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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Mechanics__in_Chemistry_(Simons_and_Nichols)/10%3A_Angular_Momentum_and_Group_Symmetries_of_Electronic_Wavefunctions/10.06%3A__Inversion_Symmetry	One more quantum number, that relating to the inversion (i) symmetry operator can be used in atomic cases because the total potential energy $V$ is unchanged when all of the electrons have their position vectors subjected to inversion (i.e., $i\textbf{r} = \textbf{-r}$). This quantum number is straightforward to determine. Because each $L, S, M_L, M_S, H$ state discussed previously consists of a few (or, in the case of configuration interaction several) symmetry adapted combinations of Slater determinant functions, the effect of the inversion operator on such a wavefunction $\Psi$ can be determined by: applying i to each orbital occupied in $\Psi$ thereby generating a ± 1 factor for each orbital (+1 for s, d, g, i, etc orbitals; -1 for p, f, h, j, etc orbitals), multiplying these $\pm 1$ factors to produce an overall sign for the character of $\Psi$ under $\hat{i}$. When this overall sign is positive, the function $\Psi$ is termed "even" and its term symbol is appended with an "e" superscript (e.g., the $^3P$ level of the $O$ atom, which has $1s^22s^22p^4$ occupancy is labeled $^3P^e$); if the sign is negative $\Psi$ is called "odd" and the term symbol is so amended (e.g., the $^3P$ level of $1s^22s^12p^1$ $B^+$ ion is labeled $^3P_o$).
Courses/Chippewa_Valley_Technical_College/CVTC_Basic_Chemistry/07%3A_Solutions/7.06%3A_Liquid-Liquid_Solutions	In 2010, a major oil spill occurred when an explosion on a drilling rig in the Gulf of Mexico released millions of gallons of crude oil into the Gulf. Oil is primarily a mixture of hydrocarbons (organic compounds composed of only carbon and hydrogen atoms). Because of its composition, oil does not dissolve in water. As a result, much of the Gulf of Mexico was contaminated, as was a great deal of shoreline in the affected area. Liquid-Liquid Solutions Nonpolar compounds do not dissolve in water. The attractive forces that operate between the particles in a nonpolar compound are weak dispersion forces. However, the nonpolar molecules are more attracted to themselves than they are to the polar water molecules. When a nonpolar liquid such as oil is mixed with water, two separate layers form because the liquids will not dissolve into each other (figure below). When another polar liquid such as ethanol is mixed with water, they completely blend and dissolve into one another. Liquids that dissolve in one another in all proportions are said to be miscible . Liquids that do not dissolve in one another are deemed immiscible . The general rule for deciding if one substance is capable of dissolving another is "like dissolves like". A nonpolar solid such as iodine will dissolve in nonpolar lighter fluid, but will not dissolve in polar water. For molecular compounds, the major factor that contributes to the material dissolving in water is the ability to form hydrogen bonds with the water solvent. Small compounds such as methanol, ethanol, acetic acid, and acetone have polar groups that can interact with the polar $\ce{H}$ of water. However, as the nonpolar portion of the molecule gets larger, solubility with water drops off. The nonpolar portion of the molecule increasingly repels the water and eventually overrides the interaction of the polar component with water. Summary Liquids that dissolve in one another in all proportions are miscible; liquids that do not dissolve in one another are immiscible. The general rule for deciding if one substance is capable of dissolving another is "like dissolves like". Nonpolar molecules are usually insoluble in water. A non-ionized molecule must be relatively polar to interact with water molecules.
Courses/Fresno_City_College/Introductory_Chemistry_Atoms_First_for_FCC/09%3A_Aqueous_Solutions/9.02%3A_Solubility_Trends/9.2.01%3A_Solutions_of_Solids_Dissolved_in_Water	Learning Objectives Explain how temperature influences solubility of solids. The solubility of a substance is the amount of that substance that is required to form a saturated solution in a given amount of solvent at a specified temperature. Solubility is often measured as the grams of solute per $100 \: \text{g}$ of solvent. The solubility of sodium chloride in water is $36.0 \: \text{g}$ per $100 \: \text{g}$ water at $20^\text{o} \text{C}$. The temperature must be specified because solubility varies with temperature. For gases, the pressure must also be specified. Solubility is specific for a particular solvent. We will consider solubility of material in water as solvent. The solubility of the majority of solid substances increases as the temperature increases. However, the effect is difficult to predict and varies widely from one solute to another. The temperature dependence of solubility can be visualized with the help of a solubility curve , a graph of the solubility vs. temperature (Figure $\PageIndex{4}$). Notice how the temperature dependence of $\ce{NaCl}$ is fairly flat, meaning that an increase in temperature has relatively little effect on the solubility of $\ce{NaCl}$. The curve for $\ce{KNO_3}$, on the other hand, is very steep and so an increase in temperature dramatically increases the solubility of $\ce{KNO_3}$. The trends for gas solubility in aqueous solution are often different than those for solids. You may notice on the graph that there are several substances—$\ce{HCl}$, $\ce{NH_3}$, and $\ce{SO_2}$— which have solubility that decreases as temperature increases. They are all gases at standard pressure. Additionally, changes in pressure can affect solubility of gases whereas they generally have little or no effect on solubility of solids. Trends of gas solubility will be discussed in the next subsection . Solubility curves can be used to determine if a given solution is saturated or unsaturated. Suppose that $80 \: \text{g}$ of $\ce{KNO_3}$ is added to $100 \: \text{g}$ of water at $30^\text{o} \text{C}$. According to the solubility curve, approximately $48 \: \text{g}$ of $\ce{KNO_3}$ will dissolve at $30^\text{o} \text{C}$. This means that the solution will be saturated since $48 \: \text{g}$ is less than $80 \: \text{g}$. We can also determine that there will be $80 - 48 = 32 \: \text{g}$ of undissolved $\ce{KNO_3}$ remaining at the bottom of the container. Now suppose that this saturated solution is heated to $60^\text{o} \text{C}$. According to the curve, the solubility of $\ce{KNO_3}$ at $60^\text{o} \text{C}$ is about $107 \: \text{g}$. Now the solution is unsaturated since it contains only the original $80 \: \text{g}$ of dissolved solute. Now suppose the solution is cooled all the way down to $0^\text{o} \text{C}$. The solubility at $0^\text{o} \text{C}$ is about $14 \: \text{g}$, meaning that $80 - 14 = 66 \: \text{g}$ of the $\ce{KNO_3}$ will re-crystallize. Summary The solubility of a solid in water usually increases with an increase in temperature. Contributions & Attributions This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality: Marisa Alviar-Agnew ( Sacramento City College ) Henry Agnew (UC Davis)
Courses/Stanford_Online_High_School/TEN2A-Acids/04%3A_Complex_Ions/4.06%3A_Qualitative_Analysis	Learning Objectives To know how to separate metal ions by selective precipitation. To understand how several common metal cations can be identified in a solution using selective precipitation. The composition of relatively complex mixtures of metal ions can be determined using qualitative analysis , a procedure for discovering the identity of metal ions present in the mixture (rather than quantitative information about their amounts). The procedure used to separate and identify more than 20 common metal cations from a single solution consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all of the metal ions are precipitated, as illustrated in Figure $\PageIndex{1}$. Group 1: Insoluble Chlorides Most metal chloride salts are soluble in water; only $\ce{Ag^{+}}$, $\ce{Pb^{2+}}$, and $\ce{Hg2^{2+}}$ form chlorides that precipitate from water. Thus the first step in a qualitative analysis is to add about 6 M $\ce{HCl}$, thereby causing $\ce{AgCl}$, $\ce{PbCl2}$, and/or $\ce{Hg2Cl2}$ to precipitate. If no precipitate forms, then these cations are not present in significant amounts. The precipitate can be collected by filtration or centrifugation. Group 2: Acid-Insoluble Sulfides Next, the acidic solution is saturated with $\ce{H2S}$ gas. Only those metal ions that form very insoluble sulfides, such as $\ce{As^{3+}}$, $\ce{Bi^{3+}}$, $\ce{Cd^{2+}}$, $\ce{Cu^{2+}}$, $\ce{Hg^{2+}}$, $\ce{Sb^{3+}}$, and $\ce{Sn^{2+}}$, precipitate as their sulfide salts under these acidic conditions. All others, such as $\ce{Fe^{2+}}$ and $\ce{Zn^{2+}}$, remain in solution. Once again, the precipitates are collected by filtration or centrifugation. Group 3: Base-Insoluble Sulfides (and Hydroxides) Ammonia or $\ce{NaOH}$ is now added to the solution until it is basic, and then $\ce{(NH4)2S}$ is added. This treatment removes any remaining cations that form insoluble hydroxides or sulfides. The divalent metal ions $\ce{Co^{2+}}$, $\ce{Fe^{2+}}$, $\ce{Mn^{2+}}$, $\ce{Ni^{2+}}$, and $\ce{Zn^{2+}}$ precipitate as their sulfides, and the trivalent metal ions $\ce{Al^{3+}}$ and $\ce{Cr^{3+}}$ precipitate as their hydroxides: $\ce{Al(OH)3}$ and $\ce{Cr(OH)3}$. If the mixture contains $\ce{Fe^{3+}}$, sulfide reduces the cation to $\ce{Fe^{2+}}$, which precipitates as $\ce{FeS}$. Group 4: Insoluble Carbonates or Phosphates The next metal ions to be removed from solution are those that form insoluble carbonates and phosphates. When $\ce{Na2CO3}$ is added to the basic solution that remains after the precipitated metal ions are removed, insoluble carbonates precipitate and are collected. Alternatively, adding $\ce{(NH4)2HPO4}$ causes the same metal ions to precipitate as insoluble phosphates. Group 5: Alkali Metals At this point, we have removed all the metal ions that form water-insoluble chlorides, sulfides, carbonates, or phosphates. The only common ions that might remain are any alkali metals ($\ce{Li^{+}}$, $\ce{Na^{+}}$, $\ce{K^{+}}$, $\ce{Rb^{+}}$, and $\ce{Cs^{+}}$) and ammonium ($\ce{NH4^{+}}$). We now take a second sample from the original solution and add a small amount of $\ce{NaOH}$ to neutralize the ammonium ion and produce $\ce{NH3}$. (We cannot use the same sample we used for the first four groups because we added ammonium to that sample in earlier steps.) Any ammonia produced can be detected by either its odor or a litmus paper test. A flame test on another original sample is used to detect sodium, which produces a characteristic bright yellow color. The other alkali metal ions also give characteristic colors in flame tests, which allows them to be identified if only one is present. Metal ions that precipitate together are separated by various additional techniques, such as forming complex ions, changing the pH of the solution, or increasing the temperature to redissolve some of the solids. For example, the precipitated metal chlorides of group 1 cations, containing $\ce{Ag^{+}}$, $\ce{Pb^{2+}}$, and $\ce{Hg2^{2+}}$, are all quite insoluble in water. Because $\ce{PbCl2}$ is much more soluble in hot water than are the other two chloride salts, however, adding water to the precipitate and heating the resulting slurry will dissolve any $\ce{PbCl2}$ present. Isolating the solution and adding a small amount of $\ce{Na2CrO4}$ solution to it will produce a bright yellow precipitate of $\ce{PbCrO4}$ if $\ce{Pb^{2+}}$ were in the original sample ( Figure $\PageIndex{2}$). As another example, treating the precipitates from group 1 cations with aqueous ammonia will dissolve any $\ce{AgCl}$ because $\ce{Ag^{+}}$ forms a stable complex with ammonia: $\ce{[Ag(NH3)2]^{+}}$. In addition, $\ce{Hg2Cl2}$ disproportionates in ammonia. \[\ce{2Hg2^{2+} \rightarrow Hg + Hg^{2+}} \nonumber \] to form a black solid that is a mixture of finely divided metallic mercury and an insoluble mercury(II) compound, which is separated from solution: \[\ce{Hg2Cl2(s) + 2NH3(aq) \rightarrow Hg(l) + Hg(NH_2)Cl(s) + NH^{+}4(aq) + Cl^{−}(aq)} \nonumber \] Any silver ion in the solution is then detected by adding $\ce{HCl}$, which reverses the reaction and gives a precipitate of white $\ce{AgCl}$ that slowly darkens when exposed to light: \[\ce{[Ag(NH3)2]^{+} (aq) + 2H^{+}(aq) + Cl^{−}(aq) \rightarrow AgCl(s) + 2NH^{+}4(aq)} \nonumber \] Similar but slightly more complex reactions are also used to separate and identify the individual components of the other groups. Summary In qualitative analysis, the identity, not the amount, of metal ions present in a mixture is determined. The technique consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all the metal ions are precipitated. Other additional steps are needed to separate metal ions that precipitate together.
Courses/Riverland_Community_College/CHEM_1000_-_Introduction_to_Chemistry_(Riverland)/04%3A_Atoms_Elements_and_the_Periodic_Table/4.12%3A_Electron_Configurations_and_the_Periodic_Table	Learning Objectives Relate the electron configurations of the elements to the shape of the periodic table. Determine the expected electron configuration of an element by its place on the periodic table. Previously, we introduced the periodic table as a tool for organizing the known chemical elements. A periodic table is shown in Figure $\PageIndex{1}$. The elements are listed by atomic number (the number of protons in the nucleus), and elements with similar chemical properties are grouped together in columns. Why does the periodic table have the structure it does? The answer is rather simple, if you understand electron configurations: the shape of the periodic table mimics the filling of the subshells with electrons. The shape of the periodic table mimics the filling of the subshells with electrons. Let us start with H and He. Their electron configurations are 1 s 1 and 1 s 2 , respectively; with He, the n = 1 shell is filled. These two elements make up the first row of the periodic table (Figure $\PageIndex{2}$) The next two electrons, for Li and Be, would go into the 2 s subshell. Figure $\PageIndex{3}$ shows that these two elements are adjacent on the periodic table. For the next six elements, the 2 p subshell is being occupied with electrons. On the right side of the periodic table, these six elements (B through Ne) are grouped together (Figure $\PageIndex{4}$). The next subshell to be filled is the 3 s subshell. The elements when this subshell is being filled, Na and Mg, are back on the left side of the periodic table (Figure $\PageIndex{5}$). Next, the 3 p subshell is filled with the next six elements (Figure $\PageIndex{6}$). Instead of filling the 3 d subshell next, electrons go into the 4 s subshell (Figure $\PageIndex{7}$). After the 4 s subshell is filled, the 3 d subshell is filled with up to 10 electrons. This explains the section of 10 elements in the middle of the periodic table (Figure $\PageIndex{8}$). ...And so forth. As we go across the rows of the periodic table, the overall shape of the table outlines how the electrons are occupying the shells and subshells. The first two columns on the left side of the periodic table are where the s subshells are being occupied. Because of this, the first two rows of the periodic table are labeled the s block . Similarly, the p block are the right-most six columns of the periodic table, the d block is the middle 10 columns of the periodic table, while the f block is the 14-column section that is normally depicted as detached from the main body of the periodic table. It could be part of the main body, but then the periodic table would be rather long and cumbersome. Figure $\PageIndex{9}$ shows the blocks of the periodic table. The electrons in the highest-numbered shell, plus any electrons in the last unfilled subshell, are called valence electrons ; the highest-numbered shell is called the valence shell . (The inner electrons are called core electrons .) The valence electrons largely control the chemistry of an atom. If we look at just the valence shell’s electron configuration, we find that in each column, the valence shell’s electron configuration is the same. For example, take the elements in the first column of the periodic table: H, Li, Na, K, Rb, and Cs. Their electron configurations (abbreviated for the larger atoms) are as follows, with the valence shell electron configuration highlighted: H: 1s1 Li: 1s22s1 Na: [Ne]3s1 K: [Ar]4s1 Rb: [Kr]5s1 Cs: [Xe]6s1 They all have a similar electron configuration in their valence shells: a single s electron. Because much of the chemistry of an element is influenced by valence electrons, we would expect that these elements would have similar chemistry— and they do . The organization of electrons in atoms explains not only the shape of the periodic table, but also the fact that elements in the same column of the periodic table have similar chemistry. The same concept applies to the other columns of the periodic table. Elements in each column have the same valence shell electron configurations, and the elements have some similar chemical properties. This is strictly true for all elements in the s and p blocks. In the d and f blocks, because there are exceptions to the order of filling of subshells with electrons, similar valence shells are not absolute in these blocks. However, many similarities do exist in these blocks, so a similarity in chemical properties is expected. Similarity of valence shell electron configuration implies that we can determine the electron configuration of an atom solely by its position on the periodic table. Consider Se, as shown in Figure $\PageIndex{10}$. It is in the fourth column of the p block. This means that its electron configuration should end in a p 4 electron configuration. Indeed, the electron configuration of Se is [Ar]4 s 2 3 d 10 4 p 4 , as expected. Example $\PageIndex{1}$: Predicting Electron Configurations From the element’s position on the periodic table, predict the valence shell electron configuration for each atom (Figure $\PageIndex{11}$). Ca Sn Solution Ca is located in the second column of the s block. We expect that its electron configuration should end with s 2 . Calcium’s electron configuration is [Ar]4 s 2 . Sn is located in the second column of the p block, so we expect that its electron configuration would end in p 2 . Tin’s electron configuration is [Kr]5 s 2 4 d 10 5 p 2 . Exercise $\PageIndex{1}$ From the element’s position on the periodic table, predict the valence shell electron configuration for each atom. Figure $\PageIndex{11}$. Ti Cl Answer a [Ar]4 s 2 3 d 2 Answer b [Ne]3 s 2 3 p 5 Summary The arrangement of electrons in atoms is responsible for the shape of the periodic table. Electron configurations can be predicted by the position of an atom on the periodic table.
Courses/Los_Angeles_Trade_Technical_College/Chem_51/zz%3A_Back_Matter/20.02%3A_Alkanes	As our country looks at the prospect of oil shortages in the future, we are searching for alternative transportation fuel sources. One very viable possibility is propane gas. Power and acceleration for propane-powered vehicles are comparable to gasoline-powered vehicles and fuel efficiency is greater. Propane has a higher octane rating than regular gasoline, leading to much longer engine life. When properly structured, propane engines can produce lower amounts of air pollution. We are seeing a growing use of propane in buses, trucks, and police cars. Maybe your next car will burn propane. Straight-Chain Alkanes Hydrocarbons A hydrocarbon is an organic compound that is made up of only carbon and hydrogen. A hydrocarbon is the simplest kind of organic molecule and is the basis for all other more complex organic compounds. Hydrocarbons can be divided into two broad categories. Aliphatic hydrocarbons are hydrocarbons that do not contain the benzene group or a benzene ring. Aromatic hydrocarbons contain one or more benzene rings. In this concept, we will discuss the aliphatic hydrocarbons. Alkanes An alkane is a hydrocarbon in which there are only single covalent bonds. The simplest alkane is methane, with the molecular formula $\ce{CH_4}$. The carbon is the central atom and makes four single covalent bonds to hydrogen atoms. The next simplest alkane is called ethane $\left( \ce{C_2H_6} \right)$ and consists of two carbon atoms with a single covalent bond between them. Each carbon is then able to bond to three hydrogen atoms. The alkane series progresses from there, increasing the length of the carbon chain by one carbon at a time. Structural formulas for ethane, propane $\left( \ce{C_3H_8} \right)$, and butane $\left( \ce{H_4H_{10}} \right)$ are shown below. These alkanes are called straight-chain alkanes because the carbon atoms are connected in one continuous chain with no branches. Naming and writing structural and molecular formulas for the straight-chain alkanes is straightforward. The name of each alkane consists of a prefix that specifies the number of carbon atoms and the ending -ane . The molecular formula follows the pattern of $\ce{C_{n}H_{2n_2}}$ where $n$ is the number of carbons in the chain. The table below lists the first ten members of the alkane series. 0 1 2 3 Table 25.2.1: First Ten Members of the Alkane Series Table 25.2.1: First Ten Members of the Alkane Series Table 25.2.1: First Ten Members of the Alkane Series Table 25.2.1: First Ten Members of the Alkane Series Name Molecular Formula Condensed Structural Formula Boiling Point $\left( ^\text{o} \text{C} \right)$ Methane $\ce{CH_4}$ $\ce{CH_4}$ -161.0 Ethane $\ce{C_2H_6}$ $\ce{CH_3CH_3}$ -88.5 Propane $\ce{C_3H_8}$ $\ce{CH_3CH_2CH_3}$ -42.0 Butane $\ce{C_4H_{10}}$ $\ce{CH_3CH_2CH_2CH_3}$ 0.5 Pentane $\ce{C_5H_{12}}$ $\ce{CH_3CH_2CH_2CH_2CH_3}$ 36.0 Hexane $\ce{C_6H_{14}}$ $\ce{CH_3CH_2CH_2CH_2CH_2CH_3}$ 68.7 Heptane $\ce{C_7H_{16}}$ $\ce{CH_3CH_2CH_2CH_2CH_2CH_2CH_3}$ 98.5 Octane $\ce{C_8H_{18}}$ $\ce{CH_3CH_2CH_2CH_2CH_2CH_2CH_2CH_3}$ 125.6 Nonane $\ce{C_9H_{20}}$ $\ce{CH_3CH_2CH_2CH_2CH_2CH_2CH_2CH_2CH_3}$ 150.7 Decane $\ce{C_{10}H_{22}}$ $\ce{CH_3CH_2CH_2CH_2CH_2CH_2CH_2CH_2CH_2CH_3}$ 174.1 Note that the table shows a variation of a structural formula called a condensed structural formula. In this formula, the covalent bonds are understood to exist between each carbon and the hydrogens associated with it, as well as between carbon atoms. This table also shows that the boiling points of the alkanes steadily increase as the length of the carbon chain increases. This is due to an increase in the strength of the intermolecular attractive forces and is a general feature of other organic molecules as well. Summary Hydrocarbon definition and classifications are given. Alkanes are defined and listed. Contributors
Courses/Sacramento_City_College/SCC%3A_Chem_420_-_Organic_Chemistry_I/09%3A_Reactions_of_Alkenes/9.03%3A__Alkene_Asymmetry_and_Markovnikov's_Rule	Learning Objective predict the products/specify the reagents for EAR of hydrohalic acids (HX) with asymmetrical alkenes using Markovnikov's Rule for Regioselectivity apply the principles of regioselectivity and stereoselectivity to the addition reactions of alkenes Addition to unsymmetrical alkenes In terms of reaction conditions and the factors affecting the rates of the reaction, there is no difference whatsoever between these alkenes and the symmetrical ones described above. The problem comes with the orientation of the addition of the hydrogen and the halogen across the double bond. Markovnikov's Rule If HCl adds to an unsymmetrical alkene like propene, there are two possible ways it could add. However, in practice, there is only one major product according to Markovnikov's Rule. Markovnikov's Rule: When HX is added to an unsymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already. Applying Markovnikov's Rule to the reaction above, the hydrogen bonds with the CH 2 group, because the CH 2 group has more hydrogens than the CH group. Notice that only the hydrogens directly attached to the carbon atoms at either end of the double bond count. Regioselectivity - a closer look If more than one reaction could occur between a set of reactants under the same conditions giving products that are constitutional isomers and if one product forms in greater amounts than the others, the overall reaction is said to be regioselective. Say three reactions could occur between the hypothetical reactants A and B under the same conditions giving the constitutionally isomeric products C , D , and E . There are two possibilities: 1. The three products form in equal amounts, i.e., of the total product 33% is C , another 33% D , the remaining 33% E . (These percentages are called relative yields of the products.) If this is what is observed, the overall reaction between A and B is not regioselective. 2. One product forms in greater amounts than the others. Say, for example, the relative yields of C , D , and E are 25%, 50%, and 25%, respectively. If this is what is observed, the overall reaction between A and B is regioselective. eg: Experimentally, 2 is the major product; 3 is the minor product. Thus, the overall reaction between 1 and HBr is regioselective toward 2 . If more than one reaction could occur between a set of reactants under the same conditions giving products that are constitutional isomers and if only one product is observed, the overall reaction is said to be 100% regioselective or regiospecific. eg: The only observed product is 5 . (Relative yields of 5 and 6 are 100% and 0%, respectively.) Thus the overall reaction between 4 and HBr is regiospecific toward 5 . Regiospecificity is merely the limiting case of regioselectivity. All regiospecific reactions are regioselective, but not all regioselective reactions are regiospecific. Exercises 1. Predict the product(s) for the following reactions: 2. In each case, suggest an alkene that would give the product shown. Answers 1. 2.
Courses/University_of_Connecticut/Chem_2444%3A_(Second_Semester_Organic_Chemistry)_UConn/03%3A_Conjugated_Pi-Systems_and_Aromaticity/3.S%3A_Benzene_and_Aromaticity_(Summary)	Concepts & Vocabulary 15.0 Introduction Aromatic compounds contain ring structures with a special type of resonance delocalization. Aromatic compounds can be drawn with alternating single and double bonds, each atom in the ring must have a p-orbital available. 15.1 Naming Aromatic Compounds Disubstituted benzene derivatives are often named using ortho (1,2), meta (1,3) and para (1,4). There are common benzene derivative names that are used by IUPAC such as toluene, phenol, benzoic acid and benzaldehyde. A benzene group that is named as a substituent is called phenyl. A benzene with a CH 2 as a substituent group is called benzyl. 15.2 Structure and Stability of Benzene Benzene does not undergo the same reactions that alkenes do, due to its aromatic stability. Aromatic molecules must have all ring atoms in the same plane to allow delocalization of the pi electrons. Heats of hydrogenation can be used to show the special stability of benzene compared to what would be expected for a theoretical cyclohexatriene molecule. 15.3 Aromaticity and the Hückel 4n + 2 Rule The four criteria for aromaticity are that the molecule must: be cyclic be planar be fully conjugated have 4n+2 π Electrons Ionic molecules and heterocyclic molecules can also be aromatic if they meet the four criteria. 15.4 Aromatic Ions Carbanions and carbocations that meet the rules for aromaticity are also aromatic. 15.5 Aromatic Heterocycles: Pyridine and Pyrrole Heterocycles that meet the rules for aromaticty are also aromatic. If a lone pair of electrons on a ring atom can result in 4n+2 π Electrons, they will be in a p-orbital. If not, they will remain in hybrid orbitals. 15.6 Polycyclic Aromatic Compounds Benzene rings can be fused together to give larger aromatic compounds with mutliple rings called polycyclic aromatic compounds (or polycyclic aromatic hydrocarbons). 15.7 Spectroscopy of Aromatic Compounds Aromatic compounds can be identified by common infrared absorptions in the 3000-3100 cm -1 and 1500-1600 cm -1 . In 1 H NMR, aromatic hydrogens appear in the 6.5-8 ppm region. Skills to Master Skill 15.1 Using IUPAC rules to name substituted benzene molecules. Skill 15.2 Use heats of hydrogenation to explain aromatic stabilization. Skill 15.3 Draw molecular orbital diagram for benzene (all 6 MO's). Skill 15.4 Use the criteria for aromaticity to determine if a molecule is aromatic or not. Skill 15.5 Determine whether lone pairs of electrons for ions and heterocycles will be in p orbitals or hybrid orbitals. Skill 15.6 Identify aromatic absorbances in infrared spectroscopy. Skill 15.7 Identify aromatic resonances in 1 H NMR spectroscopy.
Courses/South_Puget_Sound_Community_College/Chem_121%3A_Introduction_to_Chemistry/11%3A_Chapter_9_-_Gases/11.03%3A_Ideal_Gas_Law	To use the ideal gas law to describe the behavior of a gas. Calculate T, V, P, or n of the ideal gas law: PV=nRT A flat tire is not very useful. It does not cushion the rim of the wheel and creates a very uncomfortable ride. When air is added to the tire, the pressure increases as more molecules of gas are forced into the rigid tire. The amount of air that should be put into a tire depends on the pressure rating for that tire. Too little pressure and the tire will not hold its shape. Too much pressure and the tire could burst. Avogadro's Law You have learned about Avogadro's hypothesis: equal volumes of any gas at the same temperature and pressure contain the same number of molecules. It follows that the volume of a gas is directly proportional to the number of moles of gas present in the sample. Avogadro's Law states that the volume of a gas is directly proportional to the number of moles of gas, when the temperature and pressure are held constant. The mathematical expression of Avogadro's Law is: \[V = k \times n \: \: \: \text{and} \: \: \: \frac{V_1}{n_1} = \frac{V_2}{n_2}\] (Where $n$ is the number of moles of gas and $k$ is a constant). Avogadro's Law is in evidence whenever you blow up a balloon. The volume of the balloon increases as you add moles of gas to the balloon by blowing it up. If the container holding the gas is rigid rather than flexible, pressure can be substituted for volume in Avogadro's Law. Adding gas to a rigid container makes the pressure increase. There are a number of chemical reactions that require ammonia. In order to carry out a reaction efficiently, we need to know how much ammonia we have for stoichiometric purposes. Using gas laws, we can determine the number of moles present in the tank if we know the volume, temperature, and pressure of the system. Ideal Gas Law The combined gas law shows that the pressure of a gas is inversely proportional to volume and directly proportional to temperature. Avogadro's Law shows that volume or pressure is directly proportional to the number of moles of gas. Putting these laws together gives us the following equation: \[\frac{P_1 \times V_1}{T_1 \times n_1} = \frac{P_2 \times V_2}{T_2 \times n_2}\] As with the other gas laws, we can also say that $\frac{\left( P \times V \right)}{\left( T \times n \right)}$ is equal to a constant. The constant can be evaluated provided that the gas being described is considered to be ideal. The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. If we substitute in the variable $R$ for the constant, the equation becomes: \[\frac{P \times V}{T \times n} = R\] The ideal gas law is conveniently rearranged to look this way, with the multiplication signs omitted: \[PV = nRT\] The variable $R$ in the equation is called the ideal gas constant . Evaluating the Ideal Gas Constant The value of $R$, the ideal gas constant, depends on the units chosen for pressure, temperature, and volume in the ideal gas equation. It is necessary to use Kelvin for the temperature and it is conventional to use the SI unit of liters for the volume. However, pressure is commonly measured in one of three units: $\text{kPa}$, $\text{atm}$, or $\text{mm} \: \ce{Hg}$. Therefore, $R$ can have three different values. We will demonstrate how $R$ is calculated when the pressure is measured in $\text{kPa}$. Recall that the volume of $1.00 \: \text{mol}$ of any gas at STP is measured to be $22.414 \: \text{L}$. We can substitute $101.325 \: \text{kPa}$ for pressure, $22.414 \: \text{L}$ for volume, and $273.15 \: \text{K}$ for temperature into the ideal gas equation and solve for $R$. \[R = \frac{PV}{nT} = \frac{101.325 \: \text{kPa} \times 22.414 \: \text{L}}{1.000 \: \text{mol} \times 273.15 \: \text{K}} = 8.314 \: \text{kPa} \cdot \text{L/K} \cdot \text{mol}\] This is the value of $R$ that is to be used in the ideal gas equation when the pressure is given in $\text{kPa}$. The table below shows a summary of this and the other possible values of $R$. It is important to choose the correct value of $R$ to use for a given problem. Unit of $P$ Unit of $V$ Unit of $n$ Unit of $T$ Value and Unit of $R$ $\text{kPa}$ $\text{L}$ $\text{mol}$ $\text{K}$ $8.314 \: \text{J/K} \cdot \text{mol}$ $\text{atm}$ $\text{L}$ $\text{mol}$ $\text{K}$ $0.08206 \: \text{L} \cdot \text{atm/K} \cdot \text{mol}$ $\text{mm} \: \ce{Hg}$ $\text{L}$ $\text{mol}$ $\text{K}$ $62.36 \: \text{L} \cdot \text{mm} \: \ce{Hg}/\text{K} \cdot \text{mol}$ Notice that the unit for $R$ when the pressure is in $\text{kPa}$ has been changed to $\text{J/K} \cdot \text{mol}$. A kilopascal multiplied by a liter is equal to the SI unit for energy, a joule $\left( \text{J} \right)$. Example $\PageIndex{1}$ What volume is occupied by $3.760 \: \text{g}$ of oxygen gas at a pressure of $88.4 \: \text{kPa}$ and a temperature of $19^\text{o} \text{C}$? Assume the oxygen is an ideal gas. Solution Step 1: List the known quantities and plan the problem. Known $P = 88.4 \: \text{kPa}$ $T = 19^\text{o} \text{C} = 292 \: \text{K}$ Mass $\ce{O_2} = 3.760 \: \text{g}$ $\ce{O_2} = 32.00 \: \text{g/mol}$ $R = 8.314 \: \text{J/K} \cdot \text{mol}$ Unknown In order to use the ideal gas law, the number of moles of $\ce{O_2}$ $\left( n \right)$ must be found from the given mass and the molar mass. Then, use $PV = nRT$ to solve for the volume of oxygen. Step 2: Solve. \[3.760 \: \text{g} \times \frac{1 \: \text{mol} \: \ce{O_2}}{32.00 \: \text{g} \: \ce{O_2}} = 0.1175 \: \text{mol} \: \ce{O_2}\] Rearrange the ideal gas law and solve for $V$. \[V = \frac{nRT}{P} = \frac{0.1175 \: \text{mol} \times 8.314 \: \text{J/K} \cdot \text{mol} \times 292 \: \text{K}}{88.4 \: \text{kPa}} = 3.23 \: \text{L} \: \ce{O_2}\] Step 3: Think about your result. The number of moles of oxygen is far less than one mole, so the volume should be fairly small compared to molar volume $\left( 22.4 \: \text{L/mol} \right)$ since the pressure and temperature are reasonably close to standard. The result has three significant figures because of the values for $T$ and $P$. Since a joule $\left( \text{J} \right) = \text{kPa} \cdot \text{L}$, the units cancel out correctly, leaving a volume in liters. Key Takeaway The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the ideal gas law , PV = nRT . The proportionality constant, R , is called the gas constant and has the value 0.08206 (L·atm)/(K·mol), 8.3145 J/(K·mol), or 1.9872 cal/(K·mol), depending on the units used. The ideal gas law can be used to calculate any of the four properties if the other three are known.
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Non-Equilibrium_Statistical_Mechanics_(Cao)/03%3A_Hydrodynamics_and_Light_Scattering/3.01%3A_Light_Scattering	Scattering and Correlation Functions A particle or light field propagating through space can be described by a wave vector $\vec{k}$ . The direction of $k$ indicates the direction of propagation of the wave, and the magnitude of $k$ indicates the wave number, or inverse wavelength, of the wave. Scattering occurs when a propagating wave encounters a medium which alters the magnitude or direction of its wave vector. In this section, we will show that the behavior of light scattered from a medium is related to the density correlation functions of the medium. As a result, light scattering experiments can be used to probe the structure of a material. Elastic Scattering Neutron or Light Scattering In this section, we want to describe the behavior of a particle or light field that undergoes elastic scattering from a medium. This discussion could apply to x-ray, proton, neutron, or electron scattering, among others. Elastic scattering occurs when there is no transfer of energy from the particle to the scattering medium. The direction of the particle’s wave vector changes, but its wave number (or frequency) remains the same. A schematic of the scattering process is depicted in Figure 3.1. The incident particle or light field with wave vector $\overrightarrow{k_{o}}$ is scattered from the sample at point $\vec{r}$ , changing its wave vector to $\overrightarrow{k_{f}}$ . The vector $\overrightarrow{k_{f}}$ has the same magnitude as $\overrightarrow{k_{o}}$ , but a different direction. The scattered light is detected at point $\overrightarrow{r^{\prime}}$ . The scattered particle or light field can be modelled as a spherical wave. The quantum mechanical expression for this wave is \[\Psi_{s}=\frac{i}{\hbar} \int \frac{e^{i k\left\|\vec{r}-\overrightarrow{r^{\prime}}\right\|}}{\left\|\vec{r}-\overrightarrow{r^{\prime}}\right\|} e^{i \overrightarrow{k_{o}} \cdot \overrightarrow{r^{\prime}}} \rho\left(\overrightarrow{r^{\prime}}\right) d \overrightarrow{r^{\prime}}\] where $\rho(\vec{r})$ is the density of scattering agents and the integral is carried out over all scattering agents. In most light scattering experiments, the distance from the sample to the light detector is significantly larger than the size of the sample itself. In this case it is valid to make the assumption that $r>>r^{\prime}$ . Then \[e^{i k\left\|\vec{r}-\vec{r}^{\prime}\right\|} \rightarrow e^{i k r-i \overrightarrow{k_{f}} \cdot \overrightarrow{r^{\prime}}}\] and the wavefunction can be written \[\Psi_{s}=\frac{i}{\hbar} \frac{e^{i k r}}{r} \int \rho\left(\overrightarrow{r^{\prime}}\right) e^{-i\left(\overrightarrow{k_{f}}-\overrightarrow{k_{o}}\right) \cdot \overrightarrow{r^{\prime}}} d \overrightarrow{r^{\prime}}=\frac{i}{\hbar} \frac{e^{i k r}}{r} \int \rho\left(\overrightarrow{r^{\prime}}\right) e^{-i \vec{k} \cdot \vec{r}^{\prime}} d \overrightarrow{r^{\prime}}\] where $\vec{k}=\overrightarrow{k_{f}}-\overrightarrow{k_{o}}$ is the difference between the initial and scattered wave vector. We can also assume that the medium is composed of point particles, so the density is the sum over all points \[\rho(\vec{r})=\sum_{i=1}^{N} a_{i} \delta\left(\vec{r}-\vec{r}_{i}\right)=a \sum_{i=1}^{N} \delta\left(\vec{r}-\vec{r}_{i}\right)\] Then the wavefunction simplifies to \[\Psi_{s}=\frac{i}{\hbar} \frac{e^{i k r}}{r} a \sum_{i=1}^{N} e^{-i \vec{k} \cdot \overrightarrow{r_{i}}} \propto a \sum_{i=1}^{N} e^{-i \vec{k} \cdot \overrightarrow{r_{i}}}\] In light scattering experiments the measured quantity is the intensity of scattered light over the angle spanned by the detector. This quantity is called the scattering cross section $\frac{d \sigma}{d \Omega}$ , and it is proportional to the square of the wavefunction: \[I(\vec{k})=\left\|\Psi_{s}\right\|^{2} \propto \frac{1}{r^{2}} \frac{d \sigma}{d \Omega}=\frac{a^{2}}{r^{2}}\left\langle\left.\sum_{i=1}^{N} e^{-i \vec{k} \cdot \vec{r}_{i}}\right\|^{2}\right\rangle=\frac{a^{2}}{r^{2}} N S(\vec{k})\] where $S(\vec{k})$ is called the static structure factor, and is defined as: \[S(\vec{k})=\frac{1}{N}\left\langle\left\|\sum_{i=1}^{N} e^{-i \vec{k} \cdot \vec{r}_{i}}\right\|^{2}\right\rangle\] In order to find the scattering intensity, we must evaluate this term. The Static Structure Factor The static structure factor can be rewritten as: \[S(k)=\frac{1}{N}\left\langle\rho_{k} \rho_{-k}\right\rangle\] where \[\rho_{k}=\int e^{-i \vec{k} \cdot \vec{r}} \rho(\vec{r}) d \vec{r}\] and $\rho(r)$ is the local number density. For a homogeneous liquid, $\langle\rho(r)\rangle=\rho_{o}$ . To model real systems, we can simplify the calculations by expressing the density correlations as a sum of the homogeneous density $\rho_{o}$ and local fluctuations $\delta \rho$ . \[\rho(r)=\rho_{o}+\delta \rho\] Using this separation, the scattering function can be written in two pieces: \[S(k)=\frac{1}{N}\left\langle\rho_{k} \rho_{-k}\right\rangle=\rho_{0}(2 \pi)^{3} \delta(\vec{k})+\frac{1}{N}\left\langle\delta \rho_{k} \delta \rho_{-k} \mid\right\rangle\] The first term arises from the homogeneous background and is called the forward scattering. The second term gives the scattering from the density fluctuations. In an ideal gas, there is no interaction between the particles $\delta \rho=0$ , and so there is only forward scattering. The Density Correlation function It is also helpful to think about the scattering in real space. Define the density correlation function $G(\vec{r})$ as the Fourier transform of $S(k)$ into coordinate space. \[\begin{aligned} G(\vec{r})=\frac{1}{(2 \pi)^{3}} \int S(\vec{k}) e^{-i \vec{k} \cdot \vec{r}} d \vec{k} \\ =\frac{1}{N} \frac{1}{(2 \pi)^{3}} \int e^{-i \vec{k} \cdot \vec{r}}\left\langle\sum_{i} e^{i \vec{k} \cdot \vec{r}_{i}} \sum_{j} e^{-i \vec{k} \cdot \vec{r}_{j}}\right\rangle d \vec{k} \\ =\frac{1}{N} \sum_{i, j}\left\langle\delta\left(\vec{r}-\vec{r}_{i, j}\right)\right\rangle \end{aligned}\] From this, we can see that the Fourier transform of the structure factor gives probability of finding two particles separated by a vector $\vec{r}$ . We can also write the density correlation function in a slightly different form: \[\begin{aligned} G(\vec{r})=\frac{1}{N} \sum_{i, j} \int d \overrightarrow{r_{o}}\left\langle\delta\left(\overrightarrow{r_{i}}-\overrightarrow{r_{o}}-\vec{r}\right) \delta\left(\overrightarrow{r_{j}}-\overrightarrow{r_{o}}\right)\right\rangle \\ =\frac{1}{N} \int d \overrightarrow{r_{o}}\left\langle\rho\left(\overrightarrow{r_{o}}+\vec{r}\right) \rho\left(\overrightarrow{r_{o}}\right)\right\rangle=\frac{V}{N}\langle\rho(\vec{r}) \rho(0)\rangle=\frac{1}{\rho_{o}}\langle\rho(\vec{r}) \rho(0)\rangle \end{aligned}\] The Pair Distribution Function $\quad$ To better understand the physical interpretation of the structure factor and the density correlation function, we can rewrite them in terms of the pair distribution function $g(r)$ . The pair distribution function is given by: \[g(r)=\frac{1}{N^{2}} \sum_{i \neq j}\left\langle\delta\left(\vec{r}-\vec{r}_{i, j}\right)\right\rangle\] This gives the probability that, if I have a single particle $i$ , I will be able to find another particle $j$ at a distance $\vec{r}$ away. It is defined only for terms with $i \neq j$ . We can write $g(r)$ as: \[g(r)=h(r)+1\] where $h(r)$ is the pair correlation function. The Fourier transform of the pair distribution function can be written: \[\tilde{g}(\vec{k})=\tilde{h}(\vec{k})+(2 \pi)^{3} \delta(\vec{k})\] This allows us to rewrite the structure factor and the density correlation function in terms of the interactions between individual pairs of particles. To write the structure factor $S(k)$ and the density correlation function $G(r)$ in terms of the pair distribution function, separate the summations into terms with $i=j$ and terms with $i \neq j$ . The structure factor is written: \[S(k)=\frac{1}{N} \sum_{i, j}\left\langle e^{-i k r_{i}} e^{i k r_{j}}\right\rangle\] The terms with $i=j$ each contribute a value of $\frac{1}{N}$ . After taking the summation over all $N$ particles, this gives a value of 1 . \[S(k)=1+\sum_{i \neq j}\left\langle e^{-i k\left(r_{i}-r_{j}\right)}\right\rangle=1+\rho \tilde{g}(k)=1+\rho \tilde{h}(\vec{k})+(2 \pi)^{3} \delta(\vec{k}) \rho\] Now, the first two terms $1+\rho \tilde{h}(\vec{k})$ give the scattering due to the molecular structure, or fluctuations. The third term gives the forward scattering, which as we discussed earlier is the scattering that we would expect in a system with no fluctuations (an ideal gas). The density correlation function is written: \[G(\vec{r})=\frac{1}{N}\left\langle\sum_{i, j} \delta\left(r_{i, j}-r\right)\right\rangle\] when $i=j$ , we are discussing a single particle. Therefore, $r_{i, j}=0$ and each term contributes $\frac{1}{N} \delta(\vec{r})$ . After taking the summation over all $N$ particles, the $N$ cancels and we are left with $\delta(\vec{r})$ . \[\begin{aligned} & G(\vec{r})=\delta(\vec{r})+\frac{1}{N}\left\langle\sum_{i \neq j} \delta\left(r_{i, j}-r\right)\right\rangle \\ =& \delta(\vec{r})+\rho g(\vec{r})=\delta(\vec{r})+\rho(h(\vec{r})+1) \end{aligned}\] By writing the expressions for $S(\vec{k})$ and $G(\vec{r})$ in terms of $g(\vec{r})$ , their physical interpretation becomes more clear. The pair distribution function for a typical liquid and a typical solid are shown in Figure $3.3$ and Figure $3.4$ . If a particle has a radius $d$ , then clearly no other particle can be closer than distance $d$ . Therefore, for both the solid and the liquid, $g(\vec{r})$ has a value of 0 from a distance 0 to a distance $d$ . At this point, the probability rapidly increases and begins oscillating around a value of 1 . In a liquid, there is short range structure as weak intermolecular interactions form a series of solvation shells around a particle. However, these forces only act at short range, and as the distance increases the correlation decays to zero. In a solid, the structure persists throughout the sample, and therefore the oscillations do not decay. Inelastic Scattering The previous section described the behavior of a particle as it undergoes an elastic scattering event. In this section, we will address the phenomenon of inelastic scattering, which applies primarily to light fields. Inelastic scattering occurs when scattered light transfers some energy to the scattering material. While an elastic scattering event causes only a change in the direction of the wave vector, an inelastic scattering event causes both a change in the direction and the wavenumber of the scattered light. In other words, the scattered wave becomes frequency dispersed. Figure $3.5$ gives a schematic of an inelastic scattering event. Scattered Intensity To calculate the intensity of scattered light from an inelastic scattering event, we can follow a very similar process to that which we used for elastic scattering: model the scattered light as a spherical wave, and simplify it by assuming that the distance from the sample to the light detector is large compared with the size of the sample, and that the medium is composed of point particles. However, there is one major difference. Since the scattered light can transfer energy to the material, the position of the particles now depends on time. The scattered wavefunction is then: \[\Psi_{s} \propto \frac{a}{r} \sum_{i=1}^{N} e^{-i \vec{k} \cdot \vec{r}_{i}(t)}\] which gives a differential cross-section of: \[\frac{d \sigma}{d \Omega d \omega}=a^{2}\left\langle\sum_{i} e^{-i \vec{k} \cdot \overrightarrow{r_{i}}(t)} \sum_{j} e^{-i \vec{k} \cdot \overrightarrow{r_{j}}(t)}\right\rangle\] By taking the temporal Fourier transform, we can find the structure factor: \[\begin{aligned} \frac{d \sigma}{d \Omega d \omega}=a^{2} \int e^{i w t} &\left\langle\sum_{i} e^{-i \vec{k} \cdot \vec{r}_{i}(t)} \sum_{j} e^{-i \vec{k} \cdot \vec{r}_{j}(t)} d t\right\rangle \\ &=a^{2} N S(\vec{k}, \omega) \end{aligned}\] Note that the structure factor $S$ is now dependent on both the wave vector $\vec{k}$ and the frequency $\omega$ . Therefore, it is called the Dynamic Structure Factor. The Intermediate Scattering Function The intermediate scattering function is defined as the Fourier transform of the dynamic structure factor into real time. \[\begin{aligned} F(\vec{k}, t)=\frac{1}{2 \pi} \int S(\vec{k}, \omega) e^{i \omega t} d \omega \\ S(\vec{k}, \omega)=\int F(\vec{k}, t) e^{-i \omega t} d t \end{aligned}\] It is called the Intermediate scattering function is because it has one variable, the spatial dimension $k$ , expressed in Fourier space, and the other variable, the time dimension $t$ , expressed in real space. It can be expressed explicitly as: \[F(\vec{k}, t)=\frac{1}{N}\left\langle\rho_{k}(t) \rho_{-k}(0)\right\rangle\] where: \[\rho_{k}(t)=\sum_{i} e^{-i \vec{k} \cdot \overrightarrow{r_{i}}(t)}\] Note that this function looks identical to the static structural factor from section 1, except that now the density is a function of time. The Van Hove Function The Van Hove Function is defined as the Fourier transform of the intermediate scattering function into real space. \[G(\vec{r}, t)=\frac{1}{(2 \pi)^{3}} \int F(\vec{k}, t) e^{i \vec{k} \cdot \vec{r}} d \vec{k}=\frac{1}{N} \sum_{i, j}\left\langle\delta\left(\vec{r}_{i}(t)-\overrightarrow{r_{j}}(0)-\vec{r}\right)\right\rangle\] The Van Hove Function can also be expressed as \[\begin{aligned} G(\vec{r}, t)=\frac{1}{N} \int d \overrightarrow{r_{o}}\left\langle\sum_{i} \delta\left(\overrightarrow{r_{i}}(t)-\vec{r}-\overrightarrow{r_{o}}\right) \sum_{j} \delta\left(\overrightarrow{r_{j}}(0)-\overrightarrow{r_{o}}\right)\right\rangle \\ =\frac{1}{N} \int d \overrightarrow{r_{o}}\left\langle\rho\left(\vec{r}+\overrightarrow{r_{o}}, t\right) \rho\left(\overrightarrow{r_{o}}, 0\right)\right\rangle \\ =\frac{V}{N}\langle\rho(\vec{r}, t) \rho(0,0)\rangle \end{aligned}\] where $\frac{V}{N}=\rho_{o}^{-1}$ . The Van Hove function describes the fluctuation of densities at different times and positions. It can be difficult to keep track of the many functions used to describe inelastic scattering. The following table summarizes these functions and their different spatial and temporal variables. Name Symbol Spatial Dimension Temporal Dimension Dynamic Structure Factor $S(\vec{k}, \omega)$ Fourier, $\vec{k}$ Fourier, $\omega$ Intermediate Scattering Function $F(\vec{k}, t)$ Fourier, $\vec{k}$ Real, $t$ Van Hove Function $G(\vec{r}, t)$ Real, $\vec{r}$ Real, $t$ If we are only interested in the spatial structure, we can perform a sum over the temporal dimension: \[S(\vec{k})=\frac{1}{2 \pi} \int S(\vec{k}, \omega) d \omega=F(\vec{k}, 0)\] This gives the spatial structure. The density can again be expressed as the sum of a constant background $\rho_{o}$ and fluctuations $\delta \rho:$ \[\rho=\rho_{o}+\delta \rho\] Then the dynamic structure factor can be expressed as \[S(\vec{k}, w)=(2 \pi)^{4} \delta(\vec{k}) \delta(\omega) \rho_{o}+\int e^{i \omega t} \frac{1}{N}\left\langle\delta \rho_{k}(t) \delta \rho_{-k}(0)\right\rangle d t\] In the first term, $\vec{k}=0$ and $\omega=0$ . This is the forward, elastic, not scattered wave for an ideal gas. The second term gives the spectrum of density fluctuations in the fluid.
Courses/SUNY_Oneonta/Chem_322_Lecture_Content/04%3A_Arenes_Electrophilic_Aromatic_Substitution/4.E%3A_Arenes_Electrophilic_Aromatic_Substitution_(Exercises)	Problem 4-1 How many structurally different monomethyl derivatives are possible for each of the following compounds? Name each. a. naphthalene b. anthracene c. phenanthrene Problem 4-2 How many isomeric products could each of the dimethylbenzenes give on introduction of a third substituent? Name each isomer, using chlorine as the third substituent. Problem 4-3 Name each of the following compounds by the IUPAC system: a. $\ce{(C_6H_5)_2CHCl}$ b. $\ce{C_6H_5CHCl_2}$ c. $\ce{C_6H_5CCl_3}$ d. e. f. Problem 4-4 Identify the two compounds with molecular formula $\ce{C_7H_7Cl}$ from the infrared spectra shown in Figure 4-2. Figure 4-2: Infrared spectra of two isomeric compounds of formula $\ce{C_7H_7Cl}$ (see Problem 4-4) Problem 4-5 Predict the effect on the ultraviolet spectrum of a water solution of benzenamine when hydrochloric acid is added. Explain why a solution of sodium benzenoxide absorbs at longer wavelengths than a solution of benzenol (see Table 4-3 ). Problem 4-6* Estimate the chemical shifts of the protons of (a) the separate $\ce{CH_2}$ of "1,4-hexamethylenebenzene" as compared with "1,2-hexamethylenebenzene"; and (b) cyclooctatetraene (see Section 21-9A ). Problem 4-7 Establish the structures of the following benzene derivatives on the basis of their empirical formulas and NMR spectra shown in Figure 4-6. Remember that equivalent protons normally do not split each other's resonances. a. $\ce{C_8H_{10}}$ b. $\ce{C_8H_7OCl}$ c. $\ce{C_9H_{10}O_2}$ d. $\ce{C_9H_{12}}$ Figure 4-6: Proton NMR spectra of some benzene derivatives at $60 \: \text{MHz}$ with reference to TMS at $0 \: \text{ppm}$ (see Problem 4-7). Problem 4-8 Calculate from appropriate bond-energy and stabilization-energy tables ( 4-3 and 21-1 ) the heats of chlorine with benzene to give (a) chlorobenzene and (b) 5,6-dichloro-1,3-cyclohexadiene. Your answer should indicate that substitution is energetically more favorable than addition. Problem 4-9 Devise an experimental test to determine whether the following addition-elimination mechanism for bromination of benzene actually takes place. Problem 4-10 Why is nitration with ethanoyl nitrate accelerated by added fluoroboric acid, $\ce{HBF_4}$, but retarded by added hydrochloric acid? Problem 4-11 Why do fairly reactive arenes, such as benzene, methylbenzene, and ethylbenzene, react with excess nitric acid in nitromethane solution at a rate that is independent of the concentration of the arene (i.e., zero order in arene concentration)? Does this lack of dependencies on arene concentration mean that nitration of an equimolar mixture of benzene and methylbenzene would necessarily give an equimolar mixture of nitrobenzene and nitromethylbenzenes? Why or why not? Problem 4-12 Reagents, besides the molecular halogens, that effect halogen substitution include hypochlorous and hypobromous acids. They are most effective when a strong acid is present and care is taken to exclude formation of halide ions. Account for the catalytic effect of acid and the anticatalytic effect of halide ions. Problem 4-13 Arrange the following bromine-containing species in order of their expected reactivity in achieving electrophilic aromatic bromination: $\ce{HOBr}$, $\ce{Br_2}$, $\ce{Br}^\oplus$, $\ce{Br}^\ominus$, $\ce{HBr}$, $\ce{H_2OBr}^\oplus$, $\ce{BrCl}$. Problem 4-14 Aluminum chloride is a much more powerful catalyst than ferric bromide for bromination of benzene. Would you expect the combination of aluminum chloride and bromine to give much chlorobenzene in reaction with benzene? Explain. Problem 4-15 a. The bromination of benzene is catalyzed by small amounts of iodine. Devise a possible explanation for this catalytic effect. b. The kinetic expression for the bromination of naphthalene in ethanoic acid involves a term that is first order in naphthalene and second order in bromine. How can two molecules of bromine and one of naphthalene be involved in the rate-determining step of bromination? Explain why the kinetic expression simplifies to first order in naphthalene and first order in bromine in $50\%$ aqueous ethanoic acid. Problem 4-16 Write a mechanism for the alkylation of benzene with 2-propanol catalyzed by boron trifluoride. Problem 4-17 Explain how it is possible that the ratio of products isolated from equilibration of 1,2-, 1,3-, and 1,4-dimethylbenzenes is 18:58:24 if the presence of a small amount of $\ce{HF-BF_3}$, but is essentially 0:100:0 in the presence of excess $\ce{HF-BF_3}$. Notice that $\ce{HBF_4}$ is an extremely strong acid. Problem 4-18 Account for the following observations: a. 3-Methyl-2-butanol alkylates benzene in $\ce{HF}$ to give (1,1-dimethylpropyl)benzene. b. 1-Chloronorbornane will not alkylate in the presence of $\ce{AlCl_3}$. c. 1-Methylcyclopentyl cation is formed from each of the compounds shown below under the indicated conditions at low temperatures $\left( -70^\text{o} \right)$. chlorocyclohexane $\overset{\ce{SbF_5-SO_2}}{\longrightarrow}$ cyclohexene $\overset{\ce{HF-SbF_5-SO_2}}{\longrightarrow}$ cyclohexanol $\overset{\ce{FSO_3H-SbF_5}}{\longrightarrow}$ Problem 4-19 Anthraquinone can be synthesized from phthalic anhydride and benzene in two steps. The first step is catalyzed by $\ce{AlCl_3}$, the second by fuming sulfuric acid. Write mechanisms for both reactions and suggest why fuming sulfuric is required in the second step but not in the first. Problem 4-20 Suggest possible routes for the synthesis of the following compounds: a. diphenylmethane from benzoic acid and benzene b. 1-ethyl-4-methylbenzene from methylbenzene Problem 4-21 a. Substitution of a chloromethyl group, $\ce{-CH_2Cl}$, on an aromatic ring is chloromethylation and is accomplished using methanal, $\ce{HCl}$, and a metal-halide catalyst $\left( \ce{ZnCl_2} \right)$. Write reasonable mechanistic steps that could be involved in this reaction: \[\ce{C_6H_6} + \ce{CH_2O} + \ce{HCl} \overset{\ce{ZnCl_2}}{\longrightarrow} \ce{C_6H_5CH_2Cl} + \ce{CH_2O}\] b. Phenylmethyl chloride can be formed from benzene and chloromethyl methyl ether, $\ce{ClCH_2OCH_3}$, in the presence of stannic chloride, $\ce{SnCl_4}$. Write reasonable mechanistic steps, again supported by analogy, for this reaction. Notice that $\ce{SnCl_4}$ is a Lewis acid. Problem 4-22 The Gattermann reaction (not to be confused with the Gattermann-Koch aldehyde synthesis) introduces the $\ce{H-C=O}$ function into reactive aromatic compounds such as 2-naphthalenol. The necessary reagents are $\ce{HCN}$, $\ce{HCl}$, and a metal-halide catalyst ($\ce{ZnCl_2}$ or $\ce{AlCl_3}$), and the initial product must be treated with water. Write a mechanism for this reaction that is supported by analogy to other reactions discussed in this chapter. Problem 4-23 Show explicitly how an alkyl side chain of alkylbenzenesulfonates could be formed with a quaternary carbon, if the $\ce{C_{12}}$ alkane used at the start of the synthesis contained any branched-chain $\ce{C_{12}}$ isomers. Problem 4-24 Draw the structures of the intermediate cations for nitration of nitrobenzene in the 2, 3, and 4 positions. Use the structures to explain why the nitro group is meta-orienting with deactivation. Use the same kind of arguments to explain the orientation observed with $\ce{-CF_3}$, $\ce{-CHO}$, $\ce{-CH_2Cl}$, and $\ce{-NH_2}$ groups in electrophilic aromatic substitution ( Table 4-6 ). Problem 4-25* The product distribution in the bromination of methylbenzene (toluene) depends on the nature of the brominating agent. Pertinent information follows: Explain why the distribution varies with the nature of the substituting agent. Predict the product distribution of isomeric ions if $\ce{Br}^\oplus$ were to add to methylbenzene in the gas phase . Problem 4-26 Construct an energy diagram, similar to Figure 4-8 , for nitration of phenyltrimethylammonium ion in the meta and para positions. Problem 4-27 Using the rationale developed in Section 4-5, predict the major products of nitration of the following compounds. It will help to work out the Lewis structures of the substituent groups. a. phenylnitromethane, $\ce{C_6H_5CH_2NO_2}$ b. methylthiobenzene, $\ce{C_6H_5SCH_3}$ c. nitrosobenzene, $\ce{C_6H_5NO}$ d. phenyldimethylphosphine oxide, $\ce{C_6H_5PO(CH_3)_2}$ Problem 4-28 The energy diagram in Figure 4-8 represents a two-step reaction in which the first step is slower than the second. This circumstance is found in nitration and halogenation reactions. Show how this diagram would change when (a) the rate-determining step is loss of a proton from the intermediate ion, (b) the reactants rapidly form a $\pi$ complex prior to the slow step of the electrophilic attack at carbon, and (c) the rate-determining step is $\pi$-complex formation. Problem 4-29 Predict the favored position(s) of substitution in the nitration of the following compounds: a. 4-nitro-1-phenylbenzene b. 4-methylbenzenecarboxylic acid c. 3-methylbenzenecarboxylic acid d. 1,3-dibromobenzene e. 1-fluoro-3-methoxybenzene f. 1,3-dimethylbenzene Problem 4-30* a. In the nitration of para -cymene by ethanoyl nitrate in ethanoic anhydride, the observed product composition at $0^\text{o}$ is $41\%$ $5$ and $6$ , $41\%$ $3$ , $8\%$ $4$ , and $10\%$ of 4-nitromethylbenzene. Use these results to determine the relative reactivities of the para -cymene ring carbons towards $\ce{NO_2^+}$. Give your answer relative to $\ce{C_3}$ as unity ($\ce{C_3}$ is the carbon next to the isopropyl group). determine the relative reactivities based on the data obtained in Equation 4-1 . How does neglect of ipso substitution affect calculation of relative reactivities of the ring carbons? b. Write a mechanism for the solvolytic conversion of $5$ and $6$ to $3$. Problem 4-31 Draw the Kekulé-type valence-bond structures for napthalene, anthracene, and phenanthrene. Estimate the percentage of double-bond character for the 9,10 bond of phenanthrene, assuming that each of the valence-bond structures contributes equally to the hybrid structure. Problem 4-32 Devise an experiment that would establish whether the acylation of naphthalene in the 2 position in nitrobenzene solution is the result of thermodynamic control of the orientation. Problem 4-33 Predict the orientation in the following reactions: a. 1-methylnaphthalene $+ \ce{Br_2}$ b. 2-methylnaphthalene $+ \ce{HNO_3}$ c. 2-napthalenecarboxylic acid $+ \ce{HNO_3}$ Problem 4-34 Show how one can predict qualitatively the character of the 1,2 bond in acenapthylene. Problem 4-35* Explain why sodium in liquid ammonia reduces methoxybenzene (anisole) to 1-methoxy-1,4-cyclohexadiene, whereas it reduces sodium benzoate to sodium 2,5-cyclohexadienecarboxylate: Problem 4-36 Predict the Birch reduction products of the following reactions: a. anthracene $\underset{\ce{C_2H_5OH}}{\overset{\ce{Na}}{\longrightarrow}}$ b. naphthalene $\underset{\ce{NH_3} \left( l \right)}{\overset{\ce{Na}}{\longrightarrow}}$ c.* methylbenzene $\underset{\ce{NH_3} \left( l \right)}{\overset{\ce{Na}, \: \ce{C_2H_5OH}}{\longrightarrow}}$ Problem 4-37* A side reaction when reducing benzene derivatives to 1,4-cyclohexadienes with lithium or sodium in liquid ammonia is over-reduction to give cyclohexenes. Addition of ethanol greatly reduces the importance of this side reaction. Explain what role ethanol plays in preventing over-reduction. Problem 4-38 Neglecting steric-hindrance effects use the stabilization energies in Table 21-1 to explain why cis -butenedioic anhydride adds more readily to anthracene than to benzene and adds across the 9,10 positions but not the 1,4 positions of anthracene. Problem 4-39 What products would you expect to be formed in the ozonization of the following substances? Consider carefully which bonds are likely to be most reactive. a. 1,2-dimethylbenzene b. naphthalene c. acenaphthylene (see Problem 4-34) Problem 4-40* The rate of the Diels-Alder addition between cyclooctatetraene and tetracyanoethene is proportional to the tetracyanoethene concentration, $\ce{[C_2(CN)_4]}$, at low concentrations of the addends but becomes independent of $\ce{[C_2(CN)_4]}$ at high concentrations. Write a mechanism that accounts for this behavior. Problem 4-41* Write reasonable mechanisms for the different oxidation reactions of cyclooctatetraene with mercuric ethanoate in ethanoic acid, methanol, and water solutions. Notice that compounds of the type $\ce{Hg(OR)_2}$ appear to act in some cases as $^\oplus \ce{OR}$-donating agents and also that the oxide produced from cyclooctatetraene and peroxyacids ( Section 15-11C ) rearranges readily in the presence of acids to phenylethanal. Problem 4-42* The dianion $\ce{C_8H_6^{2-}}$, which corresponds to pentalene, has been prepared and appears to be reasonably stable. Why may the dianon be more stable than pentalene itself? (See Section 21-9B .) Problem 4-43* Predict which of the following compounds may have some aromatic character. Give your reasons. Problem 4-44 Write structural formulas for all of the possible isomers of $\ce{C_8H_{10}}$ that contain one benzene ring. Show how many different mononitration products each could give if no carbon skeleton rearrangements occur but nitration is possible either in the ring or side chain. Name all of the mononitration products by an accepted system. Problem 4-45 Write structural formulas (more than one may be possible) for aromatic substances that fit the following descriptions: a. $\ce{C_8H_{10}}$, which can give only one theoretically possible ring nitration product b. $\ce{C_6H_3Br3}$, which can give three theoretically possible nitration products. c. $\ce{C_6H_3Br_2Cl}$, which can give two theoretically possible nitration products. d. $\ce{C_8H_8(NO_2)_2}$, which can give only two theoretically possible different ring monobromosubstitution products. Problem 4-46 Predict the most favorable position for mononitration for each of the following substances. Indicate whether the rate is greater, or less, than for the nitration of benzene. Give your reasoning in each case. a. fluorobenzene b. trifluoromethylbenzene c. phenylethanone d. phenylmethyldimethylamine oxide, $\ce{C_6H_5CH_2} \overset{\oplus}{\ce{N}} \ce{(CH_3)_2} \overset{\ominus}{\ce{O}}$ e. diphenylmethane f. 4-bromo-1-methoxybenzene g. phenylsulfinylbenzene, $\ce{C_6H_5SOC_6H_5}$ h . 1- tert -butyl-4-methylbenzene i. diphenyliodonium nitrate, $\ce{(C_6H_5)_2} \overset{\oplus}{\ce{I}} \overset{\ominus}{\ce{N}}$ \ce{O_3}\) j. 1,3-diphenylbenzene ( meta -terphenyl) k . $\ce{N}$-(4-phenylphenyl)ethanamide Problem 4-47 Explain why the bromination of benzenamine (aniline) gives 2,4,6-tribromobenzenamine (2,4,6-tribromoaniline), whereas the nitration with mixed acids gives 3-nitrobenzenamine ( meta -nitroaniline). Problem 4-48 Explain how comparison of the following resonance structures for para substitution with the corresponding ones for meta substitution may (or may not) lead to the expectation that ortho-para orientation would be favored for the nitro, cyano, and $\ce{-CH=CHNO_2}$ groups. Problem 4-49 Starting with benzene, show how the following compounds could be prepared. Specify the required reagents and catalysts. a. 1-bromo-4-nitrobenzene b. 4-isopropyl-3-nitrobenzenesulfonic acid c. 4- tert -butylbenzenecarbaldehyde d. $\ce{C_6H_5COCH_2CH_2CO_2H}$ e. 1,2,4,5-tetrachlorocyclohexane Problem 4-50 Offer a suitable explanation of each of the following facts: a. Nitration of arenes in concentrated nitric acid is retarded by added nitrate ions and strongly accelerated by small amounts of sulfuric acid. b. Nitrobenzene is a suitable solvent to use in Friedel-Crafts acylation of benzene derivatives. c. Benzene and other arenes usually do not react with nucleophiles by either addition or substitution. d. Pyridine is almost inert to nitration with mixed nitric and sulfuric acids, a reaction the proceeds readily with benzene. Problem 4-51 Indicate the structures of the major product(s) expected in the following reactions: a. b. c. d. ($\ce{T}$ is $\ce{^3H}$, or tritium) e. f. g. h. i. Problem 4-52 Draw the structures of the products A, B, C, and D in the stepwise reaction sequences shown. a. b. Problem 4-53 The pesticide DDT is made commercially by the reaction of chlorobenzene with trichloroethanal (chloral) in the presence of an acid catalyst $\left( \ce{H_2SO_4} \right)$. Show the steps that are likely to be involved in this reaction: Problem 4-54 Hexachlorophene, the controversial germicide, is prepared from 2,4,5-trichlorobenzenol (2 moles) and methanal (1 mole) in the presence of concentrated sulfuric acid. Show the steps involved and the expected orientation of the substituents in the final product. Problem 4-55 Trifluoroperoxyethanoic acid, $\ce{CF_3C(O)O-OH}$ reacts with methoxybenzene to give 2- and 4-methoxybenzenols: Explain the nature of this reaction. What is likely to be the substituting agent? What products would you expect from trifluoroperoxyethanoic acid and fluorobenzene? Would fluorobenzene be more, or less, reactive than methoxybenzene? Problem 4-56 Ethanoic anhydride reacts with concentrated nitric acid to yield the rather unstable ethanoyl nitrate (acetyl nitrate), which is a useful nitrating agent. With mixtures of benzene and methylbenzene, ethanoyl nitrate products a mixture of nitrobenzene and 2- and 4-nitromethylbenzenes. When nitrated separately, each compound reacts at the same overall rate, but when mixed together, 25 times more nitromethylbenzene is formed than nitrobenzene. a. Write equations for the formation of ethanoyl nitrate and its use in nitration of benzene derivatives. b. Consider possible mechanisms for nitrations with ethanoyl nitrate and show how the above observations with benzene and methylbenzene alone or in mixtures can be rationalized by proper choice of the rate-determining step. Problem 4-57 4-Nitromethylbenzene-2,6-$\ce{D_2}$ is nitrated by a mixture of nitric and sulfuric acids at the same rate as ordinary 4-nitromethylbenzene under conditions in which the rate of nitration $v$ is given by $v = k \left[ \text{nitromethylbenzene} \right] \left[ \ce{NO_2^+} \right]$. (Review Section 15-6B .) a. Explain what conclusion may be drawn from this result as to the mechanism of nitration under these conditions. b. What would you expect the nitration rate of $\ce{C_6D_6}$ to be as compared with $\ce{C_6H_6}$ in the ethanoyl nitrate nitration in Problem 4-56? Problem 4-58 a. From the data of Table 21-1 estimate the overall loss in stabilization energy for the addition of chlorine to the 1,4-positions of naphthalene and to the 9,10 positions of phenanthrene. Which is likely to be the more favorable reaction? b. Predict whether anthracene is more likely to undergo electrophilic substitution at the 1,2 or 9 position. Show your reasoning. Problem 4-59 Phenanthrene is oxidized more easily than benzene or naphthalene. Chromic acid oxidation of phenanthrene forms a substance known as phenanthraquinone. Which structure, A, B, or C, would you expect to be formed most readily by oxidation of phenanthrene? Explain. Problem 4-60 Explain why the nitration and halogenation of biphenyl (phenylbenzene) goes with activation at the ortho and para positions but with deactivation at the meta position. Suggest a reason why biphenyl is more reactive than 2,2'-dimethylbiphenyl in nitration. Contributors John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
Courses/East_Tennessee_State_University/CHEM_4110%3A_Advanced_Inorganic_Chemistry/04%3A_Symmetry_and_Group_Theory/4.02%3A_Point_Groups	Click here to see a lecture on this topic . Introduction A Point Group describes all the symmetry operations that can be performed on a molecule that result in a conformation indistinguishable from the original. Point groups are used in Group Theory, the mathematical analysis of groups, to determine properties such as a molecule's molecular orbitals. Assigning Point Groups While a point group contains all of the symmetry operations that can be performed on a given molecule, it is not necessary to identify all of these operations to determine the molecule's overall point group. Instead, a molecule's point group can be determined by following a set of steps which analyze the presence (or absence) of particular symmetry elements. Steps for assigning a molecule's point group: Determine if the molecule is of high or low symmetry. If not, find the highest order rotation axis, C n . Determine whether the molecule has any C 2 axes perpendicular to the principal C n axis. If so, then there are n such C 2 axes, and the molecule is in the D set of point groups. If not, it is in either the C or S set of point groups. Determine whether the molecule has a horizontal mirror plane (σ h ) perpendicular to the principal C n axis. If so, the molecule is either in the C nh or D nh set of point groups. Determine whether the molecule has a vertical mirror plane (σ v ) containing the principal C n axis. If so, the molecule is either in the C nv or D nd set of point groups. If not, and if the molecule has n perpendicular C 2 axes, then it is part of the D n set of point groups. Determine whether there is an improper rotation axis, S 2n , collinear with the principal C n axis. If so, the molecule is in the S 2n point group. If not, the molecule is in the C n point group. Example $\PageIndex{1}$ Find the point group of benzene (C 6 H 6 ). Solution 1. Benzene is neither high nor low symmetry 2. Highest order rotation axis: C 6 3. There are 6 C 2 axes perpendicular to the principal axis 4. There is a horizontal mirror plane (σ h ) Benzene is in the D 6h point group. For Additional Practice Symmetry@Otterbein
Courses/Oregon_Institute_of_Technology/OIT_(Lund)%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)/08%3A_Nucleophilic_Substitution_Reactions/8.02%3A_Nucleophiles	What is a nucleophile? A nucleophile is an atom or functional group with a pair of electrons (usually a non-bonding, or lone pair) that can be shared. The same, however, can be said about a base: in fact, bases can act as nucleophiles, and nucleophiles can act as bases. What, then, is the difference between a base and a nucleophile? A Brønsted-Lowry, as you will recall from chapter 7, uses a lone pair of electrons to form a new bond with an acidic proton. We spent much of chapter 7 discussing how to evaluate how basic a species is. Remember that when we evaluate basicity - the strength of a base - we speak in terms of thermodynamics: where does equilibrium lie in a reference acid-base reaction? We will spend much of this section discussing how to evaluate how nucleophilic a species is - in other words, its nucleophilicity. A nucleophile shares its lone pair of electrons with an electrophile - an electron-poor atom other than a hydrogen, usually a carbon. When we evaluate nucleophilicity, we are thinking in terms of kinetics - how fast does the nucleophile react with a reference electrophile? In both laboratory and biological organic chemistry, the most common nucleophilic atoms are oxygen, nitrogen, and sulfur, and the most common nucleophilic compounds and functional groups are water/hydroxide ion, alcohols, phenols, amines, thiols, and sometimes carboxylates. In laboratory (non-biological) reactions, halide ($I^-$, $Br^-$, $Cl^-$, $F^-$) and azide ($N_3-$) anions are also commonly seen acting as nucleophiles in addition to the groups mentioned above. Carbon atoms can also be nucleophiles - enolate ions (section 7.6) are common carbon nucleophiles in biochemical reactions, while the cyanide ion ($CN^-$) is just one example of a carbon nucleophile commonly used in the laboratory. Understanding carbon nucleophiles will be critical when we study, in chapters 12 and 13, the enzyme-catalyzed reactions in which new carbon-carbon bonds are formed in the synthesis of biomolecules such as DNA and fatty acids. In the present chapter, however, we will focus on heteroatom (non-carbon) nucleophiles. Now, let's consider a number of factors that influence how nucleophilic an atom or functional group is. We'll start with protonation state. Protonation state The protonation state of a group has a very large effect on its nucleophilicity. A negatively-charged hydroxide ion is much more nucleophilic (and basic) than a water molecule. In practical terms, this means that a hydroxide nucleophile will react in an $S_N2$ reaction with chloromethane several orders of magnitude faster than will a water nucleophile. Likewise, a thiolate anion is more nucleophilic than a neutral thiol, and a neutral amine is nucleophilic, whereas an ammonium cation is not. In a non-biological context, $S_N2$ reactions tend to occur with more powerful, anionic nucleophiles, where the nucleophile can be thought of as actively displacing ('pushing') the leaving group off the carbon. $S_N1$ reactions, in contrast, tend to be solvolysis reactions, with a weak, neutral nucleophile such as water or an alcohol. Periodic trends in nucleophilicity Just as with basicity, there are predictable periodic trends associated with nucleophilicity. Moving horizontally across the second row of the periodic table, the trend in nucleophilicity parallels the trend in basicity: more nucleophilic $NH_2$- > $OH$- > $F$- less nucleophilic more nucleophilic $R$-$NH_2$ > $R$-$OH$ less nucleophilic Recall from section 7.3 that the basicity of atoms decreases as we move vertically down a column on the periodic table: thiolate ions are less basic than alkoxide ions, for example, and bromide ion is less basic than chloride ion, which in turn is less basic than fluoride ion. Recall also that this trend can be explained by considering the increasing size of the 'electron cloud' around the larger ions: the electron density inherent in the negative charge is spread around a larger volume, which tends to increase stability (and thus reduce basicity). The vertical periodic trend for nucleophilicity is somewhat more complicated that that for basicity, and depends on the solvent in which the reaction is taking place. Take the general example of the $S_N2$ reaction below: where $Nu^-$ is one of the halide ions: fluoride, chloride, bromide, or iodide, and $X$ is a common leaving group. If this reaction is occurring in a protic solvent (that is, a solvent that has a hydrogen atom bonded to an oxygen or nitrogen - water, methanol and ethanol are protic solvents), then the reaction will go fastest when iodide is the nucleophile, and slowest when fluoride is the nucleophile, reflecting the relative strength of the nucleophile. (opposite of the trend in basicity!) This is the opposite of the vertical periodic trend in basicity (section 7.3), where iodide is the least basic. What is going on here? Shouldn't the stronger base, with its more reactive unbonded valence electrons, also be the stronger nucleophile? As mentioned above, it all has to do with the solvent. Remember, we are talking now about the reaction running in a protic solvent like water. Protic solvent molecules form strong noncovalent interactions with the electron-rich nucleophile, essentially creating a 'solvent cage' of hydrogen bonds: artwork needed For the nucleophile to attack in an $S_N2$ reaction, the nucleophile-solvent hydrogen bonds must be disrupted - in other words, the nucleophilic electrons must 'escape through the bars' of the solvent cage. A weak base like iodide ion interacts weakly with the protons of the solvent, so these interactions are more readily disrupted. Furthermore, because the valence electrons on iodide ion are far from the nucleus, the electron cloud is polarizable - electron density can readily be pulled away from the nucleus, through the solvent cage and toward the electrophile. A smaller, more basic anion such as fluoride is more highly shielded by stronger interactions with the solvent molecules. The electron cloud of the fluoride ion is smaller and much less polarizable than that of an iodide ion: in water solvent, the larger iodide ion is a more powerful nucleophile than the smaller fluoride ion. The above discussion of the vertical periodic trend in nucleophilicity applies to biochemical reactions, because the biological solvent is water. The picture changes for laboratory reactions if we switch to a polar aprotic solvent , such as acetone, which is polar enough to solvate the polar and ionic compounds in the reaction but is not a hydrogen bond donor, and does not form a strong 'solvent cage' like water does. In acetone and other polar aprotic solvents, the trend in nucleophilicity is the same as the trend in basicity: fluoride is the strongest base and the strongest nucleophile. Structures of some of the most common polar aprotic solvents are shown below. These solvents are commonly used in laboratory nucleophilic substitution reactions. In biological chemistry, the most important implication of the vertical periodic trend in nucleophilicity is that thiols are more nucleophilic than alcohols. The thiol group in a cysteine amino acid residue, for example, is more nucleophilic than the alcohol group on a serine, and cysteine often acts as a nucleophile in enzymatic reactions. The thiol group on coenzyme A is another example of a nucleophile we will see often in enzymatic reactions later on. Of course, reactions with oxygen and nitrogen nucleophiles are widespread in biochemistry as well. Resonance effects on nucleophilicity Resonance effects also come into play when comparing the inherent nucleophilicity of different molecules. The reasoning involved is the same as that which we used to understand resonance effects on basicity (see section 7.3). If the electron lone pair on a heteroatom is delocalized by resonance, it is inherently less reactive - meaning less nucleophilic, and also less basic. An alkoxide ion, for example, is more nucleophilic and more basic than a carboxylate group, even though in both cases the nucleophilic atom is a negatively charged oxygen. In an alkoxide, the negative charge is localized on a single oxygen, while in the carboxylate the charge is delocalized over two oxygen atoms by resonance. The nitrogen atom on an amide is less nucleophilic than the nitrogen of an amine, due to the resonance stabilization of the nitrogen lone pair provided by the amide carbonyl group. Which amino acid has the more nucleophilic side chain - serine or tyrosine? Explain. Steric effects on nucleophilicity Steric hindrance is an important consideration when evaluating nucleophility. For example, tert-butanol is less potent as a nucleophile than methanol. The comparatively bulky methyl groups on the tertiary alcohol effectively block the route of attack by the nucleophilic oxygen, slowing the reaction down considerably (imagine trying to walk through a narrow doorway while carrying three large suitcases!). A final note: when it comes to comparing the rate of nucleophilic substitution reactions, the strength of the nucleophile only matters for $S_N2$ reactions. It is irrelevant for $S_N1$ reactions, because the rate-determining step (when the leaving group departs and a carbocation intermediate forms) does not involve the nucleophile. Video tutorial: nucleophiles Which is the better nucleophile - a cysteine side chain or a methionine side chain? A serine or a threonine? Explain. In each of the following pairs of molecules/ions, which is expected to react more rapidly with CH3Cl in acetone solvent? Explain your choice. phenolate (deprotonated phenol) or benzoate (deprotonated benzoic acid)? water or hydronium ion? trimethylamine or triethylamine? chloride anion or iodide anion? CH3NH- or CH3CH2NH2? acetate or trichloroacetate? aniline or 4-methoxyaniline? phenolate or 2,6-dimethylphenolate?
Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/15%3A_Alcohols_and_Ethers/15.04%3A_Preparation_of_Alcohols	Many of the common laboratory methods for the preparation of alcohols have been discussed in previous post or will be considered later; thus to avoid undue repetition we shall not consider them in detail at this time. Included among these methods are hydration ( Section 10-3E ) and hydroboration ( Section 11-6D ), addition of hypohalous acids to alkenes ( Section 10-4B ), $S_\text{N}1$ and $S_\text{N}2$ hydrolysis of alkyl halides ( Sections 8-4 to 8-7) and of allylic and benzylic halides ( Sections 14-3B and 14-3C ), addition of Grignard reagents to carbonyl compounds ( Section 14-12 ), and the reduction of carbonyl compounds ( Sections 16-4E and 16-5 ). These methods are summarized in Table 15-2. Some of the reactions we have mentioned are used for large-scale industrial production. For example, ethanol is made in quantity by the hydration of ethene, using an excess of steam under pressure at temperatures around $300^\text{o}$ in the presence of phosphoric acid: A dilute solution of ethanol is obtained, which can be concentrated by distillation to a constant-boiling point mixture that contains $95.6\%$ ethanol by weight. Dehydration of the remaining few percent of water to give “absolute alcohol” is achieved either by chemical means or by distillation with benzene, which results in preferential separation of the water. Ethanol also is made in large quantities by fermentation, but this route is not competitive for industrial uses with the hydration of ethene. Isopropyl alcohol and tert -butyl alcohol also are manufactured by hydration of the corresponding alkenes. Table 15-2: General Methods of Preparation of Alcohols The industrial synthesis of methyl alcohol involves hydrogenation of carbon monoxide. Although this reaction has the favorable $\Delta H^0$ value of $-28.4 \: \text{kcal mol}^{-1}$, it requires high pressures and high temperatures and a suitable catalyst; excellent conversions are achieved using zinc oxide-chromic oxide as a catalyst: Various methods of synthesis of other alcohols by reduction of carbonyl compounds will be discussed in Section 16-4E .
Courses/Brevard_College/CHE_202%3A_Organic_Chemistry_II/04%3A_Substitution_and_Elimination_reactions/4.05%3A_Factors_affecting_the_SN2_Reaction	Learning Objective determine the rate law & predict the mechanism based on its rate equation or reaction data for S N 2 reactions predict the products and specify the reagents for S N 2 reactions with stereochemistry propose mechanisms for S N 2 reactions draw and interpret Reaction Energy Diagrams for S N 2 reactions Bimolecular nucleophilic substitution (SN 2 ) reactions are concerted , meaning they are a one step process . The bond-making between the nucleophile and the electrophilic carbon occurs at the same time as the bond-breaking between the electophilic carbon and the halogen. In order of decreasing importance, the factors impacting S N 2 reaction pathways are 1) structure of the alkyl halide 2) strength of the nucleophile 3) stability of the leaving group 4) type of solvent. Structure of the alkyl halide (Substrate) and S N 2 Reaction Rates Bimolecular nucleophilic substitution (SN 2 ) reactions are concerted , meaning they are a one step process . The bond-making between the nucleophile and the electrophilic carbon occurs at the same time as the bond-breaking between the electophilic carbon and the halogen. The S N 2 transition state is very crowded with a total of five groups around the electrophilic center, the nucleophile, the leaving group, and three substituents. If each of the three substituents in this transition state were small hydrogen atoms, as illustrated in the first example below, there would be little steric repulsion between the incoming nucleophile and the electrophilic center, thereby increasing the ease at which the nucleophilic substitution reaction can occur. Remember, for the SN 2 reaction to occur, the nucleophile must be able to overlap orbitals with the electrophilic carbon center, resulting in the expulsion of the leaving group. If one of the hydrogens, however, were replaced with an R group, such as a methyl or ethyl group, there would be an increase in steric repulsion with the incoming nucleophile. If two of the hydrogens were replaced by R groups, there would be an even greater increase in steric repulsion with the incoming nucleophile. How does steric hindrance affect the rate at which an SN 2 reaction will occur? As each hydrogen is replaced by an R group, the rate of reaction is significantly diminished. This is because the addition of one or two R groups shields the backside of the electrophilic carbon impeding nucleophilic penetration. The diagram below illustrates this concept, showing that electrophilic carbons attached to three hydrogen atoms results in faster nucleophilic substitution reactions, in comparison to primary and secondary haloalkanes, which result in nucleophilic substitution reactions that occur at slower or much slower rates, respectively. Notice that a tertiary haloalkane, that which has three R groups attached, does not undergo nucleophilic substitution reactions at all. The addition of a third R group to this molecule creates a carbon that is entirely blocked. Substitutes on Neighboring Carbons Slow Nucleophilic Substitution Reactions Previously we learned that adding R groups to the electrophilic carbon results in nucleophilic substitution reactions that occur at a slower rate. What if R groups are added to neighboring carbons? It turns out that the addition of substitutes on neighboring carbons will slow nucleophilic substitution reactions as well. In the example below, 2-methyl-1-bromopropane differs from 1-bromopropane in that it has a methyl group attached to the carbon that neighbors the electrophilic carbon. The addition of this methyl group results in a significant decrease in the rate of a nucleophilic substitution reaction. If R groups were added to carbons farther away from the electrophilic carbon, we would still see a decrease in the reaction rate. However, branching at carbons farther away from the electrophilic carbon would have a much smaller effect. Strength of the Nucleophile (Nucleophilicity) In the SN2 reaction, the rate determining step for the reaction is the attack of the nucleophileto the substrate. Therefore, SN2 is easier to perform for stronger nucleophiles. There are predictable periodic trends in nucleophilicity. More electronegative elements hold their electrons more tightly, and are less able to donate them to form a new bond. For example, thiols (R-SH) are more nucleophilic than alcohols (R-OH) because oxygen has a higher electronegativity than sulfur. Also, negatively charged species are more nucleophilic than neutral molecules. For example methoxide anion CH 3 O- is more nucleophilic than methanol CH 3 OH. Table 4.5.1 shows a list of common nucleophiles and their relative nucleophilicity. 0 1 Table 4.5.1 Some common nucleophiles and their relative nucleophilicity. Table 4.5.1 Some common nucleophiles and their relative nucleophilicity. Weak Nucleophiles Strong nucleophiles H2O water R-OH alcohols R-SH thiols NH3 ammonia R-NH2 amines R-COOH carboxylic acids R-CONH2 amides HO ¯ hydroxide R-O¯ alkoxide R-S¯ thiolate NH2¯ Azanide R-NH¯ R-COO¯ carboxylate CN¯ cyanide N3¯ azide R¯ carbanions X¯ halides Resonance effects on nucleophilicity Resonance effects also come into play when comparing the inherent nucleophilicity of different molecules. The reasoning involved is the same as that which we used to understand resonance effects on basicity. If the electron lone pair on a heteroatom is delocalized by resonance, it is inherently less reactive - meaning less nucleophilic, and also less basic. An alkoxide ion, for example, is more nucleophilic and more basic than a carboxylate group, even though in both cases the nucleophilic atom is a negatively charged oxygen. In the alkoxide, the negative charge is localized on a single oxygen, while in the carboxylate the charge is delocalized over two oxygen atoms by resonance. The nitrogen atom on an amide is less nucleophilic than the nitrogen of an amine, due to the resonance stabilization of the nitrogen lone pair provided by the amide carbonyl group. The leaving group The more stable the leaving group, the lower the transition state energy, the lower the activation energy, the faster the reaction rate. Evaluating leaving group stability is analogous to determining relative acidity by evaluating conjugate base stability. The considerations are the same: identity of the atom(s) and relative position on the periodic table, resonance delocalization, and electronegativity. Orbital hybridization is rarely relevant. As Size Increases, Basicity Decreases, Leaving Group Stability Increases: In general, if we move from the top of the periodic table to the bottom of the periodic table as shown in the diagram below, the size of an atom will increase. As size increases, basicity will decrease, meaning a species will be less likely to act as a base; that is, the species will be less likely to share its electrons. When evaluating halogens as leaving groups, the same trend is significant. Fluoride has the highest electron density and is considered the worst leaving group to the point of no reactivity. As move down the column, the leaving groups have lower electron density and greater stability with iodide considered an excellent leaving group. slower reaction and requires a catalyst to overcome the alkoxides as poor leaving groups. The details of these two reactions will be studied in greater detail later in this text. Solvent Effects on an S N 2 reaction The rate of an S N 2 reaction is significantly influenced by the solvent in which the reaction takes place. The use of protic solvents (those, such as water or alcohols, with hydrogen-bond donating capability) decreases the power of the nucleophile through strong solvation. WE can view the nucleophile as being locked in a solvent cage through the strong hydrogen-bond interactions between solvent protons and the reactive lone pairs on the nucleophile. A less powerful nucleophile in turn means a slower S N 2 reaction. S N 2 reactions are faster in polar, aprotic solvents: those that lack hydrogen-bond donating capability. Below are several polar aprotic solvents that are commonly used in the laboratory: These aprotic solvents are polar but, because they do not form hydrogen bonds with the anionic nucleophile, there is a relatively weak interaction between the aprotic solvent and the nucleophile. By using an aprotic solvent we can raise the reactivity of the nucleophile. Example In each pair (A and B) below, which electrophile would be expected to react more rapidly in an S N 2 reaction with the thiol group of cysteine as the common nucleophile? Explanations to explain differences in chemical reactivity need to discuss structural and/or electrostatic differences between the reactants a) Cpd B b/c it has a more stable leaving group. The larger atomic size of S relative to O means the sulfide (CH 3 S - ) will have a lower electron density than the alkoxide (CH 3 O-). b) Cpd A b/c it has a more stable leaving group. The neutral leaving group, (CH 3 ) 2 S, is more stable than the charged sulfie leaving group (CH 3 S - ). c) Cpd B b/c the leaving group is resonance stabilized delocalizing the negative charge over two oxygen atoms. d) Cpd B b/c the leaving group has inductive electron withdrawal stabilization from the three fluorine atoms in addition to the resonance stabilzation. Exercise 1. What product(s) do you expect from the reaction of 1-bromopentane with each of the following reagents in an S N 2 reaction? a) KI b) NaOH c) CH 3 C≡C-Li d) NH 3 2. Which in the following pairs is a better nuceophile? a) (CH 3 CH 2 ) 2 N - or (CH 3 CH 2 ) 2 NH b) (CH 3 CH 2 ) 3 N or (CH 3 CH 2 ) 3 B c) H 2 O or H 2 S 3. Order the following in increasing reactivity for an S N 2 reaction. CH 3 CH 2 Br CH 3 CH 2 OTos (CH 3 CH 2 ) 3 CCl (CH 3 CH 2 ) 2 CHCl 4. Solvents benzene, ether, chloroform are non-polar and not strongly polar solvents. What effects do these solvents have on an S N 2 reaction? Answer 1. (a) - (d) 2. a) (CH 3 CH 2 ) 2 N - as there is a charge present on the nitrogen. b) (CH 3 CH 2 ) 3 N because a lone pair of electrons is present. c) H 2 O as oxygen is more electronegative. 3. 4. They will decrease the reactivity of the reaction.
Bookshelves/General_Chemistry/Chemistry_2e_(OpenStax)/17%3A_Electrochemistry/17.04%3A_Potential_Free_Energy_and_Equilibrium	Learning Objectives By the end of this section, you will be able to: Explain the relations between potential, free energy change, and equilibrium constants Perform calculations involving the relations between cell potentials, free energy changes, and equilibrium Use the Nernst equation to determine cell potentials under nonstandard conditions So far in this chapter, the relationship between the cell potential and reaction spontaneity has been described, suggesting a link to the free energy change for the reaction (see chapter on thermodynamics). The interpretation of potentials as measures of oxidant strength was presented, bringing to mind similar measures of acid-base strength as reflected in equilibrium constants (see the chapter on acid-base equilibria). This section provides a summary of the relationships between potential and the related thermodynamic properties ΔG and K. E° and ΔG° The standard free energy change of a process, $ΔG^°$, was defined in a previous chapter as the maximum work that could be performed by a system, $w_{max}$. In the case of a redox reaction taking place within a galvanic cell under standard state conditions, essentially all the work is associated with transferring the electrons from reductant-to-oxidant, $w_{elec}$: \[\Delta G^{\circ}=w_{\max }=w_{\text {elec }} \nonumber \] The work associated with transferring electrons is determined by the total amount of charge (coulombs) transferred and the cell potential: \[\Delta G^{\circ}=w_{\text {elec }}=-n F E_{\text {cell }}^{\circ} \nonumber \] where $n$ is the number of moles of electrons transferred, $F$ is Faraday’s constant , and $E^°_{cell}$ is the standard cell potential. The relation between free energy change and standard cell potential confirms the sign conventions and spontaneity criteria previously discussed for both of these properties: spontaneous redox reactions exhibit positive potentials and negative free energy changes. E° and K Combining a previously derived relation between ΔG° and K (see the chapter on thermodynamics) and the equation above relating ΔG° and E ° cell yields the following: \[\Delta G^{\circ}=-R T \ln K=-n F E_{\text {cell }}^{\circ} \label{eq1} \] with \[E_{\text {coll }}^{\circ}=\left(\frac{R T}{n F}\right) \ln K \nonumber \] This equation indicates redox reactions with large (positive) standard cell potentials will proceed far towards completion, reaching equilibrium when the majority of reactant has been converted to product. A summary of the relations between E °, Δ G ° and K is depicted in Figure $\PageIndex{1}$, and a table correlating reaction spontaneity to values of these properties is provided in Table $\PageIndex{1}$. K ΔG° E°cell Comment > 1 < 0 > 0 Reaction is spontaneous under standard conditions Products more abundant at equilibrium < 1 > 0 < 0 Reaction is non-spontaneous under standard conditions Reactants more abundant at equilibrium = 1 = 0 = 0 Reaction is at equilibrium under standard conditions Reactants and products equally abundant Example $\PageIndex{1}$: Equilibrium Constants, Standard Cell Potentials, and Standard Free Energy Changes Use data from Appendix L to calculate the standard cell potential, standard free energy change, and equilibrium constant for the following reaction at 25 °C. Comment on the spontaneity of the forward reaction and the composition of an equilibrium mixture of reactants and products. \[\ce{2 Ag^{+}(aq) + Fe(s) <=> 2 Ag(s) + Fe^{2+}(aq)} \nonumber \] Solution The reaction involves an oxidation-reduction reaction, so the standard cell potential can be calculated using the data in Appendix L. \[\begin{align} & \text { anode: } \quad && \ce{Fe(s) -> Fe^{2+}(aq) + 2 e^{-}} &&E^o_{\ce{Fe^{2+}/Fe}} =-0.447\,\text{V}\\[4pt] & \text {cathode: } \quad && \ce{2 \times (Ag^{+}(aq) + e^{-} -> Ag(s))} &&E^o_{\ce{Ag^{+}/Ag}} = +0.7996\,\text{V}\\[4pt] \hline &\text { cell: } \quad && \ce{Fe(s) + 2Ag^{+}(aq) -> Fe^{2+}(aq) + 2Ag(s)} && E^o_{\text {cell }} = +1.2247\,\text{V} \end{align} \nonumber \] with \[E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}= E^o_{\ce{Fe^{2+}/Fe}} - E^o_{\ce{Ag^{+}/Ag}} = +1.2247\,\text{V} \nonumber \] With $n = 2$, the equilibrium constant is then \[\begin{aligned} E_{\text {cell }}^{\circ} &=\frac{0.0592 V }{n} \log K \\[4pt] K&=10^{n \times E_{\text {cal }}^o / 0.0592 V } \\[4pt] &=10^{2 \times 1.247 V / 0.0592 V } \\[4pt] &=10^{42.128} \\[4pt] &=1.3 \times 10^{42} \end{aligned} \nonumber \] The standard free energy is then Equation \ref{eq1} \[\begin{aligned} & \Delta G^{\circ}=-n F E_{\text {cell }}^{\circ} \\ & =-2 \times 96,485 \frac{ C }{ mol } \times 1.247 \frac{ J }{ C }=-240.6 \frac{ kJ }{ mol } \end{aligned} \nonumber \] The reaction is spontaneous, as indicated by a negative free energy change and a positive cell potential. The K value is very large, indicating the reaction proceeds to near completion to yield an equilibrium mixture containing mostly products. Exercise $\PageIndex{1}$ What is the standard free energy change and the equilibrium constant for the following reaction at room temperature? Is the reaction spontaneous? \[\ce{Sn(s) + 2 Cu^{2+}(aq) <=> Sn^{2+}(aq) + 2 Cu^{+}(aq)} \nonumber \] Answer Spontaneous $n = 2$ $E_{\text {cell }}^{\circ}=+0.291 V$ $[\Delta G^{\circ}=-56.2 \frac{ kJ }{ mol }$ K = 6.8 10 9 . Potentials at Nonstandard Conditions: The Nernst Equation Most of the redox processes that interest science and society do not occur under standard state conditions, and so the potentials of these systems under nonstandard conditions are a property worthy of attention. Having established the relationship between potential and free energy change in this section, the previously discussed relation between free energy change and reaction mixture composition can be used for this purpose. \[\Delta G=\Delta G^{\circ}+R T \ln Q \nonumber \] Notice the reaction quotient, Q , appears in this equation, making the free energy change dependent upon the composition of the reaction mixture. Substituting the equation relating free energy change to cell potential yields the Nernst equatio n : \[\begin{align} -n F E_{\text {cell }}&=-n F E_{\text {cell }}^{\circ}+R T \ln Q \\[4pt] E_{\text {cell }}&=E_{\text {cell }}^{\circ}-\frac{R T}{n F} \ln Q \end{align} \nonumber \] This equation describes how the potential of a redox system (such as a galvanic cell) varies from its standard state value, specifically, showing it to be a function of the number of electrons transferred, n , the temperature, T , and the reaction mixture composition as reflected in Q . A convenient form of the Nernst equation for most work is one in which values for the fundamental constants (R and F) and standard temperature (298) K), along with a factor converting from natural to base-10 logarithms, have been included: \[E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0592 V }{n} \log Q \label{Nernst at room temperature} \] Example $\PageIndex{2}$: Predicting Redox Spontaneity Under Nonstandard Conditions Use the Nernst equation to predict the spontaneity of the redox reaction shown below at room temperature. \[\ce{Co(s)} + \ce{Fe^{2+}} \text{(aq, 1.94 M)} \longrightarrow \ce{Co^{2+}}\text{(aq, 0.15 M)} + \ce{Fe (s)} \nonumber \] Solution Collecting information from Appendix L and the problem, \[\begin{align} & \text { anode: } \quad && \ce{Co(s) -> Co^{2+}(aq) + 2 e^{-}} &&E^o_{\ce{Co^{2+}/Co}} =-0.28\,\text{V}\\[4pt] & \text {cathode: } \quad && \ce{Fe^{2+}(aq) + 2 e^{-} -> Fe(s)} &&E^o_{\ce{Fe^{+}/Fe}} = -0.447\,\text{V}\\[4pt] \hline &\text { cell: } \quad && \ce{Co(s) + Fe^{2+}(Aq) -> Fe(s) + Co^{2+}(aq)} && E^o_{\text {cell }} = -0.17\,\text{V} \end{align} \nonumber \] with \[E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}=-0.447 V -(-0.28 V )=-0.17 V \nonumber \] Notice the negative value of the standard cell potential indicates the process is not spontaneous under standard conditions. Substitution of the Nernst equation terms for the nonstandard conditions requires calculating the reaction quotient $Q$ \[Q =\frac{\left[\ce{Co^{2+}}\right]}{\left[\ce{Fe^{2+}}\right]}=\dfrac{0.15 M}{1.94 M}=0.077 \nonumber \] them $Q$ and $n$ are substituted into Equation \red{Nernst at room temperature} \[\begin{aligned} E_{\text {cell }} & =E_{\text {cell }}^{\circ}-\frac{0.0592 \,\text{V} }{n} \log Q \\[4pt] & =-0.17 \,\text{V} -\frac{0.0592 \,\text{V} }{2} \log 0.077 \\[4pt] & =-0.17 \,\text{V} +0.033 \,\text{V} \\[4pt] &=-0.14\,\text{V} \end{aligned} \nonumber \] The cell potential remains negative (slightly) under the specified conditions, and so the reaction remains nonspontaneous. Exercise $\PageIndex{2}$ For the cell schematic below at room temperature, identify values for n and Q , and calculate the cell potential, E cell . \[\ce{Al(s)} \| \ce{Al^{3+}}\text{(aq, 0.15 M)} \| \ce{Cu^{2+}}\text{(aq, 0.025 M)} \| \ce{Cu(s)} \nonumber \] Answer n = 6; Q = 1440; E cell = +1.97 V, spontaneous. A concentration cell is constructed by connecting two nearly identical half-cells, each based on the same half-reaction and using the same electrode, varying only in the concentration of one redox species. The potential of a concentration cell, therefore, is determined only by the difference in concentration of the chosen redox species. The example problem below illustrates the use of the Nernst equation in calculations involving concentration cells. Example $\PageIndex{3}$: Concentration Cells What is the cell potential of the concentration cell operating at room temperature described by \[\ce{Zn(s)} \| \ce{Zn^{2+}}\text{(aq, 0.10 M)} \| \ce{Zn^{2+}} \text{(aq, 0.50 M)} \| \ce{Zn(s)} \nonumber \] Solution From the information given: \[\begin{align} & \text { anode: } \quad && \ce{Zn(s) -> Zn^{2+}}\text{(aq, 0.10 M)} + \ce{2 e^{-}} &&E^o_{\ce{Co^{2+}/Co}} =-0.7618\,\text{V}\\[4pt] & \text {cathode: } \quad && \ce{Zn^{2+}}\text{(aq, 0.50 M)} + \ce{2 e^{-} -> Zn(s)} &&E^o_{\ce{Fe^{+}/Fe}} = -0,7618\,\text{V}\\[4pt] \hline &\text { cell: } \quad && \ce{\cancel{Zn(s)}} + \ce{Zn^{2+}} \text{(aq, 0.50 M)} \ce{-> Zn^{2+}}\text{(aq, 0.10 M)} + \cancel{\ce{Zn(s)}} && E^o_{\text {cell }} = 0\,\text{V} \end{align} \nonumber \] Substituting into the Nernst equation (Equation \ref{Nernst at room temperature}), \[E_{\text {cell }}=0.000\,\text{V} -\frac{0.0592\,\text{V} }{2} \log \dfrac{0.10}{0.50} = +0.021\, \text{V} \nonumber \] The positive value for cell potential indicates the overall cell reaction (see above) is spontaneous. This spontaneous reaction is one in which the zinc ion concentration in the cathode falls (it is reduced to elemental zinc) while that in the anode rises (it is produced by oxidation of the zinc anode). A greater driving force for zinc reduction is present in the cathode, where the zinc(II) ion concentration is greater ( E cathode > E anode ). Exercise $\PageIndex{1}$ The concentration cell above was allowed to operate until the cell reaction reached equilibrium at room temperature. What are the cell potential and the concentrations of zinc(II) in each half-cell for the cell now? Answer E cell = 0.000 V; [Zn 2 + ] cathode = [Zn 2 + ] anode = 0.30 M
Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/4_f-Block_Elements/The_Lanthanides/aLanthanides%3A_Properties_and_Reactions	The Lanthanides consist of the elements in the f-block of period six in the periodic table . While these metals can be considered transition metals, they have properties that set them apart from the rest of the elements. Introduction The Lanthanides were first discovered in 1787 when a unusual black mineral was found in Ytterby, Sweden. This mineral, now known as Gadolinite, was later separated into the various Lanthanide elements. In 1794, Professor Gadolin obtained yttria, an impure form of yttrium oxide, from the mineral. In 1803, Berzelius and Klaproth secluded the first Cerium compound. Later, Moseley used an x-ray spectra of the elements to prove that there were fourteen elements between Lanthanum and Hafnium. The rest of the elements were later separated from the same mineral. These elements were first classified as ‘rare earth’ due to the fact that obtained by reasonably rare minerals. However, this is can be misleading since the Lanthanide elements have a practically unlimited abundance. The term Lanthanides was adopted, originating from the first element of the series, Lanthanum. Like any other series in the periodic table, such as the Alkali metals or the Halogens , the Lanthanides share many similar characteristics. These characteristics include the following: Similarity in physical properties throughout the series Adoption mainly of the +3 oxidation state. Usually found in crystalline compounds) They can also have an oxidation state of +2 or +4, though some lanthanides are most stable in the +3 oxidation state. Adoption of coordination numbers greater than 6 (usually 8-9) in compounds Tendency to decreasing coordination number across the series A preference for more electronegative elements (such as O or F) binding Very small crystal-field effects Little dependence on ligands Ionic complexes undergo rapid ligand-exchange Electron Configuration Similarly, the Lanthanides have similarities in their electron configuration, which explains most of the physical similarities. These elements are different from the main group elements in the fact that they have electrons in the f orbital. After Lanthanum, the energy of the 4f sub-shell falls below that of the 5d sub-shell. This means that the electron start to fill the 4f sub-shell before the 5d sub-shell. The electron configurations of these elements were primarily established through experiments. The technique used is based on the fact that each line in an emission spectrum reveals the energy change involved in the transition of an electron from one energy level to another. However, the problem with this technique with respect to the Lanthanide elements is the fact that the 4f and 5d sub-shells have very similar energy levels, which can make it hard to tell the difference between the two. Another important feature of the Lanthanides is the Lanthanide Contraction , in which the 5s and 5p orbitals penetrate the 4f sub-shell. This means that the 4f orbital is not shielded from the increasing nuclear change, which causes the atomic radius of the atom to decrease that continues throughout the series. Symbol Idealized Observed Symbol.1 Idealized.1 Observed.1 La 5d16s2 5d16s2 Tb 4f85d16s2 4f9 6s2 or 4f85d16s2 Ce 4f15d16s2 4f15d16s2 Dy 4f95d16s2 4f10 6s2 Pr 4f25d16s2 4f3 6s2 Ho 4f105d16s2 4f11 6s2 Nd 4f35d16s2 4f4 6s2 Er 4f115d16s2 4f12 6s2 Pm 4f45d16s2 4f5 6s2 Tm 4f125d16s2 4f13 6s2 Sm 4f55d16s2 4f6 6s2 Yb 4f135d16s2 4f14 6s2 Eu 4f65d16s2 4f7 6s2 Lu 4f145d16s2 4f145d16s2 Gd 4f75d16s2 4f75d16s2 NaN NaN NaN Properties and Chemical Reactions One property of the Lanthanides that affect how they will react with other elements is called the basicity. Basicity is a measure of the ease at which an atom will lose electrons. In another words, it would be the lack of attraction that a cation has for electrons or anions. In simple terms, basicity refers to have much of a base a species is. For the Lanthanides, the basicity series is the following: La 3 + > Ce 3 + > Pr 3 + > Nd 3 + > Pm 3 + > Sm 3 + > Eu 3 + > Gd 3 + > Tb 3 + > Dy 3 + > Ho 3 + > Er 3 + > Tm 3 + > Yb 3 + > Lu 3+ In other words, the basicity decreases as the atomic number increases. Basicity differences are shown in the solubility of the salts and the formation of the complex species. Another property of the Lanthanides is their magnetic characteristics. The major magnetic properties of any chemical species are a result of the fact that each moving electron is a micromagnet. The species are either diamagnetic, meaning they have no unpaired electrons, or paramagnetic, meaning that they do have some unpaired electrons. The diamagnetic ions are: La 3 + , Lu 3 + , Yb 2 + and Ce 4 + . The rest of the elements are paramagnetic. Metals and their Alloys The metals have a silvery shine when freshly cut. However, they can tarnish quickly in air, especially Ce, La and Eu. These elements react with water slowly in cold, though that reaction can happen quickly when heated. This is due to their electropositive nature. The Lanthanides have the following reactions: oxidize rapidly in moist air dissolve quickly in acids reaction with oxygen is slow at room temperature, but they can ignite around 150-200 °C react with halogens upon heating upon heating, react with S, H, C and N Symbol Ionization Energy (kJ/mol) Melting Point (°C) Boiling Point (°C) La 538 920.0 3469.0 Ce 527 795.0 3468.0 Pr 523 935.0 3127.0 Nd 529 1024.0 3027.0 Pm 536 NaN NaN Sm 543 1072.0 1900.0 Eu 546 826.0 1429.0 Gd 593 1312.0 3000.0 Tb 564 1356.0 2800.0 Dy 572 1407.0 2600.0 Ho 581 1461.0 2600.0 Er 589 1497.0 2900.0 Tm 597 1545.0 1727.0 Yb 603 824.0 1427.0 Lu 523 1652.0 3327.0 Periodic Trends: Size The size of the atomic and ionic radii is determined by both the nuclear charge and by the number of electrons that are in the electronic shells. Within those shells, the degree of occupancy will also affect the size. In the Lanthanides, there is a decrease in atomic size from La to Lu. This decrease is known as the Lanthanide Contraction . The trend for the entire periodic table states that the atomic radius decreases as you travel from left to right. Therefore, the Lanthanides share this trend with the rest of the elements. Element Atomic Radius (pm) Ionic Radius (3+) Element.1 Atomic Radius (pm).1 Ionic Radius (3+).1 La 187.7 106.1 Tb 178.2 92.3 Ce 182.0 103.4 Dy 177.3 90.8 Pr 182.8 101.3 Ho 176.6 89.4 Nd 182.1 99.5 Er 175.7 88.1 Pm 181.0 97.9 Tm 174.6 89.4 Sm 180.2 96.4 Yb 194.0 85.8 Eu 204.2 95.0 Lu 173.4 84.8 Gd 180.2 93.8 NaN NaN NaN Color and Light Absorbance The color that a substance appears is the color that is reflected by the substance. This means that if a substance appears green, the green light is being reflected. The wavelength of the light determines if the light with be reflected or absorbed. Similarly, the splitting of the orbitals can affect the wavelength that can be absorbed. This is turn would be affected by the amount of unpaired electrons. Ion Unpaired Electrons Color Ion.1 Unpaired Electrons.1 Color.1 La3+ 0 Colorless Tb3+ 6.0 Pale Pink Ce3+ 1 Colorless Dy3+ 5.0 Yellow Pr3+ 2 Green Ho3+ 4.0 Pink; yellow Nd3+ 3 Reddish Er3+ 3.0 Reddish Pm3+ 4 Pink; yellow Tm3+ 2.0 Green Sm3+ 5 Yellow Yb3+ 1.0 Colorless Eu3+ 6 Pale Pink Lu3+ 0.0 Colorless Gd3+ 7 Colorless NaN NaN NaN Occurrence in Nature Each known Lanthanide mineral contains all the members of the series. However, each mineral contains different concentrations of the individual Lanthanides. The three main mineral sources are the following: Monazite: contains mostly the lighter Lanthanides. The commercial mining of monazite sands in the United States is centered in Florida and the Carolinas Xenotime: contains mostly the heavier Lanthanides Euxenite: contains a fairly even distribution of the Lanthanides In all the ores, the atoms with a even atomic number are more abundant. This allows for more nuclear stability, as explained in the Oddo-Harkins rule. The Oddo-Harkins rule simply states that the abundance of elements with an even atomic number is greater than the abundance of elements with an odd atomic number. In order to obtain these elements, the minerals must go through a separating process, known as separation chemistry. This can be done with selective reduction or oxidation. Another possibility is an ion-exchange method. The Oddo-Harkins Rule The abundance of elements with an even atomic number is greater than the abundance of elements with an odd atomic number. Applications Metals and Alloys The pure metals of the Lanthanides have little use. However, the alloys of the metals can be very useful. For example, the alloys of Cerium have been used for metallurgical applications due to their strong reducing abilities. Non-nuclear The Lanthanides can also be used for ceramic purposes. The almost glass-like covering of a ceramic dish can be created with the lanthanides. They are also used to improve the intensity and color balance of arc lights. Nuclear Like the Actinides, the Lanthanides can be used for nuclear purposes. The hydrides can be used as hydrogen-moderator carriers. The oxides can be used as diluents in nuclear fields. The metals are good for being used as structural components. The can also be used for structural-alloy-modifying components of reactors. It is also possible for some elements, such as Tm, to be used as portable x-ray sources. Other elements, such as Eu, can be used as radiation sources. Practice Problems Which elements are considered to be Lanthanides? How do the Lanthanides react with oxygen? What causes the Lanthanide Contraction? Why do Lanthanides exhibit strong electromagnetic and light properties? What do the Lanthanides have in common with the Noble Gases? Answers Elements Lanthanum (57) through Lutetium (71) on the periodic table are considered to be Lanthanides. Lanthanides tend to react with oxygen to form oxides. The reaction at room temperature can be slow while heat can cause the reaction to happen rapidly. The Lanthanide Contraction refers to the decrease in atomic size of the elements in which electrons fill the f-subshell. Since the f sub-shell is not shielded, the atomic size will decrease as the nuclear charge still increases. Lanthanides exhibit strong electromagnetic and light properties because of the presence of unpaired electrons in the f-orbitals. The majority of the Lanthanides are paramagnetic, which means that they have strong magnetic fields. Both the Lanthanides and Noble Gases tend to bind with more electronegative atoms, such as Oxygen or Fluorine. References Petrucci, Hardwood, Herring. "General Chemistry: Principles & Modern Applications". New Jersey: Macmillan Publishing Company, 2007. Moeller, Therald. The Chemistry of the Lanthanides . New York: Reinhold Publishing Corporation, 1963. Cotton, Simon. Lanthanides and Actinides . London: Macmillan Education Ltd, 1991.
Courses/Sacramento_City_College/SCC%3A_CHEM_300_-_Beginning_Chemistry/SCC%3A_CHEM_300_-_Beginning_Chemistry_(Faculty)/01%3A_The_Chemical_World/1.05%3A_A_Beginning_Chemist-_How_to_Succeed	Examples of the practical applications of chemistry are everywhere ( Figure $\PageIndex{1}$ ). Engineers need to understand the chemical properties of the substances needed to design biologically compatible implants for joint replacements; or to design roads, bridges, buildings, and nuclear reactors that do not collapse because of weakened structural materials such as steel and cement. Archeology and paleontology rely on chemical techniques to date bones and artifacts and identify their origins. Although law is not normally considered a field related to chemistry, forensic scientists use chemical methods to analyze blood, fibers, and other evidence as they investigate crimes. In particular, DNA matching—comparing biological samples of genetic material to see whether they could have come from the same person—has been used to solve many high-profile criminal cases as well as clear innocent people who have been wrongly accused or convicted. Forensics is a rapidly growing area of applied chemistry. In addition, the proliferation of chemical and biochemical innovations in industry is producing rapid growth in the area of patent law. Ultimately, the dispersal of information in all the fields in which chemistry plays a part requires experts who are able to explain complex chemical issues to the public through television, print journalism, the Internet, and popular books. Hopefully at this point you are fully convinced of how important and useful the study of chemistry can be. You may, however, still be wondering exactly what it is that a chemist does. Chemistry is the study of matter and the changes that matter undergoes. In general, chemists are interested in both characteristics that you can test and observe, like a chemical's smell or color, and characteristics that are far too small to see, like what the oxygen you breathe in or the carbon dioxide you breath out looks like under a microscope 1,000 times more powerful than any existing in the world today. Wait a minute… how can a chemist know what oxygen and carbon dioxide look like under a microscope that doesn't even exist? What happened to the scientific method? What happened to relying on observations and careful measurements ? In fact, because chemists can't see the underlying structure of different materials, they have to rely on the scientific method even more! Chemists are a lot like detectives. Suppose a detective is trying to solve a murder case—what do they do? Obviously, the detective starts by visiting the site of the crime and looking for evidence. If the murderer has left enough clues behind, the detective can piece together a theory explaining what happened. Even though the detective wasn't at the crime scene when the crime was committed and didn't actually see the murderer kill the victim, with the right evidence, the detective can be pretty sure of how the crime took place. It is the same with chemistry. When chemists go into the laboratory, they collect evidence by making measurements. Once chemists have collected enough clues from the properties that they can observe, they use that evidence to piece together a theory explaining the properties that they cannot observe—the properties that are too small to see. What kinds of properties do chemists actually measure in the laboratory? Well, you can probably guess a few. Imagine that you go to dinner at a friend's house and are served something that you don't recognize, what types of observations might you make to determine exactly what you've been given? You might smell the food. You might note the color of the food. You might try to decide whether the food is a liquid or a solid because if it's a liquid, it's probably soup or a drink. The temperature of the food could be useful if you wanted to know whether or not you had been served ice cream! You could also pick up a small amount of food with your fork and try to figure out how much it weighs—a light dessert might be something like an angel cake, while a heavy dessert is probably a pound cake. The quantity of food you have been given might be a clue too. Finally, you might want to know something about the food's texture—is it hard and granular like sugar cubes, or soft and easy to spread, like butter? Believe it or not, the observations you are likely to make when trying to identify an unknown food are very similar to the observations that a chemist makes when trying to learn about a new material. Chemists rely on smell, color, state (whether it is a solid or liquid or gas), temperature, volume, mass (which is related to weight—as will be discussed in a later section), and texture. There is, however, one property possibly used to learn about a food, but that should definitely not be used to learn about a chemical—taste! In the sections on the Atomic Theory, you will see exactly how measurements of certain properties helped early scientists to develop theories about the chemical structure of matter on a scale much smaller than they could ever hope to see. You will also learn how these theories, in turn, allow us to make predictions about new materials that humankind has not yet created. The video below gives you some important tips on how to study chemistry in this class. With practice, you too can learn to think like a chemist, and you may even enjoy it! Contributions & Attributions
Courses/Lumen_Learning/First_Year_Seminar_(Lumen)/04%3A_Module_2-_Self-Awareness_Assessment/4.01%3A_Textbook_Preface	“Success doesn’t come to you…you go to it.” This quote by Dr. Marva Collins sets the stage for the journey you are about to take. Your success, however you choose to define it, is waiting for you, and Foundations of Academic Success: Words of Wisdom ( FAS: WoW ) is your guide to your success. Some may believe that success looks like a straight and narrow line that connects the dots between where you are and where you are going, but the truth is that success looks more like a hot mess of twists and turns, curves and bumps, and hurdles and alternate pathways. Putting this textbook together was challenging because there is so much to tell you as you embark on your college journey. I have worked with college students on academic success at a number of college campuses, and have hunted for the most effective and most affordable college student academic success textbook but could never find everything I wanted to teach in one book. So, I figured the answer was to write my own textbook! Like any good research project, the outcome was not exactly what I expected. In addition to a host of true-to-life stories written by real people who have successfully navigated the journey through college, the first draft of the textbook included everything (and more) that the other similarly themed textbooks about college student academic success do. The chapters were framed by a slew of “How To” facts according to me (such as “how to efficiently take notes during a lecture and how to effectively use your preferred learning style to help you learn better”) and research-based figures according to researchers in the field of college student academic success, such as rules like “For every hour in class, successful college-level work requires about two hours of out-of-class work: reading, writing, research, labs, discussion, field work, etc.” or “A 15-credit course load is about equivalent to working a full-time job.” Once the first draft was finished, I decided to test-drive my new textbook with the students in my First Year Experience class to see what they thought. I figured, who better to give me feedback on the textbook than actual students who would use the textbook in class, right? I gave the first draft of the textbook (facts and figures and all) to my students to read, review, and reflect upon. It turned out that the pieces that my students learned the most from were the true-to-life stories. They either didn’t read or barely glanced over the facts and figures, but provided very positive feedback (and even remembered) the words of wisdom from real people who have successfully navigated the journey through college. I guess it makes sense; students love when real-life stories are infused into the activities and lessons. Plus, as a number of students told me, the facts and figures on topics such as note-taking and how many hours to study per week can be found by searching online and can vary per person. What really mattered to students were the real-life words of wisdom that you can’t find online. Thus, Foundations of Academic Success: Words of Wisdom ( FAS: WoW as I lovingly call it) emerged. I share this story because my intended outcome (to be the author of the world’s best open access college student academic success textbook) was not exactly what I expected it to be. The same is true of your journey through college, and you’ll read more about that in the stories right here in FAS: WoW . You’ll find that this is not your typical college textbook full of concrete facts and figures, nor does it tell you how to succeed. No textbook can truly do that for you—success is defined differently for everyone. The stories in FAS: WoW are relevant, relational, and reflective. The authors welcome you into their lives and offer ideas that ignite helpful discussions that will help succeed. FAS: WoW introduces you to the various aspects of student and academic life on campus and prepares you to thrive as a successful college student (since there is a difference between a college student and a successful college student). Each section of FAS: WoW is framed by self-authored, true-to-life short stories from actual State University of New York (SUNY) students, employees, and alumni. You may even know some of the authors! The advice they share includes a variety of techniques to help you cope with the demands of college. The lessons learned are meant to enlarge your awareness of self with respect to your academic and personal goals and assist you to gain the necessary skills to succeed in college. In the text, the authors tell stories about their own academic, personal, and life-career successes. When reading FAS: WoW , consider the following guiding questions: How do you demonstrate college readiness through the use of effective study skills and campus resources? How do you apply basic technological and information management skills for academic and lifelong career development? How do you demonstrate the use of critical and creative thinking skills to solve problems and draw conclusions? How do you demonstrate basic awareness of self in connection with academic and personal goals? How do you identify and demonstrate knowledge of the implications of choices related to wellness? How do you demonstrate basic knowledge of cultural diversity? After you read each story, take the time to reflect on the lessons learned from your reading and answer the guiding questions as they will help you to connect the dots between being a college student and being a successful college student. Note your areas of strength and your areas of weakness, and develop a plan to turn your weaknesses into strengths. I could go on and on (and on) about college student academic success, but what fun is the journey for you if I tell you everything now? You need to learn some stuff on your own, right? So, I will leave you to read and enjoy FAS: WoW with a list of tips that I share with college students as they embark on their journey to academic success: Early is on time , on time is late , and late is unacceptable ! Get the book(s) and read the book(s). Take notes in class and when reading for class. Know your professors (email, office location, office hours, etc.) and be familiar with what is in the course syllabus. Put your phone away in class. Emails need a salutation, a body, and a close. Don’t write the way you might text—using abbreviations and clipped sentences. Never academically advise yourself! Apply for scholarships… all of them ! Speak it into existence and keep your eyes on the prize. Enjoy the ride! Cheers, TOM Dr. Thomas C. Priester, [email protected] CC licensed content, Shared previously Foundations of Academic Success: Words of Wisdom. Authored by : Thomas C. Priester. Located at : http://textbooks.opensuny.org/foundations-of-academic-success/ . Project : Open SUNY Textbooks. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike
Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/11%3A_Liquids_and_Intermolecular_Forces/11.01%3A_A_Molecular_Comparison_of_Gases_Liquids_and_Solids	Learning Objectives To be familiar with the kinetic molecular description of liquids. The physical properties of a substance depends upon its physical state. Water vapor, liquid water and ice all have the same chemical properties, but their physical properties are considerably different. In general covalent bonds determine: molecular shape, bond energies, chemical properties, while intermolecular forces (non-covalent bonds) influence the physical properties of liquids and solids. The kinetic molecular theory of gases gives a reasonably accurate description of the behavior of gases. A similar model can be applied to liquids, but it must take into account the nonzero volumes of particles and the presence of strong intermolecular attractive forces. The state of a substance depends on the balance between the kinetic energy of the individual particles (molecules or atoms) and the intermolecular forces . The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance. The intermolecular forces are attractive forces that try to draw the particles together (Figure $\PageIndex{2}$). A discussed previously, gasses are very sensitive to temperatures and pressure. However, these also affect liquids and solids too. Heating and cooling can change the kinetic energy of the particles in a substance, and so, we can change the physical state of a substance by heating or cooling it. Increasing the pressure on a substance forces the molecules closer together, which increases the strength of intermolecular forces. Below is an overview of the general properties of the three different phases of matter. Properties of Gases A collection of widely separated molecules The kinetic energy of the molecules is greater than any attractive forces between the molecules The lack of any significant attractive force between molecules allows a gas to expand to fill its container If attractive forces become large enough, then the gases exhibit non-ideal behavior Properties of Liquids The intermolecular attractive forces are strong enough to hold molecules close together Liquids are more dense and less compressible than gasses Liquids have a definite volume, independent of the size and shape of their container The attractive forces are not strong enough, however, to keep neighboring molecules in a fixed position and molecules are free to move past or slide over one another Thus, liquids can be poured and assume the shape of their containers. Properties of Solids The intermolecular forces between neighboring molecules are strong enough to keep them locked in position Solids (like liquids) are not very compressible due to the lack of space between molecules If the molecules in a solid adopt a highly ordered packing arrangement, the structures are said to be crystalline Due to the strong intermolecular forces between neighboring molecules, solids are rigid. Cooling a gas may change the state to a liquid Cooling a liquid may change the state to a solid Increasing the pressure on a gas may change the state to a liquid Increasing the pressure on a liquid may change the state to a solid Physical Properties of Liquids In a gas, the distance between molecules, whether monatomic or polyatomic, is very large compared with the size of the molecules; thus gases have a low density and are highly compressible. In contrast, the molecules in liquids are very close together, with essentially no empty space between them. As in gases, however, the molecules in liquids are in constant motion, and their kinetic energy (and hence their speed) depends on their temperature. We begin our discussion by examining some of the characteristic properties of liquids to see how each is consistent with a modified kinetic molecular description. The properties of liquids can be explained using a modified version of the kinetic molecular theory of gases described previously. This model explains the higher density, greater order, and lower compressibility of liquids versus gases; the thermal expansion of liquids; why they diffuse; and why they adopt the shape (but not the volume) of their containers. A kinetic molecular description of liquids must take into account both the nonzero volumes of particles and the presence of strong intermolecular attractive forces. Solids and liquids have particles that are fairly close to one another, and are thus called " condensed phases " to distinguish them from gases Density : The molecules of a liquid are packed relatively close together. Consequently, liquids are much denser than gases. The density of a liquid is typically about the same as the density of the solid state of the substance. Densities of liquids are therefore more commonly measured in units of grams per cubic centimeter (g/cm 3 ) or grams per milliliter (g/mL) than in grams per liter (g/L), the unit commonly used for gases. Molecular Order: Liquids exhibit short-range order because strong intermolecular attractive forces cause the molecules to pack together rather tightly. Because of their higher kinetic energy compared to the molecules in a solid, however, the molecules in a liquid move rapidly with respect to one another. Thus unlike the ions in the ionic solids, the molecules in liquids are not arranged in a repeating three-dimensional array. Unlike the molecules in gases, however, the arrangement of the molecules in a liquid is not completely random. Compressibility : Liquids have so little empty space between their component molecules that they cannot be readily compressed. Compression would force the atoms on adjacent molecules to occupy the same region of space. Thermal Expansion : The intermolecular forces in liquids are strong enough to keep them from expanding significantly when heated (typically only a few percent over a 100°C temperature range). Thus the volumes of liquids are somewhat fixed. Notice from Table S1 (with a shorten version in Table $\PageIndex{1}$) that the density of water, for example, changes by only about 3% over a 90-degree temperature range. T (°C) Density (g/cm3) 0 0.99984 30 0.99565 60 0.98320 90 0.96535 Diffusion : Molecules in liquids diffuse because they are in constant motion. A molecule in a liquid cannot move far before colliding with another molecule, however, so the mean free path in liquids is very short, and the rate of diffusion is much slower than in gases. Fluidity : Liquids can flow, adjusting to the shape of their containers, because their molecules are free to move. This freedom of motion and their close spacing allow the molecules in a liquid to move rapidly into the openings left by other molecules, in turn generating more openings, and so forth (Figure $\PageIndex{3}$).
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/An_Introduction_to_the_Electronic_Structure_of_Atoms_and_Molecules_(Bader)/06%3A_The_Chemical_Bond/6.E%3A_Exercises	Q6.1 The element beryllium has an atomic number of four. Rationalize the following observations in terms of the valence bond theory of molecular structure. $Be_2$ does not exist except as a weakly bound van der Waals molecule. $Be$ can exhibit a valency of two in combination with a halogen, for example, BeF 2 . $BeF_2$ can undergo a further reaction with an excess of F - ions to give \[BeF_2 + 2F^- \rightarrow BeF_4^{-2} \nonumber \] In addition to explaining why this reaction occurs, predict the geometrical shape of the $BeF_4^{-2}$ ion. Q6.2 Use valence bond theory to predict the molecular formula and geometrical structure of the most stable electrically neutral hydride of phosphorus. The hydride of phosphorus can react with HI to form an ionic crystal which contains the I - ion. Explain why this reaction can occur and give the formula and geometrical structure of the positive ion which contains phosphorus and hydrogen. Q6.3 The atomic number of silicon is fourteen. What is the electronic structure of Si in its ground state? Predict the molecular formula and geometrical shape of the most stable silicon-hydrogen compound using valence bond theory. Q6.4 The element vanadium (Z = 23) forms the compound VCl 4 . Would a beam of VCl 4 molecules be deflected in an inhomogeneous magnetic field? Explain the reasoning behind your answer. 5. The CH 2 molecule may exist in two distinct forms. In the one case all the electrons are paired and the molecule does not possess a magnetic moment. In the second form the molecule exhibits a magnetism which can be shown to arise from the presence of two unpaired electrons. One of the forms of CH 2 is linear. Use valence bond theory to describe the electronic structures and geometries of both forms of CH 2 . Which of the two will possess the lower electronic energy? Q6.6 Write Lewis structures (structures in which each electron pair bond is designated by a line joining the nuclei and dots are used to designate unshared electrons in the valency shell) for H 2 O, CH 4 , CO 2 , HF, NH 4 + , H 2 O 2 . Give a discussion of the bonding of the molecules listed in part (a) in terms of valence bond theory. Denote the use of hybrid orbitals by arrows and a label as to whether they are sp , sp 2 , or sp 3 hybrids. You should predict that H 2 O and H 2 O 2 are bent molecules, that CH 4 and NH 4 + are tetrahedral and that CO 2 is linear. Q6.7 Sometimes it is possible to write a number of equivalent Lewis structures for a single species. For example, the bonding in the NO 3 - ion can be described by: Each atom in these structures is surrounded by four pairs of electrons, the first cardinal rule in writing a Lewis structure. On the average, one electron of the pair in each bond belongs to one atom. Since there are only four bonds to N and no unshared valence pairs, N on the average has but four valence electrons in these three structures. The N atom initially possessed five electrons, and a plus sign is placed at N to denote that it has, on the average, one less electron in the NO 3 - ion. The two singly-bonded oxygens have on the average seven electrons in each structure, one more than a neutral oxygen atom. This is denoted by a minus sign. The doubly-bonded O has on the average six electrons. Notice that the sum of these formal charges is minus one, the correct charge for the NO 3 - ion. The structure of the NO 3 - ion is in reality planar and symmetrical, all of the NO bonds being of equal length. This could be indicated in a single Lewis structure by indicating that the final pair of electrons in the p bond between N and one O is actually spread over all three NO bonds simultaneously: When one or more pairs of electrons are delocalized over more than two atoms, the Lewis method or the valence bond method of writing valence structures with bonds between pairs of atoms runs into difficulties. The compromise structure above correctly indicates that each NO bond in NO 3 - is stronger and shorter than a N�O single bond, but not as strong as an N=O double bond. Use the valence bond theory to account for the bonding and planar structure of the NO 3 - ion. Write Lewis structures and the corresponding valence bond structures for the CO 3 -2 ion and SO 2 . Are there full S=O or C=O double bonds in either of these molecules? Q6.8 Draw valence bond structures for benzene, C 6 H 6 . This molecule has a planar hexagonal geometry: Are there any delocalized electron pairs in the benzene molecule? Q6.9 The carbon monoxide molecule forms stable complexes with many transition metal elements. Examples are (from the first transition metal series) Cr(CO) 6 , (CO) 5 Mn�Mn(CO) 5 , Fe(CO) 5 , Ni(CO) 4 In each case the bond is formed between the metal and the unshared pair of electrons on the carbon end of carbon monoxide. The metal atom in these complexes obviously violates the octet rule, but can the electronic structures for the carbon monoxide complexes be rationalized on the basis of an expanded valency shell for the metal?
Courses/City_College_of_San_Francisco/Foundations_-_Review_Source_for_Chem_101A/04%3A_Stoichiometry_of_Chemical_Reactions/4.02%3A_Reaction_Stoichiometry	Learning Objectives Explain the concept of stoichiometry as it pertains to chemical reactions Use balanced chemical equations to derive stoichiometric factors relating amounts of reactants and products Perform stoichiometric calculations involving mass, moles, and solution molarity A balanced chemical equation provides a great deal of information in a very succinct format. Chemical formulas provide the identities of the reactants and products involved in the chemical change, allowing classification of the reaction. Coefficients provide the relative numbers of these chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. These quantitative relationships are known as the reaction’s stoichiometry , a term derived from the Greek words stoicheion (meaning “element”) and metron (meaning “measure”). In this module, the use of balanced chemical equations for various stoichiometric applications is explored. The general approach to using stoichiometric relationships is similar in concept to the way people go about many common activities. Cooking, for example, offers an appropriate comparison. Suppose a recipe for making eight pancakes calls for 1 cup pancake mix, $\dfrac{3}{4}$ cup milk, and one egg. The “equation” representing the preparation of pancakes per this recipe is \[\mathrm{1\:cup\: mix+\dfrac{3}{4}\:cup\: milk+1\: egg \rightarrow 8\: pancakes} \label{4.4.1}\] If two dozen pancakes are needed for a big family breakfast, the ingredient amounts must be increased proportionally according to the amounts given in the recipe. For example, the number of eggs required to make 24 pancakes is \[\mathrm{24\: \cancel{pancakes} \times \dfrac{1\: egg}{8\: \cancel{pancakes}}=3\: eggs} \label{4.4.2}\] Balanced chemical equations are used in much the same fashion to determine the amount of one reactant required to react with a given amount of another reactant, or to yield a given amount of product, and so forth. The coefficients in the balanced equation are used to derive stoichiometric factors that permit computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen: \[\ce{N2}(g)+\ce{3H2}(g)\rightarrow \ce{2NH3}(g) \label{4.4.3}\] This equation shows that ammonia molecules are produced from hydrogen molecules in a 2:3 ratio, and stoichiometric factors may be derived using any amount (number) unit: \[\ce{\dfrac{2NH3 \: molecules}{3H2 \: molecules}\: or \: \dfrac{2 \: doz \: NH3\: molecules}{3\: doz\:H2 \:molecules} \: or \: \dfrac{2\: mol\: NH3\: molecules}{3\: mol\: H2\: molecules}} \label{4.4.4}\] These stoichiometric factors can be used to compute the number of ammonia molecules produced from a given number of hydrogen molecules, or the number of hydrogen molecules required to produce a given number of ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation. Example $\PageIndex{1}$: Moles of Reactant Required in a Reaction How many moles of I 2 are required to react with 0.429 mol of Al according to the following equation (see Figure $\PageIndex{2}$)? \[\ce{2Al + 3I2 \rightarrow 2AlI3} \label{4.4.5}\] Solution Referring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is $\ce{\dfrac{3\: mol\: I2}{2\: mol\: Al}}$. The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor: \[\begin{align} \mathrm{mol\: I_2} &=\mathrm{0.429\: \cancel{mol\: Al}\times \dfrac{3\: mol\: I_2}{2\:\cancel{mol\: Al}}} \\[4pt] &=\mathrm{0.644\: mol\: I_2} \end{align}\] Exercise $\PageIndex{1}$ How many moles of Ca(OH) 2 are required to react with 1.36 mol of H 3 PO 4 to produce Ca 3 (PO 4 ) 2 according to the equation $\ce{3Ca(OH)2 + 2H3PO4 \rightarrow Ca3(PO4)2 + 6H2O}$ ? Answer 2.04 mol Example $\PageIndex{2}$: Number of Product Molecules Generated by a Reaction How many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation? \[\ce{C3H8 + 5O2 \rightarrow 3CO2 + 4H2O} \label{4.4.6}\] S olution The approach here is the same as for Example $\PageIndex{1}$, though the absolute number of molecules is requested, not the number of moles of molecules. This will simply require use of the moles-to-numbers conversion factor, Avogadro’s number. The balanced equation shows that carbon dioxide is produced from propane in a 3:1 ratio: \[\ce{\dfrac{3\: mol\: CO2}{1\: mol\: C3H8}} \label{4.4.7}\] Using this stoichiometric factor, the provided molar amount of propane, and Avogadro’s number, \[\mathrm{0.75\: \cancel{mol\: C_3H_8}\times \dfrac{3\: \cancel{mol\: CO_2}}{1\:\cancel{mol\:C_3H_8}}\times \dfrac{6.022\times 10^{23}\:CO_2\:molecules}{\cancel{mol\:CO_2}}=1.4\times 10^{24}\:CO_2\:molecules} \label{4.4.8}\] Exercise $\PageIndex{1}$ How many NH 3 molecules are produced by the reaction of 4.0 mol of Ca(OH) 2 according to the following equation: \[\ce{(NH4)2SO4 + Ca(OH)2 \rightarrow 2NH3 + CaSO4 + 2H2O} \label{4.4.9} \nonumber\] Answer 4.8 × 10 24 NH 3 molecules These examples illustrate the ease with which the amounts of substances involved in a chemical reaction of known stoichiometry may be related. Directly measuring numbers of atoms and molecules is, however, not an easy task, and the practical application of stoichiometry requires that we use the more readily measured property of mass. Example $\PageIndex{3}$: Relating Masses of Reactants and Products What mass of sodium hydroxide, NaOH, would be required to produce 16 g of the antacid milk of magnesia [magnesium hydroxide, Mg(OH) 2 ] by the following reaction? $\ce{MgCl2}(aq)+\ce{2NaOH}(aq)\rightarrow \ce{Mg(OH)2}(s)+\ce{2NaCl}(aq)$ S olution The approach used previously in Examples $\PageIndex{1}$ and $\PageIndex{2}$ is likewise used here; that is, we must derive an appropriate stoichiometric factor from the balanced chemical equation and use it to relate the amounts of the two substances of interest. In this case, however, masses (not molar amounts) are provided and requested, so additional steps of the sort learned in the previous chapter are required. The calculations required are outlined in this flowchart: \[\mathrm{16\:\cancel{g\: Mg(OH)_2} \times \dfrac{1\:\cancel{mol\: Mg(OH)_2}}{58.3\:\cancel{g\: Mg(OH)_2}}\times \dfrac{2\:\cancel{mol\: NaOH}}{1\:\cancel{mol\: Mg(OH)_2}}\times \dfrac{40.0\: g\: NaOH}{\cancel{mol\: NaOH}}=22\: g\: NaOH}\] Exercise $\PageIndex{3}$ What mass of gallium oxide, Ga 2 O 3 , can be prepared from 29.0 g of gallium metal? The equation for the reaction is $\ce{4Ga + 3O2 \rightarrow 2Ga2O3}$. Answer 39.0 g Example $\PageIndex{4}$: Relating Masses of Reactants What mass of oxygen gas, O 2 , from the air is consumed in the combustion of 702 g of octane, C 8 H 18 , one of the principal components of gasoline? \[\ce{2C8H18 + 25O2 \rightarrow 16CO2 + 18H2O} \nonumber\] S olution The approach required here is the same as for the Example $\PageIndex{3}$, differing only in that the provided and requested masses are both for reactant species. $\mathrm{702\:\cancel{g\:\ce{C8H18}}\times \dfrac{1\:\cancel{mol\:\ce{C8H18}}}{114.23\:\cancel{g\:\ce{C8H18}}}\times \dfrac{25\:\cancel{mol\:\ce{O2}}}{2\:\cancel{mol\:\ce{C8H18}}}\times \dfrac{32.00\: g\:\ce{O2}}{\cancel{mol\:\ce{O2}}}=2.46\times 10^3\:g\:\ce{O2}}$ Exercise $\PageIndex{4}$ What mass of CO is required to react with 25.13 g of Fe 2 O 3 according to the equation $\ce{Fe2O3 + 3CO \rightarrow 2Fe + 3CO2}$? Answer 13.22 g These examples illustrate just a few instances of reaction stoichiometry calculations. Numerous variations on the beginning and ending computational steps are possible depending upon what particular quantities are provided and sought (volumes, solution concentrations, and so forth). Regardless of the details, all these calculations share a common essential component: the use of stoichiometric factors derived from balanced chemical equations. Figure $\PageIndex{2}$ provides a general outline of the various computational steps associated with many reaction stoichiometry calculations. Airbags Airbags (Figure $\PageIndex{3}$) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, NaN 3 . When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of NaN 3 to initiate its decomposition: \[\ce{2NaN3}(s)\rightarrow \ce{3N2}(g)+\ce{2Na}(s)\] This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03–0.1 s). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function. For example, a small mass (~100 g) of NaN 3 will generate approximately 50 L of N 2 . Summary A balanced chemical equation may be used to describe a reaction’s stoichiometry (the relationships between amounts of reactants and products). Coefficients from the equation are used to derive stoichiometric factors that subsequently may be used for computations relating reactant and product masses, molar amounts, and other quantitative properties. Glossary stoichiometric factor ratio of coefficients in a balanced chemical equation, used in computations relating amounts of reactants and products stoichiometry relationships between the amounts of reactants and products of a chemical reaction
Courses/Pasadena_City_College/CHEM_001A%3A_General_Chemistry_and_Chemical_Analysis/05%3A_Introduction_to_Solutions_and_Aqueous_Reactions/5.06%3A_Representing_Aqueous_Reactions-_Molecular_Ionic_and_Complete_Ionic_Equations	0 1 NaN Howard University General Chemistry: An Atoms First Approach Unit 1: Atomic Theory Unit 2: Molecular Structure Unit 3: Stoichiometry Unit 4: Thermochem & Gases Unit 5: States of Matter Unit 6: Kinetics & Equilibria Unit 7: Electro & Thermo Chemistry Unit 8: Materials Unit 1: Atomic Theory Unit 2: Molecular Structure Unit 3: Stoichiometry Unit 4: Thermochem & Gases Unit 5: States of Matter Unit 6: Kinetics & Equilibria Unit 7: Electro & Thermo Chemistry Unit 8: Materials The chemical equations discussed in Chapter 7 showed the identities of the reactants and the products and gave the stoichiometries of the reactions, but they told us very little about what was occurring in solution. In contrast, equations that show only the hydrated species focus our attention on the chemistry that is taking place and allow us to see similarities between reactions that might not otherwise be apparent. Let’s consider the reaction of silver nitrate with potassium dichromate. As you learned in Example 9, when aqueous solutions of silver nitrate and potassium dichromate are mixed, silver dichromate forms as a red solid. The overall chemical equation A chemical equation that shows all the reactants and products as undissociated, electrically neutral compounds. for the reaction shows each reactant and product as undissociated, electrically neutral compounds: \[2AgNO_3(aq) + K_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + 2KNO_3(aq)\label{8.4.1}\] Although Equation $\ref{8.4.1}$ gives the identity of the reactants and the products, it does not show the identities of the actual species in solution. Because ionic substances such as AgNO 3 and K 2 Cr 2 O 7 are strong electrolytes, they dissociate completely in aqueous solution to form ions. In contrast, because Ag 2 Cr 2 O 7 is not very soluble, it separates from the solution as a solid. To find out what is actually occurring in solution, it is more informative to write the reaction as a complete ionic equation , showing which ions and molecules are hydrated and which are present in other forms and phases: \[2Ag^+(aq) + 2NO_3^-(aq) + 2K^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s) + 2K^+(aq) + 2NO_3^-(aq)\label{8.4.2}\] Note that K + (aq) and NO 3 − (aq) ions are present on both sides of the equation, and their coefficients are the same on both sides. These ions are called spectator ions because they do not participate in the actual reaction. Canceling the spectator ions gives the net ionic equation, which shows only those species that participate in the chemical reaction: \[2Ag^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s)\label{8.4.3}\] Both mass and charge must be conserved in chemical reactions because the numbers of electrons and protons do not change. For charge to be conserved, the sum of the charges of the ions multiplied by their coefficients must be the same on both sides of the equation. In Equation $\ref{8.4.3}$, the charge on the left side is 2(+1) + 1(−2) = 0, which is the same as the charge of a neutral Ag 2 Cr 2 O 7 formula unit. By eliminating the spectator ions, we can focus on the chemistry that takes place in a solution. For example, the overall chemical equation for the reaction between silver fluoride and ammonium dichromate is as follows: \[2AgF(aq) + (NH_4)_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + 2NH_4F(aq)\label{8.4.4}\] The complete ionic equation for this reaction is as follows: \[2Ag^+(aq) + 2F^-(aq) + 2NH_4^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s) + 2NH_4^+(aq) + 2F^-(aq)\label{8.4.5}\] Because two NH 4 + (aq) and two F − (aq) ions appear on both sides of Equation $\ref{8.4.5}$, they are spectator ions. They can therefore be canceled to give the net ionic equation (Equation $\ref{8.4.6}$), which is identical to Equation $\ref{8.4.3}$: \[2Ag^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s)\label{8.4.6}\] If we look at net ionic equations, it becomes apparent that many different combinations of reactants can result in the same net chemical reaction. For example, we can predict that silver fluoride could be replaced by silver nitrate in the preceding reaction without affecting the outcome of the reaction. Example $\PageIndex{1}$ Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous barium nitrate with aqueous sodium phosphate to give solid barium phosphate and a solution of sodium nitrate. Given: reactants and products Asked for: overall, complete ionic, and net ionic equations Strategy: Write and balance the overall chemical equation. Write all the soluble reactants and products in their dissociated form to give the complete ionic equation; then cancel species that appear on both sides of the complete ionic equation to give the net ionic equation. Solution: From the information given, we can write the unbalanced chemical equation for the reaction: \[Ba(NO_3)_2(aq) + Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + NaNO_3(aq)\] Because the product is Ba 3 (PO 4 ) 2 , which contains three Ba 2 + ions and two PO 4 3− ions per formula unit, we can balance the equation by inspection: \[ 3Ba(NO_3)_2(aq) + 2Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + 6NaNO_3(aq) \] This is the overall balanced chemical equation for the reaction, showing the reactants and products in their undissociated form. To obtain the complete ionic equation, we write each soluble reactant and product in dissociated form: \[ 3Ba^{2+}(aq) + 6NO_3^-(aq) + 6Na^+(aq) + 2PO_4^{3-}(aq) \rightarrow Ba_3(PO_4)_2(s) + 6Na^+(aq) + 6NO_3^-(aq) \] The six NO 3 − (aq) ions and the six Na + (aq) ions that appear on both sides of the equation are spectator ions that can be canceled to give the net ionic equation: \[3Ba^{2+}(aq) + 2PO_4^{3-}(aq) \rightarrow Ba_3(PO_4)_2(s)\] Exercise $\PageIndex{1}$ Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous silver fluoride with aqueous sodium phosphate to give solid silver phosphate and a solution of sodium fluoride. Answer: overall chemical equation: \[3AgF(aq) + Na_3PO_4(aq) \rightarrow Ag_3PO_4(s) + 3NaF(aq)\] complete ionic equation: \[3Ag^+(aq) + 3F^-(aq) + 3Na^+(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s) + 3Na^+(aq) + 3F^-(aq)\] net ionic equation: \[3Ag^+(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s)\] So far, we have always indicated whether a reaction will occur when solutions are mixed and, if so, what products will form. As you advance in chemistry, however, you will need to predict the results of mixing solutions of compounds, anticipate what kind of reaction (if any) will occur, and predict the identities of the products. Students tend to think that this means they are supposed to “just know” what will happen when two substances are mixed. Nothing could be further from the truth: an infinite number of chemical reactions is possible, and neither you nor anyone else could possibly memorize them all. Instead, you must begin by identifying the various reactions that could occur and then assessing which is the most probable (or least improbable) outcome. The most important step in analyzing an unknown reaction is to write down all the species—whether molecules or dissociated ions—that are actually present in the solution (not forgetting the solvent itself) so that you can assess which species are most likely to react with one another. The easiest way to make that kind of prediction is to attempt to place the reaction into one of several familiar classifications, refinements of the five general kinds of reactions introduced in Chapter 3 (acid–base, exchange, condensation, cleavage, and oxidation–reduction reactions). In the sections that follow, we discuss three of the most important kinds of reactions that occur in aqueous solutions: precipitation reactions (also known as exchange reactions), acid–base reactions, and oxidation–reduction reactions. Summary The chemical equation for a reaction in solution can be written in three ways. The overall chemical equation shows all the substances present in their undissociated forms; the complete ionic equation shows all the substances present in the form in which they actually exist in solution; and the net ionic equation is derived from the complete ionic equation by omitting all spectator ions , ions that occur on both sides of the equation with the same coefficients. Net ionic equations demonstrate that many different combinations of reactants can give the same net chemical reaction. Key Takeaway A complete ionic equation consists of the net ionic equation and spectator ions. Conceptual Problem What information can be obtained from a complete ionic equation that cannot be obtained from the overall chemical equation? Contributors Modified by Joshua Halpern ( Howard University )
Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Chemiluminescence/2%3A_Chemiluminescence_Reagents/2.01%3A_Luminol	Luminol is the common name for 5-amino-2,3-dihydro-1,4-phthalazinedione (often called 3-aminopthalhydrazide). Oxidation of luminol produces excited 3-aminophthalate, which on relaxation emits light (λ max = 425 nm) with quantum yield of ~0.01 [1] ; Information on the hazards of using luminol is available at the website of the United States National Toxicology Program . The reaction is triggered by a catalytic process, usually enzymatic, provided, for example, by heme-containing proteins, especially horseradish peroxidase (HRP, EC 1.11.1.7). In the presence of hydrogen peroxide this enzyme is converted into intermediary complexes before being regenerated. It has the distinct advantage in biological work of permitting the luminol reaction at pH as low as 8.0 to 8.5. HRP can be used as a label to detect analytes of interest and luminol chemiluminescence can be used to detect substrates of oxidase enzymes that generate hydrogen peroxide. Enzymatic catalysis is discussed fully in section B1f (ADD LINK). The catalyst may be chemical rather than enzymatic (e.g., transition metal cations or complex ions, e.g., ferricyanide, at high pH). Catalysis by metal ions is discussed fully in " The role of metal ions and metallo-complexes in Luminol chemiluminescence " by HL Nekimken. Alternatively, luminol chemiluminescence may be triggered electrochemically. Sakura [2] had proposed that luminol was oxidized at the electrode surface, after which it can react with hydrogen peroxide producing one photon per hydrogen peroxide molecule (compared with 0.5 in the HRP-catalysed reaction) giving more sensitive detection and avoiding the fragility of enzyme methods [3] . Luminol electrochemiluminescence is discussed fully in section B1d . Very many assays have been devised determining compounds by inhibiting, enhancing or catalysing luminol chemiluminescence. Detectivity reaches the sub-femtomole level but the very versatility of the chemistry limits its selectivity. This is a serious shortcoming for samples such as body fluids or natural waters are very complex; in some cases, one analyte might enhance the luminol reaction while another inhibits it and the resulting signal is a combination of effects that is difficult to interpret. The situation is rather less difficult in process analytical chemistry, where there may be one and only one expected analyte. Coupling the chemiluminescence reaction post-column with a separation step (liquid chromatography or capillary electrophoresis) (ADD LINKS) can overcome interferences and give fmol-pmol detectivity. Labelling of sample components with luminol before separation can achieve the same end. Selectivity can also be provided by coupling the luminol reaction with enzymatic reactions or with antibody detection or with recognition by molecularly imprinted polymers [3] . Many analogues of luminol have been synthesized [1] ; some of them give more intense chemiluminescence than luminol itself but only if the modifications are restricted to the benzenoid ring of the luminol molecule. Changes to the heterocyclic ring abolish chemiluminescence. Phthalic hydrazide (luminol without the amine substituent) is not chemiluminescent except in aprotic solvents. Figure B1.1 – One- and two-electron routes of primary oxidation of luminol leading to secondary oxidation and chemiluminescence. A mechanism for the oxidative chemiluminescence of luminol has been proposed by Roswell and White [1] ; some of the individual steps have been studied by Lind, Merenyi and Eriksen [4] . Figure B1.1 is a flow chart of the mechanism; the structures of the main chemical species involved in the oxidation of luminol and the abbreviations for them used in the text are shown. The model proposes two step formation of luminol diazaquinone hydroperoxide anions (LOOH – ), which spontaneously decompose (via a tricyclic endoperoxide transition state) to form dinitrogen and excited 3-aminophthalate anions that luminesce. The quantum yield of luminol oxidation by this route is high giving good analytical sensitivity. b(i) Primary oxidation of luminol The hydroperoxide intermediate (LOOH – ) is formed in aqueous solution by the primary oxidation of the luminol monoanion (LH – ) to a radical (L •– ) followed by the addition of superoxide (O 2 •– ) or by primary oxidation to diazaquinone (L) followed by addition of hydrogen peroxide anions (HO 2 – ) [5] . (a) Luminol (LH 2 ) exists in aqueous solutions at pH 10.0 as monoanions (LH – ), which undergo one-electron oxidation, e.g., by hydroxyl radicals (HO • , E 0 = +2.8 V), to form rapidly (k = 9.7 x 10 -9 dm 3 mol -1 s -1 ) diazasemiquinone radicals (L •– ): 1) LH – – e – → LH • (e.g., LH – + HO • → L •– + H 2 O) (b) Two-electron oxidation of luminol monoanion, e.g., with hydrogen peroxide, gives diazaquinone (L), 2) LH – – 2e ― → L + H + (e.g., LH – + H 2 O 2 → L + H 2 O + HO – ) Two-electron oxidation occurs at the start of the luminol-hydrogen peroxide reaction. There is no superoxide present until hydrogen peroxide, competing with luminol for the hydroxyl radical, is converted to HO 2 • , which rapidly deprotonates to O 2 •– at high pH (pK a = 4.8): 3) H 2 O 2 + HO • → O 2 •– + H 3 O + Hydroxyl radicals reacting with luminol convert the monoanions (LH – ) to L •– or LH • , depending on the pH; this is a one-electron oxidation process. Transfer of a second electron to form diazaquinone occurs only in the absence of superoxide, which otherwise would react with L •– or LH • to form luminol diazaquinone hydroperoxide anions (LOOH – ). The primary oxidation step usually determines the rate of light emission, so luminol chemiluminescence effectively measures the power of the oxidant to bring about this reaction but other factors also affect the rate of primary oxidation. Light emission from the reaction between luminol and hydrogen peroxide can be induced by the presence of cobalt(II) at concentrations low enough to be regarded as catalytic and it has been proposed that cobalt(II)-peroxide complex ions bring about the primary oxidation of luminol [6] . b(ii) Secondary oxidation of luminol In analytical luminol chemiluminescence, the initial oxidation of luminol is the rate-determining step. But chemiluminescence also depends on the availability of superoxide or hydroperoxide ions for secondary oxidation. So experiments have been performed using pulse radiolysis to bring about primary oxidation, allowing the rate of secondary oxidation to be studied in the pauses between the pulses. Protonated diazasemiquinone radicals (LH • ) formed by one-electron primary oxidation add to superoxide radicals (O 2 •– ) to form the diazaquinone hydroperoxide (LOOH – ): 1) LH •– + O 2 •– → LOOH – This reaction consumes superoxide radical anions and, in the presence of a large excess of hydrogen peroxide, the major part of LH • . LH • , however, can also recombine with itself. In the absence of superoxide, all luminol radicals are consumed by recombination, at least 80% of which is accounted for by dimerization. Diazaquinone (L), the product of two-electron primary oxidation of luminol, is converted to the peroxide by the addition of hydroperoxide anions: 2) L + HO 2 – → LOOH – b(iii). Decomposition of hydroperoxide intermediate Secondary oxidation is followed by the decomposition of the cyclic hydroperoxide intermediate to 3-aminophthalate, which emits light on relaxation to the ground state. LOOH – → 3-aminophthalate + N 2 + hν The basic peroxide adduct (LOOH – ) decomposes to form the excited state of the aminophthalate emitter, while the protonated adduct undergoes a non-chemiluminescent side reaction which forms a distinct yellow product, the so-called “dark reaction” [7] . The absorbance spectrum of LOOH – decays at the same rate as does the light emission. Emission intensity increases with pH up to a maximum at about pH = 11, reflecting increasing dissociation of H 2 O 2 into its anion and the diminishing importance of the dark reaction . Decreased light output above pH 11 reflects diminished fluorescent quantum yield (Φ FL ) of the emitter. b(iv) Determination of chemiluminescence quantum yield Lind and Merényi [8] have measured the light yield of several chemiluminescent reactions of luminol undergoing one-electron oxidation by hydroxyl radicals of radiolytic origin. Using the transfer of 100 eV from ionising particles to aqueous media, it becomes possible to calculate Φ CL by Φ CL = G(hν)/ Φ CL G OH . They propose as a standard for luminol chemiluminescence initiated by pulse radiolysis a system consisting of 10 −3 mol dm −3 aqueous luminol at pH = 10.0 saturated with 10% O 2 and 90% N 2 O. Having defined a standard with a well-defined light output, it then becomes possible to determine the chemiluminescence quantum yield of any other luminol reaction relative to the standard. This has been done by Merényi and Lind [9] by plotting integrated light intensity as a function of radiolytic dose (which has a linear relationship with hydroxyl radical concentration). The light yields and hence the relative quantum yields are obtained by comparing the slopes of the plots. B1c. Oxidants used in Luminol Chemiluminescence c(i). Hydrogen peroxide Hydrogen peroxide is analytically the most useful oxidant of luminol, but requires the catalytic effect of an electrode, a metal ion or an enzyme. For example, it reacts readily with luminol in an aqueous medium in the presence of a cobalt(II) catalyst. This reaction is a very effective bench demonstration of chemiluminescence, using equal volumes of 0.1 mol dm −3 hydrogen peroxide and 1.0 x 10 −3 mol dm −3 luminol in 0.1 mol dm − 3 carbonate buffer (pH between 10 and 11). Some metal ions, used to catalyse the oxidation of luminol e.g. iron(II), react with hydrogen peroxide: Fe 2 + + H 2 O 2 → Fe 3 + + HO • + HO – to generate hydroxyl radicals (Fenton Reaction), which have very powerful oxidizing properties and can therefore bring about the primary oxidation of luminol. But they also react with hydrogen peroxide (equation1) and hydroperoxide ions (equation 2): 1) H 2 O 2 + HO • → O 2 •– + H 3 O + 2) HO 2 – + HO • → O 2 •– + H 2 O The consumption of hydroxyl radicals in these reactions diminishes the rate of primary oxidation, but the generation of superoxide increases the rate of secondary oxidation. c(ii). Oxygen The standard reduction potential (E 0 ) of luminol radicals to monoanions (LH • + e − → LH − ) has been determined to be +0.87 V [10] . Oxidation by molecular dioxygen (E 0 = 1.229 V) is thermodynamically possible but in aqueous solutions, the reactivity is undetectably low at any pH (k = 10 −8 dm 3 mol −1 s −1 ) and so the reaction is not useful for primary oxidation. It is widely believed that air-saturated luminol solutions are indefinitely stable in the dark, even at pH = 14. In spite of this, the oxidation of luminol by dissolved oxygen in aqueous solutions is frequently reported. It is likely that what is referred to in these cases is oxidation by oxygen radicals , which may be formed from molecular dioxygen by suitable reductants such as metal ions. This phenomenon is discussed in chapter D10 along with other cases in which oxygen radicals act as chemiluminescence reagents (LINK). In dimethylsulfoxide (DMSO) solutions, luminol exists as dianions and reacts with dissolved oxygen in the presence of a strong base with intense chemiluminescence [1] . The rate constant for the reaction is about 50 dm 3 mol −1 s −1 [10] ; the rate constant for the corresponding reaction between oxygen and luminol dianions in aqueous alkali is 10 −2 dm 3 mol −1 s −1 . Because the conditions of the reaction in DMSO solution are relatively simple, the phenomenon has found great favour as a demonstration [11] , for a spatula measure of luminol in a bottle of alkaline DMSO will react on shaking at room temperature. However, while the oxidation of luminol in aqueous solution is very widely used analytically, there are no established analytical procedures making use of the reaction in dimethylsulfoxide, dimethylformamide or other organic solvents but the effect of a range of metal complexes on the reaction in DMSO has been investigated [12] [13] The emitter (3-aminophthalate ion) tautomerizes in aprotic solvents such as DMSO to a quinonoid form that gives maximum emission at 510 nm; this tautomer is favoured by the pairing of luminol anions with metal cations (e.g., Na + or K + ). If luminol is oxidized in mixed solvents, there is less emission at 425 nm (reduced ion pairing) and more at 510 nm than in aqueous media. Also in mixed solvents there is less 425 nm emission in chemiluminescence than in fluorescence because in chemiluminescence the fraction of ion-pairs is determined by the transition-state rather than by the ground-state equilibrium as in fluorescence [1] . c(iii). Higher oxidation states of manganese Permanganate ions are thermodynamically easily capable ((E 0 = 1.70 V) of oxidizing luminol. A flow injection analysis of paracetamol in pharmaceutical preparations based on inhibition of luminol-permanganate chemiluminescence has been reported [14] . A little earlier, an imaginative biosensor for urea had been fabricated, in which ammonium carbonate generated by urease-catalysed hydrolysis was used to release luminol from an anion-exchange column to react with permanganate eluted from a second column, producing chemiluminescence [15] . A steady stream of novel applications of the luminol-permanganate system followed. Oxidation of luminol by alkaline potassium permanganate produces manganate(VI) ions, which further react with luminol causing chemiluminescence. This phenomenon, termed by the authors ‘’’second chemiluminescence (SCL)’’’ has been applied in an assay for nickel(II) ions [16] . In a suitable flow injection manifold, dilute solutions of alkaline luminol and of aqueous potassium permanganate are mixed and allowed to react for a long enough time for the resulting chemiluminescence to drop to a stable minimum close to zero. The sample is then injected into the mixture and, if nickel ions are present, light emission recommences and rapidly rises to a well-defined peak before returning to baseline intensity. Optimum intensity of the second chemiluminescence was obtained by using a tenfold excess of luminol (over potassium permanganate) in 0.1 mol dm -3 aqueous sodium hydroxide and injecting sample at pH 5.10; linear relationship with nickel(II) concentration was established and the detection limit was 0.33 μg dm -3 . Numerous divalent and trivalent metal ions and nitrate ions were found to interfere with the determination. The mechanism of the luminol-manganate(VI) chemiluminescence appears to be the same as that for other luminol oxidations, with the production of excited 3-aminophthalate ion emitting at 425 nm. But oxidations both by permanganate and manganate(VI) can lead to the formation of excited manganese(II), which would be an additional source of chemiluminescence. Unfortunately, in the work described, the chemiluminescence spectrum was observed only up to 490 nm, overlooking such possible contributions to the signal. c(iv) Silver(III) A fairly stable silver(III) complex anion, diperiodatoargentate(III) (DPA), [Ag (H 2 IO 6 )(OH) 2 ] 2− , can be readily synthesized [17] . A new chemiluminescence reaction between luminol and diperiodatoargentate has been observed in alkaline aqueous solution [18] [19] . The emission of light from this reaction is strongly enhanced by iron nanoparticles and the intensity is further increased by the addition of aminophylline [20] . This forms the basis for an assay in which the chemiluminescence signal has a linear relationship with aminophylline concentration in human serum over the range 1.0 x 10 −8 to 2.0 x 10 −6 mol dm −3 . The detection limit is 9.8 x 10 −9 mol dm −3 . The relative standard deviation at 8.0 x 10 −7 mol dm −3 is 4.8% (n = 10). Penicillin antibiotics have also been found to enhance luminol-silver(III) complex chemiluminescence, which has formed the basis for a sensitive flow injection assay for these drugs in dosage forms and in urine samples. In optimized conditions, the detection limit for benzylpenicillin sodium is reported to be 70 ng cm −3 , for amoxicillin 67 ng cm −3 , for ampicillin 169 ng cm −3 and for cloxacillin sodium 64 ng cm −3 [21] . The maximum wavelength of the light emitted is about 425 nm [18] , which is the usual chemiluminescence from excited 3-aminophthalate, produced by luminol oxidation. This implies that the silver(III) complex is capable of bringing about both the primary and secondary oxidation of luminol as proposed by Shi ‘’et al.’’ who postulate one-electron primary oxidation of two luminol molecules by each diperiodatoargentate. Perhaps two-electron oxidation of one luminol molecule is more likely. The reduction potential of diperiodatoargentate(III) ion is +1.74 V [22] , high enough for two-electron oxidation to convert water into hydrogen peroxide (ε 0 = -1.763 V; Nernst equation indicates a millimolar H 2 O 2 equilibrium concentration). This provides a possible mechanism for secondary as well as primary oxidation. 2.1: B1d. Electrochemiluminescence The instrumentation of electrochemiluminescence (ECL) is dealt with in chapter D7. The resulting reaction pathways lend themselves to control of emission by varying the applied voltage or the electrode selected and are applicable to near- neutral (pH 8.0 to 8.5) aqueous solutions such as biological fluids, whereas luminol chemiluminescence usually occurs in strongly alkaline or non-aqueous solutions. It has been proposed that luminol is first oxidized at the electrode surface and, on subsequent reaction with hydrogen peroxide, the chemiluminescence quantum yield (see chapter A1 ADD LINK) is enhanced [23] [24] . Typical of the early applications is an assay of lipid hydroperoxides using ECL at a vitreous carbon electrode [25] . With applied voltages of 0.5-1.0 V, luminol monoanion loses one electron giving diazasemiquinone, which dismutes to produce diazaquinone, which reacts quantitatively with the analyte. At applied voltages above 1.0 V, the –NH2 of diazaquinone and the analyte itself are oxidized giving respectively –NH• and superoxide, which causes an interfering signal. The detection limit in optimized conditions was 0.3 nmol at S/N = 1.5. Using a voltage of 0.5-1.0 V applied to a platinum electrode, both methyl linoleate hydroperoxide (MLHP) and luminol are oxidized; the detection limit for MLHP is 0.1 nmol at S/N = 2.5. There was no emission from the closely related methyl hydroxyoctadecadienoate (a reduction product of linoleic acid hydroperoxide). The inhibition of ECL signals from luminol oxidation can be used as a method of determination of inhibitors. A recent example is the determination of melamine in dairy products and in tableware [26] . Using low voltage scan rates in phosphate buffer at high pH, ECL is observed at 1.47 V and there is a linear (r 2 =0.9911) decrease of ECL proportional to the logarithm of the melamine concentration over the range 1 to 100 ng cm −3 . The limit of detection is 0.1 ng cm −3 with high recovery. The signal arises from the reaction with luminol of reactive oxygen species (from the electrooxidation of hydroxyl ions) that are eliminated by melamine. Modification of electrodes is now a well-established way of controlling ECL and in recent years the use of nanomaterials for this purpose has grown in importance. An example involving luminol is the modification of a gold electrode by applying a composite of multi-wall carbon nanotubes and the perfluorosulfonate polymer Nafion [27] . In the course of cyclic voltammetry in carbonate buffer, three ECL peaks were obtained, up to 20 times as intense as with the unmodified electrode; in each case the emitter was identified as 3-aminophthalate anion, indicating that the improvement was due to electrode efficiency rather than to any change in the chemistry of the system. ECL immunosensors have been fabricated that have been successfully applied to the determination of human immunoglobulin G (hIgG) in serum. The primary antibody, biotin-conjugated goat anti-hIgG, is first immobilized on an electrode modified with streptavidin-coated gold nanoparticles (AuNPs). The sensors are sandwich-type immunocomplexes formed by the conjugation of hIgG to a second antibody labelled with luminol-functionalized AuNPs. ECL is generated by a double potential step in carbonate buffer containing 1.0 mmol dm −3 hydrogen peroxide. Many luminol molecules are attached to the surface of each AuNP and act as multiple sources of light emission from each antibody molecule. The amplification of ECL in this way, linked to the analyte by the biotin-streptavidin system, leads to greatly enhanced signals. The limit of detection is 1 pg cm −3 (at S/N = 3), which surpasses the performance of all previous hIgG assays [28] . The surface of a glassy carbon electrode was modified by producing on it L-cysteine reduced graphene oxide composites onto which AuNPs were self-assembled. Cholesterol oxidase (ChOx) was subsequently adsorbed on the AuNP surface to form a cholesterol biosensor with satisfactory reproducibility, stability and selectivity. The AuNPs increase the surface area of the electrode, hence permitting a higher ChOx loading, and provide a nanostructure more favourable to ECL, improving analytical performance. The linear response to cholesterol of the sensor extends over the range 3.3 x 10 −6 to 1.0 x 10 −3 mol dm −3 and the limit of detection is 1.1 x 10 −6 (at S/N = 3) [29] . A poly(luminol-3,3',5,5'-tetramethylbenzidine) copolymer manufactured by electropolymerization on screen-printed gold electrodes greatly improves the ECL of hydrogen peroxide. A cholesterol biosensor suitable for the analysis of serum samples was fabricated by immobilization of cholesterol oxidase onto the polymer. Under optimized conditions, the biosensor has a linear dynamic range of 2.4 x 10 −5 to 1.0 x 10 −3 mol dm −3 with a limit of detection of 7.3 x 10 −6 mol dm −3 . Precision (measured as relative standard deviation) was 10.3% at 5.0 x 10 − 4 mol dm −3 and the method has the additional advantages of low cost and high speed [30] .
Courses/Smith_College/CHM_222_Chemistry_II%3A_Organic_Chemistry_(2025)/10%3A_Ethers_and_Epoxides_Thiols_and_Sulfides/10.01%3A_Introduction	Objectives After completing this section, you should be able to use the terms “ether,” “diethyl ether ” and “ethyl ether ” appropriately in context. Key Terms Make certain that you can define, and use in context, the key terms below. ether ($\ce{R-O-R′}$) sulfide ($\ce{R-S-R′}$) thiol ($\ce{R-S-H}$) Study Notes As defined in the textbook, an “ether” is a substance with the general formula ($\ce{R-O-R′}$) where $\ce{R}$ and $\ce{R′}$ are alkyl, aryl, vinyl or allyl groups. However, the word “ether” is also commonly used to refer to the specific compound, $\ce{CH3-CH2-O-CH2-CH3}$, which is correctly called “diethyl ether.” Further confusion can arise because some chemists refer to “diethyl ether ” as “ethyl ether.” In this course, “ether ” will be used to refer to the class of compounds with the structure ($\ce{R-O-R′}$) ; diethyl ether will refer to the compound, $\ce{CH3-CH2-O-CH2-CH3}$; and “ethyl ether ” will not be used. Ethers and Epoxides While the general formula for ethers is R-O-R′ , keep in mind that there also cyclic ethers like tetrahydrofuran (a common organic solvent) or even epoxides which you first encounter in Section 8.7 in synthesizing diols from alkenes. Thiols and Sulfides Thiols (thio alcohols or mercaptans) and sulfides (thioethers) are the sulfur analogues of alcohols and ethers and have the general formulas of R-S-H and R-S-R′, respectively.
Courses/Madera_Community_College/Concepts_of_Physical_Science/10%3A_Elements_Are_Made_of_Atoms/10.03%3A_The_Development_of_Atomic_Theory/10.3.01%3A_Law_of_Conservation_of_Mass	Learning Objectives Recognize the law of conservation of mass. ↵ Have you ever lost a screw? The following situation happens all too often. You have taken apart a piece of equipment to clean it up. When you put the equipment back together, somehow you have an extra screw or two. Or you find out that a screw is missing that was a part of the original equipment. In either case, you know something is wrong. You expect to end up with the same amount of material that you started with, not with more or less than what you had originally. Law of Conservation of Mass By the late 1700s, chemists accepted the definition of an element as a substance that cannot be broken down into a simpler substance by ordinary chemical means. It was also clear that elements combine with one another to form more complex substances called compounds. The chemical and physical properties of these compounds are different than the properties of the elements from which they were formed. There were questions about the details of these processes. In the 1790s, a greater emphasis began to be placed on the quantitative analysis of chemical reactions. Accurate and reproducible measurements of the masses of reacting elements and the compounds they form led to the formulation of several basic laws . One of these is called the law of conservation of mass , which states that during a chemical reaction, the total mass of the products must be equal to the total mass of the reactants . In other words, mass cannot be created or destroyed during a chemical reaction, but is always conserved. This is another of the rare conservation laws we are being introduced to. LAW OF CONSERVATION OF MASS During a chemical reaction, the total mass of all products must be equal to the total mass of all reactants. As an example, consider the reaction between silver nitrate and sodium chloride. These two compounds will dissolve in water to form silver chloride and sodium nitrate. The silver chloride does not dissolve in water, so it forms a solid that we can filter off. When we evaporate the water, we can recover the sodium nitrate formed. If we react 58.5 grams of sodium chloride with 169.9 grams of silver nitrate, we start with 228.4 grams of materials. After the reaction is complete and the materials separated, we find that we have formed 143.4 grams of silver chloride and 85.0 grams of sodium nitrate, giving us a total mass of 228.4 grams for the products. So, the total mass of reactants equals the total mass of products, a proof of the law of conservation of mass. Section Summary The law of conservation of mass states that, during a chemical reaction, the total mass of the products must be equal to the total mass of the reactants. Glossary reactants substances which react during a chemical reaction. products substances which are formed during a chemical reaction. law of conservation of mass mass cannot be created or destroyed during a chemical reaction, but is always conserved.
Courses/Duke_University/CHEM_210D%3A_Modern_Applications_of_Chemistry/3%3A_Textbook-_Modern_Applications_of_Chemistry/01%3A_Primer/1.02%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/1.2.02%3A_Ionic_Bonding	Ionic bonding involves transfer of an electron from one atom (which becomes a positively charged cation ) to another (which becomes a negatively charged anion ). The two ions attract strongly to form a crystal lattice . Since ionic bonding requires that the atoms involved have unequal attraction for their valence electrons, an ionic compound must involve atoms of two quite different elements. Attraction for electrons depends on the distance of the electrons from the nucleus (which in turn depends on the amount of shielding by inner electrons). Ionic compounds generally form between metals toward the left and bottom of the periodic table, and nonmetals toward the right and top of the periodic table. The simplest example of a binary ionic compound is provided by the combination of elements number 1 (H) and number 3 (Li) in lithium hydride, LiH. On a microscopic level the formula LiH contains four electrons. In separate Li and H atoms these electrons are arranged as shown in part a of the following figure. The H atom has the electron configuration 1 s 1 , and Li is 1 s 2 2 s 1 . When the two atoms are brought close enough together, however, the striking rearrangement of the electron clouds shown part b takes place. Here the color coding shows clearly that the electron density which was associated with the 2 s orbital in the individual Li atom has been transferred to a 1 s orbital surrounding the H atom. As a result, two new microscopic species are formed. The extra electron transforms the H atom into a negative ion or anion , written H – and called the hydride ion . The two electrons left on the Li atom are not enough to balance the charge of +3 on the Li nucleus, and so removal of an electron produces a positive ion or cation , written Li + and called the lithium ion . The electron-transfer process can be summarized in of Lewis diagrams as follows: The opposite charges of Li + and H – attract each other strongly, and the ions form an ion pair (see image below) in which the two nuclei are separated by a distance of 160 pm (1.60 Å). The image above shows an ion pair of Lithium Hydride. Notice how Lithium has a strong positive charge (cation) and Hydrogen has a strong negative charge (anion). Multiple ion pairs (as seen in the image above) connect to form a crystal lattice , pictured below. All ionic solids form a crystal lattice and the shape of the lattice determines the properties and look of the solid.
Bookshelves/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.03%3A_The_Limiting_Reagent/3.3.07%3A_Geology-_Using_the_Acid_Test_to_Distinguish_the_Minerals_in_Calomine	Calamine is an obsolete name for what is now known to be a mixture of two distinct minerals: zinc carbonate (ZnCO 3 or smithsonite) and zinc silicate (Zn 4 Si 2 O 7 (OH) 2 ·H 2 O, or hemimorphite). the name "Calamine" is now used only for calomine lotion, which is a suspension of ZnO and Fe 2 O 3 [1] . Smithsonite and hemimorphite may be similar in appearance to one another, but their appearance may be quite variable depending on location, so two samples of smithsonite (or hemimorphite) may look quite different as shown in the figures: 0 1 Solid chunk of mineral with many crystalline spikes all around. This mineral has a light purple hue. Hemimorphite from Mapimi, Durango, Mexico[2] Hemimorphite[3] 0 1 Smithsonite from Tsumeb, Namibia[4] Smithsonite from Tsumeb, Namibia[5] The two minerals can only be reliably distinguished through chemical analysis. Carbonate minerals like calcite or smithsonite react with acids to efforvesce (fizz) while dissolving and producing CO 2 (see equation (1) below). This test can be done with 1 M HCl, or household vinegar (crushing the sample will help if vinegar is used). While calcite (CaCO 3 ) bubbles strongly in cold dilute acid, dolomite CaMg(CO 3 ) 2 ) and rhodochrosite (MnCO 3 ) bubble weakly. Smithsonite (along with Siderite, FeCO 3 and Magnesite, MgCO 3 ) require heating to react. Silicates, like hemimorphite, don't generally react with cold, dilute acids at all. So we could tell if a sample contained just smithsonite, because it would ocmpletely dissolve in acid. hemimorphite would not react, and a mixture of the two would partially dissolve. It would be necessary to add an excess of HCl to the sample, otherwise it might not all dissolve because there isn't enough HCl, not because it's partially hemimorphite. If we have a 100 g sample that may contain smithsonite, hemimorphite, or both, we need to add enough acid to react with the sample, assuming it's all smithsonite, just to make sure. Example 4 from Equations and Mass Relationships also illustrates the idea that one reactant in a chemical equation may be completely consumed without using up all of another. Here we want the smithsonite to be completely consumed. When it is gone, the other excess reactants have nothing to react with and they cannot be converted to products. The substance which is used up first (the smithsonite) is the limiting reagent . EXAMPLE 1 When 100.0 g of smithsonite is reacted with 100.0 g of HCl to form carbon dioxide gas, which is the limiting reagent? What mass of product will be formed? (Note: HCl is provided as a solution with a concentration of 1-5% HCl for this purpose. The mass of HCl solution would be much (20-100 times) greater than the mass of HCl given here). Solution The balanced equation ZnCO 3 + 2 HCl → ZnCl 2 + CO 2 + H 2 O (1) tells us that according to the atomic theory, 2 mol HCl are required for each mole of ZnCO 3 . That is, the stoichiometric ratio S(HCl/ZnCO 3 ) = 2 mol HCl/ 1 mol ZnCO 3 . Let us see how many moles of each we actually have $\begin{align} & n_{\text{HCl}}=\text{100}\text{.0 g}\times \frac{\text{1 mol HCl}}{\text{36.5 g}}=\text{2}\text{.74 mol HCl} \\ & n_{\text{ZnCO}_{\text{3}}}=\text{100}\text{.0 g}\times \frac{\text{1 mol ZnCO}_{\text{3}}}{\text{125}\text{.4 g}}=\text{0}\text{.798 mol ZnCO}_{\text{3}} \\ \end{align}$ Calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. We can calculate (hypothetically) how much of each reactant would be required if the other were completely consumed to demonstrate which is in excess, and which is limiting. For example, if all the HCl were to react, it would require $\frac{\text{1 mol ZnCO}_3}{\text{2 mol HCl}}~~ x ~~ \text{2.74 mol HCl} = \text{1.37 mol ZnCO}_3$ Since there is not this much ZnCO 3 present, this is impossible. HCl is in excess, and ZnCO 3 is the limiting reactant. In the table, we've crossed out this calculation, and proceeded to calculate how much HCl would be required if all the ZnCO 3 reacts (which is what happens). Unnamed: 0 ZnCO3 + HCl → ZnCl2 + CO2 + H2O m (g) 100.000 100.00 NaN NaN NaN M (g/mol) 125.100 36.50 136.300 44.000 18.000 n (mol) 0.798 2.74 NaN NaN NaN if all ZnCO3 reacts -0.798 -1.60 0.798 0.798 0.798 if all HCl reacts -2.740 -1.37 NaN NaN NaN Actual Reaction Amounts -0.798 -1.60 0.798 0.798 0.798 Actual Reaction Masses -100.000 -58.40 108.800 35.100 14.400 We use the amount of limiting reagent to calculate the amount of product formed. $\frac{\text{1 mol CO}_{2}}{\text{1 mol ZnCO}_{3}}~~ x ~~ \text{0.798 mol ZnCO}_{3} = \text{0.798 mol CO}_2$ $\text{0.798 mol CO}_2 ~~ x ~~ \frac{\text{44.0 g CO}_{2}}{\text{1 mol CO}_{2}} = \text{35.1 g CO}_2 $ When the reaction ends, 1.60 mol HCl will have reacted with 0.798 mol ZnCO 3 and there will be (2.74 – 1.60) mol HCl = 1.14 mol HCl left over. ZnCO 3 is therefore the limiting reagent. The left over HCl will ensure that if any material remains in a mineral test of a 100 g sample, that it can't be a carbonate. From this example you can begin to see what needs to be done to determine which of two reagents, X or Y, is limiting. We must compare the stoichiometric ratio S(X/Y) with the actual ratio of amounts of X and Y which were initially mixed together. In Example 1 this ratio of initial amounts $\frac{n_{\text{ZnCO}_{3}}\text{(initial)}}{n_{\text{HCl}}\text{(initial)}}=\frac{\text{0.798 mol ZnCO}_{3}}{\text{2.74 mol HCl}}=\frac{\text{0}\text{.291 mol ZnCO}_{3}}{\text{1 mol HCl}}$ was less than the stoichiometric ratio $\text{S}\left( \frac{\text{ZnCO}_{3}}{\text{HCl}} \right)=\frac{\text{1 mol ZnCO}_{3}}{\text{2 mol HCl}}~=~\frac{\text{0.5 mol ZnCO}_{3}}{\text{1 mol HCl}}$ This indicated that there was not enough Hg to react with all the bromine and mercury was the limiting reagent. The corresponding general rule, for any reagents X and Y, is $\begin{align} & \text{If}~ \frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is less than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then X is limiting}\text{.} \\ & \\ & \text{If}~\frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is greater than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then Y is limiting}\text{.} \\ \end{align}$ (Of course, when the amounts of X and Y are in exactly the stoichiometric ratio, both reagents will be completely consumed at the same time, and neither is in excess.). This general rule for determining the limiting reagent is applied in the next example. EXAMPLE 2 Iron can be obtained by reacting the ore hematite (Fe 2 O 3 ) with coke (C). The latter is converted to CO 2 . As manager of a blast furnace you are told that you have 20.5 Mg (megagrams) of Fe 2 O 3 and 2.84 Mg of coke on hand. (a) Which should you order first—another shipment of iron ore or one of coke? (b) How many megagrams of iron can you make with the materials you have? Solution a) Write a balanced equation 2Fe 2 O 3 + 3C → 3CO 2 + 4Fe The stoichiometric ratio connecting C and Fe 2 O 3 is $\text{S}\left( \frac{\text{C}}{\text{Fe}_{\text{2}}\text{O}_{\text{3}}} \right)=\frac{\text{3 mol C}}{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}=\frac{\text{1}\text{.5 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}$ The initial amounts of C and Fe 2 O 3 are calculated using appropriate molar masses $\begin{align} & \text{ }n_{\text{C}}\text{(initial)}=\text{2}\text{.84}\times \text{10}^{\text{6}}\text{g}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g}}=\text{2}\text{.36}\times \text{10}^{\text{5}}\text{mol C} \\ & \\ & n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{(initial)}=\text{20}\text{.5}\times \text{10}^{\text{6}}\text{g}\times \frac{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}{\text{159}\text{.69 g}}=\text{1}\text{.28}\times \text{10}^{\text{5}}\text{mol Fe}_{\text{2}}\text{O}_{\text{3}} \\ \end{align}$ Their ratio is $\frac{n_{\text{C}}\text{(initial)}}{n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{(initial)}}=\frac{\text{2}\text{.36}\times \text{10}^{\text{5}}\text{mol C}}{\text{1}\text{.28}\times \text{10}^{\text{5}}\text{mol Fe}_{\text{2}}\text{O}_{\text{3}}}=\frac{\text{1}\text{.84 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}$ Since this ratio is larger than the stoichiometric ratio, you have more than enough C to react with all the Fe 2 O 3 . Fe 2 O 3 is the limiting reagent, and you will want to order more of it first since it will be consumed first. b) The amount of product formed in a reaction may be calculated via an appropriate stoichiometric ratio from the amount of a reactant which was consumed . Some of the excess reactant C will be left over, but all the initial amount of Fe 2 O 3 will be consumed. Therefore we use n Fe2O3 (initial) to calculate how much Fe can be obtained $n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{ }\xrightarrow{S\text{(Fe/Fe}_{\text{2}}\text{O}_{\text{3}}\text{)}}\text{ }n_{\text{Fe}}\xrightarrow{M_{\text{Fe}}}\text{ }m_{\text{Fe}}$ $m_{\text{Fe}}=\text{1}\text{.28 }\times \text{ 10}^{\text{5}}\text{ mol Fe}_{\text{2}}\text{O}_{\text{3}}\text{ }\times \text{ }\frac{\text{4 mol Fe}}{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}\text{ }\times \text{ }\frac{\text{55}\text{.85 g}}{\text{mol Fe}}=\text{1}\text{.43 }\times \text{ 10}^{\text{7}}\text{ g Fe}$ This is 1.43 × 10 6 g, or 14.3 Mg, Fe. As you can see from the example, in a case where there is a limiting reagent, the initial amount of the limiting reagent must be used to calculate the amount of product formed . Using the initial amount of a reagent present in excess would be incorrect, because such a reagent is not entirely consumed. The concept of a limiting reagent was used by the nineteenth century German chemist Justus von Liebig (1807 to 1873) to derive an important biological and ecological law. Liebig’s law of the minimum states that the essential substance available in the smallest amount relative to some critical minimum will control growth and reproduction of any species of plant or animal life. When a group of organisms runs out of that essential limiting reagent, the chemical reactions needed for growth and reproduction must stop. Vitamins, protein, and other nutrients are essential for growth of the human body and of human populations. Similarly, the growth of algae in natural bodies of water such as Lake Erie can be inhibited by reducing the supply of nutrients such as phosphorus in the form of phosphates. It is for this reason that many states have regulated or banned the use of phosphates in detergents and are constructing treatment plants which can remove phosphates from municipal sewage before they enter lakes or streams. References ↑ en.Wikipedia.org/wiki/Calamine_lotion ↑ en.Wikipedia.org/wiki/Hemimorphite ↑ en.Wikipedia.org/wiki/Hemimorphite ↑ en.Wikipedia.org/wiki/Smithsonite ↑ en.Wikipedia.org/wiki/Smithsonite
Courses/Martin_Luther_College/Organic_Chemistry_-_MLC/02%3A_Tools_of_Organic_Chemistry/2.01%3A_Intermolecular_Forces_(IMFs)_-_Review/2.1.05%3A_Gas_Chromatography_(GC)/2.1.5F%3A_Sample_Preparation_for_Gas_Chromatography	Liquid GC Samples Fill a GC vial to the $1.5 \: \text{mL}$ mark (Figure 2.93a) with a low boiling solvent (e.g. methanol, clean acetone, diethyl ether, or dichloromethane). Add one drop of the sample to be analyzed. If you think you possibly only added a half-drop, it's probably enough. Two drops are really too much. Cap the vial and invert once or twice to dissolve the sample. If it appears like the sample drop did not dissolve fully, prepare another sample using a different solvent. Alternatively, if you have already prepared a high-field NMR sample, use one-third of this sample and dilute it further with a low-boiling solvent. Run the GC, as demonstrated by your instructor. Procedures vary at each institution. Solid GC Samples Some relatively volatile solids (never ionic solids!) can be analyzed by GC. Add one or two "specks" of solid$^{16}$ (a pile approximately $2 \: \text{mm}$ in diameter), or a very small spatula-tip of solid to a GC vial (Figure 2.93c). Then add a low boiling solid (e.g. methanol, clean acetone, diethyl ether, or dichloromethane) to the $1.5 \: \text{mL}$ mark. Cap the vial and invert several times to fully dissolve the solid. If the solid dissolves, run the GC. If the solid does not dissolve, do NOT run the GC anyway! Solids can plug the very small microliter syringes used by the instrument. If the quantity of solid does not appear to have changed at all after adding the solvent, try making another sample with a different low boiling solvent. If it appears the solid has decreased in quantity but is not fully dissolved, it is likely enough of it has dissolved to analyze. Use a pipette filter to filter the solution, and run the GC on the filtered liquid. $^{16}$The quantity of solid may need to be adjusted if the solid is "fluffy". A GC is at the ideal dilution when it produces abundances around one million using a mass spectrometer detector.
Courses/Solano_Community_College/Introductory_Chemistry_at_Solano_College_-_Updated_2023_04_03/13%3A_Solutions/13.01%3A_Solute-Solvent_Combinations	Solute-Solvent Combinations The focus of "Chapter 15: Water" was water's role in the formation of aqueous solutions. We examined the primary characteristics of a solution and how water is able to dissolve solid solutes; we differentiated between a solution, a suspension, and a colloid. There are many examples of solutions that do not involve water at all, or that involve solutes that are not solids. The table below summarizes the possible combinations of solute-solvent states, along with examples of each. Solute State Solvent State Example liquid gas water in air gas gas oxygen in nitrogen (gas mixture) solid liquid salt in water liquid liquid alcohol in water gas liquid carbon dioxide in water solid solid zinc in copper (brass alloy) liquid solid mercury in silver and tin (dental amalgam) Gas-Gas Solutions Our air is a homogenous mixture of many different gases and therefore qualifies as a solution. Approximately $78\%$ of the atmosphere is nitrogen, making it the solvent for this solution. The next major constituent is oxygen (about $21\%$), followed by the inert gas argon $\left( 0.9\% \right)$, carbon dioxide $\left( 0.03\% \right)$, and trace amounts of neon, methane, helium, and other gases. Solid-Solid Solutions Solid-solid solutions such as brass, bronze, and sterling silver are called alloys. Bronze (composed mainly of copper with added tin) was widely used in making weapons in times past, dating back to at least 2400 B.C. This metal alloy was hard and tough, but was eventually replaced by iron. Liquid-Liquid Solutions Perhaps the most familiar liquid-solid solution is dental amalgam, used to fill teeth when there is a cavity. Approximately $50\%$ of the amalgam material is liquid mercury to which a powdered alloy of silver, tin, and copper is added. Mercury is used because it binds well with the solid metal alloy. However, the use of mercury-based dental amalgam has gone under question in recent years, because of concerns regarding the toxicity of mercury. Summary
Courses/College_of_the_Canyons/Chem_151%3A_Preparatory_General_Chemistry_OER/07%3A_Chemical_Reactions/7.04%3A_Evidence_of_a_Chemical_Reaction	Learning Objectives Identify the evidence for chemical reactions. In a chemical change, new substances are formed. In order for this to occur, the chemical bonds of the substances break, and the atoms that compose them separate and rearrange themselves into new substances with new chemical bonds. When this process occurs, we call it a chemical reaction. A chemical reaction is the process in which one or more substances are changed into one or more new substances. To identify a chemical reaction, we look for a chemical change . A chemical change always produces one or more types of matter that differ from the matter present before the change. The formation of rust is a chemical change because rust is a different kind of matter than the iron, oxygen, and water present before the rust formed. The explosion of nitroglycerin is a chemical change because the gases produced are very different kinds of matter from the original substance. Other examples of chemical changes include: reactions that are performed in a lab (such as copper reacting with nitric acid), all forms of combustion (burning), and food being cooked, digested, or rotting (Figure $\PageIndex{1}$). Video$\PageIndex{1}$: Evidence of a Chemical Reaction Example $\PageIndex{1}$: Evidence of a Chemical Reaction Which of the following is a chemical reaction? Freezing liquid mercury. Adding yellow to blue to make green. Cutting a piece of paper into two pieces. Dropping a sliced orange into a vat of sodium dydroxide. Filling a balloon with natural air. Solution A, B, C, & E involve only physical changes. A sliced orange has acid (citric acid) that can react with sodium hydroxide, so the answer is D. Exercise $\PageIndex{1}$ Which of the following is a chemical reaction? Painting a wall blue. A bicycle rusting. Ice cream melting. Scratching a key across a desk. Making a sand castle. Answer B Example $\PageIndex{2}$: Evidence of a Chemical Reaction Which of the following is not a chemical reaction? Shattering glass with a baseball. Corroding metal. Fireworks exploding. Lighting a match. Baking a cake. Solution Shattering glass with a baseball results in glass broken into many pieces but no chemical change happens, so the answer is A. Exercise $\PageIndex{2}$ Which of the following is NOT a chemical reaction? Frying an egg. Slicing carrots. A Macbook falling out of a window. Creating ATP in the human body. Dropping a fizzy tablet into a glass of water. Answer B and C Summary Chemical reactions can be identified via a wide range of different observable factors including change in color, energy change (temperature change or light produced), gas production, formation of precipitate and change in properties. Contributions & Attributions This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality: Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ). Marisa Alviar-Agnew ( Sacramento City College ) Henry Agnew (UC Davis)
Courses/Saint_Francis_University/Chem_114%3A_Human_Chemistry_II_(Hargittai)/19%3A_Enzymes_and_Vitamins/19.04%3A_How_Enzymes_Work	Learning Objectives To describe the interaction between an enzyme and its substrate. Enzyme-catalyzed reactions occur in at least two steps. In the first step, an enzyme (E) and the substrate molecule or molecules (S) collide and react to form an intermediate compound called the enzyme-substrate (ES) complex . (This step is reversible because the complex can break apart into the original substrate or substrates and the free enzyme.) Once the ES complex forms, the enzyme is able to catalyze the formation of product (P), which is then released from the enzyme surface: \[E + S \leftrightarrow ES \tag{$\PageIndex{1}$} \] \[ES \rightarrow E + P \tag{$\PageIndex{2}$} \] Hydrogen bonding and other electrostatic interactions hold the enzyme and substrate together in the complex. The structural features or functional groups on the enzyme that participate in these interactions are located in a cleft or pocket on the enzyme surface. This pocket, where the enzyme combines with the substrate and transforms the substrate to product is called the active site of the enzyme (Figure $\PageIndex{1}$). Models of Enzyme-Substrate Interaction The active site of an enzyme possesses a unique conformation (including correctly positioned bonding groups) that is complementary to the structure of the substrate, so that the enzyme and substrate molecules fit together in much the same manner as a key fits into a tumbler lock. In fact, an early model describing the formation of the enzyme-substrate complex was called the lock-and-key model (Figure $\PageIndex{2}$). This model portrayed the enzyme as conformationally rigid and able to bond only to substrates that exactly fit the active site. Working out the precise three-dimensional structures of numerous enzymes has enabled chemists to refine the original lock-and-key model of enzyme actions. They discovered that the binding of a substrate often leads to a large conformational change in the enzyme, as well as to changes in the structure of the substrate or substrates. The current theory, known as the induced-fit model , says that enzymes can undergo a change in conformation when they bind substrate molecules, and the active site has a shape complementary to that of the substrate only after the substrate is bound, as shown for hexokinase in Figure $\PageIndex{3}$. After catalysis, the enzyme resumes its original structure. The structural changes that occur when an enzyme and a substrate join together bring specific parts of a substrate into alignment with specific parts of the enzyme’s active site. Amino acid side chains in or near the binding site can then act as acid or base catalysts, provide binding sites for the transfer of functional groups from one substrate to another or aid in the rearrangement of a substrate. The participating amino acids, which are usually widely separated in the primary sequence of the protein, are brought close together in the active site as a result of the folding and bending of the polypeptide chain or chains when the protein acquires its tertiary and quaternary structure. Binding to enzymes brings reactants close to each other and aligns them properly, which has the same effect as increasing the concentration of the reacting compounds. Example $\PageIndex{1}$ What type of interaction would occur between an OH group present on a substrate molecule and a functional group in the active site of an enzyme? Suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you just identified. Solution An OH group would most likely engage in hydrogen bonding with an appropriate functional group present in the active site of an enzyme. Several amino acid side chains would be able to engage in hydrogen bonding with an OH group. One example would be asparagine, which has an amide functional group. Exercise $\PageIndex{1}$ What type of interaction would occur between an COO − group present on a substrate molecule and a functional group in the active site of an enzyme? Suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you just identified. One characteristic that distinguishes an enzyme from all other types of catalysts is its substrate specificity . An inorganic acid such as sulfuric acid can be used to increase the reaction rates of many different reactions, such as the hydrolysis of disaccharides, polysaccharides, lipids, and proteins, with complete impartiality. In contrast, enzymes are much more specific. Some enzymes act on a single substrate, while other enzymes act on any of a group of related molecules containing a similar functional group or chemical bond. Some enzymes even distinguish between D- and L-stereoisomers, binding one stereoisomer but not the other. Urease, for example, is an enzyme that catalyzes the hydrolysis of a single substrate—urea—but not the closely related compounds methyl urea, thiourea, or biuret. The enzyme carboxypeptidase, on the other hand, is far less specific. It catalyzes the removal of nearly any amino acid from the carboxyl end of any peptide or protein. Enzyme specificity results from the uniqueness of the active site in each different enzyme because of the identity, charge, and spatial orientation of the functional groups located there. It regulates cell chemistry so that the proper reactions occur in the proper place at the proper time. Clearly, it is crucial to the proper functioning of the living cell. Summary A substrate binds to a specific region on an enzyme known as the active site, where the substrate can be converted to product. The substrate binds to the enzyme primarily through hydrogen bonding and other electrostatic interactions. The induced-fit model says that an enzyme can undergo a conformational change when binding a substrate. Enzymes exhibit varying degrees of substrate specificity.
Bookshelves/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/Chapter_18%3A_Acid___Base_Chemistry/18.11_Ions_as_Acids_and_Bases_(Video)	This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students . Video Topics This video discusses how to calculate the pH of a solution contains the conjugate of a weak acid or the conjugate acid of a weak base. The salts of a group one metals and the conjugate base of a weak acid are basic Chloride and bromide salts of the conjugate acid of a weak base are acidic. Link to Video Ions as Acids and Bases: https://youtu.be/XYAGNonPSow Attribution Prof. Steven Farmer ( Sonoma State University )
Courses/Woodland_Community_College/WCC%3A_Chem_10_-_Concepts_of_Chemistry/16%3A_The_Chemical_World	Template:HideTOC 16.1: The Scope of Chemistry Chemistry is the study of matter and the ways in which different forms of matter combine with each other. You study chemistry because it helps you to understand the world around you. Everything you touch or taste or smell is a chemical, and the interactions of these chemicals with each other define our universe. Chemistry forms the fundamental basis for biology and medicine. From the structure of proteins and nucleic acids, to the design, synthesis and manufacture of drugs, chemistry allows you 16.2: Chemicals Compose Ordinary Things Chemistry is the branch of science dealing with the structure, composition, properties, and the reactive characteristics of matter. Matter is anything that has mass and occupies space. Thus, chemistry is the study of literally everything around us – the liquids that we drink, the gasses we breathe, the composition of everything from the plastic case on your phone to the earth beneath your feet. Moreover, chemistry is the study of the transformation of matter. 16.3: Hypothesis, Theories, and Laws Although all of us have taken science classes throughout the course of our study, many people have incorrect or misleading ideas about some of the most important and basic principles in science. We have all heard of hypotheses, theories, and laws, but what do they really mean? Before you read this section, think about what you have learned about these terms before. What do these terms mean to you? What do you read contradicts what you thought? What do you read supports what you thought? 16.4: The Scientific Method- How Chemists Think Science is a process of knowing about the natural universe through observation and experiment. Scientists go through a rigorous process to determine new knowledge about the universe; this process is generally referred to as the scientific method. Science is broken down into various fields, of which chemistry is one. Science, including chemistry, is both qualitative and quantitative. 16.5: A Beginning Chemist- How to Succeed Most people can succeed in chemistry, but it often requires dedication, hard work, the right attitude and study habits!
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Symmetry_(Vallance)/01%3A_Chapters/1.04%3A_Symmetry_and_Physical_Properties	Carrying out a symmetry operation on a molecule must not change any of its physical properties. It turns out that this has some interesting consequences, allowing us to predict whether or not a molecule may be chiral or polar on the basis of its point group. Polarity For a molecule to have a permanent dipole moment, it must have an asymmetric charge distribution. The point group of the molecule not only determines whether the molecule may have a dipole moment, but also in which direction(s) it may point. If a molecule has a $C_n$ axis with $n > 1$, it cannot have a dipole moment perpendicular to the axis of rotation (for example, a $C_2$ rotation would interchange the ends of such a dipole moment and reverse the polarity, which is not allowed – rotations with higher values of $n$ would also change the direction in which the dipole points). Any dipole must lie parallel to a $C_n$ axis. Also, if the point group of the molecule contains any symmetry operation that would interchange the two ends of the molecule, such as a $\sigma_h$ mirror plane or a $C_2$ rotation perpendicular to the principal axis, then there cannot be a dipole moment along the axis. The only groups compatible with a dipole moment are $C_n$, $C_{nv}$ and $C_s$. In molecules belonging to $C_n$ or $C_{nv}$ the dipole must lie along the axis of rotation. Chirality One example of symmetry in chemistry that you will already have come across is found in the isomeric pairs of molecules called enantiomers. Enantiomers are non-superimposable mirror images of each other, and one consequence of this symmetrical relationship is that they rotate the plane of polarized light passing through them in opposite directions. Such molecules are said to be chiral, 2 meaning that they cannot be superimposed on their mirror image. Formally, the symmetry element that precludes a molecule from being chiral is a rotation-reflection axis $S_n$. Such an axis is often implied by other symmetry elements present in a group. For example, a point group that has $C_n$ and $\sigma_h$ as elements will also have $S_n$. Similarly, a center of inversion is equivalent to $S_2$. As a rule of thumb, a molecule definitely cannot have be chiral if it has a center of inversion or a mirror plane of any type ($\sigma_h$, $\sigma_v$ or $\sigma_d$), but if these symmetry elements are absent the molecule should be checked carefully for an $S_n$ axis before it is assumed to be chiral. 2 The word chiral has its origins in the Greek word for hand ( $\chi$$\epsilon$$\rho$$\iota$ , pronounced ‘cheri’ with a soft ch as in ‘loch’). A pair of hands is also a pair of non-superimposable mirror images, and you will often hear chirality referred to as ‘handedness’ for this reason.
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Chemical_Thermodynamics_(Supplement_to_Shepherd_et_al.)/12%3A_Fundamental_10_-_Processes/12.01%3A_Reversible_and_Irreversible_Pathways	The most common example of work in the systems discussed in this book is the work of expansion. It is also convenient to use the work of expansion to exemplify the difference between work that is done reversibly and that which is done irreversibly. The example of expansion against a constant external pressure is an example of an irreversible pathway. It does not mean that the gas cannot be re-compressed. It does, however, mean that there is a definite direction of spontaneous change at all points along the expansion. Imagine instead a case where the expansion has no spontaneous direction of change as there is no net force push the gas to seek a larger or smaller volume. The only way this is possible is if the pressure of the expanding gas is the same as the external pressure resisting the expansion at all points along the expansion. With no net force pushing the change in one direction or the other, the change is said to be reversible or to occur reversibly . The work of a reversible expansion of an ideal gas is fairly easy to calculate. If the gas expands reversibly, the external pressure ($P_{ex}$) can be replaced by a single value ($P$) which represents both the internal pressure of the gas and the external pressure. \[ dw = -PdV\] or \[ w = - \int P dV\] But now that the external pressure is not constant, $P$ cannot be extracted from the integral. Fortunately, however, there is a simple relationship that tells us how $P$ changes with changing $V$ – the equation of state ! If the gas is assumed to be an ideal gas \[ w = - \int P dV -\int \left( \dfrac{nRT}{V}\right) dV\] Constant Temperature (Isothermal) Pathways If the temperature is held constant (so that the expansion follows an isothermal pathway) the nRT term can be extracted from the integral. \[ w = -nRT \int_{V_1}^{V_2} \dfrac{dV}{V} = -nRT \ln \left( \dfrac{V_1}{V_2} \right) \label{isothermal} \] Equation \ref{isothermal} is derived for ideal gases only; a van der Waal gas would result in a different version. Example $\PageIndex{1}$: Gas Expansion What is the work done by 1.00 mol an ideal gas expanding reversibly from a volume of 22.4 L to a volume of 44.8 L at a constant temperature of 273 K? Solution : Using Equation \ref{isothermal} to calculate this \[\begin{align} w & = -(1.00 \, \cancel{mol}) \left(8.314\, \dfrac{J}{\cancel{mol}\,\cancel{ K}}\right) (273\,\cancel{K}) \ln \left( \dfrac{44.8\,L}{22.4 \,L} \right) \nonumber \\[4pt] & = -1570 \,J = 1.57 \;kJ \end{align}\] Note : A reversible expansion will always require more work than an irreversible expansion (such as an expansion against a constant external pressure) when the final states of the two expansions are the same! The work of expansion can be depicted graphically as the area under the P-V curve depicting the expansion. Comparing examples $\PageIndex{1}$ and $3.1.2$ , for which the initial and final volumes were the same, and the constant external pressure of the irreversible expansion was the same as the final pressure of the reversible expansion, such a graph looks as follows. The work is depicted as the shaded portion of the graph. It is clear to see that the reversible expansion (the work for which is shaded in both light and dark gray) exceeds that of the irreversible expansion (shaded in dark gray only) due to the changing pressure of the reversible expansion. In general, it will always be the case that the work generated by a reversible pathway connecting initial and final states will be the maximum work possible for the expansion. It should be noted (although it will be proven in a later chapter) that $\Delta U$ for an isothermal reversible process involving only P-V work is 0 for an ideal gas. This is true because the internal energy, U, is a measure of a system’s capacity to convert energy into work. In order to do this, the system must somehow store that energy. The only mode in which an ideal gas can store this energy is in the translational kinetic energy of the molecules (otherwise, molecular collisions would not need to be elastic, which as you recall, was a postulate of the kinetic molecular theory!) And since the average kinetic energy is a function only of the temperature, it (and therefore $U$) can only change if there is a change in temperature. Hence, for any isothermal process for an ideal gas, $\Delta U=0$. And, perhaps just as usefully, for an isothermal process involving an ideal gas, $q = -w$, as any energy that is expended by doing work must be replaced with heat, lest the system temperature drop. Constant Volume (Isochoric) Pathways One common pathway which processes can follow is that of constant volume. This will happen if the volume of a sample is constrained by a great enough force that it simply cannot change. It is not uncommon to encounter such conditions with gases (since they are highly compressible anyhow) and also in geological formations, where the tremendous weight of a large mountain may force any processes occurring under it to happen at constant volume. If reversible changes in which the only work that can be done is that of expansion (so-called P-V work) are considered, the following important result is obtained: \[ dU = dq + dw = dq - PdV\] However, $dV = 0$ since the volume is constant! As such, $dU$ can be expressed only in terms of the heat that flows into or out of the system at constant volume \[ dU = dq_v\] Recall that $dq$ can be found by \[ dq = \dfrac{dq}{\partial T} dT = C\, dt \label{eq1}\] This suggests an important definition for the constant volume heat capacity ($C_V$) which is \[C_V \equiv \left( \dfrac{\partial U}{\partial T}\right)_V\] When Equation \ref{eq1} is integrated the \[q = \int _{T_1}^{T_2} nC_V dt \label{isochoric}\] Example $\PageIndex{2}$: Isochoric Pathway Consider 1.00 mol of an ideal gas with $C_V = 3/2 R$ that undergoes a temperature change from 125 K to 255 K at a constant volume of 10.0 L. Calculate $\Delta U$, $q$, and $w$ for this change. Solution : Since this is a constant volume process \[w = 0 \nonumber\] Equation \ref{isochoric} is applicable for an isochoric process, \[q = \int _{T_1}^{T_2} nC_V dt \nonumber\] Assuming $C_V$ is independent of temperature: \[\begin{align} q & = nC_V \int _{T_1}^{T_2} dt \\[4pt] &= nC_V ( T_2-T_1) \\[4pt] & = (1.00 \, mol) \left( \dfrac{3}{2} 8.314\, \dfrac{J}{mol \, K}\right) (255\, K - 125 \,K) \\[4pt] & = 1620 \,J = 1.62\, kJ \end{align}\] Since this a constant volume pathway, \[ \begin{align} \Delta U & = q + \cancel{w} \\ & = 1.62 \,kJ \end{align}\] Constant Pressure (Isobaric) Pathways Most laboratory-based chemistry occurs at constant pressure. Specifically, it is exposed to the constant air pressure of the laboratory, glove box, or other container in which reactions are taking place. For constant pressure changes, it is convenient to define a new thermodynamic quantity called enthalpy . \[ H \equiv U+ pV \nonumber\] or \[\begin{align} dH &\equiv dU + d(pV) \\[4pt] &= dU + pdV + Vdp \end{align}\] For reversible changes at constant pressure ($dp = 0$) for which only P-V work is done \[\begin{align} dH & = dq + dw + pdV + Vdp \\[4pt] & = dq - \cancel{pdV} + \cancel{pdV} + \cancelto{0}{Vdp} \\ & = dq \label{heat} \end{align}\] And just as in the case of constant volume changes, this implies an important definition for the constant pressure heat capacity \[C_p \equiv \left( \dfrac{\partial H}{\partial T} \right)_p\] Example $\PageIndex{3}$: Isobaric Gas Expansion Consider 1.00 mol of an ideal gas with $C_p = 5/2 R$ that changes temperature change from 125 K to 255 K at a constant pressure of 10.0 atm. Calculate $\Delta U$, $\Delta H$, $q$, and $w$ for this change. Solution : \[q = \int_{T_1}^{T_2} nC_p dT \nonumber\] assuming $C_p$ is independent of temperature: \[ \begin{align} q & = nC_p \int _{T_1}^{T_2} dT \\ & = nC_p (T_2-T_1) \\ & = (1.00 \, mol) \left( \dfrac{5}{2} 8.314 \dfrac{J}{mol \, K}\right) (255\, K - 125\, K) = 2700\, J = 1.62\, kJ \end{align}\] So via Equation \ref{heat} (specifically the integrated version of it using differences instead of differentials) \[ \Delta H = q = 1.62 \,kJ \nonumber\] \[ \begin{align} \Delta U & = \Delta H - \Delta (pV) \\ & = \Delta H -nR\Delta T \\ & = 2700\, J - (1.00 \, mol) \left( 8.314\, \dfrac{J}{mol \, K}\right) (255\, K - 125 \,K) \\ & = 1620 \,J = 1.62\, kJ \end{align}\] Now that $\Delta U$ and $q$ are determined, then work can be calculated \[\begin{align} w & =\Delta U -q \\ & = 1.62\,kJ - 2.70\,kJ = -1.08\;kJ \end{align}\] It makes sense that $w$ is negative since this process is an gas expansion. Example $\PageIndex{4}$: Isothermal Gas Expansion Calculate $q$, $w$, $\Delta U$, and $\Delta H$ for 1.00 mol of an ideal gas expanding reversibly and isothermally at 273 K from a volume of 22.4 L and a pressure of 1.00 atm to a volume of 44.8 L and a pressure of 0.500 atm. Solution Since this is an isothermal expansion, Equation\ref{isothermal} is applicable \[ \begin{align} w & = -nRT \ln \dfrac{V_2}{V_1} \\ & = (1.00 \, mol) \left( 8.314\, \dfrac{J}{mol \, K}\right) (255\, K) \ln \left(\dfrac{44.8\,L}{22.4\,L} \right) \\ & = 1572\,J = 1.57\,kJ \\[4pt] \Delta U & = q + w \\ & = q + 1.57\,KJ \\ & = 0 \\[4pt] q &= -1.57\,kJ \end{align}\] Since this is an isothermal expansion \[\Delta H = \Delta U + \Delta (pV) = 0 + 0 \nonumber\] where $\Delta (pV) = 0$ due to Boyle’s Law ! Adiabatic Pathways An adiabatic pathway is defined as one in which no heat is transferred ($q = 0$). Under these circumstances, if an ideal gas expands, it is doing work ($w < 0$) against the surroundings (provided the external pressure is not zero!) and as such the internal energy must drop ($\Delta U <0 $). And since $\Delta U$ is negative, there must also be a decrease in the temperature ($\Delta T < 0$). How big will the decrease in temperature be and on what will it depend? The key to answering these questions comes in the solution to how we calculate the work done. If the adiabatic expansion is reversible and done on an ideal gas, \[dw = -PdV\] and \[dw = dU = nC_vdT \label{Adiabate2}\] Equating these two terms yields \[- PdV = nC_v dT\] Using the ideal gas law for an expression for $P$ ($P = nRT/V$) \[ - \dfrac{nRT}{V} dV = nC_vdT\] And rearranging to gather the temperature terms on the right and volume terms on the left yields \[\dfrac{dV}{V} = -\dfrac{C_V}{R} \dfrac{dT}{T}\] This expression can be integrated on the left between $V_1$ and $V_2$ and on the right between $T_1$ and $T_2$. Assuming that $C_v/nR$ is independent of temperature over the range of integration, it can be pulled from the integrand in the term on the right. \[ \int_{V_1}^{V_2} \dfrac{dV}{V} = -\dfrac{C_V}{R} \int_{T_1}^{T_2} \dfrac{dT}{T}\] The result is \[ \ln \left(\dfrac{V_2}{V_1} \right) = - \dfrac{C_V}{R} \ln \left( \dfrac{T_2}{T_1} \right) \] or \[ \left(\dfrac{V_2}{V_1} \right) = \left(\dfrac{T_2}{T_1} \right)^{- \frac{C_V}{R}} \] or \[ V_1T_1^{\frac{C_V}{R}} = V_2T_2^{\frac{C_V}{R}}\] or \[T_1 \left(\dfrac{V_1}{V_2} \right)^{- \frac{R} {C_V}} = T_2 \label{Eq4Alternative}\] Once $\Delta T$ is known, it is easy to calculate $w$, $\Delta U$ and $\Delta H$. Example $\PageIndex{5}$: 1.00 mol of an ideal gas (C V = 3/2 R) initially occupies 22.4 L at 273 K. The gas expands adiabatically and reversibly to a final volume of 44.8 L. Calculate $\Delta T$, $q$, $w$, $\Delta U$, and $\Delta H$ for the expansion. Solution Since the pathway is adiabatic: \[q =0 \nonumber\] Using Equation \ref{Eq4Alternative} \[ \begin{align} T_2 & = T_1 \left(\dfrac{V_1}{V_2} \right)^{- \frac{R} {C_V}} \\ & =(273\,K) \left( \dfrac{22.4\,L}{44.8\,L} \right)^{2/3} \\ & = 172\,K \end{align}\] So \[\Delta T = 172\,K - 273\,K = -101\,K \nonumber\] For calculating work, we integrate Equation \ref{Adiabate2} to get \[ \begin{align} w & = \Delta U = nC_v \Delta T \\ & = (1.00 \, mol) \left(\dfrac{3}{2} 8.314\, \dfrac{J}{mol \, K} \right) (-101\,K ) \\ & = 1.260 \,kJ \end{align}\] \[ \begin{align} \Delta H & = \Delta U + nR\Delta T \\ & = -1260\,J + (1.00 \, mol) \left(\dfrac{3}{2} 8.314\, \dfrac{J}{mol \, K} \right) (-101\,K ) \\ & = -2100\,J \end{align}\] The following table shows recipes for calculating $q$, $w$, $\Delta U$, and $\Delta H$ for an ideal gas undergoing a reversible change along the specified pathway. Pathway $q$ $w$ $\Delta U$ $\Delta H$ Isothermal $nRT \ln (V_2/V_1) $ $-nRT \ln (V_2/V_1) $ 0 0 Isochoric $C_V \Delta T$ 0 $C_V \Delta T$ $C_V \Delta T + V\Delta p$ Isobaric $C_p \Delta T$ $- p\Delta V$ $C_p \Delta T - p\Delta V$ $C_p \Delta T$ Adiabatic 0 $C_V \Delta T$ $C_V \Delta T$ $C_p \Delta T$
Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.10%3A_Electrolytic_Cells_and_Electrolysis	Learning Objectives Make sure you thoroughly understand the following essential ideas. In electrolysis, an external power source supplies the free energy required to drive a cell reaction in its non-spontaneous direction. An electrolytic cell is in this sense the opposite of a galvanic cell. In practice, the products of electrolysis are usually simpler than the reactants, hence the term electro- lysis . Transport of ions in the electrolyte in response to the potential difference between the electrodes ("drift") is largely restricted to the regions very close to the electrodes. Ionic transport in the greater part of the electrolyte is by ordinary thermal diffusion— the statistical tendency of concentrations to become uniform. When an aqueous solution is subjected to electrolysis, the oxidation or reduction of water can be a competing process and may dominate if the applied voltage is sufficiently great. Thus an attempt to electrolyze a solution of NaNO 3 will produce only H 2 and O 2 . A large number of electrolysis processes are employed by industry to refine metals and to produce both inorganic and organic products. The largest of these are the chloralkali industry (chlorine and "caustic"), and the refining of aluminum; the latter consumes approximately 5% of the electrical power generated in North America. Electrolysis refers to the decomposition of a substance by an electric current. The electrolysis of sodium and potassium hydroxides, first carried out in 1808 by Sir Humphrey Davey, led to the discovery of these two metallic elements and showed that these two hydroxides which had previously been considered un-decomposable and thus elements, were in fact compounds. Electrolysis of molten alkali halides is the usual industrial method of preparing the alkali metals: 0 1 Na+ + e– → Na(l) E° = –2.71 V Cl– → ½ Cl2(g) + e– E° = –1.36 V Na+ + Cl– → Na(l) + ½ Cl2(g) E° = –4.1 V Ions in aqueous solutions can undergo similar reactions. Thus if a solution of nickel chloride undergoes electrolysis at platinum electrodes, the reactions are 0 1 Ni2+ + 2 e– → Ni(s) E° = –0.24 V 2 Cl– → Cl2(g) + 2 e– E° = –1.36 V Ni2+ + 2 Cl– → Ni(s) + Cl2(g) E° = –1.60 v Both of these processes are carried out in electrochemical cells which are forced to operate in the "reverse", or non-spontaneous direction, as indicated by the negative for the above cell reaction. The free energy is supplied in the form of electrical work done on the system by the outside world (the surroundings). This is the only fundamental difference between an electrolytic cell and the galvanic cell in which the free energy supplied by the cell reaction is extracted as work done on the surroundings. A common misconception about electrolysis is that "ions are attracted to the oppositely-charged electrode." This is true only in the very thin interfacial region near the electrode surface. Ionic motion throughout the bulk of the solution occurs mostly by diffusion, which is the transport of molecules in response to a concentration gradient. Migration— the motion of a charged particle due to an applied electric field, is only a minor player, producing only about one non-random jump out of around 100,000 random ones for a 1 volt cm –1 electric field. Only those ions that are near the interfacial region are likely to undergo migration. Electrolysis in aqueous solutions Water is capable of undergoing both oxidation \[H_2O \rightarrow O_{2(g)} + 4 H^+ + 2 e^– \;\;\; E^o = -1.23 V\] and reduction \[2 H_2O + 2 e^– \rightarrow H_{2(g)} + 2 OH^– \;\;\; E^o = -0.83 \;V\] Thus if an aqueous solution is subjected to electrolysis, one or both of the above reactions may be able to compete with the electrolysis of the solute. For example, if we try to electrolyze a solution of sodium chloride, hydrogen is produced at the cathode instead of sodium: 0 1 2 NaN H2O + 2 e– → H2(g) + 2 OH– E =+0.41 V ([OH–] = 10-7 M) NaN Cl– → ½ Cl2(g) + e– E° = –1.36 V NaN Cl– + H2O → 2 H2(g) + ½ Cl2(g) + 2 OH– E = –0.95 V [This illustration is taken from the excellent Purdue University Chemistry site ]Reduction of Na+ (E° = –2.7 v) is energetically more difficult than the reduction of water (–1.23 V), so in aqueous solution the latter will prevail. Electrolysis of salt ("brine") is carried out on a huge scale and is the basis of the chloralkali industry. Electrolysis of water Pure water is an insulator and cannot undergo significant electrolysis without adding an electrolyte. If the object is to produce hydrogen and oxygen, the electrolyte must be energetically more difficult to oxidize or reduce than water itself. Electrolysis of a solution of sulfuric acid or of a salt such as NaNO 3 results in the decomposition of water at both electrodes: 0 1 2 NaN H2O + 2 e– → H2(g) + 2 OH– E =+0.41 V ([OH–] = 10-7 M) NaN 2 H2O → O2(g) + 4 H+ + 2 e– E° = -0.82 V NaN 2 H2O(l) → 2 H2(g) + O2(g) E = -1.23 V Electrolytic production of hydrogen is usually carried out with a dilute solution of sulfuric acid. This process is generally too expensive for industrial production unless highly pure hydrogen is required. However, it becomes more efficient at higher temperatures, where thermal energy reduces the amount of electrical energy required, so there is now some interest in developing high-temperature electrolytic processes. Most hydrogen gas is manufactured by the steam reforming of natural gas. Faraday's laws of electrolysis One mole of electric charge (96,500 coulombs), when passed through a cell, will discharge half a mole of a divalent metal ion such as Cu 2 + . This relation was first formulated by Faraday in 1832 in the form of two laws of electrolysis : The weights of substances formed at an electrode during electrolysis are directly proportional to the quantity of electricity that passes through the electrolyte. The weights of different substances formed by the passage of the same quantity of electricity are proportional to the equivalent weight of each substance. The equivalent weight of a substance is defined as the molar mass, divided by the number of electrons required to oxidize or reduce each unit of the substance. Thus one mole of V 3+ corresponds to three equivalents of this species, and will require three faradays of charge to deposit it as metallic vanadium. Most stoichiometric problems involving electrolysis can be solved without explicit use of Faraday's laws. The "chemistry" in these problems is usually very elementary; the major difficulties usually stem from unfamiliarity with the basic electrical units: current (amperes) is the rate of charge transport ; 1 amp = 1 C/sec. power (watts) is the rate of energy production or consumption; \[1 W = 1 J/sec = 1 volt-amp; 1 watt-sec = 1 J, 1 kW-h = 3600 J.\] Example $\PageIndex{1}$ A metallic object to be plated with copper is placed in a solution of CuSO 4 . To which electrode of a direct current power supply should the object be connected? What mass of copper will be deposited if a current of 0.22 amp flows through the cell for 1.5 hours? Solution Since Cu 2 + ions are being reduced, the object acts as a cathode and must be connected to the negative terminal (where the electrons come from!) The amount of charge passing through the cell is (0.22 amp) × (5400 sec) = 1200 C or (1200 C) ÷ (96500 c F –1 ) = 0.012 F Since the reduction of one mole of Cu 2 + ion requires the addition of two moles of electrons, the mass of Cu deposited will be (63.54 g mol –1 ) (0.5 mol Cu/F) (.012 F) = 0.39 g of copper Example $\PageIndex{2}$ How much electric power is required to produce 1 metric ton (1000 kg) of chlorine from brine, assuming the cells operate at 2.0 volts and assuming 100 % efficiency? Solution moles of Cl 2 produced: (10 6 g) ÷ 70 g mol –1 = 14300 mol Cl 2 faradays of charge: (2 F/mol) × (14300 mol) = 28600 F charge in coulombs: (96500 C/F) × (28600 F) = 2.76 × 10 9 C duration of electrolysis: (3600 s/h) x (24 h) = 86400 s current (rate of charge delivery): (2.76 × 10 9 amp-sec) ÷ (86400 sec) = 32300 amps power (volt-amps): (2.0 V) × (32300 amp) = 64.6 kW energy in kW-h: (64.6 kW) × (24 h) = 1550 kW-h energy in joules: (1550 kW-h) × (3.6 MJ/kW-h) = 5580 MJ (megajoules) (In the last step, recall that 1 W = 1 J/s, so 1 kW-h = 3.6 MJ) Industrial Electrolytic Processes For many industrial-scale operations involving the oxidation or reduction of both inorganic and organic substances, and especially for the production of the more active metals such as sodium, calcium, magnesium, and aluminum, the most cost-effective reducing agent is electrons supplied by an external power source. The two most economically important of these processes are described below. The chloralkali industry The electrolysis of brine is carried out on a huge scale for the industrial production of chlorine and caustic soda (sodium hydroxide). Because the reduction potential of Na + is much higher than that of water, the latter substance undergoes decomposition at the cathode, yielding hydrogen gas and OH – . 0 1 2 3 anode reactions 2 Cl– → Cl2(g) + 2 e– -1.36 V i anode reactions 4 OH– → O2(g) + 2 H2O + 4 e– -0.40 V ii cathode reactions Na+ + e– → Na(s) -2.7 V iii cathode reactions H2O + 2 e– → H2(g) + 2 OH– +.41 V iv A comparison of the E° s would lead us to predict that the reduction ( ii ) would be favored over that of ( i ). This is certainly the case from a purely energetic standpoint, but as was mentioned in the section on fuel cells, electrode reactions involving O 2 are notoriously slow (that is, they are kinetically hindered), so the anodic process here is under kinetic rather than thermodynamic control. The reduction of water ( iv ) is energetically favored over that of Na + ( iii ), so the net result of the electrolysis of brine is the production of Cl 2 and NaOH ("caustic"), both of which are of immense industrial importance: \[\ce{2 NaCl + 2 H2O -> 2 NaOH + Cl2(g) + H2(g)} \nonumber\] Since chlorine reacts with both OH – and H 2 , it is necessary to physically separate the anode and cathode compartments. In modern plants this is accomplished by means of an ion-selective polymer membrane, but prior to 1970 a more complicated cell was used that employed a pool of mercury as the cathode. A small amount of this mercury would normally find its way into the plant's waste stream, and this has resulted in serious pollution of many major river systems and estuaries and devastation of their fisheries. Electrolytic refining of aluminum Aluminum is present in most rocks and is the most abundant metallic element in the earth's crust (eight percent by weight.) However, its isolation is very difficult and expensive to accomplish by purely chemical means, as evidenced by the high E° (–1.66 V) of the Al 3 + /Al couple. For the same reason, aluminum cannot be isolated by electrolysis of aqueous solutions of its compounds, since the water would be electrolyzed preferentially. And if you have ever tried to melt a rock, you will appreciate the difficulty of electrolyzing a molten aluminum ore! Aluminum was in fact considered an exotic and costly metal until 1886, when Charles Hall (U.S.A) and Paul Hérault (France) independently developed a practical electrolytic reduction process. The Hall-Hérault process takes advantage of the principle that the melting point of a substance is reduced by admixture with another substance with which it forms a homogeneous phase. Instead of using the pure alumina ore Al 2 O 3 which melts at 2050°C, it is mixed with cryolite, which is a natural mixture of NaF and AlF 3 , thus reducing the temperature required to a more manageable 1000°C. The anodes of the cell are made of carbon (actually a mixture of pitch and coal), and this plays a direct role in the process; the carbon gets oxidized (by the oxide ions left over from the reduction of Al 3 + to CO, and the free energy of this reaction helps drive the aluminum reduction, lowering the voltage that must be applied and thus reducing the power consumption. This is important, because aluminum refining is the largest consumer of industrial electricity, accounting for about 5% of all electricity generated in North America. Since aluminum cells commonly operated at about 100,000 amperes, even a slight reduction in voltage can result in a large saving of power. The net reaction is \[\ce{2 Al_2O_3 + 3 C \rightarrow 4 Al + 3 CO_2} \nonumber\] However, large quantities of CO and of HF (from the cryolite), and hydrocarbons (from the electrodes) are formed in various side reactions, and these can be serious sources of environmental pollution.
Bookshelves/Organic_Chemistry/Organic_Chemistry_I_(Liu)/09%3A_Free_Radical_Substitution_Reaction_of_Alkanes/9.01%3A_Homolytic_and_Heterolytic_Cleavage	For the reactions we learned so far, bond breaking occurs in the way that one part of the bond takes both electrons (the electron pair) of the bond away. For example of S N 1 reaction, the leaving group Br leaves with the electron pair to form Br – and carbocation intermediate. This process is called heterolytic bond cleavage , the σ bond breaks heterolytically. As we have always been doing, an arrow with the double-barbs is used to show heterolytic cleavage, that is the transfer of electron pair specifically: There is another type of bond breaking process, in which each part of the σ bond takes one electron away, as shown below: This is called homolytic cleavage , or homolysis . The electron pair separate evenly to each part, and as a result both products contain a single electron. The species that contain one or more single electron is called radical (or free radical). Radicals are produced from homolytic cleavage. The arrow with sing-barb (like the shape of a fishhook) is used to show homolytic cleavage, that is single electron transfer specifically: Homolysis occurs mainly for non-polar bonds, heat or light (delta is the symbol for heat; hv is used to show light) is needed to provide enough energy for initiating the process. For example: Radical is another highly reactive reaction intermediate, because of the lack of octet. The substitution reaction we will learn in this chapter involves the radical intermediate.
Courses/Chippewa_Valley_Technical_College/CVTC_Basic_Chemistry/07%3A_Solutions/7.09%3A_Strong_and_Weak_Electrolytes	Car batteries are used around the world to provide the power to start car engines. One essential component of car batteries is the strong electrolyte sulfuric acid. In the battery, this material ionizes into hydrogen ions and sulfate ions. As the battery is used, the concentrations of these ions decreases. Older batteries had openings in the top where new sulfuric acid could be added to replenish the supply. Today, batteries are sealed to prevent leakage of hazardous sulfuric acid. Strong and Weak Electrolytes Some polar molecular compounds are nonelectrolytes when they are in their pure state, but become electrolytes when they are dissolved in water. Hydrogen chloride $\left( \ce{HCl} \right)$ is a gas in its pure molecular state and is a nonelectrolyte. However, when $\ce{HCl}$ is dissolved in water, it conducts a current well because the $\ce{HCl}$ molecule ionizes into hydrogen and chloride ions. \[\ce{HCl} \left( g \right) \rightarrow \ce{H^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)\nonumber \] $\ce{HCl}$ dissolved into water is called hydrochloric acid. A strong electrolyte is a solution in which a large fraction of the dissolved solute exists as ions. Ionic compounds, and some polar compounds, are completely broken apart into ions and thus conduct a current very well—this makes them strong electrolytes. Some other polar molecular compounds become electrolytes upon being dissolved into water, but do not ionize to very great extent. Gaseous nitrous acid ionizes in solution to hydrogen ions and nitrite ions, but does so very weakly. Aqueous nitrous acid is composed of only about $5\%$ ions and $95\%$ intact nitrous acid molecules. A weak electrolyte is a solution in which only a small fraction of the dissolved solute exists as ions. The equation showing the ionization of a weak electrolyte utilizes a double arrow indicating an equilibrium between the reactants and products. \[\ce{HNO_2} \left( g \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{NO_2^-} \left( aq \right)\nonumber \] Summary A strong electrolyte exists mainly as ions in solution. A weak electrolyte has only a small amount of ionization in solution.
Courses/City_College_of_San_Francisco/Chemistry_101A/Topic_E%3A_Atomic_Structure/07%3A_Electronic_Structure_of_Atoms/7.06%3A_3D_Representation_of_Orbitals	Learning Objectives To understand the 3D representation of electronic orbitals An orbital is the quantum mechanical refinement of Bohr’s orbit. In contrast to his concept of a simple circular orbit with a fixed radius, orbitals are mathematically derived regions of space with different probabilities of containing an electron. One way of representing electron probability distributions was illustrated previously for the 1 s orbital of hydrogen. Because Ψ 2 gives the probability of finding an electron in a given volume of space (such as a cubic picometer), a plot of Ψ 2 versus distance from the nucleus ( r ) is a plot of the probability density . The 1 s orbital is spherically symmetrical, so the probability of finding a 1 s electron at any given point depends only on its distance from the nucleus. The probability density is greatest at r = 0 (at the nucleus) and decreases steadily with increasing distance. At very large values of r , the electron probability density is very small but not zero. In contrast, we can calculate the radial probability (the probability of finding a 1 s electron at a distance r from the nucleus) by adding together the probabilities of an electron being at all points on a series of x spherical shells of radius r 1 , r 2 , r 3 ,…, r x − 1 , r x . In effect, we are dividing the atom into very thin concentric shells, much like the layers of an onion (Figure $\PageIndex{1a}$), and calculating the probability of finding an electron on each spherical shell. Recall that the electron probability density is greatest at r = 0 (Figure $\PageIndex{1b}$), so the density of dots is greatest for the smallest spherical shells in part (a) in Figure $\PageIndex{1}$. In contrast, the surface area of each spherical shell is equal to 4π r 2 , which increases very rapidly with increasing r (Figure $\PageIndex{1c}$). Because the surface area of the spherical shells increases more rapidly with increasing r than the electron probability density decreases, the plot of radial probability has a maximum at a particular distance (Figure $\PageIndex{1d}$). Most important, when r is very small, the surface area of a spherical shell is so small that the total probability of finding an electron close to the nucleus is very low; at the nucleus, the electron probability vanishes (Figure $\PageIndex{1d}$). Figure $\PageIndex{1}$ : Most Probable Radius for the Electron in the Ground State of the Hydrogen Atom. (a) Imagine dividing the atom’s total volume into very thin concentric shells as shown in the onion drawing. (b) A plot of electron probability density Ψ 2 versus r shows that the electron probability density is greatest at r = 0 and falls off smoothly with increasing r . The density of the dots is therefore greatest in the innermost shells of the onion. (c) The surface area of each shell, given by 4π r 2 , increases rapidly with increasing r . (d) If we count the number of dots in each spherical shell, we obtain the total probability of finding the electron at a given value of r . Because the surface area of each shell increases more rapidly with increasing r than the electron probability density decreases, a plot of electron probability versus r (the radial probability ) shows a peak. This peak corresponds to the most probable radius for the electron, 52.9 pm, which is exactly the radius predicted by Bohr’s model of the hydrogen atom. For the hydrogen atom, the peak in the radial probability plot occurs at r = 0.529 Å (52.9 pm), which is exactly the radius calculated by Bohr for the n = 1 orbit. Thus the most probable radius obtained from quantum mechanics is identical to the radius calculated by classical mechanics. In Bohr’s model, however, the electron was assumed to be at this distance 100% of the time, whereas in the Schrödinger model, it is at this distance only some of the time. The difference between the two models is attributable to the wavelike behavior of the electron and the Heisenberg uncertainty principle. Figure $\PageIndex{2}$ compares the electron probability densities for the hydrogen 1 s , 2 s , and 3 s orbitals. Note that all three are spherically symmetrical. For the 2 s and 3 s orbitals, however (and for all other s orbitals as well), the electron probability density does not fall off smoothly with increasing r . Instead, a series of minima and maxima are observed in the radial probability plots (Figure $\PageIndex{2c}$). The minima correspond to spherical nodes (regions of zero electron probability), which alternate with spherical regions of nonzero electron probability. The existence of these nodes is a consequence of changes of wave phase in the wavefunction Ψ. Figure $\PageIndex{2}$ : Probability Densities for the 1 s , 2 s , and 3 s Orbitals of the Hydrogen Atom. (a) The electron probability density in any plane that contains the nucleus is shown. Note the presence of circular regions, or nodes, where the probability density is zero. (b) Contour surfaces enclose 90% of the electron probability, which illustrates the different sizes of the 1 s , 2 s , and 3 s orbitals. The cutaway drawings give partial views of the internal spherical nodes. The orange color corresponds to regions of space where the phase of the wave function is positive, and the blue color corresponds to regions of space where the phase of the wave function is negative. (c) In these plots of electron probability as a function of distance from the nucleus ( r ) in all directions (radial probability), the most probable radius increases as n increases, but the 2 s and 3 s orbitals have regions of significant electron probability at small values of r . s Orbitals (l=0) Three things happen to s orbitals as n increases (Figure $\PageIndex{2}$): They become larger, extending farther from the nucleus. They contain more nodes. This is similar to a standing wave that has regions of significant amplitude separated by nodes, points with zero amplitude. For a given atom, the s orbitals also become higher in energy as n increases because of their increased distance from the nucleus. Orbitals are generally drawn as three-dimensional surfaces that enclose 90% of the electron density, as was shown for the hydrogen 1 s , 2 s , and 3 s orbitals in part (b) in Figure $\PageIndex{2}$. Although such drawings show the relative sizes of the orbitals, they do not normally show the spherical nodes in the 2 s and 3 s orbitals because the spherical nodes lie inside the 90% surface. Fortunately, the positions of the spherical nodes are not important for chemical bonding. p Orbitals (l=1) Only s orbitals are spherically symmetrical. As the value of l increases, the number of orbitals in a given subshell increases, and the shapes of the orbitals become more complex. Because the 2 p subshell has l = 1, with three values of m l (−1, 0, and +1), there are three 2 p orbitals. The electron probability distribution for one of the hydrogen 2 p orbitals is shown in Figure $\PageIndex{3}$. Because this orbital has two lobes of electron density arranged along the z axis, with an electron density of zero in the xy plane (i.e., the xy plane is a nodal plane), it is a $2p_z$ orbital. As shown in Figure $\PageIndex{4}$, the other two 2 p orbitals have identical shapes, but they lie along the x axis ($2p_x$) and y axis ($2p_y$), respectively. Note that each p orbital has just one angular node. The angular node has the shape of plane in the following illustrations. In each case, the phase of the wave function for each of the 2 p orbitals is positive for the lobe that points along the positive axis and negative for the lobe that points along the negative axis. It is important to emphasize that these signs correspond to the phase of the wave that describes the electron motion, not to positive or negative charges. Figure $\PageIndex{4}$ The Three Equivalent 2 p Orbitals of the Hydrogen Atom The surfaces shown enclose 90% of the total electron probability for the 2 p x , 2 p y , and 2 p z orbitals. Each orbital is oriented along the axis indicated by the subscript and a nodal plane that is perpendicular to that axis bisects each 2 p orbital. The phase of the wave function is positive (orange) in the region of space where x , y , or z is positive and negative (blue) where x , y , or z is negative. Just as with the s orbitals, the size and complexity of the p orbitals for any atom increase as the principal quantum number n increases. The shapes of the 90% probability surfaces of the 3 p , 4 p , and higher-energy p orbitals are, however, essentially the same as those shown in Figure $\PageIndex{4}$. d Orbitals (l=2) Subshells with l = 2 have five d orbitals; the first principal shell to have a d subshell corresponds to n = 3. The five d orbitals have m l values of −2, −1, 0, +1, and +2. The hydrogen 3d orbitals, shown in Figure $\PageIndex{5}$, have more complex shapes than the 2p orbitals. All five 3d orbitals contain two angular nodes . The angular nodes have the shape of a planar or conical surface where there is zero probability of finding the electron. Note the contrast as compared to one angular node for each p orbital and zero angular nodes for each s orbital. In three of the d orbitals, the lobes of electron density are oriented between the x and y, x and z, and y and z planes; these orbitals are referred to as the $3d_{xy}$, $3d_{xz}$, and $3d_{yz}$ orbitals, respectively. A fourth d orbital has lobes lying along the x and y axes; this is the $3d_{x^2−y^2}$ orbital. The fifth 3d orbital, called the $3d_{z^2}$ orbital, has a unique shape: it looks like a $2p_z$ orbital combined with an additional doughnut of electron probability lying in the xy plane. Despite its peculiar shape, the $3d_{z^2}$ orbital is mathematically equivalent to the other four and has the same energy. In contrast to p orbitals, the phase of the wave function for d orbitals is the same for opposite pairs of lobes. As shown in Figure $\PageIndex{5}$, the phase of the wave function is positive for the two lobes of the $dz^2$ orbital that lie along the z axis, whereas the phase of the wave function is negative for the doughnut of electron density in the xy plane. Like the s and p orbitals, as n increases, the size of the d orbitals increases, but the overall shapes remain similar to those depicted in Figure $\PageIndex{5}$. f Orbitals (l=3) Principal shells with n = 4 can have subshells with l = 3 and m l values of −3, −2, −1, 0, +1, +2, and +3. These subshells consist of seven f orbitals. Each f orbital has three angular nodes, so their shapes are complex. Because f orbitals are not particularly important for our purposes, we do not discuss them further, and orbitals with higher values of l are not discussed at all. Orbital Energies Although we have discussed the shapes of orbitals, we have said little about their comparative energies. We begin our discussion of orbital energies by considering atoms or ions with only a single electron (such as H or He + ). The relative energies of the atomic orbitals with n ≤ 4 for a hydrogen atom are plotted in Figure $\PageIndex{6}$; note that the orbital energies depend on only the principal quantum number n . Consequently, the energies of the 2 s and 2 p orbitals of hydrogen are the same; the energies of the 3 s , 3 p , and 3 d orbitals are the same; and so forth. Quantum mechanics predicts that in the hydrogen atom, all orbitals with the same value of n (e.g., the three 2 p orbitals) are degenerate , meaning that they have the same energy. The orbital energies obtained for hydrogen using quantum mechanics are exactly the same as the allowed energies calculated by Bohr. In contrast to Bohr’s model, however, which allowed only one orbit for each energy level, quantum mechanics predicts that there are 4 orbitals with different electron density distributions in the n = 2 principal shell (one 2 s and three 2 p orbitals), 9 in the n = 3 principal shell, and 16 in the n = 4 principal shell. The different values of l and m l for the individual orbitals within a given principal shell are not important for understanding the emission or absorption spectra of the hydrogen atom under most conditions, but they do explain the splittings of the main lines that are observed when hydrogen atoms are placed in a magnetic field. Figure $\PageIndex{6}$ shows that the energy levels become closer and closer together as the value of n increases, as expected because of the 1/ n 2 dependence of orbital energies. The energies of the orbitals in any species with only one electron can be calculated by a minor variation of Bohr’s equation, which can be extended to other single-electron species by incorporating the nuclear charge Z (the number of protons in the nucleus): \[E=−\dfrac{Z^2}{n^2}Rhc \] In general, both energy and radius decrease as the nuclear charge increases. Thus the most stable orbitals (those with the lowest energy) are those closest to the nucleus. For example, in the ground state of the hydrogen atom, the single electron is in the 1s orbital, whereas in the first excited state, the atom has absorbed energy and the electron has been promoted to one of the n = 2 orbitals. In ions with only a single electron, the energy of a given orbital depends on only n, and all subshells within a principal shell, such as the $p_x$, $p_y$, and $p_z$ orbitals, are degenerate. Summary The four chemically important types of atomic orbital correspond to values of $\ell = 0$, $1$, $2$, and $3$. Orbitals with $\ell = 0$ are s orbitals and they are spherically symmetrical with no angular nodes. The s orbitals have the greatest electron density occurring at the nucleus. Orbitals with $\ell = 1$ are p orbitals and contain a angular node (a nodal surface shaped like a plane) that includes the nucleus, giving rise to a dumbbell shape. Orbitals with $\ell = 2$ are d orbitals and have more complex shapes with two angular nodes (nodal surfaces shaped like planes or cones). Orbitals with $\ell = 3$ are f orbitals, which are still more complex. Three things happen to all orbital types (s, p, d, f) as n increases: They become larger, extending farther from the nucleus. They contain more nodes. This is similar to a standing wave that has regions of significant amplitude separated by nodes, points with zero amplitude. They become higher in energy as n increases. Because its average distance from the nucleus determines the energy of an electron, each atomic orbital with a given set of quantum numbers has a particular energy associated with it, the orbital energy . \[E=−\dfrac{Z^2}{n^2}Rhc \nonumber\] In atoms or ions with only a single electron, all orbitals with the same value of $n$ have the same energy (they are degenerate), and the energies of the principal shells increase smoothly as $n$ increases. An atom or ion with the electron(s) in the lowest-energy orbital(s) is said to be in its ground state, whereas an atom or ion in which one or more electrons occupy higher-energy orbitals is said to be in an excited state.
Courses/Community_College_of_Allegheny_County/Developmental_Studies_101_and_103/10%3A_Chapter_9%3A_Effective_Research/10.1%3A_Information_Literacy	Welcome to the Chemistry Library. This Living Library is a principal hub of the LibreTexts project , which is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning. The LibreTexts approach is highly collaborative where an Open Access textbook environment is under constant revision by students, faculty, and outside experts to supplant conventional paper-based books. Campus Bookshelves Bookshelves Learning Objects
Bookshelves/Organic_Chemistry/Organic_Chemistry_(OpenStax)/25%3A_Biomolecules_-_Carbohydrates/25.08%3A_Disaccharides	We saw in Section 25.6 that reaction of a monosaccharide with an alcohol yields a glycoside in which the anomeric –OH group is replaced by an –OR substituent. If the alcohol is itself a sugar, the glycosidic product is a disaccharide . Maltose and Cellobiose Disaccharides contain a glycosidic acetal bond between the anomeric carbon of one sugar and an –OH group at any position on the other sugar. A glycosidic bond between C1 of the first sugar and the –OH at C4 of the second sugar is particularly common. Such a bond is called a 1→4 link. The glycosidic bond to an anomeric carbon can be either α or β. Maltose, the disaccharide obtained by enzyme-catalyzed hydrolysis of starch, consists of two α - D -glucopyranose units joined by a 1→4- α -glycoside bond. Cellobiose, the disaccharide obtained by partial hydrolysis of cellulose, consists of two β - D -glucopyranose units joined by a 1→4- β -glycoside bond. Maltose and cellobiose are both reducing sugars because the anomeric carbons on their right-hand glucopyranose units have hemiacetal groups and are in equilibrium with aldehyde forms. For a similar reason, both maltose and cellobiose exhibit mutarotation of α and β anomers of the glucopyranose unit on the right. Despite the similarities of their structures, cellobiose and maltose have dramatically different biological properties. Cellobiose can’t be digested by humans and can’t be fermented by yeast. Maltose, however, is digested without difficulty and is fermented readily. Exercise $\PageIndex{1}$ Show the product you would obtain from the reaction of cellobiose with the following reagents: NaBH 4 Br 2 , H 2 O CH 3 COCl, pyridine Answer The hemiacetal ring is reduced. The hemiacetal ring is oxidized. All hydroxyl groups are acetylated. Lactose Lactose is a disaccharide that occurs naturally in both human and cow’s milk. It is widely used in baking and in commercial milk formulas for infants. Like maltose and cellobiose, lactose is a reducing sugar. It exhibits mutarotation and is a 1→4-β-linked glycoside. Unlike maltose and cellobiose, however, lactose contains two different monosaccharides— D -glucose and D -galactose—joined by a β-glycosidic bond between C1 of galactose and C4 of glucose. Sucrose Sucrose, or ordinary table sugar, is probably the most abundant pure organic chemical in the world. Whether from sugar cane (20% sucrose by weight) or sugar beets (15% by weight), and whether raw or refined, all table sugar is sucrose. Sucrose is a disaccharide that yields 1 equivalent of glucose and 1 equivalent of fructose on hydrolysis. This 1 : 1 mixture of glucose and fructose is often referred to as invert sugar because the sign of optical rotation changes, or inverts, during the hydrolysis of sucrose ([α] D = +66.5) to a glucose/ fructose mixture ([α] D = –22.0). Some insects, such as honeybees, have enzymes called invertases that catalyze the sucrose hydrolysis. Honey, in fact, is primarily a mixture of glucose, fructose, and sucrose. Unlike most other disaccharides, sucrose is not a reducing sugar and does not undergo mutarotation. These observations imply that sucrose is not a hemiacetal and that glucose and fructose must both be glycosides. This can happen only if the two sugars are joined by a glycoside link between the anomeric carbons of both sugars—C1 of glucose and C2 of fructose.
Courses/Grinnell_College/CHM_364%3A_Physical_Chemistry_2_(Grinnell_College)/04%3A_Postulates_and_Principles_of_Quantum_Mechanics/4.05%3A_Eigenfunctions_of_Operators_are_Orthogonal	Understand the properties of a Hermitian operator and their associated eigenstates Recognize that all experimental obervables are obtained by Hermitian operators Consideration of the quantum mechanical description of the particle-in-a-box exposed two important properties of quantum mechanical systems. We saw that the eigenfunctions of the Hamiltonian operator are orthogonal, and we also saw that the position and momentum of the particle could not be determined exactly. We now examine the generality of these insights by stating and proving some fundamental theorems. These theorems use the Hermitian property of quantum mechanical operators that correspond to observables, which is discuss first. Hermitian Operators Since the eigenvalues of a quantum mechanical operator correspond to measurable quantities, the eigenvalues must be real, and consequently a quantum mechanical operator must be Hermitian. To prove this, we start with the premises that $ψ$ and $φ$ are functions, $\int d\tau$ represents integration over all coordinates, and the operator $\hat {A}$ is Hermitian by definition if \[ \int \psi ^* \hat {A} \psi \,d\tau = \int (\hat {A} ^* \psi ^* ) \psi \,d\tau \label {4-37} \] This equation means that the complex conjugate of $\hat {A}$ can operate on $ψ^$ to produce the same result after integration as $\hat {A}$ operating on $φ$, followed by integration. To prove that a quantum mechanical operator $\hat {A}$ is Hermitian, consider the eigenvalue equation and its complex conjugate. \[\hat {A} \psi = a \psi \label {4-38} \] \[\hat {A}^ \psi ^* = a^* \psi ^* = a \psi ^* \label {4-39} \] Note that $a^* = a$ because the eigenvalue is real. Multiply Equation $\ref{4-38}$ and $\ref{4-39}$ from the left by $ψ^$ and $ψ$, respectively, and integrate over the full range of all the coordinates. Note that $ψ$ is normalized. The results are \[ \int \psi ^ \hat {A} \psi \,d\tau = a \int \psi ^* \psi \,d\tau = a \label {4-40} \] \[ \int \psi \hat {A}^* \psi ^* \,d \tau = a \int \psi \psi ^* \,d\tau = a \label {4-41} \] Since both integrals equal $a$, they must be equivalent. \[ \int \psi ^* \hat {A} \psi \,d\tau = \int \psi \hat {A}^* \psi ^* \,d\tau \label {4-42} \] The operator acting on the function, \[\hat {A}^* \int \psi ^* \hat {A} \psi \,d\tau = \int \psi \hat {A} ^* \psi ^* \,d\tau_* \nonumber \] produces a new function. Since functions commute, Equation $\ref{4-42}$ can be rewritten as \[ \int \psi ^* \hat {A} \psi d\tau = \int (\hat {A}^\psi ^) \psi d\tau \label{4-43} \] This equality means that $\hat {A}$ is Hermitian. Eigenfunctions of a Hermitian operator are orthogonal if they have different eigenvalues. Because of this theorem, we can identify orthogonal functions easily without having to integrate or conduct an analysis based on symmetry or other considerations. Proof $ψ$ and $φ$ are two eigenfunctions of the operator Â with real eigenvalues $a_1$ and $a_2$, respectively. Since the eigenvalues are real, $a_1^* = a_1$ and $a_2^* = a_2$. \[\hat {A} \psi = a_1 \psi \nonumber \] \[\hat {A}^* \psi ^* = a_2 \psi ^* \nonumber \] Multiply the first equation by $φ^$ and the second by $ψ$ and integrate. \[\int \psi ^ \hat {A} \psi \,d\tau = a_1 \int \psi ^* \psi \,d\tau \nonumber \] \[\int \psi \hat {A}^* \psi ^* \,d\tau = a_2 \int \psi \psi ^* \,d\tau \label {4-45} \] Subtract the two equations in Equation \ref{4-45} to obtain \[\int \psi ^\hat {A} \psi \,d\tau - \int \psi \hat {A} ^ \psi ^* \,d\tau = (a_1 - a_2) \int \psi ^* \psi \,d\tau \label {4-46} \] The left-hand side of Equation \ref{4-46} is zero because $\hat {A}$ is Hermitian yielding \[ 0 = (a_1 - a_2 ) \int \psi ^* \psi \, d\tau \label {4-47} \] If $a_1$ and $a_2$ in Equation \ref{4-47} are not equal, then the integral must be zero. This result proves that nondegenerate eigenfunctions of the same operator are orthogonal. $\square$ Two wavefunctions, $\psi_1(x)$ and $\psi_2(x)$, are said to be orthogonal if \[\int_{-\infty}^{\infty}\psi_1^\ast \psi_2 \,dx = 0. \label{4.5.1} \] Consider two eigenstates of $\hat{A}$, $\psi_a(x)$ and $\psi_{a'}(x)$, which correspond to the two different eigenvalues $a$ and $a'$, respectively. Thus, \[A\psi_a = a \psi_a \label{4.5.2} \] \[A\psi_a' = a' \psi_a' \label{4.5.3} \] Multiplying the complex conjugate of the first equation by $\psi_{a'}(x)$, and the second equation by $\psi^_{a'}(x)$, and then integrating over all $x$, we obtain \[ \int_{-\infty}^\infty (A \psi_a)^\ast \psi_{a'} dx = a \int_{-\infty}^\infty\psi_a^\ast \psi_{a'} dx, \label{ 4.5.4} \] \[ \int_{-\infty}^\infty \psi_a^\ast (A \psi_{a'}) dx = a' \int_{-\infty}^{\infty}\psi_a^\ast \psi_{a'} dx. \label{4.5.5} \] However, from Equation $\ref{4-46}$, the left-hand sides of the above two equations are equal. Hence, we can write \[(a-a') \int_{-\infty}^\infty\psi_a^\ast \psi_{a'} dx = 0. \nonumber \] By assumption, $a \neq a'$, yielding \[\int_{-\infty}^\infty\psi_a^\ast \psi_{a'} dx = 0. \nonumber \] In other words, eigenstates of an Hermitian operator corresponding to different eigenvalues are automatically orthogonal . The eigenvalues of operators associated with experimental measurements are all real . Draw graphs and use them to show that the particle-in-a-box wavefunctions for $\psi(n = 2)$ and $\psi(n = 3)$ are orthogonal to each other. Solution The two PIB wavefunctions are qualitatively similar when plotted These wavefunctions are orthogonal when \[\int_{-\infty}^{\infty} \psi(n=2) \psi(n=3) dx =0 \nonumber \] and when the PIB wavefunctions are substituted this integral becomes \[\begin{align} \int_0^L \sqrt{\dfrac{2}{L}} \sin \left( \dfrac{2n}{L}x \right) \sqrt{\dfrac{2}{L}} \sin \left( \dfrac{2n}{L}x \right) dx &= ? \\[4pt] \dfrac{2}{L} \int_0^L \sin \left( \dfrac{2}{L}x \right) \sin \left( \dfrac{3}{L}x \right) &= ? \end{align} \] We can expand the integrand using trigonometric identities to help solve the integral, but it is easier to take advantage of the symmetry of the integrand, specifically, the $\psi(n=2)$ wavefunction is even (blue curves in above figure) and the $\psi(n=3)$ is odd (purple curve). Their product (even times odd) is an odd function and the integral over an odd function is zero. Therefore $\psi(n=2)$ and $\psi(n=3)$ wavefunctions are orthogonal. This can be repeated an infinite number of times to confirm the entire set of PIB wavefunctions are mutually orthogonal as the Orthogonality Theorem guarantees. Orthogonality of Degenerate Eigenstates Consider two eigenstates of $\hat{A}$, $\psi_a$ and $\psi'_a$, which correspond to the same eigenvalue, $a$. Such eigenstates are termed degenerate . The above proof of the orthogonality of different eigenstates fails for degenerate eigenstates. Note, however, that any linear combination of $\psi_a$ and $\psi'_a$ is also an eigenstate of $\hat{A}$ corresponding to the eigenvalue $a$. Thus, even if $\psi_a$ and $\psi'_a$ are not orthogonal, we can always choose two linear combinations of these eigenstates which are orthogonal. For instance, if $\psi_a$ and $\psi'_a$ are properly normalized, we can define the overlap integral \[S= \int_{-\infty}^\infty \psi_a^\ast \psi_a' dx ,\label{ 4.5.10} \] It is easily demonstrated (but not here) that \[\psi_a'' = \frac{\vert S\vert}{\sqrt{1-\vert S\vert^2}}\left(\psi_a - S^{-1} \psi_a'\right) \label{4.5.11} \] is a properly normalized eigenstate of $\hat{A}$, corresponding to the eigenvalue $a$, which is orthogonal to $\psi_a$. It is straightforward to generalize the above argument to three or more degenerate eigenstates. Hence, we conclude that the eigenstates of a Hermitian operator are, or can be chosen to be, mutually orthogonal. Theorem: Gram-Schmidt Orthogonalization Degenerate eigenfunctions are not automatically orthogonal, but can be made so mathematically via the Gram-Schmidt Orthogonalization . The above theorem argues that if the eigenvalues of two eigenfunctions are the same then the functions are said to be degenerate and linear combinations of the degenerate functions can be formed that will be orthogonal to each other. Since the two eigenfunctions have the same eigenvalues, the linear combination also will be an eigenfunction with the same eigenvalue. The proof of this theorem shows us one way to produce orthogonal degenerate functions. Proof If $\psi_a$ and $\psi'_a$ are degenerate, but not orthogonal, we can define a new composite wavefunction $\psi_a'' = \psi'_a - S\psi_a$ where $S$ is the overlap integral: \[S= \langle \psi_a \| \psi'_a \rangle \nonumber \] then $\psi_a$ and $\psi_a'' $ will be orthogonal. \[\begin{align} \langle \psi_a \| \psi_a'' \rangle &= \langle \psi_a \| \psi'_a - S\psi_a \rangle \\[4pt] &= \cancelto{S}{\langle \psi_a \| \psi'_a \rangle} - S \cancelto{1}{\langle \psi_a \|\psi_a \rangle} \\[4pt] &= S - S =0 \end{align} \] \[ \int \psi_a^ \psi_a d\tau \nonumber \] then $\psi_a$ and $Φ$ will be orthogonal. \[ \begin{align} \int \psi_A^ Φ d\tau &= \int \psi_a^* (\psi'_a - S \psi_a ) d\tau \\[4pt] &= \int \psi_a^* \psi'_a d\tau - S \int \psi_a^\psi_a d\tau \\[4pt] &= S - S = 0 \end{align} \] $\square$ Find $N$ that normalizes $\psi$ if $\psi = N(φ_1 − Sφ_2)$ where $φ_1$ and $φ_2$ are normalized wavefunctions and $S$ is their overlap integral. \[S= \langle φ_1 \| φ_2 \rangle \nonumber \] Answer Remember that to normalize an arbitrary wavefunction, we find a constant $N$ such that $\langle \psi \| \psi \rangle = 1$. This equates to the following procedure: \[ \begin{align} \langle\psi \| \psi\rangle =\left\langle N\left(φ_{1} - Sφ_{2}\right) \| N\left(φ_{1} - Sφ_{2}\right)\right\rangle &= 1 \\[4pt] N^2\left\langle \left(φ_{1} - Sφ_{2}\right) \| \left(φ_{1}-Sφ_{2}\right)\right\rangle &=1 \\[4pt] N^2 \left[ \cancelto{1}{\langle φ_{1}\|φ_{1}\rangle} - S \cancelto{S}{\langle φ_{2}\|φ_{1}\rangle} - S \cancelto{S}{\langle φ_{1}\|φ_{2}\rangle} + S^2 \cancelto{1}{\langle φ_{2}\| φ_{2}\rangle} \right] &= 1 \\[4pt] N^2(1 - S^2 \cancel{-S^2} + \cancel{S^2})&=1 \\[4pt] N^2(1-S^2) &= 1 \end{align} \] therefore \[N = \dfrac{1}{\sqrt{1-S^2}} \nonumber \] We conclude that the eigenstates of operators are, or can be chosen to be, mutually orthogonal .
Bookshelves/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/06%3A_Solutions/6.5%3A_Polarity	So far we have considered solutions that are made up of molecules that are either polar or non-polar or ionic species that have properties that are relatively easy to predict. Many substances, however, have more complex structures that incorporate polar, ionic, and non-polar groups. For example, many biomolecules cannot be classified as exclusively polar or non-polar, but are large enough to have distinct regions of differing polarity. They are termed amphipathic. Even though the structures of proteins such as $\mathrm{RNA}$, $\mathrm{DNA}$, and other biomolecules are complex, we can use the same principles involving entropic and enthalpic effects of interacting with water to understand the interactions between biomolecules, as well as within a given biomolecule. Biomolecules are very large compared to the molecules considered in most chemistry courses, and often one part of the molecule interacts with another part of the same molecule. The intramolecular [6] interactions of biological macromolecules, together with their interactions with water, are key factors in predicting their shapes. [7] Let us begin with a relatively simple biomolecular structure. In the previous section we looked at the solubility of oils in water. Oils or fats are also known as a triglycerides. In the figure, $\mathrm{R}$ and $\mathrm{R}_{\prime}$ indicate hydrocarbon chains, which have the generic structure $\mathrm{CH}_{3}\mathrm{CnH}_{2\mathrm{n}}$, shown in the figure. If you treat an oil or fat with sodium hydroxide ($\mathrm{NaOH}$), the resulting chemical reaction leads to the formation of what is known as a fatty acid (in this example, oxygen atoms are maroon). A typical fatty acid has a long, non-polar hydrocarbon chain and one end that often contains both a polar and ionic group. The polar head of the molecule interacts with water with little or no increase in entropy, unlike a hydrocarbon, where the lack of $\mathrm{H}$-bonding interactions with water forces a more ordered shell of water molecules around the hydrocarbon molecule, leading to a decrease in entropy. On the other hand, in water the non-polar region of the molecule creates a decrease in entropy as water molecules are organized into a type of cage around it—an unfavorable outcome in terms of $\Delta \mathrm{S}$, and therefore $\Delta \mathrm{G}$ as well. So, which end of the molecule “wins”? That is do such molecules dissolve in water or not? The answer is: Both! These amphipathic molecules become arranged in such a manner that their polar groups are in contact with the water, while their non-polar regions are not. (See whether you can draw out such an arrangement, remembering to include the water molecules in your drawing.) In fact, there are several ways to produce such an arrangement, depending in part on the amount of water in the system. A standard micelle is a spherical structure with the polar heads on the outside and the non-polar tails on the inside. It is the simplest structure that can accommodate both hydrophilic and hydrophobic groups in the same molecule. If water is limiting, it is possible to get an inverted micelle arrangement, in which polar head groups (and water) are inside and the non-polar tails point outward (as shown in the figure). Other highly organized structures can form spontaneously depending on the structure of the head group and the tail. For example, lipid molecules have multiple hydrocarbon tails and carbon ring structures called sterols. That structure creates a lipid bilayer—a polar membrane made up of two lipid molecule layers that form cellular and organellar boundaries in all organisms. It should be noted that these ordered structures are possible only because dispersing the lipid molecules in water results in a substantial decrease in the disorder of the system. In fact, many ordered structures associated with living systems, such as the structure of $\mathrm{DNA}$ and proteins, are the result of entropy-driven processes, yet another counterintuitive idea. This is one of the many reasons why biological systems do not violate the laws of thermodynamics and why it is scientifically plausible that life arose solely due to natural processes! [8] Questions Questions to Answer If you had a compound that you suspected might form micelles: What structural features would you look for? How might you design an experiment to determine whether the compound would form micelles in water? What would be the experimental evidence? Why do you think some amphipathic molecules form spherical clusters (micelles or liposomes) whereas others form sheets (bilayers)? (Hint: consider the shape of the individual molecule itself.) Amphipathic molecules are often called surfactants. For example, the compounds used to disperse oil spills are surfactants. How do you think they work? Questions to Ponder If membrane formation and protein folding are entropy-driven processes, does that make the origins of life seem more or less “natural” to you? Solutions, Colloids & Emulsions So, do micelles dissolve in water? Well, micelles are not molecules but rather supramolecular assemblies composed of many distinct molecules. A glucose solution consists of isolated glucose molecules but micelles in solution consist of larger molecular aggregates. Solutions of macromolecular solutes are called colloids. These particles can be aggregates of molecules (like micelles), atoms (nanoparticles), or larger macromolecules (proteins, nucleic acids), among others. When these particles are on the order of the wavelength of visible light, they scatter the light; smaller objects do not. This is why a salt or sugar solution is translucent, whereas a colloidal dispersion of micelles or cells is cloudy. [9] This principle also explains why soap solutions are typically cloudy—they contain particles large enough to scatter the light. When the particles in a solution maintain the structure of a solid, the end result is known as a colloid. The colloid is stable because the thermal movements of these small, solid particles are suspended. As the particles get larger, the colloid becomes unstable; the influence of gravity overcomes the effects of thermal motion and the particles settle out. Before they settle out, such unstable systems are known as suspension But if the suspended particles are liquid, the system is known as an emulsion. For example, if we looked at a salad dressing made of oil and water under a microscope, we would see drops of oil suspended in water. Emulsions are often unstable, and over time the two liquid phases separate. This is why you have to shake up salad dressing just before using it. There are many colloids and emulsions in the world around us. Milk, for example, is an emulsion of fat globules and a colloid of protein (casein) micelles.
Courses/Madera_Community_College/Concepts_of_Physical_Science/06%3A_Heat_and_Energy/6.02%3A_Temperature/6.2.01%3A_Temperature_and_Thermometers	Learning Objectives Define temperature. Convert temperatures between the Celsius, Fahrenheit, and Kelvin scales. Define thermal equilibrium. State the zeroth law of thermodynamics. The concept of temperature has evolved from the common concepts of hot and cold. Human perception of what feels hot or cold is a relative one. For example, if you place one hand in hot water and the other in cold water, and then place both hands in tepid water, the tepid water will feel cool to the hand that was in hot water, and warm to the one that was in cold water. The scientific definition of temperature is less ambiguous than your senses of hot and cold. Temperature is operationally defined to be what we measure with a thermometer. (Many physical quantities are defined solely in terms of how they are measured. We shall see later how temperature is related to the kinetic energies of atoms and molecules, a more physical explanation.) Two accurate thermometers, one placed in hot water and the other in cold water, will show the hot water to have a higher temperature. If they are then placed in the tepid water, both will give identical readings (within measurement uncertainties). In this section, we discuss temperature, its measurement by thermometers, and its relationship to thermal equilibrium. Again, temperature is the quantity measured by a thermometer. MISCONCEPTION ALERT: HUMAN PERCEPTION VS. REALITY On a cold winter morning, the wood on a porch feels warmer than the metal of your bike. The wood and bicycle are in thermal equilibrium with the outside air, and are thus the same temperature. They feel different because of the difference in the way that they conduct heat away from your skin. The metal conducts heat away from your body faster than the wood does. This is just one example demonstrating that the human sense of hot and cold is not determined by temperature alone. Another factor that affects our perception of temperature is humidity. Most people feel much hotter on hot, humid days than on hot, dry days. This is because on humid days, sweat does not evaporate from the skin as efficiently as it does on dry days. It is the evaporation of sweat (or water from a sprinkler or pool) that cools us off. Any physical property that depends on temperature, and whose response to temperature is reproducible, can be used as the basis of a thermometer. Because many physical properties depend on temperature, the variety of thermometers is remarkable. For example, volume increases with temperature for most substances. This property is the basis for the common alcohol thermometer, the old mercury thermometer, and the bimetallic strip ( Figure $\PageIndex{1}$). Other properties used to measure temperature include electrical resistance and color, as shown in Figure $\PageIndex{2}$, and the emission of infrared radiation, as shown in Figure $\PageIndex{3}$. Temperature Scales Thermometers are used to measure temperature according to well-defined scales of measurement, which use pre-defined reference points to help compare quantities. The three most common temperature scales are the Fahrenheit, Celsius, and Kelvin scales. A temperature scale can be created by identifying two easily reproducible temperatures. The freezing and boiling temperatures of water at standard atmospheric pressure are commonly used. The Celsius scale (which replaced the slightly different centigrade scale) has the freezing point of water at 0ºC and the boiling point at 100ºC. Its unit is the degree Celsius (ºC). On the Fahrenheit scale (still the most frequently used in the United States), the freezing point of water is at 32ºF and the boiling point is at 212ºF. The unit of temperature on this scale is the degree Fahrenheit (ºF). Note that a temperature difference of one degree Celsius is greater than a temperature difference of one degree Fahrenheit. Only 100 Celsius degrees span the same range as 180 Fahrenheit degrees, thus one degree on the Celsius scale is 1.8 times larger than one degree on the Fahrenheit scale $180 / 100=9 / 5$. The Kelvin scale is the temperature scale that is commonly used in science. It is an absolute temperature scale defined to have 0 K at the lowest possible temperature, called absolute zero . The official temperature unit on this scale is the kelvin , which is abbreviated K, and is not accompanied by a degree sign. The freezing and boiling points of water are 273.15 K and 373.15 K, respectively. Thus, the magnitude of temperature differences is the same in units of kelvins and degrees Celsius. Unlike other temperature scales, the Kelvin scale is an absolute scale. It is used extensively in scientific work because a number of physical quantities, such as the volume of an ideal gas, are directly related to absolute temperature. The kelvin is the SI unit used in scientific work. The relationships between the three common temperature scales is shown in Figure $\PageIndex{4}$. Temperature Ranges in the Universe Figure $\PageIndex{6}$ shows the wide range of temperatures found in the universe. Human beings have been known to survive with body temperatures within a small range, from 24ºC to 44ºC (75ºF to 111ºF). The average normal body temperature is usually given as 37.0ºC (98.6ºF), and variations in this temperature can indicate a medical condition: a fever, an infection, a tumor, or circulatory problems (see Figure $\PageIndex{5}$). The lowest temperatures ever recorded have been measured during laboratory experiments: $4.5 \times 10^{-10} \mathrm{~K}$ at the Massachusetts Institute of Technology (USA), and $1.0 \times 10^{-10} \mathrm{~K}$ at Helsinki University of Technology (Finland). In comparison, the coldest recorded place on Earth’s surface is Vostok, Antarctica at 183 K (–89ºC), and the coldest place (outside the lab) known in the universe is the Boomerang Nebula, with a temperature of 1 K. MAKING CONNECTIONS: ABSOLUTE ZERO What is absolute zero? Absolute zero is the temperature at which all molecular motion has ceased. The concept of absolute zero arises from the behavior of gases. Figure $\PageIndex{7}$ shows how the pressure of gases at a constant volume decreases as temperature decreases. Various scientists have noted that the pressures of gases extrapolate to zero at the same temperature, –273.15ºC. This extrapolation implies that there is a lowest temperature. This temperature is called absolute zero . Today we know that most gases first liquefy and then freeze, and it is not actually possible to reach absolute zero. The numerical value of absolute zero temperature is –273.15ºC or 0 K. Thermal Equilibrium and the Zeroth Law of Thermodynamics Thermometers actually take their own temperature, not the temperature of the object they are measuring. This raises the question of how we can be certain that a thermometer measures the temperature of the object with which it is in contact. It is based on the fact that any two systems placed in thermal contact (meaning heat transfer can occur between them) will reach the same temperature. That is, heat will flow from the hotter object to the cooler one until they have exactly the same temperature. The objects are then in thermal equilibrium , and no further changes will occur. The systems interact and change because their temperatures differ, and the changes stop once their temperatures are the same. Thus, if enough time is allowed for this transfer of heat to run its course, the temperature a thermometer registers does represent the system with which it is in thermal equilibrium. Thermal equilibrium is established when two bodies are in contact with each other and can freely exchange energy. Furthermore, experimentation has shown that if two systems, A and B, are in thermal equilibrium with each another, and B is in thermal equilibrium with a third system C, then A is also in thermal equilibrium with C. This conclusion may seem obvious, because all three have the same temperature, but it is basic to thermodynamics. It is called the zeroth law of thermodynamics . THE ZEROTH LAW OF THERMODYNAMICS If two systems, A and B, are in thermal equilibrium with each other, and B is in thermal equilibrium with a third system, C, then A is also in thermal equilibrium with C. This law was postulated in the 1930s, after the first and second laws of thermodynamics had been developed and named. It is called the zeroth law because it comes logically before the first and second laws. An example of this law in action is seen in babies in incubators: babies in incubators normally have very few clothes on, so to an observer they look as if they may not be warm enough. However, the temperature of the air, the cot, and the baby is the same, because they are in thermal equilibrium, which is accomplished by maintaining air temperature to keep the baby comfortable. Exercise $\PageIndex{1}$ Does the temperature of a body depend on its size? Answer No, the system can be divided into smaller parts each of which is at the same temperature. We say that the temperature is an intensive quantity. Intensive quantities are independent of size. Section Summary Temperature is the quantity measured by a thermometer. Temperature is related to the average kinetic energy of atoms and molecules in a system. Absolute zero is the temperature at which there is no molecular motion. There are three main temperature scales: Celsius, Fahrenheit, and Kelvin. Temperatures on one scale can be converted to temperatures on another scale using the following equations: \[\begin{gathered} T_{\circ} \mathrm{F}=\frac{9}{5} T_{\mathrm{C}} \mathrm{C}+32 \\ T_{\mathrm{o}} \mathrm{C}=\frac{5}{9}\left(T_{\mathrm{o}}-32\right) \\ T_{\mathrm{K}}=T_{\mathrm{o}} \mathrm{C}+273.15 \\ T_{\mathrm{o}} \mathrm{C}=T_{\mathrm{K}}-273.15 \end{gathered} \nonumber\] Systems are in thermal equilibrium when they have the same temperature. Thermal equilibrium occurs when two bodies are in contact with each other and can freely exchange energy. The zeroth law of thermodynamics states that when two systems, A and B, are in thermal equilibrium with each other, and B is in thermal equilibrium with a third system, C, then A is also in thermal equilibrium with C. Glossary temperature the quantity measured by a thermometer Celsius scale temperature scale in which the freezing point of water is 0ºC and the boiling point of water is 100ºC degree Celsius unit on the Celsius temperature scale Fahrenheit scale temperature scale in which the freezing point of water is 32ºF and the boiling point of water is 212ºF degree Fahrenheit unit on the Fahrenheit temperature scale Kelvin scale temperature scale in which 0 K is the lowest possible temperature, representing absolute zero absolute zero the lowest possible temperature; the temperature at which all molecular motion ceases thermal equilibrium the condition in which heat no longer flows between two objects that are in contact; the two objects have the same temperature zeroth law of thermodynamics law that states that if two objects are in thermal equilibrium, and a third object is in thermal equilibrium with one of those objects, it is also in thermal equilibrium with the other object
Courses/National_Yang_Ming_Chiao_Tung_University/Chemistry_2/06%3A_Organic_and_Biological_Chemistry_Brown/6.06%3A_Chirality_in_Organic_Chemistry	Organic compounds, molecules created around a chain of carbon atom (more commonly known as carbon backbone), play an essential role in the chemistry of life. These molecules derive their importance from the energy they carry, mainly in a form of potential energy between atomic molecules. Since such potential force can be widely affected due to changes in atomic placement, it is important to understand the concept of an isomer, a molecule sharing same atomic make up as another but differing in structural arrangement. Stereoisomers are isomers that differ in spatial arrangement of atoms, rather than order of atomic connectivity. One of their most interesting type of isomer is the mirror-image stereoisomers, a non-superimposable set of two molecules that are mirror image of one another. The existence of these molecules are determined by concept known as chirality. Stereoisomers Molecules with the same connectivity but different arrangements of the atoms in space are called stereoisomers. There are two types of stereoisomers: geometric and optical. Geometric isomers differ in the relative position(s) of substituents in a rigid molecule. Simple rotation about a C–C σ bond in an alkene, for example, cannot occur because of the presence of the π bond. The substituents are therefore rigidly locked into a particular spatial arrangement. Thus a carbon–carbon multiple bond, or in some cases a ring, prevents one geometric isomer from being readily converted to the other. The members of an isomeric pair are identified as either cis or trans, and interconversion between the two forms requires breaking and reforming one or more bonds. Because their structural difference causes them to have different physical and chemical properties, cis and trans isomers are actually two distinct chemical compounds. Stereoisomers have the same connectivity, but different arrangements of atoms in space. Optical isomers are molecules whose structures are mirror images but cannot be superimposed on one another in any orientation. Optical isomers have identical physical properties, although their chemical properties may differ in asymmetric environments. Molecules that are nonsuperimposable mirror images of each other are said to be chiral (pronounced “ky-ral,” from the Greek cheir, meaning “hand”). Examples of some familiar chiral objects are your hands, feet, and ears. As shown in Figure $\PageIndex{1a}$, your left and right hands are nonsuperimposable mirror images. (Try putting your right shoe on your left foot—it just doesn’t work.) An achiral object is one that can be superimposed on its mirror image, as shown by the superimposed flasks in Figure $\PageIndex{1b}$. Most chiral organic molecules have at least one carbon atom that is bonded to four different groups, as occurs in the bromochlorofluoromethane molecule shown in part (a) in Figure $\PageIndex{2}$. This carbon, often designated by an asterisk in structural drawings, is called a chiral center or asymmetric carbon atom. If the bromine atom is replaced by another chlorine (Figure $\PageIndex{2b}$), the molecule and its mirror image can now be superimposed by simple rotation. Thus the carbon is no longer a chiral center. Asymmetric carbon atoms are found in many naturally occurring molecules, such as lactic acid, which is present in milk and muscles, and nicotine, a component of tobacco. A molecule and its nonsuperimposable mirror image are called enantiomers (from the Greek enantiou, meaning “opposite”). Thalidomide In the 1960’s, a drug called thalidomide was widely prescribed in the Western Europe to alleviate morning sickness in pregnant women. Thalidomide had previously been used in other countries as an antidepressant, and was believed to be safe and effective for both purposes. The drug was not approved for use in the U.S.A. It was not long, however, before doctors realized that something had gone horribly wrong: many babies born to women who had taken thalidomide during pregnancy suffered from severe birth defects. Researchers later realized the that problem lay in the fact that thalidomide was being provided as a mixture of two different isomeric forms. One of the isomers is an effective medication, the other caused the side effects. Both isomeric forms have the same molecular formula and the same atom-to-atom connectivity, so they are not constitutional isomers. Where they differ is in the arrangement in three-dimensional space about one tetrahedral, sp 3 -hybridized carbon. These two forms of thalidomide are stereoisomers . Looking for planes of symmetry in a molecule is useful, but often difficult in practice. In most cases, the easiest way to decide whether a molecule is chiral or achiral is to look for one or more stereocenters - with a few rare exceptions, the general rule is that molecules with at least one stereocenter are chiral, and molecules with no stereocenters are achiral. Carbon stereocenters are also referred to quite frequently as chiral carbons . When evaluating a molecule for chirality, it is important to recognize that the question of whether or not the dashed/solid wedge drawing convention is used is irrelevant. Chiral molecules are sometimes drawn without using wedges (although obviously this means that stereochemical information is being omitted). Conversely, wedges may be used on carbons that are not stereocenters – look, for example, at the drawings of glycine and citrate in the figure above. Just because you see dashed and solid wedges in a structure, do not automatically assume that you are looking at a stereocenter.
Bookshelves/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/06%3A_Conformational_Analysis/6.01%3A_Introduction_to_Conformation	When we smell something, the information travels to us via molecules, and almost always these are organic molecules. These molecules can be detected by a variety of organs, including noses in dogs and antennae in crickets, but no matter what organ is sensing the smell, one of the crucial factors in determining how an organism reacts to a compound is the shape of the molecule. The sense of smell depends on thousands of different receptors in the organ, working conceptually on a lock-and-key basis: a molecule with a given shape can fit into a given receptor, and when it does, a signal is sent telling the nervous system that the organism has encountered that particular type of molecule, and the organism reacts appropriately. What gives a molecule its shape? Given a structural formula, could you determine the shape of the corresponding molecule in three dimensions? Could you predict its biological activity, including not only its smell, but also a host of other behaviors linked to the shape of molecular messengers, such as anti-cancer activity or narcotic properties? These questions are at the cutting edge of biological chemistry. Although they are best answered through computer modeling, we can develop some of the qualitative ideas used in these models. Computer modeling employs "basis sets", small sets of information that the computer could apply to any molecule in order to predict its properties. In order to understand computational conformational analysis, in this chapter we will develop a very simple basis set as an example.
Courses/CSU_San_Bernardino/CHEM_2200%3A_General_Chemistry_II_(Mink)/22%3A_Appendices/22.14%3A_Answer_Key/22.14.09%3A_Chapter_9	1. The cutting edge of a knife that has been sharpened has a smaller surface area than a dull knife. Since pressure is force per unit area, a sharp knife will exert a higher pressure with the same amount of force and cut through material more effectively. 3. Lying down distributes your weight over a larger surface area, exerting less pressure on the ice compared to standing up. If you exert less pressure, you are less likely to break through thin ice. 5. 0.809 atm; 82.0 kPa 7. 2.2 10 2 kPa 9. Earth: 14.7 lb in –2 ; Venus: 1.30 × 10 3 lb in −2 11. (a) 101.5 kPa; (b) 51 torr drop 13. (a) 264 torr; (b) 35,200 Pa; (c) 0.352 bar 15. (a) 623 mm Hg; (b) 0.820 atm; (c) 83.1 kPa 17. With a closed-end manometer, no change would be observed, since the vaporized liquid would contribute equal, opposing pressures in both arms of the manometer tube. However, with an open-ended manometer, a higher pressure reading of the gas would be obtained than expected, since P gas = P atm + P vol liquid . 19. As the bubbles rise, the pressure decreases, so their volume increases as suggested by Boyle’s law. 21. (a) The number of particles in the gas increases as the volume increases. (b) temperature, pressure 23. The curve would be farther to the right and higher up, but the same basic shape. 25. About 12.5 L 27. 3.40 10 3 torr 29. 12.1 L 31. 217 L 33. 8.190 10 –2 mol; 5.553 g 35. (a) 7.24 10 –2 g; (b) 23.1 g; (c) 1.5 10 –4 g 37. 5561 L 39. 46.4 g 41. For a gas exhibiting ideal behavior: 43. (a) 1.85 L CCl 2 F 2 ; (b) 4.66 L CH 3 CH 2 F 45. 0.644 atm 47. The pressure decreases by a factor of 3. 49. 4.64 g L −1 51. 38.8 g 53. 72.0 g mol −1 55. 88.1 g mol −1 ; PF 3 57. 141 atm, 107,000 torr, 14,300 kPa 59. CH 4 : 276 kPa; C 2 H 6 : 27 kPa; C 3 H 8 : 3.4 kPa 61. Yes 63. 740 torr 65. (a) Determine the moles of HgO that decompose; using the chemical equation, determine the moles of O 2 produced by decomposition of this amount of HgO; and determine the volume of O 2 from the moles of O 2 , temperature, and pressure. (b) 0.308 L 67. (a) Determine the molar mass of CCl 2 F 2 . From the balanced equation, calculate the moles of H 2 needed for the complete reaction. From the ideal gas law, convert moles of H 2 into volume. (b) 3.72 10 3 L 69. (a) Balance the equation. Determine the grams of CO 2 produced and the number of moles. From the ideal gas law, determine the volume of gas. (b) 7.43 10 5 L 71. 42.00 L 73. (a) 18.0 L; (b) 0.533 atm 75. 10.57 L O 2 77. 5.40 10 5 L 79. XeF 4 81. 4.2 hours 83. Effusion can be defined as the process by which a gas escapes through a pinhole into a vacuum. Graham’s law states that with a mixture of two gases A and B: Both A and B are in the same container at the same temperature, and therefore will have the same kinetic energy: Therefore, 85. F 2 , N 2 O, Cl 2 , H 2 S 87. 1.4; 1.2 89. 51.7 cm 91. Yes. At any given instant, there are a range of values of molecular speeds in a sample of gas. Any single molecule can speed up or slow down as it collides with other molecules. The average speed of all the molecules is constant at constant temperature. 93. H 2 O. Cooling slows the speeds of the He atoms, causing them to behave as though they were heavier. 95. (a) The number of collisions per unit area of the container wall is constant. (b) The average kinetic energy doubles. (c) The root mean square speed increases to times its initial value; u rms is proportional to 97. (a) equal; (b) less than; (c) 29.48 g mol −1 ; (d) 1.0966 g L −1 ; (e) 0.129 g/L; (f) 4.01 10 5 g; net lifting capacity = 384 lb; (g) 270 L; (h) 39.1 kJ min −1 99. Gases C, E, and F 101. The gas behavior most like an ideal gas will occur under the conditions that minimize the chances of significant interactions between the gaseous atoms/molecules, namely, low pressures (fewer atoms/molecules per unit volume) and high temperatures (greater kinetic energies of atoms/molecules make them less susceptible to attractive forces). The conditions described in (b), high temperature and low pressure, are therefore most likely to yield ideal gas behavior. 103. SF 6 105. (a) A straight horizontal line at 1.0; (b) When real gases are at low pressures and high temperatures, they behave close enough to ideal gases that they are approximated as such; however, in some cases, we see that at a high pressure and temperature, the ideal gas approximation breaks down and is significantly different from the pressure calculated by the ideal gas equation. (c) The greater the compressibility, the more the volume matters. At low pressures, the correction factor for intermolecular attractions is more significant, and the effect of the volume of the gas molecules on Z would be a small lowering compressibility. At higher pressures, the effect of the volume of the gas molecules themselves on Z would increase compressibility (see Figure 9.35). (d) Once again, at low pressures, the effect of intermolecular attractions on Z would be more important than the correction factor for the volume of the gas molecules themselves, though perhaps still small. At higher pressures and low temperatures, the effect of intermolecular attractions would be larger. See Figure 9.35. (e) Low temperatures
Courses/Saint_Francis_University/Chem_114%3A_Human_Chemistry_II_(Muino)/12%3A_Introduction_to_Organic_Chemistry_-_Alkanes/12.08%3A_Reactions_of_Alkanes	Learning Objectives Understand the reactions of alkanes: combustion and halogenation. Alkanes are relatively stable, nonpolar molecules, that will not react with acids, bases, or oxidizing or reducing reagents. Alkanes undergo so few reactions that they are sometimes called paraffins , from the Latin parum affinis , meaning “little affinity.” However, heat or light can initiate the breaking of C–H or C–C single bonds in reactions called combustion and halogenation . Combustion Nothing happens when alkanes are merely mixed with oxygen ($O_2$) at room temperature, but when a flame or spark provides the activation energy, a highly exothermic combustion reaction proceeds vigorously. For methane (CH 4 ), the combustion reaction is as follows: \[CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + \text{heat} \label{1} \] As a consequence, alkanes are excellent fuels. For example, methane, CH 4 , is the principal component of natural gas. Butane, C 4 H 10 , used in camping stoves and lighters is an alkane. Gasoline is a liquid mixture of straight- and branched-chain alkanes, each containing from five to nine carbon atoms, plus various additives to improve its performance as a fuel. Kerosene, diesel oil, and fuel oil are primarily mixtures of alkanes with higher molecular masses. The main source of these liquid alkane fuels is crude oil, a complex mixture that is separated by fractional distillation. Fractional distillation takes advantage of differences in the boiling points of the components of the mixture (Figure $\PageIndex{1}$). You may recall that boiling point is a function of intermolecular interactions, which was discussed in an earlier chapter. If the reactants of combustion reactions are adequately mixed, and there is sufficient oxygen, the only products are carbon dioxide ($CO_2$), water ($H_2O$), and energy—heat for cooking foods, heating homes, and drying clothes. Because conditions are rarely ideal, other unwanted by-products are frequently formed. When the oxygen supply is limited, carbon monoxide ($CO$) is a by-product: \[2CH_4 + 3O_2 \rightarrow 2CO + 4H_2O\label{2} \] This reaction is responsible for dozens of deaths each year from unventilated or improperly adjusted gas heaters. (Similar reactions with similar results occur with kerosene heaters.) Halogenation In halogenation reactions, alkanes react with the halogens chlorine ($Cl_2$) and bromine ($Br_2$) in the presence of ultraviolet light or at high temperatures to yield chlorinated and brominated alkanes. For example, chlorine reacts with excess methane ($CH_4$) to give methyl chloride ($CH_3Cl$). \[CH_4 + Cl_2 \rightarrow CH_3Cl + HCl\label{12.7.3} \] With more chlorine, a mixture of products is obtained: CH 3 Cl, CH 2 Cl 2 , CHCl 3 , and CCl 4 . Fluorine ($F_2$), the lightest halogen, combines explosively with most hydrocarbons. Iodine ($I_2$) is relatively unreactive. Fluorinated and iodinated alkanes are produced by indirect methods. A wide variety of interesting and often useful compounds have one or more halogen atoms per molecule. For example, methane (CH 4 ) can react with chlorine (Cl 2 ), replacing one, two, three, or all four hydrogen atoms with Cl atoms. Several halogenated products derived from methane and ethane (CH 3 CH 3 ) are listed in Table $\PageIndex{1}$, along with some of their uses. Formula Common Name IUPAC Name Some Important Uses Derived from CH4 Derived from CH4 Derived from CH4 Derived from CH4 CH3Cl methyl chloride chloromethane refrigerant; the manufacture of silicones, methyl cellulose, and synthetic rubber CH2Cl2 methylene chloride dichloromethane laboratory and industrial solvent CHCl3 chloroform trichloromethane industrial solvent CCl4 carbon tetrachloride tetrachloromethane dry-cleaning solvent and fire extinguishers (but no longer recommended for use) CBrF3 halon-1301 bromotrifluoromethane fire extinguisher systems CCl3F chlorofluorocarbon-11 (CFC-11) trichlorofluoromethane foaming plastics CCl2F2 chlorofluorocarbon-12 (CFC-12) dichlorodifluoromethane refrigerant Derived from CH3CH3 Derived from CH3CH3 Derived from CH3CH3 Derived from CH3CH3 CH3CH2Cl ethyl chloride chloroethane local anesthetic ClCH2CH2Cl ethylene dichloride 1,2-dichloroethane solvent for rubber CCl3CH3 methylchloroform 1,1,1-trichloroethane solvent for cleaning computer chips and molds for shaping plastics Note To Your Health: Halogenated Hydrocarbons Once widely used in consumer products, many chlorinated hydrocarbons are suspected carcinogens (cancer-causing substances) and also are known to cause severe liver damage. An example is carbon tetrachloride (CCl 4 ), once used as a dry-cleaning solvent and in fire extinguishers but no longer recommended for either use. Even in small amounts, its vapor can cause serious illness if exposure is prolonged. Moreover, it reacts with water at high temperatures to form deadly phosgene (COCl 2 ) gas, which makes the use of CCl 4 in fire extinguishers particularly dangerous. Ethyl chloride, in contrast, is used as an external local anesthetic. When sprayed on the skin, it evaporates quickly, cooling the area enough to make it insensitive to pain. It can also be used as an emergency general anesthetic. Bromine-containing compounds are widely used in fire extinguishers and as fire retardants on clothing and other materials. Because they too are toxic and have adverse effects on the environment, scientists are engaged in designing safer substitutes for them, as for many other halogenated compounds. Note To Your Health: Chlorofluorocarbons and The Ozone Layer Alkanes substituted with both fluorine (F) and chlorine (Cl) atoms have been used as the dispersing gases in aerosol cans, as foaming agents for plastics, and as refrigerants. Two of the best known of these chlorofluorocarbons (CFCs) are listed in Table $\PageIndex{2}$. Chlorofluorocarbons contribute to the greenhouse effect in the lower atmosphere. They also diffuse into the stratosphere, where they are broken down by ultraviolet (UV) radiation to release Cl atoms. These in turn break down the ozone (O 3 ) molecules that protect Earth from harmful UV radiation. Worldwide action has reduced the use of CFCs and related compounds. The CFCs and other Cl- or bromine (Br)-containing ozone-destroying compounds are being replaced with more benign substances. Hydrofluorocarbons (HFCs), such as CH 2 FCF 3 , which have no Cl or Br to form radicals, are one alternative. Another is hydrochlorofluorocarbons (HCFCs), such as CHCl 2 CF 3 . HCFC molecules break down more readily in the troposphere, and fewer ozone-destroying molecules reach the stratosphere. Figure $\PageIndex{2}$: Ozone in the upper atmosphere shields Earth’s surface from UV radiation from the sun, which can cause skin cancer in humans and is also harmful to other animals and to some plants. Ozone “holes” in the upper atmosphere (the gray, pink, and purple areas at the center) are large areas of substantial ozone depletion. They occur mainly over Antarctica from late August through early October and fill in about mid-November. Ozone depletion has also been noted over the Arctic regions. The largest ozone hole ever observed occurred on 24 September 2006. Source: Image courtesy of NASA, http://ozonewatch.gsfc.nasa.gov/daily.php?date=2006-09-24 .
Bookshelves/Analytical_Chemistry/Analytical_Chemistry_Volume_I_(Harvey)/09%3A_Titrimetric_Methods/9.04%3A_Redox_Titrations	Analytical titrations using oxidation–reduction reactions were introduced shortly after the development of acid–base titrimetry. The earliest redox titration took advantage of chlorine’s oxidizing power. In 1787, Claude Berthollet introduced a method for the quantitative analysis of chlorine water (a mixture of Cl 2 , HCl, and HOCl) based on its ability to oxidize indigo, a dye that is colorless in its oxidized state. In 1814, Joseph Gay-Lussac developed a similar method to determine chlorine in bleaching powder. In both methods the end point is a change in color. Before the equivalence point the solution is colorless due to the oxidation of indigo. After the equivalence point, however, unreacted indigo imparts a permanent color to the solution. The number of redox titrimetric methods increased in the mid-1800s with the introduction of $\text{MnO}_4^-$, $\text{Cr}_2\text{O}_7^{2-}$, and I 2 as oxidizing titrants, and of Fe 2 + and $\text{S}_2\text{O}_3^{2-}$ as reducing titrants. Even with the availability of these new titrants, redox titrimetry was slow to develop due to the lack of suitable indicators. A titrant can serve as its own indicator if its oxidized and its reduced forms differ significantly in color. For example, the intensely purple $\text{MnO}_4^-$ ion serves as its own indicator since its reduced form, Mn 2 + , is almost colorless. Other titrants require a separate indicator. The first such indicator, diphenylamine, was introduced in the 1920s. Other redox indicators soon followed, increasing the applicability of redox titrimetry. Redox Titration Curves To evaluate a redox titration we need to know the shape of its titration curve. In an acid–base titration or a complexation titration, the titration curve shows how the concentration of H 3 O + (as pH) or M n + (as pM) changes as we add titrant. For a redox titration it is convenient to monitor the titration reaction’s potential instead of the concentration of one species. You may recall from Chapter 6 that the Nernst equation relates a solution’s potential to the concentrations of reactants and products that participate in the redox reaction. Consider, for example, a titration in which a titrand in a reduced state, A red , reacts with a titrant in an oxidized state, B ox . \[A_{red} + B_{ox} \rightleftharpoons B_{red} + A_{ox} \nonumber\] where A ox is the titrand’s oxidized form, B red is the titrant’s reduced form, and the stoichiometry between the two is 1:1. The reaction’s potential, E rxn , is the difference between the reduction potentials for each half-reaction. \[E_{rxn} = E_{B_{ox}/B_{red}} - E_{A_{ox}/A_{red}} \nonumber\] After each addition of titrant the reaction between the titrand and the titrant reaches a state of equilibrium. Because the potential at equilibrium is zero, the titrand’s and the titrant’s reduction potentials are identical. \[E_{B_{ox}/B_{red}} = E_{A_{ox}/A_{red}} \nonumber\] This is an important observation as it allows us to use either half-reaction to monitor the titration’s progress. Before the equivalence point the titration mixture consists of appreciable quantities of the titrand’s oxidized and reduced forms. The concentration of unreacted titrant, however, is very small. The potential, therefore, is easier to calculate if we use the Nernst equation for the titrand’s half-reaction \[E = E_{A_{ox}/A_{red}} = E_{A_{ox}/A_{red}}^{\circ} - \frac{RT}{nF}\ln{\frac{[A_{red}]}{[A_{ox}]}} \nonumber\] After the equivalence point it is easier to calculate the potential using the Nernst equation for the titrant’s half-reaction. \[E = E_{B_{ox}/B_{red}} = E_{B_{ox}/B_{red}}^{\circ} - \frac{RT}{nF}\ln{\frac{[B_{red}]}{[B_{ox}]}} \nonumber\] Although the Nernst equation is written in terms of the half-reaction’s standard state potential, a matrix-dependent formal potential often is used in its place. See Appendix 13 for the standard state potentials and formal potentials for selected half-reactions. Calculating the Titration Curve Let’s calculate the titration curve for the titration of 50.0 mL of 0.100 M Fe 2 + with 0.100 M Ce 4 + in a matrix of 1 M HClO 4 . The reaction in this case is \[\text{Fe}^{2+}(aq) + \text{Ce}^{4+}(aq) \rightleftharpoons \text{Ce}^{3+}(aq) + \text{Fe}^{3+}(aq) \label{9.1}\] Because the equilibrium constant for reaction \ref{9.1} is very large—it is approximately $6 \times 10^{15}$—we may assume that the analyte and titrant react completely. In 1 M HClO 4 , the formal potential for the reduction of Fe 3 + to Fe 2 + is +0.767 V, and the formal potential for the reduction of Ce 4 + to Ce 3 + is +1.70 V. The first task is to calculate the volume of Ce needed to reach the titration’s equivalence point. From the reaction’s stoichiometry we know that \[\text{mol Fe}^{2+} = M_\text{Fe}V_\text{Fe} = M_\text{Ce}V_\text{Ce} = \text{mol Ce}^{4+} \nonumber\] Solving for the volume of Ce 4 + gives the equivalence point volume as \[V_{eq} = V_\text{Ce} = \frac{M_\text{Fe}V_\text{Fe}}{M_\text{Ce}} = \frac{(0.100 \text{ M})(50.0 \text{ mL})}{(0.100 \text{ M})} = 50.0 \text{ mL} \nonumber\] Before the equivalence point, the concentration of unreacted Fe 2 + and the concentration of Fe 3 + are easy to calculate. For this reason we find the potential using the Nernst equation for the Fe 3 + /Fe 2 + half-reaction. \[E = +0.767 \text{ V} - 0.05916 \log{\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}} \label{9.2}\] For example, the concentrations of Fe 2 + and Fe 3 + after adding 10.0 mL of titrant are \[[\text{Fe}^{2+}] = \frac{(\text{mol Fe}^{2+})_\text{initial} - (\text{mol Ce}^{4+})_\text{added}}{\text{total volume}} = \frac{M_\text{Fe}V_\text{Fe} - M_\text{Ce}V_\text{Ce}}{V_\text{Fe} + V_\text{Ce}} \nonumber\] \[[\text{Fe}^{2+}] = \frac{(0.100 \text{ M})(50.0 \text{ mL}) - (0.100 \text{ M})(10.0 \text{ mL})}{50.0 \text{ mL} + 10.0 \text{ mL}} = 6.67 \times 10^{-2} \text{ M} \nonumber\] \[[\text{Fe}^{3+}] = \frac{(\text{mol Ce}^{4+})_\text{added}}{\text{total volume}} = \frac{M_\text{Ce}V_\text{Ce}}{V_\text{Fe} + V_\text{Ce}} \nonumber\] \[[\text{Fe}^{3+}] = \frac{(0.100 \text{ M})(10.0 \text{ mL})}{50.0 \text{ mL} + 10.0 \text{ mL}} = 1.67 \times 10^{-2} \text{ M} \nonumber\] Substituting these concentrations into Equation \ref{9.2} gives the potential as \[E = +0.767 \text{ V} - 0.05916 \log{\frac{6.67 \times 10^{-2}}{1.67 \times 10^{-2}}} = +0.731 \text{ V} \nonumber\] After the equivalence point, the concentration of Ce 3 + and the concentration of excess Ce 4 + are easy to calculate. For this reason we find the potential using the Nernst equation for the Ce 4 + /Ce 3 + half-reaction in a manner similar to that used above to calculate potentials before the equivalence point. \[E = +1.70 \text{ V} - 0.05916 \log{\frac{[\text{Ce}^{3+}]}{[\text{Ce}^{4+}]}} \label{9.3}\] For example, after adding 60.0 mL of titrant, the concentrations of Ce 3 + and Ce 4 + are \[[\text{Ce}^{3+}] = \frac{(\text{mol Fe}^{2+})_\text{initial}}{\text{total volume}} = \frac{M_\text{Fe}V_\text{Fe}}{V_\text{Fe}+V_\text{Ce}} \nonumber\] \[[\text{Ce}^{3+}] = \frac{(0.100 \text{ M})(50.0 \text{ mL})}{50.0 \text{ mL} + 60.0 \text{ mL}} = 4.55 \times 10^{-2} \text{ M} \nonumber\] \[[\text{Ce}^{4+}] = \frac{(\text{mol Ce}^{4+})_\text{added}-(\text{mol Fe}^{2+})_\text{initial}}{\text{total volume}} = \frac{M_\text{Ce}V_\text{Ce}-M_\text{Fe}V_\text{Fe}}{V_\text{Fe}+V_\text{Ce}} \nonumber\] \[[\text{Ce}^{4+}] = \frac{(0.100 \text{ M})(60.0 \text{ mL})-(0.100 \text{ M})(50.0 \text{ mL})}{50.0 \text{ mL} + 60.0 \text{ mL}} = 9.09 \times 10^{-3} \text{ M} \nonumber\] Substituting these concentrations into Equation \ref{9.3} gives a potential of \[E = +1.70 \text{ V} - 0.05916 \log{\frac{4.55 \times 10^{-2} \text{ M}}{9.09 \times 10^{-3} \text{ M}}} = +1.66 \text{ V} \nonumber\] At the titration’s equivalence point, the potential, E eq , in Equation \ref{9.2} and Equation \ref{9.3} are identical. Adding the equations together to gives \[2E_{eq} = E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} + E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ} - 0.05916 \log{\frac{[\text{Fe}^{2+}][\text{Ce}^{3+}]}{[\text{Fe}^{3+}][\text{Ce}^{4+}]}} \nonumber\] Because [Fe 2 + ] = [Ce 4 + ] and [Ce 3 + ] = [Fe 3 + ] at the equivalence point, the log term has a value of zero and the equivalence point’s potential is \[E_{eq} = \frac{E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} + E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ}}{2} = \frac{0.767 \text{ V} + 1.70 \text{ V}}{2} = +1.23 \text{ V} \nonumber\] Additional results for this titration curve are shown in Table 9.4.1 and Figure 9.4.1 . volume of Ce4+ (mL) E (V) volume of Ce4+ (mL).1 E (V).1 10.0 0.731 60.0 1.66 20.0 0.757 70.0 1.68 30.0 0.777 80.0 1.69 40.0 0.803 90.0 1.69 50.0 1.230 100.0 1.70 Calculate the titration curve for the titration of 50.0 mL of 0.0500 M Sn 2 + with 0.100 M Tl 3 + . Both the titrand and the titrant are 1.0 M in HCl. The titration reaction is \[\text{Sn}^{2+}(aq) + \text{Tl}^{3+} \rightleftharpoons \text{Tl}^+(aq) + \text{Sn}^{4+}(aq) \nonumber\] Answer The volume of Tl 3 + needed to reach the equivalence point is \[V_{eq} = V_\text{Tl} = \frac{M_\text{Sn}V_\text{Sn}}{M_\text{Tl}} = \frac{(0.050 \text{ M})(50.0 \text{ mL})}{(0.100 \text{ M})} = 25.0 \text{ mL} \nonumber\] Before the equivalence point, the concentration of unreacted Sn 2 + and the concentration of Sn 4 + are easy to calculate. For this reason we find the potential using the Nernst equation for the Sn 4 + /Sn 2 + half-reaction. For example, the concentrations of Sn 2 + and Sn 4 + after adding 10.0 mL of titrant are \[[\text{Sn}^{2+}] = \frac{(0.050 \text{ M})(50.0 \text{ mL}) - (0.100 \text{ M})(10.0 \text{ mL})}{50.0 \text{ mL} + 10.0 \text{ mL}} = 0.0250 \text{ M} \nonumber\] \[[\text{Sn}^{4+}] = \frac{(0.100 \text{ M})(10.0 \text{ mL})}{50.0 \text{ mL} + 10.0 \text{ mL}} = 0.0167 \text{ M} \nonumber\] and the potential is \[E = +0.139 \text{ V} - \frac{0.05916}{2} \log{\frac{0.0250 \text{ M}}{0.0167 \text{ M}}} = +0.134 \text{ V} \nonumber\] After the equivalence point, the concentration of Tl + and the concentration of excess Tl 3 + are easy to calculate. For this reason we find the potential using the Nernst equation for the Tl 3 + /Tl + half-reaction. For example, after adding 40.0 mL of titrant, the concentrations of Tl + and Tl 3 + are \[[\text{Tl}^{+}] = \frac{(0.050 \text{ M})(50.0 \text{ mL})}{50.0 \text{ mL} + 40.0 \text{ mL}} = 0.0278 \text{ M} \nonumber\] \[[\text{Tl}^{3+}] = \frac{(0.100 \text{ M})(40.0 \text{ mL}) - (0.050 \text{ M})(50.0 \text{ mL})}{50.0 \text{ mL} + 40.0 \text{ mL}} = 0.0167 \text{ M} \nonumber\] and the potential is \[E = +0.77 \text{ V} - \frac{0.05916}{2} \log{\frac{0.0278 \text{ M}}{0.0167 \text{ M}}} = +0.76 \text{ V} \nonumber\] At the titration’s equivalence point, the potential, E eq , potential is \[E_{eq} = \frac{0.139 \text{ V} + 0.77 \text{ V}}{2} = +0.45 \text{ V} \nonumber\] Some additional results are shown here. volume of Tl3+ (mL) E (V) volume of Tl3+ (mL).1 E (V).1 5.0 0.121 30.0 0.75 10.0 0.134 35.0 0.75 15.0 0.144 40.0 0.76 20.0 0.157 45.0 0.76 25.0 0.450 NaN NaN Sketching a Redox Titration Curve To evaluate the relationship between a titration’s equivalence point and its end point we need to construct only a reasonable approximation of the exact titration curve. In this section we demonstrate a simple method for sketching a redox titration curve. Our goal is to sketch the titration curve quickly, using as few calculations as possible. Let’s use the titration of 50.0 mL of 0.100 M Fe 2 + with 0.100 M Ce 4 + in a matrix of 1 M HClO 4 . This is the same example that we used in developing the calculations for a redox titration curve. You can review the results of that calculation in Table 9.4.1 and Figure 9.4.1 . We begin by calculating the titration’s equivalence point volume, which, as we determined earlier, is 50.0 mL. Next, we draw our axes, placing the potential, E , on the y -axis and the titrant’s volume on the x -axis. To indicate the equivalence point’s volume, we draw a vertical line that intersects the x -axis at 50.0 mL of Ce 4 + . Figure 9.4.2 a shows the result of the first step in our sketch. Before the equivalence point, the potential is determined by a redox buffer of Fe 2 + and Fe 3 + . Although we can calculate the potential using the Nernst equation, we can avoid this calculation if we make a simple assumption. You may recall from Chapter 6 that a redox buffer operates over a range of potentials that extends approximately ±(0.05916/ n ) unit on either side of $E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ}$. The potential at the buffer’s lower limit is \[E = E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} - 0.05916 \nonumber\] when the concentration of Fe 2 + is $10 \times$ greater than that of Fe 3 + . The buffer reaches its upper potential of \[E = E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} + 0.05916 \nonumber\] when the concentration of Fe 2 + is $10 \times$ smaller than that of Fe 3 + . The redox buffer spans a range of volumes from approximately 10% of the equivalence point volume to approximately 90% of the equivalence point volume. Figure 9.4.2 b shows the second step in our sketch. First, we superimpose a ladder diagram for Fe on the y -axis, using its $E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ}$ value of 0.767 V and including the buffer’s range of potentials. Next, we add points for the potential at 10% of V eq (a potential of 0.708 V at 5.0 mL) and for the potential at 90% of V eq (a potential of 0.826 V at 45.0 mL). We used a similar approach when sketching the acid–base titration curve for the titration of acetic acid with NaOH; see Chapter 9.2 for details. The third step in sketching our titration curve is to add two points after the equivalence point. Here the potential is controlled by a redox buffer of Ce 3 + and Ce 4 + . The redox buffer is at its lower limit of \[E = E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ} - 0.05916 \nonumber\] when the titrant reaches 110% of the equivalence point volume and the potential is $E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ}$ when the volume of Ce is $2 \times V_{eq}$. We used a similar approach when sketching the complexation titration curve for the titration of Mg 2 + with EDTA; see Chapter 9.3 for details. Figure 9.4.2 c shows the third step in our sketch. First, we superimpose a ladder diagram for Ce on the y -axis, using its $E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ}$ value of 1.70 V and including the buffer’s range. Next, we add points representing the potential at 110% of V eq (a value of 1.66 V at 55.0 mL) and at 200% of V eq (a value of 1.70 V at 100.0 mL). Next, we draw a straight line through each pair of points, extending the line through the vertical line that indicates the equivalence point’s volume (Figure 9.4.2 d). Finally, we complete our sketch by drawing a smooth curve that connects the three straight-line segments (Figure 9.4.2 e). A comparison of our sketch to the exact titration curve (Figure 9.4.2 f) shows that they are in close agreement. Sketch the titration curve for the titration of 50.0 mL of 0.0500 M Sn 4 + with 0.100 M Tl + . Both the titrand and the titrant are 1.0 M in HCl. The titration reaction is \[\text{Sn}^{2+}(aq) + \text{Tl}^{3+}(aq) \rightleftharpoons \text{Tl}^{+}(aq) + \text{Sn}^{4+}(aq) \nonumber\] Compare your sketch to your calculated titration curve from Exercise 9.4.1 . Answer The figure below shows a sketch of the titration curve. The two points before the equivalence point V Tl = 2.5 mL, E = +0.109 V and V Tl = 22.5 mL, E = +0.169 V are plotted using the redox buffer for Sn 4 + /Sn 2 + , which spans a potential range of +0.139 ± 0.5916/2. The two points after the equivalence point V Tl = 27.5 mL, E = +0.74 V and V Tl = 50 mL, E = +0.77 V are plotted using the redox buffer for Tl 3 + /Tl + , which spans the potential range of +0.139 ± 0.5916/2. The black dots and curve are the approximate sketch of the titration curve. The points in red are the calculations from Exercise 9.4.1 . Selecting and Evaluating the End Point A redox titration’s equivalence point occurs when we react stoichiometrically equivalent amounts of titrand and titrant. As is the case for acid–base titrations and complexation titrations, we estimate the equivalence point of a redox titration using an experimental end point. A variety of methods are available for locating a redox titration’s end point, including indicators and sensors that respond to a change in the solution conditions. Where is the Equivalence Point For an acid–base titration or a complexometric titration the equivalence point is almost identical to the inflection point on the steeply rising part of the titration curve. If you look back at Figure 9.2.2 and Figure 9.3.3 , you will see that the inflection point is in the middle of this steep rise in the titration curve, which makes it relatively easy to find the equivalence point when you sketch these titration curves. We call this a symmetric equivalence point . If the stoichiometry of a redox titration is 1:1—that is, one mole of titrant reacts with each mole of titrand—then the equivalence point is symmetric. If the titration reaction’s stoichiometry is not 1:1, then the equivalence point is closer to the top or to the bottom of the titration curve’s sharp rise. In this case we have an asymmetric equivalence point . Derive a general equation for the equivalence point’s potential when titrating Fe 2 + with $\text{MnO}_4^-$. \[5\text{Fe}^{2+}(aq) + \text{MnO}_4^-(aq) + 8\text{H}^+(aq) \rightarrow 5\text{Fe}^{3+}(aq) + \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) \nonumber\] Solution The half-reactions for the oxidation of Fe 2 + and the reduction of $\text{MnO}_4^-$ are \[\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^- \nonumber\] \[\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5 e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) \nonumber\] for which the Nernst equations are \[E = E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} - 0.5916 \log{\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}} \nonumber\] \[E = E_{\text{MnO}_4^{-}/\text{Mn}^{2+}}^{\circ} - \frac{0.5916}{5} \log{\frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^{-}][\text{H}^+]^8}} \nonumber\] Before we add together these two equations we must multiply the second equation by 5 so that we can combine the log terms; thus \[6E_{eq} = E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} + 5E_{\text{MnO}_4^{-}/\text{Mn}^{2+}}^{\circ} - 0.05916 \log{\frac{[\text{Fe}^{2+}][\text{Mn}^{2+}]}{[\text{Fe}^{3+}][\text{MnO}_4^{-}][\text{H}^+]^8}} \nonumber\] At the equivalance point we know that \[[\text{Fe}^{2+}] = 5 \times [\text{MnO}_4^-] \text{ and } [\text{Fe}^{3+}] = 5 \times [\text{Mn}^{2+}] \nonumber\] Substituting these equalities into the previous equation and rearranging gives us a general equation for the potential at the equivalence point. \[6E_{eq} = E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} + 5E_{\text{MnO}_4^{-}/\text{Mn}^{2+}}^{\circ} - 0.05916 \log{\frac{5[\text{MnO}_4^{-}][\text{Mn}^{2+}]}{5[\text{Mn}^{2+}][\text{MnO}_4^{-}][\text{H}^+]^8}} \nonumber\] \[E_{eq} = \frac{E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} + 5E_{\text{MnO}_4^{-}/\text{Mn}^{2+}}^{\circ}}{6} - \frac{0.05916}{6} \log{\frac{1}{[\text{H}^+]^8}} \nonumber\] \[E_{eq} = \frac{E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} + 5E_{\text{MnO}_4^{-}/\text{Mn}^{2+}}^{\circ}}{6} + \frac{0.05916 \times 8}{6} \log{[\text{H}^+]} \nonumber\] \[E_{eq} = \frac{E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} + 5E_{\text{MnO}_4^{-}/\text{Mn}^{2+}}^{\circ}}{6} - 0.07888 \text{pH} \nonumber\] Our equation for the equivalence point has two terms. The first term is a weighted average of the titrand’s and the titrant’s standard state potentials, in which the weighting factors are the number of electrons in their respective half-reactions. The second term shows that E eq for this titration is pH-dependent. At a pH of 1 (in H 2 SO 4 ), for example, the equivalence point has a potential of \[E_{eq} = \frac{0.768 + 5 \times 1.51}{6} - 0.07888 \times 1 = 1.31 \text{ V} \nonumber\] Figure 9.4.3 shows a typical titration curve for titration of Fe 2 + with $\text{MnO}_4^-$. Note that the titration’s equivalence point is asymmetrical. Derive a general equation for the equivalence point’s potential for the titration of U 4+ with Ce 4 + . The unbalanced reaction is \[\text{Ce}^{4+}(aq) + \text{U}^{4+}(aq) \rightarrow \text{UO}_2^{2+}(aq) + \text{Ce}^{3+}(aq) \nonumber\] What is the equivalence point’s potential if the pH is 1? Answer The two half reactions are \[\text{Ce}^{4+}(aq) + e^- \rightarrow \text{Ce}^{3+}(aq) \nonumber\] \[\text{U}^{4+}(aq) +2\text{H}_2\text{O}(l) \rightarrow \text{UO}_2^{2+}(aq)) + 4\text{H}^+(aq) +2e^- \nonumber\] for which the Nernst equations are \[E = E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ} - 0.05916 \log{\frac{[\text{Ce}^{3+}]}{[\text{Ce}^{4+}]}} \nonumber\] \[E = E_{\text{UO}_2^{2+}/\text{U}^{4+}}^{\circ} - \frac{0.05916}{2} \log{\frac{[\text{U}^{4+}]}{[\text{UO}_2^{2+}][\text{H}^+]^4}} \nonumber\] Before adding these two equations together we must multiply the second equation by 2 so that we can combine the log terms; thus \[3E = E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ} + 2E_{\text{UO}_2^{2+}/\text{U}^{4+}}^{\circ} - 0.05916 \log{\frac{[\text{Ce}^{3+}][\text{U}^{4+}]}{[\text{Ce}^{4+}][\text{UO}_2^{2+}][\text{H}^+]^4}} \nonumber\] At the equivalence point we know that \[[\text{Ce}^{3+}] = 2 \times [\text{UO}_2^{2+}] \text{ and } [\text{Ce}^{4+}] = 2 \times [\text{U}^{4+}] \nonumber\] Substituting these equalities into the previous equation and rearranging gives us a general equation for the potential at the equivalence point. \[3E = E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ} + 2E_{\text{UO}_2^{2+}/\text{U}^{4+}}^{\circ} - 0.05916 \log{\frac{2[\text{UO}_2^{2+}][\text{U}^{4+}]}{2[\text{U}^{4+}][\text{UO}_2^{2+}][\text{H}^+]^4}} \nonumber\] \[E = \frac{E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ} + 2E_{\text{UO}_2^{2+}/\text{U}^{4+}}^{\circ}}{3} - \frac{0.05916}{3} \log{\frac{1}{[\text{H}^+]^4}} \nonumber\] \[E = \frac{E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ} + 2E_{\text{UO}_2^{2+}/\text{U}^{4+}}^{\circ}}{3} + \frac{0.05916 \times 4}{3} \log{[\text{H}^+]^4} \nonumber\] \[E = \frac{E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ} + 2E_{\text{UO}_2^{2+}/\text{U}^{4+}}^{\circ}}{3} - 0.07888\text{pH} \nonumber\] At a pH of 1 the equivalence point has a potential of \[E = \frac{1.72 + 2 \times 0.327}{3} - 0.07888 \times 1 = +0.712 \text{ V} \nonumber\] Finding the End Point With an Indicator Three types of indicators are used to signal a redox titration’s end point. The oxidized and reduced forms of some titrants, such as $\text{MnO}_4^-$, have different colors. A solution of $\text{MnO}_4^-$ is intensely purple. In an acidic solution, however, permanganate’s reduced form, Mn 2 + , is nearly colorless. When using $\text{MnO}_4^-$ as a titrant, the titrand’s solution remains colorless until the equivalence point. The first drop of excess $\text{MnO}_4^-$ produces a permanent tinge of purple, signaling the end point. Some indicators form a colored compound with a specific oxidized or reduced form of the titrant or the titrand. Starch, for example, forms a dark purple complex with $\text{I}_3^-$. We can use this distinct color to signal the presence of excess $\text{I}_3^-$ as a titrant—a change in color from colorless to purple—or the completion of a reaction that consumes $\text{I}_3^-$ as the titrand— a change in color from purple to colorless. Another example of a specific indicator is thiocyanate, SCN – , which forms the soluble red-colored complex of Fe(SCN) 2+ in the presence of Fe 3 + . The most important class of indicators are substances that do not participate in the redox titration, but whose oxidized and reduced forms differ in color. When we add a redox indicator to the titrand, the indicator imparts a color that depends on the solution’s potential. As the solution’s potential changes with the addition of titrant, the indicator eventually changes oxidation state and changes color, signaling the end point. To understand the relationship between potential and an indicator’s color, consider its reduction half-reaction \[\text{In}_\text{ox} + ne^- \rightleftharpoons \text{In}_\text{red} \nonumber\] where In ox and In red are, respectively, the indicator’s oxidized and reduced forms. For simplicity, In ox and In red are shown without specific charges. Because there is a change in oxidation state, In ox and In red cannot both be neutral. The Nernst equation for this half-reaction is \[E = E_{\text{In}_\text{ox}/\text{In}_\text{red}}^{\circ} - \frac{0.05916}{n} \log{\frac{[\text{In}_\text{red}]}{[\text{In}_\text{ox}]}} \nonumber\] As shown in Figure 9.4.4 , if we assume the indicator’s color changes from that of In ox to that of In red when the ratio [In red ]/[In ox ] changes from 0.1 to 10, then the end point occurs when the solution’s potential is within the range \[E = E_{\text{In}_\text{ox}/\text{In}_\text{red}}^{\circ} \pm \frac{0.05916}{n} \nonumber\] This is the same approach we took in considering acid–base indicators and complexation indicators. A partial list of redox indicators is shown in Table 9.4.2 . Examples of an appropriate and an inappropriate indicator for the titration of Fe 2 + with Ce 4 + are shown in Figure 9.4.5 . indicator color Inox color of Inred Eo (V) indigo tetrasulfate blue colorless 0.360 methylene blue blue colorless 0.530 diphenylamine violet colorless 0.750 diphenylamine sulfonic acid red-violet colorless 0.850 tris(2,2'-bipyradine)iron pale blue red 1.120 ferroin pale blue red 1.147 tris(5-nitro-1,10-phenanthroline)iron pale blue red-violet 1.250 Other Methods for Finding the End Point Another method for locating a redox titration’s end point is a potentiometric titration in which we monitor the change in potential while we add the titrant to the titrand. The end point is found by examining visually the titration curve. The simplest experimental design for a potentiometric titration consists of a Pt indicator electrode whose potential is governed by the titrand’s or the titrant’s redox half-reaction, and a reference electrode that has a fixed potential. Other methods for locating the titration’s end point include thermometric titrations and spectrophotometric titrations. You will a further discussion of potentiometry in Chapter 11 . Representative Method 9.4.1: Determination of Total Chlorine Residual The best way to appreciate the theoretical and the practical details discussed in this section is to carefully examine a typical redox titrimetric method. Although each method is unique, the following description of the determination of the total chlorine residual in water provides an instructive example of a typical procedure. The description here is based on Method 4500-Cl B as published in Standard Methods for the Examination of Water and Wastewater , 20th Ed., American Public Health Association: Washington, D. C., 1998. Description of the Method The chlorination of a public water supply produces several chlorine-containing species, the combined concentration of which is called the total chlorine residual. Chlorine is present in a variety of chemical states, including the free residual chlorine, which consists of Cl 2 , HOCl and OCl – , and the combined chlorine residual, which consists of NH 2 Cl, NHCl 2 , and NCl 3 . The total chlorine residual is determined by using the oxidizing power of chlorine to convert I – to $\text{I}_3^-$. The amount of $\text{I}_3^-$ formed is then determined by titrating with Na 2 S 2 O 3 using starch as an indicator. Regardless of its form, the total chlorine residual is reported as if Cl 2 is the only source of chlorine, and is reported as mg Cl/L. Procedure Select a volume of sample that requires less than 20 mL of Na 2 S 2 O 3 to reach the end point. Using glacial acetic acid, acidify the sample to a pH between 3 and 4, and add about 1 gram of KI. Titrate with Na 2 S 2 O 3 until the yellow color of $\text{I}_3^-$ begins to disappear. Add 1 mL of a starch indicator solution and continue titrating until the blue color of the starch–$\text{I}_3^-$ complex disappears (Figure 9.4.6 ). Use a blank titration to correct the volume of titrant needed to reach the end point for reagent impurities. Questions 1. Is this an example of a direct or an indirect analysis? This is an indirect analysis because the chlorine-containing species do not react with the titrant. Instead, the total chlorine residual oxidizes I – to $\text{I}_3^-$, and the amount of $\text{I}_3^-$ is determined by titrating with Na 2 S 2 O 3 . 2. Why does the procedure rely on an indirect analysis instead of directly titrating the chlorine-containing species using KI as a titrant? Because the total chlorine residual consists of six different species, a titration with I – does not have a single, well-defined equivalence point. By converting the chlorine residual to an equivalent amount of $\text{I}_3^-$, the indirect titration with Na 2 S 2 O 3 has a single, useful equivalence point. Even if the total chlorine residual is from a single species, such as HOCl, a direct titration with KI is impractical. Because the product of the titration, $\text{I}_3^-$, imparts a yellow color, the titrand’s color would change with each addition of titrant, making it difficult to find a suitable indicator. 3. Both oxidizing and reducing agents can interfere with this analysis. Explain the effect of each type of interferent on the total chlorine residual. An interferent that is an oxidizing agent converts additional I – to $\text{I}_3^-$. Because this extra $\text{I}_3^-$ requires an additional volume of Na 2 S 2 O 3 to reach the end point, we overestimate the total chlorine residual. If the interferent is a reducing agent, it reduces back to I – some of the $\text{I}_3^-$ produced by the reaction between the total chlorine residual and iodide; as a result, we underestimate the total chlorine residual. Quantitative Applications Although many quantitative applications of redox titrimetry have been re- placed by other analytical methods, a few important applications continue to find relevance. In this section we review the general application of redox titrimetry with an emphasis on environmental, pharmaceutical, and industrial applications. We begin, however, with a brief discussion of selecting and characterizing redox titrants, and methods for controlling the titrand’s oxidation state. Adjusting the Titrand's Oxidation State If a redox titration is to be used in a quantitative analysis, the titrand initially must be present in a single oxidation state. For example, iron is determined by a redox titration in which Ce 4 + oxidizes Fe 2 + to Fe 3 + . Depending on the sample and the method of sample preparation, iron initially may be present in both the +2 and +3 oxidation states. Before titrating, we must reduce any Fe 3 + to Fe 2 + if we want to determine the total concentration of iron in the sample. This type of pretreatment is accomplished using an auxiliary reducing agent or oxidizing agent. A metal that is easy to oxidize—such as Zn, Al, and Ag—can serve as an auxiliary reducing agent . The metal, as a coiled wire or powder, is added to the sample where it reduces the titrand. Because any unreacted auxiliary reducing agent will react with the titrant, it is removed before we begin the titration by removing the coiled wire or by filtering. An alternative method for using an auxiliary reducing agent is to immobilize it in a column. To prepare a reduction column an aqueous slurry of the finally divided metal is packed in a glass tube equipped with a porous plug at the bottom. The sample is placed at the top of the column and moves through the column under the influence of gravity or vacuum suction. The length of the reduction column and the flow rate are selected to ensure the analyte’s complete reduction. Two common reduction columns are used. In the Jones reductor the column is filled with amalgamated zinc, Zn(Hg), which is prepared by briefly placing Zn granules in a solution of HgCl 2 . Oxidation of zinc \[\text{Zn(Hg)}(s) \rightarrow \text{Zn}^{2+}(aq) + \text{Hg}(l) + 2e^- \nonumber\] provides the electrons for reducing the titrand. In the Walden reductor the column is filled with granular Ag metal. The solution containing the titrand is acidified with HCl and passed through the column where the oxidation of silver \[\text{Ag}(s) + \text{Cl}^- (aq) \rightarrow \text{AgCl}(s) + e^- \nonumber\] provides the necessary electrons for reducing the titrand. Table 9.4.3 provides a summary of several applications of reduction columns. oxidized titrand Walden reductor Jones reducator Cr3+ — $\text{Cr}^{3+}(aq)+e^- \rightarrow \text{Cr}^{2+}(aq)$ Cu2+ $\text{Cu}^{2+}(aq)+e^- \rightarrow \text{Cu}^{+}(aq)$ $\text{Cu}^{2+}(aq)+2e^- \rightarrow \text{Cu}(s)$ Fe3+ $\text{Fe}^{3+}(aq)+e^- \rightarrow \text{Fe}^{2+}(aq)$ $\text{Fe}^{3+}(aq)+e^- \rightarrow \text{Fe}^{2+}(aq)$ TiO2+ — $\text{TiO}^{2+} (aq) + 2\text{H}^+ (aq) + e^- \rightarrow \text{Ti}^{3+} (aq) + \text{H}_2\text{O}(l)$ $\text{MoO}_4^{2+}$ $\text{MoO}_2^{2+}(aq) + e^- \rightarrow \text{MoO}_2^+(aq)$ $\text{MoO}_2^{2+}(aq) + 4\text{H}^+(aq) + 3 e^- \rightarrow \text{Mo}^{3+}(aq) + 2\text{H}_2\text{O}(l)$ $\text{VO}_4^{+}$ $\text{VO}_2^{+}(aq) + 2\text{H}^+(aq) + e^- \rightarrow \text{VO}^{2+}(aq) + \text{H}_2\text{O}(l)$ $\text{VO}_2^{+}(aq) + 4\text{H}^+(aq) + 3e^- \rightarrow \text{V}^{2+}(aq) + 2\text{H}_2\text{O}(l)$ Several reagents are used as auxiliary oxidizing agents , including ammonium peroxydisulfate, (NH 4 ) 2 S 2 O 8 , and hydrogen peroxide, H 2 O 2 . Peroxydisulfate is a powerful oxidizing agent \[\text{S}_2\text{O}_8^{2-}(aq) + 2e^- \rightarrow 2\text{SO}_4^{2-}(aq) \nonumber\] that is capable of oxidizing Mn 2 + to $\text{MnO}_4^-$, Cr 3 + to $\text{Cr}_2\text{O}_7^{2-}$, and Ce 3 + to Ce 4 + . Excess peroxydisulfate is destroyed by briefly boiling the solution. The reduction of hydrogen peroxide in an acidic solution \[\text{H}_2\text{O}_2(aq) + 2\text{H}^+(aq) + 2e^- \rightarrow 2\text{H}_2\text{O}(l) \nonumber\] provides another method for oxidizing a titrand. Excess H 2 O 2 is destroyed by briefly boiling the solution. Selecting and Standardizing a Titrant If it is to be used quantitatively, the titrant’s concentration must remain stable during the analysis. Because a titrant in a reduced state is susceptible to air oxidation, most redox titrations use an oxidizing agent as the titrant. There are several common oxidizing titrants, including $\text{MnO}_4^-$, Ce 4 + , $\text{Cr}_2\text{O}_7^{2-}$, and $\text{I}_3^-$. Which titrant is used often depends on how easily it oxidizes the titrand. A titrand that is a weak reducing agent needs a strong oxidizing titrant if the titration reaction is to have a suitable end point. The two strongest oxidizing titrants are $\text{MnO}_4^-$ and Ce 4 + , for which the reduction half-reactions are \[\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightleftharpoons \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) \nonumber\] \[\text{Ce}^{4+}(aq) + e^- \rightleftharpoons \text{Ce}^{3+}(aq) \nonumber\] A solution of Ce 4 + in 1 M H 2 SO 4 usually is prepared from the primary standard cerium ammonium nitrate, Ce(NO 3 ) 4 •2NH 4 NO 3 . When prepared using a reagent grade material, such as Ce(OH) 4 , the solution is standardized against a primary standard reducing agent such as Na 2 C 2 O 4 or Fe 2 + (prepared from iron wire) using ferroin as an indicator. Despite its availability as a primary standard and its ease of preparation, Ce 4 + is not used as frequently as $\text{MnO}_4^-$ because it is more expensive. The standardization reactions are \[\text{Ce}^{4+}(aq) + \text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + \text{Ce}^{3+}(aq) \nonumber\] \[2\text{Ce}^{4+}(aq) + \text{H}_2\text{C}_2\text{O}_4(aq) \rightarrow 2\text{Ce}^{3+}(aq) + 2\text{CO}_2(g) + 2\text{H}^+(aq) \nonumber\] A solution of $\text{MnO}_4^-$ is prepared from KMnO 4 , which is not available as a primary standard. An aqueous solution of permanganate is thermodynamically unstable due to its ability to oxidize water. \[4\text{MnO}_4^-(aq) + 2\text{H}_2\text{O}(l) \rightleftharpoons 4\text{MnO}_2(s) + 3\text{O}_2 (g) + 4\text{OH}^-(aq) \nonumber\] This reaction is catalyzed by the presence of MnO 2 , Mn 2 + , heat, light, and the presence of acids and bases. A moderately stable solution of permanganate is prepared by boiling it for an hour and filtering through a sintered glass filter to remove any solid MnO 2 that precipitates. Standardization is accomplished against a primary standard reducing agent such as Na 2 C 2 O 4 or Fe 2 + (prepared from iron wire), with the pink color of excess $\text{MnO}_4^-$ signaling the end point. A solution of $\text{MnO}_4^-$ prepared in this fashion is stable for 1–2 weeks, although you should recheck the standardization periodically. The standardization reactions are \[\text{MnO}_4^-(aq) + 5\text{Fe}^{2+}(aq) + 8\text{H}^+(aq) \rightarrow \text{Mn}^{2+}(aq) + 5\text{Fe}^{3+}(aq) + 4\text{H}_2\text{O}(l) \nonumber\] \[2\text{MnO}_4^-(aq) + 5\text{H}_2\text{C}_2\text{O}_4(aq) + 6\text{H}^+(aq) \rightarrow 2\text{Mn}^{2+}(aq) + 10\text{CO}_2(g) + 8\text{H}_2\text{O}(l) \nonumber\] Potassium dichromate is a relatively strong oxidizing agent whose principal advantages are its availability as a primary standard and its long term stability when in solution. It is not, however, as strong an oxidizing agent as $\text{MnO}_4^-$ or Ce 4 + , which makes it less useful when the titrand is a weak reducing agent. Its reduction half-reaction is \[\text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6e^- \rightleftharpoons 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) \nonumber\] Although a solution of $\text{Cr}_2\text{O}_7^{2-}$ is orange and a solution of Cr 3 + is green, neither color is intense enough to serve as a useful indicator. Diphenylamine sulfonic acid, whose oxidized form is red-violet and reduced form is colorless, gives a very distinct end point signal with $\text{Cr}_2\text{O}_7^{2-}$. Iodine is another important oxidizing titrant. Because it is a weaker oxidizing agent than $\text{MnO}_4^-$, Ce 4 + , and $\text{Cr}_2\text{O}_7^{2-}$, it is useful only when the titrand is a stronger reducing agent. This apparent limitation, however, makes I 2 a more selective titrant for the analysis of a strong reducing agent in the presence of a weaker reducing agent. The reduction half-reaction for I 2 is \[\text{I}_2(aq) + 2e^- \rightleftharpoons 2\text{I}^-(aq) \nonumber\] Because iodine is not very soluble in water, solutions are prepared by adding an excess of I – . The complexation reaction \[\text{I}_2(aq) + \text{I}^-(aq) \rightleftharpoons \text{I}_3^-(aq) \nonumber\] increases the solubility of I 2 by forming the more soluble triiodide ion, $\text{I}_3^-$. Even though iodine is present as $\text{I}_3^-$ instead of I 2 , the number of electrons in the reduction half-reaction is unaffected. \[\text{I}_3^-(aq) + 2e^-(aq) \rightleftharpoons 3\text{I}^-(aq) \nonumber\] Solutions of $\text{I}_3^-$ normally are standardized against Na 2 S 2 O 3 using starch as a specific indicator for $\text{I}_3^-$. The standardization reaction is \[\text{I}_3^-(aq) + 2\text{S}_2\text{O}_3^{2-}(aq) \rightarrow 3\text{I}^-(aq) + 2\text{S}_4\text{O}_6^{2-} (aq) \nonumber\] An oxidizing titrant such as $\text{MnO}_4^-$, Ce 4 + , $\text{Cr}_2\text{O}_7^{2-}$, and $\text{I}_3^-$, is used when the titrand is in a reduced state. If the titrand is in an oxidized state, we can first reduce it with an auxiliary reducing agent and then complete the titration using an oxidizing titrant. Alternatively, we can titrate it using a reducing titrant. Iodide is a relatively strong reducing agent that could serve as a reducing titrant except that its solutions are susceptible to the air-oxidation of I – to $\text{I}_3^-$. \[3\text{I}^-(aq) \rightleftharpoons \text{I}_3^- (aq) + 2e^- \nonumber\] A freshly prepared solution of KI is clear, but after a few days it may show a faint yellow coloring due to the presence of $\text{I}_3^-$. Instead, adding an excess of KI reduces the titrand and releases a stoichiometric amount of $\text{I}_3^-$. The amount of $\text{I}_3^-$ produced is then determined by a back titration using thiosulfate, $\text{S}_2\text{O}_3^{2-}$, as a reducing titrant. \[2\text{S}_2\text{O}_3^{2-}(aq) \rightleftharpoons \text{S}_4\text{O}_6^{2-}(aq) + 2e^- \nonumber\] Solutions of $\text{S}_2\text{O}_3^{2-}$ are prepared using Na 2 S 2 O 3 •5H 2 O and are standardized before use. Standardization is accomplished by dissolving a carefully weighed portion of the primary standard KIO 3 in an acidic solution that contains an excess of KI. The reaction between $\text{IO}_3^-$ and I – \[\text{IO}_3^-(aq) + 8\text{I}^-(aq) + 6\text{H}^+(aq) \rightarrow 3\text{I}_3^-(aq) + 3\text{H}_2\text{O}(l) \nonumber\] liberates a stoichiometric amount of I-3 . By titrating this $\text{I}_3^-$ with thiosulfate, using starch as a visual indicator, we can determine the concentration of $\text{S}_2\text{O}_3^{2-}$ in the titrant. The standardization titration is \[\text{I}_3^-(aq) + 2\text{S}_2\text{O}_3^{2-}(aq) \rightarrow 3\text{I}^-(aq) + \text{S}_4\text{O}_6^{2-}(aq) \nonumber\] which is the same reaction used to standardize solutions of $\text{I}_3^-$. This approach to standardizing solutions of $\text{S}_2\text{O}_2^{3-}$ is similar to that used in the determination of the total chlorine residual outlined in Representative Method 9.4.1 . Although thiosulfate is one of the few reducing titrants that is not readily oxidized by contact with air, it is subject to a slow decomposition to bisulfite and elemental sulfur. If used over a period of several weeks, a solution of thiosulfate is restandardized periodically. Several forms of bacteria are able to metabolize thiosulfate, which leads to a change in its concentration. This problem is minimized by adding a preservative such as HgI 2 to the solution. Another useful reducing titrant is ferrous ammonium sulfate, Fe(NH 4 ) 2 (SO 4 ) 2 •6H 2 O, in which iron is present in the +2 oxidation state. A solution of Fe 2 + is susceptible to air-oxidation, but when prepared in 0.5 M H 2 SO 4 it remains stable for as long as a month. Periodic restandardization with K 2 Cr 2 O 7 is advisable. Ferrous ammonium sulfate is used as the titrant in a direct analysis of the titrand, or, it is added to the titrand in excess and the amount of Fe 3 + produced determined by back titrating with a standard solution of Ce 4 + or $\text{Cr}_2\text{O}_7^{2-}$. Inorganic Analysis One of the most important applications of redox titrimetry is evaluating the chlorination of public water supplies. Representative Method 9.4.1 , for example, describes an approach for determining the total chlorine residual using the oxidizing power of chlorine to oxidize I – to $\text{I}_3^-$. The amount of $\text{I}_3^-$ is determined by back titrating with $\text{S}_2\text{O}_3^{2-}$. The efficiency of chlorination depends on the form of the chlorinating species. There are two contributions to the total chlorine residual—the free chlorine residual and the combined chlorine residual. The free chlorine residual includes forms of chlorine that are available for disinfecting the water supply. Examples of species that contribute to the free chlorine residual include Cl 2 , HOCl and OCl – . The combined chlorine residual includes those species in which chlorine is in its reduced form and, therefore, no longer capable of providing disinfection. Species that contribute to the combined chlorine residual are NH 2 Cl, NHCl 2 and NCl 3 . When a sample of iodide-free chlorinated water is mixed with an excess of the indicator N , N -diethyl- p -phenylenediamine (DPD), the free chlorine oxidizes a stoichiometric portion of DPD to its red-colored form. The oxidized DPD is then back-titrated to its colorless form using ferrous ammonium sulfate as the titrant. The volume of titrant is proportional to the free residual chlorine. Having determined the free chlorine residual in the water sample, a small amount of KI is added, which catalyzes the reduction of monochloramine, NH 2 Cl, and oxidizes a portion of the DPD back to its red-colored form. Titrating the oxidized DPD with ferrous ammonium sulfate yields the amount of NH 2 Cl in the sample. The amount of dichloramine and trichloramine are determined in a similar fashion. The methods described above for determining the total, free, or combined chlorine residual also are used to establish a water supply’s chlorine demand. Chlorine demand is defined as the quantity of chlorine needed to react completely with any substance that can be oxidized by chlorine, while also maintaining the desired chlorine residual. It is determined by adding progressively greater amounts of chlorine to a set of samples drawn from the water supply and determining the total, free, or combined chlorine residual. Another important example of redox titrimetry, which finds applications in both public health and environmental analysis, is the determination of dissolved oxygen. In natural waters, such as lakes and rivers, the level of dissolved O 2 is important for two reasons: it is the most readily available oxidant for the biological oxidation of inorganic and organic pollutants; and it is necessary for the support of aquatic life. In a wastewater treatment plant dissolved O 2 is essential for the aerobic oxidation of waste materials. If the concentration of dissolved O 2 falls below a critical value, aerobic bacteria are replaced by anaerobic bacteria, and the oxidation of organic waste produces undesirable gases, such as CH 4 and H 2 S. One standard method for determining dissolved O 2 in natural waters and wastewaters is the Winkler method. A sample of water is collected without exposing it to the atmosphere, which might change the concentration of dissolved O 2 . The sample first is treated with a solution of MnSO 4 and then with a solution of NaOH and KI. Under these alkaline conditions the dissolved oxygen oxidizes Mn 2 + to MnO 2 . \[2\text{Mn}^{2+}(aq) + 4\text{OH}^-(aq) + \text{O}_2(g) \rightarrow 2\text{MnO}_2(s) + 2\text{H}_2\text{O}(l) \nonumber\] After the reaction is complete, the solution is acidified with H 2 SO 4 . Under the now acidic conditions, I – is oxidized to $\text{I}_3^-$ by MnO 2 . \[\text{MnO}_2(s) + 3\text{I}^-(aq) + 4\text{H}^+(aq) \rightarrow \text{Mn}^{2+}(aq) + \text{I}_3^-(aq) + 2\text{H}_2\text{O}(l) \nonumber\] The amount of $\text{I}_3^-$ that forms is determined by titrating with $\text{S}_2\text{O}_3^{2-}$ using starch as an indicator. The Winkler method is subject to a variety of interferences and several modifications to the original procedure have been proposed. For example, $\text{NO}_2^-$ interferes because it reduces $\text{I}_3^-$ to I – under acidic conditions. This interference is eliminated by adding sodium azide, NaN 3 , which reduces $\text{NO}_2^-$ to N 2 . Other reducing agents, such as Fe 2 + , are eliminated by pretreating the sample with KMnO 4 and destroying any excess permanganate with K 2 C 2 O 4 . Another important example of redox titrimetry is the determination of water in nonaqueous solvents. The titrant for this analysis is known as the Karl Fischer reagent and consists of a mixture of iodine, sulfur dioxide, pyridine, and methanol. Because the concentration of pyridine is sufficiently large, I 2 and SO 2 react with pyridine (py) to form the complexes py•I 2 and py•SO 2 . When added to a sample that contains water, I 2 is reduced to I – and SO 2 is oxidized to SO 3 . \[\text{py}\cdot\text{I}_2 + \text{py}\cdot\text{SO}_2 + \text{H}_2\text{O} + 2\text{py} \rightarrow 2\text{py}\cdot\text{HI} + \text{py}\cdot\text{SO}_3 \nonumber\] Methanol is included to prevent the further reaction of py•SO 3 with water. The titration’s end point is signaled when the solution changes from the product’s yellow color to the brown color of the Karl Fischer reagent. Organic Analysis Redox titrimetry also is used for the analysis of organic analytes. One important example is the determination of the chemical oxygen demand (COD) of natural waters and wastewaters. The COD is a measure of the quantity of oxygen necessary to oxidize completely all the organic matter in a sample to CO 2 and H 2 O. Because no attempt is made to correct for organic matter that is decomposed biologically, or for slow decomposition kinetics, the COD always overestimates a sample’s true oxygen demand. The determination of COD is particularly important in the management of industrial wastewater treatment facilities where it is used to monitor the release of organic-rich wastes into municipal sewer systems or into the environment. A sample’s COD is determined by refluxing it in the presence of excess K 2 Cr 2 O 7 , which serves as the oxidizing agent. The solution is acidified with H 2 SO 4 , using Ag 2 SO 4 to catalyze the oxidation of low molecular weight fatty acids. Mercuric sulfate, HgSO 4 , is added to complex any chloride that is present, which prevents the precipitation of the Ag + catalyst as AgCl. Under these conditions, the efficiency for oxidizing organic matter is 95–100%. After refluxing for two hours, the solution is cooled to room temperature and the excess $\text{Cr}_2\text{O}_7^{2-}$ determined by a back titration using ferrous ammonium sulfate as the titrant and ferroin as the indicator. Because it is difficult to remove completely all traces of organic matter from the reagents, a blank titration is performed. The difference in the amount of ferrous ammonium sulfate needed to titrate the sample and the blank is proportional to the COD. Iodine has been used as an oxidizing titrant for a number of compounds of pharmaceutical interest. Earlier we noted that the reaction of $\text{S}_2\text{O}_3^{2-}$ with $\text{I}_3^-$ produces the tetrathionate ion, $\text{S}_4\text{O}_6^{2-}$. The tetrathionate ion is actually a dimer that consists of two thiosulfate ions connected through a disulfide (–S–S–) linkage. In the same fashion, $\text{I}_3^-$ is used to titrate mercaptans of the general formula RSH, forming the dimer RSSR as a product. The amino acid cysteine also can be titrated with $\text{I}_3^-$. The product of this titration is cystine, which is a dimer of cysteine. Triiodide also is used for the analysis of ascorbic acid (vitamin C) by oxidizing the enediol functional group to an alpha diketone and for the analysis of reducing sugars, such as glucose, by oxidizing the aldehyde functional group to a carboxylate ion in a basic solution. An organic compound that contains a hydroxyl, a carbonyl, or an amine functional group adjacent to an hydoxyl or a carbonyl group can be oxidized using metaperiodate, $\text{IO}_4^-$, as an oxidizing titrant. \[\text{IO}_4^-(aq) + \text{H}_2\text{O}(l) + 2e^- \rightleftharpoons \text{IO}_3^-(aq) + 2\text{OH}^-(aq) \nonumber\] A two-electron oxidation cleaves the C–C bond between the two functional groups with hydroxyl groups oxidized to aldehydes or ketones, carbonyl groups oxidized to carboxylic acids, and amines oxidized to an aldehyde and an amine (ammonia if a primary amine). The analysis is conducted by adding a known excess of $\text{IO}_4^-$ to the solution that contains the analyte and allowing the oxidation to take place for approximately one hour at room temperature. When the oxidation is complete, an excess of KI is added, which converts any unreacted $\text{IO}_4^-$ to $\text{IO}_3^-$ and $\text{I}_3^-$. \[\text{IO}_4^-(aq) + 3\text{I}^-(aq) + \text{H}_2\text{O}(l) \rightarrow \text{IO}_3^-(aq) + \text{I}_3^-(aq) + 2\text{OH}^-(aq) \nonumber\] The $\text{I}_3^-$ is then determined by titrating with $\text{S}_2\text{O}_3^{2-}$ using starch as an indicator. Quantitative Calculations The quantitative relationship between the titrand and the titrant is determined by the stoichiometry of the titration reaction. If you are unsure of the balanced reaction, you can deduce its stoichiometry by remembering that the electrons in a redox reaction are conserved. The amount of Fe in a 0.4891-g sample of an ore is determined by titrating with K 2 Cr 2 O 7 . After dissolving the sample in HCl, the iron is brought into a +2 oxidation state using a Jones reductor. Titration to the diphenylamine sulfonic acid end point requires 36.92 mL of 0.02153 M K 2 Cr 2 O 7 . Report the ore’s iron content as %w/w Fe 2 O 3 . Solution Because we are not provided with the titration reaction, we will use a conservation of electrons to deduce the stoichiometry. During the titration the analyte is oxidized from Fe 2 + to Fe 3 + , and the titrant is reduced from $\text{Cr}_2\text{O}_7^{2-}$ to Cr 3 + . Oxidizing Fe 2 + to Fe 3 + requires a single electron. Reducing $\text{Cr}_2\text{O}_7^{2-}$, in which each chromium is in the +6 oxidation state, to Cr 3+ requires three electrons per chromium, for a total of six electrons. A conservation of electrons for the titration, therefore, requires that each mole of K 2 Cr 2 O 7 reacts with six moles of Fe 2 + . The moles of K 2 Cr 2 O 7 used to reach the end point is \[(0.02153 \text{ M})(0.03692 \text{ L}) = 7.949 \times 10^{-4} \text{ mol K}_2\text{Cr}_2\text{O}_7 \nonumber\] which means the sample contains \[7.949 \times 10^{-4} \text{ mol K}_2\text{Cr}_2\text{O}_7 \times \frac{6 \text{ mol Fe}^{2+}}{\text{mol K}_2\text{Cr}_2\text{O}_7} = 4.769 \times 10^{-3} \text{ mol Fe}^{2+} \nonumber\] Thus, the %w/w Fe 2 O 3 in the sample of ore is \[4.769 \times 10^{-3} \text{ mol Fe}^{2+} \times \frac{1 \text{ mol Fe}_2\text{O}_3}{2 \text{ mol Fe}^{2+}} \times \frac{159.69 \text{g Fe}_2\text{O}_3}{\text{mol Fe}_2\text{O}_3} = 0.3808 \text{ g Fe}_2\text{O}_3 \nonumber\] \[\frac{0.3808 \text{ g Fe}_2\text{O}_3}{0.4891 \text{ g sample}} \times 100 = 77.86 \text{% w/w Fe}_2\text{O}_3 \nonumber\] Although we can deduce the stoichiometry between the titrant and the titrand in Example 9.4.2 without balancing the titration reaction, the balanced reaction \[\text{K}_2\text{Cr}_2\text{O}_7(aq) + 6\text{Fe}^{2+}(aq) + 14\text{H}^+(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 2\text{K}^+(aq) + 6\text{Fe}^{3+}(aq) + 7\text{H}_2\text{O}(l) \nonumber\] does provide useful information. For example, the presence of H + reminds us that the reaction must take place in an acidic solution. The purity of a sample of sodium oxalate, Na 2 C 2 O 4 , is determined by titrating with a standard solution of KMnO 4 . If a 0.5116-g sample requires 35.62 mL of 0.0400 M KMnO 4 to reach the titration’s end point, what is the %w/w Na 2 C 2 O 4 in the sample. Answer Because we are not provided with a balanced reaction, let’s use a conservation of electrons to deduce the stoichiometry. Oxidizing $\text{C}_2\text{O}_4^{2-}$, in which each carbon has a +3 oxidation state, to CO 2 , in which carbon has an oxidation state of +4, requires one electron per carbon or a total of two electrons for each mole of $\text{C}_2\text{O}_4^{2-}$. Reducing $\text{MnO}_4^-$, in which each manganese is in the +7 oxidation state, to Mn 2 + requires five electrons. A conservation of electrons for the titration, therefore, requires that two moles of KMnO 4 (10 moles of e - ) react with five moles of Na 2 C 2 O 4 (10 moles of e - ). The moles of KMnO 4 used to reach the end point is \[(0.0400 \text{ M KMnO}_4)(0.03562 \text{ L})=1.42 \times 10^{-3} \text{ mol KMnO}_4 \nonumber\] which means the sample contains \[1 .42 \times 10^{-3} \text{ mol KMnO}_4 \times \frac{5 \text{ mol Na}_2\text{C}_2\text{O}_4}{2 \text{ mol KMnO}_4} = 3.55 \times 10^{-3} \text{ mol Na}_2\text{C}_2\text{O}_4 \nonumber\] Thus, the %w/w Na 2 C 2 O 4 in the sample of ore is \[3.55 \times 10^{-3} \text{ mol Na}_2\text{C}_2\text{O}_4 \times \frac{134.00 \text{ g Na}_2\text{C}_2\text{O}_4}{\text{mol Na}_2\text{C}_2\text{O}_4} = 0.476 \text{ g Na}_2\text{C}_2\text{O}_4 \nonumber\] \[\frac{0.476 \text{ g Na}_2\text{C}_2\text{O}_4}{0.5116 \text{ g sample}} \times 100 = 93.0 \text{% w/w Na}_2\text{C}_2\text{O}_4 \nonumber\] As shown in the following two examples, we can easily extend this approach to an analysis that requires an indirect analysis or a back titration. A 25.00-mL sample of a liquid bleach is diluted to 1000 mL in a volumetric flask. A 25-mL portion of the diluted sample is transferred by pipet into an Erlenmeyer flask that contains an excess of KI, reducing the OCl – to Cl – and producing $\text{I}_3^-$. The liberated $\text{I}_3^-$ is determined by titrating with 0.09892 M Na 2 S 2 O 3 , requiring 8.96 mL to reach the starch indicator end point. Report the %w/v NaOCl in the sample of bleach. Solution To determine the stoichiometry between the analyte, NaOCl, and the titrant, Na 2 S 2 O 3 , we need to consider both the reaction between OCl – and I – , and the titration of $\text{I}_3^-$ with Na 2 S 2 O 3 . First, in reducing OCl – to Cl – the oxidation state of chlorine changes from +1 to –1, requiring two electrons. The oxidation of three I – to form $\text{I}_3^-$ releases two electrons as the oxidation state of each iodine changes from –1 in I – to –1⁄3 in $\text{I}_3^-$. A conservation of electrons, therefore, requires that each mole of OCl – produces one mole of $\text{I}_3^-$. Second, in the titration reaction, $\text{I}_3^-$ is reduced to I – and $\text{S}_2\text{O}_3^{2-}$ is oxidized to $\text{S}_4\text{O}_6^{2-}$. Reducing $\text{I}_3^-$ to 3I – requires two elections as each iodine changes from an oxidation state of –1⁄3 to –1. In oxidizing $\text{S}_2\text{O}_3^{2-}$ to $\text{S}_4\text{O}_6^{2-}$, each sulfur changes its oxidation state from +2 to +2.5, releasing one electron for each $\text{S}_2\text{O}_3^{2-}$. A conservation of electrons, therefore, requires that each mole of $\text{I}_3^-$ reacts with two moles of $\text{S}_2\text{O}_3^{2-}$. Finally, because each mole of OCl – produces one mole of $\text{I}_3^-$, and each mole of $\text{I}_3^-$ reacts with two moles of $\text{S}_2\text{O}_3^{2-}$, we know that every mole of NaOCl in the sample ultimately results in the consumption of two moles of Na 2 S 2 O 3 . The moles of Na 2 S 2 O 3 used to reach the titration’s end point is \[(0.09892 \text{ M})(0.00896 \text{ L}) = 8.86 \times 10^{-4} \text{ mol Na}_2\text{S}_2\text{O}_3 \nonumber\] which means the sample contains \[8.86 \times 10^{-4} \text{ mol Na}_2\text{S}_2\text{O}_3 \times \frac{1 \text{ mol NaOCl}}{\text{mol Na}_2\text{S}_2\text{O}_3} \times \frac{74.44 \text{ g NaOCl}}{\text{mol NaOCl}} = 0.03299 \text{ g NaOCl} \nonumber\] Thus, the %w/v NaOCl in the diluted sample is \[\frac{0.03299 \text{ g NaOCl}}{25.00 \text{ mL}} \times 100 = 0.132 \text{% w/v NaOCl} \nonumber\] Because the bleach was diluted by a factor of $40 \times$ (25 mL to 1000 mL), the concentration of NaOCl in the bleach is 5.28% w/v. The balanced reactions for this analysis are: \[\text{OCl}^-(aq) + 3\text{I}^-(aq) + 2\text{H}^+(aq) \rightarrow \text{I}_3^-(aq) + \text{Cl}^-(aq) + \text{H}_2\text{O}(l) \nonumber\] \[\text{I}_3^-(aq) + 2\text{S}_2\text{O}_3^{2-}(aq) \rightarrow \text{S}_4\text{O}_6^{2-}(aq) + 3\text{I}^-(aq) \nonumber\] The amount of ascorbic acid, C 6 H 8 O 6 , in orange juice is determined by oxidizing ascorbic acid to dehydroascorbic acid, C 6 H 6 O 6 , with a known amount of $\text{I}_3^-$, and back titrating the excess $\text{I}_3^-$ with Na 2 S 2 O 3 . A 5.00-mL sample of filtered orange juice is treated with 50.00 mL of 0.01023 M $\text{I}_3^-$. After the oxidation is complete, 13.82 mL of 0.07203 M Na 2 S 2 O 3 is needed to reach the starch indicator end point. Report the concentration ascorbic acid in mg/100 mL. Solution For a back titration we need to determine the stoichiometry between $\text{I}_3^-$ and the analyte, C 6 H 8 O 6 , and between $\text{I}_3^-$ and the titrant, Na 2 S 2 O 3 . The later is easy because we know from Example 9.4.3 that each mole of $\text{I}_3^-$ reacts with two moles of Na 2 S 2 O 3 . In oxidizing ascorbic acid to dehydroascorbic acid, the oxidation state of carbon changes from +2⁄3 in C 6 H 8 O 6 to +1 in C 6 H 6 O 6 . Each carbon releases 1⁄3 of an electron, or a total of two electrons per ascorbic acid. As we learned in Example 9.4.3 , reducing $\text{I}_3^-$ requires two electrons; thus, a conservation of electrons requires that each mole of ascorbic acid consumes one mole of $\text{I}_3^-$. The total moles of $\text{I}_3^-$ that react with C 6 H 8 O 6 and with Na 2 S 2 O 3 is \[(0.01023 \text{ M})(0.05000 \text{ L}) = 5.115 \times 10^{-4} \text{ mol I}_3^- \nonumber\] The back titration consumes \[0.01382 \text{ L Na}_2\text{S}_2\text{O}_3 \times \frac{0.07203 \text{ mol Na}_2\text{S}_2\text{O}_3}{\text{ L Na}_2\text{S}_2\text{O}_3} \times \frac{1 \text{ mol I}_3^-}{2 \text{ mol Na}_2\text{S}_2\text{O}_3} = 4.977 \times 10^{-4} \text{ mol I}_3^- \nonumber\] Subtracting the moles of $\text{I}_3^-$ that react with Na 2 S 2 O 3 from the total moles of $\text{I}_3^-$ gives the moles reacting with ascorbic acid. \[5.115 \times 10^{-4} \text{ mol I}_3^- - 4.977 \times 10^{-4} \text{ mol I}_3^- = 1.38 \times 10^{-5} \text{ mol I}_3^- \nonumber\] The grams of ascorbic acid in the 5.00-mL sample of orange juice is \[1.38 \times 10^{-5} \text{ mol I}_3^- \times \frac{1 \text{ mol C}_6\text{H}_8\text{O}_6}{\text{mol I}_3^-} \times \frac{176.12 \text{ g C}_6\text{H}_8\text{O}_6}{\text{mol C}_6\text{H}_8\text{O}_6} = 2.43 \times 10^{-3} \text{ g C}_6\text{H}_8\text{O}_6 \nonumber\] There are 2.43 mg of ascorbic acid in the 5.00-mL sample, or 48.6 mg per 100 mL of orange juice. The balanced reactions for this analysis are: \[\text{C}_6\text{H}_8\text{O}_6(aq) + \text{I}_3^- (aq) \rightarrow 3\text{I}^-(aq) + \text{C}_6\text{H}_6\text{O}_6(aq) + 2\text{H}^+(aq) \nonumber\] \[\text{I}_3^-(aq) + 2\text{S}_2\text{O}_3^{2-}(aq) \rightarrow \text{S}_4\text{O}_6^{2-}(aq) + 3\text{I}^-(aq) \nonumber\] A quantitative analysis for ethanol, C 2 H 6 O, is accomplished by a redox back titration. Ethanol is oxidized to acetic acid, C 2 H 4 O 2 , using excess dichromate, $\text{Cr}_2\text{O}_7^{2-}$, which is reduced to Cr 3 + . The excess dichromate is titrated with Fe 2 + , giving Cr 3 + and Fe 3 + as products. In a typical analysis, a 5.00-mL sample of a brandy is diluted to 500 mL in a volumetric flask. A 10.00-mL sample is taken and the ethanol is removed by distillation and collected in 50.00 mL of an acidified solution of 0.0200 M K 2 Cr 2 O7. A back titration of the unreacted $\text{Cr}_2\text{O}_7^{2-}$ requires 21.48 mL of 0.1014 M Fe 2 + . Calculate the %w/v ethanol in the brandy. Answer For a back titration we need to determine the stoichiometry between $\text{Cr}_2\text{O}_7^{2-}$ and the analyte, C 2 H 6 O, and between $\text{Cr}_2\text{O}_7^{2-}$ and the titrant, Fe 2 + . In oxidizing ethanol to acetic acid, the oxidation state of carbon changes from –2 in C 2 H 6 O to 0 in C 2 H 4 O 2 . Each carbon releases two electrons, or a total of four electrons per C 2 H 6 O. In reducing $\text{Cr}_2\text{O}_7^{2-}$, in which each chromium has an oxidation state of +6, to Cr 3 + , each chromium loses three electrons, for a total of six electrons per $\text{Cr}_2\text{O}_7^{2-}$. Oxidation of Fe 2 + to Fe 3 + requires one electron. A conservation of electrons requires that each mole of K 2 Cr 2 O 7 (6 moles of e – ) reacts with six moles of Fe 2 + (6 moles of e – ), and that four moles of K 2 Cr 2 O 7 (24 moles of e – ) react with six moles of C 2 H 6 O (24 moles of e – ). The total moles of K 2 Cr 2 O 7 that react with C 2 H 6 O and with Fe 2 + is \[(0.0200 \text{ M K}_2\text{Cr}_2\text{O}_7)(0.05000 \text{ L})=1.00 \times 10^{-3} \text{ mol K}_2\text{Cr}_2\text{O}_7 \nonumber\] The back titration with Fe 2 + consumes \[(0.1014 \text{ M Fe}^{2+})(0.02148 \text{ L}) \times \frac{1 \text{ mol K}_2\text{Cr}_2\text{O}_7}{6 \text{ mol Fe}^{2+}} = 3.63 \times 10^{-4} \text{ mol K}_2\text{Cr}_2\text{O}_7 \nonumber\] Subtracting the moles of K 2 Cr 2 O 7 that react with Fe 2 + from the total moles of K 2 Cr 2 O 7 gives the moles that react with the analyte. \[(1.00 \times 10^{-3} \text{ mol K}_2\text{Cr}_2\text{O}_7) - (3.63 \times 10^{-4} \text{ mol K}_2\text{Cr}_2\text{O}_7) = 6.37 \times 10^{-4} \text{ mol K}_2\text{Cr}_2\text{O}_7 \nonumber\] The grams of ethanol in the 10.00-mL sample of diluted brandy is \[6.37 \times 10^{-4} \text{ mol K}_2\text{Cr}_2\text{O}_7 \times \frac{6 \text{ mol C}_2\text{H}_6\text{O}}{4 \text{ mol K}_2\text{Cr}_2\text{O}_7} \times \frac{46.07 \text{ g C}_2\text{H}_6\text{O}}{\text{mol C}_2\text{H}_6\text{O}} = 0.0440 \text{ g C}_2\text{H}_6\text{O} \nonumber\] The %w/v C 2 H 6 O in the brandy is \[\frac{0.0440 \text{ g C}_2\text{H}_6\text{O}}{10.0 \text{ mL diluted brandy}} \times \frac{500.0 \text{ mL diluted brandy}}{5.00 \text{ mL brandy}} \times 100 = 44.0 \text{% w/v C}_2\text{H}_6\text{O} \nonumber\] Evaluation of Redox Titrimetry The scale of operations, accuracy, precision, sensitivity, time, and cost of a redox titration are similar to those described earlier in this chapter for an acid–base or a complexation titration. As with an acid–base titration, we can extend a redox titration to the analysis of a mixture of analytes if there is a significant difference in their oxidation or reduction potentials. Figure 9.4.7 shows an example of the titration curve for a mixture of Fe 2 + and Sn 2 + using Ce 4 + as the titrant. A titration of a mixture of analytes is possible if their standard state potentials or formal potentials differ by at least 200 mV.
Ancillary_Materials/Worksheets/Spectroscopy_(Worksheets)	Front Matter 1: Intro Structure Determination 2: Introduction to Absorption Spectroscopy 3: UV Spectroscopy 4: Fluorescence Spectroscopy 5: Infrared Spectroscopy 6: Mass Spectrometry 7: Carbon NMR 8: Distortionless Enhancement by Polarization Transfer (DEPT) Spectroscopy 9: Sites of Unsaturation 10: ¹H NMR 11: Combined Problems 1 12: Complex Coupling 13: 1D Spin Decoupled Experiment 14: COSY 15: TOCSY 16: Multinuclear 17: Multinuclear Spin Over Half 18: Heteronuclear Multiple Quantum Coherence (HMQC) 19: HMBC 20: Spectral Problems with 2D Spectra 21: INADEQUATE NMR Experiment 22: Nuclear Overhauser Effect (NOE) 23: Absolute Configuration 24: 3D NMR 25: Dynamic NMR Back Matter
Courses/Madera_Community_College/MacArthur_Chemistry_3A_v_1.2/06%3A_Introduction_to_Stoichiometry/6.02%3A_Mole_Ratios	Suppose that you want to add some sections to your porch. Before you go to the hardware store to buy lumber, you need to determine the unit composition (the material between two large uprights). You count how many posts, how many boards, how many rails—then you decide how many sections you want to add—before you calculate the amount of building material needed for your porch expansion. Mole Ratios Stoichiometry problems can be characterized by two things: (1) the information given in the problem, and (2) the information that is to be solved for, referred to as the unknown . The given and the unknown may both be reactants, both be products, or one may be a reactant while the other is a product. The amounts of the substances can be expressed in moles. However, in a laboratory situation, it is common to determine the amount of a substance by finding its mass in grams. The amount of a gaseous substance may be expressed by its volume. In this concept, we will focus on the type of problem where both the given and the unknown quantities are expressed in moles. Chemical equations express the amounts of reactants and products in a reaction. The coefficients of a balanced equation can represent either the number of molecules or the number of moles of each substance. The production of ammonia $\left( \ce{NH_3} \right)$ from nitrogen and hydrogen gases is an important industrial reaction called the Haber process, after German chemist Fritz Haber. \[\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightarrow 2 \ce{NH_3} \left( g \right)\] The balanced equation can be analyzed in several ways, as shown in the figure below. We see that 1 molecule of nitrogen reacts with 3 molecules of hydrogen to form 2 molecules of ammonia. This is the smallest possible relative amount of the reactants and products. To consider larger relative amounts, each coefficient can be multiplied by the same number. For example, 10 molecules of nitrogen would react with 30 molecules of hydrogen to produce 20 molecules of ammonia. The most useful quantity for counting particles is the mole. So if each coefficient is multiplied by a mole, the balanced chemical equation tells us that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. This is the conventional way to interpret any balanced chemical equation. Finally, if each mole quantity is converted to grams by using the molar mass, we can see that the law of conservation of mass is followed. $1 \: \ce{mol}$ of nitrogen has a mass of $28.02 \: \text{g}$, while $3 \: \text{mol}$ of hydrogen has a mass of $6.06 \: \text{g}$, and $2 \: \text{mol}$ of ammonia has a mass of $34.08 \: \text{g}$. \[28.02 \: \text{g} \: \ce{N_2} + 6.06 \: \text{g} \: \ce{H_2} \rightarrow 34.08 \: \text{g} \: \ce{NH_3}\] Mass and the number of atoms must be conserved in any chemical reaction. The number of molecules is not necessarily conserved. A mole ratio is a conversion factor that relates the amounts in moles of any two substances in a chemical reaction. The numbers in a conversion factor come from the coefficients of the balanced chemical equation. The following six mole ratios can be written for the ammonia forming reaction above. \[\begin{array}{ccc} \dfrac{1 \: \text{mol} \: \ce{N_2}}{3 \: \text{mol} \: \ce{H_2}} & or & \dfrac{3 \: \text{mol} \: \ce{H_2}}{1 \: \text{mol} \: \ce{N_2}} \\ \dfrac{1 \: \text{mol} \: \ce{N_2}}{2 \: \text{mol} \: \ce{NH_3}} & or & \dfrac{2 \: \text{mol} \: \ce{NH_3}}{1 \: \text{mol} \: \ce{N_2}} \\ \dfrac{3 \: \text{mol} \: \ce{H_2}}{2 \: \text{mol} \: \ce{NH_3}} & or & \dfrac{2 \: \text{mol} \: \ce{NH_3}}{3 \: \text{mol} \: \ce{H_2}} \end{array}\] In a mole ratio problem, the given substance, expressed in moles, is written first. The appropriate conversion factor is chosen in order to convert from moles of the given substance to moles of the unknown. Example $\PageIndex{1}$ How many moles of ammonia are produced if 4.20 moles of hydrogen are reacted with an excess of nitrogen? Solution Step 1: List the known quantities and plan the problem. Known 4.20 moles of hydrogen relevant mole ratios from balanced equation. Unknown mol NH 3 Plan The conversion is from $\text{mol} \: \ce{H_2}$ to $\text{mol} \: \ce{NH_3}$. The problem states that there is an excess of nitrogen, so we do not need to be concerned with any mole ratio involving $\ce{N_2}$. Choose the conversion factor that has the $\ce{NH_3}$ in the numerator and the $\ce{H_2}$ in the denominator. Step 2: Solve. \[4.20 \: \text{mol} \: \ce{H_2} \times \dfrac{2 \: \text{mol} \: \ce{NH_3}}{3 \: \text{mol} \: \ce{H_2}} = 2.80 \: \text{mol} \: \ce{NH_3}\] The reaction of $4.20 \: \text{mol}$ of hydrogen with excess nitrogen produces $2.80 \: \text{mol}$ of ammonia. Step 3: Think about your result. The result corresponds to the 3:2 ratio of hydrogen to ammonia from the balanced equation. Summary Mole ratios allow comparison of the amounts of any two materials in a balanced equation. Calculations can be made to predict how much product can be obtained from a given number of moles of reactant.
Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/10%3A_The_First_Law_of_Thermodynamics/10.14%3A_Heats_of_Reactions_Can_Be_Calculated_from_Tabulated_Heats_of_Formation	Reaction enthalpies are important, but difficult to tabulate. However, because enthalpy is a state function, it is possible to use Hess’ Law to simplify the tabulation of reaction enthalpies. Hess’ Law is based on the addition of reactions. By knowing the reaction enthalpy for constituent reactions, the enthalpy of a reaction that can be expressed as the sum of the constituent reactions can be calculated. The key lies in the canceling of reactants and products that °Ccur in the “data” reactions but not in the “target reaction. Example $\PageIndex{1}$: Find $\Delta H_{rxn}$ for the reaction \[2 CO(g) + O_2(g) \rightarrow 2 CO_2(g) \nonumber \] Given \[C(gr) + ½ O_2(g) \rightarrow CO(g) \nonumber \] with $\Delta H_1 = -110.53 \,kJ$ \[C(gr) + O_2(g) \rightarrow CO_2(g) \nonumber \] with $\Delta H_2 = -393.51\, kJ$ Solution The target reaction can be generated from the data reactions. \[ {\color{red} 2 \times} \left[ CO(g) \rightarrow C(gr) + O_2(g) \right] \nonumber \] plus \[ { \color{red} 2 \times} \left[ C(gr) + 2 O_2(g) \rightarrow 2 CO_2(g) \right] \nonumber \] equals \[2 CO(g) + O_2(g) \rightarrow 2 CO_2(g) \nonumber \] so \[{ \color{red} 2 \times} \Delta H_1 = -787.02 \, kJ \nonumber \] \[{ \color{red} 2 \times} \Delta H_2 = 221.06\, kJ \nonumber \] \[ { \color{red} 2 \times} \Delta H_1 + { \color{red} 2 \times} \Delta H_2 = -565.96 \,kJ \nonumber \] Standard Enthalpy of Formation One of the difficulties with many thermodynamic state variables (such as enthalpy) is that while it is possible to measure changes, it is impossible to measure an absolute value of the variable itself. In these cases, it is necessary to define a zero to the scale defining the variable. For enthalpy, the definition of a zero is that the standard enthalpy of formation of a pure element in its standard state is zero. All other enthalpy changes are defined relative to this standard. Thus it is essential to very carefully define a standard state. Definition: the Standard State The standard state of a substance is the most stable form of that substance at 1 atmosphere pressure and the specified temperature. Using this definition, a convenient reaction for which enthalpies can be measured and tabulated is the standard formation reaction . This is a reaction which forms one mole of the substance of interest in its standard state from elements in their standard states. The enthalpy of a standard formation reaction is the standard enthalpy of formation ($\Delta H_{f^o}$). Some examples are $NaCl(s)$: \[Na(s) + ½ Cl_2(g) \rightarrow NaCl(s) \nonumber \] with $\Delta H_f^o = -411.2\, kJ/mol$ $C_3H_8(g)$: \[3 C(gr) + 4 H_2(g) \rightarrow C_3H_8(g) \nonumber \] with $\Delta H_f^o = -103.8\, kJ/mol$ It is important to note that the standard state of a substance is temperature dependent . For example, the standard state of water at -10 °C is solid, whereas the standard state at room temperature is liquid. Once these values are tabulated, calculating reaction enthalpies becomes a snap. Consider the heat combustion ($\Delta H_c$) of methane (at 25 °C) as an example. \[CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \nonumber \] The reaction can expressed as a sum of a combination of the following standard formation reactions. \[C(gr) + 2 H_2(g) \rightarrow CH_4(g) \nonumber \] with $\Delta H_f^o = -74.6\, kJ/mol$ \[C(gr) + O_2(g) \rightarrow CO_2(g) \nonumber \] with $\Delta H_f^o = -393.5\, kJ/mol$ \[H_2(g) + ½ O_2(g) \rightarrow H_2O(l) \nonumber \] with $\Delta H_f^o = -285.8 \,kJ/mol$ The target reaction can be generated from the following combination of reactions \[{ \color{red} -1 \times} \left[ C(gr) + 2 H_2(g) \rightarrow CH_4(g)\right] \nonumber \] \[CH_4(g) \rightarrow C(gr) + 2 H_2(g) \nonumber \] with $\Delta H_f^o ={ \color{red} -1 \times} \left[ -74.6\, kJ/mol \right]= 74.6\, kJ/mol$ \[C(gr) + O_2(g) \rightarrow CO_2(g) \nonumber \] with $\Delta H_f^o = -393.5\, kJ/mol$ \[{ \color{red} 2 \times} \left[ H_2(g) + ½ O_2(g) \rightarrow H_2O(l) \right] \nonumber \] \[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \nonumber \] with $\Delta H_f^o = {\color{red} 2 \times} \left[ -285.8 \,kJ/mol \right] = -571.6\, kJ/mol$. \[CH_4(g) + 2 O_2(g) \rightarrow CO2_(g) + 2 H_2O(l) \nonumber \] with $\Delta H_c^o = -890.5\, kJ/mol$ Alternately, the reaction enthalpy could be calculated from the following relationship \[\Delta H_{rxn} = \sum_{products} \nu \cdot \Delta H_f^o - \sum_{reactants} \nu \cdot \Delta H_f^o \nonumber \] where $\nu$ is the stoichiometric coefficient of a species in the balanced chemical reaction. For the combustion of methane, this calculation is \[ \begin{align} \Delta _{rxn} & = (1\,mol) \left(\Delta H_f^o(CO_2)\right) + (2\,mol) \left(\Delta H_f^o(H_2O)\right) - (1\,mol) \left(\Delta H_f^o(CH_4)\right) \\ & = (1\,mol) (-393.5 \, kJ/mol) + (2\,mol) \left(-285.8 \, kJ/mol \right) - (1\,mol) \left(-74.6 \, kJ/mol \right) \\ & = -890.5 \, kJ/mol \end{align} \nonumber \] A note about units is in order. Note that reaction enthalpies have units of kJ, whereas enthalpies of formation have units of kJ/mol. The reason for the difference is that enthalpies of formation (or for that matter enthalpies of combustion, sublimation, vaporization, fusion, etc.) refer to specific substances and/or specific processes involving those substances. As such, the total enthalpy change is scaled by the amount of substance used. General reactions, on the other hand, have to be interpreted in a very specific way. When examining a reaction like the combustion of methane \[CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \nonumber \] with $\Delta H_{rxn} = -890.5\, kJ$. The correct interpretation is that the reaction of one mole of CH 4 (g) with two moles of O 2 (g) to form one mole of CO 2 (g) and two moles of H 2 O(l) releases 890.5 kJ at 25 °C. Ionization Reactions Ionized species appear throughout chemistry. The energy changes involved in the formation of ions can be measured and tabulated for several substances. In the case of the formation of positive ions, the enthalpy change to remove a single electron at 0 K is defined as the ionization potential . \[ M(g) \rightarrow M^+(g) + e^- \nonumber \] with $\Delta H (0 K) \equiv 1^{st} \text{ ionization potential (IP)}$ The removal of subsequent electrons requires energies called the 2 nd Ionization potential, 3 rd ionization potential, and so on. \[M^+(g) \rightarrow M^{2+}(g) + e^- \nonumber \] with $\Delta H(0 K) ≡ 2^{nd} IP$ \[M^{2+}(g) \rightarrow M^{3+}(g) + e^- \nonumber \] with $\Delta H(0 K) ≡ 3^{rd} IP$ An atom can have as many ionization potentials as it has electrons, although since very highly charged ions are rare, only the first few are important for most atoms. Similarly, the electron affinity can be defined for the formation of negative ions. In this case, the first electron affinity is defined by \[X(g) + e^- \rightarrow X^-(g) \nonumber \] with $-\Delta H(0 K) \equiv 1^{st} \text{ electron affinity (EA)}$ The minus sign is included in the definition in order to make electron affinities mostly positive. Some atoms (such as noble gases) will have negative electron affinities since the formation of a negative ion is very unfavorable for these species. Just as in the case of ionization potentials, an atom can have several electron affinities. \[X^-(g) + e^- \rightarrow X^{2-}(g) \nonumber \] with $-\Delta H(0 K) ≡ 2^{nd} EA$. \[X^{2-}(g) + e^- \rightarrow X^{3-}(g) \nonumber \] with $-\Delta H(0 K) ≡ 3^{rd} EA$. Average Bond Enthalpies In the absence of standard formation enthalpies, reaction enthalpies can be estimated using average bond enthalpies. This method is not perfect, but it can be used to get ball-park estimates when more detailed data is not available. A bond dissociation energy $D$ is defined by \[XY(g) \rightarrow X(g) + Y(g) \nonumber \] with $\Delta H \equiv D(X-Y)$ In this process, one adds energy to the reaction to break bonds, and extracts energy for the bonds that are formed. \[\Delta H_{rxn} = \sum (\text{bonds broken}) - \sum (\text{bonds formed}) \nonumber \] As an example, consider the combustion of ethanol: In this reaction, five C-H bonds, one C-C bond, and one C-O bond, and one O=O bond must be broken. Also, four C=O bonds, and one O-H bond are formed. Bond Average Bond Energy (kJ/mol) C-H 413 C-C 348 C-O 358 O=O 495 C=O 799 O-H 463 The reaction enthalpy is then given by \[ \begin{align} \Delta H_c = \, &5(413 \,kJ/mol) + 1(348\, kJ/mol) + 1(358 \,kJ/mol) \nonumber \\ & + 1(495\, kJ/mol) - 4(799 \,kJ/mol) – 2(463\, kJ/mol) \nonumber \\ =\,& -856\, kJ/mol \end{align} \nonumber \] Because the bond energies are defined for gas-phase reactants and products, this method does not account for the enthalpy change of condensation to form liquids or solids, and so the result may be off systematically due to these differences. Also, since the bond enthalpies are averaged over a large number of molecules containing the particular type of bond, the results may deviate due to the variance in the actual bond enthalpy in the specific molecule under consideration. Typically, reaction enthalpies derived by this method are only reliable to within ± 5-10%.
Courses/University_of_California_Davis/UCD_Chem_110A%3A_Physical_Chemistry__I/UCD_Chem_110A%3A_Physical_Chemistry_I_(Koski)/Text/08%3A_Multielectron_Atoms/8.02%3A_Perturbation_Theory_and_the_Variational_Method_for_Helium	Demonstrate that both perturbation theory and variational methods can be used to solve the electron structure of the helium atom. Both perturbation theory and variation method (especially the linear variational method) provide good results in approximating the energy and wavefunctions of multi-electron atoms. Below we address both approximations with respect to the helium atom. Perturbation Theory of the Helium Atom We use perturbation theory to approach the analytically unsolvable helium atom Schrödinger equation by focusing on the Coulomb repulsion term that makes it different from the simplified Schrödinger equation that we have just solved analytically. The electron-electron repulsion term is conceptualized as a correction, or perturbation, to the Hamiltonian that can be solved exactly, which is called a zero-order Hamiltonian . The perturbation term corrects the previous Hamiltonian to make it fit the new problem. In this way the Hamiltonian is built as a sum of terms, and each term is given a name. For example, we call the simplified or starting Hamiltonian, $\hat {H} ^0$, the zero order term, and the correction term $\hat {H} ^1$. \[ \hat {H} = \hat {H} ^0 + \hat {H} ^1 \label {9-17} \] The Hamilonian for the helium atom (in atomic units) is: \[\begin{align} \hat {H} ^0 &= \underbrace{-\dfrac {1}{2} \nabla ^2_1 - \dfrac {2}{r_1}}_{\text{H atom Hamiltonian}} - \underbrace{\dfrac {1}{2} \nabla ^2_2 - \dfrac {2}{r_2}}_{\text{H atom Hamiltonian}} \label {9-18} \\[4pt] \hat {H} ^1 &= \dfrac {1}{r_{12}} = \dfrac{1}{\|r_1-r_2\|} \label {9-19} \end{align} \] The expression for the first-order correction to the energy is \[ \begin{align} E^1 &= \langle \psi ^{0} \| \hat {H} ^1 \| \psi ^{0} \rangle \nonumber \\[4pt] &= \int \psi ^{0} \hat {H} ^1 \psi ^0 \,d\tau \nonumber \end{align} \label {9-28} \] Equation $\ref{9-28}$ is a general expression for the first-order perturbation energy, which provides an improvement or correction to the zero-order energy we already obtained. Hence, $E^1$ is the average interaction energy of the two electrons calculated using wavefunctions that assume there is no interaction. The solution to $\hat{H}^{0}$ (Equation \ref{9-18}) is the product of two single-electron hydrogen wavefunctions (scaled by the increased nuclear charge) since $\hat{H}^{0}$ can be separated into independent functions of each electron (i.e., Separation of Variables ). \[ \| \psi ^{0} \rangle = \| \varphi _{1s} (r_1) \varphi _{1s} (r_2) \rangle \nonumber \] So the integral in Equation $\ref{9-28}$ is \[ E^1 = \iint \varphi _{1s} (r_1) \varphi _{1s} (r_2) \dfrac {1}{r_{12}} \varphi _{1s} (r_1) \varphi _{1s} (r_2)\, d\tau _1 d\tau _2 \label {9-29} \] where the double integration symbol represents integration over all the spherical polar coordinates of both electrons $r_1, \theta _1, \varphi _1 , r_2 , \theta _2 , \varphi _2$. The evaluation of these six integrals is lengthy. When the integrals are done, the result is $E^1$ = +34.0 eV so that the total energy calculated using our second approximation method, first-order perturbation theory, is \[ E_{approx2} = E^0 + E^1 = - 74.8 eV \label {9-30} \] The new approximate value for the binding energy represents a substantial (~30%) improvement over the zero-order energy: \[E^{0} = \dfrac{2}{n^2} + \dfrac{2}{n^2} = 4\, \underbrace{E_h}_{hartrees} = 108.8\, eV \nonumber \] so the interaction of the two electrons is an important part of the total energy of the helium atom. We can continue with perturbation theory and find the additional corrections, $E^2$, $E^3$, etc. For example, \[E^0 + E^1 + E^2 = -79.2\, eV. \nonumber \] So with two corrections to the energy, the calculated result is within 0.3% of the experimental value of -79.01 eV. It takes thirteenth-order perturbation theory (adding $E^1$ through $E^{13}$ to $E^0$) to compute an energy for helium that agrees with experiment to within the experimental uncertainty. Interestingly, while we have improved the calculated energy so that it is much closer to the experimental value, we learn nothing new about the helium atom wavefunction by applying the first-order perturbation theory to the energy above. He need to expand the wavefunctions to first order perturbation theory, which requires more effort. Below, we will employ the variational method approximation to modify zero-order wavefunctions to address one of the ways that electrons are expected to interact with each other. The hartree is the atomic unit of energy (named after the British physicist Douglas Hartree) and is defined as \[E_h=2R_Hhc \nonumber \] where $R_H$ is the Rydberg constant, $h$ is the Planck constant and $c$ is the speed of light. \[\begin{align} E_h &= 4.359 \times 10^{−18} \,J \nonumber \\[4pt] &= 27.21\, eV. \nonumber \end{align} \nonumber \] The hartree is usually used as a unit of energy in atomic physics and computational chemistry. As discussed before for hydrogen emission, IR, and microwave spectroscopies, experimental measurements prefer the electronvolt ($eV$) or the wavenumber ($cm^{−1}$). Variational Method Applied to the Helium Method As discussed in Section 6.7, because of the electron-electron interactions, the Schrödinger's Equation cannot be solved exactly for the helium atom or more complicated atomic or ionic species. However, the ground-state energy of the helium atom can be estimated using approximate methods. One of these is the variational method which requires the minimizing of the following variational integral. \[\begin{align} E_{trial} &= \dfrac{\langle \psi_{trial}\| \hat{H} \| \psi_{trial} \rangle }{\langle \psi_{trial}\| \psi_{trial} \rangle}\label{7.3.1b} \\[4pt] &= \dfrac{\displaystyle \int_0^{\infty} \psi_{trial}^ \hat{H} \psi_{trial} d\tau}{\displaystyle \int_0^{\infty} \psi_{trial}^2 \, d\tau} \label{7.3.1a} \end{align} \] The five trial wavefunctions discussions below are equally "valid" trial wavefunctions that describe the probability of finding each electron (technically the wavefunction squared). What separates the "poor" approximations from the "good" approximation is whether the trial wavefunction predicts experimental results. Consequently, for all the approximations used for the rest of this TextMap, it is important to compare the theoretical results to the "true" (i.e., experimental) results. No matter how complicated an approximation is, it is only as good as the accuracy of its predicted values to experimental values. Trial Wavefunction #1: Simple Orbital Approximation with One Parameter As is clear from Equation $\ref{7.3.1b}$, the variational method approximation requires that a trial wavefunction with one or more adjustable parameters be chosen. A logical first choice for such a multi-electron wavefunction would be to assume that the electrons in the helium atom occupy two identical, but scaled, hydrogen 1s orbitals. \[ \begin{align} \| \psi (1,2) \rangle_{trial} &= \phi (1) \phi (2) \\[4pt] &= \exp\left[- \alpha (r_1 +r_2)\right] \label{7.3.2} \end{align} \] The variational energy obtained after minimizing Equation $\ref{7.3.1a}$ after substituting the trial wavefunction (Equation \ref{7.3.2}) by varying $\alpha$ is \[E_{trial} = -2.84766 \;E_h \nonumber \] and the experimentally determined ground-state energy for the helium atom is the sum of first and second ionization energies \[E_{\exp}= I_1 +I_2 = -2.90372 \;E_h \label{exp} \] The deviation of energy for the optimized trial wavefunction from the experimental value is \[ \begin{align} \left\| \dfrac{E_{trial}(\alpha)-E_{\exp}}{E_{\exp}} \right\| &= \left\| \dfrac{-2.84766 \;E_h + 2.90372 \;E_h}{-2.90372 \;E_h} \right\| \\[4pt] &= 1.93 \% \label{trial1} \end{align} \] The value of -2.8477 hartrees is within 2% of the known ground-state energy of the helium atom. The error in the calculation is attributed to the fact that the wavefunction is based on the orbital approximation and, therefore, does not adequately take electron-electron interactions into account. In other words, this wavefunction gives the electrons too much independence, given that they have like charges and tend to avoid one another. Trial Wavefunction #2: Orbital Approximation with Two Parameters Some electron-electron interactions can be built into the multi-electron wavefunction by assuming that each electron is in an orbital which is a linear combination of two different and scaled hydrogen 1s orbitals. \[\phi(r_1) = \exp(- \alpha r_1) + \exp(- \beta r_1) \label{7.3.3} \] Under the orbital approximation this assumption gives a trial wavefunction of the form \[ \begin{align} \| \psi (1,2) \rangle_{trial} &= \phi (1) \phi (2) \label{7.3.4a} \\[4pt] &= {\exp(- \alpha r_1 )\exp(- \alpha r_2)}+\exp(- \alpha r_1 )\exp(- \beta r_2)+\exp(- \beta r_1 )\exp(- \alpha r_2 )+\exp(- \beta r_1 )\exp(- \beta r_2 ) \label{7.3.4b} \end{align} \] Inspection of this trial wavefunction indicates that 50% of the time the electrons are in different orbitals, while for the first trial wavefunction the electrons were in the same orbital 100% of the time. Notice the enormous increase in the complexity of the variational expression for the energy for this trial wavefunction (Equation $\ref{7.3.1a}$). However, the calculation is very similar to that using the previous trial wavefunction. The differences are that in this case the expression for the energy is more complex and that it is being minimized simultaneously with respect to two parameters ($\alpha$ and $\beta$) rather than just one ($\alpha$). The variational energy obtained after minimizing Equation $\ref{7.3.1a}$ after substituting the trial wavefunction (Equation \ref{7.3.4b}) by varying $\alpha$ and $\beta$ is \[E_{trial} =-2.86035 \;E_h \nonumber \] The deviation of energy for the optimized trial wavefunction from the experimental value (Equation \ref{exp}) is \[ \begin{align} \left\| \dfrac{E_{trial}(\alpha, \beta)-E_{\exp}}{E_{\exp}} \right\| &= \left\| \dfrac{-2.86035 \;E_h + 2.90372 \;E_h}{-2.90372 \;E_h} \right\| \\[4pt] &= 1.49 \% \label{trial2} \end{align} \] Clearly introducing some electron-electron interactions into the trial wavefunction has improved the agreement between theory and experiment (Equation \ref{trial1} vs. \ref{trial2}). Trial Wavefunction #3: Orbital Approximation with Two Parameters The extent of electron-electron interactions can be increased further by eliminating the first and last term in the second trial wavefunction (Equation $\ref{7.3.4b}$). This yields a multi-electron wavefunction of the form, \[ \| \psi (1,2) \rangle_{trial} = \exp(- \alpha r_1 )\exp(- \beta r_2 ) + \exp(- \beta r_1 )\exp(- \alpha r_2 ) \label{7.3.5} \] This trial wavefunction places the electrons in different scaled hydrogen 1s orbitals 100% of the time this adds further improvement in the agreement with the literature value of the ground-state energy is obtained. The variational energy obtained after minimizing Equation $\ref{7.3.1a}$ after substituting the trial wavefunction (Equation \ref{7.3.5}) by varying $\alpha$ and $\beta$ is \[ E_{trial} = -2.87566 \;E_h \nonumber \] The deviation of energy for the optimized trial wavefunction from the experimental value (Equation \ref{exp}) is \[ \begin{align} \left\| \dfrac{ E_{trial} (\alpha,\beta)-E_{\exp}}{E_{\exp}} \right\| &= \left\| \dfrac{-2.87566 \;E_h + 2.90372 \;E_h}{-2.90372 \;E_h} \right\| \\[4pt] &= 0.97 \%\label{trial3} \end{align} \] This result is within 1% of the actual ground-state energy of the helium atom. Trial Wavefunction #4: Approximation with Two Parameters The third trial wavefunction, however, still rests on the orbital approximation and, therefore, does not treat electron-electron interactions adequately. Hylleraas took the calculation a step further by introducing electron-electron interactions directly into the first trial wavefunction by adding a term, $r_{12}$, involving the inter-electron separation. \[\| \psi_{trial} (1,2) \rangle = \left(\exp[- \alpha ( r_1 + r_2 )]\right) \left(1 + \beta r_{12} \right) \label{7.3.6} \] In the trial multi-electron wavefunction of Equation \ref{7.3.6}, if the electrons are far apart, then $r_{12}$ is large and the magnitude of the wavefunction increases to favor that configuration. The variational energy obtained after minimizing Equation $\ref{7.3.1a}$ after substituting the trial wavefunction (Equation \ref{7.3.6}) by varying $\alpha$ and $\beta$ is \[ E_{trial} = - 2.89112\; E_h \nonumber \] The deviation of energy for the optimized trial wavefunction from the experimental value (Equation \ref{exp}) is \[ \begin{align} \left\| \dfrac{ E_{trial} (\alpha,\beta)-E_{\exp}}{E_{\exp}} \right\| &= \left\| \dfrac{- 2.89112 \;E_h + 2.90372 \;E_h}{-2.90372 \;E_h} \right\| \\[4pt] &= 0.43 \% \label{trial4} \end{align} \] This modification of the trial wavefunction has further improved the agreement between theory and experiment to within 0.5%. Fifth Trial Wavefunction #5: Approximation with Three Parameters Chandrasakar brought about further improvement by adding Hylleraas's $r_{12}$ term to the third trial wavefunction (Equation $\ref{7.3.5}$) as shown here. \[\| \psi (1,2) \rangle_{trial} = \left[\exp(- \alpha r_1 )\exp(- \beta r_2 ) + \exp(- \beta r_1 )\exp(- \alpha r_2 ) \right][1 + \gamma r _{12} ] \label{7.3.7} \] Chandrasakar's three parameter wavefunction gives rise to a fairly complicated variational expression for ground-state energy. The variational energy obtained after minimizing Equation $\ref{7.3.1a}$ after substituting the trial wavefunction (Equation \ref{7.3.7}) by varying $\alpha$, $\beta$ and $\gamma$ is \[ E_{trial} = -2.90143 \;E_h \nonumber \] The deviation of energy for the optimized trial wavefunction from the experimental value (Equation \ref{exp}) is \[ \begin{align} \left\| \dfrac{ E_{trial} (\alpha, \beta, \gamma)-E_{\exp}}{E_{\exp}} \right\| &= \left\| \dfrac{-2.90143 + 2.90372 \;E_h}{-2.90372 \;E_h} \right\| \\[4pt] &= 0.0789 \% \label{trial5} \end{align} \] Chandrasakar's wavefunction gives a result for helium that is within 0.07% of the experimental value for the ground-state energy. Summary The purpose of this section is to examine five trial wavefunctions for the helium atom used within the Perturbation Theory and Variational method approximation. For the Variational method approximation, the calculations begin with an uncorrelated wavefunction in which both electrons are placed in a hydrogenic orbital with scale factor $\alpha$. The next four trial functions use several methods to increase the amount of electron-electron interactions in the wavefunction. As the summary of results that is appended shows this gives increasingly more favorable agreement with the experimentally determined value for the ground-state energy of the species under study. The detailed calculations show that the reason for this improved agreement with experiment is due to a reduction in electron-electron repulsion. Five variational method calculations that have been outlined above for the helium atom ($Z=2$) can be repeated for two-electron atoms (e.g., $\ce{H^-}$, $\ce{Li^+}$, $\ce{Be^{2+}}$, etc). The hydride anion is a particularly interesting case because the first two trial wavefunctions do not predict a stable ion (i.e., they are poor approximations). This indicates that electron-electron interactions is an especially important issue for atoms and ions with small nuclear charge.
Bookshelves/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/10%3A_Coordination_Chemistry_II_-_Bonding/10.03%3A_Ligand_Field_Theory/10.3.04%3A_Tetrahedral_Complexes	Tetrahedral geometry is even more common in chemistry than square planar geometry. Assessing the orbital interactions in tetrahedral geometry is somewhat more complicated, however, and it is common to proceed directly to a group theory approach. Nevertheless, let's take a look at this geometry and see what we can determine through simple observation before we see the results from a more rigorous approach. To begin, it helps to know that a tetrahedral geometry is defined as having four atoms arranged at alternating corners of a cube around a central atom. If we consider the orientation of the d orbitals, we find that they fall into two different groups. Although all five orbitals lie off-axis with respect to the ligands, some of them are pointed directly at the edges of the cube (d xy , d xz , d yz ) whereas the others point at the faces of the cube (d x 2 -y 2 , d z 2 ). The edge-touching orbitals lie a little closer to the ligands; we'll define this distance as r , which in this case is half the edge length of the cube. The face-touching orbitals are slightly farther away: based on the Pythagorean theorem, they are r 2 + r 2 = 2 r 2 = r 2 away from the ligands. Based on that simple observation, we might start to think about the dxy, dxz and dyz group as forming the antibonding orbitals upon interaction with the ligands. The dz2 and dx2-y2 would be left as non-bonding orbitals. This result would be exactly the opposite of the octahedral case. We would therefore expect an orbital splitting diagram that is exactly the inverse of the octahedral one. Maybe the splitting between orbital levels would be a little smaller, though, because of the lack of direct overlap between the ligands and the metal orbitals. After all, even the closest set of metal orbitals don't point directly at the ligands like in the octahedral case. This supposition is confirmed through a group theory approach. We can use vectors pointing along the ligand-metal axes to examine sigma bonding, as shown below. We would subject these vectors to the symmetry transformations in the tetrahedral space group, T d , shown in the table. 0 1 2 3 4 5 6 7 Td E 8C3 6C2 6S4 6σd NaN NaN A1 1 1 1 1 1 NaN x2 + y2 + z2 A2 1 1 1 -1 -1 NaN NaN E 2 -1 2 0 0 NaN (2z2 - x2 - y2, x2 - y2) T1 3 0 -1 1 -1 (Rx, Ry, Rz) NaN T2 3 0 -1 -1 1 (x, y, z) (xy, xz, yz) Γσ 4 1 0 0 2 NaN A1 + T2 Γπ 8 -1 0 0 0 NaN E + T1 + T2 That analysis leads to the reducible representation for sigma bonding, Γ σ , shown in the table. This representation reduces to the irreducible representations, A 1 + T 2 . The d orbitals represented here, the T 2 set, are indeed the d xy , d xz and d yz . The non-bonding d orbitals are the E group, corresponding to d z 2 and d x 2 -y 2 . We can go further with the group theory approach and to determine the reducible representation for pi bonding, Γ π , also shown in the table. Pi bonding is otherwise even more difficult to assess via simple inspection than was sigma bonding. The resulting representation reduces to E + T 1 + T 2 . The d orbitals represented here include the expected d z 2 and d x 2 -y 2 . Note, however, that they also include the d xy , d xz , and d yz orbitals. That means that in the presence of a pi-donor ligand, the latter set are antibonding with respect to both sigma and pi bonding. Example $\PageIndex{1}$ Demonstrate these symmetry operations on the drawing of the tetrahedron within a cube shown above. C 3 C 2 S 4 σ d Solution
Courses/Duke_University/CHEM_401L%3A_Analytical_Chemistry_Lab/CHEM_401L%3A_Analytical_Chemistry_Lab_Manual/04%3A_Determination_of_cocaine_on_bills_using_Gas_Chromatography_Mass_Spectrometry_(GC-MS)/4.04%3A_Mass_Spectrometers	Mass Spectrometry is an analytical technique that can reveal specific, characteristic, structurally related information about a compound. Sometimes the questions that attract attention to mass spectrometry are qualitative in nature. For example, "Have I successfully synthesized a C19 sterol," or "What is the structure of the by-product from my synthesis reaction," or "Is the compound that I isolated from plasma really norepinephrine". Mass spectrometry can usually answer these questions. Other times a quantitative analysis is sought and then the ideas of molar response values, peak areas and the use of internal standards, mentioned above, are included in the determination. A mass spectrometer has three essential needs: a means for producing ions, in this case (mostly) singly charged atoms; a means for separating these ions in space or in time by their mass-to-charge ratios; and a means for counting the number of ions for each mass-to-charge ratio. Figure $\PageIndex{1}$ provides a general view of a mass spectrometer, and the requirements are described briefly below. Note that the mass spectrometer is held under vacuum as this allows the ions to travel great distances without undergoing collisions that might alter their charge or energy. Sample Introduction : The compound for mass spectroscopic analysis is injected into a high vacuum where the molecules can move freely in the evacuated space. In the case of GCMS, the GC column is attached directly to the MS. Ion source : The most common means for generating ions are plasmas of various sorts, lasers, electrical sparks, and other ions. In GC-MS, positive ions are commonly produced by electron ionization (EI), electron impact on the gaseous molecules of the sample. Separating ions : The most common way of separating ions is a quadrupole magnet (described in more detail below). Counting ions: The transducer for mass spectrometry must be able to report the number of ions that emerge from the mass analyzer. In the Duke teaching labs, our mass spectrometers are quadrupole devices. They employ a combination of dc and radio frequency (rf) potentials as a mass "filter". The quadrupole typically consists of 4 cylindrical, parallel rods (10-25 cm in length) situated symmetrically in a square arrangement. Rods diagonally opposite each other are connected, in pairs, to dc and rf generators. Positive ions extracted from the ion source are accelerated into the quadrupole along the longitudinal axis of the four rods. The ions are influenced by the combined dc and rf fields. In order for an ion to reach the detector it must traverse the quadrupole without colliding with any one of the metal rods. For any level of rf/dc voltage only ions of a specific m/e avoid collision and reach the detector. The entire mass spectrum is obtained as voltages are swept from a pre-established minimum to a maximum. Quadrupole Mass Analyzers The quadrupole mass analyzer is probably the most important type of mass-selective devices; it is compact in size, low in cost, easy to use, and easy to maintain. As shown in Figure $\PageIndex{3}$, a quadrapole mass analyzer consists of four cylindrical rods, two of which are connected to the positive terminal of a variable direct current (dc) power supply and two of which are connected to the power supply's negative terminal; the two positive rods are positioned opposite of each other and the two negative rods are positioned opposite of each other. Each pair of rods is also connected to a variable alternating current (ac) source operated such that the alternating currents are 180° out-of-phase with each other. An ion beam from the source is drawn into the channel between the quadrupoles and, depending on the applied dc and ac voltages, ions with only one mass-to-charge ratio successfully travel the length of the mass analyzer and reach the transducer; all other ions collide with one of the four rods and are destroyed. To understand how a quadrupole mass analyzer achieves this separation of ions, it helps to consider the movement of an ion relative to just two of the four rods, as shown in Figure $\PageIndex{4}$ for the poles that carry a positive dc voltage. When the ion beam enters the channel between the rods, the ac voltage causes the ion to begin to oscillate. If, as in the top diagram, the ion is able to maintain a stable oscillation, it will pass through the mass analyzer and reach the transducer. If, as in the middle diagram, the ion is unable to maintain a stable oscillation, then the ion eventually collides with one of the rods and is destroyed. When the rods have a positive dc voltage, as they do here, ions with larger mass-to-charge ratios will be slow to respond to the alternating ac voltage and will pass through the transducer. The result is shown in the figure at the bottom (and repeated in Figure $\PageIndex{5}a$) where we see that ions with a sufficiently large mass-to-charge ratios successfully pass through the transducer; ions with smaller mass-to-charge ratios do not. In this case, the quadrupole mass analyzer acts as a high-pass filter. We can extend this to the behavior of the ions when they interact with rods that carry a negative dc voltage. In this case, the ions are attracted to the rods, but those ions that have a sufficiently small mass-to-charge ratio are able to respond to the alternating current's voltage and remain in the channel between the rods. The ions with larger mass-to-charge ratios move more sluggishly and eventually collide with one of the rods. As shown in Figure $\PageIndex{5}b$, in this case, the quadrupole mass analyzer acts as a low-pass filter. Together, as we see in Figure $\PageIndex{5}c$, a quadrupole mass analyzer operates as both a high-pass and a low-pass filter, allowing a narrow band of mass-to-charge ratios to pass through the transducer. By varying the applied dc voltage and the applied ac voltage, we can obtain a full mass spectrum. Quadrupole mass analyzers provide a modest mass-to-charge resolution of about 1 amu and extend to $m/z$ ratios of approximately 2000. Quadrupole mass analyzers are particularly useful for sources based on plasmas.
Courses/can/CHEM_232_-_Organic_Chemistry_II_(Puenzo)/01%3A_Conjugated_Compounds_and_Ultraviolet_Spectroscopy	Learning Objectives After you have completed Chapter 1, you should be able to fulfillall of the detailed objectives listed under each individual section. use the reactions discussed, along with those from previous chapters, when designing multi-step syntheses. use the reactions and concepts discussed to solve road-map problems. use ultraviolet-spectral data, in conjunction with other spectral data, to elucidate the structure of an unknown compound. define, and use in context, the key terms introduced. You have already studied the chemistry of compounds that contain one carbon-carbon double bond. In this chapter, you will focus your attention on compounds that contain two or more such bonds. In particular you will study the properties of those compounds that contain two carbon-carbon double bonds which are separated by one carbon-carbon single bond. These compounds are called “conjugated dienes.” To understand the properties exhibited by conjugated dienes, you must first examine their bonding in terms of the molecular orbital theory introduced in Section 1.11 . Then, you must learn how the products of a reaction are dependent on both thermodynamic and kinetic considerations. Which of these two factors is the most important can sometimes determine which of two possible products will predominate when a reaction is carried out under specific conditions. Although we shall not make extensive use of ultraviolet spectroscopy, this technique can often provide important information when conjugated compounds are being investigated. In general, ultraviolet spectroscopy is less useful than the other spectroscopic techniques introduced earlier. 1.1: Introduction 1.2: Stability of Conjugated Dienes- Molecular Orbital Theory 1.3: Electrophilic Additions to Conjugated Dienes- Allylic Carbocations 1.4: Kinetic vs. Thermodynamic Control of Reactions 1.5: The Diels-Alder Cycloaddition Reaction 1.6: Characteristics of the Diels-Alder Reaction 1.7: Structure Determination in Conjugated Systems - Ultraviolet Spectroscopy Ultraviolet spectroscopy provides much less information about the structure of molecules than do the spectroscopic techniques studied earlier (infrared spectroscopy, mass spectroscopy, and NMR spectroscopy) 1.8: Interpreting Ultraviolet Spectra- The Effect of Conjugation 1.9: Conjugation, Color, and the Chemistry of Vision 1.10: Additional Problems The "Additional Problems" section provides exercises related to conjugated compounds and ultraviolet spectroscopy, enhancing comprehension of concepts like kinetic vs. thermodynamic control, reactivity patterns, and structural analysis. These problems challenge students to apply theoretical knowledge to practical scenarios, reinforcing learning through problem-solving. 1.S: Conjugated Compounds and Ultraviolet Spectroscopy (Summary)
Courses/Grinnell_College/CHM_364%3A_Physical_Chemistry_2_(Grinnell_College)/08%3A_Multielectron_Atoms/8.01%3A_Atomic_and_Molecular_Calculations_are_Expressed_in_Atomic_Units	Demonstrate how solving electron structure problems are less cluttered by switching to atomic units instead of SI units. Atomic units (au or a.u.) form a system of natural units which is especially convenient for atomic physics calculations. Atomic units, like SI units, have a unit of mass, a unit of length, and so on. However, the use and notation is somewhat different from SI. Suppose a particle with a mass of m has 3.4 times the mass of electron. The value of mass $m$ can be written in three ways: $m=3.4\; m_e$: This is the clearest notation (but least common), where the atomic unit is included explicitly as a symbol. $m=3.4\; a.u.$: This notation is ambiguous, but is common. Here, it means that the mass $m$ is 3.4 times the atomic unit of mass. If considering a length $L$ of 3.4 times the atomic unit of length, the equation would look the same, $L= 3.4 \;a.u.$ The dimension needs to be inferred from context, which is sloppy. $m = 3.4$: This notation is similar to the previous one, and has the same dimensional ambiguity. It comes from formally setting the atomic units to 1 (Table 8.1.1 ). This article deals with "Hartree type" of atomic units, where the numerical values of the following four fundamental physical constants are all unity by definition: Dimension Name Symbol/Definition Value in SI units Value in Atomic Units mass electron rest mass $m_e$ 9.109×10−31 kg 1 charge elementary charge $e$ 1.602×10−19 C 1 action reduced Planck's constant $\hbar = \dfrac{h}{2\pi}$ 1.054×10−34 J·s 1 electric constant−1 Coulomb force constant $\displaystyle k_e = \frac{1}{4 \pi \epsilon_o}$ 8.987 x 109 kg·m3·s−2·C−2 1 Use the atomic units definitions in Table 8.1.1 to contrast the Hamiltonian for a Helium atom in Si units and in atomic units. Solution In SI units, the Hamiltonian for a Helium atom is \[ \hat {H} = -\dfrac {\hbar ^2}{2m_e} (\nabla ^2_1 + \nabla ^2_2) -\dfrac {2e^2}{4 \pi \epsilon _0 r_1} - \dfrac {2e^2}{4 \pi \epsilon _0 r_2} + \dfrac {e^2}{4 \pi \epsilon _0 r_{12}} \nonumber \] In atomic units, the same Hamiltonian \[ \hat {H} = -\dfrac {1}{2} (\nabla ^2_1 + \nabla ^2_2) - \dfrac {2}{r_1} - \dfrac {2}{r_2} + \dfrac {1}{r_{12}} \nonumber \] All the units that make the SI version of the Hamiltonian disappear to emphasize the key aspects of the operator. Atomic units are derived from certain fundamental properties of the physical world, and are free of anthropocentric considerations. It should be kept in mind that atomic units were designed for atomic-scale calculations in the present-day universe, with units normalize the reduced Planck constant and also mass and charge of the electron are set to 1, and, as a result, the speed of light in atomic units is a large value, $1/\alpha \approx 137$. For example, the orbital velocity of an electron around a small atom is of the order of 1 in atomic units. Table 8.1.2 give a few derived units. Some of them have proper names and symbols assigned, as indicated in the table. Dimension Name Symbol Expression Value in SI units Value in more common units length bohr $a_o$ $4\pi \epsilon_0 \hbar^2 / (m_\mathrm{e} e^2) = \hbar / (m_\mathrm{e} c \alpha) $ 5.291×10−11 m 0.052 nm = 0.529 Å energy hartree $E_h$ $m_\mathrm{e} e^4/(4\pi\epsilon_0\hbar)^2 = \alpha^2 m_\mathrm{e} c^2 $ 4.359×10−18 J 27.2 eV = 627.5 kcal·mol−1 time NaN NaN $\hbar / E_\mathrm{h}$ 2.418×10−17 s NaN velocity NaN NaN $ a_0 E_\mathrm{h} / \hbar = \alpha c$ 2.187×106 m·s−1 NaN Atomic units are chosen to reflect the properties of electrons in atoms. This is particularly clear from the classical Bohr model of the hydrogen atom in its ground state. The ground state electron orbiting the hydrogen nucleus has (in the classical Bohr model): Orbital velocity = 1 Orbital radius = 1 Angular momentum = 1 Orbital period = 2π Ionization energy = 1 ⁄ 2 Electric field (due to nucleus) = 1 Electrical attractive force (due to nucleus) = 1
Courses/College_of_Marin/CHEM_114%3A_Introductory_Chemistry/13%3A_Solutions/13.06%3A_Solution_Concentration-_Molarity	Template:HideTOC Learning Objectives Use molarity to determine quantities in chemical reactions. Use molarity as a conversion factor in calculations. Another way of expressing concentration is to give the number of moles of solute per unit volume of solution. Of all the quantitative measures of concentration, molarity is the one used most frequently by chemists. Molarity is defined as the number of moles of solute per liter of solution. \[\mathrm{molarity=\dfrac{number\: of\: moles\: of\: solute}{number\: of\: liters\: of\: solution}} \label{defMolarity} \] The symbol for molarity is $\text{M}$ or moles/liter. Chemists also use square brackets to indicate a reference to the molarity of a substance. For example, the expression $\left[ \ce{Ag^+} \right]$ refers to the molarity of the silver ion in solution. Solution concentrations expressed in molarity are the easiest to perform calculations with, but the most difficult to make in the lab. Such concentration units are useful for discussing chemical reactions in which a solute is a product or a reactant. Molar mass can then be used as a conversion factor to convert amounts in moles to amounts in grams. It is important to remember that “mol” in this expression refers to moles of solute and that “L” refers to liters of solution. For example, if you have 1.5 mol of NaCl dissolved in 0.500 L of solution, its molarity is \[\mathrm{\dfrac{1.5\: mol\: NaCl}{0.500\: L\: solution}=3.0\: M\: NaCl} \nonumber \] Sometimes (aq) is added when the solvent is water, as in “3.0 M NaCl (aq).” This is read as "a 3.00 molar sodium chloride solution," meaning that there are 3.00 moles of NaOH dissolved per one liter of solution. Be sure to note that molarity is calculated as the total volume of the entire solution, not just volume of solvent! The solute contributes to total volume. If the quantity of the solute is given in mass units, you must convert mass units to mole units before using the definition of molarity to calculate concentration. For example, what is the molar concentration of a solution of 22.4 g of HCl dissolved in 1.56 L? Step 1: First, convert the mass of solute to moles using the molar mass of HCl (36.5 g/mol): \[22.4\cancel{gHCl}\times \dfrac{1\: mol\: HCl}{36.5\cancel{gHCl}}=0.614\, mol\; HCl \nonumber \] Step 2: Now we can use the definition of molarity to determine a concentration: \[M \: =\: \dfrac{0.614\: mol\: HCl}{1.56L\: solution}=0.394\, M HCl \nonumber \] Before a molarity concentration can be calculated, the amount of the solute must be expressed in moles, and the volume of the solution must be expressed in liters, as demonstrated in the following example. Example $\PageIndex{1}$ A solution is prepared by dissolving $42.23 \: \text{g}$ of $\ce{NH_4Cl}$ into enough water to make $500.0 \: \text{mL}$ of solution. Calculate its molarity. Solution Steps for Problem Solving Unnamed: 1 Identify the "given" information and what the problem is asking you to "find." Given: Mass $= 42.23 \: \text{g} \: \ce{NH_4Cl}$ Volume solution $= 500.0 \: \text{mL} = 0.5000 \: \text{L}$ Find: Molarity = ? M List other known quantities. Molar mass $\ce{NH_4Cl} = 53.50 \: \text{g/mol}$ Plan the problem. 1. The mass of the ammonium chloride is first converted to moles. 2. Then the molarity is calculated by dividing by liters. Note the given volume has been converted to liters. $\mathrm{M=\dfrac{mol\: NH_4Cl}{L\: solution}} $ Cancel units and calculate. Now substitute the known quantities into the equation and solve. $\begin{align} 42.23 \ \cancel{\text{g} \: \ce{NH_4Cl}} \times \dfrac{1 \: \text{mol} \: \ce{NH_4Cl}}{53.50 \: \cancel{\text{g} \: \ce{NH_4Cl}}} &= 0.7893 \: \text{mol} \: \ce{NH_4Cl} \\ \dfrac{0.7893 \: \text{mol} \: \ce{NH_4Cl}}{0.5000 \: \text{L solution}} &= 1.579 \: \text{M} \end{align}$ Think about your result. The molarity is $1.579 \: \text{M}$, meaning that a liter of the solution would contain $1.579 \: \text{mol} \: \ce{NH_4Cl}$. Four significant figures are appropriate. Exercise $\PageIndex{1A}$ What is the molarity of a solution made when 66.2 g of C 6 H 12 O 6 are dissolved to make 235 mL of solution? Answer 1.57 M C 6 H 12 O 6 Exercise $\PageIndex{1B}$ What is the concentration, in $\text{mol/L}$, where $137 \: \text{g}$ of $\ce{NaCl}$ has been dissolved in enough water to make $500 \: \text{mL}$ of solution? Answer 4.69 M NaCl Using Molarity in Calculations Concentration can be a conversion factor between the amount of solute and the amount of solution or solvent (depending on the definition of the concentration unit). As such, concentrations can be useful in a variety of stoichiometry problems. In many cases, it is best to use the original definition of the concentration unit; it is that definition that provides the conversion factor. A simple example of using a concentration unit as a conversion factor is one in which we use the definition of the concentration unit and rearrange; we can do the calculation again as a unit conversion, rather than as a definition. Example $\PageIndex{2}$: Determining Moles of Solute, Given the Concentration and Volume of a Solution For example, suppose we ask how many moles of solute are present in 0.108 L of a 0.887 M NaCl solution. Because 0.887 M means 0.887 mol/L, we can use this second expression for the concentration as a conversion factor: Solution \[0.108\cancel{L\, solution}\times \dfrac{0.887\, mol\, NaCl}{\cancel{1L\, solution}}=0.0958\, mol\, NaCl \nonumber \] If we used the definition approach, we get the same answer, but now we are using conversion factor skills. Like any other conversion factor that relates two different types of units, the reciprocal of the concentration can be also used as a conversion factor. Example $\PageIndex{3}$: Determining Volume of a Solution, Given the Concentration and Moles of Solute Using concentration as a conversion factor, how many liters of 2.35 M CuSO 4 are needed to obtain 4.88 mol of CuSO 4 ? Solution This is a one-step conversion, but the concentration must be written as the reciprocal for the units to work out: \[4.88\cancel{mol\, CuSO_{4}}\times \dfrac{1\, L\, solution}{2.35\cancel{mol\, CuSO_{4}}}=2.08\, L\, of \, solution \nonumber \] In a laboratory situation, a chemist must frequently prepare a given volume of solutions of a known molarity. The task is to calculate the mass of the solute that is necessary. The molarity equation can be rearranged to solve for moles, which can then be converted to grams. The following example illustrates this. Example $\PageIndex{4}$ A chemist needs to prepare $3.00 \: \text{L}$ of a $0.250 \: \text{M}$ solution of potassium permanganate $\left( \ce{KMnO_4} \right)$. What mass of $\ce{KMnO_4}$ does she need to make the solution? Solution Steps for Problem Solving Unnamed: 1 Identify the "given" information and what the problem is asking you to "find." Given: Molarity $= 0.250 \: \text{M}$ Volume $= 3.00 \: \text{L}$ Find: Mass $\ce{KMnO_4} = ? \: \text{g}$ List other known quantities. Molar mass $\ce{KMnO_4} = 158.04 \: \text{g/mol}$ 0.250 mol KMnO4 to 1 L of KMnO4 solution Plan the problem. NaN Cancel units and calculate. Now substitute the known quantities into the equation and solve. $\begin{align} \text{mol} \: \ce{KMnO_4} = 0.250 \: \text{M} \: \ce{KMnO_4} \times 3.00 \: \text{L} &= 0.750 \: \text{mol} \: \ce{KMnO_4} \\ \cancel{3.00 \: L \: solution} \times \dfrac{0.250 \: \cancel{\text{mol} \: \ce{KMnO_4}}}{1\cancel{L \: solution}} \times \dfrac{158.04 \: \text{g} \: \ce{KMnO_4}}{1 \: \cancel{\text{mol} \: \ce{KMnO_4}}} &= 119 \: \text{g} \: \ce{KMnO_4} \end{align}$ Think about your result. When $119 \: \text{g}$ of potassium permanganate is dissolved into water to make $3.00 \: \text{L}$ of solution, the molarity is $0.250 \: \text{M}$. Exercise $\PageIndex{4A}$ Using concentration as a conversion factor, how many liters of 0.0444 M CH 2 O are needed to obtain 0.0773 mol of CH 2 O? Answer 1.74 L Exercise $\PageIndex{4B}$ Answer the problems below using concentration as a conversion factor. What mass of solute is present in 1.08 L of 0.0578 M H 2 SO 4 ? What volume of 1.50 M HCl solution contains 10.0 g of hydrogen chloride? Answer a 6.12 g Answer b 183 mL or 0.183L How to Indicate Concentration Square brackets are often used to represent concentration, e.g., [NaOH] = 0.50 M. Use the capital letter M for molarity, not a lower case m (this is a different concentration unit called molality ). Watch as the Flinn Scientific Tech Staff demonstrates "How To Prepare Solutions." It is important to note that there are many different ways you can set up and solve your chemistry equations. Some students prefer to answer multi-step calculations in one long step, while others prefer to work out each step individually. Neither method is necessarily better or worse than the other method—whichever makes the most sense to you is the one that you should use. In this text, we will typically use unit analysis (also called dimension analysis or factor analysis).
Courses/Saint_Francis_University/CHEM_113%3A_Human_Chemistry_I_(Muino)/01%3A_Matter_and_Measurements/1.08%3A_Measurement_and_Significant_Figures	Learning Objectives Identify the number of significant figures in a reported value. Apply the concept of significant figures to report a measurement with the proper number of digits. Scientists have established certain conventions for communicating the degree of precision of a measurement, which is dependent on the measuring device used (See Figure $\PageIndex{1}$). Imagine, for example, that you are using a meterstick to measure the width of a table. The centimeters (cm) marked on the meterstick, tell you how many centimeters wide the table is. Many metersticks also have markings for millimeters (mm), so we can measure the table to the nearest millimeter . The measurement made using millimeters is more precise, it is closer to the actual length of the table. Most metersticks do not have any smaller (or more precise) markings indicated, so you cannot report the measured width of the table any more precise than to the nearest millimeter. The concept of significant figures takes the limitation of measuring devices into account. The significant figures of a measured quantity are defined as all the digits known with certainty (those indicated by the markings on the measuring device) and the first uncertain, or estimated, digit (one digit past the smallest marking on the measuring device). It makes no sense to estimate and report any digits after the first uncertain one, so it is the last significant digit reported in a measurement. Zeros are used when needed to place the significant figures in their correct positions. Thus, zeros are sometimes counted as significant figures but are sometimes only used as placeholders (see the rules for significant figures below for more details). “Sig figs” is a common abbreviation for significant figures. Rules for Determination of Significant Figures Consider the earlier example of measuring the width of a table with a meterstick. If the table is measured and reported as being 1,357 mm wide, the number 1,357 has four significant figures. The 1 (thousands place), the 3 (hundreds place), and the 5 (tens place) are certain; the 7 (ones place) is assumed to have been estimated. It would make no sense to report such a measurement as 1,357.0 (five Sig Figs) or 1,357.00 (six Sig Figs) because that would suggest the measuring device was able to determine the width to the nearest tenth or hundredth of a millimeter, when in fact it shows only tens of millimeters and therefore the ones place was estimated. On the other hand, if a measurement is reported as 150 mm, the 1 (hundreds) and the 5 (tens) are known to be significant, but how do we know whether the zero is or is not significant? The measuring device could have had marks indicating every 100 mm or marks indicating every 10 mm. How can you determine if the zero is significant (the estimated digit), or if the 5 is significant and the zero a value placeholder? The rules for deciding which digits in a measurement are significant are as follows: Rule 1: All nonzero digits in a measurement are significant. 237 has three significant figures. 1.897 has four significant figures. Rule 2: Zeros that appear between other nonzero digits (i.e., " middle zeros ") are always significant. 39,004 has five significant figures. 5.02 has three significant figures. Rule 3: Zeros that appear in front of all of the nonzero digits are called leading zeros . Leading zeros are never significant. 0.008 has one significant figure. 0.000416 has three significant figures. Rule 4: Zeros that appear after all nonzero digits are called trailing zeros . A number with trailing zeros that lacks a decimal point may or may not be significant. 1400 is ambiguous. $1.4 \times 10^3$ has two significant figures. $1.40 \times 10^3$ three significant figures. $1.400 \times 10^3$ has four significant figures. Rule 5: Trailing zeros in a number with a decimal point are significant. This is true whether the zeros occur before or after the decimal point. 620.0 has four significant figures. 19.000 has five significant figures. It needs to be emphasized that just because a certain digit is not significant does not mean that it is not important or that it can be left out. Though the zero in a measurement of 140 may not be significant, the value cannot simply be reported as 14. An insignificant zero functions as a placeholder for the decimal point. When numbers are written in scientific notation, this becomes more apparent. The measurement 140 can be written as $1.4 \times 10^2$, with two significant figures in the coefficient or as $1.40 \times 10^3$, with three significant figures. A number less than one, such as 0.000416, can be written in scientific notation as $4.16 \times 10^{-4}$, which has 3 significant figures. In some cases, scientific notation is the only way to correctly indicate the correct number of significant figures. In order to report a value of 15,000,00 with four significant figures, it would need to be written as $1.500 \times 10^7$. Exact Quantities When numbers are known exactly, the significant figure rules do not apply. This occurs when objects are counted rather than measured. For example, a carton of eggs has 12 eggs. The actual value cannot be 11.8 eggs, since we count eggs in whole number quantities. So the 12 is an exact quantity. Exact quantities are considered to have an infinite number of significant figures; the importance of this concept will be seen later when we begin looking at how significant figures are dealt with during calculations. Numbers in many conversion factors, especially for simple unit conversions, are also exact quantities and have infinite significant figures. There are exactly 100 centimeters in 1 meter and exactly 60 seconds in 1 minute. Those values are definitions and are not the result of a measurement. Example $\PageIndex{2}$ Give the number of significant figures in each. Identify the rule for each. 5.87 0.031 52.90 00.2001 500 6 atoms Solution Explanation Answer a. All three numbers are significant (rule 1). 5.87 , three significant figures b. The leading zeros are not significant (rule 3). The 3 and the 1 are significant (rule 1) 0.031, two significant figures c. The 5, the 2 and the 9 are significant (rule 1). The trailing zero is also significant (rule 5). 52.90, four significant figures d. The leading zeros are not significant (rule 3). The 2 and the 1 are significant (rule 1) and the middle zeros are also significant (rule 2). 00.2001, four significant figures e. The number is ambiguous. It could have one, two or three significant figures. 500, ambiguous f. The 6 is a counting number. A counting number is an exact number. 6, infinite Exercise $\PageIndex{2}$ Give the number of significant figures in each. 36.7 m 0.006606 s 2,002 kg 306,490,000 people 3,800 g Answer a: three significant figures. Answer b: four significant figures. Answer c: four significant figures. Answer d: Infinite (Exact number) Answer e: Ambiguous, could be two, three or four significant figure. Summary Uncertainty exists in all measurements. The degree of uncertainty is affected in part by the quality of the measuring tool. Significant figures give an indication of the certainty of a measurement. Rules allow decisions to be made about how many digits to use in any given situation.
Courses/Harper_College/CHM_110%3A_Fundamentals_of_Chemistry/03%3A_Energy_Production/3.07%3A_Air_Pollutants	At the end of 2019, the coronavirus began spreading around the world, infecting humans and causing pneumonia known as COVID-19. The number of infections increased rapidly and in some places, healthcare systems were overrun. In an attempt to contain the spread, many countries limited transportation and closed down schools, colleges, universities and nonessential businesses. The negative impacts of the novel coronavirus pandemic are widespread and significant. However, one of the unintended consequences of the efforts to contain the spread of the disease was reduced air pollution. Many air pollutants are molecules produced from the combustion of fuels. NASA monitoring instruments measured decreased concentrations of NO 2 and CO 2 which resulted in a positive health benefit. (1,2) According to the US EPA and WHO, long-term exposure to ambient air pollution increases mortality and morbidity from cardiovascular and respiratory disease and lung cancer and decreases life expectancy. (3) The two pollutants responsible for most of the disease burden are fine particulate matter (PM 2.5 ) and ground level ozone (O 3 ). In response to poor air quality in the United States, the government established the Clean Air Act (CAA) in 1970. The law was amended in 1977 and again in 1990. This comprehensive federal law regulates air emissions from stationary sources such as factories, refineries and power plants and mobile sources such as cars, trucks and buses. CAA also authorizes the Environmental Protection Agency (EPA) to establish National Ambient Air Quality Standards (NAAQS). The goal of these standards is to protect the health and welfare of the public and regulate the emissions of hazardous air pollutants. Hazardous air pollutants are known to cause cancer and other serious health impacts. The NAAQS have been set for six common pollutants, known as the criteria air pollutants. These include ground level ozone, particulate matter carbon monoxide, lead, sulfur dioxide, and nitrogen dioxide. Four of the six criteria pollutants exist as gaseous molecules in the atmosphere. The EPA publishes the standards on their website . These six common pollutants are not the only substances that we need to worry about in the air. Soot from fires is problematic and even hazardous pollutants from industry are in our air. An example of this was ethylene oxide was being emitted from sterilization factories in Lake County IL in 2017. Ethylene oxide is cancer causing and people in the surrounding area were inhaling it. The Illinois EPA intervened and both companies were required to reduce their emissions to protest the public health. For more information visit the Lake County Health Department. Air pollution is problematic when we breathe in these compounds, they can cause all sorts of health problems from cancer, cardiovascular diseases to asthma. You can find out more here: https://www.cdc.gov/air/pollutants.htm . The EPA has also established the U.S. Air Quality Index (AQI) as a simple method for reporting air quality to the general public. Lead is the only criteria air pollutant not included in the AQI. The AQI ranges from 0-500. The lower the value the better the air quality. Additionally, the AQI is divided into six color-coded categories. Green (0-50) represents good air quality that poses little to no risk even to sensitive groups. Red (151-200) is considered unhealthy and some members of the public may experience health effects. Values above 300 are considered hazardous with everyone more likely to be affected. (4) The emission of hazardous air pollutants is not limited to human-made sources nor are people only exposed outside. Natural sources include volcanic eruptions and forest fires which are of increasing concern due their greater frequency and duration. Building materials and cleaning solvents are two fairly common sources of indoor pollutants. (1) NASA Earth Observatory. Airborne Nitrogen Dioxide Plummets Over China https ://earthobservatory.nasa.gov/images/146362/airborne-nitrogen-dioxide-plummets-over-china (accessed 19 July 2020) (2) Evans, S. Analysis: Coronavirus set to cause largest ever annual fall to CO2 emissions. https://www.carbonbrief.org/analysis-coronavirus-set-to-cause-largest-ever-annual-fall-in-co2-emissions (accessed 19 July 2020). (3) Cohen, Aaron J et al. “Estimates and 25-year trends of the global burden of disease attributable to ambient air pollution: an analysis of data from the Global Burden of Diseases Study 2015.” Lancet (London, England) vol. 389,10082 (2017): 1907-1918. doi:10.1016/S0140-6736(17)30505-6. (4) AQI basics. https://www.airnow.gov/aqi/aqi-basics/ (last accessed 19 July 2020).
Courses/Lumen_Learning/Book%3A_Microeconomics-2_(Lumen)/17%3A_Module-_Income_Distribution/17.06%3A_Reading-_Monopoly_and_Monopsony-_A_Comparison	Monopoly and Monopsony: A Comparison There is a close relationship between the models of monopoly and monopsony. A clear understanding of this relationship will help to clarify both models. Figure 14.4 compares the monopoly and monopsony equilibrium solutions. Both types of firms are price setters: The monopoly is a price setter in its product market; the monopsony is a price setter in its factor market. Both firms must change price to change quantity: The monopoly must lower its product price to sell an additional unit of output, and the monopsony must pay more to hire an additional unit of the factor. Because both types of firms must adjust prices to change quantities, the marginal consequences of their choices are not given by the prices they charge (for products) or pay (for factors). For a monopoly, marginal revenue is less than price; for a monopsony, marginal factor cost is greater than price. Figure 14.4 Monopoly and Monopsony. The graphs and the table provide a comparison of monopoly and monopsony. Both types of firms follow the marginal decision rule: A monopoly produces a quantity of the product at which marginal revenue equals marginal cost; a monopsony employs a quantity of the factor at which marginal revenue product equals marginal factor cost. Both firms set prices at which they can sell or purchase the profit-maximizing quantity. The monopoly sets its product price based on the demand curve it faces; the monopsony sets its factor price based on the factor supply curve it faces. Monopsony in the Real World Although cases of pure monopsony are rare, there are many situations in which buyers have a degree of monopsony power. A buyer has monopsony power if it faces an upward-sloping supply curve for a good, service, or factor of production. For example, a firm that accounts for a large share of employment in a small community may be large enough relative to the labor market that it is not a price taker. Instead, it must raise wages to attract more workers. It thus faces an upward-sloping supply curve and has monopsony power. Because buyers are more likely to have monopsony power in factor markets than in product markets, we shall focus on those. The next section examines monopsony power in professional sports. Monopsonies in Sports Professional sports provide a setting in which economists can test theories of wage determination in competitive versus monopsony labor markets. In their analyses, economists assume professional teams are profit-maximizing firms that hire labor (athletes and other workers) to produce a product: entertainment bought by the fans who watch their games and by other firms that sponsor the games. Fans influence revenues directly by purchasing tickets and indirectly by generating the ratings that determine television and radio advertising revenues from broadcasts of games. In a competitive system, a player should receive a wage equal to his or her MRP —the increase in team revenues the player is able to produce. As New York Yankees owner George Steinbrenner once put it, “You measure the value of a ballplayer by how many fannies he puts in the seats.” The monopsony model, however, predicts that players facing monopsony employers will receive wages that are less than their MRP s. A test of monopsony theory, then, would be to determine whether players in competitive markets receive wages equal to their MRP s and whether players in monopsony markets receive less. Since the late 1970s, there has been a major shift in the rules that govern relations between professional athletes and owners of sports teams. The shift has turned the once monopsonistic market for professional athletes into a competitive one. Before 1977, for example, professional baseball players in the United States played under the terms of the “reserve clause,” which specified that a player was “owned” by his team. Once a team had acquired a player’s contract, the team could sell, trade, retain, or dismiss the player. Unless the team dismissed him, the player was unable to offer his services for competitive bidding by other teams. Moreover, players entered major league baseball through a draft that was structured so that only one team had the right to bid for any one player. Throughout a player’s career, then, there was always only one team that could bid on him—each player faced a monopsony purchaser for his services to major league baseball. Conditions were similar in other professional sports. Many studies have shown that the salaries of professional athletes in various team sports fell far short of their MRP s while monopsony prevailed. When the reserve clauses were abandoned, players’ salaries shot up—just as economic theory predicts. Because players could offer their services to other teams, owners began to bid for their services. Profit-maximizing owners were willing to pay athletes their MRP s. Average annual salaries for baseball players rose from about $50,000 in 1975 to nearly $1.4 million in 1997. Average annual player salaries in men’s basketball rose from $109,000 in 1976 to $2.24 million in 1998. Football players worked under an almost pure form of monopsony until 1989, when a few players were allowed free agency status each year. In 1993, when 484 players were released to the market as free agents, those players received pay increases averaging more than 100%. Under the NFL collective bargaining agreement in effect in 1998, players could become unrestricted free agents if they had been playing for four years. There were 305 unrestricted free agents (out of a total player pool of approximately 1,700) that year. About half signed new contracts with their old teams while the other half signed with new teams. Table 14.1 illustrates the impact of free agency in four professional sports. Table 14.1 The Impact of Free Agency Player Salaries As Percentage of Team Revenues Unnamed: 0_level_2 MLB NBA NFL NHL 1970–73 15.9 46.1 34.4 21.3 1998 48.4 54.2 55.4 58.4 Free agency has increased player share of total revenues in each of the major men’s team sports. Table 14.1 gives player salaries as a percentage of team revenues for major league baseball (MLB), the National Basketball Association (NBA), the National Football League (NFL) and the National Hockey League (NHL) during the 1970–1973 period that players in each league worked under monopsony conditions and in 1998, when players in each league had gained the right of free agency. Given the dramatic impact on player salaries of more competitive markets for athletes, events such as the 2004–2005 lockout in hockey came as no surprise. The agreement between the owners of hockey teams and the players in 2005 to limit the total payroll of each team reinstates some of the old monopsony power of the owners. Players had a huge financial stake in resisting such attempts. Monopsony in Other Labor Markets A firm that has a dominant position in a local labor market may have monopsony power in that market. Even if a firm does not dominate the total labor market, it may have monopsony power over certain types of labor. For example, a hospital may be the only large employer of nurses in a local market, and it may have monopsony power in employing them. Colleges and universities generally pay part-time instructors considerably less for teaching a particular course than they pay full-time instructors. In part, the difference reflects the fact that full-time faculty members are expected to have more training and are expected to contribute far more in other areas. But the monopsony model suggests an additional explanation. Part-time instructors are likely to have other regular employment. A university hiring a local accountant to teach a section of accounting does not have to worry that that person will go to another state to find a better offer as a part-time instructor. For part-time teaching, then, the university may be the only employer in town—and thus able to exert monopsony power to drive the part-time instructor’s wage below the instructor’s MRP . Monopsony in Other Factor Markets Monopsony power may also exist in markets for factors other than labor. The military in different countries, for example, has considerable monopsony power in the market for sophisticated military goods. Major retailers often have some monopsony power with respect to some of their suppliers. Sears, for example, is the only wholesale buyer of Craftsman brand tools. One major development in medical care in recent years has been the emergence of managed care organizations that contract with a large number of employers to purchase medical services on behalf of employees. These organizations often have sufficient monopsony power to force down the prices charged by providers such as drug companies, physicians, and hospitals. Countries in which health care is provided by the government, such as Canada and the United Kingdom, are able to exert monopsony power in their purchase of health care services. Whatever the source of monopsony power, the expected result is the same. Buyers with monopsony power are likely to pay a lower price and to buy a smaller quantity of a particular factor than buyers who operate in a more competitive environment. KEY TAKEAWAYS In the monopsony model there is one buyer for a good, service, or factor of production. A monopsony firm is a price setter in the market in which it has monopsony power. The monopsony buyer selects a profit-maximizing solution by employing the quantity of factor at which marginal factor cost ( MFC ) equals marginal revenue product ( MRP ) and paying the price on the factor’s supply curve corresponding to that quantity. A degree of monopsony power exists whenever a firm faces an upward-sloping supply curve for a factor. Case in Point: Professional Player Salaries and Monopsony Professional athletes have not always enjoyed the freedom they have today to seek better offers from other teams. Before 1977, for example, baseball players could deal only with the team that owned their contract—one that “reserved” the player to that team. This reserve clause gave teams monopsony power over the players they employed. Similar restrictions hampered player mobility in men’s football, basketball, and hockey. Gerald Scully, an economist at the University of Texas at Dallas, estimated the impact of the reserve clause on baseball player salaries. He sought to demonstrate that the player salaries fell short of MRP . Mr. Scully estimated the MRP of players in a two-step process. First, he studied the determinants of team attendance. He found that in addition to factors such as population and income in a team’s home city, the team’s win-loss record had a strong effect on attendance. Second, he examined the player characteristics that determined win-loss records. He found that for hitters, batting average was the variable most closely associated with a team’s winning percentage. For pitchers, it was the earned-run average—the number of earned runs allowed by a pitcher per nine innings pitched. With equations that predicted a team’s attendance and its win-loss record, Mr. Scully was able to take a particular player, describe him by his statistics, and compute his MRP . Mr. Scully then subtracted costs associated with each player for such things as transportation, lodging, meals, and uniforms to obtain the player’s net MRP . He then compared players’ net MRP s to their salaries. Mr. Scully’s results, displayed in the table below, show net MRP and salaries, estimated on a career basis, for players he classified as mediocre, average, and star-quality, based on their individual statistics. For average and star-quality players, salaries fell far below net MRP , just as the theory of monopsony suggests. Unnamed: 0 Career Net MRP Career Salary Salary As % of net MRP Hitters Hitters Hitters Hitters Mediocre −$129,300 $60,800 NaN Average 906700 196200 22 Star 3139100 477200 15 Pitchers Pitchers Pitchers Pitchers Mediocre −53,600 54800 NaN Average 1119200 222500 20 Star 3969600 612500 15 The fact that mediocre players with negative net MRP s received salaries presents something of a puzzle. One explanation could be that when they were signed to contracts, these players were expected to perform well, so their salaries reflected their expected contributions to team revenues. Their actual performance fell short, so their wages exceeded their MRP s. Another explanation could be that teams paid young players more than they were expected to contribute to revenues early in their careers in hopes that they would develop into profitable members of the team. In any event, Mr. Scully found that the costs of mediocre players exceeded their estimated contribution to team revenues, giving them negative net MRP s. In 1977, a lawsuit filed by several baseball players resulted in the partial dismantling of the reserve clause. Players were given the right, after six years with a team, to declare themselves “free agents” and offer their services to other teams. Player salaries quickly rose. The accompanying table shows the pitchers that became free agents in 1977, their estimated net marginal revenue products, and their 1977 salaries. As you can see, salaries for pitchers came quite close to their net MRP s. Pitcher Net MRP 1977 Salary Doyle Alexander $166,203 $166,677 Bill Campbell $205,639 $210,000 Rollie Fingers $303,511 $332,000 Wayne Garland $282,091 $230,000 Don Gullett $340,846 $349,333 The same movement toward giving players greater freedom to deal with other teams occurred in the National Football League (NFL), the National Basketball Association (NBA), and the National Hockey League (NHL). The result in every case was the same: player salaries rose both in absolute terms and as a percentage of total team revenues. Table 14.1 gives player salaries as a percentage of total team revenues in the period 1970–73 and in 1998 for men’s baseball (MLB), basketball, football, and hockey. The greatest gains came in baseball, which had the most restrictive rules against player movement. Hockey players, too, ended up improving their salaries greatly. By 2004, their salaries totaled 75% of team revenues. The smallest gains came in basketball, where players already had options. The American Basketball Association was formed; it ultimately became part of the National Basketball Association. Basketball players also had the alternative of playing in Europe. But, the economic lesson remains clear: any weakening of the monopsony power of teams results in gains in player salaries. CC licensed content, Shared previously Principles of Microeconomics Section 14.1 . Authored by : Anonymous. Located at : http://2012books.lardbucket.org/books/microeconomics-principles-v1.0/s17-01-price-setting-buyers-the-case-.html . License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike
Courses/BridgeValley_Community_and_Technical_College/Consumer_Chemistry/01%3A_Introduction_to_Chemistry/1.06%3A_The_Scientific_Method/1.6.02%3A_Using_the_Scientific_Method	Learning Objectives To identify the components of the scientific method Classify measurements as being quantitative or qualitative. Evaluate science in the media. The Scientific Method Scientists search for answers to questions and solutions to problems by using a procedure called the scientific method. This procedure consists of making observations, formulating hypotheses, and designing experiments, which in turn lead to additional observations, hypotheses, and experiments in repeated cycles (Figure $\PageIndex{1}$). Step 1: Make observations Observations can be qualitative or quantitative. Qualitative observations describe properties or occurrences in ways that do not rely on numbers. Examples of qualitative observations include the following: the outside air temperature is cooler during the winter season, table salt is a crystalline solid, sulfur crystals are yellow, and dissolving a penny in dilute nitric acid forms a blue solution and a brown gas. Quantitative observations are measurements, which by definition consist of both a number and a unit. Examples of quantitative observations include the following: the melting point of crystalline sulfur is 115.21° Celsius, and 35.9 grams of table salt—whose chemical name is sodium chloride—dissolves in 100 grams of water at 20° Celsius. For the question of the dinosaurs’ extinction, the initial observation was quantitative: iridium concentrations in sediments dating to 66 million years ago were 20–160 times higher than normal. Step 2: Formulate a hypothesis After deciding to learn more about an observation or a set of observations, scientists generally begin an investigation by forming a hypothesis, a tentative explanation for the observation(s). The hypothesis may not be correct, but it puts the scientist’s understanding of the system being studied into a form that can be tested. For example, the observation that we experience alternating periods of light and darkness corresponding to observed movements of the sun, moon, clouds, and shadows is consistent with either of two hypotheses: Earth rotates on its axis every 24 hours, alternately exposing one side to the sun, or the sun revolves around Earth every 24 hours. Suitable experiments can be designed to choose between these two alternatives. For the disappearance of the dinosaurs, the hypothesis was that the impact of a large extraterrestrial object caused their extinction. Unfortunately (or perhaps, fortunately), this hypothesis does not lend itself to direct testing by any obvious experiment, but scientists can collect additional data that either support or refute it. Step 3: Design and perform experiments After a hypothesis has been formed, scientists conduct experiments to test its validity. Experiments are systematic observations or measurements, preferably made under controlled conditions—that is, under conditions in which a single variable changes. Step 4: Accept or modify the hypothesis A properly designed and executed experiment enables a scientist to determine whether the original hypothesis is valid. In which case he can proceed to step 5. In other cases, experiments often demonstrate that the hypothesis is incorrect or that it must be modified thus requiring further experimentation. Step 5: Development into law and/or theory More experimental data are then collected and analyzed, at which point a scientist may begin to think that the results are sufficiently reproducible (i.e., dependable) to merit being summarized in law, a verbal or mathematical description of a phenomenon that allows for general predictions. A law simply says what happens; it does not address the question of why. One example of a law, the law of definite proportions, which was discovered by the French scientist Joseph Proust (1754–1826), states that a chemical substance always contains the same proportions of elements by mass. Thus, sodium chloride (table salt) always contains the same proportion by mass of sodium to chlorine, in this case, 39.34% sodium and 60.66% chlorine by mass, and sucrose (table sugar) is always 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. Whereas a law states only what happens, a theory attempts to explain why nature behaves as it does. Laws are unlikely to change greatly over time unless a major experimental error is discovered. In contrast, a theory, by definition, is incomplete and imperfect, evolving with time to explain new facts as they are discovered. Because scientists can enter the cycle shown in Figure $\PageIndex{1}$ at any point, the actual application of the scientific method to different topics can take many different forms. For example, a scientist may start with a hypothesis formed by reading about work done by others in the field, rather than by making direct observations. A Real-World Application of the Scientific Method In 2007, my husband and I journeyed to China to adopt our daughter. Upon arrival in Beijing, I became violently ill. Due to her visa paperwork, my husband, daughter, and I were required to stay in China for two weeks. Unfortunately, I was ill the entire time. Once the two-week period was up, the three of us flew back to the United States where I continued to be sick. For the next year, I remained ill and lost a total of 30 pounds. The picture below shows me holding my daughter eight months after we returned home from China. I would like you to attempt to perform the scientific method on my situation described above. List the steps of the scientific method along with some plausible explanations. * Please have this ready to discuss in class .* Exercise $\PageIndex{1}$ Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation. Ice always floats on liquid water. Birds evolved from dinosaurs. According to Albert Einstein, mass X speed of light = energy When 10 g of ice was added to 100 mL of water at 25°C, the temperature of the water decreased to 15.5°C after the ice melted. The ingredients of Ivory soap were analyzed to see whether it really is 99.44% pure, as advertised. Solution This is a general statement of a relationship between the properties of liquid and solid water, so it is a law. This is an educated guess regarding the origin of birds, so it is a hypothesis. This is a theory that explains an explanation of events and can be disproven at any time. The temperature is measured before and after a change is made in a system, so these are quantitative observations. This is an analysis designed to test a hypothesis (in this case, the manufacturer’s claim of purity), so it is an experiment. Exercise $\PageIndex{2}$ Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation. Measured amounts of acid were added to a Rolaids tablet to see whether it really “consumes 47 times its weight in excess stomach acid.” Heat always flows from hot objects to cooler ones, not in the opposite direction. The universe was formed by a massive explosion that propelled matter into a vacuum. Michael Jordan is the greatest pure shooter ever to play professional basketball. Limestone is relatively insoluble in water but dissolves readily in dilute acid with the evolution of a gas. Gas mixtures that contain more than 4% hydrogen in air are potentially explosive. Answer a experiment Answer b law Answer c theory Answer d hypothesis Answer e qualitative observation Answer f quantitative observation Evaluating Science in the Media
Courses/American_River_College/CHEM_305%3A_Introduction_to_Chemistry_(Zumalt)/01%3A_Unit_1/1.07%3A_Significant_Figures_in_Calculations	Learning Objectives Use significant figures correctly in arithmetical operations. Rounding Before dealing with the specifics of the rules for determining the significant figures in a calculated result, we need to be able to round numbers correctly. To round a number, first decide how many significant figures the number should have. Once you know that, round to that many digits, starting from the left. If the number immediately to the right of the last significant digit is less than 5, it is dropped and the value of the last significant digit remains the same. If the number immediately to the right of the last significant digit is greater than or equal to 5, the last significant digit is increased by 1. Consider the measurement $207.518 \: \text{m}$. Right now, the measurement contains six significant figures. How would we successively round it to fewer and fewer significant figures? Follow the process as outlined in Table $\PageIndex{1}$. Number of Significant Figures Rounded Value Reasoning 6 207.518 All digits are significant 5 207.520 8 rounds the 1 up to 2 4 207.500 2 is dropped 3 208.000 5 rounds the 7 up to 8 2 210.000 8 is replaced by a 0 and rounds the 0 up to 1 1 200.000 1 is replaced by a 0 Notice that the more rounding that is done, the less reliable the figure is. An approximate value may be sufficient for some purposes, but scientific work requires a much higher level of detail. It is important to be aware of significant figures when you are mathematically manipulating numbers. For example, dividing 125 by 307 on a calculator gives 0.4071661238… to an infinite number of digits. But do the digits in this answer have any practical meaning, especially when you are starting with numbers that have only three significant figures each? When performing mathematical operations, there are two rules for limiting the number of significant figures in an answer—one rule is for addition and subtraction, and one rule is for multiplication and division. In operations involving significant figures, the answer is reported in such a way that it reflects the reliability of the least precise operation. An answer is no more precise than the least precise number used to get the answer. Multiplication and Division For multiplication or division, the rule is to count the number of significant figures in each number being multiplied or divided and then limit the significant figures in the answer to the lowest count. An example is as follows: The final answer, limited to four significant figures, is 4,094. The first digit dropped is 1, so we do not round up. Scientific notation provides a way of communicating significant figures without ambiguity. You simply include all the significant figures in the leading number. For example, the number 450 has two significant figures and would be written in scientific notation as 4.5 × 10 2 , whereas 450.0 has four significant figures and would be written as 4.500 × 10 2 . In scientific notation, all significant figures are listed explicitly. Example $\PageIndex{1}$ Write the answer for each expression using scientific notation with the appropriate number of significant figures. 23.096 × 90.300 125 × 9.000 Solution a Explanation Answer The calculator answer is 2,085.5688, but we need to round it to five significant figures. Because the first digit to be dropped (in the tenths place) is greater than 5, we round up to 2,085.6. $2.0856 \times 10^3$ b Explanation Answer The calculator gives 1,125 as the answer, but we limit it to three significant figures. $1.13 \times 10^3$ Addition and Subtraction How are significant figures handled in calculations? It depends on what type of calculation is being performed. If the calculation is an addition or a subtraction, the rule is as follows: limit the reported answer to the rightmost column that all numbers have significant figures in common. For example, if you were to add 1.2 and 4.71, we note that the first number stops its significant figures in the tenths column, while the second number stops its significant figures in the hundredths column. We therefore limit our answer to the tenths column. We drop the last digit—the 1—because it is not significant to the final answer. The dropping of positions in sums and differences brings up the topic of rounding. Although there are several conventions, in this text we will adopt the following rule: the final answer should be rounded up if the first dropped digit is 5 or greater, and rounded down if the first dropped digit is less than 5. Example $\PageIndex{2}$ 13.77 + 908.226 1,027 + 611 + 363.06 Solution a Explanation Answer The calculator answer is 921.996, but because 13.77 has its farthest-right significant figure in the hundredths place, we need to round the final answer to the hundredths position. Because the first digit to be dropped (in the thousandths place) is greater than 5, we round up to 922.00 $922.00 = 9.2200 \times 10^2$ b Explanation Answer The calculator gives 2,001.06 as the answer, but because 611 and 1027 has its farthest-right significant figure in the ones place, the final answer must be limited to the ones position. $2,001.06 = 2.001 \times 10^3$ Exercise $\PageIndex{2}$ Write the answer for each expression using scientific notation with the appropriate number of significant figures. 217 ÷ 903 13.77 + 908.226 + 515 255.0 − 99 0.00666 × 321 Answer a: $0.240 = 2.40 \times 10^{-1}$ Answer b: $1,437 = 1.437 \times 10^3$ Answer c: $156 = 1.56 \times 10^2$ Answer d: $2.14 = 2.14 \times 10^0$ Remember that calculators do not understand significant figures. You are the one who must apply the rules of significant figures to a result from your calculator. Calculations Involving Multiplication/Division and Addition/Subtraction In practice, chemists generally work with a calculator and carry all digits forward through subsequent calculations. When working on paper, however, we often want to minimize the number of digits we have to write out. Because successive rounding can compound inaccuracies, intermediate rounding needs to be handled correctly. When working on paper, always round an intermediate result so as to retain at least one more digit than can be justified and carry this number into the next step in the calculation. The final answer is then rounded to the correct number of significant figures at the very end. In the worked examples in this text, we will often show the results of intermediate steps in a calculation. In doing so, we will show the results to only the correct number of significant figures allowed for that step, in effect treating each step as a separate calculation. This procedure is intended to reinforce the rules for determining the number of significant figures, but in some cases it may give a final answer that differs in the last digit from that obtained using a calculator, where all digits are carried through to the last step. Example $\PageIndex{3}$ 2(1.008 g) + 15.99 g 137.3 s + 2(35.45 s) $ {118.7 g \over 2} - 35.5 g $ Solution a. Empty DataFrame Columns: [(Explanation, 2(1.008 g) + 15.99 g = Perform multiplication first. 2 (1.008 g 4 sig figs) = 2.016 g 4 sig figs The number with the least number of significant figures is 1.008 g; the number 2 is an exact number and therefore has an infinite number of significant figures. Then, perform the addition. 2.016 g thousandths place + 15.99 g hundredths place (least precise) = 18.006 g Round the final answer. Round the final answer to the hundredths place since 15.99 has its farthest right significant figure in the hundredths place (least precise).), (Answer, 18.01 g (rounding up))] Index: [] b. Explanation Answer 137.3 s + 2(35.45 s) = Perform multiplication first. 2(35.45 s 4 sig figs) = 70.90 s 4 sig figs The number with the least number of significant figures is 35.45; the number 2 is an exact number and therefore has an infinite number of significant figures. Then, perform the addition. 137.3 s tenths place (least precise) + 70.90 s hundredths place = 208.20 s Round the final answer. Round the final answer to the tenths place based on 137.3 s. 208.2 s c. Empty DataFrame Columns: [(Explanation, $ {118.7 g \over 2} - 35.5 g $ = Perform division first. $ {118.7 g \over 2} $ 4 sig figs = 59.35 g 4 sig figs The number with the least number of significant figures is 118.7 g; the number 2 is an exact number and therefore has an infinite number of significant figures. Perform subtraction next. 59.35 g hundredths place − 35.5 g tenths place (least precise) = 23.85 g Round the final answer. Round the final answer to the tenths place based on 35.5 g.), (Answer, 23.9 g (rounding up))] Index: [] Exercise $\PageIndex{3}$ Complete the calculations and report your answers using the correct number of significant figures. 5(1.008s) - 10.66 s 99.0 cm+ 2(5.56 cm) Answer a -5.62 s Answer b 110.2 cm Summary Rounding If the number to be dropped is greater than or equal to 5, increase the number to its left by 1 (e.g. 2.9699 rounded to three significant figures is 2.97). If the number to be dropped is less than 5, there is no change (e.g. 4.00443 rounded to four significant figures is 4.004). The rule in multiplication and division is that the final answer should have the same number of significant figures as there are in the number with the fewest significant figures. The rule in addition and subtraction is that the answer is given the same number of decimal places as the term with the fewest decimal places.
Courses/Chabot_College/Chem_12A%3A_Organic_Chemistry_Fall_2022/13%3A_Properties_and_Reactions_of_Alcohols	13.1: Introduction to Structure and Synthesis of Alcohols Alcohols, phenols, enols, and carboxylic acids all contain hydroxyl groups. It is essential to distinguish between the three functional groups. Because of the structural similarities between alcohols and phenols, their similarities and differences are emphasized. 13.2: Classification of Alcohols Alcohols are classified by the bonding pattern of the carbon bonded to the hydroxyl group. Alcohol classification is helpful in discerning patterns of reactivity. 13.3: Physical Properties of Alcohols Alcohols are the first functional group we are studying in detail that is capable of H-bonding. The effects of increased polarity and stronger intermolecular forces on the physical properties of alcohols relative to alkanes are discussed. 13.4: Spectroscopy of Alcohols The hydroxyl group plays an important role in the spectroscopy of alcohols and phenols. 13.5: Synthesis of Alcohols - Review Through the first ten chapters, we have learned to synthesize alcohols from alkyl halides via nucleophilic substitution (SN2 & SN1) and from alkenes using a variety of pathways determined by regiochemistry and stereochemistry. Gentle oxidation of the alkenes can also be used to synthesize diols. 13.6: Acidity of Alcohols and Phenols Phenols are weakly acidic (pKa = 10) because of their resonance stabilized conjugate base, phenoxide. Alcohols are considered neutral with pKa values similar to water (pKa = 14). The concepts used to predict relative acidity are explained in Chapter 1. 13.7: Reduction of the Carbonyl Group - Synthesis of 1º and 2º Alcohols Aldehydes, ketones, carboxylic acids, and esters can all be reduced to form alcohols. An important pattern of chemical reactivity is introduced when we notice that aldehydes and ketones often use the same reagents, where as carboxylic acids and esters require different reagents to create similar reactivity. 13.8: Organometallic Reagents Grignard (RMgX) and organolithium (RLi) reagents are made from alkyl halides in aprotic solvents. 13.9: Organometallic Reagents in Alcohol Synthesis Organometallic reagents can react with aldehydes, ketones, acyl halides, esters, and epoxides to synthesize alcohols with an increased number of carbon atoms in the product. Building larger organic molecules is a useful skill for multi-step synthesis. 13.10: Additional Exercises Part 1 This section has additional exercises for the key learning objectives of the chapter. 13.11: Solutions to Additional Exercises Part 1 This section has the solutions to the Additional Practice Problems in the previous section. 13.12: Reactions of Alcohols with Hydrohalic Acids Alcohols react with hydrohalic acids (HCl, HBr, and HI) to form alkyl halides via the SN1 or SN2 mechanism as determined by the structure of the alcohol. Since hydroxide is a poor leaving group, acid catalysis is required. 13.13: Reactions with Phosphorus Halides and Thionyl Chloride Because of the limited synthetic utility of reacting alcohols with hydrohalic acids to form alkyl halides, the alternative reagents thionyl chloride and phosphorous tribromide. have been developed. 13.14: Dehydration Reactions of Alcohols Alcohols can form alkenes via the E1 or E2 pathway depending on the structure of the alcohol and the reaction conditions. Markovnokov's Rule still applies and carbocation rearrangements must be considered for the E1 mechanism. 13.15: Oxidation States of Alcohols and Related Functional Groups Organic chemistry requires an expanded definition of oxidation and reduction. 13.16: Oxidation Reactions of Alcohols Alcohols can be oxidized using acidified sodium or potassium dichromate(VI) solution. This reaction has been used historically as a way of distinguishing between primary, secondary and tertiary alcohols. 13.17: Protection of Alcohols During the synthesis of complex molecules, one functional group may interfere or complete with the reagent intended for a second functional group on the same molecule. There are several methods for protecting and subsequently recovering alcohols during multiple step syntheses of complex molecules. 13.18: Additional Exercises Part 2 This section has additional exercises for the key learning objectives of the chapter. 13.19: Solutions to Additional Exercises Part 2 This section has the solutions to the additional exercises from the previous section.
Courses/University_of_California_Davis/UCD_Chem_110A%3A_Physical_Chemistry__I/UCD_Chem_110A%3A_Physical_Chemistry_I_(Koski)/Text/08%3A_Multielectron_Atoms/8.07%3A_Hartree-Fock_Calculations_Give_Good_Agreement_with_Experimental_Data	Understand how the Hartree method is expanded to include symmetrized mulit-electron determential wavefunctions via the Hartree-Fock equations. Understand how to calculate the orbital energies from HF theory. Apply HF theory with Koopman's theory to estimate ionization energies and electron affinities. The Hartree method discussed previously is useful as an introduction to the solution of many-particle system and to the concepts of self-consistency and of the self-consistent-field calculations, but its importance is confined to the history of physics. In fact the Hartree method is not just approximate, it is fundamentally wrong since its wavefunction is not antisymmetric to electron permutation! The Hartree-Fock approach discussed below is a better approach, which correctly takes into account the antisymmetric character of the trial wavefunctions. Although the Hartree equations are numerically tractable via the self-consistent field method, it is not surprising that such a crude approximation fails to capture elements of the essential physics. The Pauli exclusion principle demands that the many-body wavefunction be antisymmetric with respect to interchange of any two electron coordinates, e.g. \[\Psi(\mathbf{r}_{1},\mathbf{r}_{2}, \ldots, \mathbf{r}_{N}) = - \Psi(\mathbf{r}_{2},\mathbf{r}_{1}, \ldots, \mathbf{r}_{N}) \label{2.7} \] which clearly cannot be satisfied by the multi-electron wavefunctions of the form used in the Hartree Approximation, i.e., the orbital approximation (Equation $\ref{2.3}$). \[\Psi(\mathbf{r}_1,\mathbf{r}_2, \ldots, \mathbf{r}_N) \approx \psi_{1}(\mathbf{r}_1)\psi_{2}(\mathbf{r}_2) \ldots \psi_{N}(\mathbf{r}_N) \label{2.3} \] This indistinguishability condition can be satisfied by forming a Slater determinant of single-particle orbitals \[\Psi(\mathbf{r}_{1}, \mathbf{r}_{2}, \ldots, \mathbf{r}_{N})= \dfrac{1}{\sqrt{N}} \left \vert\psi(\mathbf{r}_{1})\psi(\mathbf{r}_{2}) \ldots \psi(\mathbf{r}_{N}) \right\vert \label{2.8} \] This decouples the electrons resulting in $N$ single-particle Hartree-Fock equations: \[\underbrace{-\dfrac{\hbar^{2}}{2m}\nabla^{2}\psi_{i}(\mathbf{r})}_{\text{kinetic energy}} + \underbrace{V_{nucleus}(\mathbf{r})\psi_{i}(\mathbf{r})}_{\text{electron-nucleus potential}} + \underbrace{V_{electron}(\mathbf{r})\psi_{i}(\mathbf{r})}_{\text{Hartree Term}} - \sum_{j} \int d\mathbf{r}^{\prime} \dfrac{\psi^{\star}_{j}(\mathbf{r}') \psi^{\star}_{i}(\mathbf{r}') \psi_{j}(\mathbf{r}) } {\left \vert \mathbf{r} - \mathbf{r}^{\prime} \right\vert} = \epsilon_{i}\psi_{i}(\mathbf{r}). \label{2.9} \] As with the Hartree equations , the first term is the kinetic energy of the $i^{th}$ electron \[-\dfrac{\hbar^{2}}{2m}\nabla^{2}\psi_{i}(\mathbf{r}) \nonumber \] and the second term is the electron-nucleus potential between the $i^{th}$ electron and nucleus \[V_{nucleus}(\mathbf{r})\psi_{i}(\mathbf{r}) \nonumber \] The third term (sometimes called the “Hartree” term) is the electrostatic potential between the $i^{th}$ electron and the average charge distribution of the other $N-1$ electrons. \[V_{electron}(\mathbf{r})\psi_{i}(\mathbf{r}) = J_{j,k} = \int \|\phi_j(r)\|^2 \|\phi_k(r’)\|^2 \dfrac{e^2}{r-r'} dr dr’ \label{8.3.9} \] These three terms are identical to Hartree Equations with the product wavefunction ansatz (i.e., orbital approximation). The fourth term of Equation $\ref{2.9}$ is not in the Hartree Equations: \[\sum_{j} \int d\mathbf{r}^{\prime} \dfrac{\psi^{\star}_{j}(\mathbf{r}') \psi^{\star}_{i}(\mathbf{r}') \psi_{j}(\mathbf{r}) } {\left \vert \mathbf{r} - \mathbf{r}^{\prime} \right\vert} \nonumber \] and is the exchange term. This term resembles the direct Coulomb term, but for the exchanged indices. It is a manifestation of the Pauli exclusion principle, and acts so as to separate electrons of the same spin. This “exchange” term acts only on electrons with the same spin and comes from the Slater determinant form of the wavefunction (Equation \ref{2.8}). Physically, the effect of exchange is for like-spin electrons to avoid each other. The exchange term adds considerably to the complexity of these equations. The Hartree-Fock Equations in Equation $\ref{2.9}$ can be recast as series of Schrödinger-like equations: \[ \hat {F} \| \varphi _i \rangle = \epsilon _i\| \varphi _i \rangle \label {8.7.2} \] where $\hat {F}$ is called the Fock operator and $ \{\| \varphi_i \rangle \}$ are the Hatree-Fock orbitals with corresponding energies $\epsilon_i$. The Fock operator is a one-electron operator and solving a Hartree-Fock equation gives the energy and Hartree-Fock orbital for one electron. For a system with 2N electrons, the variable i will range from 1 to N; i.e there will be one equation for each orbital. The reason for this is that only the spatial wavefunctions are used in Equation $\ref{8.7.2}$. Since the spatial portion of an orbital can be used to describe two electrons, each of the energies and wavefunctions found by solving Equation $\ref{8.7.2}$ will be used to describe two electrons. The nature of the Fock operator reveals how the Hartree-Fock (HF) or Self-Consistent Field (SCF) Method accounts for the electron-electron interaction in atoms and molecules while preserving the idea of independent atomic orbitals . The wavefunction written as a Slater determinant of spin-orbitals is necessary to derive the form of the Fock operator, which is \[\begin{align} \hat {F} &= \hat {H} ^0 + \sum _{j=1}^N ( 2 \hat {J} _j - \hat {K} _j ) \nonumber \\[4pt] &= -\dfrac {\hbar ^2}{2m} \nabla ^2 - \dfrac {Ze^2}{4 \pi \epsilon _0 r} + \sum _{j=1}^N (2\hat {J}_j - \hat {K} _j ) \label {8.7.3} \end{align} \] As shown by the expanded version on the far right, the first term in this equation, $\hat {H}^0$, is the familiar hydrogen-like operator that accounts for the kinetic energy of an electron and the potential energy of this electron interacting with the nucleus. The next term accounts for the potential energy of one electron in an average field created by all the other electrons in the system. The $\hat {J}$ and $\hat {K}$ operators result from the electron-electron repulsion terms in the full Hamiltonian for a multi-electron system. These operators involve the one-electron orbitals as well as the electron-electron interaction energy. The Fock operator (Equation $\ref{8.7.3}$) depends on all occupied orbitals (because of the exchange and Coulomb operators). Therefore, a specific orbital can only be determined if all the others are known. One must use iterative methods to solve the HF equations like the Self-consistent field method discussed previously for the Hartree Approximation. The exchange interaction is a quantum mechanical effect that only occurs between identical particles. Despite sometimes being called an exchange force in analogy to classical force, it is not a true force , as it lacks a force carrier. The effect is due to the wavefunction of indistinguishable particles being subject to exchange symmetry, that is, either remaining unchanged (symmetric) or changing its sign (antisymmetric) when two particles are exchanged. Both bosons and fermions can experience the exchange interaction. For fermions, it is sometimes called Pauli repulsion and related to the Pauli exclusion principle. For bosons, the exchange interaction takes the form of an effective attraction that causes identical particles to be found closer together, as in Bose–Einstein condensation. For example, the electron 1 in helium (with $Z=2$), then \[\hat {H}^0 (1) = - \dfrac {\hbar ^2}{2m} \nabla ^2_1 - \dfrac {2e^2}{4 \pi \epsilon _0 r_1} \nonumber \nonumber \] The Fock operator is couched in terms of the coordinates of the one electron whose perspective we are taking (which we will call electron 1 throughout the following discussion), and the average field created by all the other electrons in the system is built in terms of the coordinates of a generic “other electron” (which we’ll call electron 2) that is considered to occupy each orbital in turn during the summation over the $N$ spatial orbitals. The best possible one-electron wavefunctions, by definition, will give the lowest possible total energy for a multi-electron system used with the complete multielectron Hamiltonian to calculate the expectation value for the total energy of the system. These wavefunctions are called the Hartree-Fock wavefunctions and the calculated total energy is the Hartree-Fock energy of the system. As with the Hartree Equations, solving the Hartree-Fock Equations is mathematically equivalent to assuming each electron interacts only with the average charge cloud of the other electrons. This is how the electron-electron repulsion is handled. This also why this approach is also called the Self-Consistant Field (SCF) approach. Hartree-Fock Energy The Hartree-Fock equations $h_e \phi_i = \epsilon_i \phi_i$ imply that the orbital energies $\epsilon_i$ can be written as: \[ \begin{align} \epsilon_i &= \langle \phi_i \| h_e \| \phi_i \rangle \\[4pt] &= \langle \phi_i \| T + V \| \phi_i \rangle + \sum_{j({\rm occupied})} \langle \phi_i \| J_j - K_j \| \phi_i \rangle \label{8.7.6} \\[4pt] &= \langle \phi_i \| T + V \| \phi_i \rangle + \sum_{j({\rm occupied})} [ J_{i,j} - K_{i,j} ],\label{8.7.7} \end{align} \] where $T + V$ represents the kinetic ($T$) and nuclear attraction ($V$) energies, respectively. Thus, $\epsilon_i$ is the average value of the kinetic energy plus Coulombic attraction to the nuclei for an electron in $\phi_i$ plus the sum over all of the spin-orbitals occupied in $\psi$ of Coulomb minus Exchange interactions of these electrons with the electron in $\phi_i$. If $\phi_i$ is an occupied spin-orbital, the $j = i$ term $[ J_{i,i} - K_{i,i}]$ disappears in the above sum and the remaining terms in the sum represent the Coulomb minus exchange interaction of $\phi_i$ with all of the $N-1$ other occupied spin-orbitals. If $\phi_i$ is a virtual spin-orbital, this cancelation does not occur because the sum over $j$ does not include $j = i$. So, one obtains the Coulomb minus exchange interaction of $\phi_i$ with all $N$ of the occupied spin-orbitals in $\psi$. Hence the energies of occupied orbitals pertain to interactions appropriate to a total of $N$ electrons, while the energies of virtual orbitals pertain to a system with $N+1$ electrons. This difference is very important to understand and to keep in mind. To give an idea of how well HF theory can predict the ground state energies of several atoms, consider Table 8.7.1 below: Atom Hartree-Fock Energy Experiment $He$ $-5.72$ $-5.80$ $Li$ $-14.86$ $-14.96$ $Ne$ $-257.10$ $-257.88$ $Ar$ $-1053.64$ $-1055.20$ Koopmans' Theorem Koopmans' theorem states that the first ionization energy is equal to the negative of the orbital energy of the highest occupied molecular orbital. Hence, the ionization energy required to generated a cation and detached electron is represented by the removal of an electron from an orbital without changing the wavefunctions of the other electrons. This is called the "frozen orbital approximation." Let us consider the following model of the detachment or attachment of an electron in an $N$-electron system. In this model, both the parent molecule and the species generated by adding or removing an electron are treated at the single-determinant level. The Hartree-Fock orbitals of the parent molecule are used to describe both species. It is said that such a model neglects orbital relaxation (i.e., the re-optimization of the spin-orbitals to allow them to become appropriate to the daughter species). Within this model, the energy difference between the daughter and the parent can be written as follows ($\phi_k$ represents the particular spin-orbital that is added or removed: for electron detachment (vertical ionization energies) \[ E_{N-1} - E_N = - \epsilon_k \label{8.7.8} \] and for electron attachment (electron affinities) \[ E_N - E_{N+1} = - \epsilon_k .\label{8.7.9} \] The Hartree-Fock equations deal with exchange exactly; however, the equations neglect more detailed correlations due to many-body interactions. The effects of electronic correlations are not negligible; indeed the failure of Hartree-Fock theory to successfully incorporate correlation leads to one of its most celebrated failures. Let’s derive this result for the case in which an electron is added to the $N+1^{st}$ spin-orbital. The energy of the $N$-electron determinant with spin-orbitals $\phi_1$ through $f_N$ occupied is \[E_N = \sum_{i=1}^N \langle \phi_i \| T + V \| \phi_i \rangle + \sum_{i=1}^{N} [ J_{i,j} - K_{i,j} ] \nonumber \] which can also be written as \[E_N = \sum_{i=1}^N \langle \phi_i \| T + V \| \phi_i \rangle + \frac{1}{2} \sum_{i,j=1}^{N} [ J_{i,j} - K_{i,j} ].\nonumber \] Likewise, the energy of the $N+1$-electron determinant wavefunction is \[E_{N+1} = \sum_{i=1}^{N+1} \langle \phi_i \| T + V \| \phi_i \rangle + \frac{1}{2} \sum_{i,j=1}^{N+1} [ J_{i,j} - K_{i,j} ]. \nonumber \] The difference between these two energies is given by \[ \begin{align} E_{N} – E_{N+1} = &- \langle \phi_{N+1} \| T + V \| \phi_{N+1} \rangle - \frac{1}{2} \sum_{i=1}^{N+1} [ J_{i,N+1} - K_{i,N+1} ] \\[4pt] &- \frac{1}{2} \sum_{j=1}^{N+1} [ J_{N+1,j} - K_{N+1,j} ] \\[4pt] &= - \langle \phi_{N+1} \| T + V \| \phi_{N+1} \rangle - \sum_{i=1}^{N+1} [ J_{i,N+1} - K_{i,N+1} ] \\[4pt] &= - \epsilon_{N+1}. \end{align} \] That is, the energy difference is equal to minus the expression for the energy of the $N+1^{st}$ spin-orbital, which was given earlier. In the Copenhagen Interpretation, the squared modulus of the wavefunction gives the probability of finding a particle in a given place. The many-body wavefunction gives the N-particle distribution function, i.e. $\|Φ(r_1, ..., r_N )\|^2$ is the probability density that particle 1 is at $r_1$, ..., and particle $N$ is at $r_N$. However, when trying to work out the interaction between electrons, what we want to know is the probability of finding an electron at $r$, given the positions of all the other electrons $\{r_i\}$. This implies that the electron behaves quantum mechanically when we evaluate its wavefunction, but as a classical point particle when it contributes to the potential seen by the other electrons. The contributions of electron-electron interactions in N-electron systems within the Hartree and Hartree-Fock methods are shown in Figure 8.7.2 . The conditional electron probability distributions $n(r)$ of $N-1$ electrons around an electron with given spin situated at $r=0$. Within the Hartree approximation, all electrons are treated as independent, therefore $n(r)$ is structureless. However, within the Hartree-Fock approximation, the $N$-electron wavefunction reflects the Pauli exclusion principle and near the electron at $r=0$ the exchange hole can be seen where the the density of spins equal to that of the central electron is reduced. Electrons with opposite spins are unaffected (not shown). Summary So, within the limitations of the HF, frozen-orbital model, the ionization potentials (IPs) and electron affinities (EAs) are given as the negative of the occupied and virtual spin-orbital energies, respectively. This statement is referred to as Koopmans’ theorem; it is used extensively in quantum chemical calculations as a means of estimating ionization potentials (Equation $\ref{8.7.8}$) and Electron Affinities (Equation $\ref{8.7.9}$) and often yields results that are qualitatively correct (i.e., ± 0.5 eV). In general Hartree-Fock theory gives a great first order solution (99%) to describing multi-electron systems, but that last 1% is still too great for quantitatively describing many aspects of chemistry and more sophisticated approaches are necessary. These are discussed elsewhere.
Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Exemplars/Membrane_Potentials	Membrane potential is what we use to describe the difference in voltage (or electrical potential) between the inside and outside of a cell. Introduction Without membrane potentials human life would not be possible. All living cells maintain a potential difference across their membrane. Simply stated, membrane potential is due to disparities in concentration and permeability of important ions across a membrane. Because of the unequal concentrations of ions across a membrane, the membrane has an electrical charge. Changes in membrane potential elicit action potentials and give cells the ability to send messages around the body. More specifically, the action potentials are electrical signals; these signals carry efferent messages to the central nervous system for processing and afferent messages away from the brain to elicit a specific reaction or movement. Numerous active transports embedded within the cellular membrane contribute to the creation of membrane potentials, as well as the universal cellular structure of the lipid bilayer. The chemistry involved in membrane potentials reaches to many scientific disciplines. Chemically it involves molarity, concentration, electrochemistry and the Nernst equation. From a physiological standpoint, membrane potential is responsible for sending messages to and from the central nervous system. It is also very important in cellular biology and shows how cell biology is fundamentally connected with electrochemistry and physiology. The bottom line is that membrane potentials are at work in your body right now and always will be as long as you live. History The subject of membrane potential stretches across multiple scientific disciplines; Membrane Potential plays a role in the studies of Chemistry, Physiology and Biology. The culmination of the study of membrane potential came in the 19th and early 20th centuries. Early in the 20th century, a man named professor Bernstein hypothesized that there were three contributing factors to membrane potential; the permeability of the membrane and the fact that [K+] was higher inside and lower on the outside of the cell. He was very close to being correct, but his proposal had some flaws. Walther H. Nernst, notable for the development of the Nernst equation and winner of 1920 Nobel Prize in chemistry, was a major contributor to the study of membrane potential. He developed the Nernst equation to solve for the equilibrium potential for a specific ion. Goldman, Hodgkin and Katz furthered the study of membrane potential by developing the Goldman-Hodgkin-Katz equation to account for any ion that might permeate the membrane and affect its potential. The study of membrane potential utilizes electrochemistry and physiology to formulate a conclusive idea of how charges are separated across a membrane. Figure 1. Differences in concentration of ions on opposite sides of a cellular membrane produce a voltage difference called the membrane potential. The largest contributions usually come from sodium (Na + ) and chloride (Cl – ) ions which have high concentrations in the extracellular region, and potassium (K + ) ions, which along with large protein anions have high concentrations in the intracellular region. Calcium ions, which sometimes play an important role, are not shown. Membrane Potential and Cellular Biology In discussing the concept of membrane potentials and how they function, the creation of a membrane potential is essential. The lipid bilayer structure of the cellular membrane, with its lipid-phosphorous head and fatty acid tail, provides a perfect building material that creates both a hydrophobic and hydrophilic side to the cellular membrane. The membrane is often referred to as a mosaic model because of its semi-permeability and its ability to keep certain substances from entering the cell. Molecules such as water can diffuse through the cell based on concentration gradients; however, larger molecules such as glucose or nucleotides require channels. The lipid bilayer also houses the Na + /K + pump, ATPase pump, ion transporters, and voltage gated channels, and it is the site of vesicular transport. The structure regulates which ions enter and exit to determine the concentration of specific ions inside of the cell. Why is membrane potential essential to the survival of all living creatures? Animals and plants require the breakdown of organic substances through cellular respiration to generate energy. This process, which produces ATP, is dependent on the electron transport chain. Electrons travel down this path to be accepted by oxygen or other electron acceptors. The initial electrons are obtained from the breakdown of water molecules. The hydrogen build up in the extracellular fluid leaving a gradient. As per membrane potentials, when there a gradient, the molecules flow in the opposite direction. In this case, hydrogen flows back into the cell through a protein known as ATP synthase which creates ATP in the process. This action is essential to life because the number of ATP created from each glucose increases drastically. Chemical disequilibrium and membrane potentials allow bodily functions to take place. Figure 2 Transport proteins, more specifically the 'active' transport proteins, can pump ions and molecules against their concentration gradient. This is the main source of charge difference across the cellular membrane. Physiology of Membrane Potential Understanding Membrane Potential The following points should help you to understand how membrane potential works The difference between the electrical and chemical gradient is important. Electrical Gradient Opposes the chemical gradient. Represents the difference in electrical charge across the membrane Chemical Gradient Opposes the electrical gradient Represents the difference in the concentration of a specific ion across the membrane. A good example is K + . The membrane is very permeable to K + and the [K + ] inside the cell is great, therefore a positive charge is flowing out of the cell along with K + . The [K + ] inside the cell decreases causing the concentration gradient to flow towards the outside of the cell. This also causes the inside of the cell to become more electronegative increasing its electrical gradient. The Nernst equation can help us relate the numerical values of concentration to the electrical gradient. Leak Channels Channels that are always open Permit unregulated flow of ions down an electrochemical gradient. Na + /K + ATPase Pump Actively transports Na + out of the cell and K + into the cell. Helps to maintain the concentration gradient and to counteract the leak channels. Membrane Potential and Physiology of Human Nerve Cells Human nerve cells work mainly on the concept of membrane potentials. They transmit chemicals known as serotonin or dopamine through gradients. The brain receives these neurotransmitters and uses it to perform functions. Na + has a much higher concentration outside of the cell and the cell membrane is very impermeable to Na + K + has a high concentration inside the cell due to the fact that the cell membrane is very permeable to K + A - is used to refer to large ions that are found completely inside of the cell and cannot penetrate the cell membrane. Concentration (in Millimoles/ Liter) and permeability of Ions Responsible for Membrane Potential in a Resting Nerve Cell ION Extracellular Intracellular Relative Permeability Na+ 150 15 1 K+ 5 150 25-30 A- 0 65 0 Check out this YouTube video if you want to know more about how the Na + /K + pump and how the membrane potential works. www.youtube.com/watch?v=iA-Gdkje6pg How To Calculate A Membrane Potential The calculation for the charge of an ion across a membrane, The Nernst Potential, is relatively easy to calculate. The equation is as follows: (RT/zF) log([X] out /[X] in ). RT/F is approximately 61, therefore the equation can be written as (61/z) ln([X] out /[X] in ) R is the universal gas constant (8.314 J.K -1 .mol -1 ). T is the temperature in Kelvin (°K = °C + 273.15). z is the ionic charge for an ion. For example, z is +1 for K + , +2 for Mg 2 + , -1 for F - , -1 for Cl - , etc. Remember, z does not have a unit. F is the Faraday's constant (96485 C.mol -1 ). [ X ] out is the concentration of the ion outside of the species. For example the molarity outside of a neuron. [ X ] in is the concentration of the ion inside of the species. For example, the molarity inside of a neuron. The only difference in the Goldman-Hodgkin-Katz equation is that is adds together the concentrations of all permeable ions as follows (RT/zF) log([K + ] o +[Na + ] o +[Cl - ] o /[K + ] i +[Na + ] i +[Cl - ] i ) Figure 3. (Clockwise From Upper Left) 1) The charges are equal on both sides; therefore the membrane has no potential. 2)There is an unbalance of charges, giving the membrane a potential. 3) The charges line up on opposite sides of the membrane to give the membrane its potential. 4) A hypothetical neuron in the human body; a large concentration of potassium on the inside and sodium on the outside. References Kaiser, Chris A., et al. Molecular Cell Biology. 6th ed. New York. W. H. Freeman, 2007. Orians, Gordon H., et al. Life: The Science of Biology. 8th ed. Gordonville, VA. Sinaver Associates, Inc., 2008., Petrucci, Ralph H., et al. General Chemistry: Principles and Modern Applications. 9th ed. New Jersey. Pearson Education International, 2007.Fuel Cells# Sherwood, Lauralee. Human Physiology: From Cells to Systems (International Edition) . International ed ed. New York: Brooks Cole, 2009. Print. Hietler, W.J.. "Membrane Potential Tutorial." St. Andrews Biology Dept. . St. Andrews University., 13 Aug. 2007. Web. 24 May 2010. <http:/ http://www.st-andrews.ac.uk/~wjh/neurotut/ Problems 1. List the following in order from highest to lowest permeability. A - , K + , Na + 2. Which of the following statement is NOT true? The membrane potential usually requires a minimal difference of electrocharges across the membrane Membrane impermeability plays a role in membrane potentials. Membrane potential exists in all cellular structures, except for neurons. The active transports play a vital role in membrane potentials. 3. What would be the equilibrium potential for the ion K + be if [K + ] in = 5mM and [K + ] in =150mM? 4. True or false: At resting membrane potential, the inside of the membrane is slightly negatively charged while the outside is slightly positively charged. Answers: 1. K + > Na + > A - 2. Answer c.) is not true; membrane potential exists in neurons and is responsible for action potential propagation in neurons. 3. E k + = (61/z) log([K + ] out /[K + ] in ) = (61/1) log([5mM]/[150mM]) = -90mV z=1 4. True. The resting membrane potential is negative as a result of this disparity in concentration of charges. Contributors Dan Chong, Matt Klingler (UCD)
Courses/Saint_Marys_College_Notre_Dame_IN/CHEM_431%3A_Inorganic_Chemistry_(Haas)/CHEM_431_Readings/10%3A_Coordination_Chemistry_I_-_Structure_and_Isomers/10.03%3A_Isomerism	Metal complexes present a rich, interesting, and diverse structural chemistry. Major points of variation in the structural chemistry of metal complexes include coordination number and coordination geometry , which involve differences in how many ligands surround a central metal and their overall geometric arrangement. Examples of the latter include the tetrahedral and square planar geometries commonly observed for four-coordinate metal centers. Tetrahedral, square planar, and octahedral coordination serve as a backdrop to the discussion of isomerism on this page. Readers who are unfamiliar with these structures might consider reading the section on coordination geometries before this one. structural iosmerism involves differences in how potential ligand atoms are bound to metals in a complex. Possibilities for structural isomerism include linkage or ambidentate isomerism hydrate/solvate isomerism ionization isomerism coordination isomerism. Of these, linkage isomerism should always be considered when working with ambidentate ligands. As classifications, the last three forms of structural isomerism are mainly of academic interest since they represent permutations of ordinary structure patterns for solvates, salts, and coordination complexes, respectively. stereochemistry , which includes which coordination geometry is adopted for a metal with a given set of ligands in a particular oxidation state. For instance, [Ni II Cl 4 ] 2- is tetrahedral while [Pt II Cl 4 ] 2- is square planar. Since this geometry is usually fixed by the metal and ligands it is typically not a source of stereoisomeric varierty. metal-centric stereoisomerism involving possible variations in where ligands are located relative to one another around the metal. The possibilities depend on the coordination number, geometry, and number of different types of ligands. For instance, square planar complexes can exhibit cis-/trans- isomerism while tetrahedral ones cannot. Tetrahedral complexes with four different ligands exhibit R and S chirality but complexes of that type are relatively rare. The more common cases include cis and trans isomerism in square planar complexes and cis & t rans -, mer & fac -, and $\Lambda$ & $\Delta$ isomerism in octahedral ones, although some multidentate ligands present additional possibilities for stereochemical variation Stereoisomerism centered on the ligands bound to a metal center. Possibilities include stereoisomerism inherent to a ligand. This would include cases where the ligand itself is chiral or, as in the case of amines, exists as a rapidly interconverting mixture of isomers. Thus this sort of stereoisomerism is an extension of the isomerism encountered in ordinary organic and other main group compounds. The main new issues introduced by binding of such ligands to a metal center involve the creation of new possibilities for diastereoisomerism based on the stereochemistry of the metal center and the freezing out of stereocenter inversion on binding to a metal. The latter is particularly important for amines, which as free amines racemize rapidly by nitrogen inversion. or conformational isomerism involving five-membered chelate rings created when a multidentate ligand binds to a metal. This isomerism is often called chelate ring twist since individual rings can exhibit one of two conformations depending on how they twist on binding. A summary of these forms of isomerism is given in Figure $\sf{\PageIndex{1}}$. Structural isomers Structural isomerism involves different topological linkages of atoms. Differences in atomic linkages distinctive to coordination chemistry involve the linkages between metals and ligands. The main variations are: 1. Hydrate/solvate isomerism Solvate isomers differ in terms of whether a molecule acts as a ligand or whether it acts as a solvate by occupying a lattice site in the crystal. Among solvate isomers, the case in which water is the ligand or solvate is the best known. The resulting isomers are called hydrate isomers after the term for water acting as a solvate, hydrate. A well-known example that illustrates how solvate isomerism works is the series [CrCl x (H 2 O) 6-x ]Cl 3 - x ·xH 2 O, for which x = 0-2. The structures of the complex ions involved in this series are shown in Figure $\sf{\PageIndex{2}}$. 1 As can be seen from the compounds in Figure $\sf{\PageIndex{2}}$, hydrate isomers differ both in terms of whether water acts as a ligand or hydrate and in terms of whether a potential counterion acts as a counterion or ligand. Thus, in trans- [CrCl 2 (H 2 O) 4 ]Cl·2H 2 O two chlorides act as chloro ligands and four waters as aqua ligands while in [CrCl(H 2 O) 5 ]Cl 2 ·H 2 O five water molecules act as aqua ligands while only one chloride acts as a chloro ligand. In this way, solvate isomers simply represent cases in which two or more of the possible permutations for metal ligand binding among a set of solvent and counterion molecules are stable. 2. Ionization isomerism In ionization isomerism there are two or more potential ions that can act as ligands. The ionization isomers differ in terms of which of these ions act as counterions and which act as ligands. Consider, for instance, the complexes shown in Figure $\sf{\PageIndex{3}}$. These complexes differ in terms of whether the chloride or sulfate acts as a ligand, with the other acting as a counterion. 3. Coordination isomerism Coordination isomers exist in compounds containing two or more complexes, each of which possesses a different set of ligands that can in principle be swapped with a ligand of the other complex. An example involves [Co III (NH 3 ) 6 ][Cr III (ox) 3 ], depicted in Figure $\sf{\PageIndex{4A}}$. 1 Coordination isomers of this complex involve swapping the ammine ligands around the Co 3 + center for oxalato ligands surrounding the Cr 3 + center. For instance, swapping all the ammine and oxalato ligands between the metal centers gives the coordination isomer [Cr III (NH 3 ) 6 ][Co III (ox) 3 ] shown in Figure $\sf{\PageIndex{4B}}$. As classifications, hydrate/solvate, ionization, and coordination isomerism represent permutations of ordinary structure patterns for solvates, salts, and coordination complexes, respectively. As a result these forms of isomerism are rarely used as independent conceptual frameworks when thinking and talking about the structure of coordination compounds. Not so the final type of structural isomerism, linkage isomerism. That is because linkage isomerism has to do with the special capacity of ambidentate ligands to bind metals in multiple ways. 4. Linkage or ambidentate isomerism Linkage or ambidentate isomers differ in how one or more ambidentate ligands bind to metal centers in a complex. The classic example dating from the work of Jørgensen and Werner is given in Scheme $\sf{\PageIndex{I}}$. Just as alkenes exist as E and Z isomers, compounds possessing ambidentate ligands exist as one among the possible linkage isomers. As such, when working with ambidentate ligands the particular linkage isomer formed should be determined experimentally and considered when interpreting the complex's chemical and physical behavior. The main ambidentate ligands which give rise to linkage isomerism are cyanide, CN - thiocyanate, SCN - , and the O and Se analogues, OCN - and SeCN - nitrite, NO 2 - sulfite, SO 3 - nitrosyl, NO Although ambidentate ligands can bind metals in multiple ways, most exhibit a preferred binding mode ( i.e., prefer to bind metal centers in one of the possible ways). For instance, cyanide almost always binds through its carbon atom and thiocyanate almost always binds through its nitrogen ($\mu$- N ). However, the other binding mode can sometimes be formed kinetically or by using conditions that particularly favor its formation. Thus thiocyanate forms M-SCN - linkages in the presence of exceptionally soft metal centers or in hard polar solvents (in which Lewis acid groups can stablize the terminal nitrogen of a bound thiocyanate ligand). Ambidentate ligands are of significant research interest because many don't just bind to metal centers in two ways; the linkage isomers sometimes have significantly different physical properties. In addition, the possibility of two binding modes introduces the possibility of exploiting linkage isomerism to take advantage of some of these ligands: different structural chemistry in different coordination modes. In mononuclear complexes the coordination mode influences which side of the ligand faces away from the complex and might be susceptible to stabilization by interaction with solvent. Additionally, the orientation of the ligand relative to the metal center might differ between one coordination mode and another. For instance, as predicted from the minor contributor to its resonance structure, thiocyanate binds nonlinearly via its S atom and linearly via its N (Figure $\sf{\PageIndex{5}}$). Because of this the apparent steric bulk of the ligand around the metal center might differ between forms. bind two different metal centers at the same time, linking them together. Perhaps the best-known examples involve thiocyanate and cyanide. The latter forms linkages of the type M-C≡N-M'. In the dye Prussian blue these take the form Fe II -C≡N-Fe III . An example from the author's graduate research involved tetrahedral clusters containing four metal atoms linked by six cyano ligands, as shown in Figure $\sf{\PageIndex{6}}$. can be induced to change from one binding mode to another in response to a stimulus. The classic example involves the light-driven transformation of the thermodynamically more stable yellow nitro complex of pentamminecobalt(III), [Co(NH 3 ) 5 NO 2 ] 2+ to the less stable red O -nitrito complex, [Co(NH 3 ) 5 ONO] 2+ . Less stable cyanometallate coordination networks like those in the red K{Fe II [Cr III (CN) 6 ]} transform into the more stable green K{Fe II [Cr III (NC) 6 ]} form on heating, as shown in Figure $\sf{\PageIndex{7}}$. 3 Exercise $\PageIndex{1}$: Bridging ambidentate ligands and the Hard-Soft Acid- Base Principle. The hard-soft acid-base principle helps explain the preference of bridging ambidentate ligands for particular binding modes. How might the greater stability of the green K{Fe II [Cr III (NC) 6 ]} over red K{Fe II [Cr III (CN) 6 ]} be explained in terms of the hard and soft acid-base concept? Answer The greater stability of K{Fe II [Cr III (NC) 6 ]} over K{Fe II [Cr III (CN) 6 ]} reflects the greater stability of the Fe II -CN-Cr III linkages in the former over the Fe II -NC-Cr III linkages in the latter . This is consistent with the preference for hard-hard and soft-soft Lewis acid-base interactions of the Hard-Soft Acid-Base Principle. The Fe II -CN-Cr III linkages are more stable because they possess bonds between the softer Lewis acid (Fe II ) and Lewis base (the C end of CN - ) the harder Lewis acid (Cr III ) and Lewis base (the N end of CN - ) In contrast, the Fe II -NC-Cr III linkages in the less stable linkage isomer involve bonds between the softer base (the C end of CN - ) and harder acid (Cr III ) the harder base (the N end of CN - ) and softer acid (Fe II ) Stereoisomerism Optical Isomerism/Chirality Molecules of D n, C n , or C 1 symmetry with only proper rotation axes (including E = C 1 ) are chiral and exhibit optical isomerism. As described in Figure $\sf{\PageIndex{1}}$, the main sources of such optical isomerism in coordination chemistry are: Chirality inherent to an organic or main group ligand. This type of isomerism is just an extension of the sort described in undergraduate organic texts and consequently does not merit separate discussion here, other than to note that some forms of optical isomerism which are of little importance in organic chemistry lead to optical activity in coordination compounds. In particular, the nitrogen inversion process which serves to rapidly racemize chiral amines is frozen out on formation of a metal-ligand bond. Because of this chiral amine, ligands bound to a metal form non-interconvertible R and S enantiomers, as shown in Figure $\sf{\PageIndex{8}}$. Chirality arising from the symmetry of ligands about a metal center. The two common situations through which such chirality arises involve: Chirality at a tetrahedral metal center surrounded by four different ligand attachment points. Such cases are often referred to as MABCD or Mabcd where M stands for the metal and a, b, c, and d the four different ligand attachment points. An example of such a complex is given in Figure $\sf{\PageIndex{9A}}$. The chirality of such complexes is analogous to the chirality arising from a tetrahedral carbon stereocenter. There are two caveats, though. First, it is not enough to have a four-coordinate complex; the metal must possess tetrahedral symmetry since, as shown in Figure $\sf{\PageIndex{9B}}$, square planar M complexes with four different ligands are identical to their mirror images. Second, such chirality is most easily realized using macrocyclic ligands like proteins. Metal ligand bonds involving sterically unhindered monodentate complexes like that in the example of Figure $\sf{\PageIndex{9A}}$ are in general weaker and more flexible than analogous carbon-carbon bonds. Consequently, in such cases it is likely that the enantiomers would interconvert through a fluxional process, making their resolution difficult if not impossible. $\Lambda$ and $\Delta$ isomerism at octahedral metal centers surrounded by two or three chelating ligands. Specifically, Octahedral tris-chelates of formula M(L~L) 3 exhibit D 3 symmetry. cis-octahedral bis-chelates of formula M(L~L) 2 XY exhibit C 2 or C 1 symmetry: C 2 if X = Y and C 1 if X $\neq$ Y. As shown in Figure $\sf{\PageIndex{10A}}$, octahedral tris-chelates like [V(ox) 3 ] 3- are chiral and can exist as nonsuperimposable $\Lambda$ and $\Delta$ enantiomers. The chirality of these tris-chelates is analogous to that of pinwheels. As shown in Figure $\sf{\PageIndex{10B}}$, when looking directly at the head of any individual pinwheel from above, the blades will be angled from back to front going around the pinwheel in either the clockwise or counterclockwise direction. As shown in Figure $\sf{\PageIndex{10C}}$, the case for which the blades are angled in the clockwise direction is analogous to the $\Lambda$ enantiomer and the case when they are angled in the counterclockwise direction the $\Delta$. The $\Lambda$ and $\Delta$ chirality in octahedral bis-chelates is analogous to that in the tris-chelates except this time the third chelating ligand is replaced by an arbitrary pair of ligands. This lowers the symmetry of the complex to either C 2 or C 1 , as shown for the C 2 complex [CoCl 2 (en) 2 ] + in Figure $\sf{\PageIndex{11A}}$. In either case the result is that such complexes exist as $\Lambda$ and $\Delta$ enantiomers analogous to those in the metal tris-chelates, as shown in Figure $\sf{\PageIndex{11B}}$. Octahedral complexes containing ligands with denticities of four or more also exhibit $\Lambda$ and $\Delta$ chirality; it is just that in such cases the bound ligand defines multiple rings within the complex that can be defined as existing in either $\Lambda$ or $\Delta$ orientations relative to each other. The procedure for making these assignments is beyond the scope of most introductory inorganic courses but the result is that such complexes are designated as $\Lambda\Lambda$, $\Delta\Delta$, $\Delta\Lambda$, $\Delta\Delta\Delta$, etc., depending on the number of relationships involving the rings defined by the bound ligand. Interested readers should consult reference 4 for more details. Circular dichroism may be used to characterize optically active metal complexes. At the present the most definitive way to determine the absolute configuration of an optically active coordination compound is to determine its 3D molecular structure using single crystal X-ray crystallography. 5 However, as that is not always possible or convenient, the optical activity of coordination complexes is also commonly studied using circular dichroism (CD) spectroscopy. CD spectroscopy is used because, unlike most chiral organics, optically active coordination complexes possess low-energy electronic transitions that occur in the far UV and visible range. This means that it is often convenient to use the ability of a complex to refract and absorb different circular polarizations of light to derive information about both the configuration of an optically active complex and its electronic structure. The focus of this section will be to explain how the determination of absolute configuration by circular dichroism and optical rotatory dispersion work, as well as the relationship of these techniques to the more ordinary sort of polarimetry used to measure optical rotation in organic systems. Only brief notes will be made about the instrumental and electronic structure applications of these techniques since the electronic spectra of metal complexes will be discussed in Chapter 11: Coordination Chemistry III - Electronic Spectra and the instrumental aspects of CD are well-treated elsewhere . Both polarimetry and circular dichroism spectroscopy are grounded in the recognition that plane polarized light is equivalent to a superposition of left and right circularly polarized light, as shown in Figure $\sf{\PageIndex{12}}$. When plane polarized light passes through a chiral medium, its left and right circularly polarized components move at different rates ( i.e., have different indices of refraction). This causes the plane of polarization to rotate in the direction of the faster component according to the relationship \[\sf{\alpha ~=~\dfrac{n_l~-~n_r}{\lambda}} \nonumber \] where $\alpha$ is the angle of rotation, $\lambda$ the light's wavelength, and n l and n r the refractive indices experienced by left and right circularly polarized light. An example of such a rotation is shown in Figure $\sf{\PageIndex{13}}$. Chemists typically call this rotation of light by a chiral medium optical rotation. In organic chemistry it is common to measure the optical rotation at the sodium D wavelength of 589.29 nm, a much longer wavelength than most organics are able to absorb. Because of this the optical rotations commonly measured by organic chemists are largely independent of absorption and mainly serve as a characteristic physical property similar to melting points or are used to establish the purity of a mixture of enantiomers. The sort of spectroscopic data used to characterize chiral coordination compounds more commonly makes use of the relationship between absorption and optical rotation. In coordination chemistry the wavelength-dependence of optical rotation is used to discern information about compounds' electronic spectra. This measurement of the wavelength dependence of optical rotation is called optical rotatory dispersion (ORD) and the resulting plots are called optical rotatory dispersion or ORD spectra or curves. Such ORD curves are useful for characterizing the chiral environment in a coordination complex because the ORD curves exhibit a characteristic shape. This characteristic shape is because according to the Cotton Effect, the optical rotation changes sign at the absorption maximum of a chromophore in the compound. For one configuration the rotation will go from negative to positive with increasing wavelength and is called a positive cotton effect ; for the other configuration the rotation will go from positive to negative or exhibit a negative cotton effect . This behavior is summarized in the top two curves of Figure $\sf{\PageIndex{14}}$. A technique used even more often than ORD is circular dichroism (CD) , represented by the lowest spectrum in Figure $\sf{\PageIndex{14}}$. Circular dichroism arises from the differential absorption of left and right circularly polarized light by a chiral compound. This gives elliptically polarized light as shown in Figure $\sf{\PageIndex{15}}$. In circular dichroism (CD) spectroscopy a sample is irradiated with polarized light resulting in the transmission of elliptically polarized light due to differential refraction and absorption of the polarized light's left and right circularly polarized components. For achiral molecules the handedness doesn't affect absorbance, but for chiral molecules, which chiral ground and excited states the handedness of an EM wave affects can determine how readily that wave can distort the chiral ground state into the chiral excited one. For this reason chiral molecules absorb left and right circularly polarized light differently. In CD spectroscopy the magnitudes of the left and right circularly polarized components of light that is transmitted by the sample are measured and used to determine how much light of each was absorbed (just as in regular absorbance spectroscopy). The difference in absorption between left and right circularly polarized light constitutes the primary signal in CD spectroscopy. More specifically, circular dichroism spectra (CD spectra) show the difference in extinction coefficients for left and right circularly polarized light, $\Delta \epsilon$, as a function of wavelength, where \[\sf{\Delta \epsilon\ ~=~ \epsilon_l ~-~\epsilon_r} \nonumber \] where $\epsilon_\sf{l}$ and $\epsilon_\sf{r}$ are the extinction coefficients observed with left and right circularly polarized light, respectively. As with ORD spectra, the signs of the features in CD spectra like that of Δ–[Co(en) 3 ]Cl 3 given in Figure $\sf{\PageIndex{16}}$ are enantiomer-dependent. As illustrated in the lower spectrum in Figure $\sf{\PageIndex{14}}$ , the CD spectra of enantiomers mirror one another. Because of this difference in behavior it is possible in principle to determine the absolute configuration of a complex from its CD spectrum. There are several ways that ORD and CD spectra can be useful for determining the absolute configuration of a chiral complex. First, the sign of the Cotton effect and peaks in CD spectra exhibited by a given absolute configuration is often consistent across a series of compounds. Thus if a new member of the series is isolated, its configuration can be determined based on which enantiomer has the same Cotton effect. For instance, if both the $\Lambda$ isomer of a metal diimine complex like [Ru(bpy) 3 ] 2+ exhibit a positive Cotton effect and a newly resolved metal diimine exhibits a negative Cotton effect it might reasonably be inferred that the newly-resolved complex is in the opposite or $\Delta$ configuration. Second, for some systems the expected sign of the Cotton effect and CD peaks for each enantiomer may be predicted semi-empirically based on the spatial locations of substituents relative to the chromophore (the part of the molecule that changes electronic structure during the transition), although the details are beyond the scope of the present discussion. Consideration of interactions between multiple chromophores in a compound can be used to infer absolute configuration. Again, the details are beyond the scope of the present discussion. Interested readers are referred to reference 7 for details. Common patterns of Diastereomerism Geometric isomerism Geometric isomerism involves differences in the geometric placement of atoms in a compound. In coordination chemistry, geometric isomerism involves differences in the relative placement of a set of ligands about a metal center. There are two main types: cis and trans isomerism This type of isomerism has to do with how two ligands are oriented relative to one another in a square planar or octahedral complex. As shown in Figure $\sf{\PageIndex{17}}$, ligands that are next to one another with a L-M-L bond angle of 90$^{\circ}$ are said to be cis ; those on opposite sides of the metal with a L-M-L bond angle of 180$^{\circ}$ are in the trans arrangement. Simple cis and trans geometric isomers can be identified when there are two identical ligands (A) or chelating ligands with distinguishable attachment points (A~B) oriented about either a square planar or octahedral center. Examples are given in Figure $\sf{\PageIndex{18}}$. Notice from the example given in Figure $\sf{\PageIndex{18C}}$ that multidentate ligands have the potential to constrain complexes to adopt a particular geometry. Slightly more involved examples involve perturbations of the simple cases above in which there are multiple distinguishable cis and/or trans relationships. Two square planar examples are given in Figure $\sf{\PageIndex{19}}$. More complex examples of cis and trans relationships between ligands in octahedral complexes are given in the exercises that conclude this page, which also demonstrates how a systematic approach may be used to identify isomers. mer and fac isomerism This type of isomerism has to do with how three ligands are oriented relative to one another in an octahedral complex. The arrangements are represented in Figure $\sf{\PageIndex{20}}$. Since it can help to visualize the mer and fac geometry from multiple points of view, several representations are given in Figure $\sf{\PageIndex{21}}$. As shown in Figure $\sf{\PageIndex{21}}$, in fac or facial arrangements , the ligands occupy the same "face" of the octahedral coordination sphere, while in the mer or meridional geometry the ligands form a T shape in the same plane or its "meridian." 8 Octahedral complexes with three identical ligands oriented about an octahedral center can only exist in mer and fac arrangements. Consequently, such complexes can be designated as mer and fac isomers. Examples are given in Figure $\sf{\PageIndex{22}}$. Diastereomerism arising from multiple ligand-associated chiral centers Diastereomerism can also arise when two or more chiral ligands bind to a metal center. Such cases typically give rise to a complex mixture of isomers, as shown by the example in Figure $\sf{\PageIndex{23}}$. As may be seen from Figure $\sf{\PageIndex{23}}$, the differences between these isomers can arise from both changes in the stereochemical configuration of the ligand and the relationship of particular ligand centers relative to one another. For instance, the leftmost two structures in Figure $\sf{\PageIndex{23}}$ possess one R and S nitrogen center each but differ in whether the centers are oriented so that the R and S centers on the two ligands are trans or cis to one another. Diastereomerism arising from $\lambda$ and $\delta$ ring conformation isomerism. Chelate rings are formed when a chelating ligand binds a metal As may also be seen from the structures in Figure $\sf{\PageIndex{23}}$, when a multidentate ligand coordinates a metal, the ligand and metal center comprise one or more chelate rings . Examples of such chelate rings are shown in Figure $\sf{\PageIndex{24}}$. As may be inferred from the examples in Figure $\sf{\PageIndex{24}}$, chelate rings may be of different sizes, although four, six, and especially five-membered rings are particularly common. Exactly which ring sizes will be more stable for a given system depends on the coordination geometry and, due to differences in M-L bond lengths, to a lesser degree on the metal. As a result, for octahedral and square planar complexes with 90$^{\circ}$ L-M-L bond angles between cis ligands, five-membered rings tend to be especially stable (Figure $\sf{\PageIndex{25}}$). Not all metal chelates prefer to form five-membered rings. The steric requirements of chelate rings depend on both the preferred coordination geometry of the metal center and the stereochemistry of the ligand. Both of these effects are typically considered in terms of preferred L-M-L bond angles. From the perspective of the metal center, the preferred bond angle is determined by the coordination geometry. Octahedral and square planar metals prefer 90$^{\circ}$ L-M-L bond amgles, trigonal bipyramidal systems 90 and 120$^{\circ}$ L-M-L bond angles, and tetrahedral complexes 109.5$^{\circ}$ L-M-L angles. The larger preferred bond angles in tetrahedral and trigonal bipyramidal systems often require the formation of larger six or seven-membered chelate rings for maximum stability. The size of the chelate ring actually formed between a metal and ligand is determined by the ligand's structure. As the contributor of all the atoms in the chelate ring but one, the ligand directly determines the chelate ring size. More subtly, ligands naturally prefer to coordinate metals at a particular L-M-L angle, called the bite angle , as shown in Figure $\sf{\PageIndex{26A}}$. Ligands with bite angles corresponding to the ideal L-M-L angle for a metal's preferred geometry tend to form more stable complexes, although in turn ligand bite angles can cause metals with a weak coordination geometry preference to adopt the ligand's preferred geometry instead. 10 As may be seen from the values in Figure $\sf{\PageIndex{26B}}$, bite angles roughly increase with the size of the chelate ring formed but are also influenced by the types of structures used to connect the ligand's Lewis base sites. This facilitates the use of diphosphine ligands to tailor the structure and reactivity of phosphine-containing organometallic catalysts. $\lambda$ and $\delta$ isomerism involves differences in the conformation of nonplanar five-membered chelate rings As shown in Figure $\sf{\PageIndex{27}}$, some chelate ring systems are planar while others are not. Among the nonplanar chelate ring systems are ligands like en and dppe, which contain tetrahedral C, N, O, P, and S atoms. Because these atoms create kinks in the chelate ring they introduce additional opportunities for diastereomerism due to differences in the chelate ring conformation, called ring twist . Ring twist in nonplanar systems has been most extensively explored for five-membered chelates like those formed by dppe and by en and other chelating amines. Unlike the more rigid four-membered chelate rings, five-membered chelates tend to be conformationally flexible, but not so conformationally flexible that their conformers are rapidly interconverting at room temperature (at least not when the rings are substituted to introduce additional steric strain). This balance between flexibility and rigidity enables some five-membered chelate rings to exist as mixtures of distinguishable conformational isomers at room temperature. To understand the two most stable conformers that five-membered chelates rings tend to form, it is helpful to think of five-membered chelate rings as involving two components (Figure $\sf{\PageIndex{28}}$): a planar L-M-L group, where L are the atoms directly attached to the metal, M, and the remaining two ring atoms, E, which form a rigid E-E bar across the back of the chelate ring. As can be seen from Figure $\sf{\PageIndex{28}}$ , rotation about the M-L and L-E bonds causes twisting of the E-E bar relative to the L-M-L plane. The two most stable conformers produced by these motions are the $\lambda$ or $\delta$ conformations as shown in Figures $\sf{\PageIndex{29}}$ and $\sf{\PageIndex{29}}$. The relative stability of the $\lambda$ and $\delta$ conformers will depend on the configuration of any stereocenters present and steric factors. As with organic ring systems, steric effects tend to favor conformers in which bulky groups are placed in the less sterically strained equatorial positions while the configuration of the stereocenters present can serve to restrict the allowable ring twist conformations, as illustrated in Figure $\sf{\PageIndex{30}}$. A detailed treatment of such systems is beyond the scope of this page. Interested readers are referred to reference 12 for more details. In closing, it should be noted that it might seem to be making much of small effects in pointing out that one source of diastereomerism in metal complexes involves the freezing out of chelate ring conformations. However, that would be to mistake the impacts that ring conformations can have on the steric accessibility of a coordination site and shaping the course of processes that might occur there. As illustrated in Figure $\sf{\PageIndex{31}}$, a simple shift in the conformation of one ring in a square planar, pyramidal, or octahedral complex containing coplanar ethylene diamine ligands can exert a significant effect on the steric profile of the complex perpendicular to the MN 4 square plane. Because of effects like these, ring conformation isomerism plays a role in the design of stereoselective transition metal catalysts. Exercises Exercise $\PageIndex{1}$. Identify the type of isomerism What type of isomers are [CoCl(NH 3 ) 5 ](NO 3 ) and [Co(NH 3 ) 5 ( N- NO 2 )]Cl? [Co(NH 3 ) 5 -CN-Ru(NH 3 ) 5 ] 4+ and [Co(NH 3 ) 5 -NC-Ru(NH 3 ) 5 ] 4+ [Cr(CN) 5 -CN-Co(NH 3 ) 5 ] and [Co(CN) 5 -CN-Cr(NH 3 ) 5 ] cis -[Mn(en) 2 (CN) 2 ] and trans -[Mn(en) 2 (CN) 2 ] Answer a. Ionization isomers. Answer b. Linkage isomers. Answer c. Coordination isomers. Answer d. Geometric isomers, specifically cis / trans isomers. Exercise $\PageIndex{2}$. Assigning metal oxidation states in a complex Many properties of transition metal complexes depend on the metal's oxidation state. For instance, octahedral complexes of Co II lose and gain ligands rapidly octahedral complexes of Co III lose and gain ligands very slowly four-coordinate complexes are generally tetrahedral EXCEPT four-coordinate complexes of metals like Pt II , Pd II , Rh I , and Ir I , among others, are square planar For this reason it is important to be able to estimate the formal oxidation state of a metal in a complex. Fortunately, this is easy to do if you remember The sum of all atoms' oxidation states will equal the overall charge on the complex When determining the metal's oxidation state, the ligands can be treated as having an oxidation state equal to their charge - i.e., the charge they possess in the form in which they coordinate the metal - so if they need to lose a proton to bind, don't forget to account for that. Given the above, estimate the oxidation state of the metal in the following real and hypothetical complexes. Na 4 [Fe(CN) 6 ] [Cu(phen) 2 ]BF 4 [PtF 4 (NH 3 ) 2 ] [Ni(en) 2 ]SO 4 Co(acac) 3 [MnCl(O)(salen)] Answer for Na 4 [Fe(CN) 6 ] . This contains [Fe(CN) 6 ] 4- ; so O.S. Fe + 6 x (-1) (for CN - ) = -4 (the complex's charge) so O.S. Fe = +2 or Fe 2 + . Answer for [Cu(phen) 2 ]BF 4 . Since tetrafluoroborate is a monoanion, the complex is [Cu(phen) 2 ] + so O.S. Cu + 0 x 2 (for phen) = +1 (the complex's charge) so O.S. Cu = +1 or Cu + . Answer for [PtF 4 (NH 3 ) 2 ] . O.S. Pt + 4 x (-1) (for F - ) + 0 x 2 (for NH 3 ) = +0 (the complex's charge) so O.S. Pt = +4 or Pt 4 + . Answer [Ni(en) 2 ]SO 4 . The complex is [Ni(en) 2 ] + so O.S. Ni + 0 x 2 (for en) = +2 (the complex's charge) so O.S. Ni = +2 or Ni 2 + . Answer Co(acac) 3 . O.S. Co + 3 x (-1) (for acac; see table 9.2.2 ) = +0 (the complex's charge) so O.S. Co = +3 or Co 3 + . Answer [MnCl(O)(salen)] . O.S. Mn + 1 x (-1) (for Cl - ) + 1 x (-2)(for oxo) + 1 x (-2) (for salen; see table 9.2.2 ) = +0 (the complex's charge) so O.S. Mn = +5 or Mn 5 + . Exercise $\PageIndex{3}$. Drawing isomers from descriptions Draw structures that match the descriptions given, assuming that complexes in which the metal has a coordination number of six are octahedral complexes in which the metal has a coordination number of five are trigonal bipyramidal complexes in which Pt II , Pd II , or Rh I , or Ir I have a coordination number of four are square planar other complexes in which the metal has a coordination number of four will be tetrahedral mer -triammineaqua- trans- dichlorocobalt(III) ion $\Delta$-diaminebis(oxalato)manganate(III) [CoCl 4 ] 2- trans -diamminebis(ethylenediamine)Nickel(2+) tetracyanopalladate(2-) $\Lambda$-bis(ethylenediamine)cobalt(III)- μ-amido μ-hydroxo-$\Delta$-bis(ethylenediamine)cobalt(III) Answer a. mer -triammineaqua- trans- dichlorocobalt(III) ion . Answer b. $\Delta$-diaminebis(oxalato)manganate(III) . Answer c. [CoCl 4 ] 2- . Answer d. trans -diamminebis(ethylenediamine)Nickel(2+) tetracyanopalladate(2-) . Note that since the Pd in [Pd(CN) 4 ] 2- is Pd 2 + it will be square planar. Answer e. $\Lambda$-bis(ethylenediamine)cobalt(III)- μ-amido μ-hydroxo-$\Delta$-bis(ethylenediamine)cobalt(III) . . Exercise $\PageIndex{4}$. Stereoisomers of complexes containing only monodentate ligands. Draw all the stereoisomers of the following real and hypothetical complexes. You may assume that complexes in which the metal has a coordination number of six are octahedral complexes in which the metal has a coordination number of five are trigonal bipyramidal complexes in which Pt II , Pd II , or Rh I , or Ir I have a coordination number of four are square planar other complexes in which the metal has a coordination number of four will be tetrahedral [IrCl(CO)(PPh 3 ) 2 ] [CoCl 3 (NH 3 ) 3 ] [CoCl 2 (H 2 O) 2 (NH 3 ) 2 ] + [CoBrCl(H 2 O) 2 (NH 3 ) 2 ] + [CoBrClI(NH 3 ) 3 ] Answer a. [IrCl(CO)(PPh 3 ) 2 ] This is a 4-coordinate Ir I complex and, as such will be square planar. Since two of the ligands are identical, it will have cis and trans isomers. The trans isomer is famous for its ability to form adducts and is called Vaska's complex. Answer b. [CoCl 3 (NH 3 ) 3 ] This complex contains three ammine and three chloro ligands and so will have fac and mer isomers. Answer c. [CoCl 2 (H 2 O) 2 (NH 3 ) 2 ] + This is a case of an MA 2 B 2 C 2 system (A = Cl - , B = H 2 O, C = NH 3 ) involving multiple cis and trans relationships. The six stereoisomers are shown below. Since it may not be obvious how to arrive at a set of isomers like the ones above, it is worth considering how one might work through the possibilities for a system of ligands like this one. Several systems may be used to systematically identify isomers. Once approach is the Macbdef system described in Note $\sf{\PageIndex{1}}$ at the end of these exercises. The solutions presented here and in subsequent problems employ a variant of that approach. Start by fixing one set of ligands. In this case the chloro ligands were first fixed as cis to one another. Then the remaining ligands might be cis or trans to one another, although since the chloro ligands are already cis, the ammine and aqua ligands cannot both be trans at the same time. Thus for cis chloro ligands the possibilities for the others are: cis -H 2 O , cis -NH 3 , cis- -H 2 O, trans -NH 3 trans -H 2 O, cis- NH 3 Swap the configuration of the ligand set you started with. In this case it means placing the chloro ligands trans to one another. From that configuration, the possibilities for the ammine and aqua ligands are cis -H 2 O , cis -NH 3 , trans -H 2 O, trans -NH 3 Check for enantiomers and create mirror images of any chiral complexes you drew. The easiest way to do this is to assign point groups. However, it turns out that the octahedral all- cis case is D 3 and formally equivalent to the symmetry of a tris chelate, as shown below. finally, check to make sure you didn't include the same complex twice. Humans do make mistakes after all. Answer d. [CoBrCl(H 2 O) 2 (NH 3 ) 2 ] + This problem is analogous to the one above except that this complex contains a bromo and chloro ligand in place of the two chloro ligands in [CoCl 2 (H 2 O) 2 (NH 3 ) 2 ] + . Consequently, when the bromo and chloro ligands are cis to one another, additional possibilities for isomers arise based on whether ammine or aqua ligands are trans to the chloro and bromo. The result is eight isomers. Answer e. [CoBrClI(NH 3 ) 3 ] There are two possibilities for this complex - a fac -(NH 3 ) 3 and a mer- (NH 3 ) 3 arrangement. These may be taken as starting points for examining the possible permutations for the Cl - , Br - , and I - ligands. The top row presents the three possibilities within the mer- (NH 3 ) 3 arrangement; each corresponds to a different ligand trans to an ammine ligand. The bottom row presents the two possible isomers with a fac -(NH 3 ) 3 arrangement. In each, the Cl - , Br - , and I - ligands are trans to ammine ligands. However, the complexes possess C 1 symmetry and so are chiral and exist as enantiomers. Exercise $\PageIndex{5}$. Stereoisomers of complexes with bidentate ligands (ignoring ring twist). Ignoring ring twist , draw all the stereoisomers of the following real and hypothetical octahedral complexes. [CoBr 2 Cl 2 (en)] - [CoBrCl(en) 2 ] + [CoBrCl(gly) 2 ] - [Co(en) 3 ] 3+ [Co(gly) 3 ] Answer a. [CoBr 2 Cl 2 (en)] - This complex is analogous to [CoCl 2 (H 2 O) 2 (NH 3 ) 2 ] + in possessing pairs of identical ligating groups. The main difference is that in [CoBr 2 Cl 2 (en)] - the two amine ligating group of the en ligand are restricted to a cis arrangement. Under the MABCDEF system explained in Note $\PageIndex{1}$ , [CoCl 2 (H 2 O) 2 (NH 3 ) 2 ] + may be MA 2 B 2 C 2 , but this complex, [CoBr 2 Cl 2 (en)] - , is M(AA)B 2 C 2 . The isomers are: Answer b. [CoBrCl(en) 2 ] + The key to problems like this is to focus on the unique ligands, in this case Br - and Cl - . These can exist in either a cis or trans arrangement. The trans form is achiral, but in the case where Br - and Cl - are cis, $\Lambda$ and $\Delta$ enantiomers are possible. The result is three isomers. Answer c. [CoBrCl(gly) 2 ] - This case is analogous to the one above, but this time the bidentate ligand, gly, possesses distinguishable binding sites. Thus there are additional permutations involving how the gly ligands are bound relative to one another and/or whether the O or N end of the gly is bound trans to Br or Cl. The result is 11 different isomers, which are given below. Answer d. [Co(en) 3 ] 3+ As a tris-chelate of symmetric bidentate ligands, ignoring ring twist, [Co(en) 3 ] 3+ will only exhibit $\Lambda$ and $\Delta$ isomerism. Answer e. [Co(gly) 3 ] Since there are three gly ligands, each of which have carboxy and amine ends, the allowable arrangements involve whether the resulting three carboxy and three amine ends are oriented in a mer or fac arrangement. This gives two possibilities, each of which can exist as either the $\Lambda$ or $\Delta$ enantiomer, for a total of four isomers, which are given below. Exercise $\PageIndex{6}$. More stereoisomers: Now featuring linkage isomerism Ignoring ring twist , draw all chemically reasonable stereoisomers for the following real and hypothetical complexes. You may assume that complexes in which the metal has a coordination number of six are octahedral complexes in which the metal has a coordination number of five are trigonal bipyramidal complexes in which Pt II , Pd II , or Rh I , or Ir I have a coordination number of four are square planar other complexes in which the metal has a coordination number of four will be tetrahedral [Pt(en)(SCN) 2 ] [Pt(NH 3 ) 2 (SCN) 2 ] [Co(en) 2 (SCN) 2 ] + Answer a. [Pt(en)(SCN) 2 ] The Pt in this neutral complex must be Pt 2 + to balance the negative charges of the two SCN - ligands. Complexes of Pt 2 + are square planar and four coordinate, consistent with the bidentate en and two thiocyanato ligands. Of the ligands, the en ligand must bind in a cis arrangement. Consequently, the two SCN - ligands will be in a cis arrangement as well. The isomers will therefore only differ in whether the two SCN - bind $\kappa$ N or $\kappa$ S . The possibilities are: Answer b. [Pt(NH 3 ) 2 (SCN) 2 ] As with the preceding example, this will be a square planar Pt 2 + complex. The main difference is that this time the coordinated nitrogens are not constrained to adopt a cis arrangement so there will be both cis and trans isomers. Answer c. [Co(en) 2 (SCN) 2 ] + Exercise $\PageIndex{7}$. Stereoisomer free for all. Ignoring ring twist , draw all chemically reasonable stereoisomers for the following real and hypothetical complexes. You may assume that complexes in which the metal has a coordination number of six are octahedral complexes in which the metal has a coordination number of five are trigonal bipyramidal complexes in which Pt II , Pd II , or Rh I , or Ir I have a coordination number of four are square planar other complexes in which the metal has a coordination number of four will be tetrahedral [Ru(bpy)(phen)(dppe)] 2+ [CoClF(PPh 3 )(py)] - [Ni(en) 2 (NO 2 ) 2 ] [Fe(H 2 O) 3 (SCN) 3 ] [PtClF(PPh 3 )(py)] [CoBr 2 Cl(NH 3 ) 3 ] Answer a. [Ru(bpy)(phen)(dppe)] 2+ This complex is an octahedral tris-chelate containing symmetric ligands. As such it will exhibit $\Lambda$ and $\Delta$ isomers: Answer b. [CoClF(PPh 3 )(py)] - This complex has a coordination number of 4 and contains a Co 2 + ion with a d 7 electron configuration, so a tetrahedral geometry is expected. Tetrahedral complexes like this one with four different ligands are chiral and can form R and S enantiomers. Answer c. [Ni(en) 2 (NO 2 ) 2 ] This is an octahedral bis-chelate and will exist in $\Lambda$ and $\Delta$ configurations. Within each configuration the NO 2 - ligands can exist as $\kappa$ N and $\kappa$ O , leading to the following possibilities: Answer d. [Fe(H 2 O) 3 (SCN) 3 ] As an octahedral complex with three identical ligands it can exist in mer and fac configurations, with the ambidentate SCN ligand providing additional possibilities for isomerism. Answer e. [PtClF(PPh 3 )(py)] As a square planar complex with four nonidentical ligands, this complex exists as a single isomer. Answer f. [CoBr 2 Cl(NH 3 ) 3 ] As an octahedral complex with three identical ligands it will exhibit mer and fac isomerism. In addition, it will have two mer configurations that differ in terms of the cis and trans relationships between the bromo and chloro ligands. Exercise $\PageIndex{8}$ Complexes of formula Ru(TPP)py 2 have been prepared, in which TPP is tetraphenylporphyrin, which binds metals in the form given below. Draw all isomers of Ru(TPP)py 2 . Answer There is only one isomer. While normally complexes containing two identical ligands (in this case py) can exhibit cis and trans isomerism, in this case the planar tetraphenylporphyrin ring ligates the Ru 2 + ion in a square planar arrangement, as shown at left in the image below. This leaves only a pair of trans coordination sites for the chloro ligands to occupy, giving the isomer shown at right below. Exercise $\PageIndex{9}$ Draw the diastereomers formed due to the chirality of the amine nitrogen atoms in [PtCl 2 ( N , N '-dimethylethane-1,2-diamine)] which has the atomic connectivity represented below Answer The diastereomers will differ in whether the N atoms adopt an R or S configuration at the nitrogen atoms. The possible permutations are ( R,R ), (S ,S ), ( R,S ), and (S ,R ). However, as may be seen from the image below, the ( R,S ) and (S ,R ) configurations are identical (the two projections shown can be interconverted through a C 2 rotation along an axis bisecting the Cl-Pt-Cl unit). As a result there are only three unique isomers. Of these, the ( R,R ) and (S ,S ) configurations are enantiomers (they are mirror images by reflection in the PtCl 2 N 2 plane). Note that in solving problems like this one it can be helpful to keep track of possible isomers by assigning the stereochemistry at each chiral center as R or S using the Cahn-Ingold-Prelog-convention , though it is not strictly necessary to do so. Exercise $\PageIndex{10}$ Label the conformations of all chelate rings in the structure below. Answer . Appendix: The MABCDEF bookkeeping system for identifying isomers The Mabcdef or MABCDEF system is one method for identifying the number of isomers in an octahedral complex, although since it is really just a bookkeeping and organizational system it can easily be extended to other geometries as well. The M in Mabcdef stands for metal and the other letters are used to stand in for ligands. The basic approach involves classifying the ligands as A, B, C, D, E, or F in order of multiplicity. Thus for [Cr(NH 3 ) 6 ] 3+ A = NH 3 ; there are no B, C, D,E, or F ligands; and the complex is classified as MA 6 [CoCl(NH 3 ) 5 ] 2+ A = NH 3 , B = Cl - , and the complex is MA 5 B [CrCl 2 (H 2 O) 2 (NH 3 ) 2 ] + A = Cl - , B = H 2 O, C = NH 3 , and the complex is MA 2 B 2 C 2 in the classification above, multidentate ligands are typically classified before monodentate ones and are designated AA, AB, ABC, ABA, etc. based on the symmetry of the attachment points. Thus en has identical attachment amine points and is AA gly has a carboxy and amine attachment points and is AB trien is ABA CoCl 2 (en) 2 + is M(AA) 2 B 2 CoCl 2 (gly) 2 - is M(AB) 2 C 2 CoCl 2 (en)(gly) is M(AA)(AB)C 2 Systematically list out the possible trans arrangements of ligands by assigning one pair of ligands to be trans to one another. Then systematically list out the other possible trans pairs by permuting the remaining trans arrangements. It can help to organize the permutations in a table, such as that shown below for an MABCDEF complex where A and B are assigned trans . In looking at the table notice how the second set of trans permutations is systematically varied. This helps ensure that no possibility is skipped. Go through the list of isomers and remove any duplicates you generated so far. For instance, notice in the table below that the last three stereoisomers are identical with the first three ( e.g., stereoisomer 4 is identical to stereoisomer 3, 5 with 2, and 6 with 1). 0 1 2 3 4 5 Stereoisomer 1 Stereoisomer 2 Stereoisomer 3 Stereoisomer 4 (same as isomer 3) Stereoisomer 5 (same as isomer 2) Stereoisomer 6 (same as isomer 1) trans AB (fixed) trans AB (fixed) trans AB (fixed) trans AB (fixed) trans AB (fixed) trans AB (fixed) trans CD trans CE trans CF trans DE trans DF trans EF trans EF trans DF trans DE trans CF trans CE trans CD If the complex possesses multidentate ligands that demand that certain groups exist and cis pairs, then also remove any configurations which do not agree with the known binding capability of the ligand ( e.g., the two amine groups of ethylenediamine cannot be trans to one another, so if you have an en ligand, remove configurations like that from the list). Next, swap or permute the original trans pair and repeat the process you just followed. In the case of Mabcdef this gives the following results: 0 1 2 3 Isomer "fixed" trans pair Additional trans pairs Additional trans pairs 1 AB CD EF 2 AB CE DF 3 AB CF DE 4 AC BD EF 5 AC BE BF 6 AC BF DE 7 AD BC EF 8 AD BE CF 9 AD BF CE 10 AE BC DF 11 AE BD CF 12 AE BF CD 13 AF BC DE 14 AF BD CE 15 AF BE CD Since the procedure explained above only identifies isomers based on unique trans pairings, it does not identify when a configuration is chiral and corresponds to a pair of enantiomers. Any enantiomers can be identified by drawing out the complexes and either classifying their point groups or by drawing their mirror images and checking if they are superimposable. In the MABCDEF case - i.e., where all the ligands are different - all of the isomers identified have C 1 symmetry so are chiral. This means there will be a pair of enantiomers for each, giving 15 x 2 = 30 different stereoisomers. References 1. These examples are taken from http://wwwchem.uwimona.edu.jm/courses/inorgnom.html 2. Contakes S. M.; Rauchfuss, T.B. Chem . Commun . 2001, 553-554. 3. Shriver, D. F.; Shriver, S. A.; Anderson, S. E., Inorg. Chem. 1965, 4(5) , 725-730. 4. Zelewsky, A. v., Stereochemistry of Coordination Compounds Wiley, 1996. 5. In the past it was difficult to determine absolute configurations from X-ray data alone, but recent advances have made it easier to do so. 6. Pavan M. V. Raja & Andrew R. Barron "Circular Dichroism Spectroscopy and its Application for Determination of Secondary Structure of Optically Active Species" in Physical Methods in Chemistry and Nano Science https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Book%3A_Physical_Methods_in_Chemistry_and_Nano_Science_(Barron)/07%3A_Molecular_and_Solid_State_Structure/7.07%3A_Circular_Dichroism_Spectroscopy_and_its_Application_for_Determination_of_Secondary_Structure_of_Optically_Active_Species 7. Berova, N.; Bari, L. D.; Pescitelli, G., Chemical Society Reviews 2007, 36 (6), 914-931. 8. Of course, if one set of ligands occupies a meridional plane, then the other three ligands will be oriented in an equatorial one. The reason they may both be considered meridional is because if the complex were rotated, the equatorial plane would be meridional and vice versa. From that point of view both might be considered meridional planes - albeit not at the same time. 9. Bite angles are taken from Mansell, S. M., Catalytic applications of small bite-angle diphosphorus ligands with single-atom linkers. Dalton Transactions 2017, 46 (44), 15157-15174. 10. Hancock, R. D., The pyridyl group in ligand design for selective metal ion complexation and sensing. Chemical Society Reviews 2013, 42 (4), 1500-1524. 11. Sasi, D.; Ramkumar, V.; Murthy, N. N., Bite-Angle-Regulated Coordination Geometries: Tetrahedral and Trigonal Bipyramidal in Ni(II) with Biphenyl-Appended (2-Pyridyl)alkylamine N,N'-Bidentate Ligands. ACS Omega 2017, 2 (6), 2474-2481. 12. Ehnbom, A.; Ghosh, S. K.; Lewis, K. G.; Gladysz, J. A., Octahedral Werner complexes with substituted ethylenediamine ligands: a stereochemical primer for a historic series of compounds now emerging as a modern family of catalysts. Chemical Society Reviews 2016, 45 (24), 6799-6811.
Courses/National_Yang_Ming_Chiao_Tung_University/Chemical_Principles_for_Medical_Students/01%3A_Electronic_Structure_of_Atoms	Template:HideTOC In this chapter, we describe how electrons are arranged in atoms and how the spatial arrangements of electrons are related to their energies. We also explain how knowing the arrangement of electrons in an atom enables chemists to predict and explain the chemistry of an element. As you study the material presented in this chapter, you will discover how the shape of the periodic table reflects the electronic arrangements of elements. In this and subsequent chapters, we build on this information to explain why certain chemical changes occur and others do not. After reading this chapter, you will know enough about the theory of the electronic structure of atoms to explain what causes the characteristic colors of neon signs, how laser beams are created, and why gemstones and fireworks have such brilliant colors. In later chapters, we will develop the concepts introduced here to explain why the only compound formed by sodium and chlorine is NaCl, an ionic compound, whereas neon and argon do not form any stable compounds, and why carbon and hydrogen combine to form an almost endless array of covalent compounds, such as CH 4 , C 2 H 2 , C 2 H 4 , and C 2 H 6 . You will discover that knowing how to use the periodic table is the single most important skill you can acquire to understand the incredible chemical diversity of the elements. 1.1: The Wave Nature of Light Understanding the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation. A basic knowledge of the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation. A wave is a periodic oscillation by which energy is transmitted through space. All waves are periodic, repeating regularly in both space and time. Waves are characterized by several interrelated properties. 1.2: Quantized Energy and Photons Blackbody radiation is the radiation emitted by hot objects and could not be explained with classical physics. Max Planck postulated that energy was quantized and may be emitted or absorbed only in integral multiples of a small unit of energy, known as a quantum. The energy of a quantum is proportional to the frequency of the radiation; the proportionality constant h is a fundamental constant (Planck’s constant). Albert Einstein used the quantization of energy to explain the photoelectric effect 1.3: Line Spectra and the Bohr Model There is an intimate connection between the atomic structure of an atom and its spectral characteristics. Most light is polychromatic and contains light of many wavelengths. Light that has only a single wavelength is monochromatic and is produced by devices called lasers, which use transitions between two atomic energy levels to produce light in a very narrow range of wavelengths. Atoms can also absorb light of certain energies, resulting in a transition from the ground state or a lower-energy e 1.4: The Wave Behavior of Matter An electron possesses both particle and wave properties. Louis de Broglie showed that the wavelength of a particle is equal to Planck’s constant divided by the mass times the velocity of the particle. The electron in Bohr’s circular orbits could thus be described as a standing wave, one that does not move through space. Werner Heisenberg’s uncertainty principle states that it is impossible to precisely describe both the location and the speed of particles that exhibit wavelike behavior. 1.5: Quantum Mechanics and Atomic Orbitals There is a relationship between the motions of electrons in atoms and molecules and their energies that is described by quantum mechanics. Because of wave–particle duality, scientists must deal with the probability of an electron being at a particular point in space. To do so required the development of quantum mechanics, which uses wavefunctions to describe the mathematical relationship between the motion of electrons in atoms and molecules and their energies. 1.6: 3D Representation of Orbitals Orbitals with l = 0 are s orbitals and are spherically symmetrical, with the greatest probability of finding the electron occurring at the nucleus. Orbitals with values of n > 1 and l = 0 contain one or more nodes. Orbitals with l = 1 are p orbitals and contain a nodal plane that includes the nucleus, giving rise to a dumbbell shape. Orbitals with l = 2 are d orbitals and have more complex shapes with at least two nodal surfaces. l = 3 orbitals are f orbitals, which are still more complex. 1.7: Many-Electron Atoms In addition to the three quantum numbers (n, l, ml) dictated by quantum mechanics, a fourth quantum number is required to explain certain properties of atoms. This is the electron spin quantum number (ms), which can have values of +½ or −½ for any electron, corresponding to the two possible orientations of an electron in a magnetic field. This is important for chemistry because the Pauli exclusion principle implies that no orbital can contain more than two electrons (with opposite spin). 1.8: Electron Configurations Based on the Pauli principle and a knowledge of orbital energies obtained using hydrogen-like orbitals, it is possible to construct the periodic table by filling up the available orbitals beginning with the lowest-energy orbitals (the aufbau principle), which gives rise to a particular arrangement of electrons for each element (its electron configuration). Hund’s rule says that the lowest-energy arrangement of electrons is the one that places them in degenerate orbitals with parallel spins. 1.9: Electron Configurations and the Periodic Table The arrangement of atoms in the periodic table results in blocks corresponding to filling of the ns, np, nd, and nf orbitals to produce the distinctive chemical properties of the elements in the s block, p block, d block, and f block, respectively. 1.E: Electronic Structure of Atoms (Exercises) These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. 1.S: Electronic Structure of Atoms (Summary) This is the summary Module for the chapter "Electronic Structure of Atoms" in the Brown et al. General Chemistry Textmap. Thumbnail: Electron shell diagram for Sodium , the 19th element in the periodic table of elements. (CC BY-SA; 2.5; Pumbaa )
Courses/Furman_University/CHM101%3A_Chemistry_and_Global_Awareness_(Gordon)/00%3A_Front_Matter/02%3A_InfoPage	This text is disseminated via the Open Education Resource (OER) LibreTexts Project ( https://LibreTexts.org ) and like the hundreds of other texts available within this powerful platform, it is freely available for reading, printing and "consuming." Most, but not all, pages in the library have licenses that may allow individuals to make changes, save, and print this book. Carefully consult the applicable license(s) before pursuing such effects. Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new technologies to support learning. The LibreTexts mission is to unite students, faculty and scholars in a cooperative effort to develop an easy-to-use online platform for the construction, customization, and dissemination of OER content to reduce the burdens of unreasonable textbook costs to our students and society. The LibreTexts project is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being optimized by students, faculty, and outside experts to supplant conventional paper-based books. These free textbook alternatives are organized within a central environment that is both vertically (from advance to basic level) and horizontally (across different fields) integrated. The LibreTexts libraries are Powered by NICE CXOne and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. This material is based upon work supported by the National Science Foundation under Grant No. 1246120, 1525057, and 1413739. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation nor the US Department of Education. Have questions or comments? For information about adoptions or adaptions contact [email protected] . More information on our activities can be found via Facebook ( https://facebook.com/Libretexts ), Twitter ( https://twitter.com/libretexts ), or our blog ( http://Blog.Libretexts.org ). This text was compiled on 04/21/2025
Courses/South_Puget_Sound_Community_College/CHEM_110%3A_Chemical_Concepts/10%3A_Radiation_Electromagnetic_Waves_Biological_Effects_of_UV_Radiation	10.1: Radiation and Electromagnetic waves A wave is an oscillation or periodic movement that can transport energy from one point in space to another. Common examples of waves are all around us. Shaking the end of a rope transfers energy from your hand to the other end of the rope, dropping a pebble into a pond causes waves to ripple outward along the water's surface, and the expansion of air that accompanies a lightning strike generates sound waves (thunder) that can travel outward for several miles. 10.2: Content in Context You learned about different types of radiation, electromagnetic waves, frequency, and energy. In this section, we will focus on UV radiation, its harmful effects, the stratospheric ozone layer, and how to protect ourselves from the dangers of UV rays. 10.2.1: UVA, UVB, UVC and Its' Effect 10.2.2: UV Index 10.2.3: Stratospheric Ozone and Ozone Depletion 10.2.4: Sunscreen Thumbnail: CFCs & Ozone. (CC BY-SA 4.0; Nicole Leihe via Wikimedia )
Courses/Smith_College/CHM_222_Chemistry_II%3A_Organic_Chemistry_(2025)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.07%3A_Conformations_of_Ethane	Objectives After completing this section, you should be able to explain the concept of free rotation about a carbon-carbon single bond. explain the difference between conformational isomerism and structural isomerism. draw the conformers of ethane using both sawhorse representation and Newman projection. sketch a graph of energy versus bond rotation for ethane, and discuss the graph in terms of torsional strain. Key Terms Make certain that you can define, and use in context, the key terms below. conformation (conformer, conformational isomer) dihedral angle eclipsed conformation Newman projection staggered conformation strain energy torsional strain (eclipsing strain) Study Notes You should be prepared to sketch various conformers using both sawhorse representations and Newman projections. Each method has its own advantages, depending upon the circumstances. Notice that when drawing the Newman projection of the eclipsed conformation of ethane, you cannot clearly draw the rear hydrogens exactly behind the front ones. This is an inherent limitation associated with representing a 3-D structure in two dimensions. Conformational isomerism involves rotation about sigma bonds, and does not involve any differences in the connectivity of the atoms or geometry of bonding. Two or more structures that are categorized as conformational isomers, or conformers , are really just two of the exact same molecule that differ only in rotation of one or more sigma bonds. Ethane Conformations Although there are seven sigma bonds in the ethane molecule, rotation about the six carbon-hydrogen bonds does not result in any change in the shape of the molecule because the hydrogen atoms are essentially spherical. Rotation about the carbon-carbon bond, however, results in many different possible molecular conformations. In order to better visualize these different conformations, it is convenient to use a drawing convention called the Newman projection . In a Newman projection, we look lengthwise down a specific bond of interest – in this case, the carbon-carbon bond in ethane. We depict the ‘front’ atom as a dot, and the ‘back’ atom as a larger circle. The six carbon-hydrogen bonds are shown as solid lines protruding from the two carbons at 120°angles, which is what the actual tetrahedral geometry looks like when viewed from this perspective and flattened into two dimensions. Interactive Element Figure 3.6.1 : A 3D Model of Staggered Ethane. The lowest energy conformation of ethane, shown in the figure above, is called the ‘staggered’ conformation. In the staggered conformation, all of the C-H bonds on the front carbon are positioned at an angle of 60° relative to the C-H bonds on the back carbon. This angle between a sigma bond on the front carbon compared to a sigma bond on the back carbon is called the dihedral angle. In this conformation, the distance between the bonds (and the electrons in them) is maximized. Maximizing the distance between the electrons decreases the electrostatic repulsion between the electrons and results in a more stable structure. If we now rotate the front CH 3 group 60° clockwise, the molecule is in the highest energy ‘eclipsed' conformation, and the hydrogens on the front carbon are as close as possible to the hydrogens on the back carbon. This is the highest energy conformation because of unfavorable electrostatic repulsion between the electrons in the front and back C-H bonds. The energy of the eclipsed conformation is approximately 3 kcal/mol (12 kJ/mol) higher than that of the staggered conformation. Torsional strain (or eclipsing strain) is the name give to the energy difference caused by the increased electrostatic repulsion of eclipsing bonds. Another 60° rotation returns the molecule to a second eclipsed conformation. This process can be continued all around the 360° circle, with three possible eclipsed conformations and three staggered conformations, in addition to an infinite number of variations in between. We will focus on the staggered and eclipsed conformers since they are, respectively, the lowest and highest energy conformers. Unhindered (Free) Rotations Do Not Exist in Ethane The carbon-carbon bond is not completely free to rotate – the 3 kcal/mol torsional strain in ethane creates a barrier to rotation that must be overcome for the bond to rotate from one staggered conformation to another. This rotational barrier is not large enough to prevent rotation except at extremely cold temperatures. So at normal temperatures, the carbon-carbon bond is constantly rotating. However, at any given moment the molecule is more likely to be in a staggered conformation - one of the rotational ‘energy valleys’ - than in any other conformer. The potential energy associated with the various conformations of ethane varies with the dihedral angle of the bonds, as shown in Figure 3.6.2. Although the conformers of ethane are in rapid equilibrium with each other, the 3 kcal/mol energy difference leads to a substantial preponderance of staggered conformers (> 99.9%) at any given time. The animation below illustrates the relationship between ethane's potential energy and its dihedral angle Exercises 1) What is the most stable rotational conformation of ethane and explain why it is preferred over the other conformation? Solutions 1) Staggered, as there is less repulsion between the hydrogen atoms.
Courses/Nassau_Community_College/Principles_of_Chemistry/zz%3A_Back_Matter/20%3A_Glossary	Words (or words that have the same definition) The definition is case sensitive (Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages] (Optional) Caption for Image (Optional) External or Internal Link (Optional) Source for Definition (Eg. "Genetic, Hereditary, DNA ...") (Eg. "Relating to genes or heredity") NaN The infamous double helix https://bio.libretexts.org/ CC-BY-SA; Delmar Larsen Word(s) Definition Image Caption Link Source Sample Word 1 Sample Definition 1 NaN NaN NaN NaN
Bookshelves/Analytical_Chemistry/Instrumental_Analysis_(LibreTexts)/30%3A_Capillary_Electrophoresis_and_Capillary_Electrochromatography/30.03%3A_Applications_of_Capillary_Electrophoresis	There are several different forms of capillary electrophoresis, each of which has its particular advantages. Several of these methods are described briefly in this section. Capillary Zone Electrophoresis (CZE) The simplest form of capillary electrophoresis is capillary zone electrophoresis. In CZE we fill the capillary tube with a buffer and, after loading the sample, place the ends of the capillary tube in reservoirs that contain additional buffer. Usually the end of the capillary containing the sample is the anode and solutes migrate toward the cathode at a velocity determined by their respective electrophoretic mobilities and the electroosmotic flow. Cations elute first, with smaller, more highly charged cations eluting before larger cations with smaller charges. Neutral species elute as a single band. Anions are the last species to elute, with smaller, more negatively charged anions being the last to elute. We can reverse the direction of electroosmotic flow by adding an alkylammonium salt to the buffer solution. As shown in Figure 30.3.1 , the positively charged end of the alkyl ammonium ions bind to the negatively charged silanate ions on the capillary’s walls. The tail of the alkyl ammonium ion is hydrophobic and associates with the tail of another alkyl ammonium ion. The result is a layer of positive charges that attract anions in the buffer. The migration of these solvated anions toward the anode reverses the electroosmotic flow’s direction. The order of elution is exactly opposite that observed under normal conditions. Coating the capillary’s walls with a nonionic reagent eliminates the electroosmotic flow. In this form of CZE the cations migrate from the anode to the cathode. Anions elute into the source reservoir and neutral species remain stationary. Capillary zone electrophoresis provides effective separations of charged species, including inorganic anions and cations, organic acids and amines, and large biomolecules such as proteins. For example, CZE was used to separate a mixture of 36 inorganic and organic ions in less than three minutes [Jones, W. R.; Jandik, P. J. Chromatog . 1992 , 608 , 385–393]. A mixture of neutral species, of course, can not be resolved. Micellar Electrokinetic Capillary Chromatography (MEKC) One limitation to CZE is its inability to separate neutral species. Micellar electrokinetic capillary chromatography overcomes this limitation by adding a surfactant, such as sodium dodecylsulfate (Figure 30.3.2 a) to the buffer solution. Sodium dodecylsulfate, or SDS, consists of a long-chain hydrophobic tail and a negatively charged ionic functional group at its head. When the concentration of SDS is sufficiently large a micelle forms. A micelle consists of a spherical agglomeration of 40–100 surfactant molecules in which the hydrocarbon tails point inward and the negatively charged heads point outward (Figure 30.3.2 b). Because micelles have a negative charge, they migrate toward the cathode with a velocity less than the electroosmotic flow velocity. Neutral species partition themselves between the micelles and the buffer solution in a manner similar to the partitioning of solutes between the two liquid phases in HPLC. Because there is a partitioning between two phases, we include the descriptive term chromatography in the techniques name. Note that in MEKC both phases are mobile. The elution order for neutral species in MEKC depends on the extent to which each species partitions into the micelles. Hydrophilic neutrals are insoluble in the micelle’s hydrophobic inner environment and elute as a single band, as they would in CZE. Neutral solutes that are extremely hy- drophobic are completely soluble in the micelle, eluting with the micelles as a single band. Those neutral species that exist in a partition equilibrium between the buffer and the micelles elute between the completely hydro- philic and completely hydrophobic neutral species. Those neutral species that favor the buffer elute before those favoring the micelles. Micellar electrokinetic chromatography is used to separate a wide variety of samples, including mixtures of pharmaceutical compounds, vitamins, and explosives. Capillary Gel Electrophoresis (CGE) In capillary gel electrophoresis the capillary tubing is filled with a polymeric gel. Because the gel is porous, a solute migrates through the gel with a velocity determined both by its electrophoretic mobility and by its size. The ability to effect a separation using size is helpful when the solutes have similar electrophoretic mobilities. For example, fragments of DNA of varying length have similar charge-to-size ratios, making their separation by CZE difficult. Because the DNA fragments are of different size, a CGE separation is possible. The capillary used for CGE usually is treated to eliminate electroosmotic flow to prevent the gel from extruding from the capillary tubing. Samples are injected electrokinetically because the gel provides too much resistance for hydrodynamic sampling. The primary application of CGE is the separation of large biomolecules, including DNA fragments, proteins, and oligonucleotides. Capillary Electrochromatography (CEC) Another approach to separating neutral species is capillary electrochromatography. In CEC the capillary tubing is packed with 1.5–3 μm particles coated with a bonded stationary phase. Neutral species separate based on their ability to partition between the stationary phase and the buffer, which is moving as a result of the electroosmotic flow; Figure 30.3.3 provides a representative example for the separation of a mixture of hydrocarbons. A CEC separation is similar to the analogous HPLC separation, but without the need for high pressure pumps. Efficiency in CEC is better than in HPLC, and analysis times are shorter.
Courses/Smith_College/CHM_222_Chemistry_II%3A_Organic_Chemistry_(2025)/05%3A_Structure_Determination_-_Mass_Spectrometry_(reference_only)_and_Infrared_Spectroscopy/5.03%3A_Interpreting_Mass_Spectra_(reference_only)	Objectives After completing this section, you should be able to suggest possible molecular formulas for a compound, given the m/z value for the molecular ion, or a mass spectrum from which this value can be obtained. predict the relative heights of the M+·, (M + 1)+·, etc., peaks in the mass spectrum of a compound, given the natural abundance of the isotopes of carbon and the other elements present in the compound. interpret the fragmentation pattern of the mass spectrum of a relatively simple, known compound (e.g., hexane). use the fragmentation pattern in a given mass spectrum to assist in the identification of a relatively simple, unknown compound (e.g., an unknown alkane). Study Notes When interpreting fragmentation patterns, you may find it helpful to know that, as you might expect, the weakest carbon-carbon bonds are the ones most likely to break. You might wish to refer to the table of bond dissociation energies when attempting problems involving the interpretation of mass spectra. This page looks at how fragmentation patterns are formed when organic molecules are fed into a mass spectrometer, and how you can get information from the mass spectrum. The Origin of Fragmentation Patterns When the vaporized organic sample passes into the ionization chamber of a mass spectrometer, it is bombarded by a stream of electrons. These electrons have a high enough energy to knock an electron off an organic molecule to form a positive ion. This ion is called the molecular ion - or sometimes the parent ion and is often given the symbol M + or . The dot in this second version represents the fact that somewhere in the ion there will be a single unpaired electron. That's one half of what was originally a pair of electrons - the other half is the electron which was removed in the ionization process. The molecular ions are energetically unstable, and some of them will break up into smaller pieces. The simplest case is that a molecular ion breaks into two parts - one of which is another positive ion, and the other is an uncharged free radical. The uncharged free radical will not produce a line on the mass spectrum. Only charged particles will be accelerated, deflected and detected by the mass spectrometer. These uncharged particles will simply get lost in the machine - eventually, they get removed by the vacuum pump. The ion, X + , will travel through the mass spectrometer just like any other positive ion - and will produce a line on the stick diagram. All sorts of fragmentations of the original molecular ion are possible - and that means that you will get a whole host of lines in the mass spectrum. For example, the mass spectrum of pentane looks like this: Note The pattern of lines in the mass spectrum of an organic compound tells you something quite different from the pattern of lines in the mass spectrum of an element . With an element, each line represents a different isotope of that element. With a compound, each line represents a different fragment produced when the molecular ion breaks up. In the stick diagram showing the mass spectrum of pentane, the line produced by the heaviest ion passing through the machine (at m/z = 72) is due to the molecular ion . The tallest line in the stick diagram (in this case at m/z = 43) is called the base peak . This is usually given an arbitrary height of 100, and the height of everything else is measured relative to this. The base peak is the tallest peak because it represents the commonest fragment ion to be formed - either because there are several ways in which it could be produced during fragmentation of the parent ion, or because it is a particularly stable ion. Using Fragmentation Patterns This section will ignore the information you can get from the molecular ion (or ions). That is covered in three other pages which you can get at via the mass spectrometry menu. You will find a link at the bottom of the page. Example 12.2.1: Pentane Let's have another look at the mass spectrum for pentane: What causes the line at m/z = 57? How many carbon atoms are there in this ion? There cannot be 5 because 5 x 12 = 60. What about 4? 4 x 12 = 48. That leaves 9 to make up a total of 57. How about C 4 H 9 + then? C 4 H 9 + would be [CH 3 CH 2 CH 2 CH 2 ] + , and this would be produced by the following fragmentation: The methyl radical produced will simply get lost in the machine. The line at m/z = 43 can be worked out similarly. If you play around with the numbers, you will find that this corresponds to a break producing a 3-carbon ion: The line at m/z = 29 is typical of an ethyl ion, [CH 3 CH 2 ] + : The other lines in the mass spectrum are more difficult to explain. For example, lines with m/z values 1 or 2 less than one of the easy lines are often due to loss of one or more hydrogen atoms during the fragmentation process. Example 12.2.2: Pentan-3-one This time the base peak (the tallest peak - and so the commonest fragment ion) is at m/z = 57. But this is not produced by the same ion as the same m/z value peak in pentane. If you remember, the m/z = 57 peak in pentane was produced by [CH 3 CH 2 CH 2 CH 2 ] + . If you look at the structure of pentan-3-one, it's impossible to get that particular fragment from it. Work along the molecule mentally chopping bits off until you come up with something that adds up to 57. With a small amount of patience, you'll eventually find [CH 3 CH 2 CO] + - which is produced by this fragmentation: You would get exactly the same products whichever side of the CO group you split the molecular ion. The m/z = 29 peak is produced by the ethyl ion - which once again could be formed by splitting the molecular ion either side of the CO group. Peak Heights and Stability The more stable an ion is, the more likely it is to form. The more of a particular sort of ion that's formed, the higher its peak height will be. We'll look at two common examples of this. Carbocations (carbonium ions) Summarizing the most important conclusion from the page on carbocations: Order of stability of carbocations primary < secondary < tertiary Applying the logic of this to fragmentation patterns, it means that a split which produces a secondary carbocation is going to be more successful than one producing a primary one. A split producing a tertiary carbocation will be more successful still. Let's look at the mass spectrum of 2-methylbutane. 2-methylbutane is an isomer of pentane - isomers are molecules with the same molecular formula, but a different spatial arrangement of the atoms. Look first at the very strong peak at m/z = 43. This is caused by a different ion than the corresponding peak in the pentane mass spectrum. This peak in 2-methylbutane is caused by: The ion formed is a secondary carbocation - it has two alkyl groups attached to the carbon with the positive charge. As such, it is relatively stable. The peak at m/z = 57 is much taller than the corresponding line in pentane. Again a secondary carbocation is formed - this time, by: You would get the same ion, of course, if the left-hand CH 3 group broke off instead of the bottom one as we've drawn it. In these two spectra, this is probably the most dramatic example of the extra stability of a secondary carbocation. Acylium ions, [RCO] + Ions with the positive charge on the carbon of a carbonyl group, C=O, are also relatively stable. This is fairly clearly seen in the mass spectra of ketones like pentan-3-one. The base peak, at m/z=57, is due to the [CH 3 CH 2 CO] + ion. We've already discussed the fragmentation that produces this. Note The more stable an ion is, the more likely it is to form. The more of a particular ion that is formed, the higher will be its peak height. Using mass spectra to distinguish between compounds Suppose you had to suggest a way of distinguishing between pentan-2-one and pentan-3-one using their mass spectra. 0 1 2 pentan-2-one NaN CH3COCH2CH2CH3 pentan-3-one NaN CH3CH2COCH2CH3 Each of these is likely to split to produce ions with a positive charge on the CO group. In the pentan-2-one case, there are two different ions like this: [CH 3 CO] + [COCH 2 CH 2 CH 3 ] + That would give you strong lines at m/z = 43 and 71. With pentan-3-one, you would only get one ion of this kind: [CH 3 CH 2 CO] + In that case, you would get a strong line at 57. You don't need to worry about the other lines in the spectra - the 43, 57 and 71 lines give you plenty of difference between the two. The 43 and 71 lines are missing from the pentan-3-one spectrum, and the 57 line is missing from the pentan-2-one one. The two mass spectra look like this: As you've seen, the mass spectrum of even very similar organic compounds will be quite different because of the different fragmentation patterns that can occur. Provided you have a computer data base of mass spectra, any unknown spectrum can be computer analyzed and simply matched against the data base. Exercises
Courses/Riverland_Community_College/CHEM_1000_-_Introduction_to_Chemistry_(Riverland)/02%3A_Measurement_and_Problem_Solving/2.01%3A_Taking_Measurements	Learning Objectives Express quantities properly, using a number and a unit. A coffee maker’s instructions tell you to fill the coffee pot with 4 cups of water and to use 3 scoops of coffee. When you follow these instructions, you are measuring. When you visit a doctor’s office, a nurse checks your temperature, height, weight, and perhaps blood pressure (Figure $\PageIndex{1}$); the nurse is also measuring. Chemists measure the properties of matter and express these measurements as quantities. A quantity is an amount of something and consists of a number and a unit. The number tells us how many (or how much), and the unit tells us what the scale of measurement is. For example, when a distance is reported as “5 kilometers,” we know that the quantity has been expressed in units of kilometers and that the number of kilometers is 5. If you ask a friend how far they walk from home to school, and the friend answers “12” without specifying a unit, you do not know whether your friend walks 12 kilometers, 12 miles, 12 furlongs, or 12 yards. Both a number and a unit must be included to express a quantity properly. To understand chemistry, we need a clear understanding of the units chemists work with and the rules they follow for expressing numbers. The next two sections examine the rules for expressing numbers. Example $\PageIndex{1}$ Identify the number and the unit in each quantity. one dozen eggs 2.54 centimeters a box of pencils 88 meters per second Solution The number is one, and the unit is a dozen eggs. The number is 2.54, and the unit is centimeter. The number 1 is implied because the quantity is only a box. The unit is box of pencils. The number is 88, and the unit is meters per second. Note that in this case the unit is actually a combination of two units: meters and seconds. Key Take Away Identify a quantity properly with a number and a unit.
Courses/Chabot_College/Chem_12A%3A_Organic_Chemistry_Fall_2022/06%3A_Stereoisomerism/6.01%3A_Chirality	Learning Objective recognize and classify molecules as chiral or achiral and identify planes of symmetry Stereoisomers are isomers that differ in spatial arrangement of atoms, rather than order of atomic connectivity. One of their most interesting type of isomer is the mirror-image stereoisomers, a non-superimposable set of two molecules that are mirror image of one another. The existence of these molecules are determined by concept known as chirality . Introduction Organic compounds, molecules created around a chain of carbon atom (more commonly known as carbon backbone), play an essential role in the chemistry of life. These molecules derive their importance from the energy they carry, mainly in a form of potential energy between atomic molecules. Since such potential force can be widely affected due to changes in atomic placement, it is important to understand the concept of an isomer , a molecule sharing same atomic make up as another but differing in structural arrangements. This article will be devoted to a specific isomers called stereoisomers and its property of chirality (Figure 1). The concepts of steroisomerism and chirality command great deal of importance in modern organic chemistry , as these ideas helps to understand the physical and theoretical reasons behind the formation and structures of numerous organic molecules, the main reason behind the energy embedded in these essential chemicals. In contrast to more well-known constitutional isomerism, which develops isotopic compounds simply by different atomic connectivity, stereoisomerism generally maintains equal atomic connections and orders of building blocks as well as having same numbers of atoms and types of elements. What, then, makes stereoisomers so unique? To answer this question, the learner must be able to think and imagine in not just two-dimensional images, but also three-dimensional space. This is due to the fact that stereoisomers are isomers because their atoms are different from others in terms of spatial arrangement. Spatial Arrangement First and foremost, one must understand the concept of spatial arrangement in order to understand stereoisomerism and chirality. Spatial arrangement of atoms concern how different atomic particles and molecules are situated about in the space around the organic compound, namely its carbon chain. In this sense, spatial arrangement of an organic molecule are different another if an atom is shifted in any three-dimensional direction by even one degree. This opens up a very broad possibility of different molecules, each with their unique placement of atoms in three-dimensional space . Stereoisomers Stereoisomers are, as mentioned above, contain different types of isomers within itself, each with distinct characteristics that further separate each other as different chemical entities having different properties. Type called entaniomer are the previously-mentioned mirror-image stereoisomers, and will be explained in detail in this article. Another type, diastereomer, has different properties and will be introduced afterwards. The Many Synonyms of the Chiral Carbon Be aware - all of the following terms can be used to describe a chiral carbon. chiral carbon = asymmetric carbon = optically active carbon = stereo carbon Enantiomers This type of stereoisomer is the essential mirror-image, non-superimposable type of stereoisomer introduced in the beginning of the article. Figure 3 provides a perfect example; note that the gray plane in the middle demotes the mirror plane. Note that even if one were to flip over the left molecule over to the right, the atomic spatial arrangement will not be equal. This is equivalent to the left hand - right hand relationship, and is aptly referred to as 'handedness' in molecules. This can be somewhat counter-intuitive, so this article recommends the reader try the 'hand' example. Place both palm facing up, and hands next to each other. Now flip either side over to the other. One hand should be showing the back of the hand, while the other one is showing the palm. They are not same and non-superimposable. This is where the concept of chirality comes in as one of the most essential and defining idea of stereoisomerism. Chirality Chirality essentially means 'mirror-image, non-superimposable molecules', and to say that a molecule is chiral is to say that its mirror image (it must have one) is not the same as it self. Whether a molecule is chiral or achiral depends upon a certain set of overlapping conditions. Figure 4 shows an example of two molecules, chiral and achiral, respectively. Notice the distinct characteristic of the achiral molecule: it possesses two atoms of same element. In theory and reality, if one were to create a plane that runs through the other two atoms, they will be able to create what is known as bisecting plane: The images on either side of the plan is the same as the other (Figure 4). In this case, the molecule is considered 'achiral'. In other words, to distinguish chiral molecule from an achiral molecule, one must search for the existence of the bisecting plane in a molecule. All chiral molecules are deprive of bisecting plane, whether simple or complex. As a universal rule, no molecule with different surrounding atoms are achiral. Chirality is a simple but essential idea to support the concept of stereoisomerism, being used to explain one type of its kind. The chemical properties of the chiral molecule differs from its mirror image, and in this lies the significance of chilarity in relation to modern organic chemistry. Exercise 1 Identify the following as either a constitutional isomer or stereoisomer. If stereoisomer, determine if it is an enantiomer or diastereomer. Explain the reason behind the answer. Also mark chirality for each molecule. a) b) c) Solutions a) achiral b) chiral c) chiral Exercise 2 Identify the chiral centers in each of the following: Solutions References Anslyn, Eric V. and Dougherty, Dennis A. Modern Physical Organic Chemistry. Chicago, IL.: University Science. 2005 Hick, Janice M. The Physical Chemistry of Chirality. New York, N.Y.: An American Chemical Society Publication. 2001. Vollhardt, K. Peter C. and Schore, Neil E. Organic Chemistry: Structure and Function. Fifth Edition. New York, N.Y.: W. H. Freeman Company, 2007.
Ancillary_Materials/Worksheets/Worksheets%3A_Inorganic_Chemistry/Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry/12%3A_Molecular_Orbitals/12.07%3A_Experimental_Evidence	The molecular orbital picture of dioxygen differs from the Lewis picture. Both models predict an oxygen-oxygen double bond, but one model suggests unpaired electrons whereas the other indicates an electron-paired system. Often, there is experimental evidence available to check the reliability of predictions about structure. These data include measurements of bond lengths and bond strengths as well as magnetic properties. Bond dissociation energy data tell us how difficult it is to separate one atom from another in a molecule. Bond order is one of the factors that influences bond strength. Thus, measuring a bond dissociation energy is one way to confirm that dioxygen really does contain an oxygen-oxygen double bond. First, we need something to compare it to. Peroxides (such as hydrogen peroxide, H2O2, or sodium peroxide, Na2O2) probably contain oxygen-oxygen single bonds, according to their Lewis structures. These bonds are relatively weak, costing about 35 kcal/mol to break. In contrast, the bond in dioxygen costs about 70 kcal/mol to break. Its bond is about twice as strong; it is a double bond. Bond dissociation energies can be used to determine how many bonds there are between two atoms. Bond dissociation energies can be complicated to measure. They require a comparison of energy changes in numerous chemical reactions so that the energy change resulting from cleavage of a specific bond can be inferred. In contrast, infrared absorption frequencies are easy to measure. They simply require shining infrared light through a sample and measuring what frequencies of the light are absorbed by the material. (A related technique, Raman spectroscopy, gives similar information by measuring subtle changes in the frequency of laser light that is scattered off a sample). The frequencies absorbed depend on what bonds are present in the material. These frequencies vary according to two basic factors: the weights of the atoms at the ends of the bond, and the strength of the bond between them. The stronger the bond, the higher the absorption frequency. Peroxides absorb infrared light at around 800 cm-1 (this unusual frequency unit is usually pronounced "wavenumbers"). Dioxygen absorbs infrared light around 1300 cm-1 . Since the atoms at the ends of the bond in both peroxide and dioxygen are oxygens, we can be sure that this difference in frequency is not due to a difference in mass. It is due to a difference in bond strength. The bond in dioxygen is much stronger than the O-O bond in peroxide, because the former is a double bond and the latter is a single bond. Vibrational spectroscopies (IR and Raman spectroscopy) can give information about the bond order between two atoms. A third measure of bond order is found in bond length measurements. The more strongly bound two atoms are, the closer they are together. An O=O bond should be shorter than an O-O bond. Bond lengths can be measured by microwave spectroscopy (usually for gas-phase molecules), in which frequencies absorbed depend on the distance between the molecules. Alternatively, bond lengths can be measured by x-ray crystallography. X-rays can be diffracted through crystals of solid materials. The interference pattern that is produced can be mathematically decoded to produce a three-dimensional map of where all the atoms are in the material. The distances between these atoms can be measured very accurately. The O-O bond in peroxides are about 1.49 Angstroms long (an Angstrom is 10-10 m; this unit is often used for bond lengths because it is a convenient size for this task. Covalent bonds are generally one to three Angstroms long). The O-O bond in dioxygen is about 1.21 A long. The O-O bond in dioxygen is shorter and stronger than in a peroxide. Bond length data provides insight into the bond order. In addition to bond order, there is the question of electron pairing in dioxygen. The Lewis structure suggests electrons are paired in dioxygen. The molecular orbital picture suggests two unpaired electrons. Compounds with paired electrons are referred to as diamagnetic. Those with unpaired electrons are called paramagnetic. Paramagnetic substances interact strongly with magnetic fields. It turns out that oxygen does interact with a magnetic fields. A sample of liquid-phase oxygen can be held between the poles of a magnet. Oxygen has unpaired electrons. This finding is consistent with molecular orbital theory, but not with simple Lewis structures. Thus, MO theory tells us something that the Lewis picture cannot. Magnetic information, and measurements of magnetism, give us experimental evidence of spin states. We can tell if electrons are paired, unpaired, and how many unpaired spins there are. A final important source of experimental data is photoelectron spectroscopy. Photoelectron spectroscopy gives information about the electron energy levels in an atom or compound. In this technique, gas-phase molecules are subjected to high-energy electromagnetic radiation, such as ultraviolet light or X-rays. Electrons are ejected from various energy levels in the molecule, and the binding energies of electrons in those levels is determined. Thus, photoelectron spectroscopy provides verification for exactly the sort of information that quantitative molecular orbital calculations are designed to deliver. Photoelectron spectroscopy tells how much energy is needed to remove electrons from various energy levels in a molecule. This technique gives us an accurate experimental picture of the energy levels that we predict with molecular orbital calculations. Problem MO 7.1 In the previous section you were asked to draw MO diagrams for some molecules. Determine whether the molecules in problems MO6.3-6.8 are paramagnetic or diamagnetic. Problem MO7.2. Use a MO diagram to determine which species in the following pairs will have the longer bond. Give an explanation for your choice. a. N 2 or N 2 + b. N 2 or N 2 - Problem MO7.3 Use a MO diagram to determine which species in the following pairs will have the stronger bond. Give an explanation for your choice. a. O 2 or O 2 + b. O 2 or O 2 -
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/01%3A_Quantum_Fundamentals/1.68%3A_Another_Look_at_the_Wigner_Function	The Wigner function, W(x,p), is a phase space distribution function which behaves similarly to the coordinate $\left(\|\Psi(x)\|^{2}\right)$ and momentum $\left(\|\tilde{\Psi}(p)\|^{2}\right)$ distribution functions. For example, its integral over phase space is normalized. \[ \iint W(x, p) d x d p=1 \nonumber \] In phase space, position and momentum are represented by multiplicative operators, so the calculation of their expectation values has a classical appearance. This, naturally, is part of the appeal of phase space quantum mechanical calculations. \[ \langle x\rangle=\iint x W(x, p) d x d p \nonumber \] \[ \langle p\rangle=\iint p W(x, p) d x d p \nonumber \] While the Wigner function is real, unlike $\|\Psi(x)\|^{2}$ and $\|\tilde{\Psi}(p)\|^{2}$, it can take on negative values making it impossible to interpret it as a genuine probability distribution function. For this reason it is frequently referred to as a quasi-probability function, and loses some of its classical appeal. In any case, the Wigner function is redundant in the sense that it is generated from a Schrödinger coordinate or momentum wave equation. In what follows, the quantum mechanical Wigner distribution function will be rationalized by reference to familiar classical concepts, such as position, momentum and trajectory. In classical physics, a trajectory is a temporal sequence of position and momentum states. Let us try to represent a classical trajectory in a quantum mechanical formalism. Suppose a quantum mechanical object, a quon (thank you Nick Herbert), in state \|$\Psi$> moves from position x –s/2 to position x + s/2. We might represent this transition quantum mechanically as the product of two coordinate space probability amplitudes (reading from left to right). \[ \langle x- \frac{s}{2} \| \Psi\rangle\langle\Psi \| x+ \frac{s}{2} \rangle \nonumber \] Thus far we have a coordinate representation of a transition from one spatial location to another. However, a phase space description also requires a dynamic (or motional) parameter such as momentum. We can introduce momentum by first rearranging the above product of amplitudes as follows. \[ \langle\Psi \| x+ \frac{s}{2} \rangle\langle x- \frac{s}{2} \| \Psi\rangle \nonumber \] This convolution of positional states takes on the coherent character of a trajectory with the insertion of the following momentum projector (see Feynman Lectures Volume 3) coupling the two spatial states. \[ \langle x+ \frac{s}{2} \| p\rangle\langle p \| x- \frac{s}{2} \rangle \nonumber \] This gives us a quantum trajectory expressed in the following product of Dirac brackets, \[ \langle\Psi \| x+ \frac{s}{2} \rangle\langle x+ \frac{s}{2} \| p\rangle\langle p \| x- \frac{s}{2} \rangle\langle x- \frac{s}{2} \| \Psi\rangle \nonumber \] The four Dirac brackets are read now from right to left as follows: (1) is the amplitude that a particle in the state $\Psi$ has position (x - $\frac{s}{2})); (2) is the amplitude that a particle with position (x - \(\frac{s}{2}$) has momentum p; (3) is the amplitude that a particle with momentum p has position (x + $\frac{s}{2}$); (4) is the amplitude that a particle with position (x + $\frac{s}{2}$) is (still) in the state $\Psi$. Integration over s yields the Wigner distribution function, which is a superposition of all possible quantum trajectories of the state $\Psi$, which interfere constructively and destructively, providing a quasi-probability distribution in phase space. \[ \int\langle\Psi \| x+ \frac{s}{2} \rangle\langle x+ \frac{s}{2} \| p\rangle\langle p \| x- \frac{s}{2} \rangle\langle x- \frac{s}{2} \| \Psi\rangle d s = \frac{1}{h} \int \Psi(x+ \frac{s}{2})^{*} \exp \left(i \frac{p s}{\hbar}\right) \Psi(x- \frac{s}{2}) d s \nonumber \] given that \[ \langle x+ \frac{s}{2} \| p\rangle\langle p \| x- \frac{s}{2}\rangle=\frac{1}{\sqrt{h}} \exp \left(i \frac{p(x+ \frac{s}{2})}{\hbar}\right) \frac{1}{\sqrt{h}} \exp \left(-i \frac{p(x-\frac{s}{2})}{\hbar}\right)=\frac{1}{h} \exp \left(i \frac{p s}{\hbar}\right) \nonumber \] While the Wigner distribution is more than a quantum mechanical curiosity and plays an important role in current research (see references below), it is also true, as mentioned above, that it is redundant because it is generated from either a coordinate or momentum wave function. In Dan Styer’s words it is useful in exploring the quantum/classical transition, but it does not eliminate quantum weirdness – it simply repackages it (see reference 12). Having said this it should be acknowledged that the Wigner phase-space distribution has been measured for the double slit experiment using tomographic techniques (see references 17-19). Literature references to the Wigner distribution function: E. P. Wigner, “On the quantum correction for thermodynamic equilibrium,” Phys. Rev. 40 , 749 – 759 (1932). M. Hillery, R. F. O’Connell, M. O. Scully, and E. P. Wigner, “Distribution functions in physics: Fundamentals,” Phys. Rep. 106 , 121 – 167 (1984). Y. S. Kim and E. P. Wigner, “Canonical transformations in quantum mechanics,” Am. J. Phys. 58 , 439 – 448 (1990). J. Snygg, “Wave functions rotated in phase space,” Am. J. Phys. 45 , 58 – 60 (1977). J. Snygg, “Uses of operator functions to construct refined correspondence principle via the quantum mechanics of Wigner and Moyal,” Am. J. Phys. 48 , 964 – 970 (1980). N. Mukunda, “Wigner distribution for angle coordinates in quantum mechanics,” Am. J. Phys. 47 , 192 – 187 (1979). S. Stenholm, “The Wigner function: I. The physical interpretation,” Eur. J. Phys. 1, 244 – 248 (1980). G. Mourgues, J. C. Andrieux, and M. R. Feix, “Solutions of the Schrödinger equation for a system excited by a time Dirac pulse of pulse of potential. An example of the connection with the classical limit through a particular smoothing of the Wigner function,” Eur. J. Phys. 5 , 112 – 118 (1984). M. Casas, H. Krivine, and J. Martorell, “On the Wigner transforms of some simple systems and their semiclassical interpretations,” Eur. J.Phys. 12 , 105 – 111 (1991). R. A. Campos, “Correlation coefficient for incompatible observables of the quantum mechanical harmonic oscillator,” Am. J. Phys. 66 , 712 – 718 (1998). M. Belloni, M. A. Doncheski, and R. W. Robinett, “Wigner quasi-probability distribution for the infinite square well: Energy eigenstates and time-dependent wave packets,” Am. J. Phys. 72 , 1183 – 1192 (2004). D. F. Styer, et al., “Nine formulations of quantum mechanics,” Am. J. Phys. 70 , 288 – 297 (2002). H-W Lee, “Spreading of a free wave packet,” Am. J. Phys. 50 , 438 – 440 (1982). D. Home and S. Sengupta, “Classical limit of quantum mechanics,” Am. J. Phys. 51 , 265 – 267 (1983). W. H. Zurek, “Decoherence and the transition from quantum to classical,” Phys. Today 44 , 36 – 44 (October 1991). M. C. Teich and B. E. A. Saleh, “Squeezed and antibunched light,” Phys. Today 43 , 26 – 34 (June 1990). Ch. Kurtsiefer, T. Pfau, and J.Mlynek, “Measurement of the Wigner function of an ensemble of helium atoms,” Nature 386 , 150-153 (1997). M. Freyberger and W. P. Schleich, “True vision of a quantum state,” Nature 386 , 121-122 (1997). D. Leibfried, T. Pfau, and C. Monroe, “Shadows and mirrors: Reconstructing quantum states of motion,” Phys. Today 51 , 22 – 28 (April 1998). W. P. Schleich and G. Süssmann, “A jump shot at the Wigner distribution,” Phys. Today 44 , 146 – 147 (October 1991). R. A. Campos, “Correlation coefficient for incompatible observables of the quantum harmonic oscillator,” Am. J. Phys. 66 , 712 – 718 (1998). R. A. Campos, “Wigner quasiprobability distribution for quantum superpositions of coherent states, a Comment on ‘Correlation coefficient for incompatible observables of the quantum harmonic oscillator,’” Am. J. Phys. 67 , 641 – 642 (1999). C. C. Gerry and P. L. Knight, “Quantum superpositions and Schrödinger cat states in quantum optics,” Am. J. Phys. 65 , 964 – 974 (1997). K. Ekert and P. L. Knight, “Correlations and squeezing of two-mode oscillations,” Am. J. Phys. 57 , 692 – 697 (1989). W. B. Case, “Wigner functions and Weyl transforms for pedestrians,” Am. J. Phys. 76 , 937 – 946 (2008). M. G. Raymer, “Measuring the quantum mechanical wave function,” Contemp. Phys. 38 , 343 – 355 (1997). F. Rioux, “ Illuminating the Wigner function with Dirac notation ,” F. Rioux, “The Wigner distribution for the double-slit experiment,” www.users.csbsju.edu/~frioux/wigner/DBL-SLIT-NEW.pdf F. Rioux, “ Basic quantum mechanics in coordinate space, momentum space and phase space ,” F. Rioux, “ The Wigner distribution for the harmonic oscillator ,” F. Rioux, “ The Wigner distribution for the particle in a box ,” F. Rioux, “The time-dependent Wigner distribution for harmonic oscillator transitions,” F. Rioux, “The Wigner distribution distinguishes between a superposition and a mixture ,”
Courses/University_of_North_Texas/UNT%3A_CHEM_1410_-_General_Chemistry_for_Science_Majors_I/Text/01%3A_Introduction_-_The_Ambit_of_Chemistry/1.02%3A_What_Chemists_Do	What are some of the things that chemists do? Like most scientists, they observe and measure components of the natural world. Based on these observations they try to place things into useful, appropriate categories and to formulate scientific laws which summarize the results of a great many observations. Indeed, it is a fundamental belief of all science that natural events do not occur in a completely unpredictable fashion. Instead, they obey natural laws. Therefore observations and measurements made on one occasion can be duplicated by the same or another person on another occasion. Communication of such results, another important activity, affords an opportunity for the entire scientific community to test an individual’s work. Eventually a consensus is reached, and there is agreement on a new law. Like other scientists, chemists try to explain their observations and laws by means of theories or models. They constantly make use of atoms, molecules , and other very small particles. Using such theories as their guides, chemists synthesize new materials. Well over 3 million compounds are now known, and more than 9000 are in large-scale commercial production. Even a backpacker going “back to nature” takes along synthetic materials such as nylon, aluminum, and aspirin. Chemists also analyze the substances they make and those found in nature. Determining the composition of a substance is the first step in understanding its chemical properties, and detection of very small quantities of some materials in the natural world is essential in controlling air and water pollution. Another role that chemists play is in studying the processes (chemical reactions) by which one substance can be transformed into another. Will the reactions occur without prodding? If so, how quickly? Is energy given off? Can the reactions be controlled - made to occur only when we want them to? These and many other problems that interest chemists are the subject matter of this online textbook. We will return to each several times and in increasing detail. Do not forget, however, that chemistry is more than just what chemists do. Many persons in other sciences as well as in daily life are constantly doing chemistry, whether they call it by name or not. Indeed, each of us is an intricate combination of chemicals, and everything we do depends on chemical reactions. Although space limitations will prevent us from exploring all but a fraction of the applications of chemistry to other fields, we have included such excursions as often as seemed appropriate. From them we hope that you will be able to learn how to apply chemical facts and principles to the problems you will face in the future. Many of the problems are probably not even known yet, and we could not possibly anticipate them. If you have learned how to think “chemically” or “scientifically,” however, we believe that you will be better prepared to face them. Elements, Compounds and Mixtures Chemists have a unique way of characterizing the objects of their study. To a chemist, matter containing one kind of atom or molecule is considered a pure substance , so water (H 2 O) and aluminum metal (Al) are pure substances. To a chemist, aluminum is an element , the simplest kind of pure substance which contains only one kind of atom (a physicist might consider aluminum a very complex "mixture" of things like leptons, protons, quarks). To a chemist, the pure substance water is a compound (it contains two kinds of atoms bound to one another in just one kind of molecule). An environmentalist might consider water "pure" even if it contains the normal amount of dissolved oxygen and carbon dioxide, but no other "pollutants". To a chemist, water containing oxygen is no longer a pure substance, but a mixture . In pure water, the ratio of hydrogen to oxygen atoms is always 2:1, but as soon as oxygen is dissolved in the water, the ratio is no longer fixed (because oxygen does not form a new molecule by reacting with H 2 O), a sure sign of impurity to a chemist. Chemical Reactions Because the dissolved oxygen does not react with water to form a new compound, we consider "dissolving" to be a physical process (one that is not understood in terms of atoms or molecules and bond formation or breaking). Changes like dissolution, evaporation, and condensation are considered "physical changes" because, while they are clearly interesting changes, they might be studied in physics with no reference to atomic recombinations. If, however, the dissolved oxygen is used by a fish to convert food to carbon dioxide (CO 2 ) and water (H 2 O), a chemical reaction has occurred, because the process is best understood at the level of atoms, molecules, and bonds between them. The process by which the dissolved oxygen is absorbed by fish might be considered a "biological process" because knowledge of the cellular structure of the fish gills is important in its understanding. Once more, the mode of understanding is more important than the subject of study in determining whether a process is considered "chemical", "physical", or "biological". Later in this book, we will use atomic and molecular concepts like non-covalent bond formation to understand dissolution, and then it should probably be considered a chemical process, and it will become clear that no absolute distinction between physical and chemical processes is possible (or necessary). Homogeneous and Heterogeneous It is interesting that water containing ice cubes is a pure substance to a chemist (only H 2 O molecules are present), but it is heterogeneous . To a chemist, "heterogeneous" does not indicate "impurity", nor does "homogeneous" indicate purity. Oxygen (or salt) dissolved in water is a homogeneous mixture , because every part of it looks like every other part. Of course, when we say "homogeneous" we usually assume that the substance is not being observed with anything with higher magnification than a crude microscope. ADAPT $\PageIndex{1}$
Courses/can/CHEM_210%3A_General_Chemistry_I_(An_Atoms_Up_Approach)	This course is the first half of a two-semester sequence in general chemistry intended for students pursuing majors in physical sciences, biological sciences and engineering. The topics include atomic theory, stoichiometry, chemical bonding, thermochemistry, periodicity, molecular geometry, gas laws, solution stoichiometry, intermolecular forces and selected topics covering redox and acid-base reactions. The laboratory program includes gravimetric, colorimetric, and selected volumetric methods of analysis. Students are introduced to spreadsheet and graphical analysis of laboratory data and molecular modeling, and perform a variety of computer-interfaced experiments. Front Matter 1: Chemistry is the Science of Everything 2: Matter - An Introduction 3: Units, Measurements, and Conversions 4: Atoms and Elements 5: The Quantum Model of the Atom 6: Periodic Law and Periodic Properties of the Elements 7: Molecules and Compounds 8: Chemical Bonding I - The Lewis Model 9: Chemical Bonding II - Advanced Bonding Models 10: Chemical Reactions and Chemical Quantities 11: Solutions, Concentration, and Dilution 12: Aqueous Reactions 13: Thermochemistry 14: Gases 15: Gases and Gas Laws 16: Solids, Liquids, and Phase Changes Back Matter
Courses/Windward_Community_College/BIOC_141%3A_Fundamentals_of_Biochemistry_(Colmenares_and_Ashburn)/02%3A_Elements_Atoms_and_the_Periodic_Table	Just as a language has an alphabet from which words are built, chemistry has an alphabet from which matter is described. However, the chemical alphabet is larger than the one we use for spelling. You may have already figured out that the chemical alphabet consists of the chemical elements. Their role is central to chemistry, for they combine to form the millions and millions of known compounds. 2.0: Prelude to Elements, Atoms, and the Periodic Table The hardest material in the human body is tooth enamel. It has to be hard so that our teeth can serve us for a lifetime of biting and chewing; however, tough as it is, tooth enamel is susceptible to chemical attack. Acids found in some foods or made by bacteria that feed on food residues on our teeth are capable of dissolving enamel. Unprotected by enamel, a tooth will start to decay, thus developing cavities and other dental problems. 2.1: The Elements All matter is composed of elements. Chemical elements are represented by a one- or two-letter symbol. 2.2: Atomic Theory Atoms are the ultimate building blocks of all matter. The modern atomic theory establishes the concepts of atoms and how they compose matter. 2.3: The Structure of Atoms Atoms are composed of three main subatomic particles: protons, neutrons, and electrons. Protons and neutrons are grouped together in the nucleus of an atom, while electrons orbit about the nucleus. 2.4: Nuclei of Atoms Elements can be identified by their atomic number and mass number. Isotopes are atoms of the same element that have different masses. 2.5: Atomic Masses Atoms have a mass that is based largely on the number of protons and neutrons in their nucleus. 2.6: Arrangements of Electrons Electrons are organized into shells and subshells about the nucleus of an atom. 2.7: The Periodic Table The chemical elements are arranged in a chart called the periodic table. Some characteristics of the elements are related to their position on the periodic table. 2.E: Elements, Atoms, and the Periodic Table (Exercises) These are homework exercises to accompany Chapter 2 of the Ball et al. "The Basics of GOB Chemistry" Textmap. 2.S: Elements, Atoms, and the Periodic Table (Summary) To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms and ask yourself how they relate to the topics in the chapter.
Courses/can/CHEM_210%3A_General_Chemistry_I_(An_Atoms_Up_Approach)/09%3A_Chemical_Bonding_II_-_Advanced_Bonding_Models/9.02%3A_Hybridization_and_Hybrid_Orbitals_in_VBT	The localized valence bond theory uses a process called hybridization , in which atomic orbitals that are similar in energy but not equivalent are combined mathematically to produce sets of equivalent orbitals that are properly oriented to form bonds. These new combinations are called hybrid atomic orbitals because they are produced by combining ( hybridizing ) two or more atomic orbitals from the same atom. Hybridization of s and p Orbitals In BeH 2 , we can generate two equivalent orbitals by combining the 2 s orbital of beryllium and any one of the three degenerate 2 p orbitals. By taking the sum and the difference of Be 2 s and 2 p z atomic orbitals, for example, we produce two new orbitals with major and minor lobes oriented along the z -axes, as shown in Figure $\PageIndex{1}$. Because the difference A − B can also be written as A + (−B), in Figure $\PageIndex{2}$ and subsequent figures we have reversed the phase(s) of the orbital being subtracted, which is the same as multiplying it by −1 and adding. This gives us Equation \ref{9.5.1b}, where the value $\frac{1}{\sqrt{2}}$ is needed mathematically to indicate that the 2 s and 2 p orbitals contribute equally to each hybrid orbital. \[sp = \dfrac{1}{\sqrt{2}} (2s + 2p_z) \label{9.5.1a} \] and \[sp = \dfrac{1}{\sqrt{2}} (2s - 2p_z) \label{9.5.1b} \] The nucleus resides just inside the minor lobe of each orbital. In this case, the new orbitals are called sp hybrids because they are formed from one s and one p orbital. The two new orbitals are equivalent in energy, and their energy is between the energy values associated with pure s and p orbitals, as illustrated in this diagram: Because both promotion and hybridization require an input of energy, the formation of a set of singly occupied hybrid atomic orbitals is energetically uphill. The overall process of forming a compound with hybrid orbitals will be energetically favorable only if the amount of energy released by the formation of covalent bonds is greater than the amount of energy used to form the hybrid orbitals (Figure $\PageIndex{4}$). As we will see, some compounds are highly unstable or do not exist because the amount of energy required to form hybrid orbitals is greater than the amount of energy that would be released by the formation of additional bonds. The concept of hybridization also explains why boron, with a 2 s 2 2 p 1 valence electron configuration, forms three bonds with fluorine to produce BF 3 , as predicted by the Lewis and VSEPR approaches. With only a single unpaired electron in its ground state, boron should form only a single covalent bond. By the promotion of one of its 2 s electrons to an unoccupied 2 p orbital, however, followed by the hybridization of the three singly occupied orbitals (the 2 s and two 2 p orbitals), boron acquires a set of three equivalent hybrid orbitals with one electron each, as shown here: Looking at the 2 s 2 2 p 2 valence electron configuration of carbon, we might expect carbon to use its two unpaired 2 p electrons to form compounds with only two covalent bonds. We know, however, that carbon typically forms compounds with four covalent bonds. We can explain this apparent discrepancy by the hybridization of the 2 s orbital and the three 2 p orbitals on carbon to give a set of four degenerate sp 3 (“s-p-three” or “s-p-cubed”) hybrid orbitals, each with a single electron: In addition to explaining why some elements form more bonds than would be expected based on their valence electron configurations, and why the bonds formed are equal in energy, valence bond theory explains why these compounds are so stable: the amount of energy released increases with the number of bonds formed. In the case of carbon, for example, much more energy is released in the formation of four bonds than two, so compounds of carbon with four bonds tend to be more stable than those with only two. Carbon does form compounds with only two covalent bonds (such as CH 2 or CF 2 ), but these species are highly reactive, unstable intermediates that only form in certain chemical reactions. Valence bond theory explains the number of bonds formed in a compound and the relative bond strengths. The bonding in molecules such as NH 3 or H 2 O, which have lone pairs on the central atom, can also be described in terms of hybrid atomic orbitals. In NH 3 , for example, N, with a 2 s 2 2 p 3 valence electron configuration, can hybridize its 2 s and 2 p orbitals to produce four sp 3 hybrid orbitals. Placing five valence electrons in the four hybrid orbitals, we obtain three that are singly occupied and one with a pair of electrons: The three singly occupied sp 3 lobes can form bonds with three H atoms, while the fourth orbital accommodates the lone pair of electrons. Similarly, H 2 O has an sp 3 hybridized oxygen atom that uses two singly occupied sp 3 lobes to bond to two H atoms, and two to accommodate the two lone pairs predicted by the VSEPR model. Such descriptions explain the approximately tetrahedral distribution of electron pairs on the central atom in NH 3 and H 2 O. Unfortunately, however, recent experimental evidence indicates that in NH 3 and H 2 O, the hybridized orbitals are not entirely equivalent in energy, making this bonding model an active area of research. Example $\PageIndex{1}$ Use the VSEPR model to predict the number of electron pairs and molecular geometry in each compound and then describe the hybridization and bonding of all atoms except hydrogen. H 2 S CHCl 3 Given: two chemical compounds Asked for: number of electron pairs and molecular geometry, hybridization, and bonding Strategy: Using the VSEPR approach to determine the number of electron pairs and the molecular geometry of the molecule. From the valence electron configuration of the central atom, predict the number and type of hybrid orbitals that can be produced. Fill these hybrid orbitals with the total number of valence electrons around the central atom and describe the hybridization. Solution: A H 2 S has four electron pairs around the sulfur atom with two bonded atoms, so the VSEPR model predicts a molecular geometry that is bent, or V shaped. B Sulfur has a 3 s 2 3 p 4 valence electron configuration with six electrons, but by hybridizing its 3 s and 3 p orbitals, it can produce four sp 3 hybrids. If the six valence electrons are placed in these orbitals, two have electron pairs and two are singly occupied. The two sp 3 hybrid orbitals that are singly occupied are used to form S–H bonds, whereas the other two have lone pairs of electrons. Together, the four sp 3 hybrid orbitals produce an approximately tetrahedral arrangement of electron pairs, which agrees with the molecular geometry predicted by the VSEPR model. A The CHCl 3 molecule has four valence electrons around the central atom. In the VSEPR model, the carbon atom has four electron pairs, and the molecular geometry is tetrahedral. B Carbon has a 2 s 2 2 p 2 valence electron configuration. By hybridizing its 2 s and 2 p orbitals, it can form four sp 3 hybridized orbitals that are equal in energy. Eight electrons around the central atom (four from C, one from H, and one from each of the three Cl atoms) fill three sp 3 hybrid orbitals to form C–Cl bonds, and one forms a C–H bond. Similarly, the Cl atoms, with seven electrons each in their 3 s and 3 p valence subshells, can be viewed as sp 3 hybridized. Each Cl atom uses a singly occupied sp 3 hybrid orbital to form a C–Cl bond and three hybrid orbitals to accommodate lone pairs. Exercise $\PageIndex{1}$ Use the VSEPR model to predict the number of electron pairs and molecular geometry in each compound and then describe the hybridization and bonding of all atoms except hydrogen. the BF 4 − ion hydrazine (H 2 N–NH 2 ) Answer a B is sp 3 hybridized; F is also sp 3 hybridized so it can accommodate one B–F bond and three lone pairs. The molecular geometry is tetrahedral. Answer b Each N atom is sp 3 hybridized and uses one sp 3 hybrid orbital to form the N–N bond, two to form N–H bonds, and one to accommodate a lone pair. The molecular geometry about each N is trigonal pyramidal. The number of hybrid orbitals used by the central atom is the same as the number of electron pairs around the central atom. Hybridization Using d Orbitals Hybridization is not restricted to the ns and np atomic orbitals. The bonding in compounds with central atoms in the period 3 and below can also be described using hybrid atomic orbitals. In these cases, the central atom can use its valence ( n − 1) d orbitals as well as its ns and np orbitals to form hybrid atomic orbitals, which allows it to accommodate five or more bonded atoms (as in PF 5 and SF 6 ). Using the ns orbital, all three np orbitals, and one ( n − 1) d orbital gives a set of five sp 3 d hybrid orbitals that point toward the vertices of a trigonal bipyramid (part (a) in Figure $\PageIndex{7}$). In this case, the five hybrid orbitals are not all equivalent: three form a triangular array oriented at 120° angles, and the other two are oriented at 90° to the first three and at 180° to each other. Similarly, the combination of the ns orbital, all three np orbitals, and two nd orbitals gives a set of six equivalent sp 3 d 2 hybrid orbitals oriented toward the vertices of an octahedron (part (b) in Figure 9.5.6). In the VSEPR model, PF 5 and SF 6 are predicted to be trigonal bipyramidal and octahedral, respectively, which agrees with a valence bond description in which sp 3 d or sp 3 d 2 hybrid orbitals are used for bonding. Example $\PageIndex{2}$ What is the hybridization of the central atom in each species? Describe the bonding in each species. XeF 4 SO 4 2 − SF 4 Given: three chemical species Asked for: hybridization of the central atom Strategy: Determine the geometry of the molecule using the strategy in Example $\PageIndex{1}$. From the valence electron configuration of the central atom and the number of electron pairs, determine the hybridization. Place the total number of electrons around the central atom in the hybrid orbitals and describe the bonding. Solution: A Using the VSEPR model, we find that Xe in XeF 4 forms four bonds and has two lone pairs, so its structure is square planar and it has six electron pairs. The six electron pairs form an octahedral arrangement, so the Xe must be sp 3 d 2 hybridized. B With 12 electrons around Xe, four of the six sp 3 d 2 hybrid orbitals form Xe–F bonds, and two are occupied by lone pairs of electrons. A The S in the SO 4 2 − ion has four electron pairs and has four bonded atoms, so the structure is tetrahedral. The sulfur must be sp 3 hybridized to generate four S–O bonds. B Filling the sp 3 hybrid orbitals with eight electrons from four bonds produces four filled sp 3 hybrid orbitals. A The S atom in SF 4 contains five electron pairs and four bonded atoms. The molecule has a seesaw structure with one lone pair: To accommodate five electron pairs, the sulfur atom must be sp 3 d hybridized. B Filling these orbitals with 10 electrons gives four sp 3 d hybrid orbitals forming S–F bonds and one with a lone pair of electrons. Exercise $\PageIndex{2}$ What is the hybridization of the central atom in each species? Describe the bonding. PCl 4 + BrF 3 SiF 6 2 − Answer a sp 3 with four P–Cl bonds Answer a sp 3 d with three Br–F bonds and two lone pairs Answer a sp 3 d 2 with six Si–F bonds Hybridization using d orbitals allows chemists to explain the structures and properties of many molecules and ions. Like most such models, however, it is not universally accepted. Nonetheless, it does explain a fundamental difference between the chemistry of the elements in the period 2 (C, N, and O) and those in period 3 and below (such as Si, P, and S). Period 2 elements do not form compounds in which the central atom is covalently bonded to five or more atoms, although such compounds are common for the heavier elements. Thus whereas carbon and silicon both form tetrafluorides (CF 4 and SiF 4 ), only SiF 4 reacts with F − to give a stable hexafluoro dianion, SiF 6 2 − . Because there are no 2 d atomic orbitals, the formation of octahedral CF 6 2 − would require hybrid orbitals created from 2 s , 2 p , and 3 d atomic orbitals. The 3 d orbitals of carbon are so high in energy that the amount of energy needed to form a set of sp 3 d 2 hybrid orbitals cannot be equaled by the energy released in the formation of two additional C–F bonds. These additional bonds are expected to be weak because the carbon atom (and other atoms in period 2) is so small that it cannot accommodate five or six F atoms at normal C–F bond lengths due to repulsions between electrons on adjacent fluorine atoms. Perhaps not surprisingly, then, species such as CF 6 2 − have never been prepared. Example $\PageIndex{3}$: $\ce{OF4}$ What is the hybridization of the oxygen atom in OF 4 ? Is OF 4 likely to exist? Given: chemical compound Asked for: hybridization and stability Strategy: Predict the geometry of OF 4 using the VSEPR model. From the number of electron pairs around O in OF 4 , predict the hybridization of O. Compare the number of hybrid orbitals with the number of electron pairs to decide whether the molecule is likely to exist. Solution: A The VSEPR model predicts that OF 4 will have five electron pairs, resulting in a trigonal bipyramidal geometry with four bonding pairs and one lone pair. B To accommodate five electron pairs, the O atom would have to be sp 3 d hybridized. The only d orbital available for forming a set of sp 3 d hybrid orbitals is a 3 d orbital, which is much higher in energy than the 2 s and 2 p valence orbitals of oxygen. As a result, the OF 4 molecule is unlikely to exist. In fact, it has not been detected. Exercise $\PageIndex{3}$ What is the hybridization of the boron atom in $BF_6^{3−}$? Is this ion likely to exist? Answer a sp 3 d 2 hybridization; no Summary Hybridization increases the overlap of bonding orbitals and explains the molecular geometries of many species whose geometry cannot be explained using a VSEPR approach. The localized bonding model (called valence bond theory ) assumes that covalent bonds are formed when atomic orbitals overlap and that the strength of a covalent bond is proportional to the amount of overlap. It also assumes that atoms use combinations of atomic orbitals ( hybrids ) to maximize the overlap with adjacent atoms. The formation of hybrid atomic orbitals can be viewed as occurring via promotion of an electron from a filled ns 2 subshell to an empty np or ( n − 1) d valence orbital, followed by hybridization , the combination of the orbitals to give a new set of (usually) equivalent orbitals that are oriented properly to form bonds. The combination of an ns and an np orbital gives rise to two equivalent sp hybrids oriented at 180°, whereas the combination of an ns and two or three np orbitals produces three equivalent sp 2 hybrids or four equivalent sp 3 hybrids , respectively. The bonding in molecules with more than an octet of electrons around a central atom can be explained by invoking the participation of one or two ( n − 1) d orbitals to give sets of five sp 3 d or six sp 3 d 2 hybrid orbitals, capable of forming five or six bonds, respectively. The spatial orientation of the hybrid atomic orbitals is consistent with the geometries predicted using the VSEPR model.
Courses/Athabasca_University/Chemistry_350%3A_Organic_Chemistry_I/09%3A_Alkynes-_An_Introduction_to_Organic_Synthesis/9.01%3A_Naming_Alkynes	Objectives After completing this section, you should be able to provide the correct IUPAC name of an alkyne, given its Kekulé, condensed or shorthand structure. provide the correct IUPAC name of a compound containing both double and triple bonds, given its Kekulé, condensed or shorthand structure. draw the structure of a compound containing one or more triple bonds, and possibly one or more double bonds, given its IUPAC name. name and draw the structure of simple alkynyl groups, and where appropriate, use these names as part of the IUPAC system of nomenclature. Study Notes Simple alkynes are named by the same rules that are used for alkenes (see Section 7.3), except that the ending is - yne instead of - ene . Alkynes cannot exhibit E , Z (cis‑trans) isomerism; hence, in this sense, their nomenclature is simpler than that of alkenes. Alkynes are organic molecules made of the functional group carbon-carbon triple bonds and are written in the empirical formula of $C_nH_{2n-2}$. They are unsaturated hydrocarbons. Like alkenes have the suffix –ene, alkynes use the ending –yne; this suffix is used when there is only one alkyne in the molecule. Introduction Here are the molecular formulas and names of the first ten carbon straight chain alkynes . 0 1 Name Molecular Formula Ethyne C2H2 Propyne C3H4 1-Butyne C4H6 1-Pentyne C5H8 1-Hexyne C6H10 1-Heptyne C7H12 1-Octyne C8H14 1-Nonyne C9H16 1-Decyne C10H18 The more commonly used name for ethyne is acetylene, which used industrially. Naming Alkynes Like previously mentioned, the IUPAC rules are used for the naming of alkynes . Rule 1 Find the longest carbon chain that includes both carbons of the triple bond. Rule 2 Number the longest chain starting at the end closest to the triple bond. A 1-alkyne is referred to as a terminal alkyne and alkynes at any other position are called internal alkynes . For example: Rule 3 After numbering the longest chain with the lowest number assigned to the alkyne , label each of the substituents at its corresponding carbon. While writing out the name of the molecule, arrange the substituents in alphabetical order. If there are more than one of the same substituent use the prefixes di, tri, and tetra for two, three, and four substituents respectively. These prefixes are not taken into account in the alphabetical order. For example: If there is an alcohol present in the molecule, number the longest chain starting at the end closest to it, and follow the same rules. However, the suffix would be –ynol, because the alcohol group takes priority over the triple bond. When there are two triple bonds in the molecule, find the longest carbon chain including both the triple bonds. Number the longest chain starting at the end closest to the triple bond that appears first. The suffix that would be used to name this molecule would be –diyne. For example: Rule 4 Substituents containing a triple bond are called alkynyl. For example: Here is a table with a few of the alkynyl substituents : 0 1 Name Molecule Ethynyl -C≡CH 2- Propynyl -CH2C≡CH 2-Butynyl -CH3C≡CH2CH3 Rule 5 A molecule that contains both double and triple bonds is called an alkenyne. The chain can be numbered starting with the end closest to the functional group that appears first. For example: [NB If both functional groups are the exact same distance from the end the alkene takes precedence.] Reference Vollhardt, Peter, and Neil E. Schore. Organic Chemistry: Structure and Function . 5th Edition. New York: W. H. Freeman & Company, 2007. Exercises Exercise $\PageIndex{1}$ Name the following compounds: Answer A – 3,6-diethyl-4-octyne B – 3-methylbutyne C – 4-ethyl-2-heptyne D – cyclodecyne Exercise $\PageIndex{2}$ How many isomers are possible for C 5 H 8 ? Draw them. Answer 2 possible isomers Exercise $\PageIndex{3}$ Draw the following compounds: a) 4,4-dimethyl-2-pentyne b) 3-octyne c) 3-methyl-1-hexyne d) trans 3-hepten-1-yne Answer Exercise $\PageIndex{4}$ Do alkynes show cis-trans isomerism? Explain. Answer No. A triply bonded carbon atom can form only one other bond and has linear electron geometry so there are no "sides". Allkenes have two groups attached to each inyl carbon with a trigonal planar electron geometry that creates the possibility of cis-trans isomerism.
Courses/Roosevelt_University/General_Organic_and_Biochemistry_with_Problems_Case_Studies_and_Activities/09%3A_Acids_and_Bases	9.1: What are Acids and Bases? Compounds that donate a proton (a hydrogen ion) to another compound is called a Brønsted-Lowry acid. The compound that accepts the proton is called a Brønsted-Lowry base. The species remaining after a Brønsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Brønsted-Lowry base gains a proton is the conjugate acid of the base. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species. 9.2: pH and pOH The concentration of hydronium ion in a solution of an acid in water is greater than 1.0×10⁻⁷M at 25 °C. The concentration of hydroxide ion in a solution of a base in water is greater than 1.0×10⁻⁷M M at 25 °C. The concentration of H₃O⁺ in a solution can be expressed as the pH of the solution; pH=−log H₃O⁺. The concentration of OH⁻ can be expressed as the pOH of the solution: pOH=−log[OH⁻]. 9.3: Acid-Base Balance 9.4: Case Study- Navigating the Delicate pH Spectrum in Intensive Care 9.E: Exercises
Courses/Los_Angeles_Trade_Technical_College/Chem_51/15%3A_Solutions/15.07%3A_Normality	Normality expresses concentration in terms of the equivalents of one chemical species reacting stoichiometrically with another chemical species. Note that this definition makes an equivalent, and thus normality, a function of the chemical reaction. Although a solution of H 2 SO 4 has a single molarity, its normality depends on its reaction. We define the number of equivalents, n , using a reaction unit, which is the part of a chemical species participating in the chemical reaction. In a precipitation reaction, for example, the reaction unit is the charge of the cation or anion participating in the reaction; thus, for the reaction \[ \ce{Pb}^{2+}(aq) + \ce{2 I}^- (aq) \rightleftharpoons \ce{PbI}_2(s) \] n = 2 for Pb 2 + (aq) and n = 1 for 2 I - (aq) . In an acid-base reaction, the reaction unit is the number of H + ions that an acid donates or that a base accepts. For the reaction between sulfuric acid and ammonia \[\ce{H_2SO_4}(aq) + \ce{2NH_3}(aq) \rightleftharpoons \ce{2NH_4^+}(aq) + \ce{SO}_4^{2-}(aq)\] n = 2 for H 2 SO 4 (aq) because sulfuric acid donates two protons, and n = 1 for NH 3 (aq) because each ammonia accepts one proton. For a complexation reaction, the reaction unit is the number of electron pairs that the metal accepts or that the ligand donates. In the reaction between Ag + and NH 3 \[\ce{Ag^+}(aq) + \ce{2NH_3}(aq) \rightleftharpoons \ce{Ag(NH_3)2+}(aq) \] n = 2 for Ag + (aq) because the silver ion accepts two pairs of electrons, and n = 1 for NH 3 because each ammonia has one pair of electrons to donate. Finally, in an oxidation-reduction reaction the reaction unit is the number of electrons released by the reducing agent or accepted by the oxidizing agent; thus, for the reaction \[\ce{2Fe}^{3+}(aq) + \ce{Sn}^{2+}(aq) \rightleftharpoons \ce{Sn}^{4+}(aq) + \ce{2Fe}^{2+}(aq)\] $n = 1$ for $\ce{Fe^3+}(aq)$ and $n = 2$ for $\ce{Sn^2+}(aq)$. Clearly, determining the number of equivalents for a chemical species requires an understanding of how it reacts. Normality is the number of equivalent weights, $EW$, per unit volume. An equivalent weight is the ratio of a chemical species' formula weight, FW , to the number of its equivalents, $n$. \[EW = \dfrac{FW}{n}\] The following simple relationship exists between normality, N , and molarity, M . \[N = n \times M\]
Courses/University_of_Arkansas_Little_Rock/Chem_1402%3A_General_Chemistry_1_(Belford)/Worksheets/Chapter_01A_and_1B_Problem_Sets	This section has link to Worksheets that will help with Chapters 1A and 1B material. Chapters 1A Homework and 1B Homework . There is also two review worksheets for algebra. Unnamed: 0 Key Chapter 1 Worksheet 1: Algebra Review Chapter 1 Worksheet 1 Key: Algebra Review Chapter 1 Worksheet 2: Algebra Review Continue Chapter Worksheet 2 Key: Algebra Review Continue Chapter 1 Worksheet 3: Significant Digits Chapter Worksheet 3 Key: Significant Digits Chapter 1 Worksheet 4: Scientific Notation Chapter 1 Worksheet 4 Key: Scientific Notation Chapter 1 Worksheet 5: Conversion Factors Chapter 1 Worksheet 5 Key: Conversion Factors Chapter 1 Worksheet 6: Elements Chapter 1 Worksheet 6 Key: Elements
Courses/Martin_Luther_College/Organic_Chemistry_-_MLC/03%3A_Alcohols_Ethers_Thiols_Sulfides_and_Amines/3.01%3A_Alcohols_and_Phenols/3.1.06%3A_Alcohols_from_Carbonyl_Compounds-_Grignard_Reagents	Objectives After completing this section, you should be able to write an equation to illustrate the formation of a Grignard reagent. write a general equation to represent the reaction of an aldehyde or ketone with a Grignard reagent. write the detailed mechanism for the reaction of an aldehyde or ketone with a Grignard reagent. identify the product formed from the reaction of a given aldehyde or ketone with a given Grignard reagent. identify the carbonyl compound, the Grignard reagent, or both, needed to prepare a given alcohol. write the equation to describe the reaction of an ester with a Grignard reagent. identify the product formed from the reaction of a given ester with a given Grignard reagent. discuss the limitations of Grignard reagent formation, and determine whether a given compound can be used to form such a reagent. Study Notes Before you begin this section, you may wish to review Section 10.8 which discusses the formation of Grignard reagents. Link to section 10.8 Grignard reagents are among the most frequently used reagents in organic synthesis. They react with a wide variety of substrates; however, in this section, we are concerned only with those reactions that produce alcohols. Notice that in a reaction involving a Grignard reagent, not only does the functional group get changed, but the number of carbon atoms present also changes. This fact provides us with a useful method for ascending a homologous series. For example: One important route for producing an alcohol from a Grignard reagent has been omitted from the discussion in the reading. It involves the reaction of the Grignard reagent with ethylene oxide to produce a primary alcohol containing two more carbon atoms than the original Grignard reagent. As mentioned in the reading, both organolithium and Grignard reagents are good nucleophiles. They also act as strong bases in the presence of acidic protons such as −CO 2 H, −OH, −SH, −NH and terminal alkyne groups. Not only do acidic protons interfere with the nucleophilic attack on the carbonyl of these organometallic reagents, if the starting materials possess any acidic protons, reagents cannot be generated in the first place. They are also the reason these reactions must be carried out in a water-free environment. Another limitation of preparing Grignard and organolithium reagents is that they cannot already contain a carbonyl group (or other electrophilic multiple bonds like C=N, nitriles, N=O, S=O) because it would simply react with itself. A summary of the methods used to prepare alcohols from Grignard reagents is provided below. Formation of Grignard Reagents Grignard reagents (RMgX) can be prepared through the reaction of halogens with magnesium metal ( Section 10-6 ). Grignard reagents are a source of carbanion nucleophiles (R: - + MgX) which add to carbonyl compounds to yield alcohols. Ethyl ether or THF are essential for Grignard reagent formation. Lone pair electrons from two ether molecules form a complex with the magnesium in the Grignard reagent. This complex helps stabilize the organometallic and increases its ability to react. Reaction of Grignard Reagents with Carbonyls Because organometallic reagents react as their corresponding carbanion, they are excellent nucleophiles. Aldehydes, ketones, and other carbonyl containing compounds will undergo nucleophilic addition with Grignard reagents. The nucleophilic carbon in the organometallic reagents forms a C-C single bond with the electrophilic carbonyl carbon. An alkoxide ion intermediate is formed which becomes an alcohol with subsequent protonation by an acid. The type of alcohol produced depends on the number of alkyl substituents attached to the electrophilic carbonyl carbon. Reacting a Grignard reagent with formaldehyde (H 2 C=O) produces 1 o alcohols, aldehydes produce 2 o alcohols, and ketones produce 3 o alcohols. These reactions will be discussed in greater detail in Section 19.7 . General Reaction Addition to Formaldehyde gives 1 o Alcohols Addition to Aldehydes gives 2 o Alcohols Addition to Ketones gives 3 o Alcohols Predicting the Product of the Addition of Grignard Reagent to Carbonyl During the reaction, the C=O double bond in the reactant forms a C-O single bond in the product. The breaking of the C=O double bond allows for the formation of two single bonds in the product. One will be attached to the oxygen and one to the carbon which was originally in the carbonyl. The carbon will gain whatever R group was contained in the Grignard reagent and the oxygen will gain a hydrogen. Example $\PageIndex{1}$ Mechanism for the Addition of Grignard Reagents to Carbonyls The mechanism starts with the formation of a acid-base complex between + MgX and the carbonyl oxygen. The + MgBr of the Grignard reagent acts as a Lewis acid and accepts a set of lone pair electrons from the carbonyl oxygen. This gives the oxygen a positive charge which correspondingly increases the partial positive charge on the carbonyl carbon increasing its susceptibility to nucleophilic attack. Step 1: Lewis acid-base formation Step 2: Nucleophilic attack The carbanion nucleophile from the Grignard reagent adds to the electrophilic carbon of the acid-base complex forming a C-C bond. The two electrons of the C=O are pushed toward the carbonyl oxygen atom forming a tetrahedral Magnesium alkoxide intermediate. Step 3: Protonation The alkoxide intermediate is converted to an alcohol through addition of a acidic aqueous solution. The + MgX ion is also converted to HOMgX.' Grignard Reagents Convert Esters to 3 o alcohols With esters, after the first Grignard reaction, the carbonyl reforms creating a ketone which can then react with a second molecule of the Grignard. In effect, the Grignard reagent adds twice. This reaction will be discussed in greater detail in Section 21.6 . General Reaction Example $\PageIndex{2}$ Grignard Reagents Convert Epoxides to 1o alcohols Another important route for producing an alcohol from a Grignard reagent involves the reaction of the Grignard reagent with ethylene oxide to produce a primary alcohol containing two more carbon atoms than the original Grignard reagent. This reaction will be discussed in greater detail in Section 18.6 . The first step of the mechanism is shown below. With the second step following the protonation step common to the other reaction pathways studied in this section. Limitation of Organometallic Reagents Grignard and organolithium reagents are powerful bases. Because of this they cannot be prepared using halogen compounds which contain an additional functional group that has acidic hydrogens. If there are acidic hydrogens present, the organometallic reagent will act as a base and deprotonate the acidic hydrogen rather than act as a nucleophile and attack the carbonyl. A partial list of functional groups which cannot be used are: alcohols (RO H ) , carboxylic acids (RCO 2 H ), thiols (RS H ), and terminal alkynes (RCC H ). Additionally, amides (RCON H 2 ) and amines (RN H 2 ) that have NH bonds cannot be used with organometallic reagents. Planning an Alcohol Synthesis Using a Grignard Reaction The nucleophilic addition of a Grignard reagent to a carbonyl is a powerful tool in organic synthesis because if forms a C-C bond. Also, there is often more than one way to make a given target molecule. Primary alcohols have one C-C bond which can retrosynthetically cleaved. Secondary alcohols have two and tertiary alcohols have three. Worked Example $\PageIndex{1}$ What reagents are required to make the following molecule using a Grignard Reaction? Answer Analysis: Because the target molecule is a 1o alcohol there is only one C-C bond which can be cleaved to generate possible starting materials. The only possible reagents which would provide the target molecule would be formaldehyde and phenylmagnesium bromide. Worked Example $\PageIndex{2}$ What reagents are required to make the following molecule using a Grignard Reaction? Answer Analysis: Because the target molecule is an asymmetrical 2 o alcohol there are two different C-C bond cleavage points. Each of these will provide a unique set of reagents which should be considered in terms of their reactivity and availability. Pathway 1) Pathway 1 synthesis) Pathway 1 shows that propanal and propylmagnesium bromide can be reacted to create the target molecule. Pathway 2) Pathway 2 synthesis) Pathway 2 shows that butanal and ethylmagnesium bromide can be reacted to create the target molecule. Exercises Exercise $\PageIndex{1}$ 1) If allylmagnesium chloride were added to a solution of the following compound and then worked-up with acid, the product would contain a chiral center. Would the product be a racemic mixture or an enatiomerically pure product? Draw both enantiomers. 2) What combination of carbonyl compound and grignard (use MgBr) reagent would yield the following alcohols (after workup)? a) b) c) d) 3) If the following compounds were reacted with methylmagnesium bromide what would be the products? a) Cyclohexanone b) Acetophenone c) 2-Hexanone Answer 1) The result would be a racemic mixture of the following. 2) c) d) 3) a) b) c) Exercise $\PageIndex{2}$ If allylmagnesium chloride were added to a solution of the following compound and then worked-up with acid, the product would contain a chiral center. Would the product be a racemic mixture or an enatiomerically pure product? Draw both enantiomers. Answer The result would be a racemic mixture of the following. Exercise $\PageIndex{3}$ What combination of carbonyl compound and grignard (use MgBr) reagent would yield the following alcohols (after workup)? Answer Exercise $\PageIndex{4}$ Fill in the blanks of the following reaction scheme. Answer
Courses/GalwayMayo_Institute_of_Technology/Chemistry_1.1_(GMIT)/02%3A_Molecules_Compounds_and_Chemical_Equations/2.01%3A_Chemical_Bonds	Chemical bonds form when electrons can be simultaneously close to two or more nuclei, but beyond this, there is no simple, easily understood theory that would not only explain why atoms bind together to form molecules, but would also predict the three-dimensional structures of the resulting compounds as well as the energies and other properties of the bonds themselves. Unfortunately, no one theory exists that accomplishes these goals in a satisfactory way for all of the many categories of compounds that are known. Moreover, it seems likely that if such a theory does ever come into being, it will be far from simple. When we are faced with a scientific problem of this complexity, experience has shown that it is often more useful to concentrate instead on developing models . A scientific model is something like a theory in that it should be able to explain observed phenomena and to make useful predictions. But whereas a theory can be discredited by a single contradictory case, a model can be useful even if it does not encompass all instances of the phenomena it attempts to explain. We do not even require that a model be a credible representation of reality; all we ask is that be able to explain the behavior of those cases to which it is applicable in terms that are consistent with the model itself. An example of a model that you may already know about is the kinetic molecular theory of gases. Despite its name, this is really a model (at least at the level that beginning students use it) because it does not even try to explain the observed behavior of real gases. Nevertheless, it serves as a tool for developing our understanding of gases, and as a starting point for more elaborate treatments. Given the extraordinary variety of ways in which atoms combine into aggregates, it should come as no surprise that a number of useful bonding models have been developed. Most of them apply only to certain classes of compounds, or attempt to explain only a restricted range of phenomena. In this section we will provide brief descriptions of some of the bonding models; the more important of these will be treated in much more detail in later parts of this chapter. Ionic Bonding Ever since the discovery early in the 19th century that solutions of salts and other electrolytes conduct electric current, there has been general agreement that the forces that hold atoms together must be electrical in nature. Electrolytic solutions contain ions having opposite electrical charges, opposite charges attract, so perhaps the substances from which these ions come consist of positive and negatively charged atoms held together by electrostatic attraction. It turns out that this is not true generally, but a model built on this assumption does a fairly good job of explaining a rather small but important class of compounds that are called ionic solids. The most well known example of such a compound is sodium chloride, which consists of two interpenetrating lattices of Na + and Cl – ions arranged in such as way that every ion of one type is surrounded (in three dimensional space) by six ions of opposite charge. The main limitation of this model is that it applies really well only to the small class of solids composed of Group 1 and 2 elements with highly electronegative elements such as the halogens. Although compounds such as CuCl 2 dissociate into ions when they dissolve in water, the fundamental units making up the solid are more like polymeric chains of covalently-bound CuCl 2 molecules that have little ionic character. According to the ionic electrostatic model, solids such as NaCl consist of positive and negative ions arranged in a crystal lattice. Each ion is attracted to neighboring ions of opposite charge, and is repelled by ions of like charge; this combination of attractions and repulsions, acting in all directions, causes the ion to be tightly fixed in its own location in the crystal lattice. Since electrostatic forces are nondirectional, the structure of an ionic solid is determined purely by geometry: two kinds of ions, each with its own radius, will fall into whatever repeating pattern will achieve the lowest possible potential energy. Surprisingly, there are only a small number of possible structures Covalent Bonding This model originated with the theory developed by G.N. Lewis in 1916, and it remains the most widely-used model of chemical bonding. The essential element s of this model can best be understood by examining the simplest possible molecule. This is the hydrogen molecule ion H 2 + , which consists of two nuclei and one electron. First, however, think what would happen if we tried to make the even simpler molecule H 2 2 + . Since this would consist only of two protons whose electrostatic charges would repel each other at all distances, it is clear that such a molecule cannot exist; something more than two nuclei are required for bonding to occur. In the hydrogen molecule ion H 2 + we have a third particle, an electron. The effect of this electron will depend on its location with respect to the two nuclei. If the electron is in the space between the two nuclei, it will attract both protons toward itself, and thus toward each other. If the total attraction energy exceeds the internuclear repulsion, there will be a net bonding effect and the molecule will be stable. If, on the other hand, the electron is off to one side, it will attract both nuclei, but it will attract the closer one much more strongly, owing to the inverse-square nature of Coulomb's law. As a consequence, the electron will now help the electrostatic repulsion to push the two nuclei apart. We see, then, that the electron is an essential component of a chemical bond, but that it must be in the right place: between the two nuclei. Coulomb's law can be used to calculate the forces experienced by the two nuclei for various positions of the electron. This allows us to define two regions of space about the nuclei, as shown in the figure. One region, the binding region, depicts locations at which the electron exerts a net binding effect on the new nuclei. Outside of this, in the antibinding region, the electron will actually work against binding. This simple picture illustrates the number one rule of chemical bonding: chemical bonds form when electrons can be simultaneously close to two or more nuclei. It should be pointed out that this principle applies also to the ionic model; as will be explained later in this chapter, the electron that is "lost" by a positive ion ends up being closer to more nuclei (including the one from whose electron cloud it came) in the compound. The polar covalent model : A purely covalent bond can only be guaranteed when the electronegativities (electron-attracting powers) of the two atoms are identical. When atoms having different electronegativities are joined, the electrons shared between them will be displaced toward the more electronegative atom, conferring a polarity on the bond which can be described in terms of percent ionic character. The polar covalent model is thus an generalization of covalent bonding to include a very wide range of behavior. The Coulombic model : This is an extension of the ionic model to compounds that are ordinarily considered to be non-ionic. Combined hydrogen is always considered to exist as the hydride ion H – , so that methane can be treated as if it were C 4 + H –4 . This is not as bizarre as it might seem at first if you recall that the proton has almost no significant size, so that it is essentially embedded in an electron pair when it is joined to another atom in a covalent bond. This model, which is not as well known as it deserves to be, has considerable predictive power, both as to bond energies and structures. The VSEPR model : The "valence shell electron repulsion" model is not so much a model of chemical bonding as a scheme for explaining the shapes of molecules. It is based on the quantum mechanical view that bonds represent electron clouds- physical regions of negative electric charge that repel each other and thus try to stay as far apart as possible. Summary The covalent bond is formed when two atoms are able to share electrons: whereas the ionic bond is formed when the "sharing" is so unequal that an electron from atom A is completely lost to atom B, resulting in a pair of ions: The two extremes of electron sharing represented by the covalent and ionic models appear to be generally consistent with the observed properties of molecular and ionic solids and liquids. But does this mean that there are really two kinds of chemical bonds, ionic and covalent?
Courses/Madera_Community_College/MacArthur_Chemistry_3A_v_1.2/06%3A_Introduction_to_Stoichiometry/6.E%3A_Introduction_to_Stoichiometry_(Exercises)	The following questions are related to the material covered in this chapter. For additional discussion on each topic, also check the links included in each heading. 6.1: Stoichiometry Think back to the pound cake recipe. What possible conversion factors can you construct relating the components of the recipe? Think back to the pancake recipe. What possible conversion factors can you construct relating the components of the recipe? What are all the conversion factors that can be constructed from the balanced chemical reaction: \[\ce{2H2(g) + O2(g) → 2H2O(ℓ)}?\] What are all the conversion factors that can be constructed from the balanced chemical reaction N 2 (g) + 3H 2 (g) → 2NH 3 (g)? Given the chemical equation : Na(s) + H 2 O(ℓ) → NaOH(aq) + H 2 (g) Balance the equation. How many molecules of H 2 are produced when 332 atoms of Na react? Given the chemical equation: S(s) + O 2 (g) → SO 3 (g) Balance the equation. How many molecules of O 2 are needed when 38 atoms of S react? For the balanced chemical equation: 6H + (aq) + 2MnO 4 − (aq) + 5H 2 O 2 (ℓ) → 2Mn 2 + (aq) + 5O 2 (g) + 8H 2 O(ℓ) how many molecules of H 2 O are produced when 75 molecules of H 2 O 2 react? For the balanced chemical reaction 2C 6 H 6 (ℓ) + 15O 2 (g) → 12CO 2 (g) + 6H 2 O(ℓ) how many molecules of CO 2 are produced when 56 molecules of C 6 H 6 react? Given the balanced chemical equation Fe 2 O 3 (s) + 3SO 3 (g) → Fe 2 (SO 4 ) 3 how many molecules of Fe 2 (SO 4 ) 3 are produced if 321 atoms of S are reacted? For the balanced chemical equation CuO(s) + H 2 S(g) → CuS + H 2 O(ℓ) how many molecules of CuS are formed if 9,044 atoms of H react? For the balanced chemical equation Fe 2 O 3 (s) + 3SO 3 (g) → Fe 2 (SO 4 ) 3 suppose we need to make 145,000 molecules of Fe 2 (SO 4 ) 3 . How many molecules of SO 3 do we need? One way to make sulfur hexafluoride is to react thioformaldehyde, CH 2 S, with elemental fluorine: CH 2 S + 6F 2 → CF 4 + 2HF + SF 6 If 45,750 molecules of SF 6 are needed, how many molecules of F 2 are required? Construct the three independent conversion factors possible for these two reactions: 2H 2 + O 2 → 2H 2 O H 2 + O 2 → H 2 O 2 Why are the ratios between H 2 and O 2 different? The conversion factors are different because the stoichiometries of the balanced chemical reactions are different. Construct the three independent conversion factors possible for these two reactions: 2Na + Cl 2 → 2NaCl 4Na + 2Cl 2 → 4NaCl What similarities, if any, exist in the conversion factors from these two reactions? Answers \[\frac{1\, pound\, butter}{1\, pound\, flour}\] or \[\frac{1\, pound\, sugar}{1\, pound\, eggs}\] are two conversion factors that can be constructed from the pound cake recipe. Other conversion factors are also possible. 1 pound butter 1 pound flour \[\frac{2\, molecules\, H_{2}}{1\, molecule\, O_{2}}\] , \[\frac{1\, molecule\, O_{2}}{2\, molecules\, H_{2}O}\] , \[\frac{2\, molecules\, H_{2}}{2\, molecules\, H_{2}O}\] and their reciprocals are the conversion factors that can be constructed. 2 molecules H 2 1 molecule O 2 2Na(s) + 2H 2 O(ℓ) → 2NaOH(aq) + H 2 (g) 166 molecules 120 molecules 107 molecules 435,000 molecules \[\frac{2\, molecules\, H_{2}}{1\, molecule\, O_{2}}\ , \frac{1\, molecule\, O_{2}}{2\, molecules\, H_{2}O}\ , \frac{2\, molecules\, H_{2}}{2\, molecules\, H_{2}O}\] \[\frac{1\, molecules\, H_{2}}{1\, molecule\, O_{2}}\ , \frac{1\, molecule\, O_{2}}{2\, molecules\, H_{2}O_{2}}\ , \frac{1\, molecule\, H_{2}}{1\, molecule\, H_{2}O_{2}}\] 6. 2: The Mole How many atoms are present in 4.55 mol of Fe? How many atoms are present in 0.0665 mol of K? How many molecules are present in 2.509 mol of H 2 S? How many molecules are present in 0.336 mol of acetylene (C 2 H 2 )? How many moles are present in 3.55 × 10 24 Pb atoms? How many moles are present in 2.09 × 10 22 Ti atoms? How many moles are present in 1.00 × 10 23 PF 3 molecules? How many moles are present in 5.52 × 10 25 penicillin molecules? Determine the molar mass of each substance. Si SiH 4 K 2 O Determine the molar mass of each substance. Cl 2 SeCl 2 Ca(C 2 H 3 O 2 ) 2 Determine the molar mass of each substance. Al Al 2 O 3 CoCl 3 Determine the molar mass of each substance. O 3 NaI C 12 H 22 O 11 What is the mass of 4.44 mol of Rb? What is the mass of 0.311 mol of Xe? What is the mass of 12.34 mol of Al 2 (SO 4 ) 3 ? What is the mass of 0.0656 mol of PbCl 2 ? How many moles are present in 45.6 g of CO? How many moles are present in 0.00339 g of LiF? How many moles are present in 1.223 g of SF 6 ? How many moles are present in 48.8 g of BaCO 3 ? How many moles are present in 54.8 mL of mercury if the density of mercury is 13.6 g/mL? How many moles are present in 56.83 mL of O 2 if the density of O 2 is 0.00133 g/mL? Answers 2.74 × 10 24 atoms 1.511 × 10 24 molecules 5.90 mol 0.166 mol 28.086 g 32.118 g 94.195 g 26.981 g 101.959 g 165.292 g 379 g 4,222 g 1.63 mol 0.008374 mol 3.72 mol 6.3: The Mole in Chemical Reactions Express in mole terms what this chemical equation means: CH 4 + 2O 2 → CO 2 + 2H 2 O Express in mole terms what this chemical equation means. Na 2 CO 3 + 2HCl → 2NaCl + H 2 O + CO 2 How many molecules of each substance are involved in the equation in Exercise 1 if it is interpreted in terms of moles? How many molecules of each substance are involved in the equation in Exercise 2 if it is interpreted in terms of moles? For the chemical equation 2C 2 H 6 + 7O 2 → 4CO 2 + 6H 2 O what equivalents can you write in terms of moles? Use the ⇔ sign. For the chemical equation 2Al + 3Cl 2 → 2AlCl 3 what equivalents can you write in terms of moles? Use the ⇔ sign. Write the balanced chemical reaction for the combustion of C 5 H 12 (the products are CO 2 and H 2 O) and determine how many moles of H 2 O are formed when 5.8 mol of O 2 are reacted. Write the balanced chemical reaction for the formation of Fe 2 (SO 4 ) 3 from Fe 2 O 3 and SO 3 and determine how many moles of Fe 2 (SO 4 ) 3 are formed when 12.7 mol of SO 3 are reacted. For the balanced chemical equation 3Cu(s) + 2NO 3 − (aq) + 8H + (aq) → 3Cu 2 + (aq) + 4H 2 O(ℓ) + 2NO(g) how many moles of Cu 2 + are formed when 55.7 mol of H + are reacted? For the balanced chemical equation Al(s) + 3Ag + (aq) → Al 3 + (aq) + 3Ag(s) how many moles of Ag are produced when 0.661 mol of Al are reacted? For the balanced chemical reaction 4NH 3 (g) + 5O 2 (g) → 4NO(g) + 6H 2 O(ℓ) how many moles of H 2 O are produced when 0.669 mol of NH 3 react? For the balanced chemical reaction 4NaOH(aq) + 2S(s) + 3O 2 (g) → 2Na 2 SO 4 (aq) + 2H 2 O(ℓ) how many moles of Na 2 SO 4 are formed when 1.22 mol of O 2 react? For the balanced chemical reaction 4KO 2 (s) + 2CO 2 (g) → 2K 2 CO 3 (s) + 3O 2 (g) determine the number of moles of both products formed when 6.88 mol of KO 2 react. For the balanced chemical reaction 2AlCl 3 + 3H 2 O(ℓ) → Al 2 O 3 + 6HCl(g) determine the number of moles of both products formed when 0.0552 mol of AlCl 3 react. Answers One mole of CH 4 reacts with 2 mol of O 2 to make 1 mol of CO 2 and 2 mol of H 2 O. 6.022 × 10 23 molecules of CH 4 , 1.2044 × 10 24 molecules of O 2 , 6.022 × 10 23 molecules of CO 2 , and 1.2044 × 10 24 molecules of H 2 O 2 mol of C 2 H 6 ⇔ 7 mol of O 2 ⇔ 4 mol of CO 2 ⇔ 6 mol of H 2 O C 5 H 12 + 8O 2 → 5CO 2 + 6H 2 O; 4.4 mol 20.9 mol 1.00 mol 3.44 mol of K 2 CO 3 ; 5.16 mol of O 2 6.4: Mole-Mass and Mass-Mass Calculations What mass of CO 2 is produced by the combustion of 1.00 mol of CH 4 ? CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(ℓ) What mass of H 2 O is produced by the combustion of 1.00 mol of CH 4 ? CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(ℓ) What mass of HgO is required to produce 0.692 mol of O 2 ? 2HgO(s) → 2Hg(ℓ) + O 2 (g) What mass of NaHCO 3 is needed to produce 2.659 mol of CO 2 ? 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(ℓ) + CO 2 (g) How many moles of Al can be produced from 10.87 g of Ag? Al(NO 3 ) 3 (s) + 3Ag → Al + 3AgNO 3 How many moles of HCl can be produced from 0.226 g of SOCl 2 ? SOCl 2 (ℓ) + H 2 O(ℓ) → SO 2 (g) + 2HCl(g) How many moles of O 2 are needed to prepare 1.00 g of Ca(NO 3 ) 2 ? Ca(s) + N 2 (g) + 3O 2 (g) → Ca(NO 3 ) 2 (s) How many moles of C 2 H 5 OH are needed to generate 106.7 g of H 2 O? C 2 H 5 OH(ℓ) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(ℓ) What mass of O 2 can be generated by the decomposition of 100.0 g of NaClO 3 ? 2NaClO 3 → 2NaCl(s) + 3O 2 (g) What mass of Li 2 O is needed to react with 1,060 g of CO 2 ? Li 2 O(aq) + CO 2 (g) → Li 2 CO 3 (aq) What mass of Fe 2 O 3 must be reacted to generate 324 g of Al 2 O 3 ? Fe 2 O 3 (s) + 2Al(s) → 2Fe(s) + Al 2 O 3 (s) What mass of Fe is generated when 100.0 g of Al are reacted? Fe 2 O 3 (s) + 2Al(s) → 2Fe(s) + Al 2 O 3 (s) What mass of MnO 2 is produced when 445 g of H 2 O are reacted? H 2 O(ℓ) + 2MnO 4 − (aq) + Br − (aq) → BrO 3 − (aq) + 2MnO 2 (s) + 2OH − (aq) What mass of PbSO 4 is produced when 29.6 g of H 2 SO 4 are reacted? Pb(s) + PbO 2 (s) + 2H 2 SO 4 (aq) → 2PbSO 4 (s) + 2H 2 O(ℓ) If 83.9 g of ZnO are formed, what mass of Mn 2 O 3 is formed with it? Zn(s) + 2MnO 2 (s) → ZnO(s) + Mn 2 O 3 (s) If 14.7 g of NO 2 are reacted, what mass of H 2 O is reacted with it? 3NO 2 (g) + H 2 O(ℓ) → 2HNO 3 (aq) + NO(g) If 88.4 g of CH 2 S are reacted, what mass of HF is produced? CH 2 S + 6F 2 → CF 4 + 2HF + SF 6 If 100.0 g of Cl 2 are needed, what mass of NaOCl must be reacted? NaOCl + HCl → NaOH + Cl 2 Answers 44.0 g 3.00 × 10 2 g 0.0336 mol 0.0183 mol 45.1 g 507 g 4.30 × 10 3 g 163 g 76.7 g 6.5: Yields What is the difference between the theoretical yield and the actual yield? What is the difference between the actual yield and the percent yield? A worker isolates 2.675 g of SiF 4 after reacting 2.339 g of SiO 2 with HF. What are the theoretical yield and the actual yield? SiO 2 (s) + 4HF(g) → SiF 4 (g) + 2H 2 O(ℓ) A worker synthesizes aspirin, C 9 H 8 O 4 , according to this chemical equation. If 12.66 g of C 7 H 6 O 3 are reacted and 12.03 g of aspirin are isolated, what are the theoretical yield and the actual yield? C 7 H 6 O 3 + C 4 H 6 O 3 → C 9 H 8 O 4 + HC 2 H 3 O 2 A chemist decomposes 1.006 g of NaHCO 3 and obtains 0.0334 g of Na 2 CO 3 . What are the theoretical yield and the actual yield? 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(ℓ) + CO 2 (g) A chemist combusts a 3.009 g sample of C 5 H 12 and obtains 3.774 g of H 2 O. What are the theoretical yield and the actual yield? C 5 H 12 (ℓ) + 8O 2 (g) → 5CO 2 + 6H 2 O(ℓ) What is the percent yield in Exercise 3? What is the percent yield in Exercise 4? What is the percent yield in Exercise 5? What is the percent yield in Exercise 6? Answers Theoretical yield is what you expect stoichiometrically from a chemical reaction; actual yield is what you actually get from a chemical reaction. theoretical yield = 4.052 g; actual yield = 2.675 g theoretical yield = 0.635 g; actual yield = 0.0334 g 66.02% 5.26% 6 .6: Limiting Reagents The box below shows a group of nitrogen and hydrogen molecules that will react to produce ammonia, NH 3 . What is the limiting reagent? The box below shows a group of hydrogen and oxygen molecules that will react to produce water, H 2 O. What is the limiting reagent? Given the statement “20.0 g of methane is burned in excess oxygen,” is it obvious which reactant is the limiting reagent? Given the statement “the metal is heated in the presence of excess hydrogen,” is it obvious which substance is the limiting reagent despite not specifying any quantity of reactant? Acetylene (C 2 H 2 ) is formed by reacting 7.08 g of C and 4.92 g of H 2 . 2C(s) + H 2 (g) → C 2 H 2 (g) What is the limiting reagent? How much of the other reactant is in excess? Ethane (C 2 H 6 ) is formed by reacting 7.08 g of C and 4.92 g of H 2 . 2C(s) + 3H 2 (g) → C 2 H 6 (g) What is the limiting reagent? How much of the other reactant is in excess? Given the initial amounts listed, what is the limiting reagent, and how much of the other reactant is in excess? \[\underset{35.6\, g}{P_{4}O_{6}(s)}+6\underset{4.77\, g}{H_{2}O(l)}\rightarrow 4H_{3}PO_{4}\] Given the initial amounts listed, what is the limiting reagent, and how much of the other reactant is in excess? \[\underset{377\, g}{3NO_{2}(g)}+\underset{244\, g}{H_{2}O(l)}\rightarrow 2HNO_{3}(aq)+NO(g)\] To form the precipitate PbCl 2 , 2.88 g of NaCl and 7.21 g of Pb(NO 3 ) 2 are mixed in solution. How much precipitate is formed? How much of which reactant is in excess? In a neutralization reaction, 18.06 g of KOH are reacted with 13.43 g of HNO 3 . What mass of H 2 O is produced, and what mass of which reactant is in excess? Answers Nitrogen is the limiting reagent. Yes; methane is the limiting reagent. C is the limiting reagent; 4.33 g of H 2 are left over. H 2 O is the limiting reagent; 25.9 g of P 4 O 6 are left over. 6.06 g of PbCl 2 are formed; 0.33 g of NaCl is left over. 6.7: Additional Exercises How many molecules of O 2 will react with 6.022 × 10 23 molecules of H 2 to make water? The reaction is 2H 2 (g) + O 2 (g) → 2H 2 O(ℓ). How many molecules of H 2 will react with 6.022 × 10 23 molecules of N 2 to make ammonia? The reaction is N 2 (g) + 3H 2 (g) → 2NH 3 (g). How many moles are present in 6.411 kg of CO 2 ? How many molecules is this? How many moles are present in 2.998 mg of SCl 4 ? How many molecules is this? What is the mass in milligrams of 7.22 × 10 20 molecules of CO 2 ? What is the mass in kilograms of 3.408 × 10 25 molecules of SiS 2 ? What is the mass in grams of 1 molecule of H 2 O? What is the mass in grams of 1 atom of Al? What is the volume of 3.44 mol of Ga if the density of Ga is 6.08 g/mL? What is the volume of 0.662 mol of He if the density of He is 0.1785 g/L? For the chemical reaction 2C 4 H 10 (g) + 13O 2 (g) → 8CO 2 (g) + 10H 2 O(ℓ) assume that 13.4 g of C 4 H 10 reacts completely to products. The density of CO 2 is 1.96 g/L. What volume in liters of CO 2 is produced? For the chemical reaction 2GaCl 3 (s) + 3H 2 (g) → 2Ga(ℓ) + 6HCl(g) if 223 g of GaCl 3 reacts completely to products and the density of Ga is 6.08 g/mL, what volume in milliliters of Ga is produced? Calculate the mass of each product when 100.0 g of CuCl react according to the reaction 2CuCl(aq) → CuCl 2 (aq) + Cu(s) What do you notice about the sum of the masses of the products? What concept is being illustrated here? Calculate the mass of each product when 500.0 g of SnCl 2 react according to the reaction 2SnCl 2 (aq) → SnCl 4 (aq) + Sn(s) What do you notice about the sum of the masses of the products? What concept is being illustrated here? What mass of CO 2 is produced from the combustion of 1 gal of gasoline? The chemical formula of gasoline can be approximated as C 8 H 18 . Assume that there are 2,801 g of gasoline per gallon. What mass of H 2 O is produced from the combustion of 1 gal of gasoline? The chemical formula of gasoline can be approximated as C 8 H 18 . Assume that there are 2,801 g of gasoline per gallon. A chemical reaction has a theoretical yield of 19.98 g and a percent yield of 88.40%. What is the actual yield? A chemical reaction has an actual yield of 19.98 g and a percent yield of 88.40%. What is the theoretical yield? Given the initial amounts listed, what is the limiting reagent, and how much of the other reactants are in excess? \[\underset{35.0\, g}{P_{4}}+\underset{12.7\, g}{3NaOH}+\underset{9.33\, g}{3H_{2}O}\rightarrow 2Na_{2}HPO_{4}+PH_{3}\] Given the initial amounts listed, what is the limiting reagent, and how much of the other reactants are in excess? \[\underset{46.3\, g}{2NaCrO_{2}}+\underset{88.2\, g}{3NaBrO_{4}}+\underset{32.5\, g}{2NaOH}\rightarrow 3NaBrO_{3}+2Na_{2}CrO_{4}+H_{2}O\] Verify that it does not matter which product you use to predict the limiting reagent by using both products in this combustion reaction to determine the limiting reagent and the amount of the reactant in excess. Initial amounts of each reactant are given. \[\underset{26.3\, g}{C_{3}H_{8}}+\underset{21.8\, g}{5O_{2}}\rightarrow 3CO_{2}(g)+4H_{2}O(l)\] Just in case you suspect Exercise 21 is rigged, do it for another chemical reaction and verify that it does not matter which product you use to predict the limiting reagent by using both products in this combustion reaction to determine the limiting reagent and the amount of the reactant in excess. Initial amounts of each reactant are given. \[\underset{35.0\, g}{2P_{4}}+\underset{12.7\, g}{6NaOH}+\underset{9.33\, g}{6H_{2}O}\rightarrow 3Na_{2}HPO_{4}+5PH_{3}\] Answers 1.2044 × 10 24 molecules 145.7 mol; 8.77 × 10 25 molecules 52.8 mg 2.99 × 10 −23 g 39.4 mL 20.7 L 67.91 g of CuCl 2 ; 32.09 g of Cu. The two masses add to 100.0 g, the initial amount of starting material, demonstrating the law of conservation of matter. 8,632 g 17.66 g The limiting reagent is NaOH; 21.9 g of P 4 and 3.61 g of H 2 O are left over. Both products predict that O 2 is the limiting reagent; 20.3 g of C 3 H 8 are left over.
Courses/Prince_Georges_Community_College/CHEM_2000%3A_Chemistry_for_Engineers_(Sinex)/Unit_1%3A_Atomic_Structure/Chapter_1%3A_Introduction/Chapter_1.4%3A__The_Mole_and_Molar_Mass	0 1 2 NaN Prince George's Community College General Chemistry for Engineering CHM 2000 NaN Unit I: Atoms Unit II: Molecules Unit III: States of Matter Unit IV: Reactions Unit V: Kinetics & Equilibrium Unit VI: Thermo & Electrochemistry Unit VII: Nuclear Chemistry Unit I: Atoms Unit II: Molecules Unit III: States of Matter Unit IV: Reactions Unit V: Kinetics & Equilibrium Unit VI: Thermo & Electrochemistry Unit VII: Nuclear Chemistry Unit I: Atoms Unit II: Molecules Unit III: States of Matter Unit IV: Reactions Unit V: Kinetics & Equilibrium Unit VI: Thermo & Electrochemistry Unit VII: Nuclear Chemistry Learning Objective To calculate the molecular mass of a covalent compound and the formula mass of an ionic compound and to calculate the number of atoms, molecules, or formula units in a sample of a substance. Chemistry is the study of how atoms and molecules interact with each other which occurs on the atomic scale. Chemists need a way of simply determining how many molecules they have in a beaker. The mole concept, which we will introduce here, bridges that gap by relating the mass of a single atom or molecule in amu to the mass of a collection of a large number of such molecules in grams. As you learned, the mass number is the sum of the numbers of protons and neutrons present in the nucleus of an atom. The mass number is an integer that is approximately equal to the numerical value of the atomic mass. Although the atomic mass is unitless, it is assigned units called atomic mass units (amu) measured relative to the mass of a single atom of 12 C . Because a molecule or a polyatomic ion is an assembly of atoms whose identities are given in its molecular or ionic formula, we can calculate the average atomic mass of any molecule or polyatomic ion from its composition by adding together the masses of the constituent atoms. The average mass of a monatomic ion is the same as the average mass of an atom of the element because the mass of electrons is so small that it is insignificant in most calculations. This is not much help in the laboratory for the typical chemist who only has a balance to weight out chemicals but needs to know how the number of atoms or molecules. Clearly something clever is needed but first let us briefly review how to calculate molecular masses. Molecular The molecular mass The sum of the average masses of the atoms in one molecule of a substance, each multiplied by its subscript. of a substance is the sum of the average masses of the atoms in one molecule of a substance. It is calculated by adding together the atomic masses of the elements in the substance, each multiplied by its subscript (written or implied) in the molecular formula. Because the units of atomic mass are atomic mass units, the units of molecular mass are also atomic mass units. The procedure for calculating molecular masses is illustrated in Example 1. Example 1.4.1 Calculate the molecular mass of ethanol, whose condensed structural formula is CH 3 CH 2 OH. Among its many uses, ethanol is a fuel for internal combustion engines. Given: molecule Asked for: molecular mass Strategy: A Determine the number of atoms of each element in the molecule. B Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element. C Add together the masses to give the molecular mass. Solution: A The molecular formula of ethanol may be written in three different ways: CH 3 CH 2 OH (which illustrates the presence of an ethyl group, CH 3 CH 2 −, and an −OH group), C 2 H 5 OH, and C 2 H 6 O; all show that ethanol has two carbon atoms, six hydrogen atoms, and one oxygen atom. B Taking the atomic masses from the periodic table, we obtain $ 2 \times atomic;\ mass\;of\:carbon = 2\;atoms \left( {\dfrac{12.011amu}{atom}} \right) = 24.022\;amu$ $ 6 \times atomic;\ mass\;of\:hydrogen = 6\;atoms \left( {\dfrac{1.0079amu}{atom}} \right) = 6.0474\;amu$ $ 1 \times atomic;\ mass\;of\:oxygen = 1\;atoms \left( {\dfrac{15.9994amu}{atom}} \right) = 15.9994amu$ C Adding together the masses gives the molecular mass: 24.022 amu + 6.0474 amu + 15.9994 amu = 46.069 amu Alternatively, we could have used unit conversions to reach the result in one step, as described in Essential Skills 2 $ \left [ 2\; atoms\: \; \left ( \dfrac{12.011\; amu}{1\, \; atom\; C} \right ) \right ] +$ $ \left [ 6\; atoms\: H\; \left ( \dfrac{1.0079\; amu}{1\, \; atom\; H} \right ) \right ]+$ $ \left [ 1\; atoms\: O\; \left ( \dfrac{15.9994\; amu}{1\, \; atom\; O} \right ) \right ] $ The same calculation can also be done in a tabular format, which is especially helpful for more complex molecules: 0 1 2C (2 atoms)(12.011 amu/atom) = 24.022 amu 6H (6 atoms)(1.0079 amu/atom) = 6.0474 amu 1O (1 atoms)(15.9994 amu/atom) = 15.9994 amu C2H6O molecular mass of ethanol = 46.069 amu Exercise Calculate the molecular mass of trichlorofluoromethane, also known as Freon-11, whose condensed structural formula is CCl 3 F. Until recently, it was used as a refrigerant. The structure of a molecule of Freon-11 is as follows: Answer: 137.368 amu Note the Pattern Atomic mass, and molecular mass have the same units: atomic mass units. The Mole Each chemical compound has a particular combination of atoms and the ratios of the numbers of atoms of the elements present are usually small whole numbers. The problem for early chemists was to discover the quantitative relationship between the number of atoms in a chemical substance and its mass. Because the masses of individual atoms are so minuscule (on the order of 10 −23 g/atom), chemists do not measure the mass of individual atoms or molecules. In the laboratory, for example, the masses of compounds and elements used by chemists typically range from milligrams to grams, while in industry, chemicals are bought and sold in kilograms and tons. To analyze the transformations that occur between individual atoms or molecules in a chemical reaction . A process in which a substance is converted to one or more other substances with different compositions and properties , it is therefore absolutely essential for chemists to know how many atoms or molecules are contained in a measurable quantity in the laboratory—a given mass of sample. The unit that provides this link is the mole (mol). The quantity of a substance that contains the same number of units (e.g., atoms or molecules) as the number of carbon atoms in exactly 12 g of isotopically pure carbon-12. , from the Latin moles , meaning “pile” or “heap” ( not from the small subterranean animal!). Many familiar items are sold in numerical quantities that have unusual names. For example, cans of soda come in a six-pack, eggs are sold by the dozen (12), and pencils often come in a gross (12 dozen, or 144). Sheets of printer paper are packaged in reams of 500, a seemingly large number. Atoms are so small, however, that even 500 atoms are too small to see or measure by most common techniques. Any readily measurable mass of an element or compound contains an extraordinarily large number of atoms, molecules, or ions, so an extraordinarily large numerical unit is needed to count them. The mole is used for this purpose. A mole is defined as the amount of a substance that contains the number of carbon atoms in exactly 12 g of isotopically pure carbon-12. According to the most recent experimental measurements, this mass of carbon-12 contains 6.022142 × 10 23 atoms, but for most purposes 6.022 × 10 23 provides an adequate number of significant figures. Just as 1 mol of atoms contains 6.022 × 10 23 atoms, 1 mol of eggs contains 6.022 × 10 23 eggs. The number in a mole is called Avogadro’s number The number of units (e.g., atoms, molecules, or formula units) in 1 mol: 6.022142 x 10 23 , after the 19th-century Italian scientist who first proposed how to measure the number of molecules in a gas. Since the mass of the gas can also be measured on a sensitive balance, knowing both the number of molecules and their total mass allows us to simply determine the mass of a single molecule in grams. The mole provides a bridge between the atomic world (amu) and the laboratory (grams). It allows determination of the number of molecules or atoms by weighing them. The numerical value of Avogadro's number, usually written as No , is a consequence of the arbitrary value of one kilogram, a block of Pt-Ir metal called the International Prototype Kilogram, and the choice of reference for the atomic mass unit scale, one atom of carbon-12. A mole of C-12 by definition weighs exactly 12 g and Avogadro's number is determined by counting the number of atoms. It is not so easy. Avogadro's number is the fundamental constant that is least accurately determined. The definition of a mole—that is, the decision to base it on 12 g of carbon-12—is arbitrary but one arrived at after some discussion between chemists and physicists debating about whether to use naturally occurring carbon, a mixture of C-12 and C-13, or hydrogen. The important point is that 1 mol of carbon—or of anything else, whether atoms, compact discs, or houses—always has the same number of objects: 6.022 × 10 23 . In the following video, Prof. Steve Boon shows how Avogadro's hypothesis can be used to measure the molecular masses of He, N 2 and CO2. Follow along and record the measurements to get the relative masses. When we consider the behavior of gases in Unit 4, we can use the data to calculate the molecular weight of each gas. This method was, until the invention of the mass spectrometer, the best way of measuring molecular weights of gas molecules Note the Pattern One mole always has the same number of objects: 6.022 × 10 23 . To appreciate the magnitude of Avogadro’s number, consider a mole of pennies. Stacked vertically, a mole of pennies would be 4.5 × 10 17 mi high, or almost six times the diameter of the Milky Way galaxy. If a mole of pennies were distributed equally among the entire population on Earth, each person would get more than one trillion dollars. Clearly, the mole is so large that it is useful only for measuring very small objects, such as atoms. The concept of the mole allows us to count a specific number of individual atoms and molecules by weighing measurable quantities of elements and compounds. To obtain 1 mol of carbon-12 atoms, we would weigh out 12 g of isotopically pure carbon-12. Because each element has a different atomic mass, however, a mole of each element has a different mass, even though it contains the same number of atoms (6.022 × 10 23 ). This is analogous to the fact that a dozen extra large eggs weighs more than a dozen small eggs, or that the total weight of 50 adult humans is greater than the total weight of 50 children. Because of the way in which the mole is defined, for every element the number of grams in a mole is the same as the number of atomic mass units in the atomic mass of the element. For example, the mass of 1 mol of magnesium (atomic mass = 24.305 amu) is 24.305 g. Because the atomic mass of magnesium (24.305 amu) is slightly more than twice that of a carbon-12 atom (12 amu), the mass of 1 mol of magnesium atoms (24.305 g) is slightly more than twice that of 1 mol of carbon-12 (12 g). Similarly, the mass of 1 mol of helium (atomic mass = 4.002602 amu) is 4.002602 g, which is about one-third that of 1 mol of carbon-12. Using the concept of the mole, we can now restate Dalton’s theory: 1 mol of a compound is formed by combining elements in amounts whose mole ratios are small whole numbers. For example, 1 mol of water (H 2 O) has 2 mol of hydrogen atoms and 1 mol of oxygen atoms. Molar Mass The molar mass The mass in grams of 1 mol of a substance. of a substance is defined as the mass in grams of 1 mol of that substance. One mole of isotopically pure carbon-12 has a mass of 12 g. For an element, the molar mass is the mass of 1 mol of atoms of that element; for a covalent molecular compound, it is the mass of 1 mol of molecules of that compound; for an ionic compound, it is the mass of 1 mol of formula units. That is, the molar mass of a substance is the mass (in grams per mole) of 6.022 × 10 23 atoms, molecules, or formula units of that substance. In each case, the number of grams in 1 mol is the same as the number of atomic mass units that describe the atomic mass, the molecular mass, or the formula mass, respectively. Note the Pattern The molar mass of any substance is its atomic mass, molecular mass, or formula mass in grams per mole. The periodic table lists the atomic mass of carbon as 12.011 amu; the average molar mass of carbon—the mass of 6.022 × 10 23 carbon atoms—is therefore 12.011 g/mol: Substance (formula) Atomic, Molecular, or Formula Mass (amu) Molar Mass (g/mol) carbon (C) 12.011 (atomic mass) 12.011 ethanol (C2H5OH) 46.069 (molecular mass) 46.069 calcium phosphate [Ca3(PO4)2] 310.177 (formula mass) 310.177 The molar mass of naturally occurring carbon is different from that of carbon-12 and is not an integer because carbon occurs as a mixture of carbon-12, carbon-13, and carbon-14. One mole of carbon still has 6.022 × 10 23 carbon atoms, but 98.89% of those atoms are carbon-12, 1.11% are carbon-13, and a trace (about 1 atom in 10 12 ) are carbon-14. Similarly, the atomic mass of uranium is 238.03 g/mol, and the atomic mass of iodine is 126.90 g/mol. When we deal with elements such as iodine and sulfur, which occur as a diatomic molecule (I 2 ) and a polyatomic molecule (S 8 ), respectively, molar mass usually refers to the mass of 1 mol of atoms of the element—in this case I and S, not to the mass of 1 mol of molecules of the element (I 2 and S 8 ). The molar mass of ethanol is the mass of ethanol (C 2 H 5 OH) that contains 6.022 × 10 23 ethanol molecules. As you calculated in Example 1, the molecular mass of ethanol is 46.069 amu. Because 1 mol of ethanol contains 2 mol of carbon atoms (2 × 12.011 g), 6 mol of hydrogen atoms (6 × 1.0079 g), and 1 mol of oxygen atoms (1 × 15.9994 g), its molar mass is 46.069 g/mol. Similarly, the formula mass of calcium phosphate [Ca 3 (PO 4 ) 2 ] is 310.177 amu, so its molar mass is 310.177 g/mol. This is the mass of calcium phosphate that contains 6.022 × 10 23 formula units. The mole is the basis of quantitative chemistry. It provides chemists with a way to convert easily between the mass of a substance and the number of individual atoms, molecules, or formula units of that substance. Conversely, it enables chemists to calculate the mass of a substance needed to obtain a desired number of atoms, molecules, or formula units. For example, to convert moles of a substance to mass, we use the relationship (moles)(molar mass) → mass (1.4.1) or, more specifically, $ moles\left ( \dfrac{grams}{mole} \right ) = grams $ $ \left ( \dfrac{mass}{molar\; mass} \right )\rightarrow moles \tag{1.4.2} $ $ \left ( \dfrac{grams}{grams/mole} \right )=grams\left ( \dfrac{mole}{grams} \right )=moles $ Be sure to pay attention to the units when converting between mass and moles. Figure 1.4.2 is a flowchart for converting between mass; the number of moles; and the number of atoms, molecules, or formula units. The use of these conversions is illustrated in Example 3 and Example 4. Figure 1.4.2 A Flowchart for Converting between Mass; the Number of Moles; and the Number of Atoms, Molecules, or Formula Units Example 1.4.2 For 35.00 g of ethylene glycol (HOCH 2 CH 2 OH), which is used in inks for ballpoint pens, calculate the number of moles. molecules. Given: mass and molecular formula Asked for: number of moles and number of molecules Strategy: A Use the molecular formula of the compound to calculate its molecular mass in grams per mole. B Convert from mass to moles by dividing the mass given by the compound’s molar mass. C Convert from moles to molecules by multiplying the number of moles by Avogadro’s number. Solution: A The molecular mass of ethylene glycol can be calculated from its molecular formula using the method illustrated in Example 1: 0 1 2C (2 atoms)(12.011 amu/atom) = 24.022 amu 6H (6 atoms)(1.0079 amu/atom) = 6.0474 amu 2O (2 atoms)(15.9994 amu/atom) = 31.9988 amu C2H6O molecular mass of ethanol = 62.068 amu The molar mass of ethylene glycol is 62.068 g/mol B The number of moles of ethylene glycol present in 35.00 g can be calculated by dividing the mass (in grams) by the molar mass (in grams per mole): $ \; 35.00\; g\; ethylene glycol\left ( \frac{1\; mol\; ethylene\; glycol\; (g))}{62.068\; g\; ethylene\; glycol} \right )=0.5639\; mol\; ethylene\; glycol$ It is always a good idea to estimate the answer before you do the actual calculation . In this case, the mass given (35.00 g) is less than the molar mass, so the answer should be less than 1 mol. The calculated answer (0.5639 mol) is indeed less than 1 mol, so we have probably not made a major error in the calculations. C To calculate the number of molecules in the sample, we multiply the number of moles by Avogadro’s number: $ molecules\; of\; ethylene\; glycol=0.5639\; mol\left ( \dfrac{6.022\times 10^{23}}{1\; mol} \right ) $ $ = 3.396\times 10^{23}\; molecules $ Exercise For 75.0 g of CCl 3 F (Freon-11), calculate the number of moles. molecules. Answer: 0.546 mol 3.29 × 10 23 molecules Example 1.4.3 Calculate the mass of 1.75 mol of each compound. S 2 Cl 2 (common name: sulfur monochloride; systematic name: disulfur dichloride) Ca(ClO) 2 (calcium hypochlorite) Given: number of moles and molecular or empirical formula Asked for: mass Strategy: A Calculate the molecular mass of the compound in grams from its molecular formula (if covalent) or empirical formula (if ionic). B Convert from moles to mass by multiplying the moles of the compound given by its molar mass. Solution: We begin by calculating the molecular mass of S 2 Cl 2 and the formula mass of Ca(ClO) 2 . A The molar mass of S 2 Cl 2 is obtained from its molecular mass as follows: 0 1 2S (2 atoms)(32.065 amu/atom) = 64.130 amu 2Cl (2 atoms)(35.353 amu/atom) = 70.906 amu S2Cl2 molecular mass of S2Cl2 = 135.036 amu The molar mass of S 2 Cl 2 is 135.036 g/mol. The mass of 1.75 mol of S 2 Cl 2 is calculated as follows: $ moles\; S{_{2}}Cl_{2} \left [molar\; mass \dfrac{g}{mol} \right ]= mass\; S{_{2}}Cl_{2} $ $ 1.75\; mol\; S{_{2}}Cl_{2}\left ( \dfrac{135.036\; g\; S{_{2}}Cl_{2}}{1\;mol\;S{_{2}}Cl_{2}} \right )=236\;g\; S{_{2}}Cl_{2} $ B The formula mass of Ca(ClO) 2 is obtained as follows: 0 1 1Ca (1 atom )(40.078 amu/atom) = 40.078 amu 2Cl (2 atoms)(35.453 amu/atom) = 70.906 amu 2O (2 atoms)(15.9994 amu/atom) = 31.9988 amu Ca(ClO)2 formula mass of Ca(ClO)2 = 142.983 amu The molar mass of Ca(ClO) 2 is 142.983 g/mol $ moles\; Ca\left ( ClO \right )_{2}\left [ \dfrac{molar\; mass\; Ca\left ( ClO \right )_{2}}{1\; mol\; Ca\left ( ClO \right )_{2}} \right ]=mass\; Ca\left ( ClO \right )_{2} $ $ 1.75\; mol\; Ca\left ( ClO \right )_{2}\left [ \dfrac{142.983\; g Ca\left ( ClO \right )_{2}}{1\; mol\; Ca\left ( ClO \right )_{2}} \right ]=250.\; g\; Ca\left ( ClO \right )_{2} $ Exercise Calculate the mass of 0.0122 mol of each compound. Si 3 N 4 (silicon nitride), used as bearings and rollers (CH 3 ) 3 N (trimethylamine), a corrosion inhibitor Answer: 1.71 g 0.721 g Summary The molecular mass and the formula mass of a compound are obtained by adding together the atomic masses of the atoms present in the molecular formula or empirical formula, respectively; the units of both are atomic mass units (amu). The mole is a unit used to measure the number of atoms, molecules, or (in the case of ionic compounds) formula units in a given mass of a substance. The mole is defined as the amount of substance that contains the number of carbon atoms in exactly 12 g of carbon-12 and consists of Avogadro’s number (6.022 × 10 23 ) of atoms of carbon-12. The molar mass of a substance is defined as the mass of 1 mol of that substance, expressed in grams per mole, and is equal to the mass of 6.022 × 10 23 atoms, molecules, or formula units of that substance. Key Takeaway To analyze chemical transformations, it is essential to use a standardized unit of measure called the mole. Conceptual Problems Describe the relationship between an atomic mass unit and a gram. Construct a flowchart to show how you would calculate the number of moles of silicon in a 37.0 g sample of orthoclase (KAlSi 3 O 8 ), a mineral used in the manufacture of porcelain. Construct a flowchart to show how you would calculate the number of moles of nitrogen in a 22.4 g sample of nitroglycerin that contains 18.5% nitrogen by mass. Numerical Problems Derive an expression that relates the number of molecules in a sample of a substance to its mass and molecular mass. Calculate the molecular mass or formula mass of each compound. KCl (potassium chloride) NaCN (sodium cyanide) H 2 S (hydrogen sulfide) NaN 3 (sodium azide) H 2 CO 3 (carbonic acid) K 2 O (potassium oxide) Al(NO 3 ) 3 (aluminum nitrate) Cu(ClO 4 ) 2 [copper(II) perchlorate] Calculate the molecular mass or formula mass of each compound. V 2 O 4 (vanadium(IV) oxide) CaSiO 3 (calcium silicate) BiOCl (bismuth oxychloride) CH 3 COOH (acetic acid) Ag 2 SO 4 (silver sulfate) Na 2 CO 3 (sodium carbonate) (CH 3 ) 2 CHOH (isopropyl alcohol) Calculate the mass in grams of each sample. 0.520 mol of N 2 O 4 1.63 mol of C 6 H 4 Br 2 4.62 mol of (NH 4 ) 2 SO 3 Give the number of molecules or formula units in each sample. 1.30 × 10 −2 mol of SCl 2 1.03 mol of N 2 O 5 0.265 mol of Ag 2 Cr 2 O 7 Give the number of moles in each sample. 9.58 × 10 26 molecules of Cl 2 3.62 × 10 27 formula units of KCl 6.94 × 10 28 formula units of Fe(OH) 2 What is the total number of atoms in each sample? 0.431 mol of Li 2.783 mol of methanol (CH 3 OH) 0.0361 mol of CoCO 3 1.002 mol of SeBr 2 O What is the total number of atoms in each sample? 0.980 mol of Na 2.35 mol of O 2 1.83 mol of Ag 2 S 1.23 mol of propane (C 3 H 8 ) What is the total number of atoms in each sample? 2.48 g of HBr 4.77 g of CS 2 1.89 g of NaOH 1.46 g of SrC 2 O 4 Decide whether each statement is true or false and explain your reasoning. There are more molecules in 0.5 mol of Cl 2 than in 0.5 mol of H 2 . One mole of H 2 has 6.022 × 10 23 hydrogen atoms. The molecular mass of H 2 O is 18.0 amu. The formula mass of benzene is 78 amu. Complete the following table. Substance Mass (g) Number of Moles Number of Molecules or Formula Units Number of Atoms or Ions MgCl2 37.62 NaN NaN NaN AgNO3 NaN 2.84 NaN NaN BH4Cl NaN NaN 8.93 × 1025 NaN K2S NaN NaN NaN 7.69 × 1026 H2SO4 NaN 1.29 NaN NaN C6H14 11.84 NaN NaN NaN HClO3 NaN NaN 2.45 × 1026 NaN Contributors Anonymous Modified by Joshua Halpern, Scott Sinex and Scott Johnson Verifying Avogadro's Hypothesis Video from HC Communications on YouTube
Courses/Lumen_Learning/Book%3A_US_History_II_(OS_Collection)_(Lumen)/00%3A_Front_Matter/03%3A_Table_of_Contents	1: Main Body 1.1: Chapter 1 2: Instructor Resources (Materials Available with Log In) 3: The Era of Reconstruction, 1865-1877 3.1: Introduction 3.10: Assignment: Reactions to Jim Crow 3.2: Restoring the Union 3.3: Congress and the Remaking of the South, 1865–1866 3.4: Radical Reconstruction, 1867–1872 3.5: The Collapse of Reconstruction 3.6: Video: Reconstruction and 1876 3.7: Primary Source Reading: Atlanta Compromise Speech 3.8: Primary Source Reading: Souls of Black Folk 3.9: Primary Source Reading: Black Codes 4: Go West Young Man! Westward Expansion, 1840-1900 4.1: Introduction 4.2: The Loss of American Indian Life and Culture 4.3: The Impact of Expansion on Chinese Immigrants and Hispanic Citizens 4.4: The Westward Spirit 4.5: Video: Westward Expansion 4.6: Homesteading: Dreams and Realities 4.7: Making a Living in Gold and Cattle 5: Industrialization and the Rise of Big Business, 1870-1900 5.1: Introduction 5.2: Inventors of the Age 5.3: From Invention to Industrial Growth 5.4: Video: The Industrial Economy 5.5: Building Industrial America on the Backs of Labor 5.6: A New American Consumer Culture 5.7: Primary Source Reading: The Gospel of Wealth 5.8: Assignment: The Gospel of Wealth 6: The Growing Pains of Urbanization, 1870-1900 6.1: Introduction 6.2: Urbanization and Its Challenges 6.3: The African American “Great Migration” and New European Immigration 6.4: Relief from the Chaos of Urban Life 6.5: Video: Growth, Cities, and Immigration 6.6: Change Reflected in Thought and Writing 6.7: Primary Source Reading: How the Other Half Lives 6.8: Assignment: How the Other Half Lives 7: Politics in the Gilded Age, 1870-1900 7.1: Introduction 7.2: Political Corruption in Postbellum America 7.3: The Key Political Issues: Patronage, Tariffs, and Gold 7.4: Farmers Revolt in the Populist Era 7.5: Social and Labor Unrest in the 1890s 7.6: Video: Gilded Age Politics 7.7: Assignment: Social Darwinism 8: Leading the Way: The Progressive Movement, 1890-1920 8.1: Introduction 8.10: Assignment: The Jungle 8.2: The Origins of the Progressive Spirit in America 8.3: Video: The Progressive Era 8.4: Progressivism at the Grassroots Level 8.5: New Voices for Women and African Americans 8.6: Video: Women’s Suffrage 8.7: Progressivism in the White House 8.8: Video: Progressive Presidents 8.9: Primary Source Reading: The Jungle 9: Age of Empire: American Foreign Policy, 1890-1914 9.1: Introduction 9.10: Assignment: White Man’s Burden 9.2: Turner, Mahan, and the Roots of Empire 9.3: The Spanish-American War and Overseas Empire 9.4: Economic Imperialism in East Asia 9.5: Roosevelt’s “Big Stick” Foreign Policy 9.6: Taft’s “Dollar Diplomacy” 9.7: Video: American Imperialism 9.8: Assignment: The Turner Thesis 9.9: Primary Source Reading: White Man’s Burden 10: Americans and the Great War, 1914-1919 10.1: Introduction 10.2: American Isolationism and the European Origins of War 10.3: The United States Prepares for War 10.4: A New Home Front 10.5: From War to Peace 10.6: Video: America in World War I 10.7: Demobilization and Its Difficult Aftermath 10.8: Assignment: WWI Propaganda 11: The Jazz Age: Redefining the Nation, 1919-1929 11.1: Introduction 11.2: Prosperity and the Production of Popular Entertainment 11.3: Transformation and Backlash 11.4: A New Generation 11.5: Republican Ascendancy: Politics in the 1920s 11.6: Video: The Roaring 20s 11.7: Assignment: The Roaring Twenties 11.8: Primary Source Reading: A Glimpse behind the Mask of Prohibition 11.9: Primary Source Reading: Hobson Argues for Prohibition 12: Brother, Can You Spare a Dime? The Great Depression, 1929-1932 12.1: Introduction 12.2: The Stock Market Crash of 1929 12.3: President Hoover’s Response 12.4: The Depths of the Great Depression 12.5: Video: The Great Depression 12.6: Assessing the Hoover Years on the Eve of the New Deal 12.7: Primary Source Reading: Greater Security for the Average Man 12.8: Primary Source Reading: Herbert Hoover on Liberty 13: Franklin Roosevelt and the New Deal, 1932-1941 13.1: Introduction 13.2: The Rise of Franklin Roosevelt 13.3: The First New Deal 13.4: The Second New Deal 13.5: Video: The New Deal 13.6: Assignment: Perspectives on the Great Depression and the New Deal 14: Fighting the Good Fight in World War II, 1941-1945 14.1: Introduction 14.2: The Origins of War: Europe, Asia, and the United States 14.3: Primary Source Reading: Nazi Party Platform 14.4: Assignment: Why Nazis? 14.5: The Home Front 14.6: Victory in the European Theater 14.7: The Pacific Theater and the Atomic Bomb 14.8: Videos: World War II 15: Post-War Prosperity and Cold War Fears, 1945-1960 15.1: Introduction 15.10: Assignment: The Sexual Revolution 15.11: Video: Civil Rights and the 1950s 15.12: Primary Source Reading: Eisenhower’s Farewell 15.13: Assignment: Eisenhower’s Farewell 15.14: Extra Credit Assignment: Dr. Strangelove 15.2: The Challenges of Peacetime 15.3: The Cold War 15.4: Video: The Cold War 15.5: Video: The Cold War in Asia 15.6: The American Dream 15.7: Popular Culture and Mass Media 15.8: The African American Struggle for Civil Rights 15.9: Assignment: Women in the 1950s 16: Contesting Futures: America in the 1960s 16.1: Introduction 16.2: The Kennedy Promise 16.3: Lyndon Johnson and the Great Society 16.4: The Civil Rights Movement Marches On 16.5: Challenging the Status Quo 16.6: Primary Source Reading: The Black Panther Party Platform 16.7: Assignment: Black Panther Party Platform 17: Political Storms at Home and Abroad, 1968-1980 17.1: Introduction 17.2: Identity Politics in a Fractured Society 17.3: Video: The 1960s in America 17.4: Coming Apart, Coming Together 17.5: Video: The Rise of Conservatism 17.6: Vietnam: The Downward Spiral 17.7: Watergate: Nixon’s Domestic Nightmare 17.8: Jimmy Carter in the Aftermath of the Storm 17.9: Video: Ford, Carter, and the Economic Malaise 18: From Cold War to Culture Wars, 1980-2000 18.1: Introduction 18.10: Video: The Clinton Years, or the 1990s 18.2: The Reagan Revolution 18.3: Video: The Reagan Revolution 18.4: Primary Source Reading: Ronald Reagan “A Time for Choosing” 18.5: Assignment: American Conservatism 18.6: Political and Cultural Fusions 18.7: Video: George HW Bush and the End of the Cold War 18.8: A New World Order 18.9: Bill Clinton and the New Economy 19: The Challenges of the Twenty-First Century 19.1: Introduction 19.2: The War on Terror 19.3: The Domestic Mission 19.4: Video: Terrorism, War, and Bush 19.5: New Century, Old Disputes 19.6: Hope and Change 19.7: Video: Obamanation: Crash Course US History #47 20: Appendices 20.1: The Declaration of Independence 20.2: The Constitution of the United States 20.3: Presidents of the United States of America 20.4: U.S. Political Map 20.5: U.S. Topographical Map 20.6: United States Population Chart 20.7: Further Reading Back Matter Index Glossary
Courses/Lumen_Learning/Book%3A_Statistics_for_the_Social_Sciences_(Lumen)/04%3A_3-_Examining_Relationships-_Quantitative_Data/4.11%3A_Why_It_Matters-_Examining_Relationships-_Quantitative_Data	Before we begin Examining Relationships: Quantitative Data , let’s see how the new ideas in this module relate to what we learned in the previous modules, Types of Statistical Studies and Producing Data and Summarizing Data Graphically and Numerically . Recall the Big Picture: We begin a statistical investigation with a research question. The investigation proceeds with the following steps: Produce Data: Determine what to measure, then collect the data. ← Types of Statistical Studies and Producing Data Explore the Data: Analyze and summarize the data. ← Summarizing Data Graphically and Numerically, Examining Relationships: Quantitative Data Draw a Conclusion: Use the data, probability, and statistical inference to draw a conclusion about the population. Types of Statistical Studies and Producing Data focused on methods for collecting reliable data. Summarizing Data Graphically and Numerically focused on summarizing and analyzing data for a quantitative variable. In this module, we focus on summarizing and analyzing the relationship between two quantitative variables. In the Big Picture of Statistics, the material in Examining Relationships: Quantitative Data is still part of exploratory data analysis. CC licensed content, Shared previously Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution
Courses/SUNY_Oneonta/Chem_221%3A_Organic_Chemistry_I_(Bennett)/3%3AStuff_to_Review_from_General_Chemistry/09%3A_Acid-Base_Equilibria/9.06%3A_Acid-Base_Titrations	Learning Objectives Interpret titration curves for strong and weak acid-base systems Compute sample pH at important stages of a titration Explain the function of acid-base indicators As seen in the chapter on the stoichiometry of chemical reactions, titrations can be used to quantitatively analyze solutions for their acid or base concentrations. In this section, we will explore the changes in the concentrations of the acidic and basic species present in a solution during the process of a titration. Titration Curves Previously, when we studied acid-base reactions in solution, we focused only on the point at which the acid and base were stoichiometrically equivalent. No consideration was given to the pH of the solution before, during, or after the neutralization. Example $\PageIndex{1}$: Calculating pH for Titration Solutions: Strong Acid/Strong Base A titration is carried out for 25.00 mL of 0.100 M HCl (strong acid) with 0.100 M of a strong base NaOH the titration curve is shown in Figure $\PageIndex{1}$. Calculate the pH at these volumes of added base solution: 0.00 mL 12.50 mL 25.00 mL 37.50 mL Solution Since HCl is a strong acid, we can assume that all of it dissociates. The initial concentration of H 3 O + is $\ce{[H3O+]_0}=0.100\:M$. When the base solution is added, it also dissociates completely, providing OH − ions. The H 3 O + and OH − ions neutralize each other, so only those of the two that were in excess remain, and their concentration determines the pH. Thus, the solution is initially acidic (pH < 7), but eventually all the hydronium ions present from the original acid are neutralized, and the solution becomes neutral. As more base is added, the solution turns basic. The total initial amount of the hydronium ions is: \[\mathrm{n(H^+)_0=[H_3O^+]_0×0.02500\: L=0.002500\: mol} \nonumber \] Once X mL of the 0.100- M base solution is added, the number of moles of the OH − ions introduced is: \[\mathrm{n(OH^-)_0=0.100\:\mathit{M}×X\: mL×\left(\dfrac{1\: L}{1000\: mL}\right)} \nonumber \] The total volume becomes: \[V=\mathrm{(25.00\: mL+X\: mL)\left(\dfrac{1\: L}{1000\: mL}\right)} \nonumber \] The number of moles of H 3 O + becomes: \[\mathrm{n(H^+)=n(H^+)_0-n(OH^-)_0=0.002500\: mol-0.100\:\mathit{M}×X\: mL×\left(\dfrac{1\: L}{1000\: mL}\right)} \nonumber \] The concentration of H 3 O + is: \[\mathrm{[H_3O^+]=\dfrac{n(H^+)}{V}=\dfrac{0.002500\: mol-0.100\:\mathit{M}×X\: mL×\left(\dfrac{1\: L}{1000\: mL}\right)}{(25.00\: mL+X\: mL)\left(\dfrac{1\: L}{1000\: mL}\right)}} \nonumber \] \[\mathrm{=\dfrac{0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)-0.100\:\mathit{M}×X\: mL}{25.00\: mL+X\: mL}} \nonumber \] with the definition of $\mathrm{pH}$: \[\mathrm{pH=−\log([H_3O^+])} \label{phdef} \] The preceding calculations work if $\mathrm{n(H^+)_0-n(OH^-)_0>0}$ and so n(H + ) > 0. When $\mathrm{n(H^+)_0=n(OH^-)_0}$, the H 3 O + ions from the acid and the OH − ions from the base mutually neutralize. At this point, the only hydronium ions left are those from the autoionization of water, and there are no OH − particles to neutralize them. Therefore, in this case: \[\ce{[H3O+]}=\ce{[OH- ]},\:\ce{[H3O+]}=K_\ce{w}=1.0\times 10^{-14};\:\ce{[H3O+]}=1.0\times 10^{-7} \nonumber \] \[\mathrm{pH=-log(1.0\times 10^{-7})=7.00} \nonumber \] Finally, when $\mathrm{n(OH^-)_0>n(H^+)_0}$, there are not enough H 3 O + ions to neutralize all the OH − ions, and instead of $\mathrm{n(H^+)=n(H^+)_0-n(OH^-)_0}$, we calculate: $\mathrm{n(OH^-)=n(OH^-)_0-n(H^+)_0}$ In this case: \[\mathrm{[OH^-]=\dfrac{n(OH^-)}{\mathit{V}}=\dfrac{0.100\:\mathit{M}×X\: mL×\left(\dfrac{1\: L}{1000\: mL}\right)-0.002500\: mol}{(25.00\: mL+X\: mL)\left(\dfrac{1\: L}{1000\: mL}\right)}} \nonumber \] \[\mathrm{=\dfrac{0.100\:\mathit{M}×X\: mL-0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)}{25.00\: mL+X\: mL}} \nonumber \] then using the definition of $pOH$ and its relationship to $pH$ in room temperature aqueous solutios (Equation \ref{phdef}): \[\begin{align} pH &=14-pOH \nonumber \\&=14+\log([OH^-]) \nonumber\end{align} \nonumber \] Let us now consider the four specific cases presented in this problem: (a) X = 0 mL \[\mathrm{[H_3O^+]=\dfrac{n(H^+)}{\mathit{V}}=\dfrac{0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)}{25.00\: mL}=0.1\:\mathit{M}} \nonumber \] then using the definition of $pH$ (Equation \ref{phdef}): \[\begin{align} pH &= −\log(0.100) \nonumber \\ &= 1.000 \nonumber\end{align} \nonumber \] (b) X = 12.50 mL \[\mathrm{[H_3O^+]=\dfrac{n(H^+)}{\mathit{V}}=\dfrac{0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)-0.100\:\mathit{M}×12.50\: mL}{25.00\: mL+12.50\: mL}=0.0333\:\mathit{M}} \nonumber \] then using the definition of $pH$ (Equation \ref{phdef}): \[ \begin{align} pH &= −\log(0.0333) \nonumber \\ &= 1.477 \nonumber\end{align} \nonumber \] (c) X = 25.00 mL Since the volumes and concentrations of the acid and base solutions are the same: \[\mathrm{n(H^+)_0=n(OH^-)_0} \nonumber \] and \[pH = 7.000 \nonumber \] as described earlier. (d) X = 37.50 mL In this case: \[\mathrm{n(OH^-)_0>n(H^+)_0} \nonumber \] \[\mathrm{[OH^-]=\dfrac{n(OH^-)}{\mathit{V}}=\dfrac{0.100\:\mathit{M}×35.70\: mL-0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)}{25.00\: mL+37.50\: mL}=0.0200\:\mathit{M}} \nonumber \] then using the definition of $pH$ (Equation \ref{phdef}): \[ \begin{align}[pH = 14 − pOH \nonumber\\ &= 14 + \log([OH^{−}]) \nonumber \\ &= 14 + \log(0.0200) \nonumber \\ &= 12.30 \nonumber \end{align} \nonumber \] Exercise $\PageIndex{1}$ Calculate the pH for the strong acid/strong base titration between 50.0 mL of 0.100 M HNO 3 ( aq ) and 0.200 M NaOH (titrant) at the listed volumes of added base: 0.00 mL, 15.0 mL, 25.0 mL, and 40.0 mL. Answer a 0.00: 1.000 Answer b 15.0: 1.5111 Answer c 25.0: 7e. Do not delete this text first. Answer d 40.0: 12.523 In Example $\PageIndex{1}$, we calculated pH at four points during a titration. Table $\PageIndex{1}$ shows a detailed sequence of changes in the pH of a strong acid and a weak acid in a titration with NaOH. Volume of 0.100 M NaOH Added (mL) Moles of NaOH Added pH Values 0.100 M HCl1 pH Values 0.100 M $CH_3CO_2H$2 0.0 0.0 1.00 2.87 5.0 0.00050 1.18 4.14 10.0 0.00100 1.37 4.57 15.0 0.00150 1.60 4.92 20.0 0.00200 1.95 5.35 22.0 0.00220 2.20 5.61 24.0 0.00240 2.69 6.13 24.5 0.00245 3.00 6.44 24.9 0.00249 3.70 7.14 25.0 0.00250 7.00 8.72 25.1 0.00251 10.30 10.30 25.5 0.00255 11.00 11.00 26.0 0.00260 11.29 11.29 28.0 0.00280 11.75 11.75 30.0 0.00300 11.96 11.96 35.0 0.00350 12.22 12.22 40.0 0.00400 12.36 12.36 45.0 0.00450 12.46 12.46 50.0 0.00500 12.52 12.52 Titration of 25.00 mL of 0.100 M HCl (0.00250 mol of HCI) with 0.100 M NaOH. Titration of 25.00 mL of 0.100 M CH3CO2H (0.00250 mol of CH3CO2H) with 0.100 M NaOH. Titration of 25.00 mL of 0.100 M HCl (0.00250 mol of HCI) with 0.100 M NaOH. Titration of 25.00 mL of 0.100 M CH3CO2H (0.00250 mol of CH3CO2H) with 0.100 M NaOH. Titration of 25.00 mL of 0.100 M HCl (0.00250 mol of HCI) with 0.100 M NaOH. Titration of 25.00 mL of 0.100 M CH3CO2H (0.00250 mol of CH3CO2H) with 0.100 M NaOH. Titration of 25.00 mL of 0.100 M HCl (0.00250 mol of HCI) with 0.100 M NaOH. Titration of 25.00 mL of 0.100 M CH3CO2H (0.00250 mol of CH3CO2H) with 0.100 M NaOH. The simplest acid-base reactions are those of a strong acid with a strong base. Table $\PageIndex{1}$ shows data for the titration of a 25.0-mL sample of 0.100 M hydrochloric acid with 0.100 M sodium hydroxide. The values of the pH measured after successive additions of small amounts of NaOH are listed in the first column of this table, and are graphed in Figure $\PageIndex{1}$, in a form that is called a titration curve . The pH increases slowly at first, increases rapidly in the middle portion of the curve, and then increases slowly again. The point of inflection (located at the midpoint of the vertical part of the curve) is the equivalence point for the titration. It indicates when equivalent quantities of acid and base are present. For the titration of a strong acid with a strong base, the equivalence point occurs at a pH of 7.00 and the points on the titration curve can be calculated using solution stoichiometry (Table $\PageIndex{1}$ and Figure $\PageIndex{1}$). The titration of a weak acid with a strong base (or of a weak base with a strong acid) is somewhat more complicated than that just discussed, but it follows the same general principles. Let us consider the titration of 25.0 mL of 0.100 M acetic acid (a weak acid) with 0.100 M sodium hydroxide and compare the titration curve with that of the strong acid. Table $\PageIndex{1}$ gives the pH values during the titration, Figure $\PageIndex{1b}$ shows the titration curve. Although the initial volume and molarity of the acids are the same, there are important differences between the two titration curves. The titration curve for the weak acid begins at a higher value (less acidic) and maintains higher pH values up to the equivalence point. This is because acetic acid is a weak acid, which is only partially ionized. The pH at the equivalence point is also higher (8.72 rather than 7.00) due to the hydrolysis of acetate, a weak base that raises the pH: \[\ce{CH3CO2-}(aq)+\ce{H2O}(l)⇌\ce{CH3CO2H}(l)+\ce{OH-}(aq) \nonumber \] After the equivalence point, the two curves are identical because the pH is dependent on the excess of hydroxide ion in both cases. Example $\PageIndex{2}$: Titration of a Weak Acid with a Strong Base The titration curve shown in Figure $\PageIndex{1b}$ is for the titration of 25.00 mL of 0.100 M CH 3 CO 2 H with 0.100 M NaOH. The reaction can be represented as: \[\ce{CH3CO2H + OH- ⟶ CH3CO2- + H2O} \nonumber \] What is the initial pH before any amount of the NaOH solution has been added? K a = 1.8 × 10 −5 for CH 3 CO 2 H. Find the pH after 25.00 mL of the NaOH solution have been added. Find the pH after 12.50 mL of the NaOH solution has been added. Find the pH after 37.50 mL of the NaOH solution has been added. Solution (a) Assuming that the dissociated amount is small compared to 0.100 M , we find that: \[K_\ce{a}=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}}≈\ce{\dfrac{[H3O+]^2}{[CH3CO2H]_0}} \nonumber \] and \[\ce{[H3O+]}=\sqrt{K_\ce{a}×\ce{[CH3CO2H]}}=\sqrt{1.8\times 10^{-5}×0.100}=1.3\times 10^{-3} \nonumber \] \[\mathrm{pH=-\log(1.3\times 10^{-3})=2.87} \nonumber \] (b) After 25.00 mL of NaOH are added, the number of moles of NaOH and CH 3 CO 2 H are equal because the amounts of the solutions and their concentrations are the same. All of the CH 3 CO 2 H has been converted to $\ce{CH3CO2-}$. The concentration of the $\ce{CH3CO2-}$ ion is: \[\mathrm{\dfrac{0.00250\: mol}{0.0500\: L}=0.0500\: \ce{MCH3CO2-}} \nonumber \] The equilibrium that must be focused on now is the basicity equilibrium for $\ce{CH3CO2-}$: \[\ce{CH3CO2-}(aq)+\ce{H2O}(l)⇌\ce{CH3CO2H}(aq)+\ce{OH-}(aq) \nonumber \] so we must determine K b for the base by using the ion product constant for water: \[K_\ce{b}=\ce{\dfrac{[CH3CO2H][OH- ]}{[CH3CO2- ]}} \nonumber \] \[K_\ce{a}=\ce{\dfrac{[CH3CO2- ][H+]}{[CH3CO2H]}},\textrm{ so }\ce{\dfrac{[CH3CO2H]}{[CH3CO2- ]}}=\dfrac{\ce{[H+]}}{K_\ce{a}}. \nonumber \] Since K w = [H + ][OH − ]: \[\begin{align} K_\ce{b} &=\dfrac{\ce{[H+][OH- ]}}{K_\ce{a}} \\ &=\dfrac{K_\ce{w}}{K_\ce{a}} \\ &=\dfrac{1.0\times 10^{-14}}{1.8\times 10^{-5}} \\ &=5.6\times 10^{-10} \end{align} \nonumber \] Let us denote the concentration of each of the products of this reaction, CH 3 CO 2 H and OH − , as x . Using the assumption that x is small compared to 0.0500 M , $K_\ce{b}=\dfrac{x^2}{0.0500\:M}$, and then: \[x=\ce{[OH- ]}=5.3\times 10^{−6} \nonumber \] \[\ce{pOH}=-\log(5.3\times 10^{-6})=5.28 \nonumber \] \[\ce{pH}=14.00−5.28=8.72 \nonumber \] Note that the pH at the equivalence point of this titration is significantly greater than 7. (c) In (a), 25.00 mL of the NaOH solution was added, and so practically all the CH 3 CO 2 H was converted into $\ce{CH3CO2-}$. In this case, only 12.50 mL of the base solution has been introduced, and so only half of all the CH 3 CO 2 H is converted into $\ce{CH3CO2-}$. The total initial number of moles of CH 3 CO 2 H is 0.02500L × 0.100 M = 0.00250 mol, and so after adding the NaOH, the numbers of moles of CH 3 CO 2 H and $\ce{CH3CO2-}$ are both approximately equal to $\mathrm{\dfrac{0.00250\: mol}{2}=0.00125\: mol}$, and their concentrations are the same. Since the amount of the added base is smaller than the original amount of the acid, the equivalence point has not been reached, the solution remains a buffer, and we can use the Henderson-Hasselbalch equation: $\ce{pH}=\ce p K_\ce{a}+\log\ce{\dfrac{[Base]}{[Acid]}}=-\log(\mathit{K}_\ce{a})+\log\ce{\dfrac{[CH3CO2- ]}{[CH3CO2H]}}=-\log(1.8\times 10^{-5})+\log(1)$ (as the concentrations of $\ce{CH3CO2-}$ and CH 3 CO 2 H are the same) Thus: \[\ce{pH}=−\log(1.8\times 10^{−5})=4.74 \nonumber \] (the pH = the p K a at the halfway point in a titration of a weak acid) (d) After 37.50 mL of NaOH is added, the amount of NaOH is 0.03750 L × 0.100 M = 0.003750 mol NaOH. Since this is past the equivalence point, the excess hydroxide ions will make the solution basic, and we can again use stoichiometric calculations to determine the pH: $\mathrm{[OH^-]=\dfrac{(0.003750\: mol−0.00250\: mol)}{0.06250\: L}}=2.00\times 10^{−2}\:M$ So: $\mathrm{pOH=−\log(2.00\times 10^{−2})=1.70,\: and\: pH=14.00−1.70=12.30}$ Note that this result is the same as for the strong acid-strong base titration example provided, since the amount of the strong base added moves the solution past the equivalence point. Exercise $\PageIndex{2}$ Calculate the pH for the weak acid/strong base titration between 50.0 mL of 0.100 M HCOOH( aq ) (formic acid) and 0.200 M NaOH (titrant) at the listed volumes of added base: 0.00 mL, 15.0 mL, 25.0 mL, and 30.0 mL. Answer a 0.00 mL: 2.37 Answer b 15.0 mL: 3.92 Answer c 25.00 mL: 8.29 Answer d 30.0 mL: 12.097 Acid-Base Indicators Certain organic substances change color in dilute solution when the hydronium ion concentration reaches a particular value. For example, phenolphthalein is a colorless substance in any aqueous solution with a hydronium ion concentration greater than 5.0 × 10 −9 M (pH < 8.3). In more basic solutions where the hydronium ion concentration is less than 5.0 × 10 −9 M (pH > 8.3), it is red or pink. Substances such as phenolphthalein, which can be used to determine the pH of a solution, are called acid-base indicators . Acid-base indicators are either weak organic acids or weak organic bases. The equilibrium in a solution of the acid-base indicator methyl orange, a weak acid, can be represented by an equation in which we use $\ce{HIn}$ as a simple representation for the complex methyl orange molecule: \[\underbrace{\ce{HIn (aq)}}_{\ce{red}}+\ce{H2O (l)}⇌\ce{H3O^{+} (aq)}+\underbrace{\ce{In^{-} (aq)}}_{\ce{yellow}} \nonumber \] \[K_\ce{a}=\ce{\dfrac{[H3O+][In- ]}{[HIn]}}=4.0\times 10^{−4} \nonumber \] The anion of methyl orange, $\ce{In^{-}}$, is yellow, and the nonionized form, $\ce{HIn}$, is red. When we add acid to a solution of methyl orange, the increased hydronium ion concentration shifts the equilibrium toward the nonionized red form, in accordance with Le Chatelier’s principle. If we add base, we shift the equilibrium towards the yellow form. This behavior is completely analogous to the action of buffers. An indicator’s color is the visible result of the ratio of the concentrations of the two species In − and $\ce{HIn}$. If most of the indicator (typically about 60−90% or more) is present as $\ce{In^{-}}$, then we see the color of the $\ce{In^{-}}$ ion, which would be yellow for methyl orange. If most is present as $\ce{HIn}$, then we see the color of the $\ce{HIn}$ molecule: red for methyl orange. For methyl orange, we can rearrange the equation for K a and write: \[\mathrm{\dfrac{[In^-]}{[HIn]}=\dfrac{[substance\: with\: yellow\: color]}{[substance\: with\: red\: color]}=\dfrac{\mathit{K}_a}{[H_3O^+]}} \label{ABeq2} \] Equation \ref{ABeq2} shows us how the ratio of $\ce{\dfrac{[In- ]}{[HIn]}}$ varies with the concentration of hydronium ion. The above expression describing the indicator equilibrium can be rearranged: \[ \begin{align} \dfrac{[H_3O^+]}{\mathit{K}_a} &=\dfrac{[HIn]}{[In^- ]} \\[8pt] \log\left(\dfrac{[H_3O^+]}{\mathit{K}_a}\right) &= \log\left(\dfrac{[HIn]}{[In^- ]}\right) \\[8pt] \log([H_3O^+])-\log(\mathit{K}_a) &=-\log\left(\dfrac{[In^-]}{[HIn]}\right) \\[8pt] -pH+p\mathit{K}_a & =-\log\left(\dfrac{[In^-]}{[HIn]}\right) \\[8pt] pH &=p\mathit{K}_a+\log\left(\dfrac{[In^-]}{[HIn]}\right) \end {align} \nonumber \] or in general terms \[pH=p\mathit{K}_a+\log\left(\dfrac{[base]}{[acid]}\right) \label{HHeq} \] Equation \ref{HHeq} is the same as the Henderson-Hasselbalch equation , which can be used to describe the equilibrium of indicators. When [H 3 O + ] has the same numerical value as K a , the ratio of [In − ] to [HIn] is equal to 1, meaning that 50% of the indicator is present in the red form (HIn) and 50% is in the yellow ionic form (In − ), and the solution appears orange in color. When the hydronium ion concentration increases to 8 × 10 −4 M (a pH of 3.1), the solution turns red. No change in color is visible for any further increase in the hydronium ion concentration (decrease in pH). At a hydronium ion concentration of 4 × 10 −5 M (a pH of 4.4), most of the indicator is in the yellow ionic form, and a further decrease in the hydronium ion concentration (increase in pH) does not produce a visible color change. The pH range between 3.1 (red) and 4.4 (yellow) is the color-change interval of methyl orange; the pronounced color change takes place between these pH values. There are many different acid-base indicators that cover a wide range of pH values and can be used to determine the approximate pH of an unknown solution by a process of elimination. Universal indicators and pH paper contain a mixture of indicators and exhibit different colors at different pHs. Figure $\PageIndex{2}$ presents several indicators, their colors, and their color-change intervals. Titration curves help us pick an indicator that will provide a sharp color change at the equivalence point. The best selection would be an indicator that has a color change interval that brackets the pH at the equivalence point of the titration. The color change intervals of three indicators are shown in Figure $\PageIndex{3}$. The equivalence points of both the titration of the strong acid and of the weak acid are located in the color-change interval of phenolphthalein. We can use it for titrations of either strong acid with strong base or weak acid with strong base. Litmus is a suitable indicator for the HCl titration because its color change brackets the equivalence point. However, we should not use litmus for the CH 3 CO 2 H titration because the pH is within the color-change interval of litmus when only about 12 mL of NaOH has been added, and it does not leave the range until 25 mL has been added. The color change would be very gradual, taking place during the addition of 13 mL of NaOH, making litmus useless as an indicator of the equivalence point. We could use methyl orange for the HCl titration, but it would not give very accurate results: (1) It completes its color change slightly before the equivalence point is reached (but very close to it, so this is not too serious); (2) it changes color, as Figure $\PageIndex{2}$ demonstrates, during the addition of nearly 0.5 mL of NaOH, which is not so sharp a color change as that of litmus or phenolphthalein; and (3) it goes from yellow to orange to red, making detection of a precise endpoint much more challenging than the colorless to pink change of phenolphthalein. Figure $\PageIndex{2}$ shows us that methyl orange would be completely useless as an indicator for the CH 3 CO 2 H titration. Its color change begins after about 1 mL of NaOH has been added and ends when about 8 mL has been added. The color change is completed long before the equivalence point (which occurs when 25.0 mL of NaOH has been added) is reached and hence provides no indication of the equivalence point. We base our choice of indicator on a calculated pH, the pH at the equivalence point. At the equivalence point, equimolar amounts of acid and base have been mixed, and the calculation becomes that of the pH of a solution of the salt resulting from the titration. Summary A titration curve is a graph that relates the change in pH of an acidic or basic solution to the volume of added titrant. The characteristics of the titration curve are dependent on the specific solutions being titrated. The pH of the solution at the equivalence point may be greater than, equal to, or less than 7.00. The choice of an indicator for a given titration depends on the expected pH at the equivalence point of the titration, and the range of the color change of the indicator. Glossary acid-base indicator organic acid or base whose color changes depending on the pH of the solution it is in color-change interval range in pH over which the color change of an indicator takes place titration curve plot of the pH of a solution of acid or base versus the volume of base or acid added during a titration
Courses/Knox_College/Chem_322%3A_Physical_Chemisty_II/07%3A_Molecular_Spectroscopy/7.02%3A_Rotational_Transitions_Accompany_Vibrational_Transitions	Each of the modes of vibration of diatomic molecules in the gas phase also contains closely-spaced (1-10 cm -1 difference) energy states attributable to rotational transitions that accompany the vibrational transitions. A molecule’s rotation can be affected by its vibrational transition because there is a change in bond length, so these rotational transitions are expected to occur. Since vibrational energy states are on the order of 1000 cm -1 , the rotational energy states can be superimposed upon the vibrational energy states. Selection Rules Rotational and Vibration transitions (also known as rigid rotor and harmonic oscillator) of molecules help us identify how molecules interact with each other, their bond length as mentioned in previous section. In order to know each transitions, we have to consider other terms like wavenumber, force constant, quantum number, etc. There are rotational energy levels associated with all vibrational levels. From this, vibrational transitions can couple with rotational transitions to give rovibrational spectra . Rovibrational spectra can be analyzed to determine average bond length. We treat the molecule's vibrations as those of a harmonic oscillator (ignoring anharmonicity). The energy of a vibration is quantized in discrete levels and given by \[E_v=h\nu \left(v+\dfrac{1}{2} \right) \nonumber \] Where v is the vibrational quantum number and can have integer values 0, 1, 2..., and $\nu$ is the frequency of the vibration given by: \[\nu=\dfrac{1}{2\pi}\left(\dfrac{k}{\mu}\right)^\dfrac{1}{2} \nonumber \] where $k$ is the force constant and $\mu$ is the reduced mass of a diatomic molecule with atom masses $m_1$ and $m_2$, given by \[\mu=\dfrac{{m}_1{m}_2}{{m}_1+{m}_2} \label{reduced mass} \] We treat the molecule's rotations as those of a rigid rotor (ignoring centrifugal distortion from non-rigid rotor aspects). The energy of a rotation is also quantized in discrete levels given by \[ E_r=\dfrac{h^2}{8\pi^2I} J(J+1) \nonumber \] In which $I$ is the moment of inertia, given by \[{I}=\mu{r}^2 \nonumber \] where $\mu$ is the reduced mass (Equation \ref{reduced mass}) and $r$ is the equilibrium bond length. Experimentally, frequencies or wavenumbers are measured rather than energies, and dividing by $h$ or $hc$ gives more commonly seen term symbols, $F(J)$ using the rotational quantum number $J$ and the rotational constant $B$ in either frequency \[F(J)=\dfrac{E_r}{h}=\dfrac{h}{8\pi^2I} J(J+1)=BJ(J+1) \nonumber \] or wavenumbers \[\tilde{F}(J)=\dfrac{E_r}{hc}=\dfrac{h}{8\pi^2cI} J(J+1)=\tilde{B}J(J+1) \nonumber \] It is important to note in which units one is working since the rotational constant is always represented as $B$, whether in frequency or wavenumbers. Vibrational Transition Selection Rules : At room temperature, typically only the lowest energy vibrational state v= 0 is populated, so typically v 0 = 0 and ∆v = +1. The full selection rule is technically that ∆v = ±1, however here we assume energy can only go upwards because of the lack of population in the upper vibrational states. Rotational Transition Selection Rules : At room temperature, states with J≠0 can be populated since they represent the fine structure of vibrational states and have smaller energy differences than successive vibrational levels. Additionally, ∆J = ±1 since a photon contains one quantum of angular momentum and we abide by the principle of conservation of energy . This is also the selection rule for rotational transitions. These two selection rules mean that the transition ∆J = 0 (i.e. J" = 0 and J' = 0, but $\nu_0 \neq 0$ is forbidden and the pure vibrational transition is not observed in most cases. The rotational selection rule gives rise to an R-branch (when ∆J = +1) and a P-branch (when ∆J = -1). Each line of the branch is labeled R(J) or P(J), where J represents the value of the lower state Figure 13.2.1 ). R-branch When ∆J = +1, i.e. the rotational quantum number in the ground state is one more than the rotational quantum number in the excited state – R branch (in French, riche or rich). To find the energy of a line of the R-branch: \[\begin{align} \Delta{E} &=h\nu_0 +hB \left [J(J+1)-J^\prime (J^\prime{+1}) \right] \\[4pt] &=h\nu_0 +hB \left[(J+1)(J+2)-J(J+1)\right] \\[4pt] &= h\nu_0 +2hB(J+1) \end{align} \nonumber \] P-branch When ∆J = -1, i.e. the rotational quantum number in the ground state is one less than the rotational quantum number in the excited state – P branch (in French, pauvre or poor). To find the energy of a line of the P-branch: \[\begin{align} \Delta{E} &=h\nu_0 +hB \left [J(J+1)-J^\prime(J^\prime+1) \right] \\[4pt] &= h\nu_0 +hB \left [J(J-1)-J(J+1) \right] \\[4pt] &= h\nu_0 -2hBJ \end{align} \nonumber \] Q-branch When ∆J = 0, i.e. the rotational quantum number in the ground state is the same as the rotational quantum number in the excited state – Q branch (simple, the letter between P and R). To find the energy of a line of the Q-branch: \[ \begin{align} \Delta{E} &=h\nu_0 +hB[J(J+1)-J^\prime(J^\prime+1)] \\[4pt] &=h\nu_0 \end{align} \nonumber \] The Q-branch can be observed in polyatomic molecules and diatomic molecules with electronic angular momentum in the ground electronic state, e.g. nitric oxide, NO. Most diatomics, such as O 2 , have a small moment of inertia and thus very small angular momentum and yield no Q-branch. As seen in Figure 13.2.2 , the lines of the P-branch (represented by purple arrows) and R-branch (represented by red arrows) are separated by specific multiples of $B$ (i.e, $2B$), thus the bond length can be deduced without the need for pure rotational spectroscopy. The total nuclear energy of the combined rotation-vibration terms, $S(v, J)$, can be written as the sum of the vibrational energy and the rotational energy \[ S(v,J)=G(v)+F(J) \nonumber \] where $G(v)$ represents the energy of the harmonic oscillator, ignoring anharmonic components and $S(J)$ represents the energy of a rigid rotor, ignoring centrifugal distortion. From this, we can derive \[ S(v,J)=\nu_0 \left(v+\dfrac{1}{2}\right) +BJ(J+1) \nonumber \] The spectrum we expect, based on the conditions described above, consists of lines equidistant in energy from one another, separated by a value of $2B$. The relative intensity of the lines is a function of the rotational populations of the ground states, i.e. the intensity is proportional to the number of molecules that have made the transition. The overall intensity of the lines depends on the vibrational transition dipole moment. In Figure 13.2.2 , between $P(1)$ and $R(0)$ lies the zero gap, where the the first lines of both the P- and R-branch are separated by $4B$, assuming that the rotational constant B is equal for both energy levels. The zero gap is also where we would expect the Q-branch, depicted as the dotted line, if it is allowed. The relative intensity of the P- and R-branch lines depends on the thermal distribution of electrons; more specifically, they depend on the population of the lower J state. If we represent the population of the Jth upper level as N J and the population of the lower state as N 0 , we can find the population of the upper state relative to the lower state using the Boltzmann distribution: \[\dfrac{N_J}{N_0}={(2J+1)e}^\left(-\dfrac{E_r}{kT}\right) \nonumber \] (2J+1) gives the degeneracy of the Jth upper level arising from the allowed values of M J (+J to –J). As J increases, the degeneracy factor increases and the exponential factor decreases until at high J, the exponential factor wins out and N J /N 0 approaches zero at a certain level, J max . Thus, when \[ \dfrac{d}{dJ} \left( \dfrac{N_J}{N_0} \right)=0 \nonumber \] by differentiation, we obtain \[J_{max}=\left(\dfrac{kT}{2hB}\right)^\dfrac{1}{2}-\dfrac{1}{2} \nonumber \] This is the reason that rovibrational spectral lines increase in energy to a maximum as J increases, then decrease to zero as J continues to increase, as seen in Figure 13.2.2 . From this relationship, we can also deduce that in heavier molecules, $B$ will decrease because the moment of inertia will increase, and the decrease in the exponential factor is less pronounced. This results in the population distribution shifting to higher values of J. Similarly, as temperature increases, the population distribution will shift towards higher values of $J$.
Courses/Minnesota_State_Community_and_Technical_College/CHEM_1111%3A_General_Inorganic_Chemistry_I_Lab_Manual-Online_Section/02%3A_Experiments/2.07%3A_Gas_Laws-Simulations_and_Wet_Lab-Home	Learning Objectives To experiment with variables to reach gas law relationships To verify Avogadro's law To improve problem-solving skills by practicing gas-law problems Theory A gas is the state of matter that is characterized by having neither a fixed shape nor a fixed volume. Gases exert pressure, are compressible, have low densities and diffuse rapidly when mixed with other gases. On a microscopic level, the molecules (or atoms) in a gas are separated by large distances and are in constant, random motion. Four measurable properties can be used to describe a gas quantitatively: pressure ($P$), volume ($V$), temperature ($T$) and mole quantity ($n$). The relationships among these properties are summarized by the Gas Laws, as shown in the table below. 0 1 2 3 Charles’s Law $V \propto T$ $P$ and $n$ are held constant As gas temperature increases, gas volume increases. $\displaystyle \frac{V_1}{T_1}=\frac{V_2}{T_2}$ Boyle’s Law $V \propto \frac{1}{P}$ $T$ and $n$ are held constant As gas pressure increases, gas volume decreases. $\displaystyle P_1V_1=P_2V_2$ Avogadro’s Law $V \propto n$ $P$ and $T$ are held constant As the number of moles of gas increase, gas volume increases. $\displaystyle \frac{V_1}{n_1}=\frac{V_2}{n_2}$ Combined Law $V \propto \frac{T}{P}$ $n$ is held constant Obtained by combining Boyle’s Law and Charles’s Law. $\displaystyle \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$ A closer look at the Combined Law reveals that the volume of a gas depends on both the pressure and temperature. Thus, if the volumes of two gases are to be compared, they must be under the same $P$ and $T$. A commonly used set of $P$ and $T$ reference conditions is known as Standard Temperature and Pressure, or STP. Standard temperature is defined as exactly 0 °C (273 K) and standard pressure is defined as exactly 1 atm (760 mmHg). The Ideal Gas Law is obtained by combining Boyle’s Law , Charles’s Law and Avogadro’s Law together: \[PV = nRT\] Here, $P$ represents as the gas pressure (in atmospheres); $V$ is the gas volume (in Liters); $n$ is the number of moles of gas in the sample; $T$ is the gas temperature (in Kelvins). $R$ is a proportionality constant called the Gas Constant, and has a theoretical value of 0.08206 $\frac{L \cdot atm}{mol \cdot K}$. Note that the units of $R$ will allow the units of $P$, $V$, $n$ and $T$ in the Ideal Gas Law to cancel correctly. If you observe a parameter (dependent variable), say volume by changing any one of the other parameters (independent variable) like pressure, (keep the rest at constant of fixed value) you may come across some kind of data trends regarding the dependence of volume on increasing or decreasing the pressure. If you follow the results of such experiments by systematic observation and analysis you can hypothesize the relationship between the parameters. You can repeat the same experiment many times. If the trend is generalizable it could be stated as a LAW. This is how many science laws are postulated, including the simple gas laws. There are three simple gas laws studying the various parameters like temperature (T), pressure (P), and number of moles of gas particles (n) on the volume (V) of a gas. Following are the laws and the parameters they correlate · Boyle’s Law (relates Volume and Pressure: V and P ) · Charles Law (relates Volume and Temperature: V and T ) · Avogadro’s Law (relates Volume and Number of particles: V and n ) In PART A of the experiment a free online simulation on all the four parameters (V, T, P, and n) will be accessed to play around with the parameters to see how they are related, and to predict the trend in the results. In PART B of the experiment a gas evolution reaction will be followed to observe the effect of adding more particles (n) on the volume (V) of the gas to postulate the Avogadro’s law. A graphical analysis of the data will also be done. Procedure Materials Required A laptop to play the gas law simulation online, Five (16 Fl. Oz) empty water or cola bottles with narrow mouth (any type of bottles of the same size and shape with narrow mouth). Five to six medium sized balloons. Baking soda (Sodium bicarbonate), 5% Vinegar (acetic acid), spoons, 5-cups, funnel (or a paper cone to bes used as a funnel to transfer the powder to the balloon, electronic scale, graduated cylinders, water Part A: Simulations to Verify Simple Gas Laws and Ideal Gas Law The major difference between solid, liquid, and gas phases is how the particles in a phase are packed. Component particles in a gas are widely separated as if there is no interaction between them. Because of this reason, a gas can be compressed further and further down to smaller volumes (space). Most of the gases are colorless. Therefore, it is very difficult to observe them for experimental purposes. However, some parameters of gases are measurable like their volume, pressure, temperature, and number (moles) of particles present. Turn on your lap top. Access the internet and the go to the free online gas simulation website provided by your instructor Connect to the webpage, and play around with the simulation by changing volume, pressure, temperature and number of molecules by clicking and dragging the various bars after selecting one. Boyle’s law Click on the pressure button on the drag and drop bar on the right hand side. Keep the number of moles (n) and Temperature fixed at constant values. Record them. Now go ahead and drag the button on pressure bar and leave at any five different values and observe the volume reading. Record five different pressure values and the corresponding volumes on the observation sheet. This can be used to predict the relationship between the pressure and volume. This relationship is the concept behind the Boyle’s law. Charle’s law Click on the volume button on the drag and drop bar on the right hand side. Keep the number of moles (n) and Pressure fixed at constant values. Write them down. Now go ahead and drag the button on temperature (T) bar and leave at any five different values and observe the volume reading. Record all the five different Temperature values and the corresponding volumes on the observation sheet. This can be used to predict the relationship between the temperature and volume. This relationship is the concept behind the Charle’s law. Ideal gas law Click on the volume button on drag and drop bar on the right hand side. Play around with the temperature (T) and Pressure at various values. Now go ahead and . Record all the five different corresponding sets of Volume, Pressure, and Temperature values. Calculate the number of moles using the ideal gas equation. The value of Ideal gas constant, R should be taken as 0.08206 L.atm/mol.K. Part B: Wet Lab to Test Avogadro's Law Watch the following Youtube video. We will se up a similar experiment. Instead of the conical flask we will be using five identical water or cola bottles (16 Fl. Oz). Also, we will have a total of five sample instead of the four shown in the solution. The acetic acid we have is the 5% vinegar. 5% vinegar is approximately 0.83M not 1.0M as shown in the video. In order to get 0.19 mol of acetic acid in each bottle, you need to take 229 ml of the 5% (0.83M). Use the M1V1=M2V2 equation to find out the new volume 1.0 M x 190 ml =0.83 M x X ml. On solving for X you will get 229 ml as the new volume. Take five medium size water or cola bottles of the same size and shape (16 Fl. Oz). Label them as 3, 7, 12, 16, and 20 using a marker. Measure out 229 ml of 0.83M acetic (5% vinegar) into each bottle. Measure 3 g, 7 g, 12 g, 16 g, 20 g sodium bicarbonate (NaHCO 3 ) or baking soda separately in small containers. Without spilling add the powder into five similar sized balloons using a funnel, and keep them aside. Align each balloon by the side of each bottle. The balloons should be attached on the mouth of the bottles tightly while taking care not to spill the powder into the flask. Hold the balloons tight as shown in the video, as it might fly off as large amount of gas is produced. Once all the balloons are attached tightly, all the balloons should be flipped to release the powder into the bottle as shown in the video. Record the observation on the report sheet. Take a lab picture Compare the volume of the balloons with the amount of the gas particles produced. Observe the trend in the volume and enter it in the observation sheet. Following is the reaction between the acetic acid and sodium bicarbonate producing water, sodium acetate, and carbon dioxide gas. Analyze the relationship between the volume and number of moles of gas particles produced and state your hypothesis. Keep this observation and data sheet to reuse in the lab for Mole ratios and limiting reactant Report Sheet Part A: Simulations to Verify Simple Gas Laws and Ideal Gas Law Boyle's Law 0 1 2 Constant Temperature= K Constant number of moles = mol Constant Temperature= K Constant number of moles = mol Constant Temperature= K Constant number of moles = mol # Pressure ( Atm) Volume (L) 1 NaN NaN 2 NaN NaN 3 NaN NaN 4 NaN NaN 5 NaN NaN Charle' Law 0 1 2 Constant Pressure = Atm Constant number of moles = mol Constant Pressure = Atm Constant number of moles = mol Constant Pressure = Atm Constant number of moles = mol # Temperature ( K) Volume (L) 1 NaN NaN 2 NaN NaN 3 NaN NaN 4 NaN NaN 5 NaN NaN Ideal Gas Law 0 1 2 3 4 Calculate the number of mols, n for each set of data. Use the value of Ideal gas constant R as 0.08206 L.atm/mol.K Calculate the number of mols, n for each set of data. Use the value of Ideal gas constant R as 0.08206 L.atm/mol.K Calculate the number of mols, n for each set of data. Use the value of Ideal gas constant R as 0.08206 L.atm/mol.K Calculate the number of mols, n for each set of data. Use the value of Ideal gas constant R as 0.08206 L.atm/mol.K Calculate the number of mols, n for each set of data. Use the value of Ideal gas constant R as 0.08206 L.atm/mol.K # P, Pressure ( Atm) V, Volume (L) T, Temperature (K) ? mols (n) 1 NaN NaN NaN NaN 2 NaN NaN NaN NaN 3 NaN NaN NaN NaN 4 NaN NaN NaN NaN 5 NaN NaN NaN NaN Report Sheet: Part B-Wet Lab to Test Avogadro's Law Avogadro's Law 0 1 2 3 4 Constant Temperature= K (Room temperature) Constant Pressure = 1.0 Atm Constant Temperature= K (Room temperature) Constant Pressure = 1.0 Atm Constant Temperature= K (Room temperature) Constant Pressure = 1.0 Atm Constant Temperature= K (Room temperature) Constant Pressure = 1.0 Atm NaN # Grams of Sodium bicarbonate used Moles, n, of bicarbonate used Moles of Carbon dioxide produced ( assume 1:1 mol ratio) Rate the size of the inflated balloons on a scale of 1-4, 1 being the smallest 1 3.0 g mol NaN NaN 2 7.0 g mol NaN NaN 3 12.0 g mol NaN NaN 4 16.0 g mol NaN NaN 5 (Do not use this data for this lab. Reserve this for the Moles and Limiting Reactant Lab) 32.0 g mol NaN NaN Data Analysis 1. On analyzing the data in Part A simulation for Boyle's law, how are the Volume and Pressure related? Write your statement. Does this match with the theoretical statement given at the beginning of the experiment? 2. On analyzing the data in Part A simulation for Charle's law, how are the Volume and Temperature related? Write your statement. Does this match with the theoretical statement given at the beginning of the experiment? 3. On analyzing the data in Part B for Avogadro's law, how are the Volume (size of the balloons) and the number of moles of the gas produced related? Write your statement. Does this match with the theoretical statement given at the beginning of the experiment? 4. On analyzing the data in Part A simulation for the Ideal gas law, how are the number of moles related? Are you getting similar values or very different values? Explain your findings 5. Plot a graph with your independent variable in Part A Boyle's law, pressure vs. volume. Give a proper title, and label the axes with the parameters and the units. Practice Problems Exercise $\PageIndex{1}$ If 3.7L of an ideal gas at 35.8 degree Celsius and 1.5atm is heated to 373.5 K, what would be the new pressure as the gas has expanded to 7.8 L? Answer 0.59 atm Exercise $\PageIndex{2}$ Pick one of the entries from Part A-Ideal gas law simulation. Write the number of moles calculated in one of those five data points. Convert the number of moles to the mass of the gas assuming that the gas in the simulation is Carbon monoxide or CO. Answer Clue: n mol=Mass (g)/Formula mass (g/mol). Therefore, Mass (g)= n mol X Formula mass (g/mol) Contributions and Attributions Manjusha Saraswathiamma, Minnesota State Community and Technical College, Moorhead has developed this experiment to conduct this at a homeschool setting using less hazardous and cost-effective materials. The author would like to acknowledge the creators (Ohio State University) of the Youtube video used in this experiment. A portion of the Theory part is a content reuse form Chem 10 Experiment Experimental Determination of the Gas Constant (Experiment) by Santa Monica College is licensed CC BY-NC 4.0 .
Courses/SUNY_Oneonta/Chem_221%3A_Organic_Chemistry_I_(Bennett)/3%3AStuff_to_Review_from_General_Chemistry/04%3A_Thermochemistry/4.02%3A_Energy_Basics	Learning Objectives Define energy, distinguish types of energy, and describe the nature of energy changes that accompany chemical and physical changes Distinguish the related properties of heat, thermal energy, and temperature Define and distinguish specific heat and heat capacity, and describe the physical implications of both Perform calculations involving heat, specific heat, and temperature change Chemical changes and their accompanying changes in energy are important parts of our everyday world (Figure $\PageIndex{1}$). The macronutrients in food (proteins, fats, and carbohydrates) undergo metabolic reactions that provide the energy to keep our bodies functioning. We burn a variety of fuels (gasoline, natural gas, coal) to produce energy for transportation, heating, and the generation of electricity. Industrial chemical reactions use enormous amounts of energy to produce raw materials (such as iron and aluminum). Energy is then used to manufacture those raw materials into useful products, such as cars, skyscrapers, and bridges. Over 90% of the energy we use comes originally from the sun. Every day, the sun provides the earth with almost 10,000 times the amount of energy necessary to meet all of the world’s energy needs for that day. Our challenge is to find ways to convert and store incoming solar energy so that it can be used in reactions or chemical processes that are both convenient and nonpolluting. Plants and many bacteria capture solar energy through photosynthesis. We release the energy stored in plants when we burn wood or plant products such as ethanol. We also use this energy to fuel our bodies by eating food that comes directly from plants or from animals that got their energy by eating plants. Burning coal and petroleum also releases stored solar energy: These fuels are fossilized plant and animal matter. This chapter will introduce the basic ideas of an important area of science concerned with the amount of heat absorbed or released during chemical and physical changes—an area called thermochemistry . The concepts introduced in this chapter are widely used in almost all scientific and technical fields. Food scientists use them to determine the energy content of foods. Biologists study the energetics of living organisms, such as the metabolic combustion of sugar into carbon dioxide and water. The oil, gas, and transportation industries, renewable energy providers, and many others endeavor to find better methods to produce energy for our commercial and personal needs. Engineers strive to improve energy efficiency, find better ways to heat and cool our homes, refrigerate our food and drinks, and meet the energy and cooling needs of computers and electronics, among other applications. Understanding thermochemical principles is essential for chemists, physicists, biologists, geologists, every type of engineer, and just about anyone who studies or does any kind of science. Energy Energy can be defined as the capacity to supply heat or do work. One type of work ( w ) is the process of causing matter to move against an opposing force. For example, we do work when we inflate a bicycle tire—we move matter (the air in the pump) against the opposing force of the air surrounding the tire. Like matter, energy comes in different types. One scheme classifies energy into two types: potential energy , the energy an object has because of its relative position, composition, or condition, and kinetic energy , the energy that an object possesses because of its motion. Water at the top of a waterfall or dam has potential energy because of its position; when it flows downward through generators, it has kinetic energy that can be used to do work and produce electricity in a hydroelectric plant (Figure $\PageIndex{2}$). A battery has potential energy because the chemicals within it can produce electricity that can do work. Energy can be converted from one form into another, but all of the energy present before a change occurs always exists in some form after the change is completed. This observation is expressed in the law of conservation of energy: during a chemical or physical change, energy can be neither created nor destroyed, although it can be changed in form. (This is also one version of the first law of thermodynamics, as you will learn later.) When one substance is converted into another, there is always an associated conversion of one form of energy into another. Heat is usually released or absorbed, but sometimes the conversion involves light, electrical energy, or some other form of energy. For example, chemical energy (a type of potential energy) is stored in the molecules that compose gasoline. When gasoline is combusted within the cylinders of a car’s engine, the rapidly expanding gaseous products of this chemical reaction generate mechanical energy (a type of kinetic energy) when they move the cylinders’ pistons. According to the law of conservation of matter (seen in an earlier chapter), there is no detectable change in the total amount of matter during a chemical change. When chemical reactions occur, the energy changes are relatively modest and the mass changes are too small to measure, so the laws of conservation of matter and energy hold well. However, in nuclear reactions, the energy changes are much larger (by factors of a million or so), the mass changes are measurable, and matter-energy conversions are significant. This will be examined in more detail in a later chapter on nuclear chemistry. To encompass both chemical and nuclear changes, we combine these laws into one statement: The total quantity of matter and energy in the universe is fixed. Thermal Energy, Temperature, and Heat Thermal energy is kinetic energy associated with the random motion of atoms and molecules. Temperature is a quantitative measure of “hot” or “cold.” When the atoms and molecules in an object are moving or vibrating quickly, they have a higher average kinetic energy (KE), and we say that the object is “hot.” When the atoms and molecules are moving slowly, they have lower KE , and we say that the object is “cold” (Figure $\PageIndex{3}$). Assuming that no chemical reaction or phase change (such as melting or vaporizing) occurs, increasing the amount of thermal energy in a sample of matter will cause its temperature to increase. And, assuming that no chemical reaction or phase change (such as condensation or freezing) occurs, decreasing the amount of thermal energy in a sample of matter will cause its temperature to decrease. Measuring Energy and Heat Capacity Historically, energy was measured in units of calories (cal) . A calorie is the amount of energy required to raise one gram of water by 1 degree C (1 kelvin). However, this quantity depends on the atmospheric pressure and the starting temperature of the water. The ease of measurement of energy changes in calories has meant that the calorie is still frequently used. The Calorie (with a capital C), or large calorie, commonly used in quantifying food energy content, is a kilocalorie. The SI unit of heat, work, and energy is the joule. A joule (J) is defined as the amount of energy used when a force of 1 newton moves an object 1 meter. It is named in honor of the English physicist James Prescott Joule. One joule is equivalent to 1 kg m 2 /s 2 , which is also called 1 newton–meter. A kilojoule (kJ) is 1000 joules. To standardize its definition, 1 calorie has been set to equal 4.184 joules. We now introduce two concepts useful in describing heat flow and temperature change. The heat capacity ( C ) of a body of matter is the quantity of heat ( q ) it absorbs or releases when it experiences a temperature change (Δ T ) of 1 degree Celsius (or equivalently, 1 kelvin) \[C=\dfrac{q}{ΔT} \label{5.2.1} \] Heat capacity is determined by both the type and amount of substance that absorbs or releases heat. It is therefore an extensive property—its value is proportional to the amount of the substance. For example, consider the heat capacities of two cast iron frying pans. The heat capacity of the large pan is five times greater than that of the small pan because, although both are made of the same material, the mass of the large pan is five times greater than the mass of the small pan. More mass means more atoms are present in the larger pan, so it takes more energy to make all of those atoms vibrate faster. The heat capacity of the small cast iron frying pan is found by observing that it takes 18,150 J of energy to raise the temperature of the pan by 50.0 °C \[C_{\text{small pan}}=\mathrm{\dfrac{18,140\; J}{50.0\; °C} =363\; J/°C} \label{5.2.2} \] The larger cast iron frying pan, while made of the same substance, requires 90,700 J of energy to raise its temperature by 50.0 °C. The larger pan has a (proportionally) larger heat capacity because the larger amount of material requires a (proportionally) larger amount of energy to yield the same temperature change: \[C_{\text{large pan}}=\mathrm{\dfrac{90,700\; J}{50.0\;°C}=1814\; J/°C} \label{5.2.3} \] The specific heat capacity ( c ) of a substance, commonly called its “specific heat,” is the quantity of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 kelvin): \[c = \dfrac{q}{\mathrm{m\Delta T}} \label{5.2.4} \] Specific heat capacity depends only on the kind of substance absorbing or releasing heat. It is an intensive property—the type, but not the amount, of the substance is all that matters. For example, the small cast iron frying pan has a mass of 808 g. The specific heat of iron (the material used to make the pan) is therefore: \[c_\ce{iron}=\mathrm{\dfrac{18,140\; J}{(808\; g)(50.0\;°C)} = 0.449\; J/g\; °C} \label{5.2.5} \] The large frying pan has a mass of 4040 g. Using the data for this pan, we can also calculate the specific heat of iron: \[c_\ce{iron}=\mathrm{\dfrac{90,700\; J}{(4,040\; g)(50.0\;°C)}=0.449\; J/g\; °C} \label{5.2.6} \] Although the large pan is more massive than the small pan, since both are made of the same material, they both yield the same value for specific heat (for the material of construction, iron). Note that specific heat is measured in units of energy per temperature per mass and is an intensive property, being derived from a ratio of two extensive properties (heat and mass). The molar heat capacity, also an intensive property, is the heat capacity per mole of a particular substance and has units of J/mol °C (Figure $\PageIndex{7}$). Liquid water has a relatively high specific heat (about 4.2 J/g °C); most metals have much lower specific heats (usually less than 1 J/g °C). The specific heat of a substance varies somewhat with temperature. However, this variation is usually small enough that we will treat specific heat as constant over the range of temperatures that will be considered in this chapter. Specific heats of some common substances are listed in Table $\PageIndex{1}$. Substance Symbol (state) Specific Heat (J/g °C) helium He(g) 5.193 water H2O(l) 4.184 ethanol C2H6O(l) 2.376 ice H2O(s) 2.093 (at −10 °C) water vapor H2O(g) 1.864 nitrogen N2(g) 1.040 air NaN 1.007 oxygen O2(g) 0.918 aluminum Al(s) 0.897 carbon dioxide CO2(g) 0.853 argon Ar(g) 0.522 iron Fe(s) 0.449 copper Cu(s) 0.385 lead Pb(s) 0.130 gold Au(s) 0.129 silicon Si(s) 0.712 If we know the mass of a substance and its specific heat, we can determine the amount of heat, q , entering or leaving the substance by measuring the temperature change before and after the heat is gained or lost: \[\begin{align} q &= \ce{(specific\: heat)×(mass\: of\: substance)×(temperature\: change)}\label{5.2.7}\\q&=c×m×ΔT \\[4pt] &=c×m×(T_\ce{final}−T_\ce{initial})\end{align} \] In this equation, $c$ is the specific heat of the substance, m is its mass, and Δ T (which is read “delta T”) is the temperature change, T final − T initial . If a substance gains thermal energy, its temperature increases, its final temperature is higher than its initial temperature, T final − T initial has a positive value, and the value of q is positive. If a substance loses thermal energy, its temperature decreases, the final temperature is lower than the initial temperature, T final − T initial has a negative value, and the value of q is negative. Example $\PageIndex{1}$: Measuring Heat A flask containing $\mathrm{8.0 \times 10^2\; g}$ of water is heated, and the temperature of the water increases from 21 °C to 85 °C. How much heat did the water absorb? Solution To answer this question, consider these factors: the specific heat of the substance being heated (in this case, water) the amount of substance being heated (in this case, 800 g) the magnitude of the temperature change (in this case, from 21 °C to 85 °C). The specific heat of water is 4.184 J/g °C, so to heat 1 g of water by 1 °C requires 4.184 J. We note that since 4.184 J is required to heat 1 g of water by 1 °C, we will need 800 times as much to heat 800 g of water by 1 °C. Finally, we observe that since 4.184 J are required to heat 1 g of water by 1 °C, we will need 64 times as much to heat it by 64 °C (that is, from 21 °C to 85 °C). This can be summarized using the equation: \[\begin{align} q&=c×m×ΔT \\[4pt] &=c×m×(T_\ce{final}−T_\ce{initial}) \\[4pt] &=\mathrm{(4.184\:J/\cancel{g}°C)×(800\:\cancel{g})×(85−21)°C}\\[4pt] &=\mathrm{(4.184\:J/\cancel{g}°\cancel{C})×(800\:\cancel{g})×(64)°\cancel{C}}\\[4pt] &=\mathrm{210,000\: J(=210\: kJ)} \end{align} \nonumber \] Because the temperature increased, the water absorbed heat and $q$ is positive. Exercise $\PageIndex{1}$ How much heat, in joules, must be added to a $\mathrm{5.00 \times 10^2 \;g}$ iron skillet to increase its temperature from 25 °C to 250 °C? The specific heat of iron is 0.451 J/g °C. Answer $\mathrm{5.05 \times 10^4\; J}$ Note that the relationship between heat, specific heat, mass, and temperature change can be used to determine any of these quantities (not just heat) if the other three are known or can be deduced. Example $\PageIndex{2}$: Determining Other Quantities A piece of unknown metal weighs 348 g. When the metal piece absorbs 6.64 kJ of heat, its temperature increases from 22.4 °C to 43.6 °C. Determine the specific heat of this metal (which might provide a clue to its identity). Solution Since mass, heat, and temperature change are known for this metal, we can determine its specific heat using the relationship: \[\begin{align} q&=c \times m \times \Delta T \\[4pt] &=c \times m \times (T_\ce{final}−T_\ce{initial}) \end{align} \nonumber \] Substituting the known values: \[6,640\; \ce J=c \times \mathrm{(348\; g) \times (43.6 − 22.4)\; °C} \nonumber \] Solving: \[c=\mathrm{\dfrac{6,640\; J}{(348\; g) \times (21.2°C)} =0.900\; J/g\; °C} \nonumber \] Comparing this value with the values in Table $\PageIndex{1}$, this value matches the specific heat of aluminum, which suggests that the unknown metal may be aluminum. Exercise $\PageIndex{2}$ A piece of unknown metal weighs 217 g. When the metal piece absorbs 1.43 kJ of heat, its temperature increases from 24.5 °C to 39.1 °C. Determine the specific heat of this metal, and predict its identity. Answer $c = \mathrm{0.45 \;J/g \;°C}$; the metal is likely to be iron from checking Table $\PageIndex{1}$. Solar Thermal Energy Power Plants The sunlight that reaches the earth contains thousands of times more energy than we presently capture. Solar thermal systems provide one possible solution to the problem of converting energy from the sun into energy we can use. Large-scale solar thermal plants have different design specifics, but all concentrate sunlight to heat some substance; the heat “stored” in that substance is then converted into electricity. The Solana Generating Station in Arizona’s Sonora Desert produces 280 megawatts of electrical power. It uses parabolic mirrors that focus sunlight on pipes filled with a heat transfer fluid (HTF) (Figure $\PageIndex{8}$). The HTF then does two things: It turns water into steam, which spins turbines, which in turn produces electricity, and it melts and heats a mixture of salts, which functions as a thermal energy storage system. After the sun goes down, the molten salt mixture can then release enough of its stored heat to produce steam to run the turbines for 6 hours. Molten salts are used because they possess a number of beneficial properties, including high heat capacities and thermal conductivities. The 377-megawatt Ivanpah Solar Generating System, located in the Mojave Desert in California, is the largest solar thermal power plant in the world (Figure $\PageIndex{9}$). Its 170,000 mirrors focus huge amounts of sunlight on three water-filled towers, producing steam at over 538 °C that drives electricity-producing turbines. It produces enough energy to power 140,000 homes. Water is used as the working fluid because of its large heat capacity and heat of vaporization. Summary Energy is the capacity to do work (applying a force to move matter). Kinetic energy (KE) is the energy of motion; potential energy is energy due to relative position, composition, or condition. When energy is converted from one form into another, energy is neither created nor destroyed (law of conservation of energy or first law of thermodynamics). Matter has thermal energy due to the KE of its molecules and temperature that corresponds to the average KE of its molecules. Heat is energy that is transferred between objects at different temperatures; it flows from a high to a low temperature. Chemical and physical processes can absorb heat (endothermic) or release heat (exothermic). The SI unit of energy, heat, and work is the joule (J). Specific heat and heat capacity are measures of the energy needed to change the temperature of a substance or object. The amount of heat absorbed or released by a substance depends directly on the type of substance, its mass, and the temperature change it undergoes. Key Equations $q=c×m×ΔT=c×m×(T_\ce{final}−T_\ce{initial})$ Glossary calorie (cal) unit of heat or other energy; the amount of energy required to raise 1 gram of water by 1 degree Celsius; 1 cal is defined as 4.184 J endothermic process chemical reaction or physical change that absorbs heat energy capacity to supply heat or do work exothermic process chemical reaction or physical change that releases heat heat ( q ) transfer of thermal energy between two bodies heat capacity ( C ) extensive property of a body of matter that represents the quantity of heat required to increase its temperature by 1 degree Celsius (or 1 kelvin) joule (J) SI unit of energy; 1 joule is the kinetic energy of an object with a mass of 2 kilograms moving with a velocity of 1 meter per second, 1 J = 1 kg m 2 /s and 4.184 J = 1 cal kinetic energy energy of a moving body, in joules, equal to $\dfrac{1}{2}mv^2$ (where m = mass and v = velocity) potential energy energy of a particle or system of particles derived from relative position, composition, or condition specific heat capacity ( c ) intensive property of a substance that represents the quantity of heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius (or 1 kelvin) temperature intensive property of matter that is a quantitative measure of “hotness” and “coldness” thermal energy kinetic energy associated with the random motion of atoms and molecules thermochemistry study of measuring the amount of heat absorbed or released during a chemical reaction or a physical change work ( w ) energy transfer due to changes in external, macroscopic variables such as pressure and volume; or causing matter to move against an opposing force
Courses/Kutztown_University_of_Pennsylvania/CHM_320%3A_Advanced_Inorganic_Chemistry_textbook/12%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/12.03%3A_Ionic_Lattices/12.3F%3A_Structure_-_-Cristobalite_(SiO)	Silicon dioxide, $\ce{SiO2}$, also known as silica is a linear molecule that is formed by one silicon atom and two oxygen atoms with two sets of doubles bonds and 4 single bonds. Because of its main component: glass, silicon dioxide is a very common and important molecule in the construction industry. One of the forms of silicon dioxide is quartz, which is found in sand. The Structure of Silica SiO 2 is a 3 dimensional structure and comes from the tetrahedral structure, SiO 4 . Each of the Silicon atoms are connected to each other with an oxygen atom, which creates a "diamond type network". All forms of SiO 2 possess a 3 dimensional shape and has a diamond structure. The bonding angle of Si-O-Si, which is the building block of the SiO 2 molecule, is 144 degrees. These are called polymorph and in order to be stable, 3 of these polymorph are suppose to exist. This stable unit creates a a temperature for each of the different forms of SiO 2 . Forms that have (alpha) are at low temperature, while forms with (beta) are at high temperature. The structure of the different forms of SiO 2 is important because it gives each of the different forms of SiO 2 different characteristics and functions. Commercially, SiO 2 is very important in steel, electronic, and semiconductor industries because of its structure, SiO 2 is able to undergo rapid temperature changes and still maintain its shape and structure. Different Forms of Silica and Their Uses There are many different forms of SiO 2 , which mainly derived quartz glass. One form that is derived from quartz is beta-cristobalite, which is found in high temperature. Some other forms are beta-quartz, alpha-quartz, beta-tridymite, alpha-tridymite, alpha=cristobalite, and many more. The alpha and beta stands for the temperature range. Alpha molecules have low temperature while beta molecules have high temperature. References Ed. Papier, Eugene. 2000. Absorption on Silica Surfaces . Marcek Dekker. New York. Pacific Rim Conference on Ceramic and Glass Technology. 2006. Advances in Glass and Optical Materials II . American Ceramic Society. Westerville, Ohio. Housecroft, Catherine E., and Alan G. Sharpe. 2008. Inorganic Chemistry . 3rd ed. Harlow: Pearson Education.
Bookshelves/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/6%3A_Gases/6.9_Density_and_the_Molar_Mass_of_Gases_(Video)	This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students . Video Topics The ideal gas law equation can be manipulated to show the relationship between the density of a gas and the molecular weight of the gas. The equation is d = MP/RT, d is the density of the gas in g/L, M is the molar mass of the gas in g/mol, P is pressure of the gas in ATM and R is the gas law constant. The equation shows that as the density of gas increases as the molar mass increases. This videos contains a sample calculation using this equation. Link to Video Density and the Molar Mass of Gases: https://youtu.be/gnkGBsvUFVk Attribution Prof. Steven Farmer ( Sonoma State University )
Courses/Purdue/Purdue%3A_Chem_26200%3A_Organic_Chemistry_II_(Wenthold)/Course_Content	Chapter 11. Infrared Spectroscopy and Mass Spectrometry 1.The Electromagnetic Spectrum, Visible Light and Color 2.Infrared Spectroscopy 2.1Wavelength and Wavenumbers 3.Molecular Vibrations 4.Characteristic Absorptions 5.Simplified Summary of IR Stretching Frequencies 5.1Solving IR Problems 6.Introduction to Mass Spectrometry 7.Determination of the Molecular Formula by Mass Spectrometry 7.1Molecular Ions 7.2Isotopes 7.3Nitrogen Rule 8. Fragmentation Patterns in Mass Spectrometry 9. Solving MS problems 10. Solving Problems with IR and MS Chapter 12. Nuclear Magnetic Resonance 1. Theory of Nuclear Magnetic Resonance 1.1 Nuclear Spin 2. Magnetic Shielding by Electrons 3. The NMR Spectrometer 4. The Chemical Shift 5. The Number of Signals 6. Areas of Peaks 7. Spin-Spin Coupling 7.1 Complex Splitting 7.2 Stereochemical Non-equivalence of protons 8. Solving NMR spectra 8.1 Practice Problems 9. Carbon-13 NMR 9.1 Interpreting C-13 NMR 10. Integrated Structure Determination 10.1 IR and NMR 10.2 IR, MS and NMR Chapter 13. Conjugation 1. Stabilities of Dienes 2. Molecular Orbital Picture of a Conjugated System 3. Allylic Cations 4. 1,2- and 1,4-Addition to Conjugated Dienes 4.1 Kinetic versus Thermodynamic control in the Addition of HBr to Buta-1,3-diene 5. Allylic Radicals 6. Molecular Orbitals of the Allylic System 6.1 Electronic configurations of the Allyl Radical, Cation, and Anion 7. SN2 Displacement Reactions of Allylic Halides and Tosylates 8. The Diels-Alder Reaction 8.1 The Diels-Alder as an Example of a Pericyclic Reaction 9. Ultraviolet Absorption Spectroscopy 9.1 Colored Organic compounds Chapter 14. Aromaticity 1. The Discovery of Benzene 2. The Structure and Properties of Benzene 2.1 Stability of Benzene 3. Aromaticity 3.1 Huckel’s Rule 4. The Molecular Orbitals of Benzene 4.1 Frost Diagrams 5. Aromatic, Antiaromatic, and Nonaromatic Compounds 6. Examples of Aromatic Molecules 6.1 Aromatic Ions 6.2 Heterocyclic Aromatic Compounds 6.3 Polynuclear Aromatic Hydrocarbons 6.4 Aromatic Allotropes of Carbon 6.5 Fused Heterocyclic Compounds 7. Nomenclature of Benzene Derivatives Chapter 15. Reactions of Aromatic Compounds 1. Electrophilic Aromatic Substitution 1.1 Halogenation of Benzene 1.2 Nitration of Benzene 1.3 Sulfonation of Benzene 2. Friedel-Crafts Reactions 2.1 Alkylation 2.2 Acylation 3. Di- and Polysubstitution: Substitution effects on orientation 3.1 Theory of Directing effects 3.2 Activating, Ortho, Para-Directing Substituents 3.3 Deactiving, Meta-Directors 3.4 Halogen Substituents: Deactivating, but Ortho- Para-Directing 4. Nucleophilic Aromatic Substitution 4.1 Benzyne Intermediates 5. Birch Reduction of Aromatic Molecules 5.1 Regioselectivity of Birch Reduction 6. Side-Chain Reactions of Benzene Derivatives 6.1 Oxidation 6.2 Free Radical Halogenation 6.3 Substitution (S N 1 and S N 2) 6.4 Hydrogenolysis 7. Summary of Phenols 7.1 Acidity 7.2 Oxidation 7.3 Electrophilic Aromatic Substitution Chapter 16: Ketones and Aldehydes 1. Carbonyl compounds 2. Structure of the Carbonyl Group 3. Nomenclature of Ketones and Aldehydes 4. Physical Properties of Ketones and Aldehydes Review: Synthesis of Ketones and Aldehydes 5. Reactions of Ketones and Aldehydes: Introduction to Nucleophilic addition 6. Reactions of Ketones and Aldehydes: Addition of Carbon Nucleophiles 6.1 Grignard, Organolithium and other Organometallic Reagents 6.2 Cyanide: Formation of Cyanohydrins 6.3 Deprotonated Alkynes 7. The Wittig Reaction 8. Hydration of Ketones and Aldehydes 9. Formation of acetals 9.1 Use of acetals as protecting groups 10. Reactions with amines 10.1 Formation of imines 10.2 Condensations with Hydroxylamine and hydrazines 11. Oxidation of Aldehydes 12. Reductions of Ketones and Aldehydes 12.1 Metal Hydride 12.2 Other reductions Chapter 17: Carboxylic Acids 1......... Nomenclature of Carboxylic Acids 2......... Structure and Physical Properties of Carboxylic Acids 3......... Acidity of Carboxylic Acids ........... 3.1 Origins of Acidity of Carboxylic Acids ........... 3.2 Salts of Carboxylic Acids ........... 3.3 Natural sources of Carboxylic Acids 4. ...... Synthesis of Carboxylic Acids 5. ...... Condensation of Acids with Alcohols: the Fischer Esterification ........... 5.1 Esterification Using Diazomethane 6. ...... Reduction of Carboxylic Acids ........... 6.1 Alkylation of Carboxylic Acids to Form Ketones 7. ...... Synthesis of Acid Chlorides Chapter 18: Carboxylic Acid Derivatives 1. Structure and Nomenclature of Acid Derivatives 2. Physical Properties of Carboxylic Acid Derivatives 3. Substitution of Carboxylic Acid Derivatives by Nucleophilic Acyl Substitution 3.1 Substitutions of Acid Chloride 3.2 Substitutions of Acid Anhydrides 3.3 Substitutions of Esters 3.3.1 Transesterification 3.4 Amides 4. Hydrolysis of Carboxylic Acid Derivatives 5. Reduction of Acid Derivatives 6. Reactions of Acid Derivatives with Organometallic Reagents Chapter 19: Carbonyl Condensation Reactions Enols and Enolate Ions 1 Acidity of alpha-Hydrogens Alpha-Halogenation Alkylation of Enolates The Aldol Condensation of Ketones and Aldehydes 1 Base Catalyzed 2 Acid Catalyzed 3 Dehydration of Aldol Products 4 Crossed Aldol Condensations 5 Aldol Cyclizations Planning Syntheses Using Aldol Condensations Enamines The Claisen Ester Condensation 1 The Dieckmann Condensation: A Claisen Cyclization 2 Crossed Claisen Condensations Chapter 20: Amines Nomenclature of Amines Structure and Physical Properties of Amines 1 Chirality Basicity of Amines 1 Salts of Amines 2 Natural Products Review: Reactions of Amines with Ketones and Aldehydes Other Reactions of Amines 1 Alkylation of Amines by Alkyl Halides 2 The Hofmann Elimination 3 Acylation of Amines by Acid Chlorides 4 Oxidation of Amines; The Cope Elimination Synthesis and Reactions of Arenediazonium Salts Synthesis of Amines 1 Reductive Amination 2 Synthesis of Amines by Acylation-Reduction 3 Syntheses Limited to Primary Amines Chapter 21. Carbon-Carbon Bond Formation and Chiral Catalysis 1. Palladium-catalyzed Cross-Coupling Reactions 1 Heck Reaction 2 Suzuki Coupling 3 Other Reactions Stille Coupling Sonagashira Coupling Negishi Coupling 2. Alkene Metathesis Grubbs Catalyst Synthesis of Single Enantiomers 1 Asymmetric Induction Chiral Auxiliary 2 Chiral Catalysis Chapter 22. Carbohydrates Classification of Carbohydrates Fisher Projections Monosaccharides D and L Sugars Epimers Cyclic Structures of Monosaccharides Haworth Projections Anomers of Monosaccharides Glycosidic bonds Disaccharides 1 Sucrose vs HFCS Polysaccharides

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