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Bookshelves/Analytical_Chemistry/Instrumental_Analysis_(LibreTexts)/16%3A_An_Introduction_to_Infrared_Spectrometry | 16.1: Theory of Infrared Absorption Spectrometry To absorb an infrared photon, the absorbing species must experience a change in its dipole moment, which allows the oscillation in the photon's electrical field to interact with an oscillation in charge within the absorbing species. If the two oscillations have the same frequency, then absorption is possible. In this section we consider classical and quantum mechanical models for vibrational spectrscopy. 16.2: Infrared Sources and Transducers Instrumentation for IR spectroscopy requires a source of infrared radiation and a transducer for detecting the radiation after it passes through the sample. Common sources and transducers are reviewed here. 16.3: Infrared Instruments Instrumentation for infrared spectroscopy use one of three common optical benches: non-dispersive instruments, dispersive instruments, and Fourier transform instruments. As we have already examined non-dispersive and dispersive instruments in Chapter 13, and because they are no longer as common as they once were, we give them only a brief consideration here. Fourier transform instruments, which dominate the current marketplace receive more detailed treatment. |
Courses/College_of_the_Canyons/Chem_201%3A_General_Chemistry_I_OER/08%3A_Thermochemistry/8.06%3A_Constant_Pressure_Calorimeter | Because Δ H is defined as the heat flow at constant pressure, measurements made using a constant-pressure calorimeter (a device used to measure enthalpy changes in chemical processes at constant pressure) give Δ H values directly. This device is particularly well suited to studying reactions carried out in solution at a constant atmospheric pressure. A “student” version, called a coffee-cup calorimeter (Figure \(\PageIndex{1}\)), is often encountered in general chemistry laboratories. Commercial calorimeters operate on the same principle, but they can be used with smaller volumes of solution, have better thermal insulation, and can detect a change in temperature as small as several millionths of a degree (10 −6 °C). Because the heat released or absorbed at constant pressure is equal to Δ H , the relationship between heat and \(ΔH_{rxn}\) is \[ \Delta H_{rxn}=q_{rxn}=-q_{calorimater}=-mc \Delta T \label{5.5.8} \] The use of a constant-pressure calorimeter is illustrated in Example \(\PageIndex{3}\). Example \(\PageIndex{1}\) When 5.03 g of solid potassium hydroxide are dissolved in 100.0 mL of distilled water in a coffee-cup calorimeter, the temperature of the liquid increases from 23.0°C to 34.7°C. The density of water in this temperature range averages 0.9969 g/cm 3 . What is Δ H soln (in kilojoules per mole)? Assume that the calorimeter absorbs a negligible amount of heat and, because of the large volume of water, the specific heat of the solution is the same as the specific heat of pure water. Given: mass of substance, volume of solvent, and initial and final temperatures Asked for: Δ H soln Strategy: Calculate the mass of the solution from its volume and density and calculate the temperature change of the solution. Find the heat flow that accompanies the dissolution reaction by substituting the appropriate values into Equation 5.5.8. Use the molar mass of KOH to calculate Δ H soln . Solution: A To calculate Δ H soln , we must first determine the amount of heat released in the calorimetry experiment. The mass of the solution is \[ \left (100.0 \; mL\; H2O \right ) \left ( 0.9969 \; g/ \cancel{mL} \right )+ 5.03 \; g \; KOH=104.72 \; g \] The temperature change is (34.7°C − 23.0°C) = +11.7°C. B Because the solution is not very concentrated (approximately 0.9 M), we assume that the specific heat of the solution is the same as that of water. The heat flow that accompanies dissolution is thus \[ q_{calorimater}=mC_{s} \Delta T =\left ( 104.72 \; \cancel{g} \right ) \left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \cancel{^{o}C}} \right )\left ( 11.7 \; ^{o}C \right )=5130 \; J =5.13 \; lJ \] The temperature of the solution increased because heat was absorbed by the solution ( q > 0). Where did this heat come from? It was released by KOH dissolving in water. From Equation \(\ref{5.5.8}\), we see that Δ H rxn = − q calorimeter = −5.13 kJ This experiment tells us that dissolving 5.03 g of KOH in water is accompanied by the release of 5.13 kJ of energy. Because the temperature of the solution increased, the dissolution of KOH in water must be exothermic. C The last step is to use the molar mass of KOH to calculate Δ H soln —the heat released when dissolving 1 mol of KOH: \[ \Delta H_{soln}= \left ( \dfrac{5.13 \; kJ}{5.03 \; \cancel{g}} \right )\left ( \dfrac{56.11 \; \cancel{g}}{1 \; mol} \right )=-57.2 \; kJ/mol \] Exercise \(\PageIndex{1}\) A coffee-cup calorimeter contains 50.0 mL of distilled water at 22.7°C. Solid ammonium bromide (3.14 g) is added and the solution is stirred, giving a final temperature of 20.3°C. Using the same assumptions as in Example \(\PageIndex{1}\), find Δ H soln for NH 4 Br (in kilojoules per mole). Answer 16.6 kJ/mol |
Courses/Saint_Francis_University/CHEM_113%3A_Human_Chemistry_I_(Zovinka)/03%3A_Ionic_Compounds/3.02%3A_Ions_and_the_Octet_Rule | Learning Objectives Use the octet rule and electron configurations to determine if an atom will gain or lose electrons forming anions or cations. Ions are formed when an atom, usually on the left side of the periodic table, reacts with and transfers one or more electrons to another atom, usually on the right side of the periodic table. These electrons are usually lost from or gained into the valence shell , or outermost energy level (shell). Why do some atoms lose electrons and others gain electrons? How can we predict the number of electrons lost or gained? Which ions are stable and which ions do not form at all? These questions are best answered by looking at electron configurations and considering what is called the octet rule , which states that atoms gain or lose electrons to form a stable, noble gas configuration, i.e., a filled subshell containing eight electrons. Therefore, it is useful to take a closer look at electron configurations to further illustrate ion formation and electron transfer between atoms. The electron configuration for sodium shows that there are ten core electrons and one valence electron in the third energy level. When sodium loses the single valence electron, forming the cation Na + , the electron configuration is now identical to that of neon, a stable noble gas with eight valence electrons. \[\begin{array}{lcl} \ce{Na} & \rightarrow & \ce{Na^+} + \ce{e^-} \\ 1s^2 \: 2s^2 \: 2p^6 \: 3s^1 & & 1s^2 \: 2s^2 \: 2p^6 \end{array}\] Chlorine also has ten core electrons and valence electrons in the third energy level. However, chlorine has seven valence electrons, one less than the noble gas argon, which has eight valence electrons. Thus, chlorine will gain one electron, forming the anion, Cl – , and achieving a stable noble gas configuration. \[\begin{array}{lcl} \ce{Cl} + \ce{e^-} & \rightarrow & \ce{Cl^-} \\ 1s^2 \: 2s^2 \:2p^6 \: 3s^2 \: 3p^5 & & 1s^2 \: 2s^2 \: 2p^6 \: 3s^2 \: 3p^6 \end{array}\] The octet rule and the periodic table can be used to predict what ions will form; main group elements on the left side of the periodic table (metals in groups 1, 2, and 13) tend to lose electrons (form cations) to achieve the same electron configuration as the noble gas just before them in the table. The number of electrons the atom will lose depends on what group the atom is in, i.e., how many valence electrons it has. Main group elements on the right side of the periodic table (nonmetals in groups 15-17) will gain electrons to achieve the same electron configuration as the noble gas just after them in the table. Again, the number of electrons the atom will gain depends on the number of valence electrons it has and how many are needed to reach the filled subshell, eight electrons. Note Violation of the Octet Rule It is not impossible to violate the octet rule. Consider sodium: in its elemental form, it has one valence electron and is stable. It is rather reactive, however, and does not require a lot of energy to remove that electron to make the Na + ion. We could remove another electron by adding even more energy to the ion, to make the Na 2 + ion. However, that requires much more energy than is normally available in chemical reactions, so sodium stops at a 1+ charge after losing a single electron. It turns out that the Na + ion has a complete octet in its new valence shell, the n = 2 shell, which satisfies the octet rule. The octet rule is a result of trends in energies and is useful in explaining why atoms form the ions that they do. Example \(\PageIndex{1}\) Write the electron configuration of aluminum atom ( Z = 13) and underline the valence electrons . How many electrons are gained/lost to form an aluminum ion? Write the symbol and the electron configuration for an aluminum ion. Solution The electron configuration of Al atom is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 1 . Aluminum has three valence electrons in the third energy level, (3 s 2 3 p 1 ). The cation, Al 3 + , is formed when these three valence electrons are lost , leaving the configuration for the noble gas neon, 1 s 2 2 s 2 2 p 6 . Exercise \(\PageIndex{1}\) Write the electron configuration of oxygen atom ( Z = 8) and underline the valence electrons. How many electrons are gained/lost to form an oxide ion? Write the symbol and electron configuration for oxide ion. Answer The electron configuration of O atom is 1 s 2 2 s 2 2 p 4 . Oxygen has six valence electrons in the second energy level, ( 2 s 2 2 p 4 ). The anion O 2 − is formed when two electrons are gained in the valence shell. The resulting electron configuration, 1 s 2 2 s 2 2 p 6 , which is also identical to the configuration for the noble gas neon . |
Courses/Southeast_Missouri_State_University/CH185%3A_General_Chemistry_(Ragain)/03%3A_Molecules_Compounds_and_Chemical_Equations/3.05%3A_Representing_Compounds-_Chemical_Formulas_and_Molecular_Models | Chemistry is the experimental and theoretical study of materials on their properties at both the macroscopic and microscopic levels. Understanding the relationship between properties and structures/bonding is also a hot pursuit. Chemistry is traditionally divided into organic and inorganic chemistry. The former is the study of compounds containing at least one carbon-hydrogen bonds. By default, the chemical study of all other substances is called inorganic chemistry, a less well defined subject. However, the boundary between organic and inorganic compounds is not always well defined. For example, oxalic acid, H 2 C 2 O 4 , is a compound formed in plants, and it is generally considered an organic acid, but it does not contain any C-H bond. Inorganic chemistry is also closely related to other disciplines such as materials sciences, physical chemistry, thermodynamics, earth sciences, mineralogy, crystallography, spectroscopy etc. A chemical formula is a format used to express the structure of atoms. The formula tells which elements and how many of each element are present in a compound. Formulas are written using the elemental symbol of each atom and a subscript to denote the number of elements. This notation can be accredited to Swedish chemist Jons Jakob Berzeliu. The most common elements present in organic compounds are carbon, hydrogen, oxygen, and nitrogen. With carbon and hydrogen present, other elements, such as phosphorous, sulfur, silicon, and the halogens, may exist in organic compounds. Compounds that do not pertain to this rule are called inorganic compounds. Molecular Geometry and Structural Formula Understanding how atoms in a molecules are arranged and how they are bonded together is very important in giving the molecule its identity. Isomers are compounds in which two molecules can have the same number of atoms, and thus the same molecular formula, but can have completely different physical and chemical properties because of differences in structural formula. M ethylpropane and butane have the same molecular formula of C 4 H 10 , but are structurally different (methylpropane on the left, butane on the right). Polymers A polymer is formed when small molecules of identical structure, monomers, combine into a large cluster. The monomers are joined together by covalent bonds. When monomers repeat and bind, they form a polymer. While they can be comprised of natural or synthetic molecules, polymers often include plastics and rubber. When a molecule has more than one of these polymers, square parenthesis are used to show that all the elements within the polymer are multiplied by the subscript outside of the parenthesis. The subscript (shown as n in the example below) denotes the number of monomers present in the macromolecule (or polymer). Ethylene becomes the polymer polyethylene. Molecular Formula The molecular formula is based on the actual makeup of the compound. Although the molecular formula can sometimes be the same as the empirical formula, molecular compounds tend to be more helpful. However, they do not describe how the atoms are put together. Molecular compounds are also misleading when dealing with isomers, which have the same number and types of atoms (see above in molecular geometry and structural formula). Ex. Molecular Formula for Ethanol: C 2 H 6 O. Empirical Formula An empirical formula shows the most basic form of a compound. Empirical formulas show the number of atoms of each element in a compound in the most simplified state using whole numbers. Empirical formulas tend to tell us very little about a compound because one cannot determine the structure, shape, or properties of the compound without knowing the molecular formula. Usefulness of the empirical formula is decreased because many chemical compounds can have the same empirical formula. Ex. Find the empirical formula for C 8 H 16 O 2 . Answer: C 4 H 8 O (divide all subscripts by 2 to get the smallest, whole number ratio). Structural Formula A structural formula displays the atoms of the molecule in the order they are bonded. It also depicts how the atoms are bonded to one another, for example single, double, and triple covalent bond. Covalent bonds are shown using lines. The number of dashes indicate whether the bond is a single, double, or triple covalent bond. Structural formulas are helpful because they explain the properties and structure of the compound which empirical and molecular formulas cannot always represent. Ex. Structural Formula for Ethanol: Condensed Structural Formula Condensed structural formulas show the order of atoms like a structural formula but are written in a single line to save space and make it more convenient and faster to write out. Condensed structural formulas are also helpful when showing that a group of atoms is connected to a single atom in a compound. When this happens, parenthesis are used around the group of atoms to show they are together. Ex. Condensed Structural Formula for Ethanol: CH 3 CH 2 OH (Molecular Formula for Ethanol C 2 H 6 O). Line-Angle Formula Because organic compounds can be complex at times, line-angle formulas are used to write carbon and hydrogen atoms more efficiently by replacing the letters with lines. A carbon atom is present wherever a line intersects another line. Hydrogen atoms are then assumed to complete each of carbon's four bonds. All other atoms that are connected to carbon atoms are written out. Line angle formulas help show structure and order of the atoms in a compound making the advantages and disadvantages similar to structural formulas. Ex. Line-Angle Formula for Ethanol: Formulas of Inorganic Compounds Inorganic compounds are typically not of biological origin. Inorganic compounds are made up of atoms connected using ionic bonds. These inorganic compounds can be binary compounds, binary acids, or polyatomic ions. Binary compounds Binary compounds are formed between two elements, either a metal paired with a nonmetal or two nonmetals paired together. When a metal is paired with a nonmetal, they form ionic compounds in which one is a negatively charged ion and the other is positvely charged. The net charge of the compound must then become neutral. Transition metals have different charges; therefore, it is important to specify what type of ion it is during the naming of the compound. When two nonmetals are paired together, the compound is a molecular compound. When writing out the formula, the element with a positive oxidation state is placed first. Ex. Ionic Compound: BaBr 2 (Barium Bromide) Ex. Molecular Compound: N 2 O 4 (Dinitrogen Tetroxide) Binary acids Binary acids are binary compounds in which hydrogen bonds with a nonmetal forming an acid. However, there are exceptions such as NH 3 , which is a base. This is because it shows no tendency to produce a H + . Because hydrogen is positively charged, it is placed first when writing out these binary acids. Ex. HBr (Hydrobromic Acid) Polyatomic ions Polyatomic ions is formed when two or more atoms are connected with covalent bonds. Cations are ions that have are postively charged, while anions are negatively charged ions. The most common polyatomic ions that exists are those of anions. The two main polyatomic cations are Ammonium and Mercury (I). Many polyatomic ions are typically paired with metals using ionic bonds to form chemical compounds. Ex. MnO 4 - (Polyatomic ion); NaMnO 4 (Chemical Compound) Oxoacids Many acids have three different elements to form ternary compounds. When one of those three elements is oxygen, the acid is known as a oxoacid. In other words, oxacids are compounds that contain hydrogen, oxgygen, and one other element. Ex. HNO 3 (Nitric Acid) Complex Compounds Certain compounds can appear in multiple forms yet mean the same thing. A common example is hydrates: water molecules bond to another compound or element. When this happens, a dot is shown between H 2 O and the other part of the compound. Because the H 2 O molecules are embedded within the compound, the compound is not necessarily "wet". When hydrates are heated, the water in the compound evaporates and the compound becomes anhydrous. These compounds can be used to attract water such as CoCl 2 . When CoCl 2 is dry, CoCl 2 is a blue color wherease the hexahydrate (written below) is pink in color. Ex. CoCl 2 · 6 H 2 O Formulas of Organic Compounds Organic compounds contain a combination carbon and hydrogen or carbon and hydrogen with nitrogen and a few other elements, such as phosphorous, sulfur, silicon, and the halogens. Most organic compounds are seen in biological origin, as they are found in nature. Hydrocarbons Hydrocarbons are compounds that consist of only carbon and hydrogen atoms. Hydrocarbons that are bonded together with only single bonds are alkanes. The simplest example is methane (shown below). When hydrocarbons have one or more double bonds, they are called alkenes. The simplest alkene is Ethene (C 2 H 4 ) which contains a double bond between the two carbon atoms. Ex. Methane on left, Ethene on right Functional Groups Functional groups are atoms connected to carbon chains or rings of organic molecules. Compounds that are within a functional group tend to have similar properties and characteristics. Two common functional groups are hydroxyl groups and carboxyl groups. Hydroxyl groups end in -OH and are alcohols. Carboxyl groups end in -COOH, making compounds containing -COOH carboxylic acids. Functional groups also help with nomenclature by using prefixes to help name the compounds that have similar chemical properties. Ex. Hydroxyl Group on top; Carboxyl Group on bottom References Miessler, Gary L. Inorganic Chemistry. 2nd. Upper Saddle River: Prentince Hall, 1999. Munowitz, Michael. Principles of Chemistry. Norton & Company: New York, 2000. Pettrucci, Ralph H. General Chemistry: Principles and Modern Applications. 9th. Upper Saddle River: Pearson Prentice Hall, 2007. Problems Which of the following formulas are organic? HClO C 5 H 10 CO 2 What is the name of the following formula? Classify the following formulas into their appropriate functional group Acetic acid Butanol Oxalic acid What are the empirical formulas for the following compounds? C 12 H 10 O 6 CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 H 3 O What is the name of the following figure and what is the molecular formula of the following figure? Answer Key: 1. b and c. 2. Propane. 3. a. carboxyl group, b. hydroxyl group, c. carboxyl group. 4. a. C 6 H 5 O 3 , b. C 7 H 16 , c. H 3 O. 5. Methylbutane, C 5 H 12 |
Bookshelves/Analytical_Chemistry/Physical_Methods_in_Chemistry_and_Nano_Science_(Barron)/03%3A_Principles_of_Gas_Chromatography/3.05%3A_Ion_Chromatography | Ion Chromatography is a method of separating ions based on their distinct retention rates in a given solid phase packing material. Given different retention rates for two anions or two cations, the elution time of each ion will differ, allowing for detection and separation of one ion before the other. Detection methods are separated between electrochemical methods and spectroscopic methods. This guide will cover the principles of retention rates for anions and cations, as well as describing the various types of solid-state packing materials and eluents that can be used. Principles of Ion Chromatography Retention Models in Anion Chromatography The retention model for anionic chromatography can be split into two distinct models, one for describing eluents with a single anion, and the other for describing eluents with complexing agents present. Given an eluent anion or an analyte anion, two phases are observed, the stationary phase (denoted by S) and the mobile phase (denoted by M). As such, there is equilibrium between the two phases for both the eluent anions and the analyte anions that can be described by Equation \ref{1}. \[ y*[A^{x-}_{M}]\ +\ x*[E^{y-}_{S}]\ \Leftrightarrow \ y*[A^{x-}_{S}]\ +\ x*[E^{y-}_{M}] \label{1} \] This yields an equilibrium constant as given in Equation \ref{2} . \[ K_{A,E} = \frac{ [A^{x-}_{S}]^{y} [E^{y-}_{M}]^{x} \gamma ^{y} _{A^{x-}_{S} } \gamma ^{x} _{E^{y-}_{S}} }{ [A^{x-}_{M}] ^{y} [E^{y-}_{S}]^{x} \gamma ^{y} _{A^{x-}_{M}} \gamma ^{x} _{E^{y-}_{S}}} \label{2} \] Given the activity of the two ions cannot be found in the stationary or mobile phases, the activity coefficients are set to 1. Two new quantities are then introduced. The first is the distribution coefficient, D A , which is the ratio of analyte concentrations in the stationary phase to the mobile phase, Equation \ref{3} . The second is the retention factor, k 1 A , which is the distribution coefficient times the ratio of volume between the two phases, Equation \ref{4} . \[ D_{A} \ =\ \frac{[A_{S}]}{[A_{M}]} \label{3} \] \[k_{A}^{1} \ = \ D_{A} * \frac{V_{S}}{V_{M}} \label{4} \] Substituting the two quantities from Equation \ref{3} and Equation \ref{4} into Equation \ref{2} , the equilibrium constant can be written as Equation \ref{5} \[ K_{A,E} \ = (k_{A}^{1} \frac{V_{M}}{V_{S}})^{y} * (\frac{[E_{M}^{y-} ]}{[E^{y-}_{S}]})^{x} \label{5} \] Given there is usually a large difference in concentrations between the eluent and the analyte (with magnitudes of 10 greater eluent), equation 4 can be re-written under the assumption that all the solid phase packing material’s functional groups are taken up by E y- . As such, the stationary E y- can be substituted with the exchange capacity divided by the charge of E y- . This yields Equation \ref{6} \[ K_{A,E} \ = (k_{A}^{1} \frac{V_{M}}{V_{S}})^{y} * (\frac{Q}{\gamma })^{-x} [E_{M}^{y-}] \label{6} \] Solving for the retention factor Equation \ref{7} is developed. \[ z*[A^{x-}_{M}] \ +\ x*[B^{z-}_{S}] \Leftrightarrow z* [A^{x-}_{S}] \ +\ x*[B^{z-}_{M}] \label{7} \] Equation \ref{8} shows the relationship between retention factor and parameters like eluent concentration and the exchange capacity, which allows parameters of the ion chromatography to be manipulated and the retention factors to be determined. Equation \ref{9} only works for a single analyte present, but a relationship for the selectivity between two analytes [A] and [B] can easily be determined. First the equilibrium between the two analytes is determined as Equation \ref{8} \[ K_{A,B} \ = \frac{[A^{x-}_{S}]^{z} [B^{z-}_{M}]^{x}}{[A^{x-}_{M}]^{z} [B^{z-}_{S}]^{x}} \label{8} \] The equilibrium constant can be written as Equation \ref{9} (ignoring activity): \[ \alpha _{A,B} \ = \frac{[A^{x-}_{S}][B^{z-}_{M}]}{[A^{x-}_{M}][B^{z-}_{S}]} \label{9} \] The selectivity can then be determined to be Equation \ref{10} \[ \alpha _{A,B} \ = \frac{[A^{x-}_{S}][B^{z-}_{M}]}{[A^{x-}_{M}][B^{z-}_{S}]} \label{10} \] Equation \ref{10} can then be simplified into a logarithmic form as the following two equations: \[ \log \alpha _{A,B} = \frac{1}{z} log K_{A,B} \ + \frac{x-z}{z} log \frac{ k_{A}^{1} V_{M}}{V_{S}} \label{11} \] \[ \log \alpha _{A,B} = \frac{1}{x} log K_{A,B} \ + \frac{x-z}{z} log \frac{ k_{A}^{1} V_{M}}{V_{S}} \label{12} \] When the two charges are the same, it can be seen that the selectivity is only a factor of the selectivity coefficients and the charges. When the two charges are different, it can be seen that the two retention factors are dependent upon each other. In situations with a polyatomic eluent, three models are used to account for the multiple anions in the eluent. The first is the dominant equilibrium model, in which one anion is so dominant in concentration; the other eluent anions are ignored. The dominant equilibrium model works best for multivalence analytes. The second is the effective charge model, where an effective charge of the eluent anions is found, and a relationship similar to EQ is found with the effective charge. The effective charge models works best with monovalent analytes. The third is the multiple eluent species model, where Equation \ref{13} describes the retention factor: \[ \log K_{A}^{1} \ =\ C_{3} - (\frac{X_{1}}{a} + \frac{X_{2}}{b} + \frac{X_{3}}{c}) -\ log C_{P} \label{13} \] C 3 is a constant that includes the phase volume ratio between stationary, the equilibrium constant, and mobile and the exchange capacity. C p is the total concentration of the eluent species. X 1 , X 2 , X 3 , correspond to the shares of a particular eluent anion in the retention of the analyte. Retention Models of Cation Chromatography For eluents with a single cation and analytes that are alkaline earth metals, heavy metals or transition metals, a complexing agent is used to bind with the metal during chromatography. This introduces the quantity A(m) to the retention rate calculations, where A(m) is the ratio of free metal ion to the total concentration of metal. Following a similar derivation to the single anion case, Equation \ref{14} is found. \[ K_{A,E} = \ (\frac{ k_{A}^{1}}{ \alpha _{M} \phi } )^{y} * (\frac{Q}{\gamma })^{-x} [E ^{y+} _{M} ]^{x} \label{14} \] Solving for the retention coefficient, Equation \ref{15} is found. \[ k_{A}^{1} = \alpha _{M} \phi * K_{A,E} ^{\frac{1}{\gamma } } (\frac{Q}{\gamma })^{\frac{x}{y} } ([E_{M}^{y+}]^{- \frac{x}{y} } \label{15} \] From this expression, the retention rate of the cation can be determined from eluent concentration and the ratio of free metal ions to the total concentration of the metal, which itself is depended on the equilibrium of the metal ion with the complexing agent. Solid Phase Packing Materials The solid phase packing material used in the chromatography column is important to the exchange capacity of the anion or cation. There are many types of packing material, but all share a functional group that can bind either the anion or the cation complex. The functional group is mounted on a polymer surface or sphere, allowing large surface area for interaction. Packing Material for Anion Chromatography The primary functional group used for anion chromatography is the ammonium group. Amine groups are mounted on the polymer surface, and the pH is lowered to produce ammonium groups. As such, the exchange capacity is depended on the pH of the eluent. To reduce the pH dependency, the protons on the ammonium are successively replaced with alkyl groups until the all the protons are replaced and the functional group is still positively charged, but pH independent. The two packing materials used in almost all anion chromatography are trimethylamine (NMe 3 , Figure \(\PageIndex{1}\) ) and dimethylanolamine (Figure \(\PageIndex{2}\) ). Packing Material for Cation Chromatography Cation chromatography allows for the use of both organic polymer based and silica gel based packing material. In the silica gel based packing material, the most common packing material is a polymer-coated silica gel. The silicate is coated in polymer, which is held together by cross-linking of the polymer. Polybutadiene maleic acid (Figure \(\PageIndex{3}\) ) is then used to create a weakly acidic material, allowing the analyte to diffuse through the polymer and exchange. Silica gel based packing material is limited by the pH dependent solubility of the silica gel and the pH dependent linking of the silica gel and the functionalized polymer. However, silica gel based packing material is suitable for separation of alkali metals and alkali earth metals. Organic polymer based packing material is not limited by pH like the silica gel materials are, but are not suitable for separation of alkali metals and alkali earth metals. The most common functional group is the sulfonic acid group (Figure \(\PageIndex{4}\) ) attached with a spacer between the polymer and the sulfonic acid group. Detection Methods Spectroscopic Detection Methods Photometric detection in the UV region of the spectrum is a common method of detection in ion chromatography. Photometric methods limit the eluent possibilities, as the analyte must have a unique absorbance wavelength to be detectable. Cations that do not have a unique absorbance wavelength, i.e. the eluent and other contaminants have similar UV visible spectra can be complexed to for UV visible compounds. This allows detection of the cation without interference from eluents. Coupling the chromatography with various types of spectroscopy such as Mass spectroscopy or IR spectroscopy can be a useful method of detection. Inductively coupled plasma atomic emission spectroscopy is a commonly used method. Direct Conductivity Methods Direct conductivity methods take advantage of the change in conductivity that an analyte produces in the eluent, which can be modeled by Equation \ref{16} where equivalent conductivity is defined as Equation \ref{17} . \[ \Delta K \ =\frac{(\Lambda _{A} \ -\ \Lambda _{g} ) * C_{s}}{1000} \label{16} \] \[ \Lambda \ =\frac{L}{A*R} * \frac{1}{C} \label{17} \] With L being the distance between two electrodes of area A and R being the resistance the ion creates. C is the concentration of the ion. The conductivity can be plotted over time, and the peaks that appear represent different ions coming through the column as described by Equation \ref{18} \[ K_{peak} \ =\ (\Lambda _{A} \ -\ \Lambda _{g})*C_{A} \label{18} \] The values of Equivalent conductivity of the analyte and of the eluent common ions can be found in Table \(\PageIndex{1}\) Cations \(\Lambda ^{+} (S\ cm^{2} eq^{-1} ) \) Anions \(\Lambda ^{-} (S\ cm^{2} eq^{-1} ) \) \( H^{+} \) 350.0 \( OH^{-} \) 198 \( Li ^{+} \) 39.0 \( F^{-} \) 54 \( Na^{+} \) 50.0 \( Cl^{-} \) 76 \( K^{+} \) 74.0 \( Br^{-} \) 78 \( NH^{4+} \) 73.0 \( I^{-} \) 77 \( 1/2 Mg^{2+} \) 53.0 \( NO^{-}_{2} \) 72 \( 1/2 Ca^{2+} \) 60.0 \( NO^{-}_{3} \) 71 \( 1/2Sr^{2+} \) 59.0 \( HCO_{3}^{-} \) 45 \( 1/2 Ba^{2+} \) 64.0 \( 1/2 CO_{3}^{2-} \) 72 \( 1/2 Zn^{2+} \) 52.0 \( H_{2}PO_{4}^{-} \) 33 \( 1/2 Hg^{2+} \) 53.0 \( 1/2 HPO_{4}^{-} \) 57 \( 1/2 Cu^{2+} \) 55.0 \( 1/3 PO_{4}^{-} \) 69 \( 1/2 Pb ^{2+} \) 71.0 \( 1/2 SO_{4}^{2-} \) 80 \( 1/2 Co ^{2+} \) 53.0 \( CN^{-} \) 82 \( 1/3 Fe^{3+} \) 70.0 \( SCN^{-} \) 66 \( N(Et)^{4+} \) 33.0 Acetate 41 NaN NaN 1/2 Phthalate 38 NaN NaN Propionate 36 NaN NaN Benzoate 32 NaN NaN Salicylate 30 NaN NaN 1/2 Oxalate 74 Eluents The choice of eluent depends on many factors, namely, pH, buffer capacity, the concentration of the eluent, and the nature of the eluent’s reaction with the column and the packing material. Eluents in Anion Chromatography In non-suppressed anion chromatography, where the eluent and analyte are not altered between the column and the detector, there is a wide range of eluents to be used. In the non-suppressed case, the only issue that could arise is if the eluent impaired the detection ability (absorbing in a similar place in a UV-spectra as the analyte for instance). As such, there are a number of commonly used eluents. Aromatic carboxylic acids are used in conductivity detection because of their low self-conductivity. Aliphatic carboxylic acids are used for UV/visible detection because they are UV transparent. Inorganic acids can only be used in photometric detection. In suppressed anion chromatography, where the eluent and analyte are treated between the column and detection, fewer eluents can be used. The suppressor modifies the eluent and the analyte, reducing the self-conductivity of the eluent and possibly increasing the self-conductivity of the analyte. Only alkali hydroxides and carbonates, borates, hydrogen carbonates, and amino acids can be used as eluents. Eluents in Cation Chromatography The primary eluents used in cation chromatography of alkali metals and ammoniums are mineral acids such as HNO 3 . When the cation is multivalent, organic bases such as ethylenediamine (Figure \(\PageIndex{5}\) ) serve as the main eluents. If both alkali metals and alkali earth metals are present, hydrochloric acid or 2,3-diaminopropionic acid (Figure \(\PageIndex{6}\) ) is used in combination with a pH variation. If the chromatography is unsuppressed, the direct conductivity measurement of the analyte will show up as a negative peak due to the high conductivity of the H + in the eluent, but simple inversion of the data can be used to rectify this discrepancy. If transition metals or H + are the analytes in question, complexing carboxylic acids are used to suppress the charge of the analyte and to create photometrically detectable complexes, forgoing the need for direct conductivity as the detection method. |
Courses/Lumen_Learning/Book%3A_United_States_History%3A_Reconstruction_to_the_Present_(Lumen)/11%3A_Module_9-_The_Great_Depression/11.04%3A_The_Depths_of_the_Great_Depression | Learning Objectives By the end of this section, you will be able to: Identify the challenges that everyday Americans faced as a result of the Great Depression and analyze the government’s initial unwillingness to provide assistance Explain the particular challenges that African Americans faced during the crisis Identify the unique challenges that farmers in the Great Plains faced during this period From industrial strongholds to the rural Great Plains, from factory workers to farmers, the Great Depression affected millions. In cities, as industry slowed, then sometimes stopped altogether, workers lost jobs and joined breadlines, or sought out other charitable efforts. With limited government relief efforts, private charities tried to help, but they were unable to match the pace of demand. In rural areas, farmers suffered still more. In some parts of the country, prices for crops dropped so precipitously that farmers could not earn enough to pay their mortgages, losing their farms to foreclosure. In the Great Plains, one of the worst droughts in history left the land barren and unfit for growing even minimal food to live on. The country’s most vulnerable populations, such as children, the elderly, and those subject to discrimination, like African Americans, were the hardest hit. Most white Americans felt entitled to what few jobs were available, leaving African Americans unable to find work, even in the jobs once considered their domain. In all, the economic misery was unprecedented in the country’s history. STARVING TO DEATH By the end of 1932, the Great Depression had affected some sixty million people, most of whom wealthier Americans perceived as the “deserving poor.” Yet, at the time, federal efforts to help those in need were extremely limited, and national charities had neither the capacity nor the will to elicit the large-scale response required to address the problem. The American Red Cross did exist, but Chairman John Barton Payne contended that unemployment was not an “Act of God” but rather an “Act of Man,” and therefore refused to get involved in widespread direct relief efforts. Clubs like the Elks tried to provide food, as did small groups of individually organized college students. Religious organizations remained on the front lines, offering food and shelter. In larger cities, breadlines and soup lines became a common sight. At one count in 1932, there were as many as eighty-two breadlines in New York City. Despite these efforts, however, people were destitute and ultimately starving. Families would first run through any savings, if they were lucky enough to have any. Then, the few who had insurance would cash out their policies. Cash surrender payments of individual insurance policies tripled in the first three years of the Great Depression, with insurance companies issuing total payments in excess of $1.2 billion in 1932 alone. When those funds were depleted, people would borrow from family and friends, and when they could get no more, they would simply stop paying rent or mortgage payments. When evicted, they would move in with relatives, whose own situation was likely only a step or two behind. The added burden of additional people would speed along that family’s demise, and the cycle would continue. This situation spiraled downward, and did so quickly. Even as late as 1939, over 60 percent of rural households, and 82 percent of farm families, were classified as “impoverished.” In larger urban areas, unemployment levels exceeded the national average, with over half a million unemployed workers in Chicago, and nearly a million in New York City. Breadlines and soup kitchens were packed, serving as many as eighty-five thousand meals daily in New York City alone. Over fifty thousand New York citizens were homeless by the end of 1932. Children, in particular, felt the brunt of poverty. Many in coastal cities would roam the docks in search of spoiled vegetables to bring home. Elsewhere, children begged at the doors of more well-off neighbors, hoping for stale bread, table scraps, or raw potato peelings. Said one childhood survivor of the Great Depression, “You get used to hunger. After the first few days it doesn’t even hurt; you just get weak.” In 1931 alone, there were at least twenty documented cases of starvation; in 1934, that number grew to 110. In rural areas where such documentation was lacking, the number was likely far higher. And while the middle class did not suffer from starvation, they experienced hunger as well. By the time Hoover left office in 1933, the poor survived not on relief efforts, but because they had learned to be poor. A family with little food would stay in bed to save fuel and avoid burning calories. People began eating parts of animals that had normally been considered waste. They scavenged for scrap wood to burn in the furnace, and when electricity was turned off, it was not uncommon to try and tap into a neighbor’s wire. Family members swapped clothes; sisters might take turns going to church in the one dress they owned. As one girl in a mountain town told her teacher, who had said to go home and get food, “I can’t. It’s my sister’s turn to eat.” For his book on the Great Depression, Hard Times , author Studs Terkel interviewed hundreds of Americans from across the country. He subsequently selected over seventy interviews to air on a radio show that was based in Chicago. Visit Studs Terkel: Conversations with America to listen to those interviews, during which participants reflect on their personal hardships as well as on national events during the Great Depression. BLACK AND POOR: AFRICAN AMERICANS AND THE GREAT DEPRESSION Most African Americans did not participate in the land boom and stock market speculation that preceded the crash, but that did not stop the effects of the Great Depression from hitting them particularly hard. Subject to continuing racial discrimination, blacks nationwide fared even worse than their hard-hit white counterparts. As the prices for cotton and other agricultural products plummeted, farm owners paid workers less or simply laid them off. Landlords evicted sharecroppers, and even those who owned their land outright had to abandon it when there was no way to earn any income. In cities, African Americans fared no better. Unemployment was rampant, and many whites felt that any available jobs belonged to whites first. In some Northern cities, whites would conspire to have African American workers fired to allow white workers access to their jobs. Even jobs traditionally held by black workers, such as household servants or janitors, were now going to whites. By 1932, approximately one-half of all black Americans were unemployed. Racial violence also began to rise. In the South, lynching became more common again, with twenty-eight documented lynchings in 1933, compared to eight in 1932. Since communities were preoccupied with their own hardships, and organizing civil rights efforts was a long, difficult process, many resigned themselves to, or even ignored, this culture of racism and violence. Occasionally, however, an incident was notorious enough to gain national attention. One such incident was the case of the Scottsboro Boys. In 1931, nine black boys, who had been riding the rails, were arrested for vagrancy and disorderly conduct after an altercation with some white travelers on the train. Two young white women, who had been dressed as boys and traveling with a group of white boys, came forward and said that the black boys had raped them. The case, which was tried in Scottsboro, Alabama, reignited decades of racial hatred and illustrated the injustice of the court system. Despite significant evidence that the women had not been raped at all, along with one of the women subsequently recanting her testimony, the all-white jury quickly convicted the boys and sentenced all but one of them to death. The verdict broke through the veil of indifference toward the plight of African Americans, and protests erupted among newspaper editors, academics, and social reformers in the North. The Communist Party of the United States offered to handle the case and sought retrial; the NAACP later joined in this effort. In all, the case was tried three separate times. The series of trials and retrials, appeals, and overturned convictions shone a spotlight on a system that provided poor legal counsel and relied on all-white juries. In October 1932, the U.S. Supreme Court agreed with the Communist Party’s defense attorneys that the defendants had been denied adequate legal representation at the original trial, and that due process as provided by the Fourteenth Amendment had been denied as a result of the exclusion of any potential black jurors. Eventually, most of the accused received lengthy prison terms and subsequent parole, but avoided the death penalty. The Scottsboro case ultimately laid some of the early groundwork for the modern American civil rights movement. Alabama granted posthumous pardons to all defendants in 2013. The trial and conviction of nine African American boys in Scottsboro, Alabama, illustrated the numerous injustices of the American court system. Despite being falsely accused, the boys received lengthy prison terms and were not officially pardoned by the State of Alabama until 2013. Read Voices from Scottsboro for the perspectives of both participants and spectators in the Scottsboro case, from the initial trial to the moment, in 1976, when one of the women sued for slander. ENVIRONMENTAL CATASTROPHE MEETS ECONOMIC HARDSHIP: THE DUST BOWL Despite the widely held belief that rural Americans suffered less in the Great Depression due to their ability to at least grow their own food, this was not the case. Farmers, ranchers, and their families suffered more than any group other than African Americans during the Depression. From the turn of the century through much of World War I, farmers in the Great Plains experienced prosperity due to unusually good growing conditions, high commodity prices, and generous government farming policies that led to a rush for land. As the federal government continued to purchase all excess produce for the war effort, farmers and ranchers fell into several bad practices, including mortgaging their farms and borrowing money against future production in order to expand. However, after the war, prosperity rapidly dwindled, particularly during the recession of 1921. Seeking to recoup their losses through economies of scale in which they would expand their production even further to take full advantage of their available land and machinery, farmers plowed under native grasses to plant acre after acre of wheat, with little regard for the long-term repercussions to the soil. Regardless of these misguided efforts, commodity prices continued to drop, finally plummeting in 1929, when the price of wheat dropped from two dollars to forty cents per bushel. Exacerbating the problem was a massive drought that began in 1931 and lasted for eight terrible years. Dust storms roiled through the Great Plains, creating huge, choking clouds that piled up in doorways and filtered into homes through closed windows. Even more quickly than it had boomed, the land of agricultural opportunity went bust, due to widespread overproduction and overuse of the land, as well as to the harsh weather conditions that followed, resulting in the creation of the Dust Bowl. The dust storms that blew through the Great Plains were epic in scale. Drifts of dirt piled up against doors and windows. People wore goggles and tied rags over their mouths to keep the dust out. (credit: U.S. National Oceanic and Atmospheric Administration) Livestock died, or had to be sold, as there was no money for feed. Crops intended to feed the family withered and died in the drought. Terrifying dust storms became more and more frequent, as “black blizzards” of dirt blew across the landscape and created a new illness known as “dust pneumonia.” In 1935 alone, over 850 million tons of topsoil blew away. To put this number in perspective, geologists estimate that it takes the earth five hundred years to naturally regenerate one inch of topsoil; yet, just one significant dust storm could destroy a similar amount. In their desperation to get more from the land, farmers had stripped it of the delicate balance that kept it healthy. Unaware of the consequences, they had moved away from such traditional practices as crop rotation and allowing land to regain its strength by permitting it to lie fallow between plantings, working the land to death. As the Dust Bowl continued in the Great Plains, many had to abandon their land and equipment, as captured in this image from 1936, taken in Dallas, South Dakota. (credit: United States Department of Agriculture) For farmers, the results were catastrophic. Unlike most factory workers in the cities, in most cases, farmers lost their homes when they lost their livelihood. Most farms and ranches were originally mortgaged to small country banks that understood the dynamics of farming, but as these banks failed, they often sold rural mortgages to larger eastern banks that were less concerned with the specifics of farm life. With the effects of the drought and low commodity prices, farmers could not pay their local banks, which in turn lacked funds to pay the large urban banks. Ultimately, the large banks foreclosed on the farms, often swallowing up the small country banks in the process. It is worth noting that of the five thousand banks that closed between 1930 and 1932, over 75 percent were country banks in locations with populations under 2,500. Given this dynamic, it is easy to see why farmers in the Great Plains remained wary of big city bankers. For farmers who survived the initial crash, the situation worsened, particularly in the Great Plains where years of overproduction and rapidly declining commodity prices took their toll. Prices continued to decline, and as farmers tried to stay afloat, they produced still more crops, which drove prices even lower. Farms failed at an astounding rate, and farmers sold out at rock-bottom prices. One farm in Shelby, Nebraska was mortgaged at $4,100 and sold for $49.50. One-fourth of the entire state of Mississippi was auctioned off in a single day at a foreclosure auction in April 1932. Not all farmers tried to keep their land. Many, especially those who had arrived only recently, in an attempt to capitalize on the earlier prosperity, simply walked away. In hard-hit Oklahoma, thousands of farmers packed up what they could and walked or drove away from the land they thought would be their future. They, along with other displaced farmers from throughout the Great Plains, became known as Okies . Okies were an emblem of the failure of the American breadbasket to deliver on its promise, and their story was made famous in John Steinbeck’s novel, The Grapes of Wrath . Experience the Interactive Dust Bowl to see how decisions compounded to create peoples’ destiny. Click through to see what choices you would make and where that would take you. Caroline Henderson on the Dust Bowl Now we are facing a fourth year of failure. There can be no wheat for us in 1935 in spite of all our careful and expensive work in preparing ground, sowing and re-sowing our allocated acreage. Native grass pastures are permanently damaged, in many cases hopelessly ruined, smothered under by drifted sand. Fences are buried under banks of thistles and hard packed earth or undermined by the eroding action of the wind and lying flat on the ground. Less traveled roads are impassable, covered deep under by sand or the finer silt-like loam. Orchards, groves and hedge-rows cultivated for many years with patient care are dead or dying . . . Impossible it seems not to grieve that the work of hands should prove so perishable. —Caroline Henderson, Shelton, Oklahoma, 1935 Much like other farm families whose livelihoods were destroyed by the Dust Bowl, Caroline Henderson describes a level of hardship that many Americans living in Depression-ravaged cities could never understand. Despite their hard work, millions of Americans were losing both their produce and their homes, sometimes in as little as forty-eight hours, to environmental catastrophes. Lacking any other explanation, many began to question what they had done to incur God’s wrath. Note in particular Henderson’s references to “dead,” “dying,” and “perishable,” and contrast those terms with her depiction of the “careful and expensive work” undertaken by their own hands. Many simply could not understand how such a catastrophe could have occurred. CHANGING VALUES, CHANGING CULTURE In the decades before the Great Depression, and particularly in the 1920s, American culture largely reflected the values of individualism, self-reliance, and material success through competition. Novels like F. Scott Fitzgerald’s The Great Gatsby and Sinclair Lewis’s Babbit portrayed wealth and the self-made man in America, albeit in a critical fashion. In film, many silent movies, such as Charlie Chaplin’s The Gold Rush , depicted the rags-to-riches fable that Americans so loved. With the shift in U.S. fortunes, however, came a shift in values, and with it, a new cultural reflection. The arts revealed a new emphasis on the welfare of the whole and the importance of community in preserving family life. While box office sales briefly declined at the beginning of the Depression, they quickly rebounded. Movies offered a way for Americans to think of better times, and people were willing to pay twenty-five cents for a chance to escape, at least for a few hours. Even more than escapism, other films at the close of the decade reflected on the sense of community and family values that Americans struggled to maintain throughout the entire Depression. John Ford’s screen version of Steinbeck’s The Grapes of Wrath came out in 1940, portraying the haunting story of the Joad family’s exodus from their Oklahoma farm to California in search of a better life. Their journey leads them to realize that they need to join a larger social movement—communism—dedicated to bettering the lives of all people. Tom Joad says, “Well, maybe it’s like Casy says, a fella ain’t got a soul of his own, but on’y a piece of a soul—the one big soul that belongs to ever’body.” The greater lesson learned was one of the strength of community in the face of individual adversity. Another trope was that of the hard-working everyman against greedy banks and corporations. This was perhaps best portrayed in the movies of Frank Capra, whose Mr. Smith Goes to Washington was emblematic of his work. In this 1939 film, Jimmy Stewart plays a legislator sent to Washington to finish out the term of a deceased senator. While there, he fights corruption to ensure the construction of a boy’s camp in his hometown rather than a dam project that would only serve to line the pockets of a few. He ultimately engages in a two-day filibuster, standing up to the power players to do what’s right. The Depression era was a favorite of Capra’s to depict in his films, including It’s a Wonderful Life , released in 1946. In this film, Jimmy Stewart runs a family-owned savings and loan, which at one point faces a bank run similar to those seen in 1929–1930. In the end, community support helps Stewart retain his business and home against the unscrupulous actions of a wealthy banker who sought to bring ruin to his family. “Brother, Can You Spare a Dime?” They used to tell me I was building a dream, and so I followed the mob When there was earth to plow or guns to bear, I was always there, right on the job They used to tell me I was building a dream, with peace and glory ahead Why should I be standing in line, just waiting for bread? Once I built a railroad, I made it run, made it race against time Once I built a railroad, now it’s done, Brother, can you spare a dime? Once I built a tower up to the sun, brick and rivet and lime Once I built a tower, now it’s done, Brother, can you spare a dime?—Jay Gorney and “Yip” Harburg “Brother, Can You Spare a Dime?” first appeared in 1932, written for the Broadway musical New Americana by Jay Gorney, a composer who based the song’s music on a Russian lullaby, and Edgar Yipsel “Yip” Harburg, a lyricist who would go on to win an Academy Award for the song “Over the Rainbow” from The Wizard of Oz (1939). With its lyrics speaking to the plight of the common man during the Great Depression and the refrain appealing to the same sense of community later found in the films of Frank Capra, “Brother, Can You Spare a Dime?” quickly became the de facto anthem of the Great Depression. Recordings by Bing Crosby, Al Jolson, and Rudy Vallee all enjoyed tremendous popularity in the 1930s. For more on “Brother Can You Spare a Dime?” and the Great Depression, visit ArtsEdge to explore the Kennedy Center’s digital resources and learn the “Story Behind the Song.” Flying Down to Rio (1933) was the first motion picture to feature the immensely popular dance duo of Fred Astaire and Ginger Rogers. The pair would go on to star in nine more Hollywood musicals throughout the 1930s and 1940s. Finally, there was a great deal of pure escapism in the popular culture of the Depression. Even the songs found in films reminded many viewers of the bygone days of prosperity and happiness, from Al Dubin and Henry Warren’s hit “We’re in the Money” to the popular “Happy Days are Here Again.” The latter eventually became the theme song of Franklin Roosevelt’s 1932 presidential campaign. People wanted to forget their worries and enjoy the madcap antics of the Marx Brothers, the youthful charm of Shirley Temple, the dazzling dances of Fred Astaire and Ginger Rogers, or the comforting morals of the Andy Hardy series. The Hardy series—nine films in all, produced by MGM from 1936 to 1940—starred Judy Garland and Mickey Rooney, and all followed the adventures of a small-town judge and his son. No matter what the challenge, it was never so big that it could not be solved with a musical production put on by the neighborhood kids, bringing together friends and family members in a warm display of community values. All of these movies reinforced traditional American values, which suffered during these hard times, in part due to declining marriage and birth rates, and increased domestic violence. At the same time, however, they reflected an increased interest in sex and sexuality. While the birth rate was dropping, surveys in Fortune magazine in 1936–1937 found that two-thirds of college students favored birth control, and that 50 percent of men and 25 percent of women admitted to premarital sex, continuing a trend among younger Americans that had begun to emerge in the 1920s. Contraceptive sales soared during the decade, and again, culture reflected this shift. Blonde bombshell Mae West was famous for her sexual innuendoes, and her flirtatious persona was hugely popular, although it got her banned on radio broadcasts throughout the Midwest. Whether West or Garland, Chaplin or Stewart, American film continued to be a barometer of American values, and their challenges, through the decade. Section Summary The Great Depression affected huge segments of the American population—sixty million people by one estimate. But certain groups were hit harder than the rest. African Americans faced discrimination in finding employment, as white workers sought even low-wage jobs like housecleaning. Southern blacks moved away from their farms as crop prices failed, migrating en masse to Northern cities, which had little to offer them. Rural Americans were also badly hit. The eight-year drought that began shortly after the stock market crash exacerbated farmers’ and ranchers’ problems. The cultivation of greater amounts of acreage in the preceding decades meant that land was badly overworked, and the drought led to massive and terrible dust storms, creating the region’s nickname, the Dust Bowl. Some farmers tried to remain and buy up more land as neighbors went broke; others simply fled their failed farms and moved away, often to the large-scale migrant farms found in California, to search for a better life that few ever found. Maltreated by Californians who wished to avoid the unwanted competition for jobs that these “Okies” represented, many of the Dust Bowl farmers were left wandering as a result. There was very little in the way of public assistance to help the poor. While private charities did what they could, the scale of the problem was too large for them to have any lasting effects. People learned to survive as best they could by sending their children out to beg, sharing clothing, and scrounging wood to feed the furnace. Those who could afford it turned to motion pictures for escape. Movies and books during the Great Depression reflected the shift in American cultural norms, away from rugged individualism toward a more community-based lifestyle. A Open Assessments element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/ur/?p=220 Review Question What did the popular movies of the Depression reveal about American values at that time? How did these values contrast with the values Americans held before the Depression? Answer to Review Question American films in the 1930s served to both assuage the fears and frustrations of many Americans suffering through the Depression and reinforce the idea that communal efforts—town and friends working together—would help to address the hardships. Previous emphasis upon competition and individualism slowly gave way to notions of “neighbor helping neighbor” and seeking group solutions to common problems. The Andy Hardy series, in particular, combined entertainment with the concept of family coming together to solve shared problems. The themes of greed, competition, and capitalist-driven market decisions no longer commanded a large audience among American moviegoers. Glossary Dust Bowl the area in the middle of the country that had been badly overfarmed in the 1920s and suffered from a terrible drought that coincided with the Great Depression; the name came from the “black blizzard” of topsoil and dust that blew through the area Scottsboro Boys a reference to the infamous trial in Scottsboro, Alabama in 1931, where nine African American boys were falsely accused of raping two white women and sentenced to death; the extreme injustice of the trial, particularly given the age of the boys and the inadequacy of the testimony against them, garnered national and international attention CC licensed content, Shared previously US History. Authored by : P. Scott Corbett, Volker Janssen, John M. Lund, Todd Pfannestiel, Paul Vickery, and Sylvie Waskiewicz. Provided by : OpenStax College. Located at : http://openstaxcollege.org/textbooks/us-history . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/content/col11740/latest/ |
Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(LibreTexts)/03%3A_Matter_and_Energy/3.12%3A_Energy_and_Heat_Capacity_Calculations | Learning Objectives To relate heat transfer to temperature change. Heat is a familiar manifestation of transferring energy. When we touch a hot object, energy flows from the hot object into our fingers, and we perceive that incoming energy as the object being “hot.” Conversely, when we hold an ice cube in our palms, energy flows from our hand into the ice cube, and we perceive that loss of energy as “cold.” In both cases, the temperature of the object is different from the temperature of our hand, so we can conclude that differences in temperatures are the ultimate cause of heat transfer. The specific heat of a substance can be used to calculate the temperature change that a given substance will undergo when it is either heated or cooled. The equation that relates heat \(\left( q \right)\) to specific heat \(\left( c_p \right)\), mass \(\left( m \right)\), and temperature change \(\left( \Delta T \right)\) is shown below. \[q = c_p \times m \times \Delta T \nonumber \] The heat that is either absorbed or released is measured in joules. The mass is measured in grams. The change in temperature is given by \(\Delta T = T_f - T_i\), where \(T_f\) is the final temperature and \(T_i\) is the initial temperature. Every substance has a characteristic specific heat, which is reported in units of cal/g•°C or cal/g•K, depending on the units used to express Δ T . The specific heat of a substance is the amount of energy that must be transferred to or from 1 g of that substance to change its temperature by 1°. Table \(\PageIndex{1}\) lists the specific heats for various materials. Substance Specific Heat \(\left( \text{J/g}^\text{o} \text{C} \right)\) Water (l) 4.180 Water (s) 2.060 Water (g) 1.870 Ammonia (g) 2.090 Ethanol (l) 2.440 Aluminum (s) 0.897 Carbon, graphite (s) 0.709 Copper (s) 0.385 Gold (s) 0.129 Iron (s) 0.449 Lead (s) 0.129 Mercury (l) 0.140 Silver (s) 0.233 The direction of heat flow is not shown in heat = mc Δ T . If energy goes into an object, the total energy of the object increases, and the values of heat Δ T are positive. If energy is coming out of an object, the total energy of the object decreases, and the values of heat and Δ T are negative. Example \(\PageIndex{1}\) A \(15.0 \: \text{g}\) piece of cadmium metal absorbs \(134 \: \text{J}\) of heat while rising from \(24.0^\text{o} \text{C}\) to \(62.7^\text{o} \text{C}\). Calculate the specific heat of cadmium. Solution Step 1: List the known quantities and plan the problem. Known Heat \(= q = 134 \: \text{J}\) Mass \(= m = 15.0 \: \text{g}\) \(\Delta T = 62.7^\text{o} \text{C} - 24.0^\text{o} \text{C} = 38.7^\text{o} \text{C}\) Unknown \(c_p\) of cadmium \(= ? \: \text{J/g}^\text{o} \text{C}\) The specific heat equation can be rearranged to solve for the specific heat. Step 2: Solve. \[c_p = \dfrac{q}{m \times \Delta T} = \dfrac{134 \: \text{J}}{15.0 \: \text{g} \times 38.7^\text{o} \text{C}} = 0.231 \: \text{J/g}^\text{o} \text{C} \nonumber \] Step 3: Think about your result. The specific heat of cadmium, a metal, is fairly close to the specific heats of other metals. The result has three significant figures. Since most specific heats are known (Table \(\PageIndex{1}\)), they can be used to determine the final temperature attained by a substance when it is either heated or cooled. Suppose that a \(60.0 \: \text{g}\) of water at \(23.52^\text{o} \text{C}\) was cooled by the removal of \(813 \: \text{J}\) of heat. The change in temperature can be calculated using the specific heat equation: \[\Delta T = \dfrac{q}{c_p \times m} = \dfrac{813 \: \text{J}}{4.18 \: \text{J/g}^\text{o} \text{C} \times 60.0 \: \text{g}} = 3.24^\text{o} \text{C} \nonumber \] Since the water was being cooled, the temperature decreases. The final temperature is: \[T_f = 23.52^\text{o} \text{C} - 3.24^\text{o} \text{C} = 20.28^\text{o} \text{C} \nonumber \] Example \(\PageIndex{2}\) What quantity of heat is transferred when a 150.0 g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of heat flow? Solution We can use heat = mc Δ T to determine the amount of heat, but first we need to determine Δ T . Because the final temperature of the iron is 73.3°C and the initial temperature is 25.0°C, Δ T is as follows: Δ T = T final − T initial = 73.3°C − 25.0°C = 48.3°C The mass is given as 150.0 g, and Table 7.3 gives the specific heat of iron as 0.108 cal/g•°C. Substitute the known values into heat = mc Δ T and solve for amount of heat: \[\mathrm{heat=(150.0\: g)\left(0.108\: \dfrac{cal} {g\cdot {^\circ C}}\right)(48.3^\circ C) = 782\: cal} \nonumber \] Note how the gram and °C units cancel algebraically, leaving only the calorie unit, which is a unit of heat. Because the temperature of the iron increases, energy (as heat) must be flowing into the metal. Exercise \(\PageIndex{1}\) What quantity of heat is transferred when a 295.5 g block of aluminum metal is cooled from 128.0°C to 22.5°C? What is the direction of heat flow? Answer Heat leaves the aluminum block. Example \(\PageIndex{2}\) A 10.3 g sample of a reddish-brown metal gave off 71.7 cal of heat as its temperature decreased from 97.5°C to 22.0°C. What is the specific heat of the metal? Can you identify the metal from the data in Table \(\PageIndex{1}\)? Solution The question gives us the heat, the final and initial temperatures, and the mass of the sample. The value of Δ T is as follows: Δ T = T final − T initial = 22.0°C − 97.5°C = −75.5°C If the sample gives off 71.7 cal, it loses energy (as heat), so the value of heat is written as a negative number, −71.7 cal. Substitute the known values into heat = mc Δ T and solve for c : −71.7 cal = (10.3 g)( c )(−75.5°C) \(c \,\mathrm{=\dfrac{-71.7\: cal}{(10.3\: g)(-75.5^\circ C)}}\) c = 0.0923 cal/g•°C This value for specific heat is very close to that given for copper in Table 7.3. Exercise \(\PageIndex{2}\) A 10.7 g crystal of sodium chloride (NaCl) has an initial temperature of 37.0°C. What is the final temperature of the crystal if 147 cal of heat were supplied to it? Answer Summary Specific heat calculations are illustrated. |
Bookshelves/Organic_Chemistry/Organic_Chemistry_(OpenStax)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.00%3A_Why_This_Chapter | Nucleophilic substitution and base-induced elimination are two of the most widely occurring and versatile reactions in organic chemistry, both in the laboratory and in biological pathways. We’ll look at them closely in this chapter to see how they occur, what their characteristics are, and how they can be used. We’ll begin with substitution reactions. We saw in the preceding chapter that the carbon–halogen bond in an alkyl halide is polar and that the carbon atom is electron-poor. Thus, alkyl halides are electrophiles, and much of their chemistry involves polar reactions with nucleophiles and bases. Alkyl halides do one of two things when they react with a nucleophile/base such as hydroxide ion: they either undergo substitution of the X group by the nucleophile, or they undergo elimination of HX to yield an alkene. |
Courses/Fresno_City_College/Introductory_Chemistry_Atoms_First_for_FCC/09%3A_Aqueous_Solutions/9.05%3A_Colligative_Properties_and_Molality/9.5.01%3A_Freezing_Point_Depression_and_Boiling_Point_Elevation | Learning Objectives Explain what the term "colligative" means, and list the colligative properties. Indicate what happens to the boiling point and the freezing point of a solvent when a solute is added to it. Calculate boiling point elevations and freezing point depressions for a solution. People who live in colder climates have seen trucks put salt on the roads when snow or ice is forecast. Why is this done? As a result of the information you explore in this section, you will understand why these events occur. You will also learn to calculate exactly how much of an effect a specific solute can have on the boiling point or freezing point of a solution. The example given in the introduction is an example of a colligative property. Colligative properties are properties that differ based on the concentration of solute in a solvent, but not on the type of solute. What this means for the example above is that people in colder climates do not necessarily need salt to get the same effect on the roads—any solute will work. However, the higher the concentration of solute, the more these properties will change. Boiling Point Elevation Water boils at \(100^\text{o} \text{C}\) at \(1 \: \text{atm}\) of pressure, but a solution of saltwater does not . When table salt is added to water, the resulting solution has a higher boiling point than the water did by itself. The ions form an attraction with the solvent particles that prevents the water molecules from going into the gas phase. Therefore, the saltwater solution will not boil at \(100^\text{o} \text{C}\). In order for the saltwater solution to boil, the temperature must be raised about \(100^\text{o} \text{C}\). This is true for any solute added to a solvent; the boiling point will be higher than the boiling point of the pure solvent (without the solute). In other words, when anything is dissolved in water, the solution will boil at a higher temperature than pure water would. The boiling point elevation due to the presence of a solute is also a colligative property. That is, the amount of change in the boiling point is related to the number of particles of solute in a solution and is not related to the chemical composition of the solute. A \(0.20 \: \text{m}\) solution of table salt and a \(0.20 \: \text{m}\) solution of hydrochloric acid would have the same effect on the boiling point. Freezing Point Depression The effect of adding a solute to a solvent has the opposite effect on the freezing point of a solution as it does on the boiling point. A solution will have a lower freezing point than a pure solvent. The freezing point is the temperature at which the liquid changes to a solid. At a given temperature, if a substance is added to a solvent (such as water), the solute-solvent interactions prevent the solvent from going into the solid phase. The solute-solvent interactions require the temperature to decrease further in order to solidify the solution. A common example is found when salt is used on icy roadways. Salt is put on roads so that the water on the roads will not freeze at the normal \(0^\text{o} \text{C}\) but at a lower temperature, as low as \(-9^\text{o} \text{C}\). The de-icing of planes is another common example of freezing point depression in action. A number of solutions are used, but commonly a solution such as ethylene glycol, or a less toxic monopropylene glycol, is used to de-ice an aircraft. The aircrafts are sprayed with the solution when the temperature is predicted to drop below the freezing point. The freezing point depression is the difference in the freezing points of the solution from the pure solvent. This is true for any solute added to a solvent; the freezing point of the solution will be lower than the freezing point of the pure solvent (without the solute). Thus, when anything is dissolved in water, the solution will freeze at a lower temperature than pure water would. The freezing point depression due to the presence of a solute is also a colligative property. That is, the amount of change in the freezing point is related to the number of particles of solute in a solution and is not related to the chemical composition of the solute. A \(0.20 \: \text{m}\) solution of table salt and a \(0.20 \: \text{m}\) solution of hydrochloric acid would have the same effect on the freezing point. Comparing the Freezing and Boiling Point of Solutions Recall that covalent and ionic compounds do not dissolve in the same way. Ionic compounds break up into cations and anions when they dissolve. Covalent compounds typically do not break up. For example a sugar/water solution stays as sugar and water, with the sugar molecules staying as molecules. Remember that colligative properties are due to the number of solute particles in the solution. Adding 10 molecules of sugar to a solvent will produce 10 solute particles in the solution. When the solute is ionic, such as \(\ce{NaCl}\) however, adding 10 formulas of solute to the solution will produce 20 ions (solute particles) in the solution. Therefore, adding enough \(\ce{NaCl}\) solute to a solvent to produce a \(0.20 \: \text{m}\) solution will have twice the effect of adding enough sugar to a solvent to produce a \(0.20 \: \text{m}\) solution. Colligative properties depend on the number of solute particles in the solution. "\(i\)" is the number of particles that the solute will dissociate into upon mixing with the solvent. For example, sodium chloride, \(\ce{NaCl}\), will dissociate into two ions so for \(\ce{NaCl}\), \(i = 2\); for lithium nitrate, \(\ce{LiNO_3}\), \(i = 2\); and for calcium chloride, \(\ce{CaCl_2}\), \(i = 3\). For covalent compounds, \(i\) is always equal to 1. By knowing the molality of a solution and the number of particles a compound will dissolve to form, it is possible to predict which solution in a group will have the lowest freezing point. To compare the boiling or freezing points of solutions, follow these general steps: Label each solute as ionic or covalent. If the solute is ionic, determine the number of ions in the formula. Be careful to look for polyatomic ions. Multiply the original molality (\(\text{m}\)) of the solution by the number of particles formed when the solution dissolves. This will give you the total concentration of particles dissolved. Compare these values. The higher total concentration will result in a higher boiling point and a lower freezing point. Example \(\PageIndex{1}\) Rank the following solutions in water in order of increasing (lowest to highest) freezing point: \(0.1 \: \text{m} \: \ce{NaCl}\) \(0.1 \: \text{m} \: \ce{C_6H_{12}O_6}\) \(0.1 \: \text{m} \: \ce{CaI_2}\) Solution To compare freezing points, we need to know the total concentration of all particles when the solute has been dissolved. \(0.1 \: \text{m} \: \ce{NaCl}\): This compound is ionic (metal with nonmetal), and will dissolve into 2 parts. The total final concentration is: \(\left( 0.1 \: \text{m} \right) \left( 2 \right) = 0.2 \: \text{m}\) \(0.1 \: \text{m} \: \ce{C_6H_{12}O_6}\): This compound is covalent (nonmetal with nonmetal), and will stay as 1 part. The total final concentration is: \(\left( 0.1 \: \text{m} \right) \left(1 \right) = 0.1 \: \text{m}\) \(0.1 \: \text{m} \: \ce{CaI_2}\): This compound is ionic (metal with nonmetal), and will dissolve into 3 parts. The total final concentration is: \(\left( 0.1 \: \text{m} \right) \left( 3 \right) = 0.3 \: \text{m}\) Remember, the greater the concentration of particles, the lower the freezing point will be. \(0.1 \: \text{m} \: \ce{CaI_2}\) will have the lowest freezing point, followed by \(0.1 \: \text{m} \: \ce{NaCl}\), and the highest of the three solutions will be \(0.1 \: \text{m} \: \ce{C_6H_{12}O_6}\), but all three of them will have a lower freezing point than pure water. The boiling point of a solution is higher than the boiling point of a pure solvent, and the freezing point of a solution is lower than the freezing point of a pure solvent. However, the amount to which the boiling point increases or the freezing point decreases depends on the amount of solute that is added to the solvent. A mathematical equation is used to calculate the boiling point elevation or the freezing point depression. The boiling point elevation is the amount that the boiling point temperature increases compared to the original solvent. For example, the boiling point of pure water at \(1.0 \: \text{atm}\) is \(100^\text{o} \text{C}\) while the boiling point of a \(2\%\) saltwater solution is about \(102^\text{o} \text{C}\). Therefore, the boiling point elevation would be \(2^\text{o} \text{C}\). The freezing point depression is the amount that the freezing temperature decreases . Both the boiling point elevation and the freezing point depression are related to the molality of the solution. Looking at the formula for the boiling point elevation and freezing point depression, we see similarities between the two. The equation used to calculate the increase in the boiling point is: \[\Delta T_b = k_b \cdot \text{m} \cdot i \label{BP} \] Where: \(\Delta T_b =\) the amount the boiling point increases. \(k_b =\) the boiling point elevation constant which depends on the solvent (for water, this number is \(0.515^\text{o} \text{C/m}\)). \(\text{m} =\) the molality of the solution. \(i =\) the number of particles formed when that compound dissolves (for covalent compounds, this number is always 1). The following equation is used to calculate the decrease in the freezing point: \[\Delta T_f = k_f \cdot \text{m} \cdot i \label{FP} \] Where: \(\Delta T_f =\) the amount the freezing temperature decreases. \(k_f =\) the freezing point depression constant which depends on the solvent (for water, this number is \(1.86^\text{o} \text{C/m}\)). \(\text{m} =\) the molality of the solution. \(i =\) the number of particles formed when that compound dissolves (for covalent compounds, this number is always 1). Substance Chemical Formula Normal Melting Point (°C) Normal Boiling Point (°C) Kf (°C/m) Kb (°C/m) Water H2O 0.0 100.0 1.86 0.512 Diethyl Ether C4H10O -116.3 34.5 1.76 2.020 Ethanol C2H5OH -114.1 78.3 1.99 1.220 Benzene C6H6 5.5 80.1 5.10 2.530 Example \(\PageIndex{2}\): Adding Antifreeze to Protein Engines Antifreeze is used in automobile radiators to keep the coolant from freezing. In geographical areas where winter temperatures go below the freezing point of water, using pure water as the coolant could allow the water to freeze. Since water expands when it freezes, freezing coolant could crack engine blocks, radiators, and coolant lines. The main component in antifreeze is ethylene glycol, \(\ce{C_2H_4(OH)_2}\). What is the concentration of ethylene glycol in a solution of water, in molality, if the freezing point dropped by \(2.64^\text{o} \text{C}\)? The freezing point constant, \(k_f\), for water is \(1.86^\text{o} \text{C/m}\). Solution Use the equation for freezing point depression of solution (Equation \(\ref{FP}\)): \[\Delta T_f = k_f \cdot \text{m} \cdot i \nonumber \] Substituting in the appropriate values we get: \[2.64^\text{o} \text{C} = \left( 1.86^\text{o} \text{C/m} \right) \left( \text{m} \right) \left( 1 \right) \nonumber \] Solve for \(\text{m}\) by dividing both sides by \(1.86^\text{o} \text{C/m}\). \[\text{m} = 1.42 \nonumber \] Example \(\PageIndex{3}\): Adding Salt to Elevate Boiling Temperature A solution of \(10.0 \: \text{g}\) of sodium chloride is added to \(100.0 \: \text{g}\) of water in an attempt to elevate the boiling point. What is the boiling point of the solution? \(k_b\) for water is \(0.512^\text{o} \text{C/m}\). Solution Use the equation for boiling point elevation of solution (Equation \(\ref{BP}\)): \[\Delta T_b = k_b \cdot \text{m} \cdot i \nonumber \] We need to be able to substitute each variable into this equation. \(k_b = 0.52^\text{o} \text{C/m}\) \(\text{m}\): We must solve for this using stoichiometry. Given: \(10.0 \: \text{g} \: \ce{NaCl}\) and \(100.0 \: \text{g} \: \ce{H_2O}\) Find: \(\text{mol} \: \ce{NaCl}/\text{kg} \: \ce{H_2O}\). Ratios: molar mass of \(\ce{NaCl}\), \(1000 \: \text{g} = 1 \: \text{kg}\) \[\dfrac{10.0 \: \cancel{\text{g} \: \ce{NaCl}}}{100.0 \: \cancel{\text{g} \: \ce{H_2O}}} \cdot \dfrac{1 \: \text{mol} \: \ce{NaCl}}{58.45 \: \cancel{\text{g} \: \ce{NaCl}}} \cdot \dfrac{1000 \: \cancel{\text{g} \: \ce{H_2O}}}{1 \: \text{kg} \: \ce{H_2O}} = 1.71 \: \text{m} \nonumber \] For \(\ce{NaCl}\), \(i = 2\) Substitute these values into the equation \(\Delta T_b = k_b \cdot \text{m} \cdot i\). We get: \[\Delta T_b = \left( 0.512 \dfrac{^\text{o} \text{C}}{\cancel{\text{m}}} \right) \left( 1.71 \: \cancel{\text{m}} \right) \left( 2 \right) = 1.75^\text{o} \text{C} \nonumber \] Water normally boils at \(100^\text{o} \text{C}\), but our calculation shows that the boiling point increased by \(1.75^\text{o} \text{C}\). Our new boiling point is \(101.75^\text{o} \text{C}\). Note: Since sea water contains roughly 28.0 g of NaCl per liter, this saltwater solution is approximately four times more concentrated than sea water (all for a 2° C rise of boiling temperature). Summary Colligative properties are properties that are due only to the number of particles in solution, and are not related to the chemical properties of the solute. Boiling points of solutions are higher than the boiling points of the pure solvents. Freezing points of solutions are lower than the freezing points of the pure solvents. Ionic compounds split into ions when they dissolve, forming more particles. Covalent compounds stay as complete molecules when they dissolve. Vocabulary Colligative property - A property that is due only to the number of particles in solution, and not the type of the solute. Boiling point elevation - The amount that the boiling point of a solution increases from the boiling point of the pure solvent. Freezing point depression - The amount that the freezing point of a solution decreases from the freezing point of the pure solvent. |
Courses/Los_Medanos_College/Chemistry_6_and_Chemistry_7_Combined_Laboratory_Manual_(Los_Medanos_College)/01%3A_Experiments/1.02%3A_Experiment_602_Empirical_Formula_of_MgO_1_4_2 | 0 1 2 3 Student Name NaN Laboratory Date: Date Report Submitted: ___________________________ Student ID NaN Experiment Number and Title Experiment 602: Empirical Formula Experiment 60 2 : Empirical Formula Section 1: Purpose and Summary Determine the empirical formula of magnesium oxide. Calculate the mass of oxygen using weighing-by-difference. Calculate the mole of a sample from its mass. In this experiment, students will conduct the reaction between magnesium and oxygen gas. Students will determine the mass of magnesium sample before and after the reaction, and the mass of magnesium and oxygen in the product. Students will learn how to convert mass to mole of a given sample and determine empirical formula of a substance from mass and mole data. Section 2: Safety Precautions and Waste Disposal Safety Precautions: Do not look directly at the burning magnesium ribbon. The flame is bright enough to damage your eye. Use of eye protection is required for all experimental procedures. A hot crucible will break if placed directly on a cold surface. Set hot crucibles on to wire screens to cool. A hot crucible will break if splashed with water directly. Let crucibles cool prior to adding water. Waste Disposal: The solid product from the reaction can be disposed into the regular garbage can in the lab. Section 3: Procedure Part 1: Preparation of the c rucible 0 1 Heat a clean and dry porcelain crucible with lid (on a clay triangle supported on a ring stand) directly over a Bunsen burner flame for about 5 minutes. This will ensure that the crucible is clean and dry. Here is what the setup should look like: NaN Turn off the burner and allow the crucible to cool. Leave the crucible resting on the clay triangle. DO NOT put a hot crucible on the lab bench. NaN When the crucible is cool to touch, weigh the crucible and lid on the digital balance. It is important to use a balance that measures to 0.0001 grams. If you use the wrong balance, you will have to repeat this experiment. Mass of empty crucible and lid: (a)________________ grams Part 2 : Preparation of m agnesium s ample 0 1 Obtain a strip or two of magnesium ribbon, about 6.0 to 7.0 cm long. NOTE: If it is not shiny, polish it slightly with steel wool to remove any oxide coating. Use sufficient magnesium strips to mass between 0.4 and 0.6 grams on an analytical balance (a balance that measures to 0.0001 g) NaN Fold or loosely coil the magnesium ribbon(s) to make sure it sits entirely on the bottom of the crucible. NaN Put the coiled magnesium ribbon into the crucible, cover, and weigh on the digital balance. Record mass. Mass of crucible, magnesium ribbon, and lid: (b)________________ grams Part 3: Heating the m agnesium s ample 0 1 Place the crucible containing the magnesium ribbon back into the clay triangle. Put the lid close by on the lab bench, with crucible tongs to handle it with later. NaN Light the Bunsen burner and start heating the crucible and magnesium ribbon. As soon as the magnesium ribbon starts to burn, put the lid back (using crucible tongs) to put out the flame. NaN Continue heating the crucible with the lid on for a minute or so, then remove the lid again. When the magnesium ribbon starts to burn again, put the lid back to put out the flame. NaN Repeat this heating process until the magnesium ribbon no longer catches fire. NaN With the lid on, heat the crucible and sample strongly for 5 minutes. Make sure that the bottom of the crucible turns ‘red’ hot. NaN Turn off the burner, and let the crucible, lid, and sample cool on the clay triangle. NaN When the crucible is cool to touch, remove the lid, then add about 10 drops of laboratory water into the burned magnesium ribbon. Make sure to wet the entire sample, not just one spot. NaN Light the Bunsen burner and heat the crucible (without lid) with a low flame for a minute or so. This evaporates the water you just added. After the water has evaporated, increase the flame and heat it strongly for about 10 minutes. The bottom of the crucible does not need to turn ‘red’ hot during this heating. NaN Turn off the burner, and the let the crucible and the sample cool on the clay triangle. NaN When the crucible is cool to touch, weigh the crucible, product, and lid on the digital balance. Mass of crucible, product, and lid: (c)________________ grams Section 4: Calculations 0 1 From the masses you recorded in Parts 1 and 2 of this experiment, calculate the mass of magnesium ribbon. (b) – (a) (d) _________________ g Mg From the masses you recorded in Parts 2 and 3 of this experiment, calculate the mass of magnesium oxide produced from the reaction. (c) – (a) (e) _________________ g MgO From the mass of magnesium oxide (e) and the mass of magnesium ribbon (d), calculate the mass of oxygen that combined with magnesium during the reaction. (e) – (d) (f) ________________ g O Refer to a periodic table to obtain the molar mass of magnesium. Then, using this information and the mass of magnesium ribbon (d), calculate the number of moles of magnesium that you started with. Molar mass of Mg: ___________g/mol (from a Periodic Table) Show your equation here: EXAMPLE: If a student used 0.4532 g of Mg, then set up your equation like this. Use your molar mass of Mg value. Set up the equation so that the units cancel. (g) ____________ mol Mg Obtain the molar mass of oxygen from a periodic table. Using this information and the mass of oxygen you calculated (f), calculate the number of moles of oxygen that combined with the magnesium. Molar mass of O: ____________ g/mol (from a periodic table) Show your equation here: EXAMPLE: If a student measured a mass gain of 0.3005 g due to the oxygen combusted, then set up your equation like this. Use your molar mass of oxygen value. Set up the equation so that the units cancel. \(\frac{\text {0.3005 grams} \quad \text {O}}{1} \times \frac{\text {moles of O}}{\text {grams of O}}=\) mol of O (h) _______________ mol O Calculate the mole ratio between magnesium and oxygen by dividing the mol Mg (g) by the mol O (h). Show your equation here: EXAMPLE: If a student measured x moles of Mg and y moles of O, the equation should be set up as Set up the equation so that the units cancel. \(\frac{\text {x} \quad \text {moles of Mg}}{\text {y} \quad \text {moles of O}}=\) mole ratio of Mg to O in magnesium oxide (i) ________________ Using your mole ratio determined in (i), write the empirical formula of magnesium oxide. Round off the ratio to the nearest whole number. Empirical Formula: Post Lab Questions: 1. There are some experimental errors that could lead to high or low mole ratio between Mg and O. In each case below, decide whether the situation described would lead to a calculated ratio of too much oxygen, or too little oxygen, and explain your answer. (a) You forgot to do the initial drying step and proceeded right away to weighing the crucible and lid you obtained from the stockroom. (b) Your magnesium ribbon is not shiny. But you did not polish it with steel wool prior to use as indicated in the experiment. (c) You added more laboratory water than is needed in Part 3 Step 7, and you did not dry it out completely. (d) After strong heating of the crucible you removed the lid but dropped it and broke. You then obtained a new lid for the final weighing. 2. A similar experiment is performed to determine the empirical formula of an oxide of copper, and the following data were collected. Predict the empirical formula of the copper oxide from these data. Mass of crucible, cover, and copper sample 21.53 g Mass of empty crucible with cover 19.66 g Mass of crucible and cover and sample (after heating) 21.76 g How can the experiment for the determination of the empirical formula of an oxide of copper be improved? |
Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520)/06%3A_Acid-Base_Equilibrium/6.03%3A_Relative_Strengths_of_Acids_and_Bases | Learning Objectives Assess the relative strengths of acids and bases according to their ionization constants Rationalize trends in acid–base strength in relation to molecular structure Carry out equilibrium calculations for weak acid–base systems We can rank the strengths of acids by the extent to which they ionize in aqueous solution. The reaction of an acid with water is given by the general expression: \[\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq) \nonumber \] Water is the base that reacts with the acid \(\ce{HA}\), \(\ce{A^{−}}\) is the conjugate base of the acid \(\ce{HA}\), and the hydronium ion is the conjugate acid of water. A strong acid yields 100% (or very nearly so) of \(\ce{H3O+}\) and \(\ce{A^{−}}\) when the acid ionizes in water; Figure \(\PageIndex{1}\) lists several strong acids. A weak acid gives small amounts of \(\ce{H3O+}\) and \(\ce{A^{−}}\). Six Strong Acids Six Strong Acids.1 Six Strong Bases Six Strong Bases.1 \(\ce{HClO4}\) perchloric acid \(\ce{LiOH}\) lithium hydroxide \(\ce{HCl}\) hydrochloric acid \(\ce{NaOH}\) sodium hydroxide \(\ce{HBr}\) hydrobromic acid \(\ce{KOH}\) potassium hydroxide \(\ce{HI}\) hydroiodic acid \(\ce{Ca(OH)2}\) calcium hydroxide \(\ce{HNO3}\) nitric acid \(\ce{Sr(OH)2}\) strontium hydroxide \(\ce{H2SO4}\) sulfuric acid \(\ce{Ba(OH)2}\) barium hydroxide The relative strengths of acids may be determined by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, K a . For the reaction of an acid \(\ce{HA}\): \[\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq) \nonumber \] we write the equation for the ionization constant as: \[K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}} \nonumber \] where the concentrations are those at equilibrium. As noted in the section on equilibrium constants, although water is a reactant in the reaction, it is the solvent as well, so its activity has a value of 1, which does not change the value of \(K_a\). Note It is a common error to claim that the molar concentration of the solvent is in some way involved in the equilibrium law. This error is a result of a misunderstanding of solution thermodynamics. For example, it is often claimed that K a = K eq [H 2 O] for aqueous solutions. This equation is incorrect because it is an erroneous interpretation of the correct equation K a = K eq (\(\textit{a}_{H_2O}\)). Because \(\textit{a}_{H_2O}\) = 1 for a dilute solution, K a = K eq (1), or K a = K eq . The larger the \(K_a\) of an acid, the larger the concentration of \(\ce{H3O+}\) and \(\ce{A^{−}}\) relative to the concentration of the nonionized acid, \(\ce{HA}\). Thus a stronger acid has a larger ionization constant than does a weaker acid. The ionization constants increase as the strengths of the acids increase. The following data on acid-ionization constants indicate the order of acid strength: \(\ce{CH3CO2H} < \ce{HNO2} < \ce{HSO4-}\) \[ \begin{aligned} \ce{CH3CO2H}(aq) + \ce{H2O}(l) &⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \quad &K_\ce{a}=1.8×10^{−5} \\[4pt] \ce{HNO2}(aq)+\ce{H2O}(l) &⇌\ce{H3O+}(aq)+\ce{NO2-}(aq) &K_\ce{a}=4.6×10^{-4} \\[4pt] \ce{HSO4-}(aq)+\ce{H2O}(l) &⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) & K_\ce{a}=1.2×10^{−2} \end{aligned} \nonumber \] Another measure of the strength of an acid is its percent ionization . The percent ionization of a weak acid is the ratio of the concentration of the ionized acid to the initial acid concentration, times 100: \[\% \:\ce{ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}×100\% \label{PercentIon} \] Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration. Example \(\PageIndex{1}\): Calculation of Percent Ionization from pH Calculate the percent ionization of a 0.125- M solution of nitrous acid (a weak acid), with a pH of 2.09. Solution The percent ionization for an acid is: \[\ce{\dfrac{[H3O+]_{eq}}{[HNO2]_0}}×100 \nonumber \] The chemical equation for the dissociation of the nitrous acid is: \[\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{NO2-}(aq)+\ce{H3O+}(aq). \nonumber \] Since \(10^{−pH} = \ce{[H3O+]}\), we find that \(10^{−2.09} = 8.1 \times 10^{−3}\, M\), so that percent ionization (Equation \ref{PercentIon}) is: \[\dfrac{8.1×10^{−3}}{0.125}×100=6.5\% \nonumber \] Remember, the logarithm 2.09 indicates a hydronium ion concentration with only two significant figures. Exercise \(\PageIndex{1}\) Calculate the percent ionization of a 0.10 M solution of acetic acid with a pH of 2.89. Answer 1.3% ionized We can rank the strengths of bases by their tendency to form hydroxide ions in aqueous solution. The reaction of a Brønsted-Lowry base with water is given by: \[\ce{B}(aq)+\ce{H2O}(l)⇌\ce{HB+}(aq)+\ce{OH-}(aq) \nonumber \] Water is the acid that reacts with the base, \(\ce{HB^{+}}\) is the conjugate acid of the base \(\ce{B}\), and the hydroxide ion is the conjugate base of water. A strong base yields 100% (or very nearly so) of OH − and HB + when it reacts with water; Figure \(\PageIndex{1}\) lists several strong bases. A weak base yields a small proportion of hydroxide ions. Soluble ionic hydroxides such as NaOH are considered strong bases because they dissociate completely when dissolved in water. As we did with acids, we can measure the relative strengths of bases by measuring their base-ionization constant ( K b ) in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, \(\ce{B}\): \[\ce{B}(aq)+\ce{H2O}(l)⇌\ce{HB+}(aq)+\ce{OH-}(aq), \nonumber \] we write the equation for the ionization constant as: \[K_\ce{b}=\ce{\dfrac{[HB+][OH- ]}{[B]}} \nonumber \] where the concentrations are those at equilibrium. Again, we do not see water in the equation because water is the solvent and has an activity of 1. The chemical reactions and ionization constants of the three bases shown are: \[ \begin{aligned} \ce{NO2-}(aq)+\ce{H2O}(l) &⇌\ce{HNO2}(aq)+\ce{OH-}(aq) \quad &K_\ce{b}=2.17×10^{−11} \\[4pt] \ce{CH3CO2-}(aq)+\ce{H2O}(l) &⇌\ce{CH3CO2H}(aq)+\ce{OH-}(aq) &K_\ce{b}=5.6×10^{−10} \\[4pt] \ce{NH3}(aq)+\ce{H2O}(l) &⇌\ce{NH4+}(aq)+\ce{OH-}(aq) &K_\ce{b}=1.8×10^{−5} \end{aligned} \nonumber \] A table of ionization constants of weak bases appears in Table E2. As with acids, percent ionization can be measured for basic solutions, but will vary depending on the base ionization constant and the initial concentration of the solution. Consider the ionization reactions for a conjugate acid-base pair, \(\ce{HA − A^{−}}\): \[\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq) \nonumber \] with \(K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}\). \[\ce{A-}(aq)+\ce{H2O}(l)⇌\ce{OH-}(aq)+\ce{HA}(aq) \nonumber \] with \(K_\ce{b}=\ce{\dfrac{[HA][OH]}{[A- ]}}\). Adding these two chemical equations yields the equation for the autoionization for water: \[\begin{align*} \cancel{\ce{HA}(aq)}+\ce{H2O}(l)+\cancel{\ce{A-}(aq)}+\ce{H2O}(l) &⇌ \ce{H3O+}(aq)+\cancel{\ce{A-}(aq)}+\ce{OH-}(aq)+\cancel{\ce{HA}(aq)} \\[4pt] \ce{2H2O}(l) &⇌\ce{H3O+}(aq)+\ce{OH-}(aq) \end{align*} \nonumber \] As shown in the previous chapter on equilibrium, the \(K\) expression for a chemical equation derived from adding two or more other equations is the mathematical product of the other equations’ \(K\) expressions. Multiplying the mass-action expressions together and cancelling common terms, we see that: \[K_\ce{a}×K_\ce{b}=\ce{\dfrac{[H3O+][A- ]}{[HA]}×\dfrac{[HA][OH- ]}{[A- ]}}=\ce{[H3O+][OH- ]}=K_\ce{w} \nonumber \] For example, the acid ionization constant of acetic acid (CH 3 COOH) is 1.8 × 10 −5 , and the base ionization constant of its conjugate base, acetate ion (\(\ce{CH3COO-}\)), is 5.6 × 10 −10 . The product of these two constants is indeed equal to \(K_w\): \[K_\ce{a}×K_\ce{b}=(1.8×10^{−5})×(5.6×10^{−10})=1.0×10^{−14}=K_\ce{w} \nonumber \] The extent to which an acid, \(\ce{HA}\), donates protons to water molecules depends on the strength of the conjugate base, \(\ce{A^{−}}\), of the acid. If \(\ce{A^{−}}\) is a strong base, any protons that are donated to water molecules are recaptured by \(\ce{A^{−}}\). Thus there is relatively little \(\ce{A^{−}}\) and \(\ce{H3O+}\) in solution, and the acid, \(\ce{HA}\), is weak. If \(\ce{A^{−}}\) is a weak base, water binds the protons more strongly, and the solution contains primarily \(\ce{A^{−}}\) and \(\ce{H3O^{+}}\)—the acid is strong. Strong acids form very weak conjugate bases, and weak acids form stronger conjugate bases (Figure \(\PageIndex{2}\)). Figure \(\PageIndex{2}\) lists a series of acids and bases in order of the decreasing strengths of the acids and the corresponding increasing strengths of the bases. The acid and base in a given row are conjugate to each other. The first six acids in Figure \(\PageIndex{2}\) are the most common strong acids. These acids are completely dissociated in aqueous solution. The conjugate bases of these acids are weaker bases than water. When one of these acids dissolves in water, their protons are completely transferred to water, the stronger base. Those acids that lie between the hydronium ion and water in Figure \(\PageIndex{2}\) form conjugate bases that can compete with water for possession of a proton. Both hydronium ions and nonionized acid molecules are present in equilibrium in a solution of one of these acids. Compounds that are weaker acids than water (those found below water in the column of acids) in Figure \(\PageIndex{2}\) exhibit no observable acidic behavior when dissolved in water. Their conjugate bases are stronger than the hydroxide ion, and if any conjugate base were formed, it would react with water to re-form the acid. The extent to which a base forms hydroxide ion in aqueous solution depends on the strength of the base relative to that of the hydroxide ion, as shown in the last column in Figure \(\PageIndex{2}\). A strong base, such as one of those lying below hydroxide ion, accepts protons from water to yield 100% of the conjugate acid and hydroxide ion. Those bases lying between water and hydroxide ion accept protons from water, but a mixture of the hydroxide ion and the base results. Bases that are weaker than water (those that lie above water in the column of bases) show no observable basic behavior in aqueous solution. Example \(\PageIndex{2}\): The Product K a × K b = K w Use the \(K_b\) for the nitrite ion, \(\ce{NO2-}\), to calculate the \(K_a\) for its conjugate acid. Solution K b for \(\ce{NO2-}\) is given in this section as 2.17 × 10 −11 . The conjugate acid of \(\ce{NO2-}\) is HNO 2 ; K a for HNO 2 can be calculated using the relationship: \[K_\ce{a}×K_\ce{b}=1.0×10^{−14}=K_\ce{w} \nonumber \] Solving for K a , we get: \[\begin{align*} K_\ce{a} &=\dfrac{K_\ce{w}}{K_\ce{b}} \\[4pt] &=\dfrac{1.0×10^{−14}}{2.17×10^{−11}} \\[4pt] &=4.6×10^{−4} \end{align*} \nonumber \] This answer can be verified by finding the K a for HNO 2 in Table E1 Exercise \(\PageIndex{2}\) We can determine the relative acid strengths of \(\ce{NH4+}\) and \(\ce{HCN}\) by comparing their ionization constants. The ionization constant of \(\ce{HCN}\) is given in Table E1 as 4.9 × 10 −10 . The ionization constant of \(\ce{NH4+}\) is not listed, but the ionization constant of its conjugate base, \(\ce{NH3}\), is listed as 1.8 × 10 −5 . Determine the ionization constant of \(\ce{NH4+}\), and decide which is the stronger acid, \(\ce{HCN}\) or \(\ce{NH4+}\). Answer \(\ce{NH4+}\) is the slightly stronger acid ( K a for \(\ce{NH4+}\) = 5.6 × 10 −10 ). The Ionization of Weak Acids and Weak Bases Many acids and bases are weak; that is, they do not ionize fully in aqueous solution. A solution of a weak acid in water is a mixture of the nonionized acid, hydronium ion, and the conjugate base of the acid, with the nonionized acid present in the greatest concentration. Thus, a weak acid increases the hydronium ion concentration in an aqueous solution (but not as much as the same amount of a strong acid). Acetic acid (\(\ce{CH3CO2H}\)) is a weak acid. When we add acetic acid to water, it ionizes to a small extent according to the equation: \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber \] giving an equilibrium mixture with most of the acid present in the nonionized (molecular) form. This equilibrium, like other equilibria, is dynamic; acetic acid molecules donate hydrogen ions to water molecules and form hydronium ions and acetate ions at the same rate that hydronium ions donate hydrogen ions to acetate ions to reform acetic acid molecules and water molecules. We can tell by measuring the pH of an aqueous solution of known concentration that only a fraction of the weak acid is ionized at any moment (Figure \(\PageIndex{4}\)). The remaining weak acid is present in the nonionized form. For acetic acid, at equilibrium: \[K_\ce{a}=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}}=1.8 \times 10^{−5} \nonumber \] Ionization Reaction Ka at 25 °C \(\ce{HSO4- + H2O ⇌ H3O+ + SO4^2-}\) 1.2 × 10−2 \(\ce{HF + H2O ⇌ H3O+ + F-}\) 3.5 × 10−4 \(\ce{HNO2 + H2O ⇌ H3O+ + NO2-}\) 4.6 × 10−4 \(\ce{HNCO + H2O ⇌ H3O+ + NCO-}\) 2 × 10−4 \(\ce{HCO2H + H2O ⇌ H3O+ + HCO2-}\) 1.8 × 10−4 \(\ce{CH3CO2H + H2O ⇌ H3O+ + CH3CO2-}\) 1.8 × 10−5 \(\ce{HCIO + H2O ⇌ H3O+ + CIO-}\) 2.9 × 10−8 \(\ce{HBrO + H2O ⇌ H3O+ + BrO-}\) 2.8 × 10−9 \(\ce{HCN + H2O ⇌ H3O+ + CN-}\) 4.9 × 10−10 Table \(\PageIndex{1}\) gives the ionization constants for several weak acids; additional ionization constants can be found in Table E1. At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the greatest concentration. Thus, a weak base increases the hydroxide ion concentration in an aqueous solution (but not as much as the same amount of a strong base). For example, a solution of the weak base trimethylamine, (CH 3 ) 3 N, in water reacts according to the equation: \[\ce{(CH3)3N}(aq)+\ce{H2O}(l)⇌\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq) \nonumber \] This gives an equilibrium mixture with most of the base present as the nonionized amine. This equilibrium is analogous to that described for weak acids. We can confirm by measuring the pH of an aqueous solution of a weak base of known concentration that only a fraction of the base reacts with water (Figure 14.4.5). The remaining weak base is present as the unreacted form. The equilibrium constant for the ionization of a weak base, \(K_b\), is called the ionization constant of the weak base, and is equal to the reaction quotient when the reaction is at equilibrium. For trimethylamine, at equilibrium: \[K_\ce{b}=\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}} \nonumber \] The ionization constants of several weak bases are given in Table \(\PageIndex{2}\) and Table E2. Ionization Reaction Kb at 25 °C \(\ce{(CH3)2NH + H2O ⇌ (CH3)2NH2+ + OH-}\) 5.9 × 10−4 \(\ce{CH3NH2 + H2O ⇌ CH3NH3+ + OH-}\) 4.4 × 10−4 \(\ce{(CH3)3N + H2O ⇌ (CH3)3NH+ + OH-}\) 6.3 × 10−5 \(\ce{NH3 + H2O ⇌ NH4+ + OH-}\) 1.8 × 10−5 \(\ce{C6H5NH2 + H2O ⇌ C6N5NH3+ + OH-}\) 4.3 × 10−10 Example \(\PageIndex{3}\): Determination of K a from Equilibrium Concentrations Acetic acid is the principal ingredient in vinegar; that's why it tastes sour. At equilibrium, a solution contains [CH 3 CO 2 H] = 0.0787 M and \(\ce{[H3O+]}=\ce{[CH3CO2- ]}=0.00118\:M\). What is the value of \(K_a\) for acetic acid? Solution We are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction: \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber \] \[\begin{align*} K_\ce{a} &=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}} \\[4pt] &=\dfrac{(0.00118)(0.00118)}{0.0787} \\[4pt] &=1.77×10^{−5} \end{align*} \nonumber \] Exercise \(\PageIndex{3}\) What is the equilibrium constant for the ionization of the \(\ce{HSO4-}\) ion, the weak acid used in some household cleansers: \[\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \nonumber \] In one mixture of NaHSO 4 and Na 2 SO 4 at equilibrium, \(\ce{[H3O+]}\) = 0.027 M ; \(\ce{[HSO4- ]}=0.29\:M\); and \(\ce{[SO4^2- ]}=0.13\:M\). Answer \(K_a\) for \(\ce{HSO_4^-}= 1.2 ×\times 10^{−2}\) Example \(\PageIndex{4}\): Determination of K b from Equilibrium Concentrations Caffeine, C 8 H 10 N 4 O 2 is a weak base. What is the value of K b for caffeine if a solution at equilibrium has [C 8 H 10 N 4 O 2 ] = 0.050 M , \(\ce{[C8H10N4O2H+]}\) = 5.0 × 10 −3 M , and [OH − ] = 2.5 × 10 −3 M ? Solution At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction: \[\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \nonumber \] so \[K_\ce{b}=\ce{\dfrac{[C8H10N4O2H+][OH- ]}{[C8H10N4O2]}}=\dfrac{(5.0×10^{−3})(2.5×10^{−3})}{0.050}=2.5×10^{−4} \nonumber \] Exercise \(\PageIndex{4}\) What is the equilibrium constant for the ionization of the \(\ce{HPO4^2-}\) ion, a weak base: \[\ce{HPO4^2-}(aq)+\ce{H2O}(l)⇌\ce{H2PO4-}(aq)+\ce{OH-}(aq) \nonumber \] In a solution containing a mixture of \(\ce{NaH2PO4}\) and \(\ce{Na2HPO4}\) at equilibrium with: \([\ce{OH^{−}}] = 1.3 × 10^{−6} M\) \(\ce{[H2PO4^{-}]=0.042\:M}\) and \(\ce{[HPO4^{2-}]=0.341\:M}\). Answer K b for \(\ce{HPO4^2-}=1.6×10^{−7} \) Example \(\PageIndex{5}\): Determination of K a or K b from pH The pH of a 0.0516- M solution of nitrous acid, \(\ce{HNO2}\), is 2.34. What is its \(K_a\)? \[\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{NO2-}(aq) \nonumber \] Solution We determine an equilibrium constant starting with the initial concentrations of HNO 2 , \(\ce{H3O+}\), and \(\ce{NO2-}\) as well as one of the final concentrations, the concentration of hydronium ion at equilibrium. (Remember that pH is simply another way to express the concentration of hydronium ion.) We can solve this problem with the following steps in which x is a change in concentration of a species in the reaction: We can summarize the various concentrations and changes as shown here. Because water is the solvent, it has a fixed activity equal to 1. Any small amount of water produced or used up during the reaction will not change water's role as the solvent, so the value of its activity remains equal to 1 throughout the reaction. To get the various values in the ICE (Initial, Change, Equilibrium) table, we first calculate \(\ce{[H3O+]}\), the equilibrium concentration of \(\ce{H3O+}\), from the pH: \[\ce{[H3O+]}=10^{−2.34}=0.0046\:M \nonumber \] The change in concentration of \(\ce{H3O+}\), \(x_{\ce{[H3O+]}}\), is the difference between the equilibrium concentration of H 3 O + , which we determined from the pH, and the initial concentration, \(\mathrm{[H_3O^+]_i}\). The initial concentration of \(\ce{H3O+}\) is its concentration in pure water, which is so much less than the final concentration that we approximate it as zero (~0). The change in concentration of \(\ce{NO2-}\) is equal to the change in concentration of \(\ce{[H3O+]}\). For each 1 mol of \(\ce{H3O+}\) that forms, 1 mol of \(\ce{NO2-}\) forms. The equilibrium concentration of HNO 2 is equal to its initial concentration plus the change in its concentration. Now we can fill in the ICE table with the concentrations at equilibrium, as shown here: Finally, we calculate the value of the equilibrium constant using the data in the table: \[K_\ce{a}=\ce{\dfrac{[H3O+][NO2- ]}{[HNO2]}}=\dfrac{(0.0046)(0.0046)}{(0.0470)}=4.5×10^{−4} \nonumber \] Exercise \(\PageIndex{5}\) The pH of a solution of household ammonia, a 0.950- M solution of NH 3 , is 11.612. What is K b for NH 3 . Answer \(K_b = 1.8 × 10^{−5}\) Example \(\PageIndex{6}\): Equilibrium Concentrations in a Solution of a Weak Acid Formic acid, HCO 2 H, is the irritant that causes the body’s reaction to ant stings. What is the concentration of hydronium ion and the pH in a 0.534- M solution of formic acid? \[\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−4} \nonumber \] Soluti on 1. Determine x and equilibrium concentrations . The equilibrium expression is: \[\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \nonumber \] Because water is the solvent, it has a fixed activity equal to 1. Any small amount of water produced or used up during the reaction will not change water's role as the solvent, so the value of its activity remains equal to 1 throughout the reaction so we do not need to consider it when setting up the ICE table. The table shows initial concentrations (concentrations before the acid ionizes), changes in concentration, and equilibrium concentrations follows (the data given in the problem appear in color): 2. Solve for \(x\) and the equilibrium concentrations. At equilibrium: \[\begin{align*} K_\ce{a} &=1.8×10^{−4}=\ce{\dfrac{[H3O+][HCO2- ]}{[HCO2H]}} \\[4pt] &=\dfrac{(x)(x)}{0.534−x}=1.8×10^{−4} \end{align*} \nonumber \] Now solve for \(x\). Because the initial concentration of acid is reasonably large and \(K_a\) is very small, we assume that \(x << 0.534\), which permits us to simplify the denominator term as \((0.534 − x) = 0.534\). This gives: \[K_\ce{a}=1.8×10^{−4}=\dfrac{x^{2}}{0.534} \nonumber \] Solve for \(x\) as follows: \[\begin{align*} x^2 &=0.534×(1.8×10^{−4}) \\[4pt] &=9.6×10^{−5} \\[4pt] x &=\sqrt{9.6×10^{−5}} \\[4pt] &=9.8×10^{−3} \end{align*} \nonumber \] To check the assumption that \(x\) is small compared to 0.534, we calculate: \[\begin{align*} \dfrac{x}{0.534} &=\dfrac{9.8×10^{−3}}{0.534} \\[4pt] &=1.8×10^{−2} \, \textrm{(1.8% of 0.534)} \end{align*} \nonumber \] \(x\) is less than 5% of the initial concentration; the assumption is valid. We find the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration as indicated in the last line of the table: \[\begin{align*} \ce{[H3O+]} &=~0+x=0+9.8×10^{−3}\:M. \\[4pt] &=9.8×10^{−3}\:M \end{align*} \nonumber \] The pH of the solution can be found by taking the negative log of the \(\ce{[H3O+]}\), so: \[pH = −\log(9.8×10^{−3})=2.01 \nonumber \] Exercise \(\PageIndex{6}\): acetic acid Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100- M solution of acetic acid, CH 3 CO 2 H? \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−5} \nonumber \] Hint Determine \(\ce{[CH3CO2- ]}\) at equilibrium.) Recall that the percent ionization is the fraction of acetic acid that is ionized × 100, or \(\ce{\dfrac{[CH3CO2- ]}{[CH3CO2H]_{initial}}}×100\). Answer percent ionization = 1.3% The following example shows that the concentration of products produced by the ionization of a weak base can be determined by the same series of steps used with a weak acid. Example \(\PageIndex{7}\): Equilibrium Concentrations in a Solution of a Weak Base Find the concentration of hydroxide ion in a 0.25- M solution of trimethylamine, a weak base: \[\ce{(CH3)3N}(aq)+\ce{H2O}(l)⇌\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=6.3×10^{−5} \nonumber \] Solution This problem requires that we calculate an equilibrium concentration by determining concentration changes as the ionization of a base goes to equilibrium. The solution is approached in the same way as that for the ionization of formic acid in Example \(\PageIndex{6}\). The reactants and products will be different and the numbers will be different, but the logic will be the same: 1. Determine x and equilibrium concentrations . The table shows the changes and concentrations: 2. Solve for \(x\) and the equilibrium concentrations . At equilibrium: \[K_\ce{b}=\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}}=\dfrac{(x)(x)}{0.25−x=}6.3×10^{−5} \nonumber \] If we assume that x is small relative to 0.25, then we can replace (0.25 − x ) in the preceding equation with 0.25. Solving the simplified equation gives: \[x=4.0×10^{−3} \nonumber \] This change is less than 5% of the initial concentration (0.25), so the assumption is justified. Recall that, for this computation, \(x\) is equal to the equilibrium concentration of hydroxide ion in the solution (see earlier tabulation): \[\begin{align*} (\ce{[OH- ]}=~0+x=x=4.0×10^{−3}\:M \\[4pt] &=4.0×10^{−3}\:M \end{align*} \nonumber \] Then calculate pOH as follows: \[\ce{pOH}=−\log(4.3×10^{−3})=2.40 \nonumber \] Using the relation introduced in the previous section of this chapter: \[\mathrm{pH + pOH=p\mathit{K}_w=14.00}\nonumber \] permits the computation of pH: \[\mathrm{pH=14.00−pOH=14.00−2.37=11.60} \nonumber \] Check the work. A check of our arithmetic shows that \(K_b = 6.3 \times 10^{−5}\). Exercise \(\PageIndex{7}\) Show that the calculation in Step 2 of this example gives an x of 4.3 × 10 −3 and the calculation in Step 3 shows K b = 6.3 × 10 −5 . Find the concentration of hydroxide ion in a 0.0325- M solution of ammonia, a weak base with a K b of 1.76 × 10 −5 . Calculate the percent ionization of ammonia, the fraction ionized × 100, or \(\ce{\dfrac{[NH4+]}{[NH3]}}×100 \%\) Answer a \(7.56 × 10^{−4}\, M\), 2.33% Answer b 2.33% Some weak acids and weak bases ionize to such an extent that the simplifying assumption that x is small relative to the initial concentration of the acid or base is inappropriate. As we solve for the equilibrium concentrations in such cases, we will see that we cannot neglect the change in the initial concentration of the acid or base, and we must solve the equilibrium equations by using the quadratic equation. Example \(\PageIndex{8}\): Equilibrium Concentrations in a Solution of a Weak Acid Sodium bisulfate, NaHSO 4 , is used in some household cleansers because it contains the \(\ce{HSO4-}\) ion, a weak acid. What is the pH of a 0.50- M solution of \(\ce{HSO4-}\)? \[\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \hspace{20px} K_\ce{a}=1.2×10^{−2} \nonumber \] Soluti on We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of \(\ce{HSO4-}\) so that we can use \(\ce{[H3O+]}\) to determine the pH. As in the previous examples, we can approach the solution by the following steps: 1. Determine \(x\) and equilibrium concentrations . This table shows the changes and concentrations: 2. Solve for \(x\) and the concentrations . As we begin solving for \(x\), we will find this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of \(x\). At equilibrium: \[K_\ce{a}=1.2×10^{−2}=\ce{\dfrac{[H3O+][SO4^2- ]}{[HSO4- ]}}=\dfrac{(x)(x)}{0.50−x} \nonumber \] If we assume that x is small and approximate (0.50 − x ) as 0.50, we find: \[x=7.7×10^{−2} \nonumber \] When we check the assumption, we confirm: \[\dfrac{x}{\mathrm{[HSO_4^- ]_i}} \overset{?}{\le} 0.05 \nonumber \] which for this system is \[\dfrac{x}{0.50}=\dfrac{7.7×10^{−2}}{0.50}=0.15(15\%) \nonumber \] The value of \(x\) is not less than 5% of 0.50, so the assumption is not valid. We need the quadratic formula to find \(x\). The equation: \[K_\ce{a}=1.2×10^{−2}=\dfrac{(x)(x)}{0.50−x}\nonumber \] gives \[6.0×10^{−3}−1.2×10^{−2}x=x^{2+} \nonumber \] or \[x^{2+}+1.2×10^{−2}x−6.0×10^{−3}=0 \nonumber \] This equation can be solved using the quadratic formula. For an equation of the form \[ax^{2+} + bx + c=0, \nonumber \] \(x\) is given by the quadratic equation: \[x=\dfrac{−b±\sqrt{b^{2+}−4ac}}{2a} \nonumber \] In this problem, \(a = 1\), \(b = 1.2 × 10^{−3}\), and \(c = −6.0 × 10^{−3}\). Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root: \[x=7.2×10^{−2} \nonumber \] Now determine the hydronium ion concentration and the pH: \[\begin{align*} \ce{[H3O+]} &=~0+x=0+7.2×10^{−2}\:M \\[4pt] &=7.2×10^{−2}\:M \end{align*} \nonumber \] The pH of this solution is: \[\mathrm{pH=−log[H_3O^+]=−log7.2×10^{−2}=1.14} \nonumber \] Exercise \(\PageIndex{8}\) Show that the quadratic formula gives \(x = 7.2 × 10^{−2}\). Calculate the pH in a 0.010- M solution of caffeine, a weak base: \[\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=2.5×10^{−4} \nonumber \] Hint It will be necessary to convert [OH − ] to \(\ce{[H3O+]}\) or pOH to pH toward the end of the calculation. Answer pH 11.16 The Relative Strengths of Strong Acids and Bases Strong acids, such as \(\ce{HCl}\), \(\ce{HBr}\), and \(\ce{HI}\), all exhibit the same strength in water. The water molecule is such a strong base compared to the conjugate bases Cl − , Br − , and I − that ionization of these strong acids is essentially complete in aqueous solutions. In solvents less basic than water, we find \(\ce{HCl}\), \(\ce{HBr}\), and \(\ce{HI}\) differ markedly in their tendency to give up a proton to the solvent. For example, when dissolved in ethanol (a weaker base than water), the extent of ionization increases in the order \(\ce{HCl < HBr < HI}\), and so \(\ce{HI}\) is demonstrated to be the strongest of these acids. The inability to discern differences in strength among strong acids dissolved in water is known as the leveling effect of water . Water also exerts a leveling effect on the strengths of strong bases. For example, the oxide ion, O 2− , and the amide ion, \(\ce{NH2-}\), are such strong bases that they react completely with water: \[\ce{O^2-}(aq)+\ce{H2O}(l)⟶\ce{OH-}(aq)+\ce{OH-}(aq) \nonumber \] \[\ce{NH2-}(aq)+\ce{H2O}(l)⟶\ce{NH3}(aq)+\ce{OH-}(aq) \nonumber \] Thus, O 2− and \(\ce{NH2-}\) appear to have the same base strength in water; they both give a 100% yield of hydroxide ion. In the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the H-A bond strength decreases down a group in the periodic table. For group 17, the order of increasing acidity is \(\ce{HF < HCl < HBr < HI}\). Likewise, for group 16, the order of increasing acid strength is H 2 O < H 2 S < H 2 Se < H 2 Te. Across a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the H-A bond increases. Thus, the order of increasing acidity (for removal of one proton) across the second row is \(\ce{CH4 < NH3 < H2O < HF}\); across the third row, it is \(\ce{SiH4 < PH3 < H2S < HCl}\) (see Figure \(\PageIndex{6}\)). Compounds containing oxygen and one or more hydroxyl (OH) groups can be acidic, basic, or amphoteric, depending on the position in the periodic table of the central atom E, the atom bonded to the hydroxyl group. Such compounds have the general formula O n E(OH) m , and include sulfuric acid, \(\ce{O2S(OH)2}\), sulfurous acid, \(\ce{OS(OH)2}\), nitric acid, \(\ce{O2NOH}\), perchloric acid, \(\ce{O3ClOH}\), aluminum hydroxide, \(\ce{Al(OH)3}\), calcium hydroxide, \(\ce{Ca(OH)2}\), and potassium hydroxide, \(\ce{KOH}\): If the central atom, E, has a low electronegativity, its attraction for electrons is low. Little tendency exists for the central atom to form a strong covalent bond with the oxygen atom, and bond a between the element and oxygen is more readily broken than bond b between oxygen and hydrogen. Hence bond a is ionic, hydroxide ions are released to the solution, and the material behaves as a base—this is the case with Ca(OH) 2 and KOH . Lower electronegativity is characteristic of the more metallic elements; hence, the metallic elements form ionic hydroxides that are by definition basic compounds. If, on the other hand, the atom E has a relatively high electronegativity, it strongly attracts the electrons it shares with the oxygen atom, making bond a relatively strongly covalent. The oxygen-hydrogen bond, bond b , is thereby weakened because electrons are displaced toward E. Bond b is polar and readily releases hydrogen ions to the solution, so the material behaves as an acid. High electronegativities are characteristic of the more nonmetallic elements. Thus, nonmetallic elements form covalent compounds containing acidic −OH groups that are called oxyacids . Increasing the oxidation number of the central atom E also increases the acidity of an oxyacid because this increases the attraction of E for the electrons it shares with oxygen and thereby weakens the O-H bond. Sulfuric acid, H 2 SO 4 , or O 2 S(OH) 2 (with a sulfur oxidation number of +6), is more acidic than sulfurous acid, H 2 SO 3 , or OS(OH) 2 (with a sulfur oxidation number of +4). Likewise nitric acid, HNO 3 , or O 2 NOH (N oxidation number = +5), is more acidic than nitrous acid, HNO 2 , or ONOH (N oxidation number = +3). In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure \(\PageIndex{7}\)). Hydroxy compounds of elements with intermediate electronegativities and relatively high oxidation numbers (for example, elements near the diagonal line separating the metals from the nonmetals in the periodic table) are usually amphoteric. This means that the hydroxy compounds act as acids when they react with strong bases and as bases when they react with strong acids. The amphoterism of aluminum hydroxide, which commonly exists as the hydrate \(\ce{Al(H2O)3(OH)3}\), is reflected in its solubility in both strong acids and strong bases. In strong bases, the relatively insoluble hydrated aluminum hydroxide, \(\ce{Al(H2O)3(OH)3}\), is converted into the soluble ion, \(\ce{[Al(H2O)2(OH)4]-}\), by reaction with hydroxide ion: \[[\ce{Al(H2O)3(OH)3}](aq)+\ce{OH-}(aq)⇌\ce{H2O}(l)+\ce{[Al(H2O)2(OH)4]-}(aq) \nonumber \] In this reaction, a proton is transferred from one of the aluminum-bound H 2 O molecules to a hydroxide ion in solution. The \(\ce{Al(H2O)3(OH)3}\) compound thus acts as an acid under these conditions. On the other hand, when dissolved in strong acids, it is converted to the soluble ion \(\ce{[Al(H2O)6]^3+}\) by reaction with hydronium ion: \[\ce{3H3O+}(aq)+\ce{Al(H2O)3(OH)3}(aq)⇌\ce{Al(H2O)6^3+}(aq)+\ce{3H2O}(l) \nonumber \] In this case, protons are transferred from hydronium ions in solution to \(\ce{Al(H2O)3(OH)3}\), and the compound functions as a base. Summary The strengths of Brønsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons. Strong bases react with water to quantitatively form hydroxide ions. Weak bases give only small amounts of hydroxide ion. The strengths of the binary acids increase from left to right across a period of the periodic table (CH 4 < NH 3 < H 2 O < HF), and they increase down a group (HF < HCl < HBr < HI). The strengths of oxyacids that contain the same central element increase as the oxidation number of the element increases (H 2 SO 3 < H 2 SO 4 ). The strengths of oxyacids also increase as the electronegativity of the central element increases [H 2 SeO 4 < H 2 SO 4 ]. Key Equations \(K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}\) \(K_\ce{b}=\ce{\dfrac{[HB+][OH- ]}{[B]}}\) \(K_a \times K_b = 1.0 \times 10^{−14} = K_w \,(\text{at room temperature})\) \(\textrm{Percent ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}×100\) Glossary acid ionization constant ( K a ) equilibrium constant for the ionization of a weak acid base ionization constant ( K b ) equilibrium constant for the ionization of a weak base leveling effect of water any acid stronger than \(\ce{H3O+}\), or any base stronger than OH − will react with water to form \(\ce{H3O+}\), or OH − , respectively; water acts as a base to make all strong acids appear equally strong, and it acts as an acid to make all strong bases appear equally strong oxyacid compound containing a nonmetal and one or more hydroxyl groups percent ionization ratio of the concentration of the ionized acid to the initial acid concentration, times 100 |
Bookshelves/Biological_Chemistry/Medicines_by_Design_(Davis)/02%3A_Body_Heal_Thyself/2.04%3A_Our_Immune_Army | Scientist know a lot about the body's organ systems, but much more remains to be discovered. To design "smart" drugs that will seek out diseased cells and not healthy ones, researchers need to understand the body inside and out. One system in particular still puzzles scientists: the immune system. Even though researchers have accumulated vast amounts of knowledge about how our bodies fight disease using white blood cells and thousands of natural chemical weapons, a basic dilemma persists—how does the body know what to fight? The immune system constantly watches for foreign invaders and is exquisitely sensitive to any intrusion perceived as "non-self," like a transplanted organ from another person. This protection, however, can run afoul if the body slips up and views its own tissue as foreign. Autoimmune disease, in which the immune system mistakenly attacks and destroys body tissue that it believes to be foreign, can be the terrible consequence. The "Anti" Establishment Common over-the-counter medicines used to treat pain, fever, and inflammation have many uses. Here are some of the terms used to describe the particular effects of these drugs: ANTIPYRETIC—this term means fever-reducing; it comes from the Greek word pyresis , which means fire. ANTI-INFLAMMATORY—this word describes a drug's ability to reduce inflammation, which can cause soreness and swelling; it comes from the Latin word flamma , which means flame. ANALGESIC—this description refers to a medicine's ability to treat pain; it comes from the Greek word algos , which means pain. Antibodies are Y-shaped molecules of the immune system. The powerful immune army presents significant roadblocks for pharmacologists trying to create new drugs. But some scientists have looked at the immune system through a different lens. Why not teach the body to lunch an attack on its own diseased cells? Many researchers are pursuing immunotherapy as a way to treat a wide range of health problems, especially cancer. With advances in biotechnology, researchers are now able to tailor-produce in the lab modified forms of antibodies—out immune system's front-line agents. Antibodies are spectacularly specific proteins that seek out and mark for destruction anything they do not recognize as belonging to body. Scientists have learned how to join antibody-making cells with cells that grow and divide continuously. This strategy creates cellular "factories" that work around the clock to produce large quantities of specialized molecules, called monoclonal antibodies, that attach to and destroy single kinds of targets. Recently, researchers have also figured out how to produce monoclonal antibodies in the egg whites of chickens. This may reduce production costs of these increasingly important drugs. Doctors are already using therapeutic monoclonal antibodies to attack tumors. A drug called Rituxan ® was the first therapeutic antibody approved by the Food and Drug Administration to treat cancer. This monoclonal antibody targets a unique tumor "fingerprint" on the surface of immune cells, called B cells, in a blood cancer called non-Hodgkin's lymphoma. Another therapeutic antibody for cancer, Herceptin ® , latches onto breast cancer cell receptors that signal growth to either mask the receptors from view or lure immune cells to kill the cancer cells. Herceptin's actions prevent breast cancer from spreading to other organs. Researchers are also investigating a new kind of "vaccine" as therapy for diseases such as cancer. The vaccines are not designed to prevent cancer, but rather to treat the disease when it has already taken hold in the body. Unlike the targeted-attack approach of antibody therapy, vaccines aim to recruit the entire immune system to fight off a tumor. Scientists are conducing clinical trials of vaccines against cancer to evaluate the effectiveness of this treatment approach. The body machine has a tremendously complex collection of chemical signals that are relayed back and forth through the blood and into and out of cells. While scientists are hopeful that future research will point the way toward getting a sick body to heal itself, it is likely that there will always be a need for medicines to speed recovery from the many illnesses that plague humankind. A Shock to the System A body-wide syndrome caused by an infection called sepsis is a leading cause of death in hospital intensive care units, striking 750,000 people every year and killing more than 215,000. Sepsis is a serious public health problem, causing more deaths annually than heart disease. The most severe form of sepsis occurs when bacteria leak into the blood-stream, spilling their poisons and leading to a dangerous condition called septic shock. Blood pressure plunges dangerously low, the heart has difficulty pumping enough blood, and body temperature climbs or falls rapidly. In many cases, multiple organs fail and the patient dies. Despite the obvious public health importance of finding effective ways to treat sepsis, researchers have been frustratingly unsuccessful. Kevin Tracey of the North Shore-Long Island Jewish Research Institute in Manhasset, New York, has identified an unusual suspect in the deadly crime of sepsis: the nervous system. Tracey and his coworkers have discovered an unexpected link between cytokines, the chemical weapons released by the immune system during sepsis, and a major nerve that controls critical body function such as heart rate and digestion. In animal studies, Tracy found that electrically stimulating this nerve, called the vagus nerve , significantly lowered blood levels of TNF, a cytokine that is produced when the body senses the presence of bacteria in the blood. Further research has led Tracey to conclude that production of the neurotransmitter acetylcholine underlies the inflammation-blocking response. Tracey is investigating whether stimulating the vagus nerve can be used as a component of therapy for sepsis and as a treatment for other immune disorders. |
Courses/Tennessee_State_University/CHEM_4210%3A_Inorganic_Chem_II_(Siddiquee)/03%3A_d-Block_Metal_Chemistry-_Coordination_Complexes/3.04%3A_Angular_Overlap/3.4.02%3A_Pi_Acceptors_in_the_Angular_Overlap_Model | So far, we have considered only the effects of sigma donation on the d orbital splitting diagram using the angular overlap model. When we looked at ligand field theory, we saw that pi-donor and pi-acceptor effects produced significant changes in these diagrams. We generally see these effects in any ligands that have orbitals that can accept electron density from the metal (back-bonding). The paradigm of a pi acceptor is carbon monoxide, of course. Similar effects can be found in related ligands in which the donor atoms participate in pi bonding with another atom in the ligand, thus making a pi* orbital available for back-bonding. In addition, back-bonding is a feature of phosphine ligands and some N-heterocyclic carbenes. We usually think about this interaction as illustrated below. The empty ligand orbital approaches so that it is perpendicular to the bond axis, allowing overlap with the filled metal d orbital. The interaction lowers the energy of the d electrons, which become bonding in nature, and raises the energy of the empty ligand orbital. We can use the same ligand positions for pi acceptors that we already used for sigma donors. The orbitals will be oriented differently than the sigma donor orbitals but they will approach from the same directions.. As before, we can use the results of calculations of the strengths of these interactions based on the amount of overlap; we don't need to know exactly how the numbers in the table below came about. This time, the maximum overlap occurs between a dxz orbital and a p orbital approaching in position 1, perpendicular to the bond axis. Several other combinations will be equally strong. This time, the stabilization is expressed in terms of eπ rather that eσ. The amount of energy in this case is somewhat smaller than in sigma donation because of a lesser degree of metal-ligand orbital overlap. Ligand Positions dz2 dx2-y2 dxy dxz dyz 1 0 0 0 1 1 2 0 0 1 1 0 3 0 0 1 0 1 4 0 0 1 1 0 5 0 0 1 0 1 6 0 0 0 1 1 7 2/3 2/3 2/9 2/9 2/9 8 2/3 2/3 2/9 2/9 2/9 9 2/3 2/3 2/9 2/9 2/9 10 2/3 2/3 2/9 2/9 2/9 11 0 3/4 1/4 1/4 3/4 12 0 3/4 1/4 1/4 3/4 These interactions modify the picture we built previously for simple sigma donors. In the new interaction diagram, a second set of ligand p orbitals is destabilized by the pi interaction. At the same time, some of the d orbitals are stabilized by the additional interaction. This modification is illustrated below for octahedral geometry. Problems 1. Use the table of pi interactions to calculate orbital energy stabilization or destabilization for the following geometries. a) trigonal planar ML3 b) square planar ML4 c) trigonal bipyramidal ML5 Solutions 1. a) Positions 2, 11, 12. dz2: 0 dx2-y2: - (0 + 3/4 + 3/4) eπ = 6/4 = - 3/2 eπ dxy: - (1 + 1/4 + 1/4) eσ = 6/4 = - 3/2 eπ dxz: - (1 + 1/4 + 1/4) eσ = 6/4 = - 3/2 eπ dyz: - (0 + 3/4 + 3/4) eσ = 6/4 = - 3/2 eπ Ligand in position 2: (0 + 0 + 0 + 1 + 1) eπ = 2eπ Ligand in position 11: (0 + 3/4 + 1/4 + 1/4 + 3/4) eπ = 2eπ Ligand in position 12: (0 + 3/4 + 1/4 + 1/4 + 3/4) eπ = 2eπ b) Positions 2, 3, 4, 5. dz2: 0 dx2-y2: 0 dxy: - (1 + 1 + 1 + 1) = -4 eπ dxz: - (1 + 0 + 1 + 0) = -2 eπ dyz: - (0 + 1 + 0 + 1) = -2 eπ Ligand in position 2: (0 + 0 + 1 + 1 + 0) eπ = 2 eπ Ligand in position 3: (0 + 0 + 1 + 0 + 1) eπ = 2 eπ Ligand in position 4: (0 + 0 + 1 + 1 + 0) eπ = 2 eπ Ligand in position 5: (0 + 0 + 1 + 0 + 1) eπ = 2 eπ c) Positions 1, 2, 6, 11, 12. dz2: 0 dx2-y2: - (0 + 0 + 0 + 3/4 + 3/4) eπ = -3/2 eπ dxy: - (0 + 1 + 0 + 1/4 + 1/4) eπ = -3/2 eπ dxz: - (1 + 1 + 1 + 1/4 + 1/4) eπ = -7/2 eπ dyz: - (1 + 0 + 1 + 3/4 + 4/4) eπ = -7/2 eπ Ligand in position 1: (0 + 0 + 0 + 1 + 1) eπ = 2 eπ Ligand in position 2: (0 + 0 + 1 + 1 + 0) eπ = 2 eπ Ligand in position 6: (0 + 0 + 0 + 1 + 1) eπ = 2 eπ Ligand in position 11: (0 + 3/4 + 1/4 + 1/4 + 3/4) eπ = 2 eπ Ligand in position 12: (0 + 3/4 + 1/4 + 1/4 + 3/4) eπ = 2 eπ |
Courses/University_of_Connecticut/Organic_Chemistry_-_Textbook_for_Chem_2443/11%3A_Reactions_of_Alcohols | 11.1: Alcohols as Acids and Bases 11.2: Reactions of Alcohols with Base- Preparation of Alkoxides 11.3: Synthesis of Alcohols by Nucleophilic Substitution 11.4: Synthesis of Alcohols- Oxidation-Reduction Relation Between Alcohols and Carbonyl Compounds 11.5: Organometallic Reagents- Sources of Nucleophilic Carbon for Alcohol Synthesis 11.6: Organometallic Reagents in the Synthesis of Alcohols 11.7: Keys to Success- An Introduction to Synthetic Strategy 11.8: Williamson Ether Synthesis 11.9: Synthesis of Ethers- Alcohols and Mineral Acids 11.10: Videos The OER Remixer is a self-service tool to rapidly assemble a LibreText from existing sources. This tutorial will include both an explanation of the User Interface as well as a walkthrough of how to do basic tasks. |
Courses/SUNY_Oneonta/Chem_221%3A_Organic_Chemistry_I_(Bennett)/3%3AStuff_to_Review_from_General_Chemistry/01%3A_Atoms_Molecules_and_Ions/1.03%3A_Evolution_of_Atomic_Theory | Learning Objectives Outline milestones in the development of modern atomic theory Summarize and interpret the results of the experiments of Thomson, Millikan, and Rutherford Describe the three subatomic particles that compose atoms Introduce the term isotopes In the two centuries since Dalton developed his ideas, scientists have made significant progress in furthering our understanding of atomic theory. Much of this came from the results of several seminal experiments that revealed the details of the internal structure of atoms. Here, we will discuss some of those key developments, with an emphasis on application of the scientific method, as well as understanding how the experimental evidence was analyzed. While the historical persons and dates behind these experiments can be quite interesting, it is most important to understand the concepts resulting from their work. Atomic Theory after the Nineteenth Century If matter were composed of atoms, what were atoms composed of? Were they the smallest particles, or was there something smaller? In the late 1800s, a number of scientists interested in questions like these investigated the electrical discharges that could be produced in low-pressure gases, with the most significant discovery made by English physicist J. J. Thomson using a cathode ray tube. This apparatus consisted of a sealed glass tube from which almost all the air had been removed; the tube contained two metal electrodes. When high voltage was applied across the electrodes, a visible beam called a cathode ray appeared between them. This beam was deflected toward the positive charge and away from the negative charge, and was produced in the same way with identical properties when different metals were used for the electrodes. In similar experiments, the ray was simultaneously deflected by an applied magnetic field, and measurements of the extent of deflection and the magnetic field strength allowed Thomson to calculate the charge-to-mass ratio of the cathode ray particles. The results of these measurements indicated that these particles were much lighter than atoms (Figure \(\PageIndex{1}\)). Based on his observations, here is what Thomson proposed and why: The particles are attracted by positive (+) charges and repelled by negative (−) charges, so they must be negatively charged (like charges repel and unlike charges attract); they are less massive than atoms and indistinguishable, regardless of the source material, so they must be fundamental, subatomic constituents of all atoms. Although controversial at the time, Thomson’s idea was gradually accepted, and his cathode ray particle is what we now call an electron , a negatively charged, subatomic particle with a mass more than one thousand-times less that of an atom. The term “electron” was coined in 1891 by Irish physicist George Stoney, from “ electr ic i on .” In 1909, more information about the electron was uncovered by American physicist Robert A. Millikan via his “oil drop” experiments. Millikan created microscopic oil droplets, which could be electrically charged by friction as they formed or by using X-rays. These droplets initially fell due to gravity, but their downward progress could be slowed or even reversed by an electric field lower in the apparatus. By adjusting the electric field strength and making careful measurements and appropriate calculations, Millikan was able to determine the charge on individual drops (Figure \(\PageIndex{2}\)). Looking at the charge data that Millikan gathered, you may have recognized that the charge of an oil droplet is always a multiple of a specific charge, 1.6 \(\times\) 10 −19 C. Millikan concluded that this value must therefore be a fundamental charge—the charge of a single electron—with his measured charges due to an excess of one electron (1 times 1.6 \(\times\) 10 −19 C), two electrons (2 times 1.6 \(\times\) 10 −19 C), three electrons (3 times 1.6 \(\times\) 10 −19 C), and so on, on a given oil droplet. Since the charge of an electron was now known due to Millikan’s research, and the charge-to-mass ratio was already known due to Thomson’s research (1.759 \(\times\) 10 11 C/kg), it only required a simple calculation to determine the mass of the electron as well. \[\mathrm{Mass\: of\: electron=1.602\times 10^{-19}\:\cancel{C}\times \dfrac{1\: kg}{1.759\times 10^{11}\:\cancel{C}}=9.107\times 10^{-31}\:kg} \tag{2.3.1} \] Scientists had now established that the atom was not indivisible as Dalton had believed, and due to the work of Thomson, Millikan, and others, the charge and mass of the negative, subatomic particles—the electrons—were known. However, the positively charged part of an atom was not yet well understood. In 1904, Thomson proposed the “plum pudding” model of atoms, which described a positively charged mass with an equal amount of negative charge in the form of electrons embedded in it, since all atoms are electrically neutral. A competing model had been proposed in 1903 by Hantaro Nagaoka , who postulated a Saturn-like atom, consisting of a positively charged sphere surrounded by a halo of electrons (Figure \(\PageIndex{3}\)). The next major development in understanding the atom came from Ernest Rutherford , a physicist from New Zealand who largely spent his scientific career in Canada and England. He performed a series of experiments using a beam of high-speed, positively charged alpha particles (α particles) that were produced by the radioactive decay of radium; α particles consist of two protons and two neutrons (you will learn more about radioactive decay in the chapter on nuclear chemistry). Rutherford and his colleagues Hans Geiger (later famous for the Geiger counter) and Ernest Marsden aimed a beam of α particles, the source of which was embedded in a lead block to absorb most of the radiation, at a very thin piece of gold foil and examined the resultant scattering of the α particles using a luminescent screen that glowed briefly where hit by an α particle. What did they discover? Most particles passed right through the foil without being deflected at all. However, some were diverted slightly, and a very small number were deflected almost straight back toward the source (Figure \(\PageIndex{4}\)). Rutherford described finding these results: “It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you” 1 (p. 68). Here is what Rutherford deduced: Because most of the fast-moving α particles passed through the gold atoms undeflected, they must have traveled through essentially empty space inside the atom. Alpha particles are positively charged, so deflections arose when they encountered another positive charge (like charges repel each other). Since like charges repel one another, the few positively charged α particles that changed paths abruptly must have hit, or closely approached, another body that also had a highly concentrated, positive charge. Since the deflections occurred a small fraction of the time, this charge only occupied a small amount of the space in the gold foil. Analyzing a series of such experiments in detail, Rutherford drew two conclusions: The volume occupied by an atom must consist of a large amount of empty space. A small, relatively heavy, positively charged body, the nucleus , must be at the center of each atom. This analysis led Rutherford to propose a model in which an atom consists of a very small, positively charged nucleus, in which most of the mass of the atom is concentrated, surrounded by the negatively charged electrons, so that the atom is electrically neutral (Figure \(\PageIndex{5}\)). After many more experiments, Rutherford also discovered that the nuclei of other elements contain the hydrogen nucleus as a “building block,” and he named this more fundamental particle the proton , the positively charged, subatomic particle found in the nucleus. With one addition, which you will learn next, this nuclear model of the atom, proposed over a century ago, is still used today. Another important finding was the discovery of isotopes. During the early 1900s, scientists identified several substances that appeared to be new elements, isolating them from radioactive ores. For example, a “new element” produced by the radioactive decay of thorium was initially given the name mesothorium. However, a more detailed analysis showed that mesothorium was chemically identical to radium (another decay product), despite having a different atomic mass. This result, along with similar findings for other elements, led the English chemist Frederick Soddy to realize that an element could have types of atoms with different masses that were chemically indistinguishable. These different types are called isotopes —atoms of the same element that differ in mass. Soddy was awarded the Nobel Prize in Chemistry in 1921 for this discovery. One puzzle remained: The nucleus was known to contain almost all of the mass of an atom, with the number of protons only providing half, or less, of that mass. Different proposals were made to explain what constituted the remaining mass, including the existence of neutral particles in the nucleus. As you might expect, detecting uncharged particles is very challenging, and it was not until 1932 that James Chadwick found evidence of neutrons , uncharged, subatomic particles with a mass approximately the same as that of protons. The existence of the neutron also explained isotopes: They differ in mass because they have different numbers of neutrons, but they are chemically identical because they have the same number of protons. This will be explained in more detail later in this chapter. Summary Although no one has actually seen the inside of an atom, experiments have demonstrated much about atomic structure. Thomson’s cathode ray tube showed that atoms contain small, negatively charged particles called electrons. Millikan discovered that there is a fundamental electric charge—the charge of an electron. Rutherford’s gold foil experiment showed that atoms have a small, dense, positively charged nucleus; the positively charged particles within the nucleus are called protons. Chadwick discovered that the nucleus also contains neutral particles called neutrons. Soddy demonstrated that atoms of the same element can differ in mass; these are called isotopes. Footnotes Ernest Rutherford, “The Development of the Theory of Atomic Structure,” ed. J. A. Ratcliffe, in Background to Modern Science , eds. Joseph Needham and Walter Pagel, (Cambridge, UK: Cambridge University Press, 1938), 61–74. Accessed September 22, 2014, https://ia600508.us.archive.org/3/it...e032734mbp.pdf . Glossary alpha particle (α particle) positively charged particle consisting of two protons and two neutrons electron negatively charged, subatomic particle of relatively low mass located outside the nucleus isotopes atoms that contain the same number of protons but different numbers of neutrons neutron uncharged, subatomic particle located in the nucleus proton positively charged, subatomic particle located in the nucleus nucleus massive, positively charged center of an atom made up of protons and neutrons |
Bookshelves/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/21_Electrochemistry/21.1_Redox_Reactions_(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 During a redox reaction, one species is oxidized and another is reduced. This involves the transfer of electrons. Loss of electrons means oxidation. The oxidation half reaction produces electrons. The species that loses the electrons is called the reducing agent. Gain of electrons means reduction. The reduction half reaction uses electrons. The species that gains the electrons is called the oxidizing agent. Link to Video Redox Reactions: https://youtu.be/1v3yaaR_nHc Attribution Prof. Steven Farmer ( Sonoma State University ) |
Courses/University_of_Ottawa/OU%3A_CHM1721B_-_Chimie_Organique_I/Partie_II%3A_Reactivite_Chimique/Unit_6%3A_%CF%80_bonds_as_Electrophiles_and_Nucleophiles/R%C3%A9actions_des_alc%C3%A8nes/7.08%3A_Hydroboration-Oxidation%3A_A_Stereospecific_Anti-Markovnikov_Hydration | Objectives After completing this section, you should be able to identify hydroboration (followed by oxidation) as a method for bringing about the (apparently) non-Markovnikov addition of water to an alkene. write an equation for the formation of a trialkylborane from an alkene and borane. write an equation for the oxidation of a trialkylborane to an alcohol. draw the structure of the alcohol produced by the hydroboration, and subsequent oxidation, of a given alkene. determine whether a given alcohol should be prepared by oxymercuration-demercuration or by hydroboration-oxidation, and identify the alkene and reagents required to carry out such a synthesis. write the detailed mechanism for the addition of borane to an alkene, and explain the stereochemistry and regiochemistry of the reaction. Key Terms Make certain that you can define, and use in context, the key term below. hydroboration Study Notes The two most important factors influencing organic reactions are polar (or electronic) effects and steric effects. Hydroboration-Oxidation is a two step pathway used to produce alcohols. The reaction proceeds in an Anti-Markovnikov manner, where the hydrogen (from BH 3 or BHR 2 ) attaches to the more substituted carbon and the boron attaches to the least substituted carbon in the alkene bouble bond. Furthermore, the borane acts as a lewis acid by accepting two electrons in its empty p orbital from an alkene that is electron rich. This process allows boron to have an electron octet. A very interesting characteristic of this process is that it does not require any activation by a catalyst. The Hydroboration mechanism has the elements of both hydrogenation and electrophilic addition and it is a stereospecific ( syn addition), meaning that the hydroboration takes place on the same face of the double bond, this leads cis stereochemistry. Introduction Hydroboration-oxidation of alkenes has been a very valuable laboratory method for the stereoselectivity and regioselectivity of alkenes. An Additional feature of this reaction is that it occurs without rearrangement. The Borane Complex First off it is very imporatnt to understand little bit about the structure and the properties of the borane molecule. Borane exists naturally as a very toxic gas and it exists as dimer of the general formula B 2 H 6 (diborane). Additionally, the dimer B 2 H 6 ignites spontaneously in air. Borane is commercially available in ether and tetrahydrofuran (THF), in these solutions the borane can exist as a lewis acid-base complex, which allows boron to have an electron octet. B H 3 → B 2 H 6 The Mechanism Step #1 Part #1: Hydroboration of the alkene. In this first step the addittion of the borane to the alkene is initiated and prceeds as a concerted reaction because bond breaking and bond formation occurs at the same time. This part consists of the vacant 2p orbital of the boron electrophile pairing with the electron pair of the ? bondof the nucleophile. Transition state * Note that a carbocation is not formed. Therefore, no rearrangement takes place . Part #2: The Anti Markovnikov addition of Boron. The boron adds to the less substituted carbon of the alkene, which then places the hydrogen on the more substituted carbon. Both, the boron and the hydrogen add simultaneously on the same face of the double bond (syn addition). Oxidation of the Trialkylborane by Hydrogen Peroxide Step #2 Part #1: the first part of this mechanism deals with the donation of a pair of electrons from the hydrogen peroxide ion. the hydrogen peroxide is the nucleophile in this reaction because it is the electron donor to the newly formed trialkylborane that resulted from hydroboration. Part 2: In this second part of the mechanism, a rearrangement of an R group with its pair of bonding electrons to an adjacent oxygen results in the removal of a hydroxide ion. Two more of these reactions with hydroperoxide will occur in order give a trialkylborate Part 3: This is the final part of the Oxidation process. In this part the trialkylborate reacts with aqueous NaOH to give the alcohol and sodium borate. If you need additional visuals to aid you in understanding the mechanism, click on the outside links provided here that will take you to other pages and media that are very helpful as well. Stereochemistry of hydroboration The hydroboration reaction is among the few simple addition reactions that proceed cleanly in a syn fashion. As noted above, this is a single-step reaction. Since the bonding of the double bond carbons to boron and hydrogen is concerted, it follows that the geometry of this addition must be syn. Furthermore, rearrangements are unlikely inasmuch as a discrete carbocation intermediate is never formed. These features are illustrated for the hydroboration of α-pinene. Since the hydroboration procedure is most commonly used to hydrate alkenes in an anti-Markovnikov fashion, we also need to know the stereoselectivity of the second oxidation reaction, which substitutes a hydroxyl group for the boron atom. Independent study has shown this reaction takes place with retention of configuration so the overall addition of water is also syn. The hydroboration of α-pinene also provides a nice example of steric hindrance control in a chemical reaction. In the less complex alkenes used in earlier examples the plane of the double bond was often a plane of symmetry, and addition reagents could approach with equal ease from either side. In this case, one of the methyl groups bonded to C-6 (colored blue in the equation) covers one face of the double bond, blocking any approach from that side. All reagents that add to this double bond must therefore approach from the side opposite this methyl. Outside links http://en.wikipedia.org/wiki/Hydroboration-oxidation http://bcs.whfreeman.com/vollhardtschore4e/cat_010/ch12/12010-03.htm http://www.chemhelper.com/hydroboration.html http://www.cartage.org.lb/en/themes/...roboration.htm http://www.organic-chemistry.org/nam...oboration.shtm References Vollhardt, Peter, and Neil Shore. Organic Chemistry: Structure and Function . 5th. New York: W.H. Freeman and Company, 2007. Foote, S. Christopher, and William H. Brown. Organic Chemistry . 5th. Belmont, CA: Brooks/Cole Cengage Learning, 2005. Bruice, Paula Yurkanis. Oragnic Chemistry . 5th. CA. Prentice Hall, 2006. Bergbreiter E. David , and David P. Rainville. Stereochemistry of hydroboration-oxidation of terminal alkenes. J. Org. Chem. , 1976 , 41 (18), pp 3031–3033 Ilich, Predrag-Peter; Rickertsen, Lucas S., and Becker Erienne. Polar Addition to C=C Group: Why Is Anti-Markovnikov Hydroboration-Oxidation of Alkenes Not "Anti-"? Journal of Chemical Education., 2006, v83, n11, pg 1681-1685 Problems What are the products of these following reactions? #1. #2. #3. Draw the structural formulas for the alcohols that result from hydroboration-oxidation of the alkenes shown. #4. #5. (E)-3-methyl-2-pentene If you need clarification or a reminder on the nomenclature of alkenes refer to the link below on naming the alkenes. Answers #1. #2. #3. #4. #5. Exercises Questions Q8.5.1 Write out the reagents or products (A–D) shown in the following reaction schemes. Solutions S8.5.1 Contributors Dr. Dietmar Kennepohl FCIC (Professor of Chemistry, Athabasca University ) Prof. Steven Farmer ( Sonoma State University ) Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) Jim Clark ( Chemguide.co.uk ) William Reusch, Professor Emeritus ( Michigan State U. ), Virtual Textbook of Organic Chemistry |
Courses/Chandler_Gilbert_Community_College/Fundamental_Organic_ala_Mech/04%3A_Functional_Groups/4.06%3A_Naming_Aromatic_Compounds | Simple aromatic hydrocarbons come from two main sources: coal and petroleum. Coal is an enormously complex mixture consisting primarily of large arrays of benzene-like rings joined together. Thermal breakdown of coal occurs when heated to 1000 °C in the absence of air, and a mixture of volatile products called coal tar boils off. Fractional distillation of coal tar yields benzene, toluene, xylene (dimethylbenzene), naphthalene, and a host of other aromatic compounds (Figure \(\PageIndex{1}\)). Unlike coal, petroleum contains few aromatic compounds and consists largely of alkanes (see Chapter 3 Chemistry Matters ). During petroleum refining, however, aromatic molecules are formed when alkanes are passed over a catalyst at about 500 °C under high pressure. Aromatic substances, more than any other class of organic compounds, have acquired a large number of nonsystematic names. IUPAC rules discourage the use of most such names but do allow some of the more widely used ones to be retained (Figure \(\PageIndex{1}\)). Thus, methylbenzene is known commonly as toluene; hydroxybenzene as phenol; aminobenzene as aniline; and so on. Structure Name NaN Toluene (bp 111 °C) NaN Phenol (mp 43 °C) NaN Aniline (bp 184 °C) NaN Acetophenone (mp 21 °C) NaN Benzaldehyde (bp 178 °C) NaN Benzoic acid (mp 122 °C) NaN ortho-Xylene (bp 144 °C) NaN Styrene (bp 145 °C) Monosubstituted benzenes are named systematically in the same manner as other hydrocarbons, with - benzene as the parent name. Thus, C 6 H 5 Br is bromobenzene, C 6 H 5 NO 2 is nitrobenzene, and C 6 H 5 CH 2 CH 2 CH 3 is propylbenzene. Alkyl-substituted benzenes are sometimes referred to as arenes and are named in different ways depending on the size of the alkyl group. If the alkyl substituent is smaller than the ring (six or fewer carbons), the arene is referred to as an alkyl-substituted benzene. If the alkyl substituent is larger than the ring (seven or more carbons), the compound is referred to as a phenyl-substituted alkane. The name phenyl , pronounced fen -nil and sometimes abbreviated as Ph or Ф (Greek phi ), is used for the –C 6 H 5 unit when the benzene ring is considered a substituent. The word is derived from the Greek pheno (“I bear light”), commemorating the discovery of benzene by Michael Faraday in 1825 from the oily residue left by the illuminating gas used in London street lamps. In addition, the name benzyl is used for the C 6 H 5 CH 2 – group. Disubstituted benzenes are named using the prefixes ortho ( o ) , meta ( m ) , or para ( p ) . An ortho-disubstituted benzene has its two substituents in a 1,2 relationship on the ring, a meta-disubstituted benzene has its two substituents in a 1,3 relationship, and a para-disubstituted benzene has its substituents in a 1,4 relationship. The ortho, meta, para system of nomenclature is also useful when discussing reactions. For example, we might describe the reaction of bromine with toluene by saying, “Reaction occurs at the para position,” meaning at the position para to the methyl group already present on the ring. As with cycloalkanes (Section 4.1), benzenes with more than two substituents are named by choosing a point of attachment as carbon 1 and numbering the substituents on the ring so that the second substituent has as low a number as possible. If ambiguity still exists, number so that the third or fourth substituent has as low a number as possible, until a point of difference is found. The substituents are listed alphabetically when writing the name. Note in the second and third examples shown that - phenol and - toluene are used as the parent names rather than - benzene . Any of the monosubstituted aromatic compounds shown in Figure \(\PageIndex{1}\) can serve as a parent name, with the principal substituent (–OH in phenol or –CH 3 in toluene) attached to C1 on the ring. Exercise \(\PageIndex{1}\) Tell whether the following compounds are ortho-, meta-, or para-disubstituted: Answer Meta Para Ortho Exercise \(\PageIndex{2}\) Answer m-Bromochlorobenzene (3-Methylbutyl)benzene p-Bromoaniline 2,5-Dichlorotoluene 1-Ethyl-2,4-dinitrobenzene 1,2,3,5-Tetramethylbenzene Exercise \(\PageIndex{3}\) Draw structures corresponding to the following IUPAC names: p-Bromochlorobenzene p-Bromotoluene m-Chloroaniline 1-Chloro-3,5-dimethylbenzene Answer |
Courses/Chabot_College/Introduction_to_General_Organic_and_Biochemistry/11%3A_Acids_and_Bases/11.03%3A_Acid_Dissociation_Constants | Learning Objectives Write the acid dissociation constant (\(K_a\)) expression. Determine the relative strength of an acid using the (\(K_a\)) value. The ionization for a general weak acid, \(\ce{HA}\), can be written as follows: \[\ce{HA} \left( aq \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{A^-} \left( aq \right) \nonumber \] Because the acid is weak, an equilibrium expression can be written. An acid ionization constant \(\left( K_\text{a} \right)\) is the equilibrium constant for the ionization of an acid. \[K_\text{a} = \frac{\left[ \ce{H^+} \right] \left[ \ce{A^-} \right]}{\left[ \ce{HA} \right]} \nonumber \] The acid ionization represents the fraction of the original acid that has been ionized in solution. Therefore, the numerical value of \(K_\text{a}\) is a reflection of the strength of the acid. Weak acids with relatively higher \(K_\text{a}\) values are stronger than acids with relatively lower \(K_\text{a}\) values. Because strong acids are essentially \(100\%\) ionized, the concentration of the acid in the denominator is nearly zero and the \(K_\text{a}\) value approaches infinity. For this reason, \(K_\text{a}\) values are generally reported for weak acids only. The table below is a listing of acid ionization constants for several acids. Note that polyprotic acids have a distinct ionization constant for each ionization step, with each successive ionization constant being smaller than the previous one. Name of Acid Ionization Equation \(K_\text{a}\) Sulfuric acid \(\ce{H_2SO_4} \rightleftharpoons \ce{H^+} + \ce{HSO_4^-}\) \(\ce{HSO_4} \rightleftharpoons \ce{H^+} + \ce{SO_4^{2-}}\) very large \(1.3 \times 10^{-2}\) Hydrofluoric acid \(\ce{HF} \rightleftharpoons \ce{H^+} + \ce{F^-}\) \(7.1 \times 10^{-4}\) Nitrous acid \(\ce{HNO_2} \rightleftharpoons \ce{H^+} + \ce{NO_2^-}\) \(4.5 \times 10^{-4}\) Benzoic acid \(\ce{C_6H_5COOH} \rightleftharpoons \ce{H^+} + \ce{C_6H_5COO^-}\) \(6.5 \times 10^{-5}\) Acetic acid \(\ce{CH_3COOH} \rightleftharpoons \ce{H^+} + \ce{CH_3COO^-}\) \(1.8 \times 10^{-5}\) Carbonic acid \(\ce{H_2CO_3} \rightleftharpoons \ce{H^+} + \ce{HCO_3^-}\) \(\ce{HCO_3^-} \rightleftharpoons \ce{H^+} + \ce{CO_3^{2-}}\) \(4.2 \times 10^{-7}\) \(4.8 \times 10^{-11}\) Hydrofluoric acid \(HF_{(aq)}\) reacts directly with glass (very few chemicals react with glass). Hydrofluoric acid is used in glass etching. |
Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Molecular_and_Atomic_Spectroscopy/01_In_Class_Problems%3A_Molecular_Spectroscopy/04_In-class_Questions%3A__Molecular_Luminescence | Energy level diagrams for organic molecules Draw an energy level diagram for a typical organic compound with \(\pi\) and \(\pi\) * orbitals and indicate which orbitals are filled and which are empty. Now consider the electron spin possibilities for the ground and excited state. Are there different possible ways to orient the spins (if so, these represent different spin states). Do you think these different spin states have different energies? Which one do you expect to be lower in energy? If the spin state is defined as (2S + 1) where S represents the total electronic spin for the system, try to come up with names for the ground and possible excited states for the system that are based on their spin state. Draw a diagram of the energy levels for such a molecule. Draw arrows for the possible transitions that could occur for the molecule. What do you expect for the lifetime of an electron in the T 1 state? Why is phosphorescence emission weak in most substances? Which transition (\(\pi\) *- \(\pi\) or \(\pi\) *- n) would have a higher fluorescent intensity? Justify your answer. Instrumental considerations for luminescence measurements What would constitute the basic instrumental design of a fluorescence spectrophotometer? What would be the difference between an excitation and emission spectrum in fluorescence spectroscopy? Draw representative examples of the excitation and emission spectrum for a molecule. Describe a way to measure the phosphorescence spectrum of a species that is not compromised by the presence of any fluorescence emission. If performing quantitative analysis in fluorescence spectroscopy, which wavelengths would you select from the spectra you drew in the problem above? Which method is more sensitive, absorption or fluorescence spectroscopy? Variables that influence fluorescence measurements What variables influence fluorescence measurements? For each variable, describe its relationship to the intensity of fluorescence emission. Consider the reaction shown below for the dissociation of 2-naphthol. This reaction may be either slow (slow exchange) or fast (fast exchange) on the time scale of fluorescence spectroscopy. Draw the series of spectra that would result for an initial concentration of 2-naphthol of 10 -6 M if the pH was adjusted to 2, 8.5, 9.5, 10.5, and 13 and slow exchange occurred. Draw the spectra at the same pH when the exchange rate is fast. Devise a procedure that might allow you to determine the pKa of a weak acid such as 2-naphthol. Which compound will have a higher quantum yield: anthracene or diphenylmethane? |
Courses/Mt._San_Antonio_College/Chem_10_-_Chemistry_for_Allied_Health_Majors_(1st_semester)/11%3A_Nuclear_Chemistry | Most chemists pay little attention to the nucleus of an atom except to consider the number of protons it contains because that determines an element’s identity. However, in nuclear chemistry, the composition of the nucleus and the changes that occur there are very important. Applications of nuclear chemistry may be more widespread than you realize. Many people are aware of nuclear power plants and nuclear bombs, but nuclear chemistry also has applications ranging from smoke detectors to medicine, from the sterilization of food to the analysis of ancient artifacts. In this chapter, we will examine some of the basic concepts of nuclear chemistry and some of the nuclear reactions that are important in our everyday lives. 11.1: Nuclear Reactions Nuclear reactions are very different from chemical reactions. In chemical reactions, atoms become more stable by participating in a transfer of electrons or by sharing electrons with other atoms. In nuclear reactions, it is the nucleus of the atom that gains stability by undergoing a change of some kind. The energies that are released in nuclear reactions are many orders of magnitude greater than the energies involved in chemical reactions. 11.2: The Discovery and Nature of Radioactivity In 1896, Henri Becquerel found that a uranium compound placed near a photographic plate made an image on the plate and reasoned that the compound was emitting some kind of radiation. Further investigations showed that the radiation was a combination of particles and electromagnetic rays, with its ultimate source as the atomic nucleus. These emanations were ultimately called, collectively, radioactivity. The major types of radioactivity include alpha particles, beta particles, and gamma rays. 11.3: Stable and Unstable Isotopes In nuclear reactions, it is the nucleus of the atom that gains stability by undergoing a change of some kind. A radioisotope is an isotope of an element that is unstable and undergoes radioactive decay. The energies that are released in nuclear reactions are many orders of magnitude greater than the energies involved in chemical reactions. Unlike chemical reactions, nuclear reactions are not noticeably affected by changes in environmental conditions, such as temperature or pressure. 11.4: Nuclear Decay Unstable nuclei spontaneously emit radiation in the form of particles and energy. This generally changes the number of protons and/or neutrons in the nucleus, resulting in a more stable nuclide. One type of a nuclear reaction is radioactive decay, a reaction in which a nucleus spontaneously disintegrates into a slightly lighter nucleus, accompanied by the emission of particles, energy, or both. 11.5: Radioactive Half-Life Natural radioactive processes are characterized by a half-life, the time it takes for half of the material to decay radioactively. The amount of material left over after a certain number of half-lives can be easily calculated. 11.6: Detecting and Measuring Radiation We previously used mass to indicate the amount of radioactive substance present. However, this is only one of several units used to express amounts of radiation. Some units describe the number of radioactive events occurring per unit time, while others express the amount of a person's exposure to radiation. 11.7: Artificial Transmutation Although the conversion of one element to another is the basis of natural radioactive decay, it is also possible to convert one element to another artificially. The conversion of one element to another is the process of transmutation. 11.8: Nuclear Fission and Nuclear Fusion Nuclear energy comes from tiny mass changes in nuclei as radioactive processes occur. In fission, large nuclei break apart and release energy; in fusion, small nuclei merge together and release energy. |
Bookshelves/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/22%3A_Organic_Chemistry/22.13%3A_Amines | Learning Objectives Name the typical reactions that take place with amines. Describe heterocyclic amines. Recall that ammonia (NH 3 ) acts as a base because the nitrogen atom has a lone pair of electrons that can accept a proton. Amines also have a lone electron pair on their nitrogen atoms and can accept a proton from water to form substituted ammonium (NH 4 + ) ions and hydroxide (OH − ) ions: As a specific example, methylamine reacts with water to form the methylammonium ion and the OH − ion. Nearly all amines, including those that are not very soluble in water, will react with strong acids to form salts soluble in water. Amine salts are named like other salts: the name of the cation is followed by the name of the anion. Example \(\PageIndex{1}\) What are the formulas of the acid and base that react to form [CH 3 NH 2 CH 2 CH 3 ] + CH 3 COO − ? Solution The cation has two groups—methyl and ethyl—attached to the nitrogen atom. It comes from ethylmethylamine (CH 3 NHCH 2 CH 3 ). The anion is the acetate ion. It comes from acetic acid (CH 3 COOH). Exercise \(\PageIndex{1}\) What are the formulas of the acid and base that react to form \(\ce{(CH3CH2CH2)3NH^{+}I^{−}}\)? To Your Health: Amine Salts as Drugs Salts of aniline are properly named as anilinium compounds, but an older system, still in use for naming drugs, identifies the salt of aniline and hydrochloric acid as “aniline hydrochloride.” These compounds are ionic—they are salts—and the properties of the compounds (solubility, for example) are those characteristic of salts. Many drugs that are amines are converted to hydrochloride salts to increase their solubility in aqueous solution. Heterocyclic Amines Looking back at the various cyclic hydrocarbons discussed previously, we see that all the atoms in the rings of these compounds are carbon atoms. In other cyclic compounds, called heterocyclic compounds (Greek heteros , meaning “other”), nitrogen, oxygen, sulfur, or some other atom is incorporated in the ring. Many heterocyclic compounds are important in medicine and biochemistry. Some compose part of the structure of the nucleic acids, which in turn compose the genetic material of cells and direct protein synthesis. Many heterocyclic amines occur naturally in plants. Like other amines, these compounds are basic. Such a compound is an alkaloid , a name that means “like alkalis.” Many alkaloids are physiologically active, including the familiar drugs caffeine, nicotine, and cocaine. To Your Health: Three Well-Known Alkaloids Caffeine is a stimulant found in coffee, tea, and some soft drinks. Its mechanism of action is not well understood, but it is thought to block the activity of adenosine, a heterocyclic base that acts as a neurotransmitter, a substance that carries messages across a tiny gap (synapse) from one nerve cell (neuron) to another cell. The effective dose of caffeine is about 200 mg, corresponding to about two cups of strong coffee or tea. Nicotine acts as a stimulant by a different mechanism; it probably mimics the action of the neurotransmitter acetylcholine. People ingest this drug by smoking or chewing tobacco. Its stimulant effect seems transient, as this initial response is followed by depression. Nicotine is highly toxic to animals. It is especially deadly when injected; the lethal dose for a human is estimated to be about 50 mg. Nicotine has also been used in agriculture as a contact insecticide. Cocaine acts as a stimulant by preventing nerve cells from taking up dopamine, another neurotransmitter, from the synapse. High levels of dopamine are therefore available to stimulate the pleasure centers of the brain. The enhancement of dopamine action is thought to be responsible for cocaine’s “high” and its addictive properties. After the binge, dopamine is depleted in less than an hour. This leaves the user in a pleasureless state and (often) craving more cocaine. Cocaine is used as the salt cocaine hydrochloride and in the form of broken lumps of the free (unneutralized) base, which is called crack cocaine . \[\underbrace{\ce{C17H21O4N}}_{\text{cocaine (freebase)}} + \ce{HCl ->} \underbrace{\ce{C17H21O4NH^{+}Cl^{-}}}_{\text{cocaine hydrochloride}} \nonumber \] Because it is soluble in water, cocaine hydrochloride is readily absorbed through the watery mucous membranes of the nose when it is snorted. Crack cocaine is more volatile than cocaine hydrochloride. It vaporizes at the temperature of a burning cigarette. When smoked, cocaine reaches the brain in 15 s. Summary Amines are bases; they react with acids to form salts. Salts of aniline are properly named as anilinium compounds, but an older system is used to name drugs: the salts of amine drugs and hydrochloric acid are called “hydrochlorides.” Heterocyclic amines are cyclic compounds with one or more nitrogen atoms in the ring. |
Bookshelves/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/01%3A_Chemistry/1.04%3A_Chemistry_-_A_Study_of_Matter_and_Its_Changes | Learning Objectives Separate physical from chemical properties. Label a change as chemical or physical. C h emistry is the study of matter—what it consists of, what its properties are, and how it changes. Matter is anything that has mass and takes up space—that is, anything that is physic ally real. Some things are easily identified as matter—the screen on which you are reading this book, for example. Others are not so obvious. Because we move so easily through air, we sometimes forget that it, too, is matter. Because of this, chemistry is a science that has its fingers in just about everything. Even a description of the ingredients in a cake, and how those ingredients change when the cake is baked, is chemistry! Physical and Chemical Properties All matter has physical and chemical properties. Physical properties are characteristics that scientists can measure without changing the composition of the sample under study, such as mass, color, and volume (the amount of space occupied by a sample). Chemical properties describe the characteristic ability of a substance to react to form new substances; they include its flammability and susceptibility to corrosion. All samples of a pure substance have the same chemical and physical properties. For example, pure copper is always a reddish-brown solid (a physical property) and always dissolves in dilute nitric acid to produce a blue solution and a brown gas (a chemical property). Physical Property A physical property is a characteristic of a substance that can be observed or measured without changing the identity of the substance. Silver is a shiny metal that conducts electricity very well. It can be molded into thin sheets—a property called malleability. Salt is dull and brittle and conducts electricity when it has been dissolved into water, which it does quite easily. Physical properties of matter include color, hardness, malleability, solubility, electrical conductivity, density, melting point, and boiling point. In Table \(\PageIndex{1}\), notice that corn oil has a lower mass to volume ratio than water. This means that when added to water, corn oil will “float.” Substance Density at 25°C (g/cm3) blood 1.035 body fat 0.918 whole milk 1.030 corn oil 0.922 mayonnaise 0.910 honey 1.420 Hardness helps determine how an element (especially a metal) might be used. Many elements are fairly soft (silver and gold, for example) while others (such as titanium, tungsten, and chromium) are much harder. Carbon is an interesting example of hardness. In graphite, (the "lead" found in pencils) the carbon is very soft, while the carbon in a diamond is roughly seven times as hard. Melting and boiling points are somewhat unique identifiers, especially of compounds. In addition to giving some idea of the compound's identity, important information can be obtained about the purity of the material. Chemical Property A chemical property of matter describes its "potential" to undergo some chemical change or reaction by virtue of its composition; as well as what elements, electrons, and bonding are present to give the potential for chemical change. It is quite difficult to define a chemical property without using the word "change". Eventually, students of chemistry should be able to look at the formula of a compound and state a chemical property. For example, hydrogen has the potential to ignite and explode, given the right conditions—this is a chemical property. Metals in general have the chemical property of reacting with acids. Zinc reacts with hydrochloric acid to produce hydrogen gas—this is a chemical property. A chemical property of iron is that it is capable of combining with oxygen to form iron oxide, the chemical name of rust (Figure \(\PageIndex{2}\)). The more general term for rusting and other similar processes is corrosion. Other terms that are commonly used in descriptions of chemical changes are burn , rot , explode , decompose , and ferment . Chemical properties are very useful in identifying substances. However, unlike physical properties, chemical properties can only be observed as the substance is in the process of being changed into a different substance. Physical Properties Chemical Properties Gallium metal melts at 30 oC Iron metal rusts. Mercury is a very dense liquid. A green banana turns yellow when it ripens. Gold is shiny. A dry piece of paper burns. Example \(\PageIndex{1}\) Which of the following is a chemical property of iron? Iron corrodes in moist air. Density = 7.874 g/cm 3 Iron is soft when pure. Iron melts at 1808 K. Solution "Iron corrodes in air" is the only chemical property of iron from the list. Exercise \(\PageIndex{1A}\) Which of the following is a physical property of matter? corrosiveness pH (acidity) density flammability Answer 3. Exercise \(\PageIndex{1B}\) Which of the following is a chemical property? flammability melting point boiling point density Answer 1. Physical and Chemical Changes Change is happening all around us all of the time. Just as chemists have classified elements and compounds, they have also classified types of changes. Changes are either classified as physical or chemical changes. Chemists learn a lot about the nature of matter by studying the changes that matter can undergo. Chemists make a distinction between two different types of changes that they study—physical changes and chemical changes. Physical Change Physical changes are changes in which no bonds are broken or formed. This means that the same types of compounds or elements that were there at the beginning of the change are there at the end of the change. Because the ending materials are the same as the beginning materials, the properties (such as color, boiling point, etc) will also be the same. Physical changes involve moving molecules around, but not changing them. Some types of physical changes include: Changes of state (changes from a solid to a liquid or a gas and vice versa). Separation of a mixture. Physical deformation (cutting, denting, stretching). Making solutions (special kinds of mixtures). As an ice cube melts, its shape changes as it acquires the ability to flow. However, its composition does not change. Melting is an example of a physical change. A physical change is a change to a sample of matter in which some properties of the material change, but the identity of the matter does not. When liquid water is heated, it changes to water vapor. However, even though the physical properties have changed, the molecules are exactly the same as before. Each water molecule still contains two hydrogen atoms covalently bonded to one oxygen atom. When you have a jar containing a mixture of pennies and nickels, and you sort the mixture so that you have one pile of pennies and another pile of nickels, you have not altered the identity of either the pennies or the nickels—you've merely separated them into two groups. This would be an example of a physical change. Similarly, if you have a piece of paper, you don't change it into something other than a piece of paper by ripping it up. What was paper before you started tearing is still paper when you're done. Again, this is an example of a physical change. Physical changes can further be classified as reversible or irreversible. The melted ice cube may be refrozen, so melting is a reversible physical change. Physical changes that involve a change of state are all reversible. Other changes of state include vaporization (liquid to gas), freezing (liquid to solid), and condensation (gas to liquid). Dissolution is also a reversible physical change. When salt is dissolved into water, the salt is said to have entered the aqueous state. The salt may be regained by boiling off the water, leaving the salt behind. Chemical Change Chemical changes occur when bonds are broken and/or formed between molecules or atoms. This means that one substance with a certain set of properties (such as melting point, color, taste, etc.) is turned into a different substance with different properties. Chemical changes are frequently harder to reverse than physical changes. One good example of a chemical change is burning a candle. The act of burning paper actually results in the formation of new chemicals (carbon dioxide and water, to be exact) from the burning of the wax. Another example of a chemical change is what occurs when natural gas is burned in a furnace. This time, on the left there is a molecule of methane, \(\ce{CH_4}\), and two molecules of oxygen, \(\ce{O_2}\); on the right are two molecules of water, \(\ce{H_2O}\), and one molecule of carbon dioxide, \(\ce{CO_2}\). In this case, not only has the appearance changed, but the structure of the molecules has also changed. The new substances do not have the same chemical properties as the original ones. Therefore, this is a chemical change. We can't actually see molecules breaking and forming bonds, although that's what defines chemical changes. We have to make other observations to indicate that a chemical change has happened. Some of the evidence for chemical change will involve the energy changes that occur in chemical changes, but some evidence involves the fact that new substances with different properties are formed in a chemical change. Observations that help to indicate chemical change include: Temperature changes (temperature increase or decrease). Light given off. Unexpected color changes (a substance with a different color is made, rather than just mixing the original colors together). Bubbles are formed (but the substance is not boiling—you made a substance that is a gas at the temperature of the beginning materials, instead of a liquid). Different smell or taste (do not taste your chemistry experiments, though!). A solid forms if two clear liquids are mixed (look for floaties —technically called a precipitate). Example \(\PageIndex{2}\) Label each of the following changes as a physical or chemical change. Give evidence to support your answer. Boiling water. A nail rusting. A green solution and colorless solution are mixed. The resulting mixture is a solution with a pale green color. Two colorless solutions are mixed. The resulting mixture has a yellow precipitate. Solution Physical: boiling and melting are physical changes. When water boils, no bonds are broken or formed. The change could be written as: \(\ce{H_2O} \left( l \right) \rightarrow \ce{H_2O} \left( g \right)\) Chemical: The dark grey nail changes color to form an orange flaky substance (the rust); this must be a chemical change. Color changes indicate chemical change. The following reaction occurs: \(\ce{Fe} + \ce{O_2} \rightarrow \ce{Fe_2O_3}\) Physical: because none of the properties changed, this is a physical change. The green mixture is still green and the colorless solution is still colorless. They have just been combined. No color change, or other evidence of chemical change, occurred. Chemical: the formation of a precipitate and the color change from colorless to yellow indicates a chemical change. Exercise \(\PageIndex{2}\) Label each of the following changes as a physical or chemical change. A mirror is broken. An iron nail corroded in moist air. Copper metal is melted. A catalytic converter changes nitrogen dioxide to nitrogen gas and oxygen gas. Answer 1: physical change Answer 2: chemical change Answer 3: physical change Answer 4: chemical change Summary A physical property is a characteristic of a substance that can be observed or measured without changing the identity of the substance. Physical properties include color, density, hardness, and melting and boiling points. A chemical property describes the ability of a substance to undergo a specific chemical change. To identify a chemical property, we look for a chemical change. Physical changes are changes that do not alter the identity of a substance. Chemical changes are changes that occur when one substance is turned into another substance. Chemical changes are frequently harder to reverse than physical changes. Observations that indicate a chemical change occurred include color change, temperature change, light given off, formation of bubbles, and formation of a precipitate. |
Courses/Furman_University/CHM101%3A_Chemistry_and_Global_Awareness_(Gordon)/01%3A_Introduction_to_Chemistry/1.05%3A_Applying_Properties_-_MSDS_SDS | Learning Objectives Review physical and chemical properties. Know the purpose of a MSDS/SDS. Explain the bolded words on this page. What is a MSDS/SDS? A Material Safety Data Sheet (MSDS) or Safety Data Sheet (SDS) is designed to provide both workers and emergency personnel with the proper procedures for handling or working with a particular substance. MSDS's include information such as physical data (melting point, boiling point, flash point, etc.), toxicity, health effects, first aid, reactivity, storage, disposal, protective equipment, and spill/leak procedures. These are of particular use if a spill or other accident occurs. The MSDS's are available from many sources such as the chemistry store, health and safety offices, material producing companies, etc. Any business or industry that has chemicals on site is required to have access to MSDS files for employers to be able to access at any time. If an employee is concerned about their exposure to chemicals, they can peruse an MSDS to obtain helpful information. Likewise, one could research household chemicals that could be dangerous to humans and animals. Many substances found around the home (pesticides, fuels, and cleaners) can be hazardous or harmless. Other Terms Used in MSDS/SDS Aspects of toxicity and how chemicals will affect a body can also be found in this resource. Along with physical and chemical properties, you should be familiar with the terms below that could also be found in an MSDS/SDS document. Mutagen: able to alter DNA and could result in carcinogenesis (a chemical example would be nitrosamine) Carcinogen: able to cause cancer (this can get complicated, animal and human carcinogens both exist). There are very few chemicals classified as being human carcinogens. (benzene, radon, asbestos) Neurotoxin: affects the nervous system (chemical examples include lead and mercury) Teratogen : affects developing fetus or embryo (pharmaceutical examples would be thalidomide and Accutane) Corrosive : can chemically destroy things easily (acids/bases are examples) Flammable/Combustible : able to catch on fire (examples would include gasoline and acetone) Volatile : a liquid that can transform into gas easily (examples would include gasoline and acetone) MSDS/SDS's can be several pages long. A complete MSDS/SDS guide for acetic acid (also known as vinegar) can be viewed by clicking on this here . Although this chemical might appear dangerous, it used in the production of vinegar, ketchup, and barbeque sauce. After looking over vinegar's entire MSDS/SDS sheet, write down any questions for your instructor regarding unfamiliar terminology. Exercise \(\PageIndex{1}\) After looking at the MSDS for acetic acid, answer the questions below: Is this chemical a carcinogen? Does acetic acid react with metals? What odor does acetic acid produce? If this chemical is ingested, should you induce vomiting to remove it from the body? Does this chemical have a color? Answer a Acetic acid is not a carcinogen. It will not alter cause cancer. Answer b Acetic acid reacts with metals. This is a chemical property of this substance. Answer c Acetic acid smells like vinegar. Answer d Acetic acid is very corrosive and will eat biological tissues. If swallowed, vomiting should not be induced. Answer e This chemical is colorless Figure \(\PageIndex{3}\): IARC's carcinogenic classifications. Andy Brunning @ Compound Interest. |
Bookshelves/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/03%3A_An_Introduction_to_Organic_Compounds%3A_Nomenclature_Physical_Properties_and_Representation_of_Structure/3.10%3A_Conformers_of_Cyclohexane | Cycloalkanes are very important in components of food, pharmaceutical drugs, and much more. However, to use cycloalkanes in such applications, we must know the effects, functions, properties, and structures of cycloalkanes. Cycloalkanes are alkanes that are in the form of a ring; hence, the prefix cyclo-. Stable cycloalkanes cannot be formed with carbon chains of just any length. Recall that in alkanes, carbon adopts the sp 3 tetrahedral geometry in which the angles between bonds are 109.5°. For some cycloalkanes to form, the angle between bonds must deviate from this ideal angle, an effect known as angle strain . Additionally, some hydrogen atoms may come into closer proximity with each other than is desirable (become eclipsed), an effect called torsional strain . These destabilizing effects, angle strain and torsional strain are known together as ring strain . The smaller cycloalkanes, cyclopropane and cyclobutane, have particularly high ring strains because their bond angles deviate substantially from 109.5° and their hydrogens eclipse each other. Cyclopentane is a more stable molecule with a small amount of ring strain, while cyclohexane is able to adopt the perfect geometry of a cycloalkane in which all angles are the ideal 109.5° and no hydrogens are eclipsed; it has no ring strain at all. Cycloalkanes larger than cyclohexane have ring strain and are not commonly encountered in organic chemistry. Ring Strain and the Structures of Cycloalkanes There are many forms of cycloalkanes, such as cyclopropane, cyclobutane, cyclopentane, cyclohexane, among others. The process of naming cycloalkanes is the same as naming alkanes but the addition of the prefix cyclo- is required. Cyclobutane is in a form of a square, which is highly unfavorable and unstable (this will be explained soon). There are different drawings for cyclobutane, but they are equivalent to each other. Cyclobutane can reduce the ring string by puckering the square cyclobutane. Cyclopentane takes the shape of a pentagon and cyclohexane is in the shape of a hexagon. Chair Conformation of Cyclohexane - Equitorial and Axial There are two ways to draw cyclohexane because it can be in a hexagon shape or in a different conformational form called the chair conformation and the boat conformation. The chair conformation drawing is more favored than the boat because of the energy, the steric hindrance, and a new strain called the transannular strain . The boat conformation is not the favored conformation because it is less stable and has a steric repulsion between the two H's, shown with the pink curve. This is known as the transannular strain, which means that the strain results from steric crowding of two groups across a ring. The boat is less stable than the chair by 6.9 kcal/mol. The boat conformation, however, is flexible, and when we twist one of the C-C bonds, it reduces the transannular strain. When we twist the C-C bond in a boat, it becomes a twisted boat. Some Conformations of Cyclohexane Rings. (William Reusch, MSU) Although there are multiple ways to draw cyclohexane, the most stable and major conformer is the chair because is has a lower activation barrier from the energy diagram. Conformational Energy Profile of Cyclohexane. (William Reusch, MSU). The transition state structure is called a half chair . This energy diagram shows that the chair conformation is lower in energy; therefore, it is more stable. The chair conformation is more stable because it does not have any steric hindrance or steric repulsion between the hydrogen bonds. By drawing cyclohexane in a chair conformation, we can see how the H's are positioned. There are two positions for the H's in the chair conformation, which are in an axial or an equitorial formation. This is how a chair conformation looks, but you're probably wondering which H's are in the equitorial and axial form. Here are more pictures to help. These are hydrogens in the axial form. These hydrogens are in an equitorial form. Of these two positions of the H's, the equitorial form will be the most stable because the hydrogen atoms, or perhaps the other substituents, will not be touching each other. This is the best time to build a chair conformation in an equitorial and an axial form to demonstrate the stability of the equitorial form. Ring Strain Cycloalkanes tend to give off a very high and non-favorable energy, and the spatial orientation of the atoms is called the ring strain. When atoms are close together, their proximity is highly unfavorable and causes steric hindrance. The reason we do not want ring strain and steric hindrance is because heat will be released due to an increase in energy; therefore, a lot of that energy is stored in the bonds and molecules, causing the ring to be unstable and reactive. Another reason we try to avoid ring strain is because it will affect the structures and the conformational function of the smaller cycloalkanes. One way to determine the presence of ring strain is by its heat of combustion. By comparing the heat of combustion with the value measured for the straight chain molecule, we can determine the stability of the ring. There are two types of strain, which are eclipsing/torsional strain and bond angle strain. Bond angle strain causes a ring to have a poor overlap between the atoms, resulting in weak and reactive C-C bonds. An eclipsed spatial arrangement of the atoms on the cycloalkanes results in high energy. With so many cycloalkanes, which ones have the highest ring strain and are very unlikely to stay in its current form? The figures below show cyclopropane, cyclobutane, and cyclopentane, respectively. Cyclopropane is one of the cycloalkanes that has an incredibly high and unfavorable energy, followed by cyclobutane as the next strained cycloalkane. Any ring that is small (with three to four carbons) has a significant amount of ring strain; cyclopropane and cyclobutane are in the category of small rings. A ring with five to seven carbons is considered to have minimal to zero strain, and typical examples are cyclopentane, cyclohexane, and cycloheptane. However, a ring with eight to twelve carbons is considered to have a moderate strain, and if a ring has beyond twelve carbons, it has minimal strain. There are different types of ring strain: Transannular strain isdefined as the crowding of the two groups in a ring. Eclipsing strain, also known as torsional strain, is intramolecular strain due to the bonding interaction between two eclipsed atoms or groups. Bond angle strain is present when there is a poor overlap between the atoms. There must be an ideal bond angle to achieve the maximum bond strength and that will allow the overlapping of the atomic/hybrid orbitals. Cyclohexane Most of the time, cyclohexane adopts the fully staggered, ideal angle chair conformation . In the chair conformation, if any carbon-carbon bond were examined, it would be found to exist with its substituents in the staggered conformation and all bonds would be found to possess an angle of 109.5°. Cyclohexane in the chair conformation. (William Reusch, MSU). In the chair conformation, hydrogen atoms are labeled according to their location. Those hydrogens which exist above or below the plane of the molecule (shown with red bonds above) are called axial . Those hydrogens which exist in the plane of the molecule (shown with blue bonds above) are called equatorial . Although the chair conformation is the most stable conformation that cyclohexane can adopt, there is enough thermal energy for it to also pass through less favorable conformations before returning to a different chair conformation. When it does so, the axial and equatorial substituents change places. The passage of cyclohexane from one chair conformation to another, during which the axial substituents switch places with the equatorial substituents, is called a ring flip . Methylcyclohexane Methylcyclohexane is cyclohexane in which one hydrogen atom is replaced with a methyl group substituent. Methylcyclohexane can adopt two basic chair conformations: one in which the methyl group is axial, and one in which it is equatorial. Methylcyclohexane strongly prefers the equatorial conformation. In the axial conformation, the methyl group comes in close proximity to the axial hydrogens, an energetically unfavorable effect known as a 1,3-diaxial interaction (Figure 3). Thus, the equatorial conformation is preferred for the methyl group. In most cases, if the cyclohexane ring contains a substituent, the substituent will prefer the equatorial conformation. Methylcyclohexane in the chair conformation. (William Reusch, MSU). References Bachrach, Steven M. Computational Organic Chemistry . Wiley- Interscience. (2007). McMurray, John E. Organic Chemistry Sixth Edition. Brooks Cole. (2003). Vollhardt and Schore. Organic Chemistry Structure and Function Fifth Edition. New York. (2007). Problems Trans- 1,2-dimethylcyclopropane is more stable than cis-1,2-dimethylcyclopropane. Why? Drawing a picture of the two will help your explanation. Out of all the cycloalkanes, which one has the most ring strain and which one is strain free? Explain. Which of these chair conformations are the most stable and why? 3. What does it mean when people say "increase in heat leads to increase in energy" and how does that statement relate to ring strains? Why is that the bigger rings have lesser strains compared to smaller rings? Answers The cis isomer suffers from steric hindrance and has a larger heat of combustion. Cyclopropane- ring strain. Cyclohexane chair conformation- ring strain free. Top one is more stable because it is in an equitorial conformation. When assembling it with the OChem Molecular Structure Tool Kit, equitorial formation is more spread out. When there is an increase in heat there will be an increase of energy released therefore there will be a lot of energy stored in the bond and molecule making it unstable. Smaller rings are more compacts, which leads to steric hindrance and the angles for these smaller rings are harder to get ends to meet. Bigger rings tend to have more space and that the atoms attached to the ring won't be touching each other as much as atoms attached to the smaller ring. Contributors Jonathan Mooney (McGill University) |
Courses/University_of_North_Texas/UNT%3A_CHEM_1410_-_General_Chemistry_for_Science_Majors_I/Text/01%3A_Introduction_-_The_Ambit_of_Chemistry/1.09%3A_Density | The terms heavy and light are commonly used in two different ways. We refer to weight when we say that an adult is heavier than a child. On the other hand, something else is alluded to when we say that oak is heavier than balsa wood. A small shaving of oak would obviously weigh less than a roomful of balsa wood, but oak is heavier in the sense that a piece of given size weighs more than the same-size piece of balsa. What we are actually comparing is the mass per unit volume , that is, the density . In order to determine these densities, we might weigh a cubic centimeter of each type of wood. If the oak sample weighed 0.71 g and the balsa 0.15 g, we could describe the density of oak as 0.71 g cm –3 and that of balsa as 0.15 g cm –3 . (Note that the negative exponent in the units cubic centimeters indicates a reciprocal. Thus 1 cm –3 = 1/cm 3 and the units for our densities could be written as \(\frac{\text{g}}{\text{cm}^\text{3}}\), g/cm 3 , or g cm –3 . In each case the units are read as grams per cubic centimeter, the per indicating division.) We often abbreviate "cm 3 " as "cc", and 1 cm 3 = 1 mL exactly by definition. In general it is not necessary to weigh exactly 1 cm 3 of a material in order to determine its density. We simply measure mass and volume and divide volume into mass: \[\text{Density} = \dfrac{\text{mass}}{\text{volume}} \nonumber \] or \[\rho = \dfrac{m}{V} \quad \label{1} \] where \(ρ\) is the density, \(m\) is the mass, and \(V\) volume. Example \(\PageIndex{1}\): Density of Aluminum Calculate the density of a piece of aluminum whose mass is 37.42 g and which, when submerged, increases the water level in a graduated cylinder by 13.9 ml; an aluminum cylinder of mass 25.07 g, radius 0.750 cm, and height 5.25 cm. Solution a Since the submerged metal displaces its own volume, \[ \begin{align*} \text{Density} &= \rho =\dfrac{m}{V} \\[4pt] &= \dfrac{\text{37.42 g}}{\text{13.9 ml}} \\[4pt] &= \text{2.69 g}/\text{ml or 2.69 g ml}^{-\text{1}} \end{align*} \nonumber \] b The volume of the cylinder must be calculated first, using the formula \[ \begin{align*} \text{V} &= \pi r^\text{2} h\\[4pt] &= \text{3.142}\times\text{(0.750 cm)}^\text{2}\times\text{5.25 cm} \\[4pt] &= \text{9.278 718 8 cm}^\text{3} \end{align*} \nonumber \] Then \[ \rho = \dfrac{m}{V} = \dfrac{\text{25.07 g}}{\text{9.278 718 8 cm}^\text{3}} = \begin{cases} 2.70 \dfrac{\text{g}}{\text{cm}^\text{3}} \\ \text{2.70 g cm}^{-\text{3}} \\ \text{2.70 g}/ \text{cm}^\text{3} \end{cases} \nonumber \] which are all acceptable alternatives. Note that unlike mass or volume ( extensive properties ), the density of a substance is independent of the size of the sample ( intensive property ). Thus density is a property by which one substance can be distinguished from another. A sample of pure aluminum can be trimmed to any desired volume or adjusted to have any mass we choose, but its density will always be 2.70 g/cm 3 at 20°C. The densities of some common pure substances are listed below. Tables and graphs are designed to provide a maximum of information in a minimum of space. When a physical quantity (number × units) is involved, it is wasteful to keep repeating the same units. Therefore it is conventional to use pure numbers in a table or along the axes of a graph. A pure number can be obtained from a quantity if we divide by appropriate units. For example, when divided by the units gram per cubic centimeter, the density of aluminum becomes a pure number 2.70: \[\dfrac{\text{Density of aluminum}}{\text{1 g cm}^{-\text{3}}} = \dfrac{\text{2.70 g cm}^{-\text{3}}}{\text{1 g cm}^{-\text{3}}} = \text{2.70} \nonumber \] Therefore, a column in a table or the axis of a graph is conveniently labeled in the following form: Quantity/units This indicates the units that must be divided into the quantity to yield the pure number in the table or on the axis. This has been done in the second column of the table. Substance Density / g cm-3 Helium gas 0.000 16 Dry air 0.001 185 Gasoline 0.66-0.69 (varies) Kerosene 0.82 Benzene 0.880 Water 1.000 Carbon tetrachloride 1.595 Magnesium 1.74 Salt 2.16 Aluminum 2.70 Iron 7.87 Copper 8.96 Silver 10.5 Lead 11.34 Uranium 19.05 Gold 19.32 Converting Densities In our exploration of density , notice that chemists may express densities differently depending on the subject. The density of pure substances may be expressed in kg/m 3 in some journals which insist on strict compliance with SI units; densities of soils may be expressed in lb/ft 3 in some agricultural or geological tables; the density of a cell may be expressed in mg/µL; and other units are in common use. It is easy to transform densities from one set of units to another, by multiplying the original quantity by one or more unity factors : Example \(\PageIndex{2}\): Density of Water Convert the density of water, 1 g/cm 3 to (a) lb/cm 3 and (b) lb/ft 3 Solution a The equality \(\text{454 g} = \text{1 lb}\) can be used to write two unity factors, \[ \dfrac{\text{454 g}}{\text{1 lb}}\nonumber \] or \[\dfrac{\text{1 lb}}{\text{454 g}} \nonumber \] The given density can be multiplied by one of the unity factors to get the desired result. The correct conversion factor is chosen so that the units cancel: \[ \text{1} \dfrac{\text{g}}{\text{cm}^\text{3}} \times \dfrac{\text{1 lb}}{\text{454 g}} = \text{0.002203} \dfrac{\text{lb}}{\text{cm}^\text{3}}\nonumber \] b Similarly, the equalities \(\text{2.54 cm} = \text{1 inch}\), and \(\text{12 inches} = \text{1 ft}\) can be use to write the unity factors: \[ \dfrac{\text{2.54 cm}}{\text{1 in}} \text{, } \dfrac{\text{1 in}}{\text{2.54 cm}} \text{, } \dfrac{\text{12 in}}{\text{1 ft}} \text{ and } \dfrac{\text{1 ft}}{\text{12 in}} \nonumber \] In order to convert the cm 3 in the denominator of 0.002203 to in 3 , we need to multiply by the appropriate unity factor three times, or by the cube of the unity factor: \[ \text{0.002203} \dfrac{\text{g}}{\text{cm}^\text{3}} \times \dfrac{\text{2.54 cm}}{\text{1 in}} \times \dfrac{\text{2.54 cm}}{\text{1 in}} \times \dfrac{\text{2.54 cm}}{\text{1 in}}\nonumber \] or \[ \text{0.002203} \dfrac{\text{g}}{\text{cm}^\text{3}} \times \left(\dfrac{\text{2.54 cm}}{\text{1 in}}\right)^\text{3} = \text{0.0361 lb}/ \text{in}^\text{3}\nonumber \] This can then be converted to lb/ft 3 : \[ \text{0.0361 lb}/ \text{in}^\text{3}\times \left(\dfrac{\text{12 in}}{\text{1 ft}}\right)^\text{3} = \text{62.4 lb}/\text{ft}^\text{3}\nonumber \] It is important to notice that we have used conversion factors to convert from one unit to another unit of the same parameter |
Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/24%3A_Organonitrogen_Compounds_II_-_Amides_Nitriles_and_Nitro_Compounds/24.02%3A_Amides_as_Acids_and_Bases | Acidity Amides with \(\ce{N-H}\) bonds are weakly acidic, the usual \(K_a\) being about \(10^{-16}\): Nonetheless, amides clearly are far more acidic than ammonia \(\left( K_a \sim 10^{-33} \right)\), and this difference reflects a substantial degree of stabilization of the amide anion. However, amides still are very weak acids (about as weak as water) and, for practical purposes, are regarded as neutral compounds. Where there are two carbonyl groups to stabilize the amide anion, as in the 1,2-benzenedicarboximide (phthalimide) anion ( Section 18-10C ), the acidity increases markedly and imides can be converted to their conjugate bases with concentrated aqueous hydroxide ion. We have seen how imide salts can be used for the synthesis of primary amines. ( Gabriel synthesis , Section 23-9D and Table 23-6). Basicity The degree of basicity of amides is very much less than that of aliphatic amines. For ethanamide, \(K_b\) is about \(10^{-15}\) (\(K_a\) of the conjugate acid is \(\sim 10\)): The proton can become attached either to nitrogen or to oxygen, and the choice between the assignments is not an easy one. Of course, nitrogen is intrinsically more basic than oxygen; but formation of the \(\ce{N}\)-conjugate acid would cause loss of all the amide stabilization energy. Addition to oxygen actually is favored, but amides are too weakly basic for protonation to occur to any extent in water solution. |
Courses/Mendocino_College/Introduction_to_Chemistry_(CHM_200)/01%3A_Introduction_to_Chemistry_and_Measurement/1.07%3A_Temperature_and_Density | Learning Objective Learn about the various temperature scales that are commonly used in chemistry. Define density and use it as a conversion factor. There are other units in chemistry that are important, and we will cover others over the course of the entire book. One of the fundamental quantities in science is temperature. Temperature is a measure of the average amount of energy of motion, or kinetic energy , a system contains. Temperatures are expressed using scales that use units called degrees, and there are several temperature scales in use. In the United States, the commonly used temperature scale is the Fahrenheit scale (symbolized by °F and spoken as "degrees Fahrenheit"). On this scale, the freezing point of liquid water (the temperature at which liquid water turns to solid ice) is 32°F, and the boiling point of water (the temperature at which liquid water turns to steam) is 212°F. Science also uses other scales to express temperature. The Celsius scale (symbolized by °C and spoken as "degrees Celsius") is a temperature scale where 0°C is the freezing point of water and 100°C is the boiling point of water; the scale is divided into 100 divisions between these two landmarks and extended higher and lower. By comparing the Fahrenheit and Celsius scales, a conversion between the two scales can be determined: \[\begin{align} \ce{^{\circ}C} &= \ce{(^{\circ}F-32)\times 5/9} \label{eq1} \\[4pt] \ce{ ^{\circ}F} &= \left(\ce{^{\circ}C \times 9/5 } \right)+32 \label{eq2} \end{align} \] Example \(\PageIndex{1}\): Conversions What is 98.6 °F in degrees Celsius? What is 25.0 °C in degrees Fahrenheit? Solution Using Equation \ref{eq1}, we have \[\begin{align*} ^{\circ}C &=(98.6-32)\times \dfrac{5}{9} \\[4pt] &= 66.6\times \dfrac{5}{9} \\[4pt] &= 37.0^{\circ}C \end{align*}\nonumber \] Using Equation \ref{eq2}, we have \[\begin{align*} ^{\circ}F &= \left(25.0\times \dfrac{9}{5}\right)+32 \\[4pt] &= 45.0+32 \\[4pt] &= 77.0^{\circ}F \end{align*}\nonumber \] Exercise \(\PageIndex{1}\) Convert 0 °F to degrees Celsius. Convert 212 °C to degrees Fahrenheit. Answer a −17.8 °C Answer b 414 °F The fundamental unit of temperature (another fundamental unit of science, bringing us to four) in SI is the kelvin (K). The Kelvin temperature scale (note that the name of the scale capitalizes the word Kelvin , but the unit itself is lowercase) uses degrees that are the same size as the Celsius degree, but the numerical scale is shifted up by 273.15 units. That is, the conversion between the Kelvin and Celsius scales is as follows: \[K = {^{\circ}C + 273.15}\nonumber \] For most purposes, it is acceptable to use 273 instead of 273.15. Note that the Kelvin scale does not use the word degrees ; a temperature of 295 K is spoken of as "two hundred ninety-five kelvins" and not "two hundred ninety-five degrees Kelvin." The reason that the Kelvin scale is defined this way is because there exists a minimum possible temperature called absolute zero (zero kelvins). The Kelvin temperature scale is set so that 0 K is absolute zero, and temperature is counted upward from there. Normal room temperature is about 295 K, as seen in the following example. Example \(\PageIndex{2}\): Room Temperature If normal room temperature is 72.0°F, what is room temperature in degrees Celsius and kelvin? Solution First, we use the formula to determine the temperature in degrees Celsius: \[\begin{align*} ^{\circ}C &= (72.0-32)\times \dfrac{5}{9} \nonumber \\[4pt] &= 40.0\times \dfrac{5}{9} \nonumber \\[4pt] &= 22.2^{\circ}C \end{align*}\nonumber \] Then we use the appropriate formula above to determine the temperature in the Kelvin scale: \[\begin{align*} K &= 22.2^{\circ}C+273.15 \nonumber \\[4pt] &= 295.4K \end{align*}\nonumber \] So, room temperature is about 295 K. Exercise \(\PageIndex{2}\) What is 98.6°F on the Kelvin scale? Answer 310.2 K Figure \(\PageIndex{1}\) compares the three temperature scales. Note that science uses the Celsius and Kelvin scales almost exclusively; virtually no practicing chemist expresses laboratory-measured temperatures with the Fahrenheit scale. In fact, the United States is one of the few countries in the world that still uses the Fahrenheit scale on a daily basis. The other two countries are Liberia and Myanmar (formerly Burma). People driving near the borders of Canada or Mexico may pick up local radio stations on the other side of the border that express the daily weather in degrees Celsius, so do not get confused by their weather reports. Density Density is a physical property that is defined as a substance's mass divided by its volume: \[density= \dfrac{mass}{volume}\Rightarrow d= \dfrac{m}{v}\nonumber \] Density is usually a measured property of a substance, so its numerical value affects the significant figures in a calculation. Notice that density is defined in terms of two dissimilar units, mass and volume. That means that density overall has derived units, just like velocity. Common units for density include g/mL, g/cm 3 , g/L, kg/L, and even kg/m 3 . Densities for some common substances are listed in Table \(\PageIndex{1}\). Substance Density (g/mL or g/cm3) water 1.0 gold 19.3 mercury 13.6 air 0.0012 cork 0.22–0.26 aluminum 2.7 iron 7.87 Because of how it is defined, density can act as a conversion factor for switching between units of mass and volume. For example, suppose you have a sample of aluminum that has a volume of 7.88 cm 3 . How can you determine what mass of aluminum you have without measuring it? You can use the volume to calculate it. If you multiply the given volume by the known density (from Table \(\PageIndex{1}\)), the volume units will cancel and leave you with mass units, telling you the mass of the sample: \[7.88\,\cancel{cm^{3}}\times \dfrac{2.7\,g}{\cancel{cm^{3}}}= 21\, g \text{ of aluminium} \nonumber \nonumber \] where we have limited our answer to two significant figures. Example \(\PageIndex{3}\): Mercury What is the mass of 44.6 mL of mercury? Solution Use the density from Table \(\PageIndex{1}\) "Densities of Some Common Substances" as a conversion factor to go from volume to mass: \[44.6\,\cancel{mL}\times \dfrac{13.6\,g}{\cancel{mL}}= 607\,g \nonumber \nonumber \] The mass of the mercury is 607 g. Exercise \(\PageIndex{3}\) What is the mass of 25.0 cm 3 of iron? Answer 197 g Density can also be used as a conversion factor to convert mass to volume—but care must be taken. We have already demonstrated that the number that goes with density normally goes in the numerator when density is written as a fraction. Take the density of gold, for example: \[d=19.3\,g/mL =\dfrac{19.3\,g}{mL} \nonumber \nonumber \] Although this was not previously pointed out, it can be assumed that there is a 1 in the denominator: \[d=19.3\,g/mL =\dfrac{19.3\,g}{mL} \nonumber \nonumber \] That is, the density value tells us that we have 19.3 grams for every 1 milliliter of volume, and the 1 is an exact number. When we want to use density to convert from mass to volume, the numerator and denominator of density need to be switched—that is, we must take the reciprocal of the density. In so doing, we move not only the units, but also the numbers: \[\dfrac{1}{d}= \dfrac{1\,mL}{19.3\,g} \nonumber \nonumber \] This reciprocal density is still a useful conversion factor, but now the mass unit will cancel and the volume unit will be introduced. Thus, if we want to know the volume of 45.9 g of gold, we would set up the conversion as follows: \[45.9\,\cancel{g}\times \dfrac{1\,mL}{19.3\cancel{g}}= 2.38\,mL \nonumber \nonumber \] Note how the mass units cancel, leaving the volume unit, which is what we are looking for. Example \(\PageIndex{4}\): Calculating Volume from Density A cork stopper from a bottle of wine has a mass of 3.78 g. If the density of cork is 0.22 g/cm 3 , what is the volume of the cork? Solution To use density as a conversion factor, we need to take the reciprocal so that the mass unit of density is in the denominator. Taking the reciprocal, we find: \[\dfrac{1}{d}= \dfrac{1\,cm^{3}}{0.22\,g} \nonumber \nonumber \] We can use this expression as the conversion factor. So \[3.78\,\cancel{g}\times \dfrac{1\,cm^{3}}{0.22\,\cancel{g}}= 17.2\,cm^{3} \nonumber \nonumber \] Exercise \(\PageIndex{4}\) What is the volume of 3.78 g of gold? Answer 0.196 cm 3 Care must be used with density as a conversion factor. Make sure the mass units are the same, or the volume units are the same, before using density to convert to a different unit. Often, the unit of the given quantity must be first converted to the appropriate unit before applying density as a conversion factor. Food and Drink Application: Cooking Temperatures Because degrees Fahrenheit is the common temperature scale in the United States, kitchen appliances, such as ovens, are calibrated in that scale. A cool oven may be only 150°F, while a cake may be baked at 350°F and a chicken roasted at 400°F. The broil setting on many ovens is 500°F, which is typically the highest temperature setting on a household oven. People who live at high altitudes, typically 2,000 ft above sea level or higher, are sometimes urged to use slightly different cooking instructions on some products, such as cakes and bread, because water boils at a lower temperature the higher in altitude you go, meaning that foods cook slower. For example, in Cleveland water typically boils at 212°F (100°C), but in Denver, the Mile-High City, water boils at about 200°F (93.3°C), which can significantly lengthen cooking times. Good cooks need to be aware of this. A meat thermometer with a dial. Notice the markings for Fahrenheit (outer scale) and Celsius (inner scale) temperatures. Recipes for cooking food in an oven can use very different numbers, depending on the country you're in. (CC BY2.0 Bev Sykes ) At the other end is pressure cooking. A pressure cooker is a closed vessel that allows steam to build up additional pressure, which increases the temperature at which water boils. A good pressure cooker can get to temperatures as high as 252°F (122°C); at these temperatures, food cooks much faster than it normally would. Great care must be used with pressure cookers because of the high pressure and high temperature. (When a pressure cooker is used to sterilize medical instruments, it is called an autoclave .) Other countries use the Celsius scale for everyday purposes. Therefore, oven dials in their kitchens are marked in degrees Celsius. It can be confusing for US cooks to use ovens abroad—a 425°F oven in the United States is equivalent to a 220°C oven in other countries. These days, many oven thermometers are marked with both temperature scales. Key Takeaways Chemistry uses the Celsius and Kelvin scales to express temperatures. A temperature on the Kelvin scale is the Celsius temperature plus 273.15. The minimum possible temperature is absolute zero and is assigned 0 K on the Kelvin scale. Density relates the mass and volume of a substance. Density can be used to calculate volume from a given mass or mass from a given volume. |
Courses/Riverland_Community_College/CHEM_1000_-_Introduction_to_Chemistry_(Riverland)/09%3A_Chemical_Equilibrium/9.04%3A_The_Path_of_a_Reaction_and_the_Effect_of_a_Catalyst | The addition of a catalyst to an equilibrium system is a final stress factor. We consider how adding a catalyst affects the following: \[\ce{N_2(g) + O_2(g) \leftrightharpoons 2NO(g)} \nonumber \] Adding a catalyst to this, or any other equilibrium system, will not affect the position of an equilibrium. A catalyst speeds up both the forward and the reverse reactions, so there is no uneven change in reaction rates. Generally, a catalyst will help a reaction to reach the point of equilibrium sooner , but it will not affect the equilibrium otherwise. |
Courses/Erie_Community_College/ECC%3A_Introduction_to_General_Organic_and_Biochemistry_(Sorrentino)/Text/01%3A_Chemistry_Matter_and_Measurement/1.03%3A_The_Classification_of_Matter | Learning Objectives Use physical and chemical properties, including phase, to describe matter. Identify a sample of matter as an element, a compound, or a mixture. Part of understanding matter is being able to describe it. One way chemists describe matter is to assign different kinds of properties to different categories. Physical and Chemical Properties The properties that chemists use to describe matter fall into two general categories. Physical properties are characteristics that describe matter. They include characteristics such as size, shape, color, and mass. These characteristics can be observed or measured without changing the identity of the matter in question. Chemical properties are characteristics that describe how matter changes its chemical structure or composition. An example of a chemical property is flammability—a material’s ability to burn—because burning (also known as combustion) changes the chemical composition of a material. The observation of chemical properties involves a chemical change of the matter in question, resulting in matter with a different identity and different physical and chemical properties. Elements and Compounds Any sample of matter that has the same physical and chemical properties throughout the sample is called a substance . There are two types of substances. A substance that cannot be broken down into chemically simpler components is called an element . Aluminum, which is used in soda cans and is represented by the symbol Al, is an element. A substance that can be broken down into chemically simpler components (because it consists of more than one element) is called a compound . Water is a compound composed of the elements hydrogen and oxygen and is described by the chemical formula, H 2 O. Today, there are about 118 elements in the known universe. In contrast, scientists have identified tens of millions of different compounds to date. Sometimes the word pure is used to describe a substance, but this is not absolutely necessary. By definition, any single substance, element or compound is pure . The smallest part of an element that maintains the identity of that element is called an atom . Atoms are extremely tiny; to make a line of iron atoms that is 1 inch long, you would need approximately 217 million iron atoms. The smallest part of a compound that maintains the identity of that compound is called a molecule . Molecules are composed of two or more atoms that are attached together and behave as a unit. Scientists usually work with millions and millions of atoms and molecules at a time. When a scientist is working with large numbers of atoms or molecules at a time, the scientist is studying the macroscopic viewpoint of the universe. However, scientists can also describe chemical events on the level of individual atoms or molecules, which is referred to as the microscopic viewpoint . We will see examples of both macroscopic and microscopic viewpoints throughout this book (Figure \(\PageIndex{2}\)). Mixtures A material composed of two or more substances is a mixture . In a mixture, the individual substances maintain their chemical identities. Many mixtures are obvious combinations of two or more substances, such as a mixture of sand and water. Such mixtures are called heterogeneous mixtures . In some mixtures, the components are so intimately combined that they act like a single substance (even though they are not). Mixtures with a consistent or uniform composition throughout are called homogeneous mixtures (or solutions). For example, when sugar is dissolved in water to form a liquid solution, the individual properties of the components cannot be distinguished. Other examples or homogenous mixtures include solid solutions, like the metal alloy steel, and gaseous solutions, like air which is a mixture of mainly nitrogen and oxygen. Example \(\PageIndex{1}\) How would a chemist categorize each example of matter? saltwater soil water oxygen Answer a Saltwater acts as if it were a single substance even though it contains two substances—salt and water. Saltwater is a homogeneous mixture , or a solution. Answer b Soil is composed of small pieces of a variety of materials, so it is a h eterogeneous mixture . Answer c Water is a substance ; more specifically, because water is composed of hydrogen and oxygen, it is a compound . Answer d Oxygen, a substance , is an element . Exercise \(\PageIndex{2}\) How would a chemist categorize each example of matter? breakfast coffee hydrogen an egg Answer a homogeneous mixture or solution Answer b element Answer c heterogeneous mixture Phases or Physical States of Matter All matter can be further classified by one of three physical states or phases , solid, liquid or gas. These three descriptions each imply that the matter has certain physical properties when in these states. A solid has a definite shape and a definite volume. Liquids ordinarily have a definite volume but not a definite shape; they take the shape of their containers. Gases have neither a definite shape nor a definite volume, and they expand to fill their containers. We encounter matter in each phase every day; in fact, we regularly encounter water in all three phases: ice (solid), water (liquid), and steam (gas) (Figure \(\PageIndex{2}\)). We know from our experience with water that substances can change from one phase to another if the conditions are right. Typically, varying the temperature of a substance (and, less commonly, the pressure exerted on it) can cause a phase change , a physical process in which a substance changes from one phase to another (Figure \(\PageIndex{5}\)). Phase changes are identified by particular names depending on what phases are involved, as summarized in Table \(\PageIndex{1}\). Change Name solid to liquid melting, fusion solid to gas sublimation liquid to gas boiling, evaporation liquid to solid solidification, freezing gas to liquid condensation gas to solid deposition Figure \(\PageIndex{3}\) illustrates the relationships between the different ways matter can be classified. Concept Review Exercises Explain the differences between the physical properties of matter and the chemical properties of matter. What is the difference between a heterogeneous mixture and a homogeneous mixture? Give an example of each. Give at least two examples of a phase change and state the phases involved in each. Answers Physical properties describe the existence of matter, and chemical properties describe how substances change into other substances. A heterogeneous mixture is obviously a mixture, such as dirt; a homogeneous mixture behaves like a single substance, such as saltwater. solid to liquid (melting) and liquid to gas (boiling) (answers will vary) Key Takeaways Matter can be described with both physical properties and chemical properties. Matter can be identified as an element, a compound, or a mixture Exercise \(\PageIndex{3}\) Does each statement refer to a chemical property or a physical property? Balsa is a very light wood. If held in a flame, magnesium metal burns in air. Mercury has a density of 13.6 g/mL. Human blood is red. Answer physical property chemical property physical property physical property Exercise \(\PageIndex{4}\) Does each statement refer to a chemical property or a physical property? The elements sodium and chlorine can combine to make table salt. The metal tungsten does not melt until its temperature exceeds 3,000°C. The ingestion of ethyl alcohol can lead to disorientation and confusion. The boiling point of isopropyl alcohol, which is used to sterilize cuts and scrapes, is lower than the boiling point of water Answer chemical property physical property chemical property physical property Exercise \(\PageIndex{5}\) Define element . How does it differ from a compound? Answer An element is a substance that cannot be broken down into chemically simpler components. Compounds can be broken down into simpler substances. Exercise \(\PageIndex{6}\) Define compound . How does it differ from an element? Answer A compound is composed of two or more elements combined in a fixed ratio. An element is the simplest chemical substance. Exercise \(\PageIndex{7}\) Give two examples of a heterogeneous mixture. Answer a salt and pepper mix and a bowl of cereal (answers will vary) Exercise \(\PageIndex{8}\) Give two examples of a homogeneous mixture. Answer vinegar and rubbing alcohol (answers will vary) Exercise \(\PageIndex{9}\) Identify each substance as an element, a compound, a heterogeneous mixture, or a solution. xenon, a substance that cannot be broken down into chemically simpler components blood, a substance composed of several types of cells suspended in a salty solution called plasma water, a substance composed of hydrogen and oxygen Answer element heterogeneous mixture compound Exercise \(\PageIndex{10}\) Identify each substance as an element, a compound, a heterogeneous mixture, or a solution. sugar, a substance composed of carbon, hydrogen, and oxygen hydrogen, the simplest chemical substance dirt, a combination of rocks and decaying plant matter Answer compound element heterogeneous mixture Exercise \(\PageIndex{11}\) Identify each substance as an element, a compound, a heterogeneous mixture, or a solution. air, primarily a mixture of nitrogen and oxygen ringer’s lactate, a standard fluid used in medicine that contains salt, potassium, and lactate compounds all dissolved in sterile water tartaric acid, a substance composed of carbon, hydrogen, and oxygen Answer heterogeneous mixture solution compound Exercise \(\PageIndex{12}\) What word describes each phase change? solid to liquid liquid to gas solid to gas Answer melting or fusion boiling or evaporation sublimation Exercise \(\PageIndex{13}\) What word describes each phase change? liquid to solid gas to liquid gas to solid Answer freezing condensation deposition |
Courses/Kenyon_College/Chemistry_231_and_232_-_Kenyon_College_(Getzler_Hofferberth_and_Hunsen)/24%3A_Carbohydrates%3A_Polyfunctional_Compounds_in_Nature/24.2%3A_Conformations_and__Cyclic__Forms__of_Sugars | Objectives After completing this section, you should be able to determine whether a given monosaccharide will exist as a pyranose or furanose. draw the cyclic pyranose form of a monosaccharide, given its Fischer projection. draw the Fischer projection of a monosaccharide, given its cyclic pyranose form. draw, from memory, the cyclic pyranose form of D‑glucose. determine whether a given cyclic pyranose form represents the D or L form of the monosaccharide concerned. describe the phenomenon known as mutarotation. explain, through the use of chemical equations, exactly what happens at the molecular level during the mutarotation process. Key Terms Make certain that you can define, and use in context, the key terms below. alpha anomer anomer anomeric centre beta anomer furanose mutarotation pyranose Study Notes If necessary, before you attempt to study this section, review the formation of hemiacetals discussed in Section 19.10. As noted above, the preferred structural form of many monosaccharides may be that of a cyclic hemiacetal. Five and six-membered rings are favored over other ring sizes because of their low angle and eclipsing strain. Cyclic structures of this kind are termed furanose (five-membered) or pyranose (six-membered), reflecting the ring size relationship to the common heterocyclic compounds furan and pyran shown on the right. Ribose, an important aldopentose, commonly adopts a furanose structure, as shown in the following illustration. By convention for the D-family, the five-membered furanose ring is drawn in an edgewise projection with the ring oxygen positioned away from the viewer. The anomeric carbon atom (colored red here) is placed on the right. The upper bond to this carbon is defined as beta, the lower bond then is alpha. The cyclic pyranose forms of various monosaccharides are often drawn in a flat projection known as a Haworth formula, after the British chemist, Norman Haworth. As with the furanose ring, the anomeric carbon is placed on the right with the ring oxygen to the back of the edgewise view. In the D-family, the alpha and beta bonds have the same orientation defined for the furanose ring (beta is up & alpha is down). These Haworth formulas are convenient for displaying stereochemical relationships, but do not represent the true shape of the molecules. We know that these molecules are actually puckered in a fashion we call a chair conformation. Examples of four typical pyranose structures are shown below, both as Haworth projections and as the more representative chair conformers. The anomeric carbons are colored red. The size of the cyclic hemiacetal ring adopted by a given sugar is not constant, but may vary with substituents and other structural features. Aldolhexoses usually form pyranose rings and their pentose homologs tend to prefer the furanose form, but there are many counter examples. The formation of acetal derivatives illustrates how subtle changes may alter this selectivity. A pyranose structure for D-glucose is drawn in the rose-shaded box on the left. Acetal derivatives have been prepared by acid-catalyzed reactions with benzaldehyde and acetone. As a rule, benzaldehyde forms six-membered cyclic acetals, whereas acetone prefers to form five-membered acetals. The top equation shows the formation and some reactions of the 4,6-O-benzylidene acetal, a commonly employed protective group. A methyl glycoside derivative of this compound (see below) leaves the C-2 and C-3 hydroxyl groups exposed to reactions such as the periodic acid cleavage, shown as the last step. The formation of an isopropylidene acetal at C-1 and C-2, center structure, leaves the C-3 hydroxyl as the only unprotected function. Selective oxidation to a ketone is then possible. Finally, direct di-O-isopropylidene derivatization of glucose by reaction with excess acetone results in a change to a furanose structure in which the C-3 hydroxyl is again unprotected. However, the same reaction with D-galactose, shown in the blue-shaded box, produces a pyranose product in which the C-6 hydroxyl is unprotected. Both derivatives do not react with Tollens' reagent. This difference in behavior is attributed to the cis-orientation of the C-3 and C-4 hydroxyl groups in galactose, which permits formation of a less strained five-membered cyclic acetal, compared with the trans-C-3 and C-4 hydroxyl groups in glucose. Derivatizations of this kind permit selective reactions to be conducted at different locations in these highly functionalized molecules. Anomers of Simple Sugars: Mutarotation of Glucose Figure 1: Cyclization of D-Glucose. D-Glucose can be represented with a Fischer projection (a) or three dimensionally (b). By reacting the OH group on the fifth carbon atom with the aldehyde group, the cyclic monosaccharide (c) is produced. When a straight-chain monosaccharide, such as any of the structures shown in Figure 1, forms a cyclic structure, the carbonyl oxygen atom may be pushed either up or down, giving rise to two stereoisomers, as shown in Figure 2. The structure shown on the left side of Figure 2, with the OH group on the first carbon atom projected downward, represent what is called the alpha (α) form . The structures on the right side, with the OH group on the first carbon atom pointed upward, is the beta (β) form . These two stereoisomers of a cyclic monosaccharide are known as anomers; they differ in structure around the anomeric carbon—that is, the carbon atom that was the carbonyl carbon atom in the straight-chain form. Figure 2: Monosaccharides. In an aqueous solution, monosaccharides exist as an equilibrium mixture of three forms. The interconversion between the forms is known as mutarotation , which is shown for D-glucose (a) and D-fructose (b). It is possible to obtain a sample of crystalline glucose in which all the molecules have the α structure or all have the β structure. The α form melts at 146°C and has a specific rotation of +112°, while the β form melts at 150°C and has a specific rotation of +18.7°. When the sample is dissolved in water, however, a mixture is soon produced containing both anomers as well as the straight-chain form, in dynamic equilibrium (part (a) of Figure 2). You can start with a pure crystalline sample of glucose consisting entirely of either anomer, but as soon as the molecules dissolve in water, they open to form the carbonyl group and then reclose to form either the α or the β anomer. The opening and closing repeats continuously in an ongoing interconversion between anomeric forms and is referred to as mutarotation (Latin mutare , meaning “to change”). At equilibrium, the mixture consists of about 36% α-D-glucose, 64% β-D-glucose, and less than 0.02% of the open-chain aldehyde form. The observed rotation of this solution is +52.7°. Even though only a small percentage of the molecules are in the open-chain aldehyde form at any time, the solution will nevertheless exhibit the characteristic reactions of an aldehyde. As the small amount of free aldehyde is used up in a reaction, there is a shift in the equilibrium to yield more aldehyde. Thus, all the molecules may eventually react, even though very little free aldehyde is present at a time. Commonly, (e.g., in Figures 1 and 2) the cyclic forms of sugars are depicted using a convention first suggested by Walter N. Haworth, an English chemist. The molecules are drawn as planar hexagons with a darkened edge representing the side facing toward the viewer. The structure is simplified to show only the functional groups attached to the carbon atoms. Any group written to the right in a Fischer projection appears below the plane of the ring in a Haworth projection, and any group written to the left in a Fischer projection appears above the plane in a Haworth projection. The difference between the α and the β forms of sugars may seem trivial, but such structural differences are often crucial in biochemical reactions. This explains why we can get energy from the starch in potatoes and other plants but not from cellulose, even though both starch and cellulose are polysaccharides composed of glucose molecules linked together. Summary Monosaccharides that contain five or more carbons atoms form cyclic structures in aqueous solution. Two cyclic stereoisomers can form from each straight-chain monosaccharide; these are known as anomers. In an aqueous solution, an equilibrium mixture forms between the two anomers and the straight-chain structure of a monosaccharide in a process known as mutarotation. Exercises Exercise 25.5.1 Draw the following in their most stable chair conformation: α-D-galactopyranose and α-D-mannopyranose. Which is expected to be the more stable? Answer Because the both have two axial OH's their chair conformations should be roughly the same stability. Exercise 25.5.2 Draw the two chair conformations of the sugar called mannose, being sure to clearly show each non-hydrogen substituent as axial or equatorial. Predict which conformation is likely to be more stable, and explain why. Answer Exercise 25.5.3 Draw the cyclic structure of α-D-altrose. Answer Exercise 25.5.4 Draw the cyclic structure for β-D-galactose. Identify the anomeric carbon. Answer To identify the structure, we should first start with the Fischer projection of D-galactose. Since it is an aldohexose, we will start with the pyranose ring. The beta anomer was requested, so the OH on the anomeric carbon (C1) is cis to C6. Since C6 is top face (pointing up), the OH will be top face. Carbons 2, 3, and 4 are then arranged based on the Fischer projection arrangement at those carbons (C2 right, C3 left, and C4 left). Exercise 25.5.5 Given that the aldohexose D-mannose differs from D-glucose only in the configuration at the second carbon atom, draw the cyclic structure for α-D-mannose. Answer Exercise 25.5.6 Draw the cyclic structure for β-D-glucose. Identify the anomeric carbon. Answer |
Courses/Widener_University/CHEM_145%3A_FA22_Van_Bramer/08%3A_Chemical_Bonding_and_Molecular_Geometry/8.04%3A_Lewis_Symbols_and_Structures | Learning Objectives Write Lewis symbols for neutral atoms and ions Draw Lewis structures depicting the bonding in simple molecules Thus far, we have discussed the various types of bonds that form between atoms and/or ions. In all cases, these bonds involve the sharing or transfer of valence shell electrons between atoms. In this section, we will explore the typical method for depicting valence shell electrons and chemical bonds, namely Lewis symbols and Lewis structures. Lewis Symbols We use Lewis symbols to describe valence electron configurations of atoms and monatomic ions. A Lewis symbol consists of an elemental symbol surrounded by one dot for each of its valence electrons: Lewis symbols can be used to illustrate the formation of cations from atoms, as shown here for sodium and calcium: Likewise, they can be used to show the formation of anions from atoms, as shown here for chlorine and sulfur: Lewis Structures We also use Lewis symbols to indicate the formation of covalent bonds, which are shown in Lewis structures , drawings that describe the bonding in molecules and polyatomic ions. For example, when two chlorine atoms form a chlorine molecule, they share one pair of electrons: The Lewis structure indicates that each Cl atom has three pairs of electrons that are not used in bonding (called lone pairs ) and one shared pair of electrons (written between the atoms). A dash (or line) is usually used to indicate a shared pair of electrons: In the Lewis model, a single shared pair of electrons is a single bond . Each Cl atom interacts with eight valence electrons total: the six in the lone pairs and the two in the single bond. The Octet The other halogen molecules (F 2 , Br 2 , I 2 , and At 2 ) form bonds like those in the chlorine molecule: one single bond between atoms and three lone pairs of electrons per atom. This allows each halogen atom to have a noble gas electron configuration. The tendency of main group atoms to form enough bonds to obtain eight valence electrons is known as the octet rule . The number of bonds that an atom can form can often be predicted from the number of electrons needed to reach an octet (eight valence electrons); this is useful for the nonmetals of the second period of the periodic table (C, N, O, and F). For example, each atom of a group 14 element has four electrons in its outermost shell and therefore requires four more electrons to reach an octet. These four electrons can be gained by forming four covalent bonds, as illustrated here for carbon in CCl 4 (carbon tetrachloride) and silicon in SiH 4 (silane). Because hydrogen only needs two electrons to fill its valence shell, it is an exception to the octet rule and only needs to form one bond. The transition elements and inner transition elements also do not follow the octet rule since they have d and f electrons involved in their valence shells. Group 15 elements such as nitrogen have five valence electrons in the atomic Lewis symbol: one lone pair and three unpaired electrons. To obtain an octet, these atoms form three covalent bonds, as in NH 3 (ammonia). Oxygen and other atoms in group 16 obtain an octet by forming two covalent bonds: Double and Triple Bonds As previously mentioned, when a pair of atoms shares one pair of electrons, we call this a single bond. However, a pair of atoms may need to share more than one pair of electrons in order to achieve the requisite octet. A double bond forms when two pairs of electrons are shared between a pair of atoms, as between the carbon and oxygen atoms in CH 2 O (formaldehyde) and between the two carbon atoms in C 2 H 4 (ethylene): A triple bond forms when three electron pairs are shared by a pair of atoms, as in carbon monoxide (CO) and the cyanide ion (CN – ): Writing Lewis Structures with an Octet For very simple molecules and molecular ions, we can write the Lewis structures by merely pairing up the unpaired electrons on the constituent atoms. See these examples: For more complicated molecules and molecular ions, it is helpful to follow the step-by-step procedure outlined here: Determine the total number of valence (outer shell) electrons among all the atoms. For cations, subtract one electron for each positive charge. For anions, add one electron for each negative charge. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom. (Generally, the least electronegative element should be placed in the center.) Connect each atom to the central atom with a single bond (one electron pair). Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen), completing an octet around each atom. Place all remaining electrons on the central atom. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible. Let us determine the Lewis structures of SiH 4 , \(\ce{CHO2-}\), NO + , and OF 2 as examples in following this procedure: Determine the total number of valence (outer shell) electrons in the molecule or ion. For a molecule, we add the number of valence electrons on each atom in the molecule: \(\begin{align} &\phantom{+}\ce{SiH4}\\ &\phantom{+}\textrm{Si: 4 valence electrons/atom × 1 atom = 4}\\ &\underline{\textrm{+H: 1 valence electron/atom × 4 atoms = 4}}\\ &\hspace{271px}\textrm{= 8 valence electrons} \end{align}\) For a negative ion , such as \(\ce{CHO2-}\), we add the number of valence electrons on the atoms to the number of negative charges on the ion (one electron is gained for each single negative charge): \(\ce{CHO2-}\\ \textrm{C: 4 valence electrons/atom × 1 atom} \hspace{6px}= \phantom{1}4\\ \textrm{H: 1 valence electron/atom × 1 atom} \hspace{12px}= \phantom{1}1\\ \textrm{O: 6 valence electrons/atom × 2 atoms = 12}\\ \underline{+\hspace{100px}\textrm{1 additional electron} \hspace{9px}= \phantom{1}1}\\ \hspace{264px}\textrm{= 18 valence electrons}\) For a positive ion , such as NO + , we add the number of valence electrons on the atoms in the ion and then subtract the number of positive charges on the ion (one electron is lost for each single positive charge) from the total number of valence electrons: \(\ce{NO+}\\ \textrm{N: 5 valence electrons/atom × 1 atom} = \phantom{−}5\\ \textrm{O: 6 valence electron/atom × 1 atom}\hspace{5px} = \phantom{−}6\\ \underline{\textrm{+ −1 electron (positive charge)} \hspace{44px}= −1}\\ \hspace{260px}\textrm{= 10 valence electrons}\) Since OF 2 is a neutral molecule, we simply add the number of valence electrons: \(\phantom{+ }\ce{OF2}\\ \phantom{+ }\textrm{O: 6 valence electrons/atom × 1 atom} \hspace{10px}= 6\\ \underline{\textrm{+ F: 7 valence electrons/atom × 2 atoms} = 14}\\ \hspace{280px}\textrm{= 20 valence electrons}\) Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom and connecting each atom to the central atom with a single (one electron pair) bond. (Note that we denote ions with brackets around the structure, indicating the charge outside the brackets:) When several arrangements of atoms are possible, as for \(\ce{CHO2-}\), we must use experimental evidence to choose the correct one. In general, the less electronegative elements are more likely to be central atoms. In \(\ce{CHO2-}\), the less electronegative carbon atom occupies the central position with the oxygen and hydrogen atoms surrounding it. Other examples include P in POCl 3 , S in SO 2 , and Cl in \(\ce{ClO4-}\). An exception is that hydrogen is almost never a central atom. As the most electronegative element, fluorine also cannot be a central atom. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen) to complete their valence shells with an octet of electrons. There are no remaining electrons on SiH 4 , so it is unchanged: Place all remaining electrons on the central atom. For SiH 4 , \(\ce{CHO2-}\), and NO + , there are no remaining electrons; we already placed all of the electrons determined in Step 1. For OF 2 , we had 16 electrons remaining in Step 3, and we placed 12, leaving 4 to be placed on the central atom: Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible. SiH 4 : Si already has an octet, so nothing needs to be done. \(\ce{CHO2-}\): We have distributed the valence electrons as lone pairs on the oxygen atoms, but the carbon atom lacks an octet: NO + : For this ion, we added eight outer electrons, but neither atom has an octet. We cannot add any more electrons since we have already used the total that we found in Step 1, so we must move electrons to form a multiple bond: This still does not produce an octet, so we must move another pair, forming a triple bond: In OF 2 , each atom has an octet as drawn, so nothing changes. Example \(\PageIndex{1}\): Writing Lewis Structures NASA’s Cassini-Huygens mission detected a large cloud of toxic hydrogen cyanide (HCN) on Titan, one of Saturn’s moons. Titan also contains ethane (H 3 CCH 3 ), acetylene (HCCH), and ammonia (NH 3 ). What are the Lewis structures of these molecules? Solution Calculate the number of valence electrons. HCN: (1 × 1) + (4 × 1) + (5 × 1) = 10 H 3 CCH 3 : (1 × 3) + (2 × 4) + (1 × 3) = 14 HCCH: (1 × 1) + (2 × 4) + (1 × 1) = 10 NH 3 : (5 × 1) + (3 × 1) = 8 Draw a skeleton and connect the atoms with single bonds. Remember that H is never a central atom: Where needed, distribute electrons to the terminal atoms: HCN: six electrons placed on N H 3 CCH 3 : no electrons remain HCCH: no terminal atoms capable of accepting electrons NH 3 : no terminal atoms capable of accepting electrons Where needed, place remaining electrons on the central atom: HCN: no electrons remain H 3 CCH 3 : no electrons remain HCCH: four electrons placed on carbon NH 3 : two electrons placed on nitrogen Where needed, rearrange electrons to form multiple bonds in order to obtain an octet on each atom: HCN: form two more C–N bonds H 3 CCH 3 : all atoms have the correct number of electrons HCCH: form a triple bond between the two carbon atoms NH 3 : all atoms have the correct number of electrons Exercise \(\PageIndex{1}\) Both carbon monoxide, CO, and carbon dioxide, CO 2 , are products of the combustion of fossil fuels. Both of these gases also cause problems: CO is toxic and CO 2 has been implicated in global climate change. What are the Lewis structures of these two molecules? Answer Beyond the Octet Many covalent molecules have central atoms that do not have eight electrons in their Lewis structures. These molecules fall into three categories: Odd-electron molecules have an odd number of valence electrons, and therefore have an unpaired electron. Electron-deficient molecules have a central atom that has fewer electrons than needed for a noble gas configuration. Hypervalent molecules have a central atom that has more electrons than needed for a noble gas configuration. Odd-electron Molecules We call molecules that contain an odd number of electrons free radicals . Nitric oxide, NO, is an example of an odd-electron molecule; it is produced in internal combustion engines when oxygen and nitrogen react at high temperatures. To draw the Lewis structure for an odd-electron molecule like NO, we follow the same five steps we would for other molecules, but with a few minor changes: Determine the total number of valence (outer shell) electrons . The sum of the valence electrons is 5 (from N) + 6 (from O) = 11. The odd number immediately tells us that we have a free radical, so we know that not every atom can have eight electrons in its valence shell. Draw a skeleton structure of the molecule . We can easily draw a skeleton with an N–O single bond: N–O Distribute the remaining electrons as lone pairs on the terminal atoms . In this case, there is no central atom, so we distribute the electrons around both atoms. We give eight electrons to the more electronegative atom in these situations; thus oxygen has the filled valence shell: Place all remaining electrons on the central atom . Since there are no remaining electrons, this step does not apply. Rearrange the electrons to make multiple bonds with the central atom in order to obtain octets wherever possible. We know that an odd-electron molecule cannot have an octet for every atom, but we want to get each atom as close to an octet as possible. In this case, nitrogen has only five electrons around it. To move closer to an octet for nitrogen, we take one of the lone pairs from oxygen and use it to form a NO double bond. (We cannot take another lone pair of electrons on oxygen and form a triple bond because nitrogen would then have nine electrons:) Electron-deficient Molecules We will also encounter a few molecules that contain central atoms that do not have a filled valence shell. Generally, these are molecules with central atoms from groups 2 and 13 and outer atoms that are hydrogen or other atoms that do not form multiple bonds. For example, in the Lewis structures of beryllium dihydride, BeH 2 , and boron trifluoride, BF 3 , the beryllium and boron atoms each have only four and six electrons, respectively. It is possible to draw a structure with a double bond between a boron atom and a fluorine atom in BF 3 , satisfying the octet rule, but experimental evidence indicates the bond lengths are closer to that expected for B–F single bonds. This suggests the best Lewis structure has three B–F single bonds and an electron deficient boron. The reactivity of the compound is also consistent with an electron deficient boron. However, the B–F bonds are slightly shorter than what is actually expected for B–F single bonds, indicating that some double bond character is found in the actual molecule. An atom like the boron atom in BF 3 , which does not have eight electrons, is very reactive. It readily combines with a molecule containing an atom with a lone pair of electrons. For example, NH 3 reacts with BF 3 because the lone pair on nitrogen can be shared with the boron atom: Hypervalent Molecules Elements in the second period of the periodic table ( n = 2) can accommodate only eight electrons in their valence shell orbitals because they have only four valence orbitals (one 2 s and three 2 p orbitals). Elements in the third and higher periods ( n ≥ 3) have more than four valence orbitals and can share more than four pairs of electrons with other atoms because they have empty d orbitals in the same shell. Molecules formed from these elements are sometimes called hypervalent molecules .Table \(\PageIndex{5}\) shows the Lewis structures for two hypervalent molecules, PCl 5 and SF 6 . In some hypervalent molecules, such as IF 5 and XeF 4 , some of the electrons in the outer shell of the central atom are lone pairs: When we write the Lewis structures for these molecules, we find that we have electrons left over after filling the valence shells of the outer atoms with eight electrons. These additional electrons must be assigned to the central atom. Example \(\PageIndex{2}\): Octet Rule Violations Xenon is a noble gas, but it forms a number of stable compounds. We examined \(\ce{XeF4}\) earlier. What are the Lewis structures of \(\ce{XeF2}\) and \(\ce{XeF6}\)? Solution We can draw the Lewis structure of any covalent molecule by following the six steps discussed earlier. In this case, we can condense the last few steps, since not all of them apply. Step 1: Calculate the number of valence electrons: \(\ce{XeF2}\): 8 + (2 × 7) = 22 \(\ce{XeF6}\): 8 + (6 × 7) = 50 Step 2: Draw a skeleton joining the atoms by single bonds. Xenon will be the central atom because fluorine cannot be a central atom: Step 3: Distribute the remaining electrons. XeF 2 : We place three lone pairs of electrons around each F atom, accounting for 12 electrons and giving each F atom 8 electrons. Thus, six electrons (three lone pairs) remain. These lone pairs must be placed on the Xe atom. This is acceptable because Xe atoms have empty valence shell d orbitals and can accommodate more than eight electrons. The Lewis structure of XeF 2 shows two bonding pairs and three lone pairs of electrons around the Xe atom: XeF 6 : We place three lone pairs of electrons around each F atom, accounting for 36 electrons. Two electrons remain, and this lone pair is placed on the Xe atom: Exercise \(\PageIndex{2}\): interhalogens The halogens form a class of compounds called the interhalogens, in which halogen atoms covalently bond to each other. Write the Lewis structures for the interhalogens \(\ce{BrCl3}\) and \(\ce{ICl4-}\). Answer Summary Valence electronic structures can be visualized by drawing Lewis symbols (for atoms and monatomic ions) and Lewis structures (for molecules and polyatomic ions). Lone pairs, unpaired electrons, and single, double, or triple bonds are used to indicate where the valence electrons are located around each atom in a Lewis structure. Most structures—especially those containing second row elements—obey the octet rule, in which every atom (except H) is surrounded by eight electrons. Exceptions to the octet rule occur for odd-electron molecules (free radicals), electron-deficient molecules, and hypervalent molecules. Glossary double bond covalent bond in which two pairs of electrons are shared between two atoms free radical molecule that contains an odd number of electrons hypervalent molecule molecule containing at least one main group element that has more than eight electrons in its valence shell Lewis structure diagram showing lone pairs and bonding pairs of electrons in a molecule or an ion Lewis symbol symbol for an element or monatomic ion that uses a dot to represent each valence electron in the element or ion lone pair two (a pair of) valence electrons that are not used to form a covalent bond octet rule guideline that states main group atoms will form structures in which eight valence electrons interact with each nucleus, counting bonding electrons as interacting with both atoms connected by the bond single bond bond in which a single pair of electrons is shared between two atoms triple bond bond in which three pairs of electrons are shared between two atoms |
Courses/San_Francisco_State_University/General_Physical_Chemistry_I_(Gerber)/08%3A_Phase_Equilibria/8.01%3A_A_Phase_Diagram_Summarizes_the_Solid-Liquid-Gas_Behavior_of_a_Substance | A good map will take you to your destination with ease, provided you know how to read it. A map is an example of a diagram , a pictorial representation of a body of knowledge. In science they play a considerable role. Next to plots and tables diagrams are an important means of making information and/or theoretical knowledge accessible. Constructing them takes quite a bit of thought. You want to represent as much of what you know and give as accurate a picture of it without conveying anything incorrect. If the drawing can be made to scale that makes it quite a bit more powerful, but this is not strictly necessary. A remark like not to scale or schematically does need to be given if applicable. A good caption or description is essential. Thermodynamic Stability and Fluctuations There are different kinds of equilibrium, besides the stable equilibrium that represents an absolute minimum in the \(G\) function. Of course \(G\) is potentially a function of a great number of variables, but let us look at a diagram in which \(G\) is shown as a function of only one unspecified variable. You could think of the density, the mole fraction of one of the components of a mixture or an applied electrical field or whatever, but the argument is general. Figure 23.1.1 is helpful to point out that besides a stable equilibrium (\(A\)) there can also be an metastable equilibrium (B) or an indifferent equilibrium (\(C\)). The local derivatives of \(G\) (versus all variables of which we only show one) are zero in all three cases, which means that changes in the variables are not spontaneous . For a labile equilibrium (\(D\)) the opposite is true. Any small deviation will make the system role down hill. (Note that the second derivative has the opposite sign compared to cas A) and B)) A labile equilibrium is seldom or never observed except in a circus where artists delight in balancing objects on their heads (because you pay for it).. This usually requires continuous small corrections to maintain the precarious balance. All other points in our diagram represent state of instability because locally \(dG\) is not zero and a spontaneous process can take place. The fact that \(dG=0\) in the equilibrium points does not mean small deviations from the minimum cannot happen at times. We have seen e.g. that the Boltzmann distribution was simply the most likely distribution. The most likely one is the one that has the highest number of realizations W. Another way of saying that is that it is the one with the highest entropy S. A slightly less likely distribution may occur from time to time by chance. It will have a little less entropy, but the same \(\langle E \rangle\). That means it will have a slightly higher \(G\) (\(G=H-TS\)). From time to time therefore \(G\) will fluctuate a bit. Such fluctuations are very small for large systems, but they are of greater relative importance for small systems (like a nanoparticle). (Statistical averaging works best on large ensembles.) The fluctuations in \(G\) mean that small fluctuations in its variables like density etc. can also occur. They are usually kept in check, because \(dG\) is no longer zero when moving away from the equilibrium state. This drives the system back to the minimum spontaneously. You could picture the system wobbling a bit around in its G-well. This holds for stable and metastable equilibria alike. In the indifferent case (\(C\)) however the derivative is zero (or very close to zero) for a range of neighboring values of some variable. In contrast to \(A\) and \(B\) also the second derivative is zero. This means that there is little penalty to much larger deviations in the variable. If this variable is the density the system becomes milky and shows opalescence a strong scattering of light because the refractive index depends on the strongly fluctuating density. This is observed near critical points and is called critical opalescence . Unary Phase Diagrams A unary phase diagram summarizes the equilibrium states of a single pure substance. We will see that we can also look at mixtures of two components (binary diagrams) or more (ternary, quaternary, quinary, senary etc.). Usually a phase diagram only maps out stable equilibria, but occasionally metastable ones may be given too (e.g., with a dashed line). Liquid-Vapor Equilibrium Curve We have seen that the Gibbs function \(G\) depends strongly (logarithmically) on pressure for a gas, but only slightly (and linearly) for a liquid. The two curves intersect in a point representing the equilibrium vapor pressure of the liquid. At lower pressures the vapor is more stable, at higher ones the liquid. (For a solid the same holds as for the liquid). This means that except at the intersection point we only will observe one phase. Figure 23.1.2 : A cylinder with one substance It is important to stress that this holds in the absence of other matter , e.g., when we put only water into an evacuated cylinder (Figure 23.1.2 ). We may get three cases: we compress the cylinder until it only contains the liquid under hydrostatic pressure (\(P>P_{eq}\)) we expand the cylinder until all water has vaporized (\(P<p_{eq}\))> we let part of the water evaporate, just enough so that the space above the liquid is filled with an equilibrium vapor pressure (\(P=P_{eq}\)) At room temperature \(P_{eq}\) for water is only about 15 Torr. If we apply 1 bar -or let the atmosphere do the job- we will only have liquid water. If other gases are present, e.g., air, we must distinguish between the total pressure (e.g., 1 bar) and the equilibrium vapor pressure which will now be the partial pressure. In a cylinder with water and one bar of air just enough water will evaporate to establish equilibrium. The evaporation will be limited to the gas-liquid interface unless the partial pressure equals the total pressure. Then the liquid will boil . (Do allow the volume to expand, though., why? If the volume is constant the pressure builds up and boiling will stop. If we consider the set of equilibrium pressures as a function of temperature and plot that in a P vs. T diagram we have one component of our phase diagram. Gas-solid Equilibrium Curve For solids the situation is similar as the \(G(P)\) curve is once again an almost flat straight line. The intersection with the logarithmic curve for the gas will define an equilibrium pressure for gas-solid co-existence. Generally vapor pressures above solids are quite small, but not negligible. As for liquids we can construct a line representing the equilibrium pressures for sublimation as function of temperature and add it to the phase diagram. Liquid-solid Equilibrium Curve The \(\text{solid} \rightleftharpoons \text{liquid}\) equilibrium is known as melting or freezing is not very dependent on pressure. Usually melting points increase a little bit with pressure, although water is a peculiar exception. It expands upon freezing and the melting point goes down (a bit) with pressure. In our diagram this will represent an almost vertical line leaning a little forwards for most substances, but backwards for water and a few others. Putting the Curve Together The three lines come together in the triple point , the only point where all three phases are at equilibrium with each other. For water, its temperature is only 0.01 K different from the normal melting point (273.16 K) and its pressure is only 4.58 Torr. The intersection points with a line representing atmospheric pressure give the melting and boiling points at that pressure. A typical unary diagram If the triple point lies above the line that represents atmospheric pressure this implies that a liquid is never observed. On earth CO 2 is such a substance. The intersection of the solid-vapor equilibrium line with the 1 bar line represents a state where the solid will 'boil' (evaporate from inside out). This is known as the sublimation point . The melting points at P=1 bar are known as the standard melting point , the only slightly different one at 760 Torr = 1 atm is called normal melting point . The same goes for boiling and sublimation points. There is nothing magical about \(P=1 \,bar\). It just happens to be the pressure of our home planet. On a planet with higher atmospheric pressures \(CO_2\) may well be a liquid and on such a planet, all boiling points will be quite different (higher than on earth). The melting points will also differ, but only slightly so. On Mars where atmospheric pressure is much lower water can not occur in liquid form, much like carbon dioxide on earth - it sublimes. We should also realize that in a closed container (glass ampoule, hermetically sealed DSC pan), we can observe melting points at only very slightly different temperature values, but we will not see a boiling effect. Why? To see a boiling point the container must be open to the ( constant! ) 1 bar pressure of the earths atmosphere that defines it and causes the boiling phenomenon. If the ampoule is sealed it will generate its own (autogenous) pressure, depending on what you put in, how much of it in relation to the volume, how volatile it is and the temperature. The autogenous pressure does not interfere with the melting point much (the melting line is almost vertical), but as \(P\) changes with temperature you may never reach boiling conditions. In DSC experiments it is possible to observe boiling points only if the pan has been carefully perforated with a hole of known size. It must be big enough that the pressure inside the pan does not build up above atmospheric, but small enough that it does not cause premature loss of mass during the run. The latter spoils the calculation of the intensive value (per mole, per gram) of the heat of vaporization. The liquid evaporation line ends in a point that we have encountered before: the critical point \(T_C\). As temperature increases the liquid and vapor phases in equilibrium with each other start to resemble each other more and more and at \(T_c\) they coalesce. At this point the liquid-gas equilibrium becomes indifferent with respect to density and large fluctuations occur leading to critical opalescence. Notice that there is a relationship of dimensionality between the objects in the diagram and the number of phases present: 2 D planes: one phase 1 D curves: two phases 0 D point: three phases As you see the sum is always three. Number of moles So far we have typically considered one substance at the time, but for chemists it is imperative to deal with more than one because we are typically changing one into the other in our reactions. This means that the number of moles \(n\), that we often simply set equal to one now becomes an important variable in its own right. Besides we will actually have two (or more) of them: the number of moles of component one and the one for the other component. This makes n a much less trivial variable. This is already the case at a simple melting point, say when ice melts, because we are dealing with changing quantities of ice and water: \[n_{ice} + n_{water} = n_{total} \nonumber \] If all we do is turn water into ice or vice versa, we have \(dn_{total}=0\), so that: \[dn_{ice} =- dn_{water} \nonumber \] To deal with changing n's, we need to expand our mathematical notation a bit. Partial variables So far we have simply divided our thermodynamic functions if they were extensive by the number of moles and arrived at intensive molar values: \[G_{molar} = \dfrac{G}{ n} \nonumber \] \[V_{molar} = \dfrac{V}{n} \nonumber \] We have written such intensive molar values by writing a bar over the symbol G or V. We should note that scaling the function this way departs from the assumption that the function \(G\) depends on the variable n as a straight line that passes through the origin. If we have the same pure compound in two phases, like ice and water we can still apply this principle and write: \[G_{molar}^{ice} = \dfrac{G^{ice}}{n^{ice}} \nonumber \] \[V_{molar}^{ice} = \dfrac{V^{ice}}{n^{ice}} \nonumber \] \[G_{molar}^{water} = \dfrac{G^{water}}{n^{water}} \nonumber \] \[V_{molar}^{water} = \dfrac{V^{water}}{n^{water}} \nonumber \] If we have a mixture of two substances present as \(n_1\) and \(n_2\) moles the dependency need not be linear on either if the two substances interact with each other. This is also true for function like the volume of a liquid mixture. In the presence of interactions volumes do not have to be linearly additive. We can define a partial molar value of e.g. for the volume: \[V_{partial molar,1} = \dfrac{\partial V}{\partial n_1} \nonumber \] at \(n_2\) = constant \[V_{partial molar,2} = \dfrac{\partial V}{\partial n_2} \nonumber \] at \(n_1\) = constant The notation of putting a bar over the \(V\) symbol is used for these partial quantities as well. Partial molar volumes have been measured for many binary systems. They are functions of the composition (mole fraction) as well as the temperature and to a lesser extent the pressure. The partial molar Gibbs free energy The partial molar Gibbs free energy (\(\left(\dfrac{∂G}{∂n_i}\right)_{P,T}\), all other n's) is denoted with \(μ\) and is called the thermodynamic potential. When numbers of moles can change we can write the corresponding change in \(G\) as: \[dG = -SdT + VdP + \sum_i^N \left( \dfrac{\partial G}{\partial n_1} \right)_{P,T,n_{j\neq i}}dn_i \nonumber \] \[dG = -SdT + VdP + \sum_i^N μ_idn_i \nonumber \] over \(N\) phases in the system. As you can see we are adding a set of conjugate variables \(μ_in_i\) for each phase \(i\). If we are considering a pure component (but in different modifications, like ice and steam), we can still write: \[ μ_i = \left( \dfrac{\partial G}{\partial n_1} \right)_{P,T,n_{j\neq i}} = \dfrac{G_i}{n_i} \nonumber \] As soon as we are dealing with mixtures we really do have derivatives. |
Courses/University_of_the_Pacific/Book%3A_Molecular_Quantum_Mechanics_(Dutoi)/2%3A_Principles_of_Quantum_Theory/2.2%3A_Formal_Development | A scientific theory is a conceptual framework for rationalizing sets of observations. The structure of a theory is to forward a set of fundamental postulates, which, if taken as truths lead to results that match observation. For example, Dalton's atomic theory posits such things as matter consisting of immutable atoms, where each atom is identified as belonging to an element, and the atoms of a given element are all identical. You can see that the discovery of isotopes required modification of that last point, for example, but the atomic theory as proposed by Dalton still describes much of chemistry. Theories are generally not entirely right or wrong, but they have varying levels of usefulness. Theories are rarely (if ever) complete. The following is a set of formal postulates of the non-relativistic quantum theory. We have seen that the birth of quantum theory was intertwined with the birth of the theory of relativity, but the relativistic theory of quantum mechanics is still a work in progress (an area of considerable intellectual tumult, if general relativity is considered). However, there are systems for which the speed of light is effectively infinite compared to system velocities (\(p/m = v \ll c\)), and the non-relativistic theory suffices. (Similarly, very massive fast objects are handled well with relativity and no quantum mechanics). The majority of chemistry is described well without relativity (core electrons of ever heavier elements being an exception). It is worth noting before we start that, while there is consensus on the minimal content of the postulates of quantum theory, their exact organization, including the number of postulates, varies from text to text. More abstract ways of looking at the theory can sometimes summarize what might be considered multiple postulates as one. 1. Postulate 1: Wavefunction for the state The state of a physical system ( e.g., a particle ) is represented by a wavefunction, which, along with the particle identities, contains all physical information about the system. This postulate has already been thoroughly discussed qualitatively in this text. We simply take a moment to draw an analogy between the wavefunction and the combined position and momentum of a classical particle. The position and momentum together define the physical state of a fundamental (internally structureless) classical particle of known identity. With only this information, and knowledge of the surrounding field in which it moves (perhaps time-dependent), the state of the particle at any later time may be predicted. There is nothing else to know about the particle itself. Similarly, there is no inquiry that one can make of a composite system of many particles that cannot be answered by knowing all the positions, momenta, and identities of all of the particles. In quantum mechanics, the state of a particle is fully described by a complex wave on its spatial coordinates. For a composite system, the state of the whole system is a wave on the space of the coordinates of all of the particles. There is nothing to know about the system that is not accessible through the total wave function at some given time, and knowledge of the particle identities. With additional knowledge of the surrounding external field, even its future state (or its history) can be determined. There are a few formal details, however. Mainly, there are conditions on acceptable wavefunctions. Wavefunctions must be single-valued, continuous, and also have continuous first and second derivatives. They may not diverge, and they must be square integrable (the integral of the square modulus must be finite). All of the functions shown in the figure [#] below are inadmissible. Two of the conditions, (c) and (e) in the figure [#] above, can be relaxed in practice, though not formally. Case (c) indicates a wavefunction with a "kink," which can be thought of as a place where the "local" wavelength goes suddenly to zero (opposite of having a very long wavelength, with nearly zero curvature). This would be interpreted as having infinite local kinetic energy at that point. In fact, this can happen if potentials are introduced having isolated points at which they diverge. However, no real physical systems have truly infinite potentials. The Coulomb potential diverges for volumeless point particles, but that is an idealization. Regarding case (e), wavefunctions that cannot be normalized will sometimes be introduced into a calculation, and this is a useful technique, but they do not represent the state of any actual system. Rather, they are idealizations that can then be summed together (generally integrated over) to produce proper wavefunctions. Such non-normalizable functions also represent mathematically limiting cases of very broad particle waves. 2. Postulate 2: Superposition principle A linear combination (superposition) of any number of valid wavefunctions for a system is itself a valid wave function for that system. A linear combination is a generalization of a sum, where (potentially complex) coefficients may multiply the functions that are summed or integrated together, with each function appearing only to the to the first power (i.e., linear, not squared, etc.). One of the most fundamental aspects of waves is that they exhibit interference patterns. The simplest example was already encountered in the double-slit experiment. The total wave may be constructed as the superposition (sum) of the waves that would result if only one or the other slit were open, as illustrated in the figure [#] below. The resulting interference pattern gives rise to the observable undulation in light intensity across the screen on the other side, which one would not see if only one slit were open; there would be no superposition or consequent interference. This superposition principle was hinted at already at the end of Postulate 1. It is essentially the second half of asserting that matter moves as waves. It may be considered an implicit consequence of Postulate 1, as opposed to a separate postulate. However, it is an important principle with its own oft-used name, and so we have discussed it separately. 3. Postulate 3: Operators for physical properties Any physical property can be associated with a complete set of orthogonal wavefunctions, which each have a well-defined, real value for that property. This implies that these states are eigenfunctions of a self-adjoint (Hermitian) linear operator corresponding to that property, where the allowed property values are the eigenvalues associated with each eigenfunction. a. complete sets of states This postulate has a rather complex structure. Let us take each concept in turn. First, we will illustrate what is meant by a "complete" set by examining those wavefunctions that we know to have either well-defined momentum or position in 1D, which are illustrated in the figure [#] above. All such functions of either kind build a complete set of orthogonal functions. To be complete means that any admissible wavefunction (in 1D) may be constructed as a superposition of the members of that set (even if, by themselves , they do not meet all admissibility criteria). For example, though we do not delve into the mathematical details, the restrictions we have put on an admissible wavefunction \(\psi(x)\) assure us that it may be expressed as \[ \psi(x)=\int_{-\infty}^{\infty} \text{d} \bar{\nu}~\phi(\bar{\nu}) \, \text{e}^{\text{i} 2 \pi \bar{\nu} x} \label{fourier}\] for some equally well behaved choice of \(\phi(\bar{\nu})\). Mathematicians call \(\phi(\bar{\nu})\) the Fourier transform of \(\psi(x)\). We see that integration has the effect of “summing” all possible plane-wave functions (with different \(\bar{\nu}\)) giving different amplitude \(\phi(\bar{\nu})\) to each. The fact that any admissible \(\psi(x)\) in 1D may be written as such is what we mean by saying that the set of all plane waves \(\big\{\text{e}^{\text{i}2\pi\bar{\nu}x}\big\}\) is complete. Note that it is not problematic that the plane waves themselves are inadmissible wavefunctions; all this means is that the set is bigger than it needs to be, in order to be complete in the sense required here. The plane waves can also build infinitely many inadmissible wavefunctions, in addition to the admissible ones. Now let us illustrate the completeness of the definite-position functions. The theory of such mathematical objects is quite advanced, but for the purposes of this text, the following definition will suffice for the Dirac delta function \(\delta(x)\), which is centered at the origin: A \(\delta\)-function located elsewhere, at \(x_0\), may simply be written as \(\delta(x-x_0)\), a translation of the function in eqn. [not numbered, found in “figure [#]” above]. Let us then consider the mechanics of the following identity \[\begin{align} \psi(x) &=\int_{-\infty}^{\infty} \text{d} x^{\prime} ~ \psi(x^{\prime}) \, \delta(x-x^{\prime}) \nonumber\\ &=\lim _{\varepsilon \rightarrow 0} \int_{x-\varepsilon}^{x+\varepsilon} \text{d} x^{\prime}\psi(x^{\prime}) \, \delta(x-x^{\prime}) \nonumber\\ &=\psi(x) \, \lim _{\varepsilon \rightarrow 0} \int_{x-\varepsilon}^{x+\varepsilon} \text{d} x^{\prime} \, \delta(x-x^{\prime}) \nonumber\\ &=\psi(x) \, \lim _{\varepsilon \rightarrow 0} \int_{+\varepsilon}^{-\varepsilon} -\text{d} u~\delta(u) \nonumber\\ &=\psi(x) \, \lim _{\varepsilon \rightarrow 0} \int_{-\varepsilon}^{+\varepsilon} \text{d} u~\delta(u) \nonumber\\&= \psi(x) \end{align}\] This identity rests on the fact that, since the \(\delta\)-function is infinitely narrow, the integration bounds can be brought to the point where only a single value of \(\psi(x^{\prime})\) matters, and then the unit area of the \(\delta\)-function “plucks out” only this value. (Technically, this identity defines the \(\delta\)-function, and that gives rise to the interpretation of the function as we have drawn it.) The mechanics of the identity aside, it reads analogously to the Fourier transform of eqn. \ref{fourier}; any function \(\psi(x)\) may be built by “summing" the \(\delta\)-functions at every point in space. It makes good sense that the amplitude for each position function is the wavefunction itself at that same point. Finally, in analogy to the plane waves, the set of \(\delta\)-functions is not only complete, but it could also be used to build inadmissible wavefunctions, so this set of functions is also bigger than complete (for the purpose of building admissible 1D wavefunctions). In the concrete context of 1D states that we already understand to have either definite momentum (the plane waves) or definite position (the \(\delta\)-functions), we have illustrated that either such set is complete. So, the properties of both momentum and position are associated with complete sets of states, which correspond to its allowed values. b. orthogonal states The discussion of the first sentence of the third postulate has thus far only left the word “orthogonal” to be addressed. The physical interpretation of orthogonality is that the waves corresponding to different values of some physical property are completely distinct from one another. Geometrically, it means that one function contains no component of another function to which it is said to be orthogonal, in analogy to what is meant by orthogonal vectors. This is most intuitive for the set of position functions; clearly a \(\delta\)-function centered at one point will have no amplitude in building a \(\delta\)-function centered at a different point. Mathematically, if two functions in 1D, \(f(x)\) and \(g(x)\), are normalizable, they are said to also be orthogonal if \[ \int_{-\infty}^{\infty} \text{d}x\; f^{*\!}(x) \, g(x)=0 \] Where the asterisk (\({}^{*\!}\)) denotes the taking of a complex conjugate. One might clearly see the physical interpretation that, if \(f(x)\) and \(g(x)\) are very different from each other ( e.g. , very far from each other), then this integral will go to zero for these "distinct" functions, but it certainly would not if they were the same, or even quite similar. The word "orthogonal" is used because of the relationship of this integral to the formula for the dot product of vectors (when the integration is written as a Riemann sum ); it is, in fact a rigorous generalization of orthogonality to "spaces" of functions. In the case where the functions are not normalizable (say, for plane waves or \(\delta\)-functions), the following holds for orthogonal functions in 1D \[ 0= \lim _{L \rightarrow \infty} \frac{\int_{-L}^{L} \text{d}x\; f^{*\!}(x)\,g(x)}{\int_{-L} ^{L}\text{d}x~|f(x)|^{2}}=\lim _{L \rightarrow \infty} \frac{\int_{-L}^{L} \text{d}x\; f^{*\!}(x) \, g(x)}{\int_{-L}^{L} \text{d}x~|g(x)|^{2}} \] In this case, the denominator diverges, but the numerator does not. c. operators We are now in a position to handle the second sentence of this postulate concerning Hermitian operators. We will start by demonstrating that we that we can find "operators" whose "eigenfunction–eigenvalue" pairs are the 1D definite-position and definite-momentum wavefunctions just introduced. In the course of doing so, our purpose is to introduce the concepts of linear operators, eigenfunctions, and eigenvalues. First, we discuss the concept of operators, specifically, linear operators. An operation on a mathematical function is anything that is done to a function which produces another function on that same set of coordinates (such as multiplication or differentiation). From any operation, we can abstract an operator . For example, those shown in the table [#] above. The last of these provides a good example of how operators themselves have their own set of manipulations and equalities, which can be gleaned by always assuming that there is some arbitrary function to the right of them. In this text, we will follow a typical convention of distinguishing an operator by a “hat,” when referenced symbolically. For example (for operators that will be discussed more shortly) \[\begin{align} \hat{x} = x\cdot \quad\quad\quad &(\text{position along the x axis, multiplication by} ~x) \\ \hat{p}_\text{x} = -\text{i} \hbar \frac{\partial}{\partial x} \quad\quad &(\text{momentum in the x direction}) \end{align}\] Operators may be linear or nonlinear, depending on whether they obey a generalized distributive property ( i.e. , depending on how they act on linear combinations of functions). For example, in considering the examples, \[\begin{align} \frac{\text{d}}{\text{d}x}\big(f(x)+g(x)\big) &= \frac{\text{d}}{\text{d}x} f(x)+\frac{\text{d}}{\text{d}x} g(x) \\ \sqrt{f(x)+g(x)} &\neq \sqrt{f(x)}+\sqrt{g(x)} \text{,}\end{align}\] we say that the differential in the first example is a linear operator because it can be distributed over a linear combination of the functions \(f(x)\) and \(g(x)\), whereas this is not a valid manipulation for the square-root operator in the second example, meaning that it does not obey the definition of a linear operator. All operators corresponding to properties in quantum mechanics are linear operators. An important note is in order: although the operators we will meet will all have this “distributive” property, they will not usually have a commutative property. For example, \[ \hat{x}\,\hat{p}_\text{x} ~\neq~ \hat{p}_\text{x}\,\hat {x}. \] We say that \(\hat{x}\) and \(\hat{p}_\text{x}\) do not commute . d. eigenfunctions and eigenvalues With this basic concept of operators in hand, let us now discuss one of their important properties, which is that an operator may have eigenfunctions, each with an associated eigenvalue. An eigenvalue–eigenfunction pair for some operator \(\hat{a}\) on 1D functions is a constant \(a_i\) and a function \(f_{i}(x)\) that satisfy \[ \hat{a}\, f_{i}(x)=a_{i}\, f_{i}(x) \] Clearly, only functions with a special relationship to \(\hat{a}\) have this property. There are generally infinitely many such functions (a complete set, in fact) for a given \(\hat{a}\), hence the index \(i\). Let us now show that the 1D plane waves illustrated in the figure [#] above, which we associate with well-defined momenta, are indeed eigenfunctions of the operator that we have just asserted to be momentum operator, \(\hat{p}_\text{x}\), and have eigenvalues equal to the momentum associated with that wave. Recall that the relationship should be \(p=h / \lambda = h\bar{\nu}\), and that complex waves are necessary to define the direction of travel, which is critical for momentum. Inserting these presumed eigenfunctions into the eigenvalue equation for \(\hat{p}_\text{x}\) we verify this property as \[\begin{align} \hat{p}_\text{x} \, \text{e}^{\pm \text{i} 2 \pi \bar{\nu} x} &=-\text{i} \hbar \frac{\partial}{\partial x} \text{e}^{\pm \text{i} 2 \pi \bar{\nu} x} \nonumber\\ &=(-\text{i} \hbar)(\pm \text{i} 2 \pi \bar{\nu}) \, \text{e}^{\pm \text{i} 2 \pi \bar{\nu} x} \nonumber\\ &=\pm h\bar{\nu}\, \text{e}^{\pm \text{i} 2 \pi \bar{\nu} x} \nonumber\\ &=p_\text{x}\, \text{e}^{\pm \text{i} 2 \pi \bar{\nu} x} \end{align}\] We can also show that the functions with definite position in 1D, the \(\delta\)-functions, are eigenfunctions of the position operator \(\hat{x}=x\cdot\), with eigenvalue corresponding to the position. Since the full theory of these advanced mathematical objects is beyond the scope of this text, we will illustrate this only graphically, as shown in the figure [#] below. We have now seen that, indeed, for position and momentum, there do exist operators, \(\hat{x}\) and \(\hat{p}_\text{x}\), whose eigenfunctions are states with eigenvalues that are the associated property for that state. The final qualifier to address is that the operators are Hermitian. An operator \(\hat{a}\) in 1D is said to be Hermitian if, for two normalizable functions \(f(x)\) and \(g(x)\), it is true that \[ \int_{-\infty}^{\infty} \text{d}x\;f^{*\!}(x) \, \hat{a} \, g(x) = \bigg[\int_{-\infty}^{\infty} \text{d}x\;g^{*\!}(x) \, \hat{a} \, f(x)\bigg]^* \text{.}\] This can be illustrated by writing \(f(x)\) and \(g(x)\) as linear combinations the eigenfunctions of \(\hat{a}\), which are presumed to build a complete set (possibly an overcomplete set) of orthogonal functions, having real eigenvalues (corresponding to the values of physical properties). This demonstration is left as an exercise for the reader. In many texts, the discussion of the third postulate is reversed from how it is done here. A reader were versed well in linear algebra might recognize that the second second sentence in the third postulate is a consequence of the first one. The existence of a complete set of orthogonal functions, each with an associated real value, is all that is needed to construct a Hermitian linear operator. We could, therefore, have started with the statement that every property corresponds to a Hermitian linear operator on the space of valid wavefunctions, and then proceed to invoke the known mathematical properties of Hermitian operators ( i.e. , that they have as their eigenfunctions a complete set of orthogonal functions, whose eigenvalues are real) to complete the discussion of having states with well-defined values for the property under discussion. However, given the flow of the text up until the introduction of this postulate, we have preferred to begin with states and then move on to operators. Our discussion of the physical content of the third postulate would, in fact, have been complete, without ever mentioning the connection to operators. However, discussion of the fourth postulate would be very cumbersome without it. Indeed, this is the utility of reaching for higher levels of mathematical abstraction within physics, and, should we have foregone this discussion, we would have also found ourselves uncomfortably distant from the operator language ubiquitously used, so much so that we have even given this third postulate the short title of “Operators for physical properties.” 4. Postulate 4: Measurement probabilities content 5. Postulate 5: Time evolution content |
Courses/American_River_College/CHEM_305%3A_Introduction_to_Chemistry_(Zumalt)/05%3A_Unit_5/5.22%3A_Functional_Groups | Learning Objectives To know the major classes of organic compounds and identify important functional groups. You were previously introduced to several structural units that chemists use to classify organic compounds and predict their reactivities. These functional groups, which determine the chemical reactivity of a molecule under a given set of conditions, can consist of a single atom (such as Cl) or a group of atoms (such as CO 2 H). The major families of organic compounds are characterized by their functional groups. Figure \(\PageIndex{1}\) summarizes five families introduced in earlier chapters, gives examples of compounds that contain each functional group, and lists the suffix or prefix used in the systematic nomenclature of compounds that contain each functional group. The first family listed in Figure \(\PageIndex{1}\) is the hydrocarbons. These include alkanes, with the general molecular formula C n H 2n +2 where n is an integer; alkenes, represented by C n H 2n ; alkynes, represented by C n H 2n−2 ; and arenes. Halogen-substituted alkanes, alkenes, and arenes form a second major family of organic compounds, which include the alkyl halides and the aryl halides. Oxygen-containing organic compounds, a third family, may be divided into two main types: those that contain at least one C–O bond, which include alcohols, phenols (derivatives of benzene), and ethers, and those that contain a carbonyl group (C=O), which include aldehydes, ketones, and carboxylic acids. Carboxylic acid derivatives, the fourth family listed, are compounds in which the OH of the –CO 2 H functional group is replaced by either an alkoxy (–OR) group, producing an ester, or by an amido (–NRR′, where R and R′ can be H and/or alkyl groups), forming an amide. Nitrogen-containing organic compounds, the fifth family, include amines; nitriles, which have a C≡N bond; and nitro compounds, which contain the –NO 2 group. The systematic nomenclature of organic compounds indicates the positions of substituents using the lowest numbers possible to identify their locations in the carbon chain of the parent compound. If two compounds have the same systematic name, then they are the same compound. Although systematic names are preferred because they are unambiguous, many organic compounds are known by their common names rather than their systematic names. Common nomenclature uses the prefix form—for a compound that contains no carbons other than those in the functional group, and acet—for those that have one carbon atom in addition [two in the case of acetone, (CH 3 ) 2 C=O]. Thus methanal and ethanal, respectively, are the systematic names for formaldehyde and acetaldehyde. Recall that in the systematic nomenclature of aromatic compounds, the positions of groups attached to the aromatic ring are indicated by numbers, starting with 1 and proceeding around the ring in the direction that produces the lowest possible numbers. For example, the position of the first CH 3 group in dimethyl benzene is indicated with a 1, but the second CH 3 group, which can be placed in any one of three positions, produces 1,2-dimethylbenzene, 1,3-dimethylbenzene, or 1,4-dimethylbenzene (Figure \(\PageIndex{2}\)). In common nomenclature, in contrast, the prefixes ortho-, meta-, and para- are used to describe the relative positions of groups attached to an aromatic ring. If the CH 3 groups in dimethylbenzene, whose common name is xylene, are adjacent to each other, the compound is commonly called ortho-xylene, abbreviated o-xylene. If they are across from each other on the ring, the compound is commonly called para-xylene or p-xylene. When the arrangement is intermediate between those of ortho- and para- compounds, the name is meta-xylene or m-xylene. We begin our discussion of the structure and reactivity of organic compounds by exploring structural variations in the simple saturated hydrocarbons known as alkanes. These compounds serve as the scaffolding to which the various functional groups are most often attached. Summary Functional groups determine the chemical reactivity of an organic molecule. Functional groups are structural units that determine the chemical reactivity of a molecule under a given set of conditions. Organic compounds are classified into several major categories based on the functional groups they contain. In the systematic names of organic compounds, numbers indicate the positions of functional groups in the basic hydrocarbon framework. Many organic compounds also have common names, which use the prefix form—for a compound that contains no carbons other than those in the functional group and acet—for those that have one additional carbon atom. Conceptual Problems Can two substances have the same systematic name and be different compounds? Is a carbon–carbon multiple bond considered a functional group? |
Courses/City_College_of_San_Francisco/Chemistry_101A/Topic_B%3A_Reactions_in_Aqueous_Solution/04%3A_Reactions_in_Aqueous_Solution/4.2%3A_Precipitation_and_Solubility_Rules | Learning Objectives Predict the solubility of common inorganic compounds by using solubility rules Write simple net ionic equations for precipitation reactions. Precipitation Reactions and Solubility Rules A precipitation reaction is one in which dissolved substances react to form one (or more) solid products. Many reactions of this type involve the exchange of ions between ionic compounds in aqueous solution and are sometimes referred to as double displacement , double replacement , or metathesis reactions. These reactions are common in nature and are responsible for the formation of coral reefs in ocean waters and kidney stones in animals. They are used widely in industry for production of a number of commodity and specialty chemicals. Precipitation reactions also play a central role in many chemical analysis techniques, including spot tests used to identify metal ions and gravimetric methods for determining the composition of matter (see the last module of this chapter). The extent to which a substance may be dissolved in water, or any solvent, is quantitatively expressed as its solubility , defined as the maximum concentration of a substance that can be achieved under specified conditions. Substances with relatively large solubilities are said to be soluble . A substance will precipitate when solution conditions are such that its concentration exceeds its solubility. Substances with relatively low solubilities are said to be insoluble , and these are the substances that readily precipitate from solution. More information on these important concepts is provided in the text chapter on solutions. For purposes of predicting the identities of solids formed by precipitation reactions, one may simply refer to patterns of solubility that have been observed for many ionic compounds (Table \(\PageIndex{1}\)). All compounds that contain the cations Na + , K + or NH 4 + , or the anions NO 3 – or C 2 H 3 O 2 – are soluble in water. It is best to memorize these. Beyond this, solubilities are normally classified using the anion in the compound. Here are the rules that you will use in Chem 101A: An anion from this column... ...combined with a cation from this column produces a soluble compound (a precipitate will NOT form) ... combined with a cation from this column produces an insoluble compound (a precipitate will form) NO3–, C2H3O2– All NaN Cl–, Br–, I– Most Ag+ Pb2+ Hg22+ SO42– Most Ag+ Pb2+ Hg22+ Ca2+ Sr2+ Ba2+ (the heavier IIA elements) OH– Na+ K+ (NH4+ reacts with OH–: see below) Ba2+ All others (note: Ag+ forms an oxide product, rather than hydroxide product) CO32–, PO43– Na+ K+ NH4+ All others S2– Na+ K+ NH4+ Mg2+ Ca2+ Sr2+ Ba2+ (group IIA) All others (the reactions of sulfide with 3+ ions are not simple precipitations: you do not need to know these) Note that the reaction of Ag + with OH – produces Ag 2 O (and water), not AgOH. This is a “quirk” of the chemistry of silver ions. The net ionic equation for this reaction is: \[\ce{2Ag^+}(aq)+\ce{2OH-}(aq)\rightarrow \ce{Ag2O}(s)+\ce{H2O}(l)\] Note that ammonia (NH 3 ) dissolves in water to produce a small concentration of hydroxide ions (discussed in a later section.) The resulting hydroxide ions can participate in precipitation reactions. Here is an example, using Mg 2+ : \[\ce{Mg^2+}(aq)+\ce{2NH3}(aq)+\ce{2H2O}(l)\rightarrow \ce{Mg(OH)2}(s)+\ce{2NH4^+}(aq)\] Note that the above equation is written in terms of the major species in solution (NH 3 and H 2 O) as opposed to the minor species (NH 4 + and OH-). A vivid example of precipitation is observed when solutions of potassium iodide and lead nitrate are mixed, resulting in the formation of solid lead iodide. This observation is consistent with the solubility guidelines given above: The only insoluble combination among all those possible is lead and iodide. The net ionic equation representing this reaction is: \[\ce{Pb^2+}(aq)+\ce{2I-}(aq)\rightarrow \ce{PbI2}(s)\] Lead iodide is a bright yellow solid that was formerly used as an artist’s pigment known as iodine yellow (Figure \(\PageIndex{1}\)). The properties of pure PbI 2 crystals make them useful for fabrication of X-ray and gamma ray detectors. The solubility guidelines in Table \(\PageIndex{1}\) may be used to predict whether a precipitation reaction will occur when solutions of soluble ionic compounds are mixed together. One merely needs to identify all the ions present in the solution and then consider if possible cation/anion pairing could result in an insoluble compound. For example, mixing solutions of silver nitrate and sodium chloride will yield a solution containing Ag + , \(\ce{NO3-}\), Na + , and Cl − ions. Aside from the two ionic compounds originally present in the solutions, AgNO 3 and NaCl, two additional ionic compounds may be derived from this collection of ions: NaNO 3 and AgCl. The solubility guidelines indicate all nitrate salts are soluble but that AgCl is an insoluble combination. A precipitation reaction, therefore, is predicted to occur, as described by the following equation: \[\ce{Ag+}(aq)+\ce{Cl-}(aq)\rightarrow \ce{AgCl}(s)\hspace{20px}\ce{(net\: ionic)}\] Example \(\PageIndex{1}\): Predicting Precipitation Products Predict the result of mixing reasonably concentrated solutions of the following ionic compounds. If precipitation is expected, write a balanced net ionic equation for the reaction. potassium sulfate and barium nitrate lithium chloride and silver acetate lead nitrate and ammonium carbonate Solution (a) The two possible products for this combination are KNO 3 and BaSO 4 . The solubility guidelines indicate BaSO 4 is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is \[\ce{Ba^2+}(aq)+\ce{SO4^2-}(aq)\rightarrow \ce{BaSO4}(s)\] (b) The two possible products for this combination are LiC 2 H 3 O 2 and AgCl. The solubility guidelines indicate AgCl is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is \[\ce{Ag+}(aq)+\ce{Cl-}(aq)\rightarrow \ce{AgCl}(s)\] (c) The two possible products for this combination are PbCO 3 and NH 4 NO 3 . The solubility guidelines indicate PbCO 3 is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is \[\ce{Pb^2+}(aq)+\ce{CO3^2-}(aq)\rightarrow \ce{PbCO3}(s)\] Exercise \(\PageIndex{1}\) Which solution could be used to precipitate the barium ion, Ba 2+ , in a water sample: sodium chloride, sodium hydroxide, or sodium sulfate? What is the formula for the expected precipitate? Answer sodium sulfate, BaSO 4 Glossary insoluble of relatively low solubility; dissolving only to a slight extent precipitate insoluble product that forms from reaction of soluble reactants precipitation reaction reaction that produces one or more insoluble products; when reactants are ionic compounds, sometimes called double-displacement or metathesis salt ionic compound that can be formed by the reaction of an acid with a base that contains a cation and an anion other than hydroxide or oxide soluble of relatively high solubility; dissolving to a relatively large extent solubility the extent to which a substance may be dissolved in water, or any solvent |
Bookshelves/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/07%3A_The_Crystalline_Solid_State/7.01%3A_Molecular_Orbitals_and_Band_Structure/7.1.04%3A_Semiconductors-_Band_Gaps_Colors_Conductivity_and_Doping | Semiconductors, as we noted above, are somewhat arbitrarily defined as insulators with band gap energy < 3.0 eV (~290 kJ/mol). This cutoff is chosen because, as we will see, the conductivity of undoped semiconductors drops off exponentially with the band gap energy and at 3.0 eV it is very low. Also, materials with wider band gaps (e.g. SrTiO 3 , E gap = 3.2 eV) do not absorb light in the visible part of the spectrum There are a number of places where we find semiconductors in the periodic table: Early transition metal oxides and nitrides, especially those with d 0 electron counts such as TiO 2 , TaON, and WO 3 Oxides of later 3d elements such as Fe 2 O 3 , NiO, and Cu 2 O Layered transition metal chalcogenides with d 0 , d 2 and d 6 electron counts including TiS 2 , ZrS 2 , MoS 2 , WSe 2 , and PtS 2 d 10 copper and sliver halides, e.g., CuI, AgBr, and AgI Zincblende- and wurtzite-structure compounds of the p-block elements, especially those that are isoelectronic with Si or Ge, such as GaAs and CdTe. While these are most common, there are other p-block semiconductors that are not isoelectronic and have different structures, including GaS, PbS, and Se. 0 A 2" wafer cut from a GaAs single crystal. GaAs, like many p-block semiconductors, has the zincblende structure. The p-block octet semiconductors are by far the most studied and important for technological applications, and are the ones that we will discuss in detail. Zincblende- and wurtzite-structure semiconductors have 8 valence electrons per 2 atoms. These combinations include 4-4 (Si, Ge, SiC,…), 3-5 (GaAs, AlSb, InP,…), 2-6 (CdSe, HgTe, ZnO,…), and 1-7 (AgCl, CuBr,…) semiconductors. Other variations that add up to an octet configuration are also possible, such as Cu I In III Se 2 , which has the chalcopyrite structure, shown at the right. 0 The chalcopyrite structure is adopted by ABX2 octet semiconductors such as CuIInIIISe2 and CdIISnIVP2. The unit cell is doubled relative to the parent zincblende structure because of the ordered arrangement of cations. Each anion (yellow) is coordinated by two cations of each type (blue and red). How does the band gap energy vary with composition ? There are two important trends (1) Going down a group in the periodic table, the gap decreases : C (diamond) > Si > Ge > α-Sn E gap (eV): 5.4 1.1 0.7 0.0 This trend can be understood by recalling that E gap is related to the energy splitting between bonding and antibonding orbitals . This difference decreases (and bonds become weaker) as the principal quantum number increases. (2) For isoelectronic compounds, increasing ionicity results in a larger band gap . Ge < GaAs < ZnSe 0.7 1.4 2.8 eV Sn < InSb < CdTe < AgI 0.0 0.2 1.6 2.8 eV This trend can also be understood from a simple MO picture, as we discussed in Ch. 2. As the electronegativity difference Δχ increases, so does the energy difference between bonding and antibonding orbitals. The band gap is a very important property of a semiconductor because it determines its color and conductivity. Many of the applications of semiconductors are related to band gaps: Narrow gap materials (Hg x Cd 1 - x Te, VO 2 , InSb, Bi 2 Te 3 ) are used as infrared photodetectors and thermoelectrics (which convert heat to electricity). Wider gap materials (Si, GaAs, GaP, GaN, CdTe, CuIn x Ga 1-x Se 2 ) are used in electronics, light-emitting diodes, and solar cells. 0 Color wheel showing the colors and wavelengths of emitted light. Semiconductor solid solutions such as GaAs 1-x P x have band gaps that are intermediate between the end member compounds, in this case GaAs and GaP (both zincblende structure). Often, there is a linear relation between composition and band gap, which is referred to as Vegard's Law . This "law" is often violated in real materials, but nevertheless offers useful guidance for designing materials with specific band gaps. For example, red and orange light-emitting diodes (LED's) are made from solid solutions with compositions of GaP 0.40 As 0.60 and GaP 0.65 As 0.35 , respectively. Increasing the mole fraction of the lighter element (P) results in a larger band gap, and thus a higher energy of emitted photons. Colors of semiconductors The color of absorbed and emitted light both depend on the band gap of the semiconductor. Visible light covers the range of approximately 390-700 nm, or 1.8-3.1 eV. The color of emitted light from an LED or semiconductor laser corresponds to the band gap energy and can be read off the color wheel shown at the right. 0 Fe2O3 powder is reddish orange because of its 2.2 eV band gap The color of absorbed light includes the band gap energy, but also all colors of higher energy (shorter wavelength), because electrons can be excited from the valence band to a range of energies in the conduction band. Thus semiconductors with band gaps in the infrared (e.g., Si, 1.1 eV and GaAs, 1.4 eV) appear black because they absorb all colors of visible light. Wide band gap semiconductors such as TiO 2 (3.0 eV) are white because they absorb only in the UV. Fe 2 O 3 has a band gap of 2.2 eV and thus absorbs light with λ < 560 nm. It thus appears reddish-orange (the colors of light reflected from Fe 2 O 3 ) because it absorbs green, blue, and violet light. Similarly, CdS (E gap = 2.6 eV) is yellow because it absorbs blue and violet light. Electrons and holes in semiconductors Pure (undoped) semiconductors can conduct electricity when electrons are promoted, either by heat or light, from the valence band to the conduction band. The promotion of an electron (e - ) leaves behind a hole (h + ) in the valence band. The hole, which is the absence of an electron in a bonding orbital, is also a mobile charge carrier, but with a positive charge. The motion of holes in the lattice can be pictured as analogous to the movement of an empty seat in a crowded theater. An empty seat in the middle of a row can move to the end of the row (to accommodate a person arriving late to the movie) if everyone moves over by one seat. Because the movement of the hole is in the opposite direction of electron movement, it acts as a positive charge carrier in an electric field. The opposite process of excitation, which creates an electron-hole pair, is their recombination. When a conduction band electron drops down to recombine with a valence band hole, both are annihilated and energy is released. This release of energy is responsible for the emission of light in LEDs. 0 An electron-hole pair is created by adding heat or light energy E > Egap to a semiconductor (blue arrow). The electron-hole pair recombines to release energy equal to Egap (red arrow). At equilibrium, the creation and annihilation of electron-hole pairs proceed at equal rates. This dynamic equilibrium is analogous to the dissociation-association equilibrium of H + and OH - ions in water. We can write a mass action expression: \(n \times p = K_{eq} = n_{i}^{2}\) where n and p represent the number density of electrons and holes, respectively, in units of cm -3 . The intrinsic carrier concentration, n i , is equal to the number density of electrons or holes in an undoped semiconductor, where n = p = n i . Note the similarity to the equation for water autodissociation: \([H^{+}][OH^{-}] = K_{w}\) By analogy, we will see that when we increase n (e.g., by doping), p will decrease, and vice-versa, but their product will remain constant at a given temperature. Temperature dependence of the carrier concentration. Using the equations \(K_{eq} = e^{(\frac{- \Delta G^{o}}{RT})} \) and \(\Delta G^{o} = \Delta H^{o} - T \Delta S^{o}\), we can write: \[ n \times p = n_{i}^{2} = e^{(\frac{\Delta S^{o}} {R})} e^{(\frac{- \Delta H^{o}}{RT})}\] The entropy change for creating electron hole pairs is given by: \[\Delta S^{o} = R ln (N_{V}) + R ln (N_{V}) = R ln (N_{C}N_{V})\] where N V and N C are the effective density of states in the valence and conduction bands, respectively. and thus we obtain \[n_{i}^{2} = N_{C}N_{V} e^{({- \Delta H^{o}}{RT})}\] Since the volume change is negligible, \(\Delta H^{o} \approx \Delta E^{o}\), and therefore \(\frac {\Delta H^{o}}{R} \approx \frac{E_{gap}}{k}\), from which we obtain \[n_{i}^{2} = N_{C}N_{V} e^{(\frac{-E_{gap}}{kT})}\] and finally \[\mathbf{n= p = n_{i} = (N_{C}N_{V})^{\frac{1}{2}} e^{(\frac{-E_{gap}}{2kT})}}\] For pure Si (E gap = 1.1 eV) with N ≈ 10 22 /cm 3 , we can calculate from this equation a carrier density n i of approximately 10 10 /cm 3 at 300 K. This is about 12 orders of magnitude lower than the valence electron density of Al, the element just to the left of Si in the periodic table. Thus we expect the conductivity of pure semiconductors to be many orders of magnitude lower than those of metals. Conductivity of intrinsic semiconductors The conductivity (σ) is the product of the number density of carriers (n or p), their charge (e), and their mobility (µ). Recall from Chapter 6 that µ is the ratio of the carrier drift velocity to the electric field and has units of cm 2 /Volt-second. Typically electrons and holes have somewhat different mobilities (µ e and µ h , respectively) so the conductivity is given by: \[\sigma = ne\mu_{e} + pe\mu_{h}\] For either type of charge carrier, we recall from Ch. 6 that the mobility μ is given by: \[\mu = \frac{v_{drift}}{E} = \frac{e\tau}{m}\] where e is the fundamental unit of charge, τ is the scattering time, and m is the effective mass of the charge carrier. Taking an average of the electron and hole mobilities, and using n = p, we obtain \[\mathbf{\sigma= \sigma_{o} e^{(\frac{-E_{gap}}{2kT})}}, \: where \: \sigma_{o} = 2(N_{C}N_{V})^{\frac{1}{2}}e\mu\] By measuring the conductivity as a function of temperature, it is possible to obtain the activation energy for conduction, which is E gap /2. This kind of plot, which resembles an Arrhenius plot, is shown at the right for three different undoped semiconductors. The slope of the line in each case is -E gap /2k. 0 Plots of ln(σ) vs. inverse temperature for intrinsic semiconductors Ge (Egap = 0.7 eV), Si (1.1 eV) and GaAs (1.4 eV). The slope of the line is -Egap/2k. Doping of semiconductors. Almost all applications of semiconductors involve controlled doping , which is the substitution of impurity atoms, into the lattice. Very small amounts of dopants (in the parts-per-million range) dramatically affect the conductivity of semiconductors. For this reason, very pure semiconductor materials that are carefully doped - both in terms of the concentration and spatial distribution of impurity atoms - are needed. n- and p-type doping . In crystalline Si, each atom has four valence electrons and makes four bonds to its neighbors. This is exactly the right number of electrons to completely fill the valence band of the semiconductor. Introducing a phosphorus atom into the lattice (the positively charged atom in the figure at the right) adds an extra electron, because P has five valence electrons and only needs four to make bonds to its neighbors. The extra electron, at low temperature, is bound to the phosphorus atom in a hydrogen-like molecular orbital that is much larger than the 3s orbital of an isolated P atom because of the high dielectric constant of the semiconductor. In silicon, this "expanded" Bohr radius is about 42 Å, i.e., 80 times larger than in the hydrogen atom. The energy needed to ionize this electron – to allow it to move freely in the lattice - is only about 40–50 meV, which is not much larger the thermal energy (26 meV) at room temperature. Therefore the Fermi level lies just below the conduction band edge, and a large fraction of these extra electrons are promoted to the conduction band at room temperature, leaving behind fixed positive charges on the P atom sites. The crystal is n-doped , meaning that the majority carrier (electron) is n egatively charged. Alternatively, boron can be substituted for silicon in the lattice, resulting in p-type doping, in which the majority carrier (hole) is p ositively charged. Boron has only three valence electrons, and "borrows" one from the Si lattice, creating a positively charged hole that exists in a large hydrogen-like orbital around the B atom. This hole can become delocalized by promoting an electron from the valence band to fill the localized hole state. Again, this process requires only 40–50 meV, and so at room temperature a large fraction of the holes introduced by boron doping exist in delocalized valence band states. The Fermi level (the electron energy level that has a 50% probability of occupancy at zero temperature) lies just above the valence band edge in a p-type semiconductor. 0 n- and p-type doping of semiconductors involves substitution of electron donor atoms (light orange) or acceptor atoms (blue) into the lattice. These substitutions introduce extra electrons or holes, respectively, which are easily ionized by thermal energy to become free carriers. The Fermi level of a doped semiconductor is a few tens of mV below the conduction band (n-type) or above the valence band (p-type). As noted above, the doping of semiconductors dramatically changes their conductivity. For example, the intrinsic carrier concentration in Si at 300 K is about 10 10 cm -3 . The mass action equilibrium for electrons and holes also applies to doped semiconductors, so we can write: \[n \times p = n_{i}^{2} = 10^{20} cm^{-6} \: at \: 300K\] If we substitute P for Si at the level of one part-per-million, the concentration of electrons is about 10 16 cm -3 , since there are approximately 10 22 Si atoms/cm 3 in the crystal. According to the mass action equation, if n = 10 16 , then p = 10 4 cm -3 . There are three consequences of this calculation: The density of carriers in the doped semiconductor (10 16 cm -3 ) is much higher than in the undoped material (~10 10 cm -3 ), so the conductivity is also many orders of magnitude higher. The activation energy for conduction is only 40–50 meV, so the conductivity does not change much with temperature (unlike in the intrinsic semiconductor) The minority carriers (in this case holes) do not contribute to the conductivity, because their concentration is so much lower than that of the majority carrier (electrons). Similarly, for p-type materials, the conductivity is dominated by holes, and is also much higher than that of the intrinsic semiconductor. Chemistry of semiconductor doping. Sometimes it is not immediately obvious what kind of doping (n- or p-type) is induced by "messing up" a semiconductor crystal lattice. In addition to substitution of impurity atoms on normal lattice sites (the examples given above for Si), it is also possible to dope with vacancies - missing atoms - and with interstitials - extra atoms on sites that are not ordinarily occupied. Some simple rules are as follows: For substitutions, adding an atom to the right in the periodic table results in n-type doping, and an atom to the left in p-type doping. For example, when TiO 2 is doped with Nb on some of the Ti sites, or with F on O sites, the result is n-type doping. In both cases, the impurity atom has one more valence electron than the atom for which it was substituted. Similarly, substituting a small amount of Zn for Ga in GaAs, or a small amount of Li for Ni in NiO, results in p-type doping. Anion vacancies result in n-type doping, and cation vacancies in p-type doping. Examples are anion vacancies in CdS 1-x and WO 3-x , both of which give n-type semiconductors, and copper vacancies in Cu 1-x O, which gives a p-type semiconductor. Interstitial cations (e.g. Li) donate electrons to the lattice resulting in n-type doping. Interstitial anions are rather rare but would result in p-type doping. Sometimes, there can be both p- and n-type dopants in the same crystal, for example B and P impurities in a Si lattice, or cation and anion vacancies in a metal oxide lattice. In this case, the two kinds of doping compensate each other, and the doping type is determined by the one that is in higher concentration. A dopant can also be present on more than one site. For example, Si can occupy both the Ga and As sites in GaAs, and the two substitutions compensate each other. Si has a slight preference for the Ga site, however, resulting in n-type doping. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Mathematical_Methods_in_Chemistry_(Levitus)/08%3A_Calculus_in_More_than_One_Variable | Chapter Objectives Review the concept of partial derivative. Review the properties of partial derivatives. Be able to use the properties of partial derivatives in the context of physical chemistry problems. Review the concept of double and triple integrals. Learn the concept of equation of state. Understand the concept of a van der Waals gas from the molecular point of view. Learn about phase transitions and critical phenomena. 8.1: Functions of Two Independent Variables A function of two independent variables, z=f(x,y) , defines a surface in three-dimensional space. For a function of two or more variables, there are as many independent first derivatives as there are independent variables. 8.2: The Equation of State An equation of state is an expression relating the density of a fluid with its temperature and pressure. Note that the density is related to the number of moles and the volume, so it takes care of these two variables together. There is no single equation of state that predicts the behavior of all substances under all conditions. 8.3: The Chain Rule The chain rule allow us to create these ‘universal ’ relationships between the derivatives of different coordinate systems. 8.4: Double and Triple Integrals We can extend the idea of a definite integral to more dimensions. 8.5: Real Gases We have already mentioned some thermodynamic variables, but in order to make more connections between chemistry and math we need to introduce some concepts that we need to start discussing real gases. 8.6: Problems Thumbnail: Surface \(Σ\) with closed boundary \(∂Σ\). \(\vec{F}\) could be the \(\vec{E}\) or \(\vec{B}\) fields. \(n\) is the unit normal. (Public Domain; Maschen). |
Courses/SUNY_Oneonta/Chem_221%3A_Organic_Chemistry_I_(Bennett)/2%3ALab_Textbook_(Nichols)/02%3A_Chromatography/2.05%3A_Gas_Chromatography_(GC)/2.5E%3A_GC_Parameters | There are many factors that affect the separation (and/or retention times) in gas chromatography, including the type of column, sample concentration, oven temperature, and flow rate of carrier gas. The factors controllable by a student are described in this section, and are related to concentration and temperature. Dilution GC runs must be conducted on very dilute samples. A dilute sample enables adequate stationary-mobile phase equilibrium inside the column, resulting in narrow Gaussian-shaped peaks. If samples are too concentrated, the peak shapes will often be broad and flattened on top (Figure 2.87a), indicating the column (and/or detector) has been overwhelmed. Appropriate abundances (the y-axis on GC spectra) using a mass-spectrometer detector should be in the low millions (e.g. 1,500,000 TIC's or so, see Figure 2.87b). A GC should never be run on undiluted liquid. Overly concentrated samples do not only cause peaks to broaden and possibly overlap, they can also be harmful to detectors. Mass spectrometers require very small quantities of liquid, or over time, their filaments degrade. To prepare a routine GC sample, \(1.5 \: \text{mL}\) of a low-boiling solvent (e.g. methanol, clean acetone, diethyl ether, or dichloromethane) is added to a GC vial (Figure 2.88) along with one drop of compound. the instrument then injects roughly \(1 \: \mu \text{L}\) of this dilute solution into the instrument, which is further diverted internally, resulting in very small quantities entering the narrow capillary column and detector. The normal function of a GC instrument requires tiny quantities of material, which is a major advantage to this technique. Solvent Delay (With Mass Spectrometers) A properly prepared GC sample contains a small amount of compound dissolved in a solvent, and yet GC spectra (when paired with a mass spectrometer detector) often do not show any solvent peaks. The reason is that GC methods often include a " solvent delay ", where the detector is turned off until a certain amount of time has passed after injection. Even with the dilution inside the GC-MS instrument, the quantity of solvent reaching the detector would overwhelm and eventually degrade it. Therefore, the detector is left inactive until the solvent has passed through the column, and activated afterwards. If you look at any GC-MS spectrum, the retention time (x-axis) never starts at zero minutes, which represents the time the sample is injected on the column. In the GC spectrum of hexanes, the spectrum starts at 1.40 minutes, (as is indicated with an arrow in Figure 2.89). The solvent would have eluted from the column before this time. The GC solvent must be chosen such that the solvent will elute before the compound of interest (the solvent must have a significantly lower boiling point than the compound), so that the solvent delay may be set between elution of the solvent and sample. For example, methanol (b.p. \(65^\text{o} \text{C}\)) would have been an unsuitable solvent for the sample of hexanes (b.p. \(68^\text{o} \text{C}\)). If the solvent delay was set such that the hexane signals could be detected, the GC spectrum would have been overwhelmed by signals from the methanol, as the two compounds have very similar boiling points and would elute rather closely. If the detector was activated after the methanol had eluted, the instrument would likely miss detection of the closely eluting hexanes. Therefore, a significantly lower boiling solvent was used (diethyl ether, b.p. \(35^\text{o} \text{C}\)) to produce the GC spectrum of hexanes in Figure 2.89, and this solvent would have eluted before the solvent delay of 1.40 minutes. Oven Temperature The temperature of the GC oven is an easily adjusted variable (although it should be adjusted by an instructor, not student). The oven temperature has a dramatic effect on retention times, much in a similar way that changing the mobile phase in TLC has on \(R_f\). To demonstrate, a sample containing heptane, octane, nonane, and decane was run using three GC methods, differing only in their starting temperature (\(65^\text{o} \text{C}\), \(45^\text{o} \text{C}\), or \(35^\text{o} \text{C}\)). The order of elution of the compounds was the same in all methods (Figure 2.90), but the components eluted faster with a hotter oven temperature (had lower retention times). This result can be explained in that the compounds were more able to overcome their intermolecular attraction with the stationary phase (the coating of the GC column) when there was more available thermal energy. This resulted in compounds spending less time adhering to the stationary phase, and less time retained by the column. Using a Temperature Ramp A GC can be run at one oven temperature the entire time (an "isothermal" run), or the method may include temperature programming. In a programmed method, the temperature may remain fixed for a certain length of time, and then may include a " ramp ", where the temperature is increased at a steady rate during the run. A graphical display of a ramp generated by the GC instrument is shown in Figure 2.91. Temperature programming is useful because the longer a compound stays in the column, the wider it elutes as diffusion broadens the sample (analogously to how spots broaden during elution in TLC). Figure 2.92a shows an isothermal GC run of four compounds, and the peak shape significantly broadens as the retention time increases. Figure 2.92b shows a GC run of the same four compounds eluted using a temperature ramp. The ramp allows the compounds to elute faster, and for peak shape to sharpen and improve. The optimal ramp for each situation often takes experimentation to achieve the best peak shape and separation of components. |
Courses/Woodland_Community_College/Chem_2B%3A_Introductory_Chemistry_II/09%3A_Energy_Metabolism/9.07%3A_Stage_II_of_Lipid_Catabolism | Learning Objectives To describe the reactions needed to completely oxidize a fatty acid to carbon dioxide and water. Like glucose, the fatty acids released in the digestion of triglycerides and other lipids are broken down in a series of sequential reactions accompanied by the gradual release of usable energy. Some of these reactions are oxidative and require nicotinamide adenine dinucleotide (NAD + ) and flavin adenine dinucleotide (FAD). The enzymes that participate in fatty acid catabolism are located in the mitochondria, along with the enzymes of the citric acid cycle, the electron transport chain, and oxidative phosphorylation. This localization of enzymes in the mitochondria is of the utmost importance because it facilitates efficient utilization of energy stored in fatty acids and other molecules. Fatty acid oxidation is initiated on the outer mitochondrial membrane. There the fatty acids, which like carbohydrates are relatively inert, must first be activated by conversion to an energy-rich fatty acid derivative of coenzyme A called fatty acyl-coenzyme A (CoA). The activation is catalyzed by acyl-CoA synthetase . For each molecule of fatty acid activated, one molecule of coenzyme A and one molecule of adenosine triphosphate (ATP) are used, equaling a net utilization of the two high-energy bonds in one ATP molecule (which is therefore converted to adenosine monophosphate [AMP] rather than adenosine diphosphate [ADP]): The fatty acyl-CoA diffuses to the inner mitochondrial membrane, where it combines with a carrier molecule known as carnitine in a reaction catalyzed by carnitine acyltransferase . The acyl-carnitine derivative is transported into the mitochondrial matrix and converted back to the fatty acyl-CoA. Steps in the β-Oxidation of Fatty Acids Further oxidation of the fatty acyl-CoA occurs in the mitochondrial matrix via a sequence of four reactions known collectively as β-oxidation because the β-carbon undergoes successive oxidations in the progressive removal of two carbon atoms from the carboxyl end of the fatty acyl-CoA (Figure \(\PageIndex{1}\)). The first step in the catabolism of fatty acids is the formation of an alkene in an oxidation reaction catalyzed by acyl-CoA dehydrogenase . In this reaction, the coenzyme FAD accepts two hydrogen atoms from the acyl-CoA, one from the α-carbon and one from the β-carbon, forming reduced flavin adenine dinucleotide (FADH 2 ). The FADH 2 is reoxidized back to FAD via the electron transport chain that supplies energy to form 1.5–2 molecules of ATP. Next, the trans -alkene is hydrated to form a secondary alcohol in a reaction catalyzed by enoyl-CoA hydratase . The enzyme forms only the L-isomer. The secondary alcohol is then oxidized to a ketone by β-hydroxyacyl-CoA dehydrogenase , with NAD + acting as the oxidizing agent. The reoxidation of each molecule of NADH to NAD + by the electron transport chain furnishes 2.5–3 molecules of ATP. The final reaction is cleavage of the β-ketoacyl-CoA by a molecule of coenzyme A. The products are acetyl-CoA and a fatty acyl-CoA that has been shortened by two carbon atoms. The reaction is catalyzed by thiolase . The shortened fatty acyl-CoA is then degraded by repetitions of these four steps, each time releasing a molecule of acetyl-CoA. The overall equation for the β-oxidation of palmitoyl-CoA (16 carbon atoms) is as follows: Because each shortened fatty acyl-CoA cycles back to the beginning of the pathway, β-oxidation is sometimes referred to as the fatty acid spiral. The fate of the acetyl-CoA obtained from fatty acid oxidation depends on the needs of an organism. It may enter the citric acid cycle and be oxidized to produce energy, it may be used for the formation of water-soluble derivatives known as ketone bodies, or it may serve as the starting material for the synthesis of fatty acids. For more information about the citric acid cycle, see Section 20.4 . Looking Closer: Ketone Bodies In the liver, most of the acetyl-CoA obtained from fatty acid oxidation is oxidized by the citric acid cycle. However, some of the acetyl-CoA is used to synthesize a group of compounds known as ketone bodies : acetoacetate, β-hydroxybutyrate, and acetone. Two acetyl-CoA molecules combine, in a reversal of the final step of β-oxidation, to produce acetoacetyl-CoA. The acetoacetyl-CoA reacts with another molecule of acetyl-CoA and water to form β-hydroxy-β-methylglutaryl-CoA, which is then cleaved to acetoacetate and acetyl-CoA. Most of the acetoacetate is reduced to β-hydroxybutyrate, while a small amount is decarboxylated to carbon dioxide and acetone. The acetoacetate and β-hydroxybutyrate synthesized by the liver are released into the blood for use as a metabolic fuel (to be converted back to acetyl-CoA) by other tissues, particularly the kidney and the heart. Thus, during prolonged starvation, ketone bodies provide about 70% of the energy requirements of the brain. Under normal conditions, the kidneys excrete about 20 mg of ketone bodies each day, and the blood levels are maintained at about 1 mg of ketone bodies per 100 mL of blood. In starvation, diabetes mellitus, and certain other physiological conditions in which cells do not receive sufficient amounts of carbohydrate, the rate of fatty acid oxidation increases to provide energy. This leads to an increase in the concentration of acetyl-CoA. The increased acetyl-CoA cannot be oxidized by the citric acid cycle because of a decrease in the concentration of oxaloacetate, which is diverted to glucose synthesis. In response, the rate of ketone body formation in the liver increases further, to a level much higher than can be used by other tissues. The excess ketone bodies accumulate in the blood and the urine, a condition referred to as ketosis . When the acetone in the blood reaches the lungs, its volatility causes it to be expelled in the breath. The sweet smell of acetone, a characteristic of ketosis, is frequently noticed on the breath of severely diabetic patients. Because two of the three kinds of ketone bodies are weak acids, their presence in the blood in excessive amounts overwhelms the blood buffers and causes a marked decrease in blood pH (to 6.9 from a normal value of 7.4). This decrease in pH leads to a serious condition known as acidosis . One of the effects of acidosis is a decrease in the ability of hemoglobin to transport oxygen in the blood. In moderate to severe acidosis, breathing becomes labored and very painful. The body also loses fluids and becomes dehydrated as the kidneys attempt to get rid of the acids by eliminating large quantities of water. The lowered oxygen supply and dehydration lead to depression; even mild acidosis leads to lethargy, loss of appetite, and a generally run-down feeling. Untreated patients may go into a coma. At that point, prompt treatment is necessary if the person’s life is to be saved. ATP Yield from Fatty Acid Oxidation The amount of ATP obtained from fatty acid oxidation depends on the size of the fatty acid being oxidized. For our purposes here. we’ll study palmitic acid, a saturated fatty acid with 16 carbon atoms, as a typical fatty acid in the human diet. Calculating its energy yield provides a model for determining the ATP yield of all other fatty acids. The breakdown by an organism of 1 mol of palmitic acid requires 1 mol of ATP (for activation) and forms 8 mol of acetyl-CoA. Recall from Table 20.4.1 that each mole of acetyl-CoA metabolized by the citric acid cycle yields 10 mol of ATP. The complete degradation of 1 mol of palmitic acid requires the β-oxidation reactions to be repeated seven times. Thus, 7 mol of NADH and 7 mol of FADH 2 are produced. Reoxidation of these compounds through respiration yields 2.5–3 and 1.5–2 mol of ATP, respectively. The energy calculations can be summarized as follows: 1 mol of ATP is split to AMP and 2Pi −2 ATP 8 mol of acetyl-CoA formed (8 × 12) 96 ATP 7 mol of FADH2 formed (7 × 2) 14 ATP 7 mol of NADH formed (7 × 3) 21 ATP Total 129 ATP The number of times β-oxidation is repeated for a fatty acid containing n carbon atoms is n/2 – 1 because the final turn yields two acetyl-CoA molecules. The combustion of 1 mol of palmitic acid releases a considerable amount of energy: \[C_{16}H_{32}O_2 + 23O_2 → 16CO_2 + 16H_2O + 2,340\; kcal \nonumber \] The percentage of this energy that is conserved by the cell in the form of ATP is as follows: \[\mathrm{\dfrac{energy\: conserved}{total\: energy\: available}\times100=\dfrac{(129\: ATP)(7.4\: kcal/ATP)}{2,340\: kcal}\times100=41\%} \nonumber \] The efficiency of fatty acid metabolism is comparable to that of carbohydrate metabolism, which we calculated to be 42%. For more information about the efficiency of fatty acid metabolism, see II of Carbohydrate Catabolism" data-cke-saved-href="/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20:_Energy_Metabolism/20.05:_Stage_II_of_Carbohydrate_Catabolism" href="/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20:_Energy_Metabolism/20.05:_Stage_II_of_Carbohydrate_Catabolism" data-quail-id="91">Section 20.5. The oxidation of fatty acids produces large quantities of water. This water, which sustains migratory birds and animals (such as the camel) for long periods of time. Summary Fatty acids, obtained from the breakdown of triglycerides and other lipids, are oxidized through a series of reactions known as β-oxidation. In each round of β-oxidation, 1 molecule of acetyl-CoA, 1 molecule of NADH, and 1 molecule of FADH 2 are produced. The acetyl-CoA, NADH, and FADH 2 are used in the citric acid cycle, the electron transport chain, and oxidative phosphorylation to produce ATP. |
Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.07%3A_Optical_Activity_and_Racemic_Mixtures | Learning Objective define and explain the lack of optical activity of racemic mixtures determine the percent composition of an enantiomeric mixture from polarimetry data and the for specific rotation formula Racemic Mixtures (Racimates) A racemic mixture is a 50:50 mixture of two enantiomers. Racemic mixtures were an interesting experimental discovery because two optically active samples can be combined in a 1:1 ratio to create an optically INACTIVE sample. Polarimetry is used to measure optical activity. The history and theoretical foundation are discussed below. Identifying and distinguishing enantiomers is inherently difficult, since their physical and chemical properties are largely identical. Fortunately, a nearly two hundred year old discovery by the French physicist Jean-Baptiste Biot has made this task much easier. This discovery disclosed that the right- and left-handed enantiomers of a chiral compound perturb plane-polarized light in opposite ways. This perturbation is unique to chiral molecules, and has been termed optical activity . Polarimetry Plane-polarized light is created by passing ordinary light through a polarizing device, which may be as simple as a lens taken from polarizing sun-glasses. Such devices transmit selectively only that component of a light beam having electrical and magnetic field vectors oscillating in a single plane. The plane of polarization can be determined by an instrument called a polarimeter , shown in the diagram below. Monochromatic (single wavelength) light, is polarized by a fixed polarizer next to the light source. A sample cell holder is located in line with the light beam, followed by a movable polarizer (the analyzer) and an eyepiece through which the light intensity can be observed. In modern instruments an electronic light detector takes the place of the human eye. In the absence of a sample, the light intensity at the detector is at a maximum when the second (movable) polarizer is set parallel to the first polarizer (α = 0º). If the analyzer is turned 90º to the plane of initial polarization, all the light will be blocked from reaching the detector. Chemists use polarimeters to investigate the influence of compounds (in the sample cell) on plane polarized light. Samples composed only of achiral molecules (e.g. water or hexane), have no effect on the polarized light beam. However, if a single enantiomer is examined (all sample molecules being right-handed, or all being left-handed), the plane of polarization is rotated in either a clockwise (positive) or counter-clockwise (negative) direction, and the analyzer must be turned an appropriate matching angle, α , if full light intensity is to reach the detector. In the above illustration, the sample has rotated the polarization plane clockwise by +90º, and the analyzer has been turned this amount to permit maximum light transmission. The observed rotations (\(\alpha\)) of enantiomers are opposite in direction. One enantiomer will rotate polarized light in a clockwise direction, termed dextrorotatory or (+), and its mirror-image partner in a counter-clockwise manner, termed levorotatory or (–). The prefixes dextro and levo come from the Latin dexter , meaning right, and laevus , for left, and are abbreviated d and l respectively. If equal quantities of each enantiomer are examined , using the same sample cell, then the magnitude of the rotations will be the same, with one being positive and the other negative. To be absolutely certain whether an observed rotation is positive or negative it is often necessary to make a second measurement using a different amount or concentration of the sample. In the above illustration, for example, α might be –90º or +270º rather than +90º. If the sample concentration is reduced by 10%, then the positive rotation would change to +81º (or +243º) while the negative rotation would change to –81º, and the correct α would be identified unambiguously. Since it is not always possible to obtain or use samples of exactly the same size, the observed rotation is usually corrected to compensate for variations in sample quantity and cell length. Thus it is common practice to convert the observed rotation, α, to a specific rotation , by the following formula: \[[\alpha]_D = \dfrac{\alpha}{l c} \tag{5.3.1}\] where \([\alpha]_D\) is the specific rotation \(l\) is the cell length in dm \(c\) is the concentration in g/ml \(D\) designates that the light used is the 589 line from a sodium lamp Compounds that rotate the plane of polarized light are termed optically active . Each enantiomer of a stereoisomeric pair is optically active and has an equal but opposite-in-sign specific rotation. Specific rotations are useful in that they are experimentally determined constants that characterize and identify pure enantiomers. For example, the lactic acid and carvone enantiomers discussed earlier have the following specific rotations. 0 1 Carvone from caraway: [α]D = +62.5º this isomer may be referred to as (+)-carvone or d-carvone Carvone from spearmint: [α]D = –62.5º this isomer may be referred to as (–)-carvone or l-carvone Lactic acid from muscle tissue: [α]D = +2.5º this isomer may be referred to as (+)-lactic acid or d-lactic acid Lactic acid from sour milk: [α]D = –2.5º this isomer may be referred to as (–)-lactic acid or l-lactic acid A 50:50 mixture of enantiomers has no observable optical activity. Such mixtures are called racemates or racemic modifications, and are designated (±). When chiral compounds are created from achiral compounds, the products are racemic unless a single enantiomer of a chiral co-reactant or catalyst is involved in the reaction. The addition of HBr to either cis- or trans-2-butene is an example of racemic product formation (the chiral center is colored red in the following equation). 0 1 2 CH3CH=CHCH3 + HBr NaN (±) CH3CH2CHBrCH3 Chiral organic compounds isolated from living organisms are usually optically active, indicating that one of the enantiomers predominates (often it is the only isomer present). This is a result of the action of chiral catalysts we call enzymes, and reflects the inherently chiral nature of life itself. Chiral synthetic compounds, on the other hand, are commonly racemates, unless they have been prepared from enantiomerically pure starting materials. There are two ways in which the condition of a chiral substance may be changed: 1. A racemate may be separated into its component enantiomers. This process is called resolution . 2. A pure enantiomer may be transformed into its racemate. This process is called racemization . Enantiomeric Excess The "optical purity" is a comparison of the optical rotation of a pure sample of unknown stereochemistry versus the optical rotation of a sample of pure enantiomer. It is expressed as a percentage. If the sample only rotates plane-polarized light half as much as expected, the optical purity is 50%. Because R and S enantiomers have equal but opposite optical activity, it naturally follows that a 50:50 racemic mixture of two enantiomers will have no observable optical activity. If we know the specific rotation for a chiral molecule, however, we can easily calculate the ratio of enantiomers present in a mixture of two enantiomers, based on its measured optical activity. When a mixture contains more of one enantiomer than the other, chemists often use the concept of enantiomeric excess (ee) to quantify the difference. Enantiomeric excess can be expressed as: For example, a mixture containing 60% R enantiomer (and 40% S enantiomer) has a 20% enantiomeric excess of R: ((60-50) x 100) / 50 = 20 %. Example The specific rotation of ( S )-carvone is (+)61°, measured 'neat' (pure liquid sample, no solvent). The optical rotation of a neat sample of a mixture of R and S carvone is measured at (-)23°. Which enantiomer is in excess, and what is its ee? What are the percentages of ( R )- and ( S )-carvone in the sample? Solution The observed rotation of the mixture is levorotary (negative, counter-clockwise), and the specific rotation of the pure S enantiomer is given as dextrorotary (positive, clockwise), meaning that the pure R enantiomer must be levorotary, and the mixture must contain more of the R enantiomer than of the S enantiomer. Rotation ( R / S Mix) = [Fraction( S ) × Rotation ( S )] + [Fraction( R ) × Rotation ( R )] Let Fraction ( S ) = x, therefore Fraction ( R ) = 1 – x Rotation ( R / S Mix) = x[Rotation ( S )] + (1 – x)[Rotation ( R )] –23 = x(+61) + (1 – x)(–61) Solve for x: x = 0.3114 and (1 – x) = 0.6885 Therefore the percentages of ( R )- and ( S )-carvone in the sample are 68.9% and 31.1%, respectively. ee = [(% more abundant enantiomer – 50) × 100]/50 = [68.9 – 50) × 100]/50 = 37.8% Chiral molecules are often labeled according to the sign of their specific rotation, as in (S)-(+)-carvone and (R)-(-)-carvone, or (±)-carvone for the racemic mixture. However, there is no relationship whatsoever between a molecule's R/S designation and the sign of its specific rotation. Without performing a polarimetry experiment or looking in the literature, we would have no idea that (-)-carvone has the R configuration and (+)-carvone has the S configuration. Separation of Chiral Compounds As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution. Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. For example, if a racemic mixture of a chiral alcohol is reacted with a enantiomerically pure carboxylic acid, the result is a mixture of diastereomers: in this case, because the pure (R) entantiomer of the acid was used, the product is a mixture of (R-R) and (R-S) diastereomeric esters, which can, in theory, be separated by their different physical properties. Subsequent hydrolysis of each separated ester will yield the 'resolved' (enantiomerically pure) alcohols. The used in this technique are known as ' Exercise 1 A sample with a concentration of 0.3 g/mL was placed in a cell with a length of 5 cm. The resulting rotation at the sodium D line was +1.52°. What is the [α] D ? Solution 5 cm = 0.5 dm [α] D = α/(c x l) = +1.52/(0.3 x 0.5) = +10.1° |
Bookshelves/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.14%3A_The_Direction_of_Proton_Transfer | Brønsted-Lowry acidity is a special case of Lewis acidity. In Lewis acidity, an electron donor shares electrons with an electron acceptor, forming a bond. In some cases, the electron acceptor is a proton. If the proton accepts electrons from the donor, but does not relinquish its bond to its previous partner, the interaction is called a hydrogen bond. If the proton exchanges a new bond with the donor for an old bond with its previous partner -- if it releases a pair of electrons to its partned as it accepts a pair of electrons from the new donor -- the event is called proton transfer. The compound that provided the proton is called a Brønsted-Lowry acid. The compound that donated the new bond to the proton is called the Brønsted-Lowry base. Because the proton allowed its former partner to take the pair of electrons from their former bond, that partner becomes a Lewis base. In a proton transfer, the proton moves from one Lewis base to another. The proton could conceivably move back to its original partner, however. The original partner could simply donate its pair of electrons to the proton again. It would displace the new partner and win the proton back. This situation is called reversibility. Reactions can often move back and forth. In order to convey this idea, when illustrating or writing about these reactions, a pair of opposing arrows are used to show that the reaction can go from left to right as written, as well as from right to left. In a reversible reaction, the change that occurs in the reaction can be undone. The reaction can move forwards and backwards. In most cases, the reaction settles out on one side or the other. Either the reaction goes mostly forward or it goes mostly backward. The point where the reaction settles is termed the equilibrium. At equilibrium, the reactants that come together may mostly be converted into new products. Looking at an equation or diagram of the reaction, , the equilibrium is said to "lie to the right", because the products of the reaction are usually written on the right hand side of the reaction arrow or equilibrium arrow. Some reactions "lie to the left", meaning very little of the original reactants are ultimately converted to the products show. The equilibrium is the balance established between the products formed and the original reactants in a reversible reaction. In a proton transfer, the equilibrium is determined by how tightly the proton is held by each Brønsted acid. The proton will simply remain bonded to whichever compound binds it more tightly. If the difference in binding is great, the equilibrium will lie far to the left or far to the right. If the difference in proton binding is small, there will be a mixture, in which the proton could be in either position. Remember, most reactions involve zillions of molecules. There is plenty of room for mixtures. We can predict where the proton will end up by looking at pKas. A higher pKa means the proton is more tightly held. By comparing the pKa's of the Brønsted acids on both sides of the equation, we can determine which compound will retain the proton. The equilibrium will lie towards the compound with the higher pKa. This idea is illustrated in the equilibrium between hydronium ion and ammonium ion. As another example, if hydrogen chloride is dissolved in water, the HCl may give up its proton to the water. Water has a lone pair and can act as a base. However, in doing so, the water will form hydronium ion, H 3 O + . Hydronium ion is Brønsted acidic and can provide a proton to something else that has a lone pair, such as a chloride ion. This reaction could go back and forth. Where will it settle out? HCl has a pKa of -8. Hydronium ion has a pKa of about -1.7. The equilibrium in the reaction described above lies to the right, towards the hydronium ion produced when the hydrogen chloride dissociates. The proton will remain on the oxygen. Exercise \(\PageIndex{1}\) Write an equation for the proton transfer reactions that could happen in each of the following mixtures. Use structures in your equations. Predict the position of the equilibrium in each case. a) HF plus water b) CH 3 CO 2 H plus ammonia c) phenol (C 6 H 5 OH) plus sodium carbonate d) HCN plus acetonitrile Answer a Answer b Answer c Answer d Exercise \(\PageIndex{2}\) Use curved arrows to show the proton transfer reaction between the following compounds. Predict the products for these proton transfer reactions. Use pKa to determine whether each reaction is reactant favored OR mixture OR product favored. Answer |
Bookshelves/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/19%3A_Buffers_and_Titrations/19.7.1_Initial_pH_for_a_Strong_Acid_Strong_Base_Titration_(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 is the first of a series of theoretical calculation for a titration of a strong acid with a strong base. You start with 50.0 mL of a 0.100 M solution of HCl. What is the initial pH? Because HCl is a strong acid we can say: {HCl} = {H 3 O + } = 0.100 M So the pH can be found directly pH = -Log{0.100 M} = 1.00 Link to Video Initial pH for a Strong Acid/Strong Base Titration: https://youtu.be/wp3QFchYasM Attribution Prof. Steven Farmer ( Sonoma State University ) |
Courses/Pasadena_City_College/CHEM_001A%3A_General_Chemistry_and_Chemical_Analysis/22%3A_Organic_Chemistry | Template:HideTOC 22.1: Fragrances and Odor 22.2: Carbon- Why It Is Unique 22.3: Hydrocarbons- Compounds Containing Only Carbon and Hydrocarbon 22.4: Alkanes- Saturated Hydrocarbons Simple alkanes exist as a homologous series, in which adjacent members differ by a CH2 unit. 22.5: Alkenes and Alkynes As noted before, alkenes are hydrocarbons with carbon-to-carbon double bonds (R2C=CR2) and alkynes are hydrocarbons with carbon-to-carbon triple bonds (R–C≡C–R). Collectively, they are called unsaturated hydrocarbons because they have fewer hydrogen atoms than does an alkane with the same number of carbon atoms, as is indicated in the following general formulas: 22.6: Hydrocarbon Reactions The alkanes and cycloalkanes, with the exception of cyclopropane, are probably the least chemically reactive class of organic compounds. Alkanes contain strong carbon-carbon single bonds and strong carbon-hydrogen bonds. The carbon-hydrogen bonds are only very slightly polar. Alkanes can be burned, alkanes can react with some of the halogens, breaking carbon-hydrogen bonds, and alkanes can crack by breaking the carbon-carbon bonds. 22.7: Aromatic Hydrocarbons Aromatic hydrocarbons contain ring structures with delocalized π electron systems. 22.8: Functional Groups Functional groups are atoms or small groups of atoms (two to four) that exhibit a characteristic reactivity. A particular functional group will almost always display its characteristic chemical behavior when it is present in a compound. Because of their importance in understanding organic chemistry, functional groups have characteristic names that often carry over in the naming of individual compounds incorporating specific groups 22.9: Alcohols In the IUPAC system, alcohols are named by changing the ending of the parent alkane name to -ol. Alcohols are classified according to the number of carbon atoms attached to the carbon atom that is attached to the OH group. Alcohols have higher boiling points than do ethers and alkanes of similar molar masses because the OH group allows alcohol molecules to engage in hydrogen bonding. Alcohols of four or fewer carbon atoms are soluble in water because the alcohol molecules engage in hydrogen bond 22.10: Aldehydes and Ketones The common names of aldehydes are taken from the names of the corresponding carboxylic acids: formaldehyde, acetaldehyde, and so on. The polar carbon-to-oxygen double bond causes aldehydes and ketones to have higher boiling points than those of ethers and alkanes of similar molar masses but lower than those of comparable alcohols that engage in intermolecular hydrogen bonding. Aldehydes are readily oxidized to carboxylic acids, whereas ketones resist oxidation. 22.11: Carboxylic Acids and Esters Aldehydes and ketones are characterized by the presence of a carbonyl group (C=O), and their reactivity can generally be understood by recognizing that the carbonyl carbon contains a partial positive charge ( δ+ ) and the carbonyl oxygen contains a partial negative charge ( δ− ). Aldehydes are typically more reactive than ketones. 22.12: Ethers To give ethers common names, simply name the groups attached to the oxygen atom, followed by the generic name ether. If both groups are the same, the group name should be preceded by the prefix di-. Ether molecules have no OH group and thus no intermolecular hydrogen bonding. Ethers therefore have quite low boiling points for a given molar mass. Ether molecules have an oxygen atom and can engage in hydrogen bonding with water molecules. An ether molecule has about the same solubility in water as 22.13: Amines An amine is a derivative of ammonia in which one, two, or all three hydrogen atoms are replaced by hydrocarbon groups. Amines are classified as primary, secondary, or tertiary by the number of hydrocarbon groups attached to the nitrogen atom. Primary and secondary amines have higher boiling points than those of alkanes or ethers of similar molar mass because they can engage in intermolecular hydrogen bonding. Amines are bases; they react with acids to form salts. |
Courses/Pasadena_City_College/Chem_2A_(Ku)_Textbook/02%3A_Atoms_and_Elements | 2.1: Chemistry - The Central Science Chemistry is the study of matter and how it behaves. The scientific method is the general process by which we learn about the natural universe. 2.2: What is Matter? 2.2.1: In Your Room 2.2.2: What is Matter? 2.2.3: Classifying Matter According to Its State—Solid, Liquid, and Gas 2.2.4: Classifying Matter According to Its Composition 2.2.5: Differences in Matter- Physical and Chemical Properties 2.2.6: Changes in Matter - Physical and Chemical Changes 2.3: What are atoms? 2.3.1: Cutting Aluminum until you get Atoms 2.3.2: Indivisible - The Atomic Theory 2.3.3: The Nuclear Atom 2.3.4: The Properties of Protons, Neutrons, and Electrons 2.3.5: Elements- Defined by Their Number of Protons 2.3.6: Looking for Patterns - The Periodic Table 2.3.7: Ions - Losing and Gaining Electrons 2.3.8: Isotopes - When the Number of Neutrons Varies 2.3.9: Atomic Mass - The Average Mass of an Element’s Atoms |
Courses/Nassau_Community_College/CHE200_-_Introduction_to_Organic_Chemistry_(Resch)/13%3A_Reactions_at_the_-Carbon_Part_I/13.02%3A_Review_of_Acidity_at_the_-Carbon | Let's review what we learned in section 7.6 about the acidity of a proton on an a-carbon and the structure of the relevant conjugate base, the enolate ion. Remember that this acidity can be explained by the fact that the negative charge on the enolate conjugate base is delocalized by resonance to both the \(\alpha\)-carbon and the carbonyl oxygen. The \(\alpha\)-carbon on the enolate is \(sp^2\)-hybridized with trigonal planar geometry, as are the carbonyl carbon and oxygen atoms (now would be a good time to go back to section 2.1, section 2.2, and section 2.3 to review, if necessary, the geometry of p-bonding in conjugated systems). \(pK_a\) of a typical a-proton in aqueous solution is approximately 18-20: acidic, but only weakly so. Recall from section 7.8, however, that the effective \(pK_a\) of a functional group on an enzyme-bound molecule can be altered dramatically by the 'microenvironment' of the active site. In order to lower the \(pK_a\) of an \(\alpha\)-proton, an enzyme catalyzing a reaction that begins with an a-proton abstraction step must further stabilize the negative charge that develops on the oxygen atom of the (enolate) conjugate base. Different enzymes have evolved different strategies for accomplishing this task: in some cases, a metal cation (often \(Zn^{+2}\)) is bound in the active site to provide a stabilizing ion-ion interaction. In other cases, stabilization is provided by a proton-donating group positioned near the oxygen. As a third possibility, the active site architecture sometimes provides one or more stabilizing hydrogen bond donor groups. In most of the mechanism illustrations in this chapter where an enolate intermediate is depicted, stabilizing metal ions or hydrogen bond interactions will not be explicitly drawn, for the sake of clarity. However, whenever you see an enolate intermediate in an enzyme-catalyzed reaction, you should remember that there are stabilizing interactions in play inside the active site. |
Courses/University_of_North_Carolina_Charlotte/CHEM_2141%3A__Survey_of_Physical_Chemistry/08%3A_Optional-_Special_topics/8.01%3A_Chemistry_of_Cooking_(Rodriguez-Velazquez)/8.1.02%3A_Flour/8.1.2.08%3A_Flour_in_Baking | Flour forms the foundation for bread, cakes, and pastries. It may be described as the skeleton, which supports the other ingredients in a baked product. This applies to both yeast and chemically leavened products. The strength of flour is represented in protein (gluten) quality and quantity. This varies greatly from flour to flour. The quality of the protein indicates the strength and stability of the flour, and the result in bread making depends on the method used to develop the gluten by proper handling during the fermentation. Gluten is a rubber-like substance that is formed by mixing flour with water. Before it is mixed it contains two proteins. In wheat, these two proteins are gliadin and glutenin . Although we use the terms protein and gluten interchangeably, gluten only develops once the flour is moistened and mixed. The protein in the flour becomes gluten. Hard spring wheat flours are considered the best for bread making as they have a larger percentage of good quality gluten than soft wheat flours. It is not an uncommon practice for mills to blend hard spring wheat with hard winter wheat for the purpose of producing flour that combines the qualities of both. Good bread flour should have about 13% gluten. Storing Flour Flour should be kept in a dry, well-ventilated storeroom at a fairly uniform temperature. A temperature of about 21°C (70°F) with a relative humidity of 60% is considered ideal. Flour should never be stored in a damp place. Moist storerooms with temperatures greater than 23°C (74°F) are conducive to mould growth, bacterial development, and rapid deterioration of the flour. A well-ventilated storage room is necessary because flour absorbs and retains odors. For this reason, flour should not be stored in the same place as onions, garlic, coffee, or cheese, all of which give off strong odors. Flour Tests Wheat that is milled and blended with modern milling methods produce flours that have a fairly uniform quality all year round and, if purchased from a reliable mill, they should not require any testing for quality. The teacher, student, and professional baker, however, should be familiar with qualitative differences in flours and should know the most common testing methods. Flours are mainly tested for: Color Absorption Gluten strength Baking quality Other tests, done in a laboratory, are done for: Albumen Starch Sugar Dextrin Mineral and fat content Color The color of the flour has a direct bearing on baked bread, providing that fermentation has been carried out properly. The addition of other ingredients to the dough, such as brown sugar , malt, molasses, salt, and colored margarine, also affects the color of bread. To test the color of the flour, place a small quantity on a smooth glass, and with a spatula, work until a firm smooth mass about 5 cm (2 in.) square is formed. The thickness should be about 2 cm (4/5 in.) at the back of the plate to a thin film at the front. The test should be made in comparison with a flour of known grade and quality, both flours being worked side by side on the same glass. A creamy white color indicates a hard flour of good gluten quality. A dark or greyish color indicates a poor grade of flour or the presence of dirt. Bran specks indicate a low grade of flour. After making a color comparison of the dry samples, dip the glass on an angle into clean water and allow to partially dry. Variations in color and the presence of bran specks are more easily identified in the damp samples. Absorption Flours are tested for absorption because different flours absorb different amounts of water and therefore make doughs of different consistencies. The absorption ability of a flour is usually between 55% and 65%. To determine the absorption factor, place a small quantity of flour (100 g/4 oz.) in a bowl. Add water gradually from a beaker containing a known amount of water. As the water is added, mix with a spoon until the dough reaches the desired consistency. You can knead the dough by hand for final mixing and determination of consistency. Weigh the unused water. Divide the weight of the water used by the weight of the flour used. The result is the absorption ability in percentage. For example: Weight of flour used 100 g (4 oz.) Weight of water used 60 g (2.7 oz.) T therefore absorption = 6/10 or 60% Prolonged storage in a dry place results in a natural moisture loss in flour and has a noticeable effect on the dough. For example, a sack of flour that originally weighed 40 kg (88 lb.) with a moisture content of 14% may be reduced to 39 kg (86 lb.) during storage. This means that 1 kg (2 lb.) of water is lost and must be made up when mixing. The moisture content of the wheat used to make the flour is also important from an economic standpoint. Hard wheat flour absorbs more liquid than soft flour. Good hard wheat flour should feel somewhat granular when rubbed between the thumb and fingers. A soft, smooth feeling indicates a soft wheat flour or a blend of soft and hard wheat flour. Another indicator is that hard wheat flour retains its form when pressed in the hollow of the hand and falls apart readily when touched. Soft wheat flour tends to remain lumped together after pressure. Gluten Strength The gluten test is done to find the variation of gluten quality and quantity in different kinds of flour. Hard flour has more gluten of better quality than soft flour. The gluten strength and quality of two different kinds of hard flour may also vary with the weather conditions and the place where the wheat is grown. The difference may be measured exactly by laboratory tests, or roughly assessed by the variation of gluten balls made from different kinds of hard flours. For example, to test the gluten in hard flour and all-purpose flour, mix 250 g (9 oz.) of each in separate mixing bowls with enough water to make each dough stiff. Mix and develop each dough until smooth. Let the dough rest for about 10 minutes. Wash each dough separately while kneading it under a stream of cold water until the water runs clean and all the starch is washed out. (Keep a flour sieve in the sink to prevent dough pieces from being washed down the drain.) What remains will be crude gluten. Shape the crude gluten into round balls, then place them on a paper-lined baking pan and bake at 215°C (420°F) for about one hour. The gluten ball made from the hard flour will be larger than the one made from all-purpose flour. This illustrates the ability of hard flour to produce a greater volume because of its higher gluten content. Ash Content Ash or mineral content of flour is used as another measurement of quality. Earlier in the chapter, we talked about extraction rates as an indicator of how much of the grain has been refined. Ash content refers to the amount of ash that would be left over if you were to burn 100 g of flour. A higher ash content indicates that the flour contains more of the germ, bran, and outer endosperm. Lower ash content means that the flour is more highly refined (i.e., a lower extraction rate). Baking Quality The final and conclusive test of any flour is the kind of bread that can be made from it. The baking test enables the baker to check on the completed loaf that can be expected from any given flour. Good volume is related to good quality gluten; poor volume to young or green flour. Flour that lacks stability or power to hold during the entire fermentation may result in small, flat bread. Flour of this type may sometimes respond to an increase in the amount of yeast. More yeast shortens the fermentation time and keeps the dough in better condition during the pan fermentation period. |
Ancillary_Materials/Worksheets/Worksheets%3A_Analytical_Chemistry_II/Calibration_Curves_(Oxley) | Submit an Excel spreadsheet with your work. Student Learning Outcomes After this exercise, students will be able to: Prepare a linear calibration curve from experimental data. Perform a linear fit, including regression statistics, in Excel. Use a calibration curve to determine the quantity of an unknown. Discuss the limitations of a calibration curve. The purpose of this exercise is to introduce calibration curves in the context of quantitative chemical analysis. The concentration of chloride in a water sample can be determined by a technique called ion chromatography. The instrument response is peak area (no units). A series of standard chloride solutions were prepared by diluting a 30 ppm Cl - standard by 2x, 5x, 10x, and 50x in ultra-pure water. The peak area of the undiluted and diluted solutions were 6.642, 3.250, 1.308, 0.634, 0.061, respectively. Prepare a calibration curve for Cl - . Perform a linear least squares fit to the data, and display the equation and R 2 value on the plot. Determine the concentration of Cl - in a sample yielding a peak area of 1.443. Determine the uncertainty in the slope and intercept of the calibration curve. Five water samples from the same location were analyzed, yielding peak areas of 1.684, 1.454, 1.338, 1.472, and 1.443. Report the concentration of Cl - as the 95% confidence interval. Which source of error – the calibration curve or the sample reproducibility – dominates the total error in the analysis? |
Courses/Howard_University/General_Chemistry%3A_An_Atoms_First_Approach/Unit_3%3A_Stoichiometry/Chapter_7%3A_Stoichiometry/Chapter_7.7%3A_Industrially_Important_Chemicals | 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 Learning Objectives To appreciate the scope of the chemical industry and its contributions to modern society. It isn’t easy to comprehend the scale on which the chemical industry must operate to supply the huge amounts of chemicals required in modern industrial societies. Figure 7.7.1 lists the names and formulas of the chemical industry’s “top 25” for 2002—the 25 chemicals produced in the largest quantity in the United States that year—along with the amounts produced, in billions of pounds. To put these numbers in perspective, consider that the 88.80 billion pounds of sulfuric acid produced in the United States in 2002 has a volume of 21.90 million cubic meters (2.19 × 10 7 m 3 ), enough to fill the Pentagon, probably the largest office building in the world, about 22 times. Figure 7.7.1 Top 25 Chemicals Produced in the United States in 2002* According to Figure 7.7.1 , 11 of the top 15 compounds produced in the United States are inorganic, and the total mass of inorganic chemicals produced is almost twice the mass of organic chemicals. Yet the diversity of organic compounds used in industry is such that over half of the top 25 compounds (13 out of 25) are organic. Why are such huge quantities of chemical compounds produced annually? They are used both directly as components of compounds and materials that we encounter on an almost daily basis and indirectly in the production of those compounds and materials. The single largest use of industrial chemicals is in the production of foods: 7 of the top 15 chemicals are either fertilizers (ammonia, urea, and ammonium nitrate) or used primarily in the production of fertilizers (sulfuric acid, nitric acid, nitrogen, and phosphoric acid). Many of the organic chemicals on the list are used primarily as ingredients in the plastics and related materials that are so prevalent in contemporary society. Ethylene and propylene, for example, are used to produce polyethylene and polypropylene, which are made into plastic milk bottles, sandwich bags, indoor-outdoor carpets, and other common items. Vinyl chloride, in the form of polyvinylchloride, is used in everything from pipes to floor tiles to trash bags. Though not listed in Figure 7.7.1 , butadiene and carbon black are used in the manufacture of synthetic rubber for tires, and phenol and formaldehyde are ingredients in plywood, fiberglass, and many hard plastic items. We do not have the space in this text to consider the applications of all these compounds in any detail, but we will return to many of them after we have developed the concepts necessary to understand their underlying chemistry. Instead, we conclude this chapter with a brief discussion of petroleum refining as it relates to gasoline and octane ratings and a look at the production and use of the topmost industrial chemical, sulfuric acid. Petroleum The petroleum that is pumped out of the ground at locations around the world is a complex mixture of several thousand organic compounds, including straight-chain alkanes, cycloalkanes, alkenes, and aromatic hydrocarbons with four to several hundred carbon atoms. The identities and relative abundances of the components vary depending on the source. So Texas crude oil is somewhat different from Saudi Arabian crude oil. In fact, the analysis of petroleum from different deposits can produce a “fingerprint” of each, which is useful in tracking down the sources of spilled crude oil. For example, Texas crude oil is “sweet,” meaning that it contains a small amount of sulfur-containing molecules, whereas Saudi Arabian crude oil is “sour,” meaning that it contains a relatively large amount of sulfur-containing molecules. Gasoline Petroleum is converted to useful products such as gasoline in three steps: distillation, cracking, and reforming. Recall from Chapter 1 that distillation separates compounds on the basis of their relative volatility, which is usually inversely proportional to their boiling points. Part (a) in Figure 7.7.2 shows a cutaway drawing of a column used in the petroleum industry for separating the components of crude oil. The petroleum is heated to approximately 400°C (750°F), at which temperature it has become a mixture of liquid and vapor. This mixture, called the feedstock , is introduced into the refining tower. The most volatile components (those with the lowest boiling points) condense at the top of the column where it is cooler, while the less volatile components condense nearer the bottom. Some materials are so nonvolatile that they collect at the bottom without evaporating at all. Thus the composition of the liquid condensing at each level is different. These different fractions, each of which usually consists of a mixture of compounds with similar numbers of carbon atoms, are drawn off separately. Part (b) in Figure 7.7.2 shows the typical fractions collected at refineries, the number of carbon atoms they contain, their boiling points, and their ultimate uses. These products range from gases used in natural and bottled gas to liquids used in fuels and lubricants to gummy solids used as tar on roads and roofs. Figure 7.7.2 The Distillation of Petroleum (a) This is a diagram of a distillation column used for separating petroleum fractions. (b) Petroleum fractions condense at different temperatures, depending on the number of carbon atoms in the molecules, and are drawn off from the column. The most volatile components (those with the lowest boiling points) condense at the top of the column, and the least volatile (those with the highest boiling points) condense at the bottom. The economics of petroleum refining are complex. For example, the market demand for kerosene and lubricants is much lower than the demand for gasoline, yet all three fractions are obtained from the distillation column in comparable amounts. Furthermore, most gasolines and jet fuels are blends with very carefully controlled compositions that cannot vary as their original feedstocks did. To make petroleum refining more profitable, the less volatile, lower-value fractions must be converted to more volatile, higher-value mixtures that have carefully controlled formulas. The first process used to accomplish this transformation is cracking A process in petroleum refining in which the larger and heavier hydrocarbons in kerosene and higher-boiling-point fractions are heated to high temperatures, causing the carbon–carbon bonds to break (“crack”), thus producing a more volatile mixture. , in which the larger and heavier hydrocarbons in the kerosene and higher-boiling-point fractions are heated to temperatures as high as 900°C. High-temperature reactions cause the carbon–carbon bonds to break, which converts the compounds to lighter molecules similar to those in the gasoline fraction. Thus in cracking, a straight-chain alkane with a number of carbon atoms corresponding to the kerosene fraction is converted to a mixture of hydrocarbons with a number of carbon atoms corresponding to the lighter gasoline fraction. The second process used to increase the amount of valuable products is called reforming The second process used in petroleum refining, which is the chemical conversion of straight-chain alkanes to either branched-chain alkanes or mixtures of aromatic hydrocarbons. ; it is the chemical conversion of straight-chain alkanes to either branched-chain alkanes or mixtures of aromatic hydrocarbons. Using metals such as platinum brings about the necessary chemical reactions. The mixtures of products obtained from cracking and reforming are separated by fractional distillation. Octane Ratings The quality of a fuel is indicated by its octane rating A measure of a fuel’s ability to burn in a combustion engine without knocking or pinging (indications of premature combustion). The higher the octane rating, the higher quality the fuel. , which is a measure of its ability to burn in a combustion engine without knocking or pinging. Knocking and pinging signal premature combustion ( Figure 7.7.3 ), which can be caused either by an engine malfunction or by a fuel that burns too fast. In either case, the gasoline-air mixture detonates at the wrong point in the engine cycle, which reduces the power output and can damage valves, pistons, bearings, and other engine components. The various gasoline formulations are designed to provide the mix of hydrocarbons least likely to cause knocking or pinging in a given type of engine performing at a particular level. Figure 7.7.3 The Burning of Gasoline in an Internal Combustion Engine (a) Normally, fuel is ignited by the spark plug, and combustion spreads uniformly outward. (b) Gasoline with an octane rating that is too low for the engine can ignite prematurely, resulting in uneven burning that causes knocking and pinging. The octane scale was established in 1927 using a standard test engine and two pure compounds: n -heptane and isooctane (2,2,4-trimethylpentane). n -Heptane, which causes a great deal of knocking on combustion, was assigned an octane rating of 0, whereas isooctane, a very smooth-burning fuel, was assigned an octane rating of 100. Chemists assign octane ratings to different blends of gasoline by burning a sample of each in a test engine and comparing the observed knocking with the amount of knocking caused by specific mixtures of n -heptane and isooctane. For example, the octane rating of a blend of 89% isooctane and 11% n -heptane is simply the average of the octane ratings of the components weighted by the relative amounts of each in the blend. Converting percentages to decimals, we obtain the octane rating of the mixture: \[ 0.89(100) + 0.11(0) = 89 \] A gasoline that performs at the same level as a blend of 89% isooctane and 11% n -heptane is assigned an octane rating of 89; this represents an intermediate grade of gasoline. Regular gasoline typically has an octane rating of 87; premium has a rating of 93 or higher. As shown in Figure 7.7.4 , many compounds that are now available have octane ratings greater than 100, which means they are better fuels than pure isooctane. In addition, antiknock agents, also called octane enhancers, have been developed. One of the most widely used for many years was tetraethyllead [(C 2 H 5 ) 4 Pb], which at approximately 3 g/gal gives a 10–15-point increase in octane rating. Since 1975, however, lead compounds have been phased out as gasoline additives because they are highly toxic. Other enhancers, such as methyl t -butyl ether (MTBE), have been developed to take their place. They combine a high octane rating with minimal corrosion to engine and fuel system parts. Unfortunately, when gasoline containing MTBE leaks from underground storage tanks, the result has been contamination of the groundwater in some locations, resulting in limitations or outright bans on the use of MTBE in certain areas. As a result, the use of alternative octane enhancers such as ethanol, which can be obtained from renewable resources such as corn, sugar cane, and, eventually, corn stalks and grasses, is increasing. Figure 7.7.4 The Octane Ratings of Some Hydrocarbons and Common Additives Example 12 You have a crude (i.e., unprocessed or straight-run ) petroleum distillate consisting of 10% n -heptane, 10% n -hexane, and 80% n -pentane by mass, with an octane rating of 52. What percentage of MTBE by mass would you need to increase the octane rating of the distillate to that of regular-grade gasoline (a rating of 87), assuming that the octane rating is directly proportional to the amounts of the compounds present? Use the information presented in Figure 7.7.4 . Given: composition of petroleum distillate, initial octane rating, and final octane rating Asked for: percentage of MTBE by mass in final mixture Strategy: A Define the unknown as the percentage of MTBE in the final mixture. Then subtract this unknown from 100% to obtain the percentage of petroleum distillate. B Multiply the percentage of MTBE and the percentage of petroleum distillate by their respective octane ratings; add these values to obtain the overall octane rating of the new mixture. C Solve for the unknown to obtain the percentage of MTBE needed. Solution: A The question asks what percentage of MTBE will give an overall octane rating of 87 when mixed with the straight-run fraction. From Figure 7.7.4 , the octane rating of MTBE is 116. Let x be the percentage of MTBE, and let 100 − x be the percentage of petroleum distillate. B Multiplying the percentage of each component by its respective octane rating and setting the sum equal to the desired octane rating of the mixture (87) times 100 gives \begin{matrix} final\; octane\; rating\; of\; mixture & = &87\left ( 100 \right ) \\ & = &52\left ( 100-x \right )+116x \\ & = &5200-52x+116x \\ & = &5200+64x \end{matrix} C Solving the equation gives x = 55%. Thus the final mixture must contain 55% MTBE by mass. To obtain a composition of 55% MTBE by mass, you would have to add more than an equal mass of MTBE (actually 0.55/0.45, or 1.2 times) to the straight-run fraction. This is 1.2 tons of MTBE per ton of straight-run gasoline, which would be prohibitively expensive. Thus there are sound economic reasons for reforming the kerosene fractions to produce toluene and other aromatic compounds, which have high octane ratings and are much cheaper than MTBE. Exercise As shown in Figure 7.7.4 , toluene is one of the fuels suitable for use in automobile engines. How much toluene would have to be added to a blend of the petroleum fraction in this example containing 15% MTBE by mass to increase the octane rating to that of premium gasoline (93)? Answer: The final blend is 56% toluene by mass, which requires a ratio of 56/44, or 1.3 tons of toluene per ton of blend. Sulfuric Acid Sulfuric acid is one of the oldest chemical compounds known. It was probably first prepared by alchemists who burned sulfate salts such as FeSO 4 ·7H 2 O, called green vitriol from its color and glassy appearance (from the Latin vitrum , meaning “glass”). Because pure sulfuric acid was found to be useful for dyeing textiles, enterprising individuals looked for ways to improve its production. By the mid-18th century, sulfuric acid was being produced in multiton quantities by the lead-chamber process , which was invented by John Roebuck in 1746. In this process, sulfur was burned in a large room lined with lead, and the resulting fumes were absorbed in water. Production The production of sulfuric acid today is likely to start with elemental sulfur obtained through an ingenious technique called the Frasch process , which takes advantage of the low melting point of elemental sulfur (115.2°C). Large deposits of elemental sulfur are found in porous limestone rocks in the same geological formations that often contain petroleum. In the Frasch process, water at high temperature (160°C) and high pressure is pumped underground to melt the sulfur, and compressed air is used to force the liquid sulfur-water mixture to the surface ( Figure 7.7.5 ). The material that emerges from the ground is more than 99% pure sulfur. After it solidifies, it is pulverized and shipped in railroad cars to the plants that produce sulfuric acid, as shown here. Transporting sulfur. A train carries elemental sulfur through the White Canyon of the Thompson River in British Columbia, Canada. Figure 7.7.5 Extraction of Elemental Sulfur from Underground Deposits In the Frasch process for extracting sulfur, very hot water at high pressure is injected into the sulfur-containing rock layer to melt the sulfur. The resulting mixture of liquid sulfur and hot water is forced up to the surface by compressed air. An increasing number of sulfuric acid manufacturers have begun to use sulfur dioxide (SO 2 ) as a starting material instead of elemental sulfur. Sulfur dioxide is recovered from the burning of oil and gas, which contain small amounts of sulfur compounds. When not recovered, SO 2 is released into the atmosphere, where it is converted to an environmentally hazardous form that leads to acid rain ( Chapter 8 ). If sulfur is the starting material, the first step in the production of sulfuric acid is the combustion of sulfur with oxygen to produce SO 2 . Next, SO 2 is converted to SO 3 by the contact process , in which SO 2 and O 2 react in the presence of V 2 O 5 to achieve about 97% conversion to SO 3 . The SO 3 can then be treated with a small amount of water to produce sulfuric acid. Usually, however, the SO 3 is absorbed in concentrated sulfuric acid to produce oleum , a more potent form called fuming sulfuric acid . Because of its high SO 3 content (approximately 99% by mass), oleum is cheaper to ship than concentrated sulfuric acid. At the point of use, the oleum is diluted with water to give concentrated sulfuric acid ( very carefully because dilution generates enormous amounts of heat). Because SO 2 is a pollutant, the small amounts of unconverted SO 2 are recovered and recycled to minimize the amount released into the air. Uses Two-thirds of the sulfuric acid produced in the United States is used to make fertilizers, most of which contain nitrogen, phosphorus, and potassium (in a form called potash ). In earlier days, phosphate-containing rocks were simply ground up and spread on fields as fertilizer, but the extreme insolubility of many salts that contain the phosphate ion (PO 4 3− ) limits the availability of phosphorus from these sources. Sulfuric acid serves as a source of protons (H + ions) that react with phosphate minerals to produce more soluble salts containing HPO 4 2− or H 2 PO 4 − as the anion, which are much more readily taken up by plants. In this context, sulfuric acid is used in two principal ways: (1) the phosphate rocks are treated with concentrated sulfuric acid to produce “superphosphate,” a mixture of 32% CaHPO 4 and Ca(H 2 PO 4 ) 2 ·H 2 O, 50% CaSO 4 ·2H 2 O, approximately 3% absorbed phosphoric acid, and other nutrients; and (2) sulfuric acid is used to produce phosphoric acid (H 3 PO 4 ), which can then be used to convert phosphate rocks to “triple superphosphate,” which is largely Ca(H 2 PO 4 ) 2 ·H 2 O. Sulfuric acid is also used to produce potash, one of the other major ingredients in fertilizers. The name potash originally referred to potassium carbonate (obtained by boiling wood ash es with water in iron pot s), but today it also refers to compounds such as potassium hydroxide (KOH) and potassium oxide (K 2 O). The usual source of potassium in fertilizers is actually potassium sulfate (K 2 SO 4 ), which is produced by several routes, including the reaction of concentrated sulfuric acid with solid potassium chloride (KCl), which is obtained as the pure salt from mineral deposits. Summary Many chemical compounds are prepared industrially in huge quantities and used to produce foods, fuels, plastics, and other such materials. Petroleum refining takes a complex mixture of naturally occurring hydrocarbons as a feedstock and, through a series of steps involving distillation, cracking , and reforming , converts them to mixtures of simpler organic compounds with desirable properties. A major use of petroleum is in the production of motor fuels such as gasoline. The performance of such fuels in engines is described by their octane rating , which depends on the identity of the compounds present and their relative abundance in the blend. Sulfuric acid is the compound produced in the largest quantity in the industrial world. Much of the sulfur used in the production of sulfuric acid is obtained via the Frasch process, in which very hot water forces liquid sulfur out of the ground in nearly pure form. Sulfuric acid is produced by the reaction of sulfur dioxide with oxygen in the presence of vanadium(V) oxide (the contact process), followed by absorption of the sulfur trioxide in concentrated sulfuric acid to produce oleum. Most sulfuric acid is used to prepare fertilizers. Key Takeaway Many chemical compounds are prepared industrially in huge quantities to prepare the materials we need and use in our daily lives. Conceptual Problems Describe the processes used for converting crude oil to transportation fuels. If your automobile engine is knocking, is the octane rating of your gasoline too low or too high? Explain your answer. Tetraethyllead is no longer used as a fuel additive to prevent knocking. Instead, fuel is now marketed as “unleaded.” Why is tetraethyllead no longer used? If you were to try to extract sulfur from an underground source, what process would you use? Describe briefly the essential features of this process. Why are phosphate-containing minerals used in fertilizers treated with sulfuric acid? Answer Phosphate salts contain the highly-charged PO 4 3− ion, salts of which are often insoluble. Protonation of the PO 4 3− ion by strong acids such as H 2 SO 4 leads to the formation of the HPO 4 2− and H 2 PO 4 − ions. Because of their decreased negative charge, salts containing these anions are usually much more soluble, allowing the anions to be readily taken up by plants when they are applied as fertilizer. Numerical Problem In Example 12, the crude petroleum had an overall octane rating of 52. What is the composition of a solution of MTBE and n -heptane that has this octane rating? Contributors Anonymous Modified by Joshua Halpern |
Courses/Saint_Francis_University/CHEM_113%3A_Human_Chemistry_I_(Zovinka)/07%3A_Chemical_Reactions_-_Energy_Rates_and_Equilibrium/7.04%3A_Why_Do_Chemical_Reactions_Occur_Free_Energy | Learning Outcomes Describe the meaning of a spontaneous reaction in terms of enthalpy and entropy changes. Define free energy. Determine the spontaneity of a reaction based on the value of its change in free energy at high and low temperatures. Spontaneous Reactions A spontaneous reaction is a reaction that favors the formation of products at the conditions under which the reaction is occurring . A roaring bonfire (see Figure \(\PageIndex{1}\) below) is an example of a spontaneous reaction. A fire is exothermic , which means a decrease in the energy of the system as energy is released to the surroundings as heat. The products of a fire are composed mostly of gases such as carbon dioxide and water vapor, so the entropy of the system increases during most combustion reactions. This combination of a decrease in energy and an increase in entropy means that combustion reactions occur spontaneously. A nonspontaneous reaction is a reaction that does not favor the formation of products at the given set of conditions . In order for a reaction to be nonspontaneous, one or both of the driving forces must favor the reactants over the products. In other words, the reaction is endothermic, is accompanied by a decrease in entropy, or both. Out atmosphere is composed primarily of a mixture of nitrogen and oxygen gases. One could write an equation showing these gases undergoing a chemical reaction to form nitrogen monoxide. \[\ce{N_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{NO} \left( g \right)\] Fortunately, this reaction is nonspontaneous at normal temperatures and pressures, it is a highly endothermic reaction. However, nitrogen monoxide is capable of being produced at very high temperatures, and this reaction has been observed to occur as a result of lightning strikes. One must be careful not to confuse the term spontaneous with the notion that a reaction occurs rapidly. A spontaneous reaction is one in which product formation is favored, even if the reaction is extremely slow. You do not have to worry about a piece of paper on your desk suddenly bursting into flames, although its combustion is a spontaneous reaction. What is missing is the required activation energy to get the reaction started. If the paper were to be heated to a high enough temperature, it would begin to burn, at which point the reaction would proceed spontaneously until completion. Entropy as a Driving Force An example of a very simple spontaneous process is that of a melting ice cube. Energy is transferred from the room to the ice cube, causing it to change from the solid to the liquid state. \[\ce{H_2O} \left( s \right) + 6.01 \: \text{kJ} \rightarrow \ce{H_2O} \left( l \right)\] The solid state of water, ice, is highly ordered because its molecules are fixed in place. The melting process frees the water molecules from their hydrogen-bonded network and allows them a greater degree of movement. Water is more disordered than ice. The change from the solid to the liquid state of any substance corresponds to an increase in the disorder of the system. The tendency in nature for systems to proceed toward a state of greater disorder or randomness is called entropy , which is symbolized by S , and expressed in units of Joules per mole-kelvin, \(\mathrm{J/(mol\cdot K)}\). Larger values of S indicate that the particles in a substance have more disorder or randomness. In the example above, the particles in the ice cube (solid water) have lower freedom of motion, they are less random. As the ice melts to liquid water, the particles become more disordered and entropy increases. If the liquid water was heated further, the particles would even gain more freedom of motion, become more disordered, and eventually change to a gas, which has even higher entropy. Chemical reactions also tend to proceed in such a way as to increase the total entropy of the system, measured by the entropy change (\(\Delta S\)) between reactants and products. How can you tell if a certain reaction shows an increase or a decrease in entropy? The states of the reactants and produces provide certain clues. The general cases below illustrate entropy at the molecular level. For a given substance, the entropy of the liquid state is greater than the entropy of the solid state. Likewise, the entropy of the gas is greater than the entropy of the liquid. Therefore, entropy increases in processes in which solid or liquid reactants form gaseous products. Entropy also increases when solid reactants form liquid products. Entropy increases when a substance is broken up into multiple parts. The process of dissolving increases entropy because the solute particles become separated from one another when a solution is formed. Entropy increases as temperature increases. An increase in temperature means that the particles of the substance have greater kinetic energy. The faster moving particles have more disorder than particles that are moving more slowly at a lower temperature. Entropy generally increases in reactions in which the total number of product molecules is greater than the total number of reactant molecules. An exception to this rule is when nongaseous products are formed from gaseous reactants. The examples below will serve to illustrate how the entropy change in a reaction can be predicted. \[\ce{Cl_2} \left( g \right) \rightarrow \ce{Cl_2} \left( l \right)\] The entropy is decreasing because a gas is becoming a liquid. \[\ce{CaCO_3} \left( s \right) \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right)\] The entropy is increasing because a gas is being produced, and the number of molecules is increasing. \[\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightarrow 2 \ce{NH_3} \left( g \right)\] The entropy is decreasing because four total reactant molecules are forming two total product molecules. All are gases. \[\ce{AgNO_3} \left( aq \right) + \ce{NaCl} \left( aq \right) \rightarrow \ce{NaNO_3} \left( aq \right) + \ce{AgCl} \left( s \right)\] The entropy is decreasing because a solid is formed from aqueous reactants. \[\ce{H_2} \left( g \right) + \ce{Cl_2} \left( g \right) \rightarrow 2 \ce{HCl} \left( g \right)\] The entropy change is unknown (but likely not zero) because there are equal numbers of molecules on both sides of the equation, and all are gases. Gibbs Free Energy Many chemical reactions and physical processes release energy that can be used to do other things. When the fuel in a car is burned, some of the released energy is used to power the vehicle. Free energy is energy that is available to do work . Spontaneous reactions release free energy as they proceed. The determining factors for spontaneity of a reaction depend on both the enthalpy and entropy changes that occur for the system. The free energy change (\(\Delta G\)) of a reaction is a mathematical combination of the enthalpy change and the entropy change. \[\Delta G^\text{o} = \Delta H^\text{o} - T \Delta S^\text{o}\] The symbol for free energy is \(G\), in honor of American scientist Josiah Gibbs (1839 - 1903), who made many contributions to thermodynamics. The change in Gibbs free energy is equal to the change in enthalpy minus the mathematical product of the change in entropy multiplied by the Kelvin temperature. Each thermodynamic quantity in the equation is for substances in their standard states, as indicated by the \(^\text{o}\) superscripts. A spontaneous reaction is one that releases free energy, and so the sign of \(\Delta G\) must be negative. Since both \(\Delta H\) and \(\Delta S\) can be either positive or negative, depending on the characteristics of the particular reaction, there are four different possible combinations. The outcomes for \(\Delta G\) based on the signs of \(\Delta H\) and \(\Delta S\) are outlined in the table below. Recall that \(- \Delta \text{H}\) indicates that the reaction is exothermic and a \(+ \Delta \text{H}\) means the reaction is endothermic. For entropy, \(+ \Delta \text{S}\) means the entropy is increasing and the system is becoming more disordered. A \(- \Delta \text{S}\) means that entropy is decreasing and the system is becoming less disordered (more ordered). A process that releases free energy, (\(-\Delta G\)), is said to be exergonic . Processes that require free energy (\(+\Delta G\)), are endergonic . These terms are used when considering chemical reactions that occur in living systems. \(\Delta H\) \(\Delta S\) \(\Delta G\) negative positive always negative positive positive negative at higher temperatures, positive at lower temperatures negative negative negative at lower temperatures, positive at higher temperatures positive negative always positive Keep in mind that the temperature in the Gibbs free energy equation is the Kelvin temperature, so it can only have a positive value. When \(\Delta H\) is negative and \(\Delta S\) is positive, the sign of \(\Delta G\) will always be negative, and the reaction will be spontaneous at all temperatures. This corresponds to both driving forces being in favor of product formation. When \(\Delta H\) is positive and \(\Delta S\) is negative, the sign of \(\Delta G\) will always be positive, and the reaction can never be spontaneous. This corresponds to both driving forces working against product formation. When one driving force favors the reaction, but the other does not, it is the temperature that determines the sign of \(\Delta G\). Consider first an endothermic reaction (positive \(\Delta H\)) that also displays an increase in entropy (positive \(\Delta S\)). It is the entropy term that favors the reaction. Therefore, as the temperature increases, the \(T \Delta S\) term in the Gibbs free energy equation will begin to predominate and \(\Delta G\) will become negative. A common example of a process which falls into this category is the melting of ice (see figure below). At a relatively low temperature (below \(273 \: \text{K}\)), the melting is not spontaneous because the positive \(\Delta H\) term "outweighs" the \(T \Delta S\) term. When the temperature rises above \(273 \: \text{K}\), the process becomes spontaneous because the larger \(T\) value has tipped the sign of \(\Delta G\) over to being negative. When the reaction is exothermic (negative \(\Delta H\)) but undergoes a decrease in entropy (negative \(\Delta S\)), it is the enthalpy term which favors the reaction. In this case, a spontaneous reaction is dependent upon the \(T \Delta S\) term being small relative to the \(\Delta H\) term, so that \(\Delta G\) is negative. The freezing of water is an example of this type of process. It is spontaneous only at a relatively low temperature. Above \(273. \: \text{K}\), the larger \(T \Delta S\) value causes the sign of \(\Delta G\) to be positive, and freezing does not occur. |
Bookshelves/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/Appendix_M%3A_Half-Lives_for_Several_Radioactive_Isotopes | Half-Lives for Several Radioactive Isotopes Isotope Half-Life1 Type of Emission2 Isotope.1 Half-Life3 Type of Emission4 614C614C 5730 y (β−)(β−) 83210Bi83210Bi 5.01 d (β−)(β−) 713N713N 9.97 m (β+)(β+) 83212Bi83212Bi 60.55 m (αorβ−)(αorβ−) 915F915F 4.1 ×× 10−22 s (p)(p) 84210Po84210Po 138.4 d (α)(α) 1124Na1124Na 15.00 h (β−)(β−) 84212Po84212Po 3 ×× 10−7 s (α)(α) 1532P1532P 14.29 d (β−)(β−) 84216Po84216Po 0.15 s (α)(α) 1940K1940K 1.27 ×× 109 y (βorE.C.)(βorE.C.) 84218Po84218Po 3.05 m (α)(α) 2649Fe2649Fe 0.08 s (β+)(β+) 85215At85215At 1.0 ×× 10−4 s (α)(α) 2660Fe2660Fe 2.6 ×× 106 y (β−)(β−) 85218At85218At 1.6 s (α)(α) 2760Co2760Co 5.27 y (β−)(β−) 86220Rn86220Rn 55.6 s (α)(α) 3787Rb3787Rb 4.7 ×× 1010 y (β−)(β−) 86222Rn86222Rn 3.82 d (α)(α) 3890Sr3890Sr 29 y (β−)(β−) 88224Ra88224Ra 3.66 d (α)(α) 49115In49115In 5.1 ×× 1015 y (β−)(β−) 88226Ra88226Ra 1600 y (α)(α) 53131I53131I 8.040 d (β−)(β−) 88228Ra88228Ra 5.75 y (β−)(β−) 58142Ce58142Ce 5 ×× 1015 y (α)(α) 89228Ac89228Ac 6.13 h (β−)(β−) 81208Tl81208Tl 3.07 m (β−)(β−) 90228Th90228Th 1.913 y (α)(α) 82210Pb82210Pb 22.3 y (β−)(β−) 90232Th90232Th 1.4 ×× 1010 y (α)(α) 82212Pb82212Pb 10.6 h (β−)(β−) 90233Th90233Th 22 m (β−)(β−) 82214Pb82214Pb 26.8 m (β−)(β−) 90234Th90234Th 24.10 d24.10 d (β−)(β−) 83206Bi83206Bi 6.243 d (E.C.)(E.C.) 91233Pa91233Pa 27 d (β−)(β−) 92233U92233U 1.59 ×× 105 y (α)(α) 96242Cm96242Cm 162.8 d (α)(α) 92234U92234U 2.45 ×× 105 y (α)(α) 97243Bk97243Bk 4.5 h (αorE.C.)(αorE.C.) 92235U92235U 7.03 ×× 108 y (α)(α) 99253Es99253Es 20.47 d (α)(α) 92238U92238U 4.47 ×× 109 y (α)(α) 100254Fm100254Fm 3.24 h (αorS.F.)(αorS.F.) 92239U92239U 23.54 m (β−)(β−) 100255Fm100255Fm 20.1 h (α)(α) 93239Np93239Np 2.3 d (β−)(β−) 101256Md101256Md 76 m (αorE.C.)(αorE.C.) 94239Pu94239Pu 2.407 ×× 104 y (α)(α) 102254No102254No 55 s (α)(α) 94239Pu94239Pu 6.54 ×× 103 y (α)(α) 103257Lr103257Lr 0.65 s (α)(α) 94241Pu94241Pu 14.4 y (αorβ−)(αorβ−) 105260Ha105260Ha 1.5 s (αorS.F.)(αorS.F.) 95241Am95241Am 432.2 y (α)(α) 106263Sg106263Sg 0.8 s (αorS.F.)(αorS.F.) Table M1 Footnotes 1 y = years, d = days, h = hours, m = minutes, s = seconds 2 E.C. = electron capture, S.F. = Spontaneous fission 3 y = years, d = days, h = hours, m = minutes, s = seconds 4 E.C. = electron capture, S.F. = Spontaneous fission |
Courses/Sacramento_City_College/SCC%3A_CHEM_300_-_Beginning_Chemistry/SCC%3A_CHEM_300_-_Beginning_Chemistry_(Faculty)/11%3A_Gases/11.08%3A_Avogadro%E2%80%99s_Law-_Volume_and_Moles | 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. How much air 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 (or number of particles) of gas when the temperature and pressure are held constant. The mathematical expression of Avogadro's Law is: \[V = k \times n \nonumber \] or \[\dfrac{V_1}{n_1} = \dfrac{V_2}{n_2} \nonumber \] 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. Example \(\PageIndex{1}\) A balloon has been filled to a volume of \(1.90 \: \text{L}\) with \(0.0920 \: \text{mol}\) of helium gas. If \(0.0210 \: \text{mol}\) of additional helium is added to the balloon while the temperature and pressure are held constant, what is the new volume of the balloon? Solution Steps for Problem Solving Unnamed: 1 Identify the "given" information and what the problem is asking you to "find." Given: \(V_1 = 1.90 \: \text{L}\) \(n_1 = 0.0920 \: \text{mol}\) Find: \(V_2 = ? \: \text{L}\) List other known quantities. Note that the final number of moles has to be calculated by adding the original number of moles to the moles of added helium. \(n_2 = 0.0920 + 0.0210 = 0.1130 \: \text{mol}\) Plan the problem. First, rearrange the equation algebraically to solve for \(V_2\). \[V_2 = \dfrac{V_1 \times n_2}{n_1} \nonumber \] Calculate. Now substitute the known quantities into the equation and solve. \[V_2 = \dfrac{1.90 \: \text{L} \times 0.1130 \: \cancel{\text{mol}}}{0.0920 \: \cancel{\text{mol}}} = 2.33 \: \text{L} \nonumber \] Think about your result. Since a relatively small amount of additional helium was added to the balloon, its volume increases slightly. Exercise \(\PageIndex{1}\) A 12.8 L volume of gas contains .000498 moles of oxygen gas. At constant temperature and pressure, what volume does .0000136 moles of the gas fill? Answer 0.350 L Summary Calculations for relationships between volume and number of moles of a gas can be performed using Avogadro's Law. |
Ancillary_Materials/Worksheets/Worksheets%3A_Inorganic_Chemistry/Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Chem_315)/315_Workbook/18b%3A_ComplexI | Name: ______________________________ Section: _____________________________ Student ID#:__________________________ Complex I (NADH-quinone oxidoreductase) Electron Transfer in Complex I Complex I is located in the inner mitochondrial membrane in eukaryotes. The electrons from NADH (produced in the TCA cycle) begin to be shuttled through small steps to capture the energy. This section will examine the mechanisms of electron transfer by the peripheral domain, proton transfer by the membrane domain and how their coupling can drive proton transport. The net reaction of Complex I is the oxidation of NADH and the reduction of ubiquinone. Net reaction: NADH + H + + UQ -> NAD + + UQH 2 Use the table of reduction potentials and the Faraday equation to calculate ΔG for this reaction. The standard free-energy change Δ G ° ́ is related the reduction potential Δ E ́ 0 by: ΔG 0' = - n FDE’ 0 F (Faraday’s constant) = 96,485 J/V mol ΔE’ 0 = difference in reduction potentials between donor and acceptor n = number of electrons transferred Complex I A cartoon model of the complex shows redox active cofactors in complex I. Complex I contains at least 46 different proteins and a number of prosthetic groups: a flavin redox agent (FMN), a ubiquinone redox agent (UQ) and some iron sulfide clusters (FeS). The hydrophilic or peripheral domain catalyzes electron transfer while the membrane domain is involved in active transport of protons . Nature 465 (2010) 441-447. doi:10.1038/nature09066Nature 465 (2010) 428-429. Journal Biological Chemistry 284 , (2009) 29773–29783. Journal Biological Chemistry 286 (2011) 18056–18065 Label the membrane-bound part of the complex. Label the peripheral part of the complex (i.e. the other part). Electron transfer occurs more readily in polar environments. What amino acid side chains do you think are present in the peripheral region? Follow the Electron Transfer through Complex I Complex 1 and Close up views of FMN (a flavin redox agent) and Fe/S clusters are shown below where N2, N3, N4, N5, and N6 are tetranuclear FeS clusters. (N7 is found in some bacteria). Sketch in the possible site for NADH and ubiquinone (UQ) binding Indicate a possible path for flow of electrons from NADH to UQ. Given this path, what must be the relative relationships of the standard reduction potentials for FMN, the Fe/S clusters and ubiquinone for electrons to flow along the path. Are these inner sphere or outer sphere electron transfers? Why are there so many transfers? Transfer 1: Electron transfer from NADH to FMN The ultimate goal is to transfer electrons in baby steps in order to release energy incrementally, rather than all at once. The electron transfer chain uses Fe(II)/Fe(III) couples. The first step in this process is the reaction of NADH with FMN. The oxidation of NAD+ has been covered earlier: FMN is a flavin redox cofactor like FAD). How many electrons can FADH2 transfer at once? [ 1 OR 2 OR either] How many electrons can NADH transfer at once? [ 1 OR 2 OR either] Why doesn’t NADH transfer electrons directly to the iron? Assume that the FeS clusters can only accept one electron at a time. Why is a flavin redox center required as an intermediary? Flavin cofactors are usually tightly bound to enzyme and are not released and rebound to the enzyme during the catalytic cycle. Why is it necessary for a flavin to be bound to the enzyme in Complex I? Transfer 2: Electron Transfer from FMNH2 to Fe/S Clusters of Complex I The electrons are transferred from FMNH2 to the FeS cluster. The tetranuclear Fe/S cluster is based on the cubanestructure with Fe and S occupying alternating corners of a square in a tetrahedral geometry. Each Fe is also coordinated to thiolate anions. The actual structure is a distorted cube as shown below, along with that of the binuclear cluster, whose bond angle also deviate from those in a tetrahedron. What is the coordination number of the irons in the cubane ring? Assuming that all Fe are in the III oxidation state, what is the overall charge on the cluster? If one electron is added, what is the overall charge? Electrons are transferred from one FeS cluster to another in this pathway. Are these outer sphere or inner sphere transfers? EPR studies suggest that not all of these electron transfers are exergonic, but rather the chain is like a roller coaster. Why does the system use a mixture of endergonic and exergonic electron transfers? Transfer 3: Electron Transfer from FeS clusters to UQ The Fe/S Clusters transfer single electrons to UQ one at a time. The reduction of the ubiquinone (UQ) is shown below while the transfer of electrons from the FeS clusters to UQ is on the following page. Add the electrons to the UQ templates to show how UQ could be reduced in single electron transfers and protonation steps. Studies have shown that the ubiquinone radical is found only when bound to Complex I and not free in the membrane. This is true at many different overall ratios of UQ/UQH2. Why is this biologically important? Mechansim of e- and H+ flow from FeS clusters to UQ Electrons and protons flow from the bottom of each box to the top. Eventually there will be an increase in 2 e - and 2 H + to be transferred to the UQ. Berrisford and Sazanov, JBC, 284, 29773, 2009. Complete the diagram with correct charges and electron flow. Complex 1 is done with its redox task; the UQH 2 moves on to the next complex... Summary: Proton Transfer by the Membrane Domain in Complex I In the catalytic cycle of Complex I (earlier), protons are produced AND moved from inside mitochondria to a space between the membranes. How many protons are moved across the membrane for each catalytic cycle of Complex I? Is this active transport or passive diffusion? If this is active, what is fueling this transport? Is this with or against the concentration gradient? (i.e. antiporter or synporter?) Application Problems Complex I is inhibited by more than 60 different families of compoounds. They include the classic Complex I inhibitor rotenone and many other synthetic insecticides/acaricides. The classes include: Class I/A (the prototype of which is Piericidin A), Class II/B (the prototype of which is Rotenone) and Class C (the prototype of which is Capsaicin). They appear to bind at the same site. From the structure of the 3 prototypes, what are the characteristics of the pharmacophore, the “ideal binding ligand”? Where do they likely bind? How “promiscuous” is the binding site? ManydevastatingneurologicaldiseaseslikeParkinson’sDiseaseareassociated with defects in Complex I. In addition to major problems with oxidative ATP production, reactive oxygen species (ROS) increase. The major sites for generation of ROS are Complex 1 and Complex III. Given the locations of the electron carriers at the periphery and internal within the protein complex, which electron carriers might most readily leak electrons to dioxygen? What ROS is likely to form in the process? InhibitorsmightblockaccessofUQorconformationalchangesnecessaryfor final reduction of the ubiqinone free radical. Class A inhibitors dramatically increase ROS production. The actual site of ROS production in Complex is a bit controversial. One possible electron donor to dioxygen is FMN. Why is this a likely candidate? Mutants that lack N2 iron-sulfur cluster showed ROS production. Is this consistent with FMN site involvement in ROS production? Basedonthestandardmidpointreductionpotentials,whichmightbethebest candidate to promote dioxygen conversion to superoxide (not to water), FMN or N2? Aparticularinhibitor,DPI,inhibitedROSproductioninthepresenceofexcess NADH but enhanced it in the presence of high UQH 2 (which could allow reverse electron transport. Does this suggest FMN or N2 involvement in ROS production? DPI inhibition requires NADH. The higher the NADH/NAD + ratios, the high the inhibition. At high NADH levels the FeS clusters stay oxidized. Must the protein be reduced or oxidized for interaction with DPI? IfcomplexIisisolatedfromthemitochondriaandinsertedintoartificial membranes, superoxide is produced in a slow step (along with a bit of H 2 O 2 ). In a fast process, the following disproportionation (self-redox) can occur: O 2 - + O 2 - + 4 H + -> H 2 O 2 + O 2 , allowing superoxide production to be monitored by measuring peroxide levels. As above, the rate of superoxide production depends on the NADH/NAD+ ratio. Hence what must be bound at the active site for superoxide production to occur? In intact mitochondria, if a Q site inhibitor like rotenone is added, superoxide production increases. What would Q site inhibitors do to the levels of NADH? In submitochondrial preparations,normal Complex I activity occurs (which leads to formation of a sustained proton gradient). Also reverse electron transport, powered by an artificial proton gradient can occur, which leads to the reduction of NAD+ can occur (see diagram below). A summary of finding on superoxide production by Complex I is given below: Superoxide production is inhibited flavin site inhibitors but not Q site inhibitors. Reverse electron transport leads to NAD + and O 2 reduction Reverse electron transport superoxide production is inhibited by both flavin and Q site inhibitors Based on these findings, which site (the flavin or Q site), is involved in superoxide production. |
Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Molecular_and_Atomic_Spectroscopy/04_Learning_Objectives/06_Atomic_Spectroscopy | After completing this unit the student will be able to: Compare and contrast the advantages of flame, furnace and inductively coupled plasma atomization sources. Justify why continuum radiation sources are usually not practical to use for atomic absorption spectroscopy. Describe the design of a hollow cathode lamp and justify the reasons for a hollow cathode configuration and low pressure of argon filler gas. Devise an instrumental procedure to account for flame noise in atomic absorption spectroscopy. Devise an instrumental procedure to account for molecular absorption and scatter from particulate matter in atomic absorption spectroscopy. Describe three possible strategies that can be used to overcome the problem of non-volatile metal complexes. Devise a procedure to overcome excessive ionization of an analyte. Devise a procedure to account for matrix effects. |
Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/End_Creek%3A_Spotted_Frogs_and_Aquatic_Snails_in_Wetlands__A_Water_Quality_Investigation/02_Qualitative_Investigation_of_Cations_and_Anions | Adapted from “A Study of Onondaga Lake Water” – Chemistry in Context: Applying Chemistry to Society Lab Manual The End Creek Wetland Restoration Project is located 15 km north of La Grande, in Union County, Oregon. The approximately 550 acre restored wetland embodies many qualities of high value to the community. The wetland is the only natural grassland/floodplain area of any size at the north end of the Grande Ronde Valley. It forms a natural corridor between the Blue Mountains and the Grande Ronde River as it transects through extensive, uninterrupted agricultural lands. Consequently, it protects a rare island of habitat for mammals, fish, amphibians, and reptiles and is an important harbor for threatened, endangered, or sensitive plant and animal species. The wet meadows and ponds provide stopover sites for migratory waterfowl traveling north and south through the intermountain west, and nesting and feeding habitat for those that stay. Drained and converted to agricultural land by the end of the last century and operated as a dairy farm until the mid-50s, the property became part of the Wetland Reserve Program in 2005. In 2006 plans began to restore the creek to its original sinuosity with the goals of improving fish habitat and restoring original vegetation. Creek restoration helped rebuild the water table and feed natural ponds. In 2008 Columbia Spotted Frogs were discovered in one of the ponds and a breeding population study began. However, over the following years this population has dwindled down and research shows no correlation to water quality parameters. We will therefore focus our research on aquatic snails. Three families of aquatic snails were identified but their distribution differs between ponds and seasonally. In this investigation, you will learn how to identify some metal ions and anions that are connected to water quality. The presence of these ions may play a role as to why certain ponds provide a more suitable habitat for certain types of aquatic snails. The ultimate goal will be for you to determine whether any significant difference exists in the water composition across three different ponds that may justify differences in snail populations. The question to be answered is: Are there differences in water chemistry that influence snail family distributions in the ponds at End Creek? Background You will be investigating the presence of positive and negative ions in water samples using a variety of chemical and physical tests. Presence of ionic species in solution will influence the conductivity . Higher concentrations of ions will result in a higher conductivity. Flame tests can be used to qualitatively identify the presence of specific ions. Much like in fireworks, when metal ions are heated in a flame, electrons from lower energy levels absorb energy and are promoted to higher levels. When these electrons relax back to the ground state, they emit light of characteristic wavelength. By observing the color emitted by known solutions of metal ions, the presence of metals can be determined in an unknown sample. You will also make use of chemical reactions to identify ions based on the formation of insoluble salts or highly colored compounds. You will first investigate different reagents and determine which reagent is best suited for the identification of a specific ion. Once you have developed your own analytical scheme on a set of known positive and negative ions, you will use the same chemical reactions to analyze water sampled at End Creek and determine whether there are differences in the chemical composition of the three ponds. Overview Test the electrical conductivity of water from the three End Creek ponds. Use the flame test to test for presence of Na + , Ca 2 + and Fe 3 + in standard samples. Use the flame test to test for presence of Na + , Ca 2 + and Fe 3 + in the End Creek ponds. Conduct chemical tests on samples of known composition to identify best reagents to analyze for presence of Ca 2 + and Fe 3 + . Once you have identified the best reagents, test the End Creek pond water for presence of Ca 2 + and Fe 3 + . Conduct chemical tests on samples of known composition to identify best reagents to analyze for presence of Cl - , SO 4 2 - and PO 4 3 - . Once you have identified the best reagents, test the End Creek pond water for presence of Cl - , SO 4 2 - and PO 4 3 - . Materials Needed Chemicals Equipment Dropper bottles with dilute solutions of: Sodium chloride Calcium chloride Iron (III) chloride Sodium oxalate Potassium thiocyanate Sodium sulfate Sodium hydrogen phosphate Silver nitrate Barium nitrate Plastic well plates Small stirring rods Disposable pipets Ni-Cr wire for flame test Conductivity tester Bunsen burner Experimental Procedure I. Test for conductivity Place a small amount of water from each pond in a well plate well. Immerse the two leads of the conductivity tester in each water sample making sure the two wires do not touch. Record your observations. II. Flame test for metal ions Using separate, clean wells, place a few drops each of sodium chloride, calcium chloride, and iron (III) chloride. Light the Bunsen burner and adjust the flame so to obtain a bright blue inner cone (instructor will demonstrate). Wash the Ni-Cr wire with deionized water and bring wire into the flame. If the wire is clean, no color should be emitted. Dip the wire in the calcium chloride solution and position it right above the inner blue cone where the hottest part of the flame is located. Observe the color emitted from calcium and record your observations. Clean the wire and then repeat flame test with sodium chloride and iron (III) chloride. Compare the colors and record your observations. Using clean wells, place a few drops each of the pond water samples. Make sure your Ni-Cr wire is clean! Dip the wire in each water sample and bring the wire into the flame. Do you observe any color? Record your results. III. Chemical test for positive ions In this investigation you will discover which reagent is best suited to analyze a water sample for presence of sodium, calcium and iron ions. Available reagents are sodium oxalate (Na 2 C 2 O 4 ) and potassium thiocyanate (KSCN). A suitable reagent is one that will uniquely react with the analyte of interest. Signs of chemical reactions include change of color, precipitation, evolution of gasses, etc. Using the well plate, standard solutions of sodium chloride, calcium chloride and iron (III) chloride, devise a strategy to identify the most suitable reagent that can be used to identify presence of sodium, calcium and iron ions. Jot down an experiment design and check with the instructor before proceeding. Once you have identified the best reagents for the identification of metal ions in water, test the three pond water samples. Record your observations. Based on these results, decide whether any metal ions are present in any of the three ponds. Clean up: empty any contents of the well plates into proper disposal bottle located under the hood. Rinse well plate and shake to remove any residual water. IV. Chemical test for negative ions In this investigation you will discover which reagent is best suited to analyze a water sample for presence of anions such as sulfate, chloride and phosphate. Available reagents are silver nitrate (AgNO 3 ), barium nitrate (Ba(NO 3 ) 2 ) and calcium chloride (CaCl 2 ). Again, look for signs of chemical reactions which may include change of color, precipitation, evolution of gasses, etc. Using the well plate and standard solutions of sodium sulfate, sodium chloride and sodium hydrogen phosphate, devise a strategy to identify the most suitable reagent that can be used to identify presence of sulfate, chloride and phosphate ions in water. Jot down an experiment design and check with the instructor before proceeding. Once you have identified the best reagents for the identification of anions in water, test the three pond water samples. Record your observations. Based on these results, decide whether any anions are present in any of the three ponds. Clean up: empty any contents of the well plates into proper disposal bottle located under the hood. Rinse well plate. V. Questions to be answered after completing experiment What do the results of the conductivity test indicate about what is present in the End Creek water samples? Is there a difference in the conductivity of the three samples? What color do sodium, calcium and iron ions produce in the flame? Based on the results from the flame test, which of these ions are present in the pond water samples? Describe which reagent you chose, respectively, for the analysis of calcium and iron ions and write the balanced chemical reaction for each test. Based on your results, which metal ions are present in the pond water? Research the solubility rules for anions and cations. Was any reagent suitable for testing for sodium ion? Based on solubility rules, why would it be hard to devise a test for sodium based on the formation of a precipitate? Describe which reagent you chose respectively for the analysis of sulfate, chloride and phosphate and write the balanced chemical reaction for each test. Based on your results, which anions are present in the pond water? Review all the results gathered through this exploration. Based on these results, are there any differences in water composition among the ponds at End Creek? Summarize your conclusions. |
Courses/University_of_Kansas/CHEM_130%3A_General_Chemistry_I_(Sharpe_Elles)/07%3A_Reactions_in_Aqueous_Solution/7.05%3A__Stoichiometry_of_Reactions_in_Solution | Learning Objectives To balance equations that describe reactions in solution. To solve quantitative problems involving the stoichiometry of reactions in solution. Quantitative calculations involving reactions in solution are carried out in the same manner as we discussed previously. Instead of masses , however, we use volumes of solutions of known concentration to determine the number of moles of reactants. Whether we are dealing with volumes of solutions of reactants or masses of reactants, the coefficients in the balanced chemical equation tell us the number of moles of each reactant needed and the number of moles of each product that can be produced. Calculating Moles from Volume An expanded version of the flowchart for stoichiometric calculations illustrated previously is shown in Figure \(\PageIndex{1}\). We can use the balanced chemical equation for the reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products to determine the amounts of other species. Either the masses or the volumes of solutions of reactants and products can be used to determine the amounts of other species in a balanced chemical equation. Note the Pattern The balanced chemical equation for a reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products can be used in stoichiometric calculations. Example \(\PageIndex{1}\) Gold is extracted from its ores by treatment with an aqueous cyanide solution, which causes a reaction that forms the soluble [Au(CN) 2 ] − ion. Gold is then recovered by reduction with metallic zinc according to the following equation: \( Zn(s) + 2[Au(CN)_2]^-(aq) \rightarrow [Zn(CN)_4]^{2-}(aq) + 2Au(s) \notag \) What mass of gold would you expect to recover from 400.0 L of a 3.30 × 10 −4 M solution of [Au(CN) 2 ] − ? Given: chemical equation and molarity and volume of reactant Asked for: mass of product Strategy: A Check the chemical equation to make sure it is balanced as written; balance if necessary. Then calculate the number of moles of [Au(CN) 2 ] − present by multiplying the volume of the solution by its concentration. B From the balanced chemical equation, use a mole ratio to calculate the number of moles of gold that can be obtained from the reaction. To calculate the mass of gold recovered, multiply the number of moles of gold by its molar mass. Solution The equation is balanced as written, so we can proceed to the stoichiometric calculation. We can adapt the flowchart fro above for this particular problem as follows: As indicated in the strategy, we start by calculating the number of moles of [Au(CN) 2 ] − present in the solution from the volume and concentration of the [Au(CN) 2 ] − solution: \( \begin{align} moles\: [Au(CN)_2 ]^-& = V_L M_{mol/L} \notag \\ & = 400 .0\: \cancel{L} \left( \dfrac{3 .30 \times 10^{4-}\: mol\: [Au(CN)_2 ]^-} {1\: \cancel{L}} \right) = 0 .132\: mol\: [Au(CN)_2 ]^- \notag \end{align} \) B Because the coefficients of gold and the [Au(CN) 2 ] − ion are the same in the balanced chemical equation, if we assume that Zn(s) is present in excess, the number of moles of gold produced is the same as the number of moles of [Au(CN) 2 ] − we started with (i.e., 0.132 mol of Au). The problem asks for the mass of gold that can be obtained, so we need to convert the number of moles of gold to the corresponding mass using the molar mass of gold: \( \begin{align} mass\: of\: Au &= (moles\: Au)(molar\: mass\: Au) \notag \\ &= 0 .132\: \cancel{mol\: Au} \left( \dfrac{196 .97\: g\: Au} {1\: \cancel{mol\: Au}} \right) = 26 .0\: g\: Au \notag \end{align} \) At a 2011 market price of over $1400 per troy ounce (31.10 g), this amount of gold is worth $1170. \( 26 .0\: \cancel{g\: Au} \times \dfrac{1\: \cancel{troy\: oz}} {31 .10\: \cancel{g}} \times \dfrac{\$1400} {1\: \cancel{troy\: oz\: Au}} = \$1170 \notag \) Exercise \(\PageIndex{1}\) What mass of solid lanthanum(III) oxalate nonahydrate [La 2 (C 2 O 4 ) 3 ·9H 2 O] can be obtained from 650 mL of a 0.0170 M aqueous solution of LaCl 3 by adding a stoichiometric amount of sodium oxalate? Answer 3.89 g Limiting Reactants in Solutions The concept of limiting reactants applies to reactions that are carried out in solution as well as to reactions that involve pure substances. If all the reactants but one are present in excess, then the amount of the limiting reactant may be calculated as illustrated in Example \(\PageIndex{2}\). Example \(\PageIndex{2}\) Because the consumption of alcoholic beverages adversely affects the performance of tasks that require skill and judgment, in most countries it is illegal to drive while under the influence of alcohol. In almost all US states, a blood alcohol level of 0.08% by volume is considered legally drunk. Higher levels cause acute intoxication (0.20%), unconsciousness (about 0.30%), and even death (about 0.50%). The Breathalyzer is a portable device that measures the ethanol concentration in a person’s breath, which is directly proportional to the blood alcohol level. The reaction used in the Breathalyzer is the oxidation of ethanol by the dichromate ion: \( 3CH_3 CH_2 OH(aq) + \underset{yellow-orange}{2Cr_2 O_7^{2 -}}(aq) + 16H ^+ (aq) \underset{H_2 SO_4 (aq)}{\xrightarrow{\hspace{10px} Ag ^+\hspace{10px}} } 3CH_3 CO_2 H(aq) + \underset{green}{4Cr^{3+}} (aq) + 11H_2 O(l) \) When a measured volume (52.5 mL) of a suspect’s breath is bubbled through a solution of excess potassium dichromate in dilute sulfuric acid, the ethanol is rapidly absorbed and oxidized to acetic acid by the dichromate ions. In the process, the chromium atoms in some of the Cr 2 O 7 2 − ions are reduced from Cr 6 + to Cr 3 + . In the presence of Ag + ions that act as a catalyst, the reaction is complete in less than a minute. Because the Cr 2 O 7 2 − ion (the reactant) is yellow-orange and the Cr 3 + ion (the product) forms a green solution, the amount of ethanol in the person’s breath (the limiting reactant) can be determined quite accurately by comparing the color of the final solution with the colors of standard solutions prepared with known amounts of ethanol. A Breathalyzer ampul before (a) and after (b) ethanol is added. When a measured volume of a suspect’s breath is bubbled through the solution, the ethanol is oxidized to acetic acid, and the solution changes color from yellow-orange to green. The intensity of the green color indicates the amount of ethanol in the sample. A typical Breathalyzer ampul contains 3.0 mL of a 0.25 mg/mL solution of K 2 Cr 2 O 7 in 50% H 2 SO 4 as well as a fixed concentration of AgNO 3 (typically 0.25 mg/mL is used for this purpose). How many grams of ethanol must be present in 52.5 mL of a person’s breath to convert all the Cr 6 + to Cr 3 + ? Given: volume and concentration of one reactant Asked for: mass of other reactant needed for complete reaction Strategy: A Calculate the number of moles of Cr 2 O 7 2 − ion in 1 mL of the Breathalyzer solution by dividing the mass of K 2 Cr 2 O 7 by its molar mass. B Find the total number of moles of Cr 2 O 7 2 − ion in the Breathalyzer ampul by multiplying the number of moles contained in 1 mL by the total volume of the Breathalyzer solution (3.0 mL). C Use the mole ratios from the balanced chemical equation to calculate the number of moles of C 2 H 5 OH needed to react completely with the number of moles of Cr 2 O 7 2 − ions present. Then find the mass of C 2 H 5 OH needed by multiplying the number of moles of C 2 H 5 OH by its molar mass. Solution A In any stoichiometry problem, the first step is always to calculate the number of moles of each reactant present. In this case, we are given the mass of K 2 Cr 2 O 7 in 1 mL of solution, which we can use to calculate the number of moles of K 2 Cr 2 O 7 contained in 1 mL: \( \dfrac{moles\: K_2 Cr_2 O_7} {1\: mL} = \dfrac{(0 .25\: \cancel{mg}\: K_2 Cr_2 O_7 )} {mL} \left( \dfrac{1\: \cancel{g}} {1000\: \cancel{mg}} \right) \left( \dfrac{1\: mol} {294 .18\: \cancel{g}\: K_2 Cr_2 O_7} \right) = 8.5 \times 10 ^{-7}\: moles \) B Because 1 mol of K 2 Cr 2 O 7 produces 1 mol of Cr 2 O 7 2 − when it dissolves, each milliliter of solution contains 8.5 × 10 −7 mol of Cr 2 O 7 2 − . The total number of moles of Cr 2 O 7 2 − in a 3.0 mL Breathalyzer ampul is thus \( moles\: Cr_2 O_7^{2-} = \left( \dfrac{8 .5 \times 10^{-7}\: mol} {1\: \cancel{mL}} \right) ( 3 .0\: \cancel{mL} ) = 2 .6 \times 10^{-6}\: mol\: Cr_2 O_7^{2–} \) C The balanced chemical equation tells us that 3 mol of C 2 H 5 OH is needed to consume 2 mol of Cr 2 O 7 2 − ion, so the total number of moles of C 2 H 5 OH required for complete reaction is \( moles\: of\: C_2 H_5 OH = ( 2.6 \times 10 ^{-6}\: \cancel{mol\: Cr_2 O_7 ^{2-}} ) \left( \dfrac{3\: mol\: C_2 H_5 OH} {2\: \cancel{mol\: Cr _2 O _7 ^{2 -}}} \right) = 3 .9 \times 10 ^{-6}\: mol\: C _2 H _5 OH \) As indicated in the strategy, this number can be converted to the mass of C 2 H 5 OH using its molar mass: \( mass\: C _2 H _5 OH = ( 3 .9 \times 10 ^{-6}\: \cancel{mol\: C _2 H _5 OH} ) \left( \dfrac{46 .07\: g} {\cancel{mol\: C _2 H _5 OH}} \right) = 1 .8 \times 10 ^{-4}\: g\: C _2 H _5 OH \) Thus 1.8 × 10 −4 g or 0.18 mg of C 2 H 5 OH must be present. Experimentally, it is found that this value corresponds to a blood alcohol level of 0.7%, which is usually fatal. Exercise \(\PageIndex{2}\) The compound para -nitrophenol (molar mass = 139 g/mol) reacts with sodium hydroxide in aqueous solution to generate a yellow anion via the reaction Because the amount of para -nitrophenol is easily estimated from the intensity of the yellow color that results when excess NaOH is added, reactions that produce para -nitrophenol are commonly used to measure the activity of enzymes, the catalysts in biological systems. What volume of 0.105 M NaOH must be added to 50.0 mL of a solution containing 7.20 × 10 −4 g of para -nitrophenol to ensure that formation of the yellow anion is complete? Answer 4.93 × 10 −5 L or 49.3 μL In examples above, the identity of the limiting reactant was apparent: [Au(CN) 2 ] − , LaCl 3 , ethanol, and para -nitrophenol. When the limiting reactant is not apparent, we can determine which reactant is limiting by comparing the molar amounts of the reactants with their coefficients in the balanced chemical equation, just as we did in an earlier chapter. The only difference is that now we use the volumes and concentrations of solutions of reactants rather than the masses of reactants to calculate the number of moles of reactants. Example \(\PageIndex{3}\) When aqueous solutions of silver nitrate and potassium dichromate are mixed, an exchange reaction occurs, and silver dichromate is obtained as a red solid. The overall chemical equation for the reaction is as follows: \(2AgNO_3(aq) + K_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + 2KNO_3(aq) \) What mass of Ag 2 Cr 2 O 7 is formed when 500 mL of 0.17 M K 2 Cr 2 O 7 are mixed with 250 mL of 0.57 M AgNO 3 ? Given: balanced chemical equation and volume and concentration of each reactant Asked for: mass of product Strategy: A Calculate the number of moles of each reactant by multiplying the volume of each solution by its molarity. B Determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation. C Use mole ratios to calculate the number of moles of product that can be formed from the limiting reactant. Multiply the number of moles of the product by its molar mass to obtain the corresponding mass of product. Solution A The balanced chemical equation tells us that 2 mol of AgNO 3 (aq) reacts with 1 mol of K 2 Cr 2 O 7 (aq) to form 1 mol of Ag 2 Cr 2 O 7 (s) (Figure 8.3.2). The first step is to calculate the number of moles of each reactant in the specified volumes: \( moles\: K_2 Cr_2 O_7 = 500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .17\: mol\: K_2 Cr_2 O_7} {1\: \cancel{L}} \right) = 0 .085\: mol\: K_2 Cr_2 O_7 \) \( moles\: AgNO_3 = 250\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .57\: mol\: AgNO_3} {1\: \cancel{L}} \right) = 0 .14\: mol\: AgNO_3 \) B Now we can determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient: \( K_2 Cr_2 O_7: \: \dfrac{0 .085\: mol} {1\: mol} = 0 .085 \) \( AgNO_3: \: \dfrac{0 .14\: mol} {2\: mol} = 0 .070 \) Because 0.070 < 0.085, we know that AgNO 3 is the limiting reactant. C Each mole of Ag 2 Cr 2 O 7 formed requires 2 mol of the limiting reactant (AgNO 3 ), so we can obtain only 0.14/2 = 0.070 mol of Ag 2 Cr 2 O 7 . Finally, we convert the number of moles of Ag 2 Cr 2 O 7 to the corresponding mass: \( mass\: of\: Ag_2 Cr_2 O_7 = 0 .070\: \cancel{mol} \left( \dfrac{431 .72\: g} {1 \: \cancel{mol}} \right) = 30\: g \: Ag_2 Cr_2 O_7 \) The Ag + and Cr 2 O 7 2− ions form a red precipitate of solid Ag 2 Cr 2 O 7 , while the K + and NO 3 − ions remain in solution. (Water molecules are omitted from molecular views of the solutions for clarity.) Exercise \(\PageIndex{3}\) Aqueous solutions of sodium bicarbonate and sulfuric acid react to produce carbon dioxide according to the following equation: \(2NaHCO_3(aq) + H_2SO_4(aq) \rightarrow 2CO_2(g) + Na_2SO_4(aq) + 2H_2O(l)\) If 13.0 mL of 3.0 M H 2 SO 4 are added to 732 mL of 0.112 M NaHCO 3 , what mass of CO 2 is produced? Answer 3.4 g Summary Quantitative calculations that involve the stoichiometry of reactions in solution use volumes of solutions of known concentration instead of masses of reactants or products. The coefficients in the balanced chemical equation tell how many moles of reactants are needed and how many moles of product can be produced. Contributors Anonymous Modified by Joshua Halpern ( Howard University ) Thumbnail from Wikimedia |
Bookshelves/Environmental_Chemistry/Toxicology_MSDT/6%3A_Principles_of_Toxicology/Section_3%3A_Toxic_Effects/3.4%3A_Organ_Specific_Toxic_Effects | Organ Specific Toxic Effects Toxic effects that pertain to specific organs and organ systems include: Figure \(\PageIndex{1}\). Organ -specific toxic effects pertain to specific organs and organ systems (Image Source: Adapted from iStock Photos, ©) Blood and Cardiovascular/Cardiac Toxicity Blood and Cardiovascular/Cardiac Toxicity results from xenobiotics acting directly on cells in circulating blood, bone marrow , and the heart. Examples of blood and cardiovascular/cardiac toxicity are: Hypoxia due to carbon monoxide binding of hemoglobin preventing transport of oxygen. Decrease in circulating leukocytes due to chloramphenicol damage to bone marrow cells . Leukemia due to benzene damage of bone marrow cells . Arteriosclerosis due to cholesterol accumulation in arteries and veins. Death of normal cells in and around the heart as a result of exposure to drugs used to treat cancer . Dermal Toxicity Dermal Toxicity can occur when a toxicant comes into direct contact with the skin or is distributed to it internally. Effects range from mild irritation to severe changes, such as irreversible damage, hypersensitivity , and skin cancer . Examples of dermal toxicity include: Dermal irritation from skin exposure to gasoline. Dermal corrosion from skin exposure to sodium hydroxide (lye). Dermal itching, irritation , and sometimes painful rash from poison ivy, caused by urushiol . Skin cancer due to ingestion of arsenic or skin exposure to UV light. Epigenetic Alterations Epigenetics is an emerging area in toxicology . In the field of genetics, epigenetics involves studying how external or environmental factors can switch genes on and off and change the programming of cells . More specifically, epigenetics refers to stable changes in the programming of gene expression which can alter the phenotype without changing the DNA sequence (genotype). Epigenetic modifications include DNA methylation, covalent modifications of histone tails, and regulation by non-coding RNAs, among others. Toxicants are examples of factors that can alter genetic programming. In the past, toxicology studies have assessed toxicity without measuring its impact at the level where gene expression occurs. Exogenous agents could cause long-term toxicity that continues after the initial exposure has disappeared, and such toxicities remain undetected by current screening methods. Thus, a current challenge in toxicology is to develop screening methods that would detect epigenetic alterations caused by toxicants . Research is being done to assess epigenetic changes caused by toxicants . For example, the National Institutes of Health (NIH) National Institute of Environmental Health Sciences (NIEHS) Environmental Epigenetics program provides funding for a variety of research projects that use state-of-the-art technologies to analyze epigenetic changes caused by environmental exposures. NIEHS-supported researchers use animals, cell cultures, and human tissue samples to pinpoint how epigenetic changes can lead to harmful health effects and can potentially be passed down to the next generation. Eye Toxicity Eye Toxicity results from direct contact with or internal distribution to the eye. Because the cornea and conjunctiva are directly exposed to toxicants , conjunctivitis and corneal erosion may be observed following occupational exposure to chemicals . Many household items can cause conjunctivitis. Chemicals in the circulatory system can distribute to the eye and cause corneal opacity, cataracts, and retinal and optic nerve damage. For example: Acids and strong alkalis may cause severe corneal corrosion . Corticosteroids may cause cataracts. Methanol (wood alcohol) may damage the optic nerve. Hepatotoxicity Hepatotoxicity is toxicity to the liver, bile duct, and gall bladder. Because of its extensive blood supply and significant role in metabolism , the liver is particularly susceptible to xenobiotics Thus, it is exposed to high doses of the toxicant or its toxic metabolites . The primary forms of hepatotoxicity are: Steatosis — lipid accumulation in the hepatocytes. Chemical hepatitis — inflammation of the liver. Hepatic necrosis — death of the hepatocytes. Intrahepatic cholestasis — backup of bile salts into the liver cells . Hepatic cancer — cancer of the liver. Cirrhosis — chronic fibrosis , often due to alcohol. Hypersensitivity — immune reaction resulting in hepatic necrosis . Related Resource: LiverTox ® Immunotoxicity Immunotoxicity is toxicity of the immune system. It can take several forms: Hypersensitivity ( allergy and autoimmunity ) Immunodeficiency Uncontrolled proliferation ( leukemia and lymphoma) The normal function of the immune system is to recognize and defend against foreign invaders. This is accomplished by production of cells that engulf and destroy the invaders or by antibodies that inactivate foreign material. Examples include: Contact dermatitis due to exposure to poison ivy. Systemic lupus erythematosus ("lupus") in workers exposed to hydrazine. Immunosuppression by cocaine. Leukemia induced by benzene. Figure \(\PageIndex{7}\) . Bone (which contains bone marrow ) and spleen, both components of the immune system, which recognizes and defends against foreign invaders (Image Source: iStock Photos, ©) Nephrotoxicity The kidney is highly susceptible to toxicants because a high volume of blood flows through the organ and it filters large amounts of toxins which can concentrate in the kidney tubules. Nephrotoxicity is toxicity to the kidneys. It can result in systemic toxicity causing: Decreased ability to excrete body wastes. Inability to maintain body fluid and electrolyte balance. Decreased synthesis of essential hormones (for example, erythropoietin, which increases the rate of blood cell production). Neurotoxicity Neurotoxicity represents toxicant damage to cells of the central nervous system (brain and spinal cord) and the peripheral nervous system (nerves outside the CNS ). The primary types of neurotoxicity are: Neuronopathies ( neuron injury) Axonopathies ( axon injury) Demyelination (loss of axon insulation) Interference with neurotransmission Reproductive Toxicity Reproductive Toxicity involves toxicant damage to either the male or female reproductive system . Toxic effects may cause: Decreased libido and impotence. Infertility. Interrupted pregnancy (abortion, fetal death, or premature delivery). Infant death or childhood morbidity. Altered sex ratio and multiple births. Chromosome abnormalities and birth defects. Childhood cancer . Respiratory Toxicity Respiratory Toxicity relates to effects on the upper respiratory system (nose, pharynx, larynx, and trachea) and the lower respiratory system (bronchi, bronchioles , and lung alveoli ). The primary types of respiratory toxicity are: Pulmonary irritation Asthma/bronchitis Reactive airway disease Emphysema Allergic alveolitis Fibrotic lung disease Pneumoconiosis Lung cancer Knowledge Check Knowledge Check 1. Toxic effects are primarily categorized into two general types: Systemic or organ-specific effects Carcinogenic or teratogenic effects Hepatic or nephrotoxic effects Answer Systemic or organ-specific effects - This is the correct answer. Toxic effects are broadly categorized as either systemic or organ- specific effects. 2. What is the main difference between acute and chronic toxicity? Different organs are involved Acute toxicity occurs only after a single dose, whereas chronic toxicity occurs with multiple doses Acute toxicity appears within hours or days of an exposure, whereas chronic toxicity takes many months or years to become a recognizable clinical disease Acute toxicity is less likely to lead to death than is chronic toxicity Answer Acute toxicity appears within hours or days of an exposure, whereas chronic toxicity takes many months or years to become a recognizable clinical disease - This is the correct answer. 3. Police respond to a 911 call in which two people are found dead in an enclosed bedroom heated by an unvented kerosene stove. There was no sign of trauma or violence. A likely cause of death is: Excess oxygen generated by the combustion of kerosene Acute toxicity due to uncombusted kerosene fumes Acute toxicity due to carbon monoxide poisoning Answer Acute toxicity due to carbon monoxide poisoning - This is the correct answer. The victims most likely died as a result of acute toxicity from exposure to carbon monoxide. 4. Genetic toxicity can result in: Gene mutation Changes in the structure and/or number of chromosomes Epigenetic alterations All of the above Answer All of the above - This is the correct answer. Genetic toxicity can cause gene mutations, changes in chromosome structure (aberration), increases or decreases in the number of chromosomes (aneuploidy or polyploidy), and changes to genetic programming (epigenetic alterations). |
Courses/Pasadena_City_College/CHEM_001A%3A_General_Chemistry_and_Chemical_Analysis/20%3A_Electrochemistry/20.04%3A_Standard_Reduction_Potentials | Learning Objectives To use redox potentials to predict whether a reaction is spontaneous. To balance redox reactions using half-reactions. In a galvanic cell, current is produced when electrons flow externally through the circuit from the anode to the cathode because of a difference in potential energy between the two electrodes in the electrochemical cell. In the Zn/Cu system, the valence electrons in zinc have a substantially higher potential energy than the valence electrons in copper because of shielding of the s electrons of zinc by the electrons in filled d orbitals. Hence electrons flow spontaneously from zinc to copper(II) ions, forming zinc(II) ions and metallic copper. Just like water flowing spontaneously downhill, which can be made to do work by forcing a waterwheel, the flow of electrons from a higher potential energy to a lower one can also be harnessed to perform work. Because the potential energy of valence electrons differs greatly from one substance to another, the voltage of a galvanic cell depends partly on the identity of the reacting substances. If we construct a galvanic cell similar to the one in part (a) in Figure \(\PageIndex{1}\) but instead of copper use a strip of cobalt metal and 1 M Co 2 + in the cathode compartment, the measured voltage is not 1.10 V but 0.51 V. Thus we can conclude that the difference in potential energy between the valence electrons of cobalt and zinc is less than the difference between the valence electrons of copper and zinc by 0.59 V. The measured potential of a cell also depends strongly on the concentrations of the reacting species and the temperature of the system. To develop a scale of relative potentials that will allow us to predict the direction of an electrochemical reaction and the magnitude of the driving force for the reaction, the potentials for oxidations and reductions of different substances must be measured under comparable conditions. To do this, chemists use the standard cell potential (E° cell ), defined as the potential of a cell measured under standard conditions—that is, with all species in their standard states (1 M for solutions, concentrated solutions of salts (about 1 M) generally do not exhibit ideal behavior, and the actual standard state corresponds to an activity of 1 rather than a concentration of 1 M. Corrections for non ideal behavior are important for precise quantitative work but not for the more qualitative approach that we are taking here. 1 atm for gases, pure solids or pure liquids for other substances) and at a fixed temperature, usually 25°C. Measured redox potentials depend on the potential energy of valence electrons, the concentrations of the species in the reaction, and the temperature of the system. Measuring Standard Electrode Potentials It is physically impossible to measure the potential of a single electrode: only the difference between the potentials of two electrodes can be measured (this is analogous to measuring absolute enthalpies or free energies ; recall that only differences in enthalpy and free energy can be measured.) We can, however, compare the standard cell potentials for two different galvanic cells that have one kind of electrode in common. This allows us to measure the potential difference between two dissimilar electrodes. For example, the measured standard cell potential (E°) for the Zn/Cu system is 1.10 V, whereas E° for the corresponding Zn/Co system is 0.51 V. This implies that the potential difference between the Co and Cu electrodes is 1.10 V − 0.51 V = 0.59 V. In fact, that is exactly the potential measured under standard conditions if a cell is constructed with the following cell diagram: \[Co_{(s)} ∣ Co^{2+}(aq, 1 M)∥Cu^{2+}(aq, 1 M) ∣ Cu (s)\;\;\; E°=0.59\; V \label{20.4.1} \] This cell diagram corresponds to the oxidation of a cobalt anode and the reduction of Cu 2 + in solution at the copper cathode. All tabulated values of standard electrode potentials by convention are listed for a reaction written as a reduction, not as an oxidation, to be able to compare standard potentials for different substances ( Table P1 ). The standard cell potential (E° cell ) is therefore the difference between the tabulated reduction potentials of the two half-reactions, not their sum: \[E°_{cell} = E°_{cathode} − E°_{anode} \label{20.4.2} \] In contrast, recall that half-reactions are written to show the reduction and oxidation reactions that actually occur in the cell, so the overall cell reaction is written as the sum of the two half-reactions. According to Equation \(\ref{20.4.2}\), when we know the standard potential for any single half-reaction, we can obtain the value of the standard potential of many other half-reactions by measuring the standard potential of the corresponding cell. Although it is impossible to measure the potential of any electrode directly, we can choose a reference electrode whose potential is defined as 0 V under standard conditions. The standard hydrogen electrode (SHE) is universally used for this purpose and is assigned a standard potential of 0 V. It consists of a strip of platinum wire in contact with an aqueous solution containing 1 M H + . The [H + ] in solution is in equilibrium with H 2 gas at a pressure of 1 atm at the Pt-solution interface (Figure \(\PageIndex{2}\)). Protons are reduced or hydrogen molecules are oxidized at the Pt surface according to the following equation: \[2H^+_{(aq)}+2e^− \rightleftharpoons H_{2(g)} \label{20.4.3} \] One especially attractive feature of the SHE is that the Pt metal electrode is not consumed during the reaction. Figure \(\PageIndex{3}\) shows a galvanic cell that consists of a SHE in one beaker and a Zn strip in another beaker containing a solution of Zn 2 + ions. When the circuit is closed, the voltmeter indicates a potential of 0.76 V. The zinc electrode begins to dissolve to form Zn 2 + , and H + ions are reduced to H 2 in the other compartment. Thus the hydrogen electrode is the cathode, and the zinc electrode is the anode. The diagram for this galvanic cell is as follows: \[Zn_{(s)}∣Zn^{2+}_{(aq)}∥H^+(aq, 1 M)∣H_2(g, 1 atm)∣Pt_{(s)} \label{20.4.4} \] The half-reactions that actually occur in the cell and their corresponding electrode potentials are as follows: cathode: \[2H^+_{(aq)} + 2e^− \rightarrow H_{2(g)}\;\;\; E°_{cathode}=0 V \label{20.4.5} \] anode: \[Zn_{(s)} \rightarrow Zn^{2+}_{(aq)}+2e^−\;\;\; E°_{anode}=−0.76\; V \label{20.4.6} \] overall: \[Zn_{(s)}+2H^+_{(aq)} \rightarrow Zn^{2+}_{(aq)}+H_{2(g)} \label{20.4.7} \] We then use Equation \ref{20.4.2} to calculate the cell potential \[\begin{align*} E°_{cell} &=E°_{cathode}−E°_{anode}\\[4pt] &=0.76\; V \end{align*} \nonumber \] Although the reaction at the anode is an oxidation, by convention its tabulated E° value is reported as a reduction potential. The potential of a half-reaction measured against the SHE under standard conditions is called the standard electrode potential for that half-reaction. In this example, the standard reduction potential for Zn 2 + (aq) + 2e − → Zn(s) is −0.76 V, which means that the standard electrode potential for the reaction that occurs at the anode, the oxidation of Zn to Zn 2 + , often called the Zn/Zn 2 + redox couple, or the Zn/Zn 2 + couple, is −(−0.76 V) = 0.76 V. We must therefore subtract E° anode from E° cathode to obtain \[E°_{cell}: 0 \,V − (−0.76\, V) = 0.76\, V \nonumber \] Because electrical potential is the energy needed to move a charged particle in an electric field, standard electrode potentials for half-reactions are intensive properties and do not depend on the amount of substance involved. Consequently, E° values are independent of the stoichiometric coefficients for the half-reaction, and, most important, the coefficients used to produce a balanced overall reaction do not affect the value of the cell potential. E° values do NOT depend on the stoichiometric coefficients for a half-reaction, because it is an intensive property. The Standard Hydrogen Electrode (SHE): The Standard Hydrogen Electrode (SHE)(opens in new window) [youtu.be] Standard Electrode Potentials To measure the potential of the Cu/Cu 2 + couple, we can construct a galvanic cell analogous to the one shown in Figure \(\PageIndex{3}\) but containing a Cu/Cu 2 + couple in the sample compartment instead of Zn/Zn 2 + . When we close the circuit this time, the measured potential for the cell is negative (−0.34 V) rather than positive. The negative value of \(E°_{cell}\) indicates that the direction of spontaneous electron flow is the opposite of that for the Zn/Zn 2 + couple. Hence the reactions that occur spontaneously, indicated by a positive \(E°_{cell}\), are the reduction of Cu 2 + to Cu at the copper electrode. The copper electrode gains mass as the reaction proceeds, and H 2 is oxidized to H + at the platinum electrode. In this cell, the copper strip is the cathode, and the hydrogen electrode is the anode. The cell diagram therefore is written with the SHE on the left and the Cu 2 + /Cu couple on the right: \[Pt_{(s)}∣H_2(g, 1 atm)∣H^+(aq, 1\; M)∥Cu^{2+}(aq, 1 M)∣Cu_{(s)} \label{20.4.8} \] The half-cell reactions and potentials of the spontaneous reaction are as follows: Cathode: \[Cu^{2+}{(aq)} + 2e^− \rightarrow Cu_{(g)}\;\;\; E°_{cathode} = 0.34\; V \label{20.4.9} \] Anode: \[H_{2(g)} \rightarrow 2H^+_{(aq)} + 2e^−\;\;\; E°_{anode} = 0\; V \label{20.4.10} \] Overall: \[H_{2(g)} + Cu^{2+}_{(aq)} \rightarrow 2H^+_{(aq)} + Cu_{(s)} \label{20.4.11} \] We then use Equation \ref{20.4.2} to calculate the cell potential \[\begin{align*} E°_{cell} &= E°_{cathode}− E°_{anode} \\[4pt] &= 0.34\; V \end{align*} \nonumber \] Thus the standard electrode potential for the Cu 2 + /Cu couple is 0.34 V. Electrode Potentials and ECell: Electrode and Potentials and Ecell(opens in new window) [youtu.be] Balancing Redox Reactions Using the Half-Reaction Method In Section 4.4 , we described a method for balancing redox reactions using oxidation numbers. Oxidation numbers were assigned to each atom in a redox reaction to identify any changes in the oxidation states. Here we present an alternative approach to balancing redox reactions, the half-reaction method, in which the overall redox reaction is divided into an oxidation half-reaction and a reduction half-reaction, each balanced for mass and charge. This method more closely reflects the events that take place in an electrochemical cell, where the two half-reactions may be physically separated from each other. We can illustrate how to balance a redox reaction using half-reactions with the reaction that occurs when Drano, a commercial solid drain cleaner, is poured into a clogged drain. Drano contains a mixture of sodium hydroxide and powdered aluminum, which in solution reacts to produce hydrogen gas: \[Al_{(s)} + OH^−_{(aq)} \rightarrow Al(OH)^−_{4(aq)} + H_{2(g)} \label{20.4.12} \] In this reaction, \(Al_{(s)}\) is oxidized to Al 3 + , and H + in water is reduced to H 2 gas, which bubbles through the solution, agitating it and breaking up the clogs. The overall redox reaction is composed of a reduction half-reaction and an oxidation half-reaction. From the standard electrode potentials listed Table P1 , we find the corresponding half-reactions that describe the reduction of H + ions in water to H 2 and the oxidation of Al to Al 3 + in basic solution: reduction: \[2H_2O_{(l)} + 2e^− \rightarrow 2OH^−_{(aq)} + H_{2(g)} \label{20.4.13} \] oxidation: \[Al_{(s)} + 4OH^−_{(aq)} \rightarrow Al(OH)^−_{4(aq)} + 3e^− \label{20.4.14} \] The half-reactions chosen must exactly reflect the reaction conditions, such as the basic conditions shown here. Moreover, the physical states of the reactants and the products must be identical to those given in the overall reaction, whether gaseous, liquid, solid, or in solution. In Equation \(\ref{20.4.13}\), two H + ions gain one electron each in the reduction; in Equation \(\ref{20.4.14}\), the aluminum atom loses three electrons in the oxidation. The charges are balanced by multiplying the reduction half-reaction (Equation \(\ref{20.4.13}\)) by 3 and the oxidation half-reaction (Equation \(\ref{20.4.14}\)) by 2 to give the same number of electrons in both half-reactions: reduction: \[6H_2O_{(l)} + 6e^− \rightarrow 6OH^−_{(aq)} + 3H_{2(g)} \label{20.4.15} \] oxidation: \[2Al_{(s)} + 8OH^−_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 6e^− \label{20.4.16} \] Adding the two half-reactions, \[6H_2O_{(l)} + 2Al_{(s)} + 8OH^−_{(aq)} \rightarrow 2Al(OH)^−{4(aq)} + 3H_{2(g)} + 6OH^−_{(aq)} \label{20.4.17} \] Simplifying by canceling substances that appear on both sides of the equation, \[6H_2O_{(l)} + 2Al_{(s)} + 2OH^−_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{20.4.18} \] We have a −2 charge on the left side of the equation and a −2 charge on the right side. Thus the charges are balanced, but we must also check that atoms are balanced: \[2Al + 8O + 14H = 2Al + 8O + 14H \label{20.4.19} \] The atoms also balance, so Equation \(\ref{20.4.18}\) is a balanced chemical equation for the redox reaction depicted in Equation \(\ref{20.4.12}\). The half-reaction method requires that half-reactions exactly reflect reaction conditions, and the physical states of the reactants and the products must be identical to those in the overall reaction. We can also balance a redox reaction by first balancing the atoms in each half-reaction and then balancing the charges. With this alternative method, we do not need to use the half-reactions listed in Table P1 , but instead focus on the atoms whose oxidation states change, as illustrated in the following steps: Step 1: Write the reduction half-reaction and the oxidation half-reaction. For the reaction shown in Equation \(\ref{20.4.12}\), hydrogen is reduced from H + in OH − to H 2 , and aluminum is oxidized from Al° to Al 3 + : reduction: \[OH^−_{(aq)} \rightarrow H_{2(g)} \label{20.4.20} \] oxidation: \[Al_{(s)} \rightarrow Al(OH)^−_{4(aq)} \label{20.4.21} \] Step 2: Balance the atoms by balancing elements other than O and H. Then balance O atoms by adding H 2 O and balance H atoms by adding H + . Elements other than O and H in the previous two equations are balanced as written, so we proceed with balancing the O atoms. We can do this by adding water to the appropriate side of each half-reaction: reduction: \[OH^−_{(aq)} \rightarrow H_{2(g)} + H_2O_{(l)} \label{20.4.22} \] oxidation: \[Al_{(s)} + 4H_2O_{(l)} \rightarrow Al(OH)^−_{4(aq)} \label{20.4.23} \] Balancing H atoms by adding H + , we obtain the following: reduction: \[OH^−_{(aq)} + 3H^+_{(aq)} \rightarrow H_{2(g)} + H_2O_{(l)} \label{20.4.24} \] oxidation: \[Al_{(s)} + 4H_2O_{(l)} \rightarrow Al(OH)^−_{4(aq)} + 4H^+_{(aq)} \label{20.4.25} \] We have now balanced the atoms in each half-reaction, but the charges are not balanced. Step 3: Balance the charges in each half-reaction by adding electrons. Two electrons are gained in the reduction of H + ions to H 2 , and three electrons are lost during the oxidation of Al° to Al 3 + : reduction: \[OH^−_{(aq)} + 3H^+_{(aq)} + 2e^− \rightarrow H_{2(g)} + H_2O_{(l)} \label{20.4.26} \] oxidation: \[Al_{(s)} + 4H_2O_{(l)} \rightarrow Al(OH)^−_{4(aq)} + 4H^+_{(aq)} + 3e^− \label{20.4.27} \] Step 4: Multiply the reductive and oxidative half-reactions by appropriate integers to obtain the same number of electrons in both half-reactions. In this case, we multiply Equation \(\ref{20.4.26}\) (the reductive half-reaction) by 3 and Equation \(\ref{20.4.27}\) (the oxidative half-reaction) by 2 to obtain the same number of electrons in both half-reactions: reduction: \[3OH^−_{(aq)} + 9H^+_{(aq)} + 6e^− \rightarrow 3H_{2(g)} + 3H_2O_{(l)} \label{20.4.28} \] oxidation: \[2Al_{(s)} + 8H_2O_{(l)} \rightarrow 2Al(OH)^−_{4(aq)} + 8H^+_{(aq)} + 6e^− \label{20.4.29} \] Step 5: Add the two half-reactions and cancel substances that appear on both sides of the equation. Adding and, in this case, canceling 8H + , 3H 2 O, and 6e − , \[2Al_{(s)} + 5H_2O_{(l)} + 3OH^−_{(aq)} + H^+_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{20.4.30} \] We have three OH − and one H + on the left side. Neutralizing the H + gives us a total of 5H 2 O + H 2 O = 6H 2 O and leaves 2OH − on the left side: \[2Al_{(s)} + 6H_2O_{(l)} + 2OH^−_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{20.4.31} \] Step 6: Check to make sure that all atoms and charges are balanced. Equation \(\ref{20.4.31}\) is identical to Equation \(\ref{20.4.18}\), obtained using the first method, so the charges and numbers of atoms on each side of the equation balance. Example \(\PageIndex{1}\) In acidic solution, the redox reaction of dichromate ion (\(\ce{Cr2O7^{2−}}\)) and iodide (\(\ce{I^{−}}\)) can be monitored visually. The yellow dichromate solution reacts with the colorless iodide solution to produce a solution that is deep amber due to the presence of a green \(\ce{Cr^{3+}(aq)}\) complex and brown \(\ce{I2(aq)}\) ions (Figure \(\PageIndex{4}\)): \[\ce{Cr2O7^{2−}(aq) + I^{−}(aq) -> Cr^{3+}(aq) + I2(aq)} \nonumber \] Balance this equation using half-reactions. Given: redox reaction and Table P1 Asked for: balanced chemical equation using half-reactions Strategy: Follow the steps to balance the redox reaction using the half-reaction method. Solution From the standard electrode potentials listed in Table P1 , we find the half-reactions corresponding to the overall reaction: reduction: \[\ce{Cr2O7^{2−}(aq) + 14H^{+}(aq) + 6e^{−} -> 2Cr^{3+}(aq) + 7H2O(l)} \nonumber \] oxidation: \[\ce{2I^{−}(aq) -> I2(aq) + 2e^{−}} \nonumber \] Balancing the number of electrons by multiplying the oxidation reaction by 3, oxidation: \[\ce{6I^{−}(aq) -> 3I2(aq) + 6e^{−}} \nonumber \] Adding the two half-reactions and canceling electrons, \[\ce{Cr2O^{2−}7(aq) + 14H^{+}(aq) + 6I^{−}(aq) -> 2Cr^{3+}(aq) + 7H2O(l) + 3I2(aq)} \nonumber \] We must now check to make sure the charges and atoms on each side of the equation balance: \[\begin{align*} (−2) + 14 + (−6) &= +6 \\[4pt] +6 &\overset{\checkmark}{=} +6 \end{align*} \nonumber \] and atoms \[\ce{2Cr + 7O + 14H + 6I} \overset{\checkmark}{=} \ce{2Cr + 7O + 14H + 6I} \nonumber \] Both the charges and atoms balance, so our equation is balanced. We can also use the alternative procedure, which does not require the half-reactions listed in Table P1 . Step 1: Chromium is reduced from \(\ce{Cr^{6+}}\) in \(\ce{Cr2O7^{2−}}\) to \(\ce{Cr^{3+}}\), and \(\ce{I^{−}}\) ions are oxidized to \(\ce{I2}\). Dividing the reaction into two half-reactions, reduction: \[Cr_2O^{2−}_{7(aq)} \rightarrow Cr^{3+}_{(aq)} \nonumber \] oxidation: \[I^−_{(aq)} \rightarrow I_{2(aq)} \nonumber \] Step 2: Balancing the atoms other than oxygen and hydrogen, reduction: \[Cr_2O^{2−}_{7(aq)} \rightarrow 2Cr^{3+}_{(aq)} \nonumber \] oxidation: \[2I^−_{(aq)} \rightarrow I_{2(aq)} \nonumber \] We now balance the O atoms by adding H 2 O—in this case, to the right side of the reduction half-reaction. Because the oxidation half-reaction does not contain oxygen, it can be ignored in this step. reduction: \[Cr_2O^{2−}_{7(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 7H_2O_{(l)} \nonumber \] Next we balance the H atoms by adding H + to the left side of the reduction half-reaction. Again, we can ignore the oxidation half-reaction. reduction: \[Cr_2O^{2−}_{7(aq)} + 14H^+_{(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 7H_2O_{(l)} \nonumber \] Step 3: We must now add electrons to balance the charges. The reduction half-reaction (2Cr +6 to 2Cr +3 ) has a +12 charge on the left and a +6 charge on the right, so six electrons are needed to balance the charge. The oxidation half-reaction (2I − to I 2 ) has a −2 charge on the left side and a 0 charge on the right, so it needs two electrons to balance the charge: reduction: Cr 2 O 7 2 − (aq) + 14H + (aq) + 6e − → 2Cr 3 + (aq) + 7H 2 O(l) oxidation: 2I − (aq) → I 2 (aq) + 2e − Step 4: To have the same number of electrons in both half-reactions, we must multiply the oxidation half-reaction by 3: oxidation: 6I − (aq) → 3I 2 (s) + 6e − Step 5: Adding the two half-reactions and canceling substances that appear in both reactions, \[\ce{Cr2O7^{2−}(aq) + 14H^{+}(aq) + 6I^{−}(aq) → 2Cr^{3+}(aq) + 7H2O(l) + 3I2(aq)} \nonumber \] Step 6: This is the same equation we obtained using the first method. Thus the charges and atoms on each side of the equation balance. Exercise \(\PageIndex{1}\) Copper is found as the mineral covellite (\(\ce{CuS}\)). The first step in extracting the copper is to dissolve the mineral in nitric acid (\(\ce{HNO3}\)), which oxidizes sulfide to sulfate and reduces nitric acid to \(\ce{NO}\): \[\ce{CuS(s) + HNO3(aq) \rightarrow NO(g) + CuSO4(aq)} \nonumber \] Balance this equation using the half-reaction method. Answer \[\ce{3CuS(s) + 8HNO3(aq) -> 8NO(g) + 3CuSO4(aq) + 4H2O(l)} \nonumber \] Calculating Standard Cell Potentials The standard cell potential for a redox reaction (E° cell ) is a measure of the tendency of reactants in their standard states to form products in their standard states; consequently, it is a measure of the driving force for the reaction, which earlier we called voltage. We can use the two standard electrode potentials we found earlier to calculate the standard potential for the Zn/Cu cell represented by the following cell diagram: \[ Zn{(s)}∣Zn^{2+}(aq, 1 M)∥Cu^{2+}(aq, 1 M)∣Cu_{(s)} \label{20.4.32} \] We know the values of E° anode for the reduction of Zn 2 + and E° cathode for the reduction of Cu 2 + , so we can calculate \(E°_{cell}\): cathode: \[Cu^{2+}_{(aq)} + 2e^− \rightarrow Cu_{(s)} \;\;\; E°_{cathode} = 0.34\; V \label{20.4.33} \] anode: \[Zn_{(s)} \rightarrow Zn^{2+}(aq, 1 M) + 2e^−\;\;\; E°_{anode} = −0.76\; V \label{20.4.34} \] overall: \[Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)} \label{20.4.35} \] \[E°_{cell} = E°_{cathode} − E°_{anode} = 1.10\; V \nonumber \] This is the same value that is observed experimentally. If the value of \(E°_{cell}\) is positive, the reaction will occur spontaneously as written. If the value of \(E°_{cell}\) is negative, then the reaction is not spontaneous, and it will not occur as written under standard conditions; it will, however, proceed spontaneously in the opposite direction. As we shall see in Section 20.9 , this does not mean that the reaction cannot be made to occur at all under standard conditions. With a sufficient input of electrical energy, virtually any reaction can be forced to occur. Example \(\PageIndex{2}\) and its corresponding exercise illustrate how we can use measured cell potentials to calculate standard potentials for redox couples. A positive \(E°_{cell}\) means that the reaction will occur spontaneously as written. A negative \(E°_{cell}\) means that the reaction will proceed spontaneously in the opposite direction. Example \(\PageIndex{2}\) A galvanic cell with a measured standard cell potential of 0.27 V is constructed using two beakers connected by a salt bridge. One beaker contains a strip of gallium metal immersed in a 1 M solution of GaCl 3 , and the other contains a piece of nickel immersed in a 1 M solution of NiCl 2 . The half-reactions that occur when the compartments are connected are as follows: cathode: Ni 2 + (aq) + 2e − → Ni(s) anode: Ga(s) → Ga 3 + (aq) + 3e − If the potential for the oxidation of Ga to Ga 3 + is 0.55 V under standard conditions, what is the potential for the oxidation of Ni to Ni 2 + ? Given: galvanic cell, half-reactions, standard cell potential, and potential for the oxidation half-reaction under standard conditions Asked for: standard electrode potential of reaction occurring at the cathode Strategy: Write the equation for the half-reaction that occurs at the anode along with the value of the standard electrode potential for the half-reaction. Use Equation \(\ref{20.4.2}\) to calculate the standard electrode potential for the half-reaction that occurs at the cathode. Then reverse the sign to obtain the potential for the corresponding oxidation half-reaction under standard conditions. Solution A We have been given the potential for the oxidation of Ga to Ga 3 + under standard conditions, but to report the standard electrode potential, we must reverse the sign. For the reduction reaction Ga 3 + (aq) + 3e − → Ga(s), E° anode = −0.55 V. B Using the value given for \(E°_{cell}\) and the calculated value of E° anode , we can calculate the standard potential for the reduction of Ni 2 + to Ni from Equation \(\ref{20.4.2}\): \[\begin{align*} E°_{cell} &= E°_{cathode} − E°_{anode} \\[4pt] 0.27\, V &= E^o°_{cathhode} − (−0.55\, V) \\[4pt] E^°_{cathode} &= −0.28 \,V \end{align*} \nonumber \] This is the standard electrode potential for the reaction Ni 2 + (aq) + 2e − → Ni(s). Because we are asked for the potential for the oxidation of Ni to Ni 2 + under standard conditions, we must reverse the sign of E° cathode . Thus E° = −(−0.28 V) = 0.28 V for the oxidation. With three electrons consumed in the reduction and two produced in the oxidation, the overall reaction is not balanced. Recall, however, that standard potentials are independent of stoichiometry. Exercise \(\PageIndex{2}\) A galvanic cell is constructed with one compartment that contains a mercury electrode immersed in a 1 M aqueous solution of mercuric acetate \(\ce{Hg(CH_3CO_2)_2}\) and one compartment that contains a strip of magnesium immersed in a 1 M aqueous solution of \(\ce{MgCl2}\). When the compartments are connected, a potential of 3.22 V is measured and the following half-reactions occur: cathode: \(\ce{Hg^{2+} (aq) + 2e^{−} → Hg(l)}\) anode: \(\ce{Mg(s) → Mg^{2+}(aq) + 2e^{−}}\) If the potential for the oxidation of \(\ce{Mg}\) to \(\ce{Mg^{2+}}\) is 2.37 V under standard conditions, what is the standard electrode potential for the reaction that occurs at the cathode? Answer 0.85 V Reference Electrodes and Measuring Concentrations When using a galvanic cell to measure the concentration of a substance, we are generally interested in the potential of only one of the electrodes of the cell, the so-called indicator electrode , whose potential is related to the concentration of the substance being measured. To ensure that any change in the measured potential of the cell is due to only the substance being analyzed, the potential of the other electrode, the reference electrode , must be constant. You are already familiar with one example of a reference electrode: the SHE. The potential of a reference electrode must be unaffected by the properties of the solution, and if possible, it should be physically isolated from the solution of interest. To measure the potential of a solution, we select a reference electrode and an appropriate indicator electrode. Whether reduction or oxidation of the substance being analyzed occurs depends on the potential of the half-reaction for the substance of interest (the sample) and the potential of the reference electrode. The potential of any reference electrode should not be affected by the properties of the solution to be analyzed, and it should also be physically isolated. There are many possible choices of reference electrode other than the SHE. The SHE requires a constant flow of highly flammable hydrogen gas, which makes it inconvenient to use. Consequently, two other electrodes are commonly chosen as reference electrodes. One is the silver–silver chloride electrode , which consists of a silver wire coated with a very thin layer of AgCl that is dipped into a chloride ion solution with a fixed concentration. The cell diagram and reduction half-reaction are as follows: \[Cl^−_{(aq)}∣AgCl_{(s)}∣Ag_{(s)} \label{20.4.36} \] \[AgCl_{(s)}+e^− \rightarrow Ag_{(s)} + Cl^−_{(aq)} \nonumber \] If a saturated solution of KCl is used as the chloride solution, the potential of the silver–silver chloride electrode is 0.197 V versus the SHE. That is, 0.197 V must be subtracted from the measured value to obtain the standard electrode potential measured against the SHE. A second common reference electrode is the saturated calomel electrode (SCE) , which has the same general form as the silver–silver chloride electrode. The SCE consists of a platinum wire inserted into a moist paste of liquid mercury (Hg 2 Cl 2 ; called calomel in the old chemical literature) and KCl. This interior cell is surrounded by an aqueous KCl solution, which acts as a salt bridge between the interior cell and the exterior solution (part (a) in Figure \(\PageIndex{5}\). Although it sounds and looks complex, this cell is actually easy to prepare and maintain, and its potential is highly reproducible. The SCE cell diagram and corresponding half-reaction are as follows: \[Pt_{(s)} ∣ Hg_2Cl_{2(s)}∣KCl_{(aq, sat)} \label{20.4.37} \] \[Hg_2Cl_{2(s)} + 2e^− \rightarrow 2Hg_{(l)} + 2Cl^−{(aq)} \label{20.4.38} \] At 25°C, the potential of the SCE is 0.2415 V versus the SHE, which means that 0.2415 V must be subtracted from the potential versus an SCE to obtain the standard electrode potential. One of the most common uses of electrochemistry is to measure the H + ion concentration of a solution. A glass electrode is generally used for this purpose, in which an internal Ag/AgCl electrode is immersed in a 0.10 M HCl solution that is separated from the solution by a very thin glass membrane (part (b) in Figure \(\PageIndex{5}\). The glass membrane absorbs protons, which affects the measured potential. The extent of the adsorption on the inner side is fixed because [H + ] is fixed inside the electrode, but the adsorption of protons on the outer surface depends on the pH of the solution. The potential of the glass electrode depends on [H + ] as follows (recall that pH = −log[H + ]): \[E_{glass} = E′ + (0.0591\; V \times \log[H^+]) = E′ − 0.0591\; V \times pH \label{20.4.39} \] The voltage E′ is a constant that depends on the exact construction of the electrode. Although it can be measured, in practice, a glass electrode is calibrated; that is, it is inserted into a solution of known pH, and the display on the pH meter is adjusted to the known value. Once the electrode is properly calibrated, it can be placed in a solution and used to determine an unknown pH. Ion-selective electrodes are used to measure the concentration of a particular species in solution; they are designed so that their potential depends on only the concentration of the desired species (part (c) in Figure \(\PageIndex{5}\)). These electrodes usually contain an internal reference electrode that is connected by a solution of an electrolyte to a crystalline inorganic material or a membrane, which acts as the sensor. For example, one type of ion-selective electrode uses a single crystal of Eu-doped \(LaF_3\) as the inorganic material. When fluoride ions in solution diffuse to the surface of the solid, the potential of the electrode changes, resulting in a so-called fluoride electrode. Similar electrodes are used to measure the concentrations of other species in solution. Some of the species whose concentrations can be determined in aqueous solution using ion-selective electrodes and similar devices are listed in Table \(\PageIndex{1}\). Species Type of Sample H+ laboratory samples, blood, soil, and ground and surface water NH3/NH4+ wastewater and runoff water K+ blood, wine, and soil CO2/HCO3− blood and groundwater F− groundwater, drinking water, and soil Br− grains and plant extracts I− milk and pharmaceuticals NO3− groundwater, drinking water, soil, and fertilizer Summary Redox reactions can be balanced using the half-reaction method. The standard cell potential is a measure of the driving force for the reaction. \(E°_{cell} = E°_{cathode} − E°_{anode} \nonumber \] The flow of electrons in an electrochemical cell depends on the identity of the reacting substances, the difference in the potential energy of their valence electrons, and their concentrations. The potential of the cell under standard conditions (1 M for solutions, 1 atm for gases, pure solids or liquids for other substances) and at a fixed temperature (25°C) is called the standard cell potential (E° cell ). Only the difference between the potentials of two electrodes can be measured. By convention, all tabulated values of standard electrode potentials are listed as standard reduction potentials. The overall cell potential is the reduction potential of the reductive half-reaction minus the reduction potential of the oxidative half-reaction (E° cell = E° cathode − E° anode ). The potential of the standard hydrogen electrode (SHE) is defined as 0 V under standard conditions. The potential of a half-reaction measured against the SHE under standard conditions is called its standard electrode potential. The standard cell potential is a measure of the driving force for a given redox reaction. All E° values are independent of the stoichiometric coefficients for the half-reaction. Redox reactions can be balanced using the half-reaction method, in which the overall redox reaction is divided into an oxidation half-reaction and a reduction half-reaction, each balanced for mass and charge. The half-reactions selected from tabulated lists must exactly reflect reaction conditions. In an alternative method, the atoms in each half-reaction are balanced, and then the charges are balanced. Whenever a half-reaction is reversed, the sign of E° corresponding to that reaction must also be reversed. If \(E°_{cell}\) is positive, the reaction will occur spontaneously under standard conditions. If \(E°_{cell}\) is negative, then the reaction is not spontaneous under standard conditions, although it will proceed spontaneously in the opposite direction. The potential of an indicator electrode is related to the concentration of the substance being measured, whereas the potential of the reference electrode is held constant. Whether reduction or oxidation occurs depends on the potential of the sample versus the potential of the reference electrode. In addition to the SHE, other reference electrodes are the silver–silver chloride electrode; the saturated calomel electrode (SCE); the glass electrode, which is commonly used to measure pH; and ion-selective electrodes, which depend on the concentration of a single ionic species in solution. Differences in potential between the SHE and other reference electrodes must be included when calculating values for E°. |
Courses/South_Puget_Sound_Community_College/CHEM_110%3A_Chemical_Concepts/06%3A_Covalent_Bonds_Covalent_Compounds/6.04%3A_Molecular_Geometry | Learning Objective Determine the shape of simple molecules. Molecules have shapes. There is an abundance of experimental evidence to that effect—from their physical properties to their chemical reactivity. Small molecules—molecules with a single central atom—have shapes that can be easily predicted. The basic idea in molecular shapes is called valence shell electron pair repulsion ( VSEPR ). VSEPR says that electron pairs, being composed of negatively charged particles, repel each other to get as far away from one another as possible. VSEPR makes a distinction between electron group geometry , which expresses how electron groups (bonds and nonbonding electron pairs) are arranged, and molecular geometry , which expresses how the atoms in a molecule are arranged. However, the two geometries are related. There are two types of electron groups : any type of bond—single, double, or triple—and lone electron pairs. When applying VSEPR to simple molecules, the first thing to do is to count the number of electron groups around the central atom. Remember that a multiple bond counts as only one electron group. Any molecule with only two atoms is linear. A molecule whose central atom contains only two electron groups orients those two groups as far apart from each other as possible—180° apart. When the two electron groups are 180° apart, the atoms attached to those electron groups are also 180° apart, so the overall molecular shape is linear. Examples include BeH 2 and CO 2 : The two molecules, shown in the figure below in a "ball and stick" model. A molecule with three electron groups orients the three groups as far apart as possible. They adopt the positions of an equilateral triangle—120° apart and in a plane. The shape of such molecules is trigonal planar . An example is BF 3 : Some substances have a trigonal planar electron group distribution but have atoms bonded to only two of the three electron groups. An example is GeF 2 : From an electron group geometry perspective, GeF 2 has a trigonal planar shape, but its real shape is dictated by the positions of the atoms. This shape is called bent or angular . A molecule with four electron groups about the central atom orients the four groups in the direction of a tetrahedron, as shown in Figure \(\PageIndex{1}\) Tetrahedral Geometry. If there are four atoms attached to these electron groups, then the molecular shape is also tetrahedral . Methane (CH 4 ) is an example. This diagram of CH 4 illustrates the standard convention of displaying a three-dimensional molecule on a two-dimensional surface. The straight lines are in the plane of the page, the solid wedged line is coming out of the plane toward the reader, and the dashed wedged line is going out of the plane away from the reader. NH 3 is an example of a molecule whose central atom has four electron groups, but only three of them are bonded to surrounding atoms. Although the electron groups are oriented in the shape of a tetrahedron, from a molecular geometry perspective, the shape of NH 3 is trigonal pyramidal . H 2 O is an example of a molecule whose central atom has four electron groups, but only two of them are bonded to surrounding atoms. Although the electron groups are oriented in the shape of a tetrahedron, the shape of the molecule is bent or angular . A molecule with four electron groups about the central atom, but only one electron group bonded to another atom, is linear because there are only two atoms in the molecule. Double or triple bonds count as a single electron group. The Lewis electron dot diagram of formaldehyde (CH 2 O) is shown in Figure \(\PageIndex{9}\). The central C atom has three electron groups around it because the double bond counts as one electron group. The three electron groups repel each other to adopt a trigonal planar shape. (The lone electron pairs on the O atom are omitted for clarity.) The molecule will not be a perfect equilateral triangle because the C–O double bond is different from the two C–H bonds, but both planar and triangular describe the appropriate approximate shape of this molecule. Table \(\PageIndex{1}\) summarizes the shapes of molecules based on the number of electron groups and surrounding atoms. Number of Electron Groups on Central Atom Number of Bonding Groups Number of Lone Pairs Electron Geometry Molecular Shape 2 2 0 linear linear 3 3 0 trigonal planar trigonal planar 3 2 1 trigonal planar bent 4 4 0 tetrahedral tetrahedral 4 3 1 tetrahedral trigonal pyramidal 4 2 2 tetrahedral bent Example \(\PageIndex{1}\) What is the approximate shape of each molecule? PCl 3 NOF Solution The first step is to draw the Lewis structure of the molecule. For \(\ce{PCl3}\), the electron dot diagram is as follows: The lone electron pairs on the Cl atoms are omitted for clarity. The P atom has four electron groups with three of them bonded to surrounding atoms, so the molecular shape is trigonal pyramidal. The electron dot diagram for \(\ce{NOF}\) is as follows: The N atom has three electron groups on it, two of which are bonded to other atoms. The molecular shape is bent. Exercise \(\PageIndex{1}\) What is the approximate molecular shape of \(\ce{CH2Cl2}\)? Answer Tetrahedral Exercise \(\PageIndex{2}\) Ethylene (\(\ce{C2H4}\)) has two central atoms. Determine the geometry around each central atom and the shape of the overall molecule. (Hint: hydrogen is a terminal atom.) Answer Trigonal planar about both central C atoms. Summary The approximate shape of a molecule can be predicted from the number of electron groups and the number of surrounding atoms. |
Bookshelves/Environmental_Chemistry/Green_Chemistry_and_the_Ten_Commandments_of_Sustainability_(Manahan)/04%3A_Compounds-_Safer_Materials_for_a_Safer_World/4.04%3A_Covalent_Bonds_in_H2_and_Other_Molecules | Lewis symbols can be used to show how some atoms of elements on the left side of the table with only one or two outer-shell electrons can lose those electrons to form cations such as Na + or Ca 2 + . It is also easily seen that atoms from groups near the right side of the periodic table can accept one or two electrons to gain stable octets and become anions such as Cl - or O 2- . But, it is difficult to impossible to take more than two electrons away from an atom to form cations with charges greater than +2 or to add 3 or more electrons to form anions with charges of -3 or even more negative, although ions such as Al 3 + and N 3- do exist. So atoms of elements in the middle of the periodic table and the nonmetals on the right have a tendency to share electrons in covalent bonds, rather than becoming ions. It is readily visualized how mutually attracting ions of opposite charge are held together in a crystalline lattice. Shared electrons in covalent bonds act to reduce the forces of repulsion between the positively charged nuclei of the atoms that they join together. That is most easily seen for the case of the hydrogen molecule, H 2 . The nuclei of H atoms consist of single protons, and the two H atom nuclei in the H 2 molecule repel each other. However, their 2 shared electrons compose a cloud of negative charge between the two H nuclei, shielding the nuclei from each other’s repelling positive charge and enabling the molecule to exist as a covalently bound molecule (Figure 4.7). |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.04%3A_Half-lives | The half-life of a reaction (\(t_{1/2}\)), is the amount of time needed for a reactant concentration to decrease by half compared to its initial concentration. Its application is used in chemistry and medicine to predict the concentration of a substance over time. The concepts of half life plays a key role in the administration of drugs into the target, especially in the elimination phase, where half life is used to determine how quickly a drug decrease in the target after it has been absorbed in units of time (e.g., s, min., day, etc.) or elimination rate constant \(k\) in units of 1/time (e.g., min -1 , hr -1 , day -1 , etc.). It is important to note that the half-life is varied between different type of reactions. The following section will go over different type of reaction, as well as how its half-life reaction are derived. The last section will talk about the application of half-life in the elimination phase of pharmacokinetics. Zero-Order Kinetics In zero-order kinetics, the rate of a reaction does not depend on the substrate concentration. In other words, saturating the amount of substrate does not speed up the rate of the reaction. Below is a graph of time (\(t\)) vs. concentration (\([A]\)) in a zero order reaction, several observation can be made: the slope of this plot is a straight line with negative slope equal negative \(k\), the half-life of zero order reaction decreases as the concentration decreases. We learn that the zero-order kinetic rate law is as followed, where \([A]\) is the current concentration, \([A]_o\) is the initial concentration, and \(k\) is the reaction constant and \(t\) is time: \[ [A]= [A]_o - kt \label{1} \] We need to isolate \(t_{1/2}\) when \[[A]=\dfrac{[A]_o}{2} \nonumber \] Substituting into Equation \ref{1} \[\begin{align} \dfrac{[A]_o}{2} &= [A]_o - kt_{1/2} \nonumber \\[4pt] kt_{1/2} &= [A]_o - \dfrac{[A]_o}{2} \nonumber \\[4pt] t_{1/2} &= \dfrac{[A]_o}{2k} \label{2} \end{align} \] Equation \ref{2} show the half-life for a zero-order reaction depends on both the initial concentration and rate constant. First-Order Kinetics In First order reactions , the graph represents the half-life is different from zero order reaction in a way that the slope continually decreases as time progresses until it reaches zero. We can also easily see that the length of half-life will be constant, independent of concentration. For example, it takes the same amount of time for the concentration to decrease from one point to another point. In order to solve the half life of first order reactions, we recall that the rate law of a first order reaction was: \[[A]=[A]_o e^{-kt} \label{4} \] We need to isolate \(t_{1/2}\) when \[[A]=\dfrac{[A]_o}{2} \nonumber \] Substituting into Equation \ref{4} \[ \begin{align} \dfrac{[A]_0}{2} &=[A]_o e^{-kt_{1/2}} \nonumber \\[4pt] \dfrac{1}{2} &= e^{-kt_{1/2}} \nonumber \\[4pt] \ln \dfrac{1}{2} &= -kt_{1/2} \nonumber \\[4pt] t_{1/2} &= \dfrac{\ln 2}{k} \nonumber \\[4pt] &\approx \dfrac{0.693}{k} \label{5} \end{align} \] Equation \ref{5} shows that for first-order reactions, the half-life depends solely on the reaction rate constant, \(k\). We can visually see this on the graph for first order reactions when we note that the amount of time between one half life and the next are the same. Another way to see it is that the half life of a first order reaction is independent of its initial concentration. Second-Order Kinetics Half-life of second order reactions shows concentration \([A]\) vs. time (\(t\)), which is similar to first order plots in that their slopes decrease to zero with time. However, second order reactions decrease at a much faster rate as the graph shows. We can also note that the length of half-life increase while the concentration of substrate constantly decreases, unlike zero and first order reaction. In order to solve for half life of second order reactions we need to remember that the rate law of a second order reaction is: \[\dfrac{1}{[A]} = kt + \dfrac{1}{[A]_0} \label{6} \] As in zero-order and first-order reactions, we need to isolate \(t_{1/2}\) when \[[A]=\dfrac{[A]_o}{2} \nonumber \] Substituting into Equation \ref{6} \[ \begin{align} \dfrac{2}{[A]_0} &= kt_{1/2} + \dfrac{1}{[A]_0}\nonumber \\[4pt] -kt_{1/2} &= \dfrac{1}{[A]_0} - \dfrac{2}{[A]_0} \nonumber \\[4pt] t_{1/2} &= \dfrac{1}{k[A]_0} \label{7} \end{align} \] Equation \ref{7} shows that for second-order reactions, the half-life depends on both the initial concentration and the rate constant. Example \(\PageIndex{1}\): Pharmacokinetics A following example is given below to illustrate the role of half life in pharmacokinetics to determine the drugs dosage interval. The therapeutic range of drug A is 15-30 mg/L. Its half life in the target in 5 hours. Once the drug is metabolized in the target, its concentration will decrease over time. To ensure its maximal effect of the drug in the target, the administration will be monitored so that the minimum serum concentration will never go lower than 15 mg/L and the maximum serum concentration will never exceed 30 mg/L. As a result, it is important to administer drug A to the target every 5 hours to ensure its effective therapeutic range. Another important application of half life in pharmacokinetics is that half-life tells how tightly drugs bind to each ligands before it is undergoing decay (\(k_s\)). The smaller the value of \(k_s\), the higher the affinity binding of drug to its target ligand, which is an important aspect of drug design Exercise \(\PageIndex{1}\) Examine the following graph and answer What is the therapeutic range of drug B? From the graph, estimate the dosage interval of drug B to ensure its maximum effect? The patient forgot to take the drug at the end of the dosage interval, he decided to take double the amount of drug B at the end of the next dosage interval. Will the drug still be in its therapeutic range? Answer Looking at the graph, we can see the therapeutic range is the amplitude of the graph, which is 5-15 mg/L The dosage interval is the half-life of the drug, looking at the graph, the half-life is 10 hours. Even though it will get in the therapeutic range, such practice is not recommended. Exercise \(\PageIndex{2}\) A patient is treating with \(\ce{^{32}P}\). How long does it takes for the radioactivity to decay by 90%? The half-life of the material is 15 days. Answer If we want the product to decay by 90%, that means 10% is left non-decayed, so \[\dfrac{[A]_t}{[A]_o} = 0.1 \nonumber \] From ln([A] t /[A] o ) = -kt, plug in value of k and [A] t /[A] o we then have t = 50 days Exercise \(\PageIndex{3}\) In first order half life, what is the best way to determine the rate constant \(k\)? Why? Answer The best way to determine rate constant \(k\) in half-life of first order is to determine half-life by experimental data. The reason is half-life in first order order doesn't depend on initial concentration. Exercise \(\PageIndex{4}\) In a first order reaction, \(\ce{A -> B}\). The half-life is 10 days. Determine its rate constant \(k\)? How much time required for this reaction to be at least 50% and 60% complete? Answer This is a direct application of Equation \ref{7}. The rate constant, \(k\), will be equal to \[k=\dfrac{\ln 2}{t_{1/2}} \nonumber \] so \(k = 0.0693 \,day^{-1}\). For the reaction to be 50% complete, that will be exactly the half-life of the reaction at 10 days. For the reaction to be 60% complete, using the similar equation derived from question 4, we have \[\dfrac{[A]_t}{[A]_o} = 0.4 \nonumber \] \[t = 13.2\, days \nonumber \] References Chang, Raymond. Physical Chemistry for the Biosciences . Sausalito,CA: University Science Books. Pages 312-319. 2005 Bauer, Larry. Applied Clinical Pharmacokinetics. New York City, New York: McGraw-Hill. 2008 Mozayani, Ashraf and Raymond, Lionel. Handbook of Drug Interactions: A Clinical and Forensic Guide. Totowa, New Jersy: Humana Press. 2003 |
Courses/Providence_College/CHM_331_Advanced_Analytical_Chemistry_1/03%3A_Evaluating_Analytical_Data | When we use an analytical method we make three separate evaluations of experimental error. First, before we begin the analysis we evaluate potential sources of errors to ensure they will not adversely effect our results. Second, during the analysis we monitor our measurements to ensure that errors remain acceptable. Finally, at the end of the analysis we evaluate the quality of the measurements and results, and compare them to our original design criteria. This chapter provides an introduction to sources of error, to evaluating errors in analytical measurements, and to the statistical analysis of data. 3.1: Characterizing Measurements and Results One way to characterize data from multiple measurements/runs is to assume that the measurements are randomly scattered around a central value that provides the best estimate of expected, or “true” value. We describe the distribution of these results by reporting its central tendency and its spread. 3.2: Characterizing Experimental Errors Two essential questions arise from any set of data. First, does our measure of central tendency agree with the expected result? Second, why is there so much variability in the individual results? The first of these questions addresses the accuracy of our measurements and the second addresses the precision of our measurements. In this section we consider the types of experimental errors that affect accuracy and precision. 3.3: Propagation of Uncertainty A propagation of uncertainty allows us to estimate the uncertainty in a result from the uncertainties in the measurements used to calculate the result. Derivation of the general equation for any function and rectangular solid example added by J. Breen 3.4: The Distribution of Measurements and Results To compare two samples to each other, we need more than measures of their central tendencies and their spreads based on a small number of measurements. We need also to know how to predict the properties of the broader population from which the samples were drawn; in turn, this requires that we understand the distribution of samples within a population. 3.5: Statistical Analysis of Data A confidence interval is a useful way to report the result of an analysis because it sets limits on the expected result. In the absence of determinate error, a confidence interval based on a sample’s mean indicates the range of values in which we expect to find the population’s mean. In this section we introduce a general approach to the statistical analysis of data. Specific statistical tests are presented in the next section. 3.6: Statistical Methods for Normal Distributions The most common distribution for our results is a normal distribution. Because the area between any two limits of a normal distribution curve is well defined, constructing and evaluating significance tests is straightforward. The Median/Mad methods appearing in the section on outliers was added by J. Breen. 3.7: Detection Limits The International Union of Pure and Applied Chemistry (IUPAC) defines a method’s detection limit as the smallest concentration or absolute amount of analyte that has a signal significantly larger than the signal from a suitable blank. 3.8: Using Excel and R to Analyze Data Although the calculations in this chapter are relatively straightforward, it can be tedious to work problems using nothing more than a calculator. Both Excel and R include functions for many common statistical calculations. In addition, R provides useful functions for visualizing your data. 3.9: Problems End-of-chapter problems to test your understanding of topics in this chapter. 3.10: Additional Resources A compendium of resources to accompany topics in this chapter. 3.11: Chapter Summary and Key Terms Summary of chapter's main topics and a list of key terms introduced in this chapter. |
Bookshelves/Organic_Chemistry/Organic_Chemistry_(OpenStax)/06%3A_An_Overview_of_Organic_Reactions/6.14%3A_Summary | All chemical reactions, whether in the laboratory or in living organisms, follow the same chemical rules. To understand both organic and biological chemistry, it’s necessary to know not just what occurs but also why and how chemical reactions take place. In this chapter, we’ve taken a brief look at the fundamental kinds of organic reactions, we’ve seen why reactions occur, and we’ve seen how reactions can be described. There are four common kinds of reactions: addition reactions take place when two reactants add together to give a single product; elimination reactions take place when one reactant splits apart to give two products; substitution reactions take place when two reactants exchange parts to give two new products; and rearrangement reactions take place when one reactant undergoes a reorganization of bonds and atoms to give an isomeric product. A full description of how a reaction occurs is called its mechanism . There are two general kinds of mechanisms by which most reactions take place: radical mechanisms and polar mechanisms. Polar reactions, the more common type, occur because of an attractive interaction between a nucleophilic (electron-rich) site in one molecule and an electrophilic (electron-poor) site in another molecule. A bond is formed in a polar reaction when the nucleophile donates an electron pair to the electrophile. This transfer of electrons is indicated by a curved arrow showing the direction of electron travel from the nucleophile to the electrophile. Radical reactions involve species that have an odd number of electrons. A bond is formed when each reactant donates one electron. The energy changes that take place during reactions can be described by considering both rates (how fast the reactions occur) and equilibria (how much the reactions occur). The position of a chemical equilibrium is determined by the value of the free-energy change (Δ G ) for the reaction, where Δ G = Δ H − T Δ S . The enthalpy term (Δ H ) corresponds to the net change in strength of chemical bonds broken and formed during the reaction; the entropy term (Δ S ) corresponds to the change in the amount of molecular randomness during the reaction. Reactions that have negative values of Δ G release energy, are said to be exergonic , and have favorable equilibria. Reactions that have positive values of Δ G absorb energy, are said to be endergonic , and have unfavorable equilibria. A reaction can be described pictorially using an energy diagram that follows the reaction course from reactants through transition state to product. The transition state is an activated complex occurring at the highest-energy point of a reaction. The amount of energy needed by reactants to reach this high point is the activation energy, Δ G ‡ . The higher the activation energy, the slower the reaction. Many reactions take place in more than one step and involve the formation of a reaction intermediate . An intermediate is a species that lies at an energy minimum between steps on the reaction curve and is formed briefly during the course of a reaction. |
Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Biological_Mass_Spectrometry%3A_Proteomics/Instructors_Manual/Section_3%3A_Peptide_Mass_Mapping/Section_3A._MALDI-TOF_Mass_Spectrometry | Matrix-assisted laser desorption ionization ( MALDI ) is a technique that generates gas-phase peptide and protein ions. Proteins are nonvolatile and cannot simply be heated to make the gaseous form. (Think about a steak on the grill. The steak chars but does not vaporize.) Instead proteins are desorbed (released from a surface) by mixing the protein with a matrix molecule and irradiating the mixture with a pulse from an ultraviolet laser. The following figure shows the structure of various molecules used as the MALDI matrix. Discussion Question 1. What structural features do these molecules have in common? Why are these common features important for a MALDI matrix? A. They have aromatic rings so they have strong absorption in the ultraviolet region of the spectrum. The have a carboxylic acid group so they can transfer a proton to the biomolecule giving it a positive charge. The α-cyano-4-hydroxcinnamic acid (CCA) molecule is a MALDI matrix used for peptide and protein analysis. The matrix is combined with a mixture of peptides and placed on a metal plate. A pulsed nitrogen laser with a wavelength of 337 nm is used to irradiate the peptides and matrix. The absorption spectrum for a peptide and the CCA matrix is shown. (a) (b) Figure. (a) Absorbance spectrum of a peptide. (b) Absorbance spectrum of CCA matrix. Discussion Questions 1. Does the peptide absorb the laser light? Does the matrix absorb the laser light? A. No, the peptide does not absorb the laser light. The peptide absorbs strongly at 220 nm and 280 nm. The matrix does absorb the laser light. 2. Propose a reason that the laser is matched to the absorption of the matrix (CCA) and not the peptide. What would happen to the peptide if it absorbed a high energy pulse of laser light? A. It would break bonds in the peptide causing fragmentation. In the MALDI process, the laser is directed at the mixture of CCA matrix and peptides. The matrix absorbs the laser light, but the peptides do not. The matrix heats up and “explodes” from the surface carrying the peptide with it. (It is similar to skydiving. The airplane (matrix) carries the people (peptides) into the air and when they jump, they are temporarily flying.) After absorbing light the matrix is in an excited state and transfers a proton (H + ) to the peptide giving it a positive charge. The invention of MALDI enabled the mass spectrometric analysis of large nonvolatile biomolecules and was so important that its inventor Koichi Tanaka of Shimadzu Corporation was awarded one-third of the 2002 Nobel Prize in Chemistry. The MALDI process is depicted in the figure below. Figure. The MALDI process for creating gas-phase ions of peptides. (Figure from Wikimedia Commons) The purple color represents the MALDI matrix and the green color represents peptides. Energy from the laser is absorbed by the matrix, causing the matrix to desorb from the surface. The matrix carries the peptides into the gas phase and gives them a charge by donating a proton. Molecular ions of the peptides are created because peptides do not absorb laser light which would break bonds within the molecule. An example of a typical mass spectrum of a mixture of proteins ionized by the MALDI method is shown below. T he matrix is 2,5-dihydroxybenzoic acid (DHB) and the mixture contains three proteins; ubiquitin, cytochrome C and equine myoglobin. Figure. MALDI-TOF mass spectrum of three proteins; ubiquitin, cytochrome C, and equine myoglobin. Figure adapted from King’s College London http://www.kcl.ac.uk/innovation/research/corefacilities/smallrf/mspec/cemsw/instr/maldi-tof-ms.aspx Discussion Questions 1. Examine the mass spectrum of the mixture of three proteins obtained using MALDI as the ionization method. Would you classify MALDI as a “soft” (little to no fragmentation) ionization technique or a “hard” (fragmentation) ionization technique? A. Soft ionization method. Fragmentation is not evident in this mass spectrum. 2. What is the identity of Peak 1 in the mass spectrum? A. Cytochrome C+ 2 H with a +2 charge. The m/z ratio of cytochrome C+H with a +1 charge is 12400. If cytochrome C accepts 2 protons from the matrix, then the m/z ratio is 6200. |
Courses/College_of_Marin/CHEM_114%3A_Introductory_Chemistry/10%3A_Chemical_Bonding/10.02%3A_Representing_Valence_Electrons_with_Dots | Template:HideTOC Learning Objective Draw a Lewis electron dot diagram for an atom or a monatomic ion. In almost all cases, chemical bonds are formed by interactions of valence electrons in atoms. To facilitate our understanding of how valence electrons interact, a simple way of representing those valence electrons would be useful. A Lewis electron dot diagram (or electron dot diagram, or a Lewis diagram, or a Lewis structure) is a representation of the valence electrons of an atom that uses dots around the symbol of the element. The number of dots equals the number of valence electrons in the atom. These dots are arranged to the right and left and above and below the symbol, with no more than two dots on a side. (The order in which the positions are used does not matter.) For example, the Lewis electron dot diagram for hydrogen is simply \[\mathbf{H}\mathbf{\cdot} \nonumber \] Because the side is not important, the Lewis electron dot diagram could also be drawn as follows: \[\mathbf{\dot{H}}\; \; or\; \mathbf{\cdot}\mathbf{H}\; \; \; or\; \; \; \mathbf{\underset{.}H} \nonumber \] The electron dot diagram for helium, with two valence electrons, is as follows: \[\mathbf{He}\mathbf{:} \nonumber \] By putting the two electrons together on the same side, we emphasize the fact that these two electrons are both in the 1 s subshell; this is the common convention we will adopt, although there will be exceptions later. The next atom, lithium, has an electron configuration of 1 s 2 2 s 1 , so it has only one electron in its valence shell. Its electron dot diagram resembles that of hydrogen, except the symbol for lithium is used: \[\mathbf{Li}\mathbf{\cdot} \nonumber \] Beryllium has two valence electrons in its 2 s shell, so its electron dot diagram is like that of helium: \[\mathbf{Be}\mathbf{:} \nonumber \] The next atom is boron. Its valence electron shell is 2 s 2 2 p 1 , so it has three valence electrons. The third electron will go on another side of the symbol: \[\mathbf{\dot{B}}\mathbf{:} \nonumber \] Again, it does not matter on which sides of the symbol the electron dots are positioned. For carbon, there are four valence electrons, two in the 2 s subshell and two in the 2 p subshell. As usual, we will draw two dots together on one side, to represent the 2 s electrons. However, conventionally, we draw the dots for the two p electrons on different sides. As such, the electron dot diagram for carbon is as follows: \[\mathbf{\cdot \dot{C}}\mathbf{:} \nonumber \] With N, which has three p electrons, we put a single dot on each of the three remaining sides: \[\mathbf{\cdot}\mathbf{\dot{\underset{.}N}}\mathbf{:} \nonumber \] For oxygen, which has four p electrons, we now have to start doubling up on the dots on one other side of the symbol. When doubling up electrons, make sure that each side has no more than two electrons. \[\mathbf{\cdot}\mathbf{\ddot{\underset{.}O}}\mathbf{:} \nonumber \] Fluorine and neon have seven and eight dots, respectively: \[\mathbf{:}\mathbf{\ddot{\underset{.}F}}\mathbf{:} \nonumber \] \[\mathbf{:}\mathbf{\ddot{\underset{.\: .}Ne}}\mathbf{:} \nonumber \] With the next element, sodium, the process starts over with a single electron because sodium has a single electron in its highest-numbered shell, the n = 3 shell. By going through the periodic table, we see that the Lewis electron dot diagrams of atoms will never have more than eight dots around the atomic symbol. Example \(\PageIndex{1}\): Lewis Dot Diagrams What is the Lewis electron dot diagram for each element? aluminum selenium Solution The valence electron configuration for aluminum is 3 s 2 3 p 1 . So it would have three dots around the symbol for aluminum, two of them paired to represent the 3 s electrons: \[\dot{Al:} \nonumber \] The valence electron configuration for selenium is 4 s 2 4 p 4 . In the highest-numbered shell, the n = 4 shell, there are six electrons. Its electron dot diagram is as follows: \[\mathbf{\cdot }\mathbf{\dot{\underset{.\: .}Se}}\mathbf{:} \nonumber \] Exercise \(\PageIndex{1}\) What is the Lewis electron dot diagram for each element? phosphorus argon Answer a \[\mathbf{\cdot }\mathbf{\dot{\underset{.}P}}\mathbf{:} \nonumber \] Answer b \[\mathbf{:}\mathbf{\ddot{\underset{.\, .}Ar}}\mathbf{:} \nonumber \] Summary Lewis electron dot diagrams use dots to represent valence electrons around an atomic symbol. Lewis electron dot diagrams for ions have less (for cations) or more (for anions) dots than the corresponding atom. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/25%3A_Solutions_II_-_Nonvolatile_Solutes/25.04%3A_Osmotic_Pressure_can_Determine_Molecular_Masses | Some membrane materials are permeable for some molecules, but not for others. This is often a matter of the size of the molecules, but it can also be a question of solubility of the molecule in the barrier material. Many biological membranes have semipermeable properties and osmosis is therefore an important biological process. Figure 25.4.1 shows a simple osmotic cell. Both compartments contain water, but the one on the right also contains a solute whose molecules (represented by green circles) are too large to pass through the membrane. Many artificial and natural substances are capable of acting as semi-permeable membranes. For example, the walls of most plant and animal cells fall into this category. If solvent molecules can pass through the membrane, but solute molecules (or ions) cannot, solvent molecules will spontaneously migrate across the membrane to increase the solution's volume and thus reduce its concentration. If the solution is ideal, this process is in many ways analogous to the spontaneous increase in volume of a gas allowed to expand against vacuum. Of course the volume of the 'solute-gas' is limited by the availability of solvent and, if done under gravity in a U-shaped tube, by the build up of hydrostatic pressure. This pressure is known as the osmotic pressure \(\Pi\). At equilibrium we can write: \[μ^*(T,P) = μ^{sln}(T,P+ Π,a_1) \nonumber \] \[μ^*(T,P) = μ^*(T,P+ Π)+ RT\ln a_1 \nonumber \] From Gibbs energy (\(dG\)) in its natural variables (\(P,T\)) we know that: \[ \left(\dfrac{∂G}{∂P} \right)_{T,x} = V \nonumber \] Taking the partial versus \(x_1\) we get: \[ \left( \dfrac{∂μ^*}{∂P} \right)_{T,x_{j}} = \bar{V}^*_{1} \nonumber \] This means we can integrate over the molar volume to convert \(μ^*(T,P+ Π)\) to a different pressure: \[μ^*(T,P+\Pi) = μ^*(T,P) + \int_P^{P+ \Pi} \bar{V}^*_{1} dP \nonumber \] Thus we get: \[μ^*(T,P) = μ^*(T,P+ \Pi)+ RT\ln a_1 \nonumber \] \[μ^*(T,P) = μ^*(T,P)+ \Pi\bar{V}^*_{1}+ RT\ln a_1 \nonumber \] Once again using the ideal approximation: \[\ln a_1 \ln x_1 ≈ -x_2 \nonumber \] we get: \[RT x_2 = \Pi\bar{V}^*_{1} \nonumber \] \[x_2 = \dfrac{n_2}{n_1+n_2}≈\dfrac{n_2}{n_1} \nonumber \] The combination gives an expression involving the molarity: \[\Pi=RTc \nonumber \] Where \(c\) is the molar concentration. Osmosis can be used in reverse, if we apply about 30 bar to sea water we can obtain fresh water on the other side of a suitable membrane. This process is used in some places, but better membranes would be desirable and they easily get clogged. The resulting water is not completely salt-free and this means that if used for agriculture the salt may accumulate on the field over time. Determining Molar Masses Both melting point depression and boiling point elevation only facilitate the determination of relatively small molar weights. The need for such measurements is no longer felt because we now have good techniques to determine the structure of most small to medium size molecules. For polymers this is a different matter. They usually have a molecular weight (mass) distribution and determining it is an important topic of polymer science. Osmometry is still of some practical usefulness. It is also colligative and able to measure up to about 8000 daltons. Many polymers are much bigger than that. Their mass distribution is usually determined by different means. The polymers is dissolved and led over a chromatographic column usually based on size-exclusion. The effluent is then probed as function of the elution time by a combination of techniques: UV absorption (determine the monomer concentration) Low Angle Laser Light Scattering (LALLS) and/or Viscometry The latter two provide information on the molar mass distribution but they give a different moment of that distribution. The combination of techniques gives an idea not only of how much material there is of a given molar mass but also of the linearity or degree of branching of the chains. Nevertheless melting point depression is still used in a somewhat different application. When a slightly impure solid is melted its melting point in depressed. Also the melting process is not sudden but takes place over the whole trajectory from typically a lower eutectic temperature up to the depressed melting point (the liquid line in the phase diagram). In organic synthesis the melting behavior is often used as a first convenient indication of purity. In a differential scanning calorimetry (DSC) experiment the melting peak becomes progressively skewed towards lower temperatures at higher impurity levels. The shape of the curve can be modeled with a modified version of the melting point depression expression. This yields a value for the total impurity level in the solid. This technique is used in the pharmaceutical industry for quality control purposes. |
Courses/Honolulu_Community_College/CHEM_100%3A_Chemistry_and_Society/16%3A_Nuclear_Chemistry/16.04%3A_Natural_Radioactivity | Learning Objectives Know the different sources of background radiation. Describe the biological impact of ionizing radiation List common sources of radiation exposure in the US. Background Radiation We are all exposed to a small amount of radiation in our daily lives. This background radiation comes from naturally occurring sources and from human-produced radiation. Exposure to x-rays and nuclear medicine isotopes, ground sources, and cosmic radiation account for almost half of the background exposure of the average American. Radon gas, formed from the decay of uranium and thorium isotopes, is responsible for a little over half the total amount of background radiation. See the table below for background sources. 0 1 Table \(\PageIndex{1}\) Sources of Background Radiation Table \(\PageIndex{1}\) Sources of Background Radiation radon \(54\%\) consumer products \(3\%\) nuclear medicine \(4\%\) cosmic radiation \(8\%\) terrestrial \(8\%\) internal \(11\%\) x-rays \(11\%\) other \(1\%\) The Problem of Radon For many people, one of the largest sources of exposure to radiation is from radon gas (Rn-222). Radon-222 is an α emitter with a half–life of 3.82 days. It is one of the products of the radioactive decay series of U-238, which is found in trace amounts in soil and rocks. The radon gas that is produced slowly escapes from the ground and gradually seeps into homes and other structures above. Since it is about eight times more dense than air, radon gas accumulates in basements and lower floors, and slowly diffuses throughout buildings (Figure \(\PageIndex{1}\)). Radon is found in buildings across the country, with amounts depending on where you live. The average concentration of radon inside houses in the US (1.25 pCi/L) is about three times the levels found in outside air, and about one in six houses have radon levels high enough that remediation efforts to reduce the radon concentration are recommended. Exposure to radon increases one’s risk of getting cancer (especially lung cancer), and high radon levels can be as bad for health as smoking a carton of cigarettes a day. Radon is the number one cause of lung cancer in nonsmokers and the second leading cause of lung cancer overall. Radon exposure is believed to cause over 20,000 deaths in the US per year. Measuring Radiation Exposure Several different devices are used to detect and measure radiation, including Geiger counters, scintillation counters (scintillators), and radiation dosimeters (Figure \(\PageIndex{2}\)). Probably the best-known radiation instrument, the Geiger counter (also called the Geiger-Müller counter) detects and measures radiation. Radiation causes the ionization of the gas in a Geiger-Müller tube. The rate of ionization is proportional to the amount of radiation. A scintillation counter contains a scintillator—a material that emits light (luminesces) when excited by ionizing radiation—and a sensor that converts the light into an electric signal. Radiation dosimeters also measure ionizing radiation and are often used to determine personal radiation exposure. Commonly used types are electronic, film badge, thermoluminescent, and quartz fiber dosimeters. Radiation Damage to Cells The increased use of radioisotopes has led to increased concerns over the effects of these materials on biological systems (such as humans). All radioactive nuclides emit high-energy particles or electromagnetic waves. When this radiation encounters living cells, it can cause heating, break chemical bonds, or ionize molecules. The most serious biological damage results when these radioactive emissions fragment or ionize molecules. For example, alpha and beta particles emitted from nuclear decay reactions possess much higher energies than ordinary chemical bond energies. When these particles strike and penetrate matter, they produce ions and molecular fragments that are extremely reactive. The damage this does to biomolecules in living organisms can cause serious malfunctions in normal cell processes, taxing the organism’s repair mechanisms and possibly causing illness or even death (Figure \(\PageIndex{3}\)). Radiation can harm either the whole body (somatic damage) or eggs and sperm (genetic damage). Its effects are more pronounced in cells that reproduce rapidly, such as the stomach lining, hair follicles, bone marrow, and embryos. This is why patients undergoing radiation therapy often feel nauseous or sick to their stomach, lose hair, have bone aches, and so on, and why particular care must be taken when undergoing radiation therapy during pregnancy. Ionizing vs. Nonionizing Radiation There is a large difference in the magnitude of the biological effects of nonionizing radiation (for example, light and microwaves) and ionizing radiation , emissions energetic enough to knock electrons out of molecules (for example, α and β particles, γ rays, X-rays, and high-energy ultraviolet radiation) (Figure \(\PageIndex{4}\)). Energy absorbed from nonionizing radiation speeds up the movement of atoms and molecules, which is equivalent to heating the sample. Although biological systems are sensitive to heat (as we might know from touching a hot stove or spending a day at the beach in the sun), a large amount of nonionizing radiation is necessary before dangerous levels are reached. Ionizing radiation, however, may cause much more severe damage by breaking bonds or removing electrons in biological molecules, disrupting their structure and function. The damage can also be done indirectly, by first ionizing H 2 O (the most abundant molecule in living organisms), which forms a H 2 O + ion that reacts with water, forming a hydronium ion and a hydroxyl radical: Effects of Long-term Radiation Exposure on the Human Body The effects of radiation depend on the type, energy, and location of the radiation source, and the length of exposure. As shown in Figure \(\PageIndex{6}\), the average person is exposed to background radiation, including cosmic rays from the sun and radon from uranium in the ground; radiation from medical exposure, including CAT scans, radioisotope tests, X-rays, and so on; and small amounts of radiation from other human activities, such as airplane flights (which are bombarded by increased numbers of cosmic rays in the upper atmosphere), radioactivity from consumer products, and a variety of radionuclides that enter our bodies when we breathe (for example, carbon-14) or through the food chain (for example, potassium-40, strontium-90, and iodine-131). A short-term, sudden dose of a large amount of radiation can cause a wide range of health effects, from changes in blood chemistry to death. Short-term exposure to tens of rems of radiation will likely cause very noticeable symptoms or illness; a dose of about 500 rems is estimated to have a 50% probability of causing the death of the victim within 30 days of exposure. Exposure to radioactive emissions has a cumulative effect on the body during a person’s lifetime, which is another reason why it is important to avoid any unnecessary exposure to radiation. Health effects of short-term exposure to radiation are shown in Table \(\PageIndex{2}\). Exposure (rem) Health Effect Time to Onset (without treatment) 5–10 changes in blood chemistry — 50 nausea hours 55 fatigue — 70 vomiting — 75 hair loss 2–3 weeks 90 diarrhea — 100 hemorrhage — 400 possible death within 2 months 1000 destruction of intestinal lining — NaN internal bleeding — NaN death 1–2 weeks 2000 damage to central nervous system — NaN loss of consciousness; minutes NaN death hours to days It is impossible to avoid some exposure to ionizing radiation. We are constantly exposed to background radiation from a variety of natural sources, including cosmic radiation, rocks, medical procedures, consumer products, and even our own atoms. We can minimize our exposure by blocking or shielding the radiation, moving farther from the source, and limiting the time of exposure. Summary Background radiation is defined and different sources of background radiation are listed. We are constantly exposed to radiation from a variety of naturally occurring and human-produced sources. Various devices, including Geiger counters, scintillators, and dosimeters, are used to detect and measure radiation, and monitor radiation exposure. Ionizing radiation is the most harmful because it can ionize molecules or break chemical bonds, which damages the molecule and causes malfunctions in cell processes. It can also create reactive hydroxyl radicals that damage biological molecules and disrupt physiological processes. |
Courses/UWMilwaukee/CHE_125%3A_GOB_Introductory_Chemistry/11%3A_Energy_and_Chemical_Processes/11.01%3A_Prelude_to_Energy_and_Chemical_Processes | Metabolism is the collective term for the chemical reactions that occur in cells and provide energy to keep cells alive. Some of the energy from metabolism is in the form of heat, and some animals use this heat to regulate their body temperatures. Such warm-blooded animals are called endotherms . In endotherms, problems with metabolism can lead to fluctuations in body temperature. When humans get sick, for instance, our body temperatures can rise higher than normal; we develop a fever. When food is scarce (especially in winter), some endotherms go into a state of controlled decreased metabolism called hibernation . During hibernation, the body temperatures of these endotherms actually decrease. In hot weather or when feverish, endotherms will pant or sweat to rid their bodies of excess heat. Endotherm Body Temperature (°F) Body Temperature (°C) bird up to 110 up to 43.5 cat 101.5 38.6 dog 102 38.9 horse 100.0 37.8 human 98.6 37.0 pig 102.5 39.2 Ectotherms , sometimes called cold-blooded animals, do not use the energy of metabolism to regulate body temperature. Instead, they depend on external energy sources, such as sunlight. Fish, for example, will seek out water of different temperatures to regulate body temperature. The amount of energy available is directly related to the metabolic rate of the animal. When energy is scarce, ectotherms may also hibernate. The connection between metabolism and body temperature is a reminder that energy and chemical reactions are intimately related. A basic understanding of this relationship is especially important when those chemical reactions occur within our own bodies. |
Courses/Fresno_City_College/Introductory_Chemistry_Atoms_First_for_FCC/07%3A_Solids_Liquids_and_Phase_Changes/7.01%3A_Properties_of_Liquids_and_Solids | Learning Objectives Describe the solid and liquid phases. Solids and liquids are collectively called condensed phases because their particles are in virtual contact. The two states share little else, however. Solids In the solid state, the individual particles of a substance are in fixed positions with respect to each other because there is not enough thermal energy to overcome the intermolecular interactions between the particles. As a result, solids have a definite shape and volume. Most solids are hard, but some (like waxes) are relatively soft. Many solids composed of ions can also be quite brittle. Solids usually have their constituent particles arranged in a regular, three-dimensional array of alternating positive and negative ions called a crystal. The effect of this regular arrangement of particles is sometimes visible macroscopically, as shown in Figure \(\PageIndex{1}\). Some solids, especially those composed of large molecules, cannot easily organize their particles in such regular crystals and exist as amorphous (literally, “without form”) solids. Glass is one example of an amorphous solid. Liquids If the particles of a substance have enough energy to partially overcome intermolecular interactions, then the particles can move about each other while remaining in contact. This describes the liquid state. In a liquid, the particles are still in close contact, so liquids have a definite volume. However, because the particles can move about each other rather freely, a liquid has no definite shape and takes a shape dictated by its container. Gases If the particles of a substance have enough energy to completely overcome intermolecular interactions, then the particles can separate from each other and move about randomly in space. Like liquids, gases have no definite shape, but unlike solids and liquids, gases have no definite volume either. The change from solid to liquid usually does not significantly change the volume of a substance. However, the change from a liquid to a gas significantly increases the volume of a substance, by a factor of 1,000 or more. Figure \(\PageIndex{3}\) shows the differences among solids, liquids, and gases at the molecular level, while Table \(\PageIndex{1}\) lists the different characteristics of these states. Characteristic Solid Liquid Gas shape definite indefinite indefinite volume definite definite indefinite relative intermolecular interaction strength strong moderate weak relative particle positions in contact and fixed in place in contact but not fixed not in contact, random positions Example \(\PageIndex{1}\) What state or states of matter does each statement describe? This state has a definite volume. This state has no definite shape. This state allows the individual particles to move about while remaining in contact. Solution This statement describes either the liquid state or the solid state. This statement describes either the liquid state or the gas state. This statement describes the liquid state. Exercise \(\PageIndex{1}\) What state or states of matter does each statement describe? This state has individual particles in a fixed position with regard to each other. This state has individual particles far apart from each other in space. This state has indefinite shape. Answer a solid Answer b gas Answer c liquid or gas Looking Closer: Water, the Most Important Liquid Earth is the only known body in our solar system that has liquid water existing freely on its surface; life on Earth would not be possible without the presence of liquid water. Water has several properties that make it a unique substance among substances. It is an excellent solvent; it dissolves many other substances and allows those substances to react when in solution. In fact, water is sometimes called the universal solvent because of this ability. Water has unusually high melting and boiling points (0°C and 100°C, respectively) for such a small molecule. The boiling points for similar-sized molecules, such as methane (BP = −162°C) and ammonia (BP = −33°C), are more than 100° lower. Though a liquid at normal temperatures, water molecules experience a relatively strong intermolecular interaction that allows them to maintain the liquid phase at higher temperatures than expected. Unlike most substances, the solid form of water is less dense than its liquid form, which allows ice to float on water. In colder weather, lakes and rivers freeze from the top, allowing animals and plants to continue to live underneath. Water also requires an unusually large amount of energy to change temperature. While 100 J of energy will change the temperature of 1 g of Fe by 230°C, this same amount of energy will change the temperature of 1 g of H 2 O by only 100°C. Thus, water changes its temperature slowly as heat is added or removed. This has a major impact on weather, as storm systems like hurricanes can be impacted by the amount of heat that ocean water can store. Water’s influence on the world around us is affected by these properties. Isn’t it fascinating that such a small molecule can have such a big impact? Key Takeaway Solids and liquids are phases that have their own unique 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: Marisa Alviar-Agnew ( Sacramento City College ) Henry Agnew (UC Davis) |
Courses/Oregon_Tech_PortlandMetro_Campus/OT_-_PDX_-_Metro%3A_General_Chemistry_I/07%3A_Ideal_Gas_Behavior/7.01%3A_Temperature_and_Pressure/7.1.01%3A_Practice_Problems-_Temperature_and_Pressure | PROBLEM \(\PageIndex{1}\) Convert the boiling temperature of gold, 2966 °C, into degrees Fahrenheit and kelvin. Answer 5371 °F 3239 K PROBLEM \(\PageIndex{2}\) Convert the temperature of scalding water, 54 °C, into degrees Fahrenheit and kelvin. Answer 129.2 °F 327 K Click here to see a video solution PROBLEM \(\PageIndex{3}\) Convert the temperature of the coldest area in a freezer, −10 °F, to degrees Celsius and kelvin. Answer −23 °C 250 K PROBLEM \(\PageIndex{4}\) Convert the temperature of dry ice, −77 °C, into degrees Fahrenheit and kelvin. Answer -106.6 °F 196 K Click here to see a video solution PROBLEM \(\PageIndex{5}\) Convert the boiling temperature of liquid ammonia, −28.1 °F, into degrees Celsius and kelvin. Answer −33.4 °C 239.8 K PROBLEM \(\PageIndex{6}\) The label on a pressurized can of spray disinfectant warns against heating the can above 130 °F. What are the corresponding temperatures on the Celsius and kelvin temperature scales? Answer 90 °C 363 K Click here to see a video solution PROBLEM \(\PageIndex{7}\) The weather in Europe was unusually warm during the summer of 1995. The TV news reported temperatures as high as 45 °C. What was the temperature on the Fahrenheit scale? Answer 113 °F PROBLEM \(\PageIndex{8}\) Why do some small bridges have weight limits that depend on how many wheels or axles the crossing vehicle has? Answer Pressure is defined as force per unit area. Similarly to the example in the text of the elephant and the figure skater, the more wheels or axles on the vehicle, the more area the weight is spread over, causing less pressure to be exerted on any one point on the bridge. PROBLEM \(\PageIndex{9}\) A typical barometric pressure in Redding, California, is about 750 mm Hg. Calculate this pressure in atm and kPa Answer 0.987 atm 99.97 kPa PROBLEM \(\PageIndex{10}\) A typical barometric pressure in Denver, Colorado, is 615 mm Hg. What is this pressure in atmospheres and kilopascals? Answer 0.809 atm 82.0 kPa Click here to see a video solution PROBLEM \(\PageIndex{11}\) Canadian tire pressure gauges are marked in units of kilopascals. What reading on such a gauge corresponds to 32 psi? Answer 2.2 × 10 2 kPa PROBLEM \(\PageIndex{12}\) During the Viking mission landings on Mars, the atmospheric pressure was determined to be on the average about 6.50 millibars (1 bar = 0.987 atm). What is that pressure in torr and kPa? Answer 4.88 torr 0.650 kPa Click here to see a video solution PROBLEM \(\PageIndex{13}\) The pressure of the atmosphere on the surface of the planet Venus is about 88.8 atm. Compare that pressure in psi to the normal pressure on earth at sea level in psi. Answer Earth: 14.7 psi Venus: 13.1 × 10 3 psi PROBLEM \(\PageIndex{14}\) Consider this scenario and answer the following questions: On a mid-August day in the northeastern United States, the following information appeared in the local newspaper: atmospheric pressure at sea level 29.97 in. Hg, 1013.9 mbar. What was the pressure in kPa? The pressure near the seacoast in the northeastern United States is usually reported near 30.0 in. Hg. During a hurricane, the pressure may fall to near 28.0 in. Hg. Calculate the drop in pressure in torr. Answer a 101.5 kPa Answer b 51 torr drop PROBLEM \(\PageIndex{15}\) The pressure of a sample of gas is measured at sea level is 264 mmHg. Determine the pressure of the gas in: torr Pa bar Answer a 264 torr Answer b 35197 Pa Answer c 0.352 bar Contributors 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] ). Adelaide Clark, Oregon Institute of Technology |
Courses/University_of_Georgia/CHEM_3212%3A_Physical_Chemistry_II/06%3A_Entropy_Part_I/6.0S%3A_6.S%3A_The_Second_Law_(Summary) | Learning Objectives After mastering the material presented in this chapter, one will be able to: Describe a Carnot engine and derive a relationship for its efficiency of converting heat into work, in terms of the two temperatures at which the engine operates. Define entropy and be able to calculate entropy changes for systems (and the surroundings) undergoing changes which are definable as following various pathways, including constant temperature, constant pressure, constant volume, and adiabatic pathways. Relate entropy to disorder in a crystal based on the number of equivalent orientations a single formula unit may take within the crystal. State the Third Law of Thermodynamics, and use it to calculate total entropies for substances at a given temperature. Understand how isothermal compressibility differs from adiabatic compressibility and relate that difference to the measurement of the speed of sound waves traveling through a gas medium. Vocabulary and Concepts adiabatic compressibility Carnot cycle Clausius theorem criterion for spontaneity Debye Extrapolation efficiency entropy heat engine isentropic second law of thermodynamics speed of sound spontaneous spontaneous process Third Law Entropy Third Law of Thermodynamics References Biba, E. (2010, February 26). What is Time? One Physicist Hunts for the Ultimate theory. Wired . Retrieved from http://www.wired.com/2010/02/what-is-time/ Clausius, R. (1879). The Mechanical Theory of Heat. London: McMillan and Company. Doc, T. (n.d.). Isaac Newton . Retrieved April 2, 2016, from FamousScientists.org: http://www.famousscientists.org/isaac-newton/ Mendoza, E. (2016). Sadi Carnot: French engineer and physicist . (Encyclopedia Brittanica, Inc.) Retrieved March 30, 2016, from Encyclopædia Britannica Online: http://www.britannica.com/biography/...ench-scientist Newton, I. (1723). Philosophiae naturalis principia mathematica. Amstelodamum. O'Connor, J. J., & Robertson, E. F. (1998, September). Retrieved March 30, 2016, from www-groups.dcs.st-and.ac.uk/~...Boltzmann.html O'Connor, J. J., & Robertson, E. F. (1999). Pierre-Simon Laplace . Retrieved April 2, 2016, from School of Mathematics and Statistics, University of St Andrews, Scotland: www-groups.dcs.st-and.ac.uk/~...s/Laplace.html Yamashita, I., Tojo, T., Kawaji, H., Atake, T., Linnard, Y., & Richet, P. (2001). Low-temperature heat capacity of sodium borosilicate glasses at temperatures from 13 K to 300 K. The Journal of Chemical Thermodynamics, 33 (5), 535-53. doi:10.1006/jcht.2000.0744 |
Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/01%3A_Introduction_to_Chemistry/1.09%3A_Materials | How does chemistry affect the clothing that we wear? Chemistry research is often full of surprises. One such surprise came to Stephanie Kwolek of the DuPont chemical company when she was working on materials known as polymers. These chemicals had been around for a while and were being used for new types of textiles. Kwolek was looking for a strong and rigid petroleum product. She came up with a material that did not look like your average polymer; she played a hunch and had it made into threads. This new material had stiffness about nine times that of any known polymers of the time. Further research and development led to the production of Kevlar, a material now widely used in body armor. Kevlar also found a range of application in racing sails, car tires, brakes, and fire-resistant clothing worn by firefighters. Materials Electronics Chemists are involved in the design and production of new materials. Some of the materials that chemists have helped discover or develop in recent years include polymers, ceramics, adhesives, coatings, and liquid crystals. Liquid crystals are used in electronic displays, as in watches and calculators. The silicon-based computer chip has revolutionized modern society, and chemists have played a key role in the design and continued improvement of these chips. The calculator shown below uses both a liquid crystal display and chips inside the device. Superconductors Many chemists are currently working in the field of superconductivity. Superconductors are materials that are able to conduct electricity with \(100\%\) efficiency, meaning that no energy is lost during the electrical transmission, as happens with conventional conduction materials like copper cable. The challenge is to design materials that can act as superconductors at normal temperatures, as opposed to only being able to superconduct at very low temperatures. Clothing The fibers that compose the materials for our clothing are either natural or human-made. Silk and cotton are examples of natural fibers. Silk is produced by the silkworm and cotton is grown as a plant. Human-made fabrics include nylon, orlon, and a number of other polymers. These materials are made from hydrocarbons found in petroleum products. Synthetic polymers are also used in shoes, rain gear, and camping items. Synthetic fabrics tend to be lighter than the natural ones, and can be treated to be more water-resistant and durable. Materials originally developed as textiles are finding a wide variety of other uses. Nylon is found in a number of plastic utensils. Taking advantage of its strength and light weight, nylon is a component of ropes, fishing nets, tents, and parachutes. Summary Chemists produce materials for electronics, superconductors, textiles, and other applications. Review Use the link below to answer the following questions: Who developed Kevlar? Where are liquid crystals used? What is a superconductor? What are synthetic polymers made from? |
Courses/Lumen_Learning/Book%3A_US_History_I_(AY_Collection)_(Lumen)/08%3A_A_New_Nation/08.13%3A_Conclusion | A grand debate over political power engulfed the young United States. The Constitution ensured that there would be a strong federal government capable of taxing, waging war, and making law, but it could never resolve the young nation’s many conflicting constituencies. The Whiskey Rebellion proved that the nation could stifle internal dissent but exposed a new threat to liberty. Hamilton’s banking system provided the nation with credit but also constrained frontier farmers. The Constitution’s guarantee of religious liberty conflicted with many popular prerogatives. Dissension only deepened, and as the 1790s progressed, Americans became bitterly divided over political parties and foreign wars. During the ratification debates, Alexander Hamilton had written of the wonders of the Constitution. “A nation, without a national government,” he wrote, would be “an awful spectacle.” But, he added, “the establishment of a Constitution, in time of profound peace, by the voluntary consent of a whole people, is a prodigy,” a miracle that should be witnessed “with trembling anxiety.” Anti-Federalists had grave concerns about the Constitution, but even they could celebrate the idea of national unity. By 1795, even the staunchest critics would have grudgingly agreed with Hamilton’s convictions about the Constitution. Yet these same individuals could also take the cautions in Washington’s 1796 farewell address to heart. “There is an opinion,” Washington wrote, “that parties in free countries are useful checks upon the administration of the government and serve to keep alive the spirit of liberty.” This, he conceded, was probably true, but in a republic, he said, the danger was not too little partisanship, but too much. “A fire not to be quenched,” Washington warned, “it demands a uniform vigilance to prevent its bursting into a flame, lest, instead of warming, it should consume.” For every parade, thanksgiving proclamation, or grand procession honoring the unity of the nation, there was also some political controversy reminding American citizens of how fragile their union was. And as party differences and regional quarrels tested the federal government, the new nation increasingly explored the limits of its democracy. CC licensed content, Shared previously American Yawp. Located at : http://www.americanyawp.com/index.html . Project : American Yawp. License : CC BY-SA: Attribution-ShareAlike |
Courses/Smith_College/Organic_Chemistry_(LibreTexts)/19%3A_Aldehydes_and_Ketones-_Nucleophilic_Addition_Reactions/19.14%3A_Conjugate_Nucleophilic_Addition_to_alpha_beta-unsaturated_Aldehydes_and_Ketones | Objectives After completing this section, you should be able to explain how the carbonyl group which is present in α , β ‑unsaturated aldehydes and ketones activates the carbon‑carbon double bond so that it is susceptible to attack by nucleophiles. write equations to illustrate the addition of amines, water and lithium diorganocopper reagents to α , β ‑unsaturated aldehydes and ketones. identify the product formed from the reaction of a given primary or secondary amine with a given α , β ‑unsaturated aldehyde or ketone. identify the aldehyde or ketone, the primary or secondary amine, or both, needed to prepare a given β ‑amino aldehyde or ketone. identify the product formed from the reaction of an α , β ‑unsaturated aldehyde or ketone with water. identify the product formed from the reaction of a given α , β ‑unsaturated aldehyde or ketone with a given lithium diorganocopper reagent. identify the α , β ‑unsaturated aldehyde or ketone, the lithium diorganocopper reagent, or both, needed to prepare a given product through a conjugate addition reaction. Study Notes At first this section may appear to contain a considerable amount of information, but you should realize that much of the material presented is really repetition. Essentially we see how three different nucleophilic reagents, primary and secondary amines, water and lithium diorganocoppers can add across a carbon‑carbon double bond when the latter is conjugated to the carbonyl group of an aldehyde or ketone. Note that the first reagent can also react directly with the carbonyl group of an aldehyde or ketone when there is no conjugated carbon‑carbon double bond present, but that the third reagent, lithium dialkylcopper, cannot do so. You may be confused about the designation of the product from the conjugate addition to an α , β ‑unsaturated aldehyde or ketone as a 1,4 adduct. You can more clearly understand this name if you recognize that the proton added in the second step of the reaction first adds to the oxygen of the enolate ion to produce an enol. The latter then tautomerizes to the more stable keto form. One of the largest and most diverse classes of reactions involves nucleophilic additions to a carbonyl group. As discussed in Section 19.4 , carbonyl carbons are electrophilic due to bond polarity created by resonance. Previously in this chapter, we have discussed a nucleophilic addition to carbonyl carbons called a 1,2 addition . During 1,2 addition the nucleophile adds to the carbonyl carbon which is defined as the one position. Subsequently, hydrogen adds to the carbonyl oxygen which considered the two position. Overall an atom is added in both the 1 and 2 position justifying the reaction name, 1,2 addition. Basic Reaction of 1,2 Addition An important functional group is created when an alkene is placed in conjugation with a carbonyl. These conjugated carbonyl are called enones or α, β-unsaturated carbonyls . The term α commonly refers to the carbon adjacent to a carbonyl and β referrers to the next carbon in the chain. Conjugation transmits the electrophilic character of the carbonyl carbon to the β-carbon of the α, β-unsaturated carbonyl double bond. The resonance structure shown below shows that the electronegative oxygen atom in α, β-unsaturated carbonyls pulls electrons away from the β carbon making it more electrophilic than a typical alkene carbon. From this resonance form, it should be clear that nucleophiles may attack either at the carbonyl carbon or at the β-alkene carbon. These two modes of reaction are referred to as 1,2-addition and 1,4-addition respectively. A 1,4-addition is also called a conjugate addition . Basic Reaction of 1,4 Conjugate Addition In 1,4 addition, a nucleophile is added to the carbon β to the carbonyl while a hydrogen is added to the carbon α to the carbonyl. Overall, the carbonyl is unaffected by the nucleophilic addition. It is important to note that this reaction only occurs because the alkene is conjugated with a carbonyl. The utility of 1,4 conjugate addition is shown by the wide variety of nucleophiles which can be added to a α, β unsaturated carbonyls. General Mechanism for 1,4 Conjugate Addition The mechanism starts with the nucleophile attacking the electrophilic β carbon forming a single bond. The two electrons from the alkene pi bond are pushed onto the electronegative carbonyl oxygen creating an enolate. In the next step the the enolate is protonated to form an enol. If the original nucleophile was neutral, this addition will cause it to become positively charge. A proton transfer will occur making the nucleophile neutral and turning the enolate into an enol. If the original nucleophile was negatively charged this protonation is accomplished by the subsequent addition of a proton source. The product of the second step of the mechanism shows why the reaction is called a 1,4 addition. The nucleophile bonds to the β alkene carbon which is considered the one position and the hydrogen adds to the carbonyl oxygen which is in the four position. Overall, addition occurs in the one and four position. In the final step of the mechanism, the enol undergoes a rearranges to form a carbonyl during a process called tautomerization. Tautomerization causes the hydrogen to move from the oxygen to the β-carbon. The tautomerization process will be discussed in greater detail in Section 22.3 . Step 1: Nucleophilic attack Step 2: Protonation Step 3: Tautomerization Predicting the Products of a 1,4 Conjugate Addition In total, 1,4 addition occurs across the alkene bond of the α, β unsaturated carbonyl. The alkene pi bonds is broken to form two single bonds, one on the α-carbon and one on the β-carbon. During 1,4 addition, the α-carbon of the α, β unsaturated carbonyl forms a bond with a hydrogen while the β-carbon forms a bond to the nucleophile. Remember that neutral nucleophiles typically lose a hydrogen during 1,4 addition. Nucleophiles Which add 1,4 to α, β Unsaturated Carbonyls Hydroxide Alcohols Thiols 1 o Amines 2 o Amines Cyanides HBr Examples The synthesis of 3-( N -Methylamino)cyclohexanone The Synthesis of 5-Hydroxy-3-heptenone 1,2 vs. 1,4 Addition Whether 1,2 or 1,4-addition occurs to a α, β unsaturated carbonyl depends on multiple variables but is mostly determined by the nature of the nucleophile. During the addition of a nucleophile there is a competition between the formation 1,2 and 1,4 addition products. If the nucleophile is a strong base, such as Grignard reagents or metal hydrides, both the 1,2 and 1,4 reactions are irreversible and therefor are under kinetic control. Since 1,2-additions to the carbonyl group are fast, we would expect to find a predominance of 1,2-products from these reactions. If the nucleophile is a weak base, such as, water, alcohols or amines, then the possible 1,2 addition is usually reversible. This means the competition between 1,2 and 1,4 addition is under thermodynamic control. In this most cases, the 1,4-addition product dominates because the stable carbonyl group is retained. Nucleophiles Which add 1,2 to α, β-Unsaturated Carbonyls Metal Hydrides (LiAlH 4 ) Grignard Reagents Organolithium Reagents Example Gilman Reagents Another important reaction exhibited by organometallic reagents is metal exchange. Organolithium reagents react with cuprous iodide to give a lithium diorganocopper reagent , which often is referred to as a Gilman reagent . Remember that organolithium reagents are formed by a reaction of lithium metal with an organohalide. Lithium diorganocopper reagents are considered a source of carbanion like nucleophiles similar to Grignard and Organolithium reagents. However, the reactivity of lithium diorganocuprate reagents is slightly different and this difference will be exploited in different situations. Diorganocuprate reagents are made from the reaction of two equivalents of an organolithium reagent and copper (I) iodide (CuI). The created lithium diorganocuprate reagent acts as a source of R: - Lithium Diorganocopper (Gilman Reagent) General Reaction Example Formation of Lithium Dimethylcopper Reaction of 1,4 Conjugate Addition of Gilman Reagents to α, β Unsaturated Ketones Lithium diorganocopper reagents, R 2 CuLi, undergo 1,4 conjugate addition when reacted with α, β Unsaturated ketones. Using lithium diorganocopper reagents allows for a wide range of organic groups to undergo this 1,4 conjugate addition including alkyl, aryl, and alkenyl groups. Because a C-C single bond is formed this reaction is an excellent method for adding to the carbon framework of a ketone. Example Mechanism for the 1,4 Conjugate Addition of Lithium Diorganocopper Reagents to α, β-Unsaturated Ketone This mechanism is only slightly different than the general mechanism for 1,4 conjugate addition described above. The mechanism starts with the nucleophilc diorganocopper anion (R 2 Cu - ) adding to the electrophilc β alkene carbon forming a Cu-C bond. The R group from the diorganocopper is then transferred to the β alkene carbon with elimination of a neutral organocopper species (RCu). Protonation of the enolate ion followed by tautomerization creates the final 1,4 addition product. Synthesis of Ketones using 1,4 Conjugate Addition of Lithium Diorganocopper Reagents to α, β Unsaturated Ketones When using retrosynthetic analysis to plan the synthesis of a ketone, remember this reaction allows for the formation of a C-C bond between third and fourth carbon away from a ketone. The carbonyl group in the target molecule will become a α, β unsaturated carbonyl in the probable reactant. To determine the structure of a possible reactant, start by cleaving the C-C between the third and fourth carbon away from the carbonyl to create two fragments. The take the fragment which retains the carbonyl and remove a hydrogen from the second carbon and connect the second and third carbon with a double bond. The creates the required α, β unsaturated carbonyl reactant. The other fragment will become the lithium diorganocopper reagent. Remember the lithium diorganocopper reagent contains two of the required R group. Note! If the fourth carbon away from the ketone in the target molecule is tertiary or quarternary there may be multiple bonds which could be retrosynthetically cleaved. Something Extra: Why do Metals Have a Smell When They are Nonvolatile? If you have ever encountered a metallic scent when counting coins or handling metals, especially copper, you were probably being fooled. The metallic smell could not have come from the metal because metals are non-volatile and do not evaporate. This means the odor molecules associated with metal scent must come from a different source. The volatile molecules that create the metallic smell are produced by a chemical reaction between skin oils and the metal itself. These volatile molecules, primarily aldehydes and ketones, create a sensory illusion that the metal is producing the smell, even though metals have no natural odor. The main molecule is the α, β Unsaturated Ketone, 1-Octen-3-one, which is described as smelling like mushrooms or metal. This is an illusion because the molecule doesn’t smell like metals; metals smells like this molecule. Because we have smelled 1-Octen-3-one coming from metals for so long, we mistake its smell with the metal itself. This kind of chemical reaction also explain why blood produces a metallic smell when it comes into contact with skin. Blood contains iron in hemoglobin, and the same chemical reaction produces odor molecules when blood touches skin. Structure of 1-Octen-3-one Exercise \(\PageIndex{1}\) How would you make the following molecule using a 1,4 Conjugate Addition of a Gilman Reagent to a α, β Unsaturated Ketone? Answer Exercise \(\PageIndex{2}\) Draw the bond-line structures for the products of the reactions below. a) b) Answer A b) Exercise \(\PageIndex{3}\) Specify the reagents needed to perform the following chemical transformation. Answer |
Courses/Modesto_Junior_College/Chemistry_143_-_Bunag/Chemistry_143_-_Introductory_Chemistry_(Bunag)/05%3A_Electrons_in_Atoms/5.13%3A_Hund's_Rule_and_Orbital_Filling_Diagrams | Have you ever wondered what those load limit signs mean on a bridge? The sign above signifies that nothing over five tons is allowed because it will do damage to the structure. There are limits to the amount of weight that a bridge can support, there are limits to the number of people that can safely occupy a room, and there are limits to what can go into an electron orbital. Hund's Rule The last of the three rules for constructing electron arrangements requires electrons to be placed one at a time in a set of orbitals within the same sublevel. This minimizes the natural repulsive forces that one electron has for another. Hund's rule states that orbitals of equal energy are each occupied by one electron before any orbital is occupied by a second electron, and that each of the single electrons must have the same spin. The figure below shows how a set of three \(p\) orbitals is filled with one, two, three, and four electrons. Orbital Filling Diagrams An orbital filling diagram is the more visual way to represent the arrangement of all the electrons in a particular atom. In an orbital filling diagram, the individual orbitals are shown as circles (or squares) and orbitals within a sublevel are drawn next to each other horizontally. Each sublevel is labeled by its principal energy level and sublevel. Electrons are indicated by arrows inside the circles. An arrow pointing upwards indicates one spin direction, while a downward pointing arrow indicates the other direction. The orbital filling diagrams for hydrogen, helium, and lithium are shown in the figure below. According to the Aufbau process, sublevels and orbitals are filled with electrons in order of increasing energy. Since the \(s\) sublevel consists of just one orbital, the second electron simply pairs up with the first electron as in helium. The next element is lithium and necessitates the use of the next available sublevel, the \(2s\). The filling diagram for carbon is shown in the figure below. There are two \(2p\) electrons for carbon and each occupies its own \(2p\) orbital. Oxygen has four \(2p\) electrons. After each \(2p\) orbital has one electron in it, the fourth electron can be placed in the first \(2p\) orbital with a spin opposite that of the other electron in that orbital. Summary Hund's rule specifies the order of electron filling within a set of orbitals. Orbital filling diagrams are a way of indicating electron locales in orbitals. Review State Hund’s rule. What is an orbital filling diagram? Is the diagram in figure below correct? Explain your answer. Is the diagram in figure below correct? Explain your answer. |
Bookshelves/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/14%3A_Intermolecular_Forces___Phase_Changes/14.3.1_Surface_Tension_Viscosity_and_Melting_Point_(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 Surface tension: A molecules in the interior of a liquid have more IMF interactions than molecules on the surface of a liquid. Molecules in the interior of a liquid are more stable than those on the surface. Surface tension is created because molecules on the surface would rather be in the interior. IMF ↑ Surface Tension ↑. Viscosity: Intermolecular forces create internal friction in a liquid. IMF ↑ Viscosity ↑. As the intermolecular forces of a liquid increases the viscosity tends to increase. IMF ↑ Melting Point ↑. Link to Video Surface Tension, Viscosity, & Melting Point: https://youtu.be/OgKDGrdTRRM Attribution Prof. Steven Farmer ( Sonoma State University ) |
Courses/Pasadena_City_College/PCC_Chemistry_2A/07%3A_Solutions/7.01%3A_Intermolecular_Forces/7.1.01%3A_Intramolecular_forces_and_intermolecular_forces | There are electrostatic interaction between charges or partial charges, i.e., the same charges attract each other, and opposite charges repel each other, as illustrated in Fig. 3.9.1.There are two types of electrostatic forces in compounds or molecules, intramolecular forces that exist between the bonded atoms of a compound or a molecule, and intermolecular forces that exist between molecules as described below. Intramolecular forces Intramolecular forces are the chemical bonds holding the atoms together in the molecules. The three major types of chemical bonds are the metallic bond, the ionic bond, and the covalent bond. Metallic bond Metals exist as a collection of many atoms as +ions arranged in a well-defined 3D arrangement called crystal lattice with some of the outermost electrons roaming around in the whole piece of the metal, forming a sea of electrons around the metal atoms, as illustrated in Fig. 3.9.2. The attraction between +ions and the sea of free moving electrons is the metallic bond that holds the atoms together in a piece of metal. The metallic bond is usually the strongest type of chemical bond. Ionic bond When the electronegativity difference between bonded atoms is large, i.e., more than 1.9 in most cases, the bonding electrons completely transfer from a more electropositive atom to a more electronegative atom creating a cation and an anion, respectively. There is the electrostatic interaction between cation and anion, i.e., the same charges attract each other, and opposite charges repel each other, as illustrated in Fig. 3.9.1. The cations and anions orient themselves in a 3D crystal lattice in such a way that attractive interactions maximize and the repulsive interactions minimize, as illustrated in Fig. 3.9.3. Ionic bonds are usually weaker than metallic bonds but stronger there the other types of bonds. Covalent bond When the electronegativity difference between bonded atoms is moderate to zero, i.e., usually less than 1.9, the bonding electrons are shared between the bonded atoms, as illustrated in Fig. 3.9.4. The attractive force between the bonding electrons and the nuclei is the covalent bond that holds the atoms together in the molecules. The covalent bond is usually weaker than the metallic and the ionic bonds but much stronger than the intermolecular forces. Criteria to predict the type of chemical bond Metals tend to have lower electronegativity and nonmetals have higher electronegativity. When the electronegativity difference between the bonded atoms is large, usually more than 1.9, the bond is ionic. Generally, a bond between a metal and a nonmetal is ionic. When the electronegativity difference is low, usually less than 1.9, the bond is either metallic or covalent. Nonmetals tend to make a covalent bond with each other. Nonmetals also have higher electronegativities. So, when the average electronegativity of the bonded atom is high and the electronegativity difference between them is low, they tend to make a covalent bond. Metals tend to make the metallic bond with each other. Metals also tend to have lower electronegativity values. So, when the average electronegativity of the bonded atom is low and the electronegativity difference between them is also low, they tend to make a metallic bond. Fig. 3.9.5 illustrates the criteria to predict the type of chemical bond based on the electronegativity difference. Keep in mind that there is no sharp boundary between metallic, ionic, and covalent bonds based on the electronegativity differences or the average electronegativity values. Intermolecular forces Intermolecular forces are the electrostatic interactions between molecules. The intermolecular forces are usually much weaker than the intramolecular forces, but still, they play important role in determining the properties of the compounds. The major intermolecular forces include dipole-dipole interaction, hydrogen bonding, and London dispersion forces. Dipole-dipole interactions Polar molecules have permanent dipoles, one end of the molecule is partial positive (δ+) and the other is partial negative (δ-). The polar molecules have electrostatic interactions with each other through their δ+ and δ- ends called dipole-dipole interactions, though these interactions are weaker than ionic bonds. The polar molecules orient in a way to maximize the attractive forces between the opposite charges and minimize the repulsive forces between the same charges, as illustrated in Fig. 3.9.6. Hydrogen bonds Hydrogen bonding is a dipole-dipole interaction when the dipole is a hydrogen bond to O, N, or F, e.g. in water molecules as illustrated in Fig. 3.9.7. Although hydrogen bond is a dipole-dipole interaction, it is distinguished from the usual dipole-dipole interactions because of the following special features. The electronegativity difference between H and O, N, or F is usually more than other polar bonds. The charge density on hydrogen is higher than the δ+ ends of the rest of the dipoles because of the smaller size of hydrogen. The δ+ Hydrogen can penetrate in less accessible spaces to interact with the δ- O, N, or F of the other molecule because of its small size. A hydrogen bond is usually stronger than the usual dipole-dipole interactions. Hydrogen bonding is the most common and essential intermolecular interaction in biomolecules. For example, two strands of DNA molecules are held together through hydrogen bonding, as illustrated in Fig. 3.9.8. Proteins also acquire structural features needed for their functions mainly through hydrogen bonding. London dispersion forces It may appear that the nonpolar molecules should not have intermolecular interactions. Practically, there are intermolecular interactions called London dispersion forces, in all the molecules, including the nonpolar molecules. The electron cloud around atoms is not all the time symmetrical around the nuclei. It temporarily sways to one side or the other, generating a transient dipole . The transient dipole induces a dipole in the neighboring. A transient dipole-induced dipole interaction, called London dispersion force or wander Wall’s force , is established between the neighboring molecules as illustrated in Fig. 3.9.9. Although London dispersion forces are transient, they keep re-appearing randomly distributed in space and time. London dispersion forces are not unique to nonpolar molecules, they are present in all types of molecules, but these are the only intramolecular forces present in the nonpolar molecules. |
Courses/Chabot_College/Introduction_to_General_Organic_and_Biochemistry/06%3A_Ionic_and_Molecular_Compounds/6.19%3A_Polar_Molecules | Learning Objectives Recognize bond characteristics of covalent compounds: bond length and bond polarity. Use electronegativity values to predict bond polarity. If there is only one bond in the molecule, the bond polarity determines the molecular polarity . Any diatomic molecule in which the two atoms are the same element must be a nonpolar molecule. A diatomic molecule that consists of a polar covalent bond, such as HF, is a polar molecule where one end of the molecule is slightly positive, while the other end is slightly negative. The two electrically charged regions on either end of the molecule are called poles, similar to a magnet having a north and a south pole. Hence, a molecule with two poles has a dipole moment . For molecules with more than two atoms, the molecular geometry must also be taken into account when determining if the molecule is polar or nonpolar. The figure below shows a comparison between carbon dioxide and water. Carbon dioxide \(\left( \ce{CO_2} \right)\) is a linear molecule with carbon in the center and two oxygens at the terminal ends. The oxygen atoms are more electronegative than the carbon atom, so there are two individual dipoles pointing outward from the \(\ce{C}\) atom to each \(\ce{O}\) atom. However, since the dipoles are of equal strength and pointing in opposite directions, they cancel out and the overall molecular polarity of \(\ce{CO_2}\) is zero (no net dipole), therefore \(\ce{CO_2}\) is a nonpolar molecule . Water has a bent molecular structure because it has four electron groups, two bonded groups and two lone electron groups on the central oxygen atom. The individual O–H bond dipoles point from the slightly positive \(\ce{H}\) atoms toward the more electronegative \(\ce{O}\) atom. Because of the bent shape, the dipoles, which are equal in strength, both point towards the oxygen atom and will not cancel each other out, therefore, the water molecule is polar. In the figure below, you can see that the oxygen end of the molecule is slightly negative and the hydrogen side is slightly positive, there is a separation of charge throughout the whole molecule, a net dipole (shown in blue) that points upward. Some other molecules are shown in the figure below. Notice that a tetrahedral molecule such as \(\ce{CH_4}\) is nonpolar. However, if one of the peripheral \(\ce{H}\) atoms is replaced with another atom that has a different electronegativity, the molecule becomes polar. A trigonal planar molecule \(\left( \ce{BF_3} \right)\) may be nonpolar if all three peripheral atoms are the same, but a trigonal pyramidal molecule \(\left( \ce{NH_3} \right)\) is polar. To summarize, to be polar, a molecule must: Contain at least one polar covalent bond. Have a molecular structure such that the sum of the vectors of each bond dipole moment do not cancel. Steps to Identify Polar Molecules Draw the Lewis structure. Figure out the geometry (using VSEPR theory). Visualize or draw the geometry. Find the net dipole moment (you don't have to actually do calculations if you can visualize it). If the net dipole moment is zero, it is non-polar. Otherwise, it is polar. Properties of Polar Molecules Polar molecules tend to align when placed in an electric field with the positive end of the molecule oriented toward the negative plate and the negative end toward the positive plate (Figure \(\PageIndex{4}\)). We can use an electrically charged object to attract polar molecules, but nonpolar molecules are not attracted. Also, polar solvents are better at dissolving polar substances, and nonpolar solvents are better at dissolving nonpolar substances. While molecules can be described as "polar covalent" or "ionic", it must be noted that this is often a relative term, with one molecule simply being more polar or less polar than another. However, the following properties are typical of such molecules. Polar molecules tend to: have higher melting points than nonpolar molecules have higher boiling points than nonpolar molecules be more soluble in water (dissolve better) than nonpolar molecules have lower vapor pressures than nonpolar molecules Interactive Element: Molecular Polarity The following simulation is very useful in exploring more about bond and molecular polarity. Explore on your own or use the exercise and examples below. Example \(\PageIndex{1}\): Polarity Simulations Open the molecule polarity simulation (above) and select the “Three Atoms” tab at the top. This should display a molecule ABC with three electronegativity adjustors. You can display or hide the bond moments, molecular dipoles, and partial charges at the right. Turning on the Electric Field will show whether the molecule moves when exposed to a field. Use the electronegativity controls to determine how the molecular dipole will look for the starting bent molecule if: A and C are very electronegative and B is in the middle of the range. A is very electronegative, and B and C are not. Solution Molecular dipole moment points immediately between A and C. Molecular dipole moment points along the A–B bond, toward A. Exercise \(\PageIndex{1}\) Determine the partial charges that will give the largest possible bond dipoles. Answer The largest bond moments will occur with the largest partial charges. The two solutions above represent how unevenly the electrons are shared in the bond. The bond moments will be maximized when the electronegativity difference is greatest. The controls for A and C should be set to one extreme, and B should be set to the opposite extreme. Although the magnitude of the bond moment will not change based on whether B is the most electronegative or the least, the direction of the bond moment will. |
Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/17%3A_Additional_Aspects_of_Acid-Base_Equilibria/17.E%3A_Exercises | 17.1: Common-Ion Effect in Acid-Base Equilibria Questions The solubility product K sp for bismuth sulfide \(\ce{Bi2S3}\) is \(1.6 \times 10^{-72}\) at 25 °C. What is the molar solubility of bismuth sulfide in a solution that is 0.0010 M in sodium sulfide \(\ce{Na2S}\)? John poured 1.0 mL of 0.10 M \(\ce{NaCl}\), 1.0 mL of 0.10 M \(\ce{KOH}\), and 1.0 mL 0.20 \(\ce{HCl}\) solutions together and then he made the total volume to be 100.0 mL. What is the \(\ce{[Cl- ]}\) in the final solution? The K sp for \(\ce{AgCl}\) is \(1.0 \times 10^{-10}\). From which of the following solutions would silver chloride precipitate? A: A solution 0.10 M in \(\ce{Ag+}\) and 1.00 M in \(\ce{Cl-}\) B: A solution \(1.0\times 10^{-5}\; M\) in \(\ce{Ag+}\) and 0.20 M in \(\ce{Cl-}\) C: A solution \(1.0 \times 10^{-7}\; M\) in \(\ce{Ag+}\) and 1.0E-7 M in \(\ce{Cl-}\) A only B only C only A and B A, B, and C Substance Ksp magnesium hydroxide \(1.2\times 10^{-11}\) magnesium carbonate \(1.6\times 10^{-5}\) magnesium fluoride \(6.4\times 10^{-9}\) Addition of which of the following substances will cause the precipitation of a salt from one liter of a \(1\times 10^{-4}\; M\) \(\ce{Mg^2+}\) solution? \(1\times 10^{-4}\) mole NaOH \(1\times 10^{-1}\) mole nitric acid \(1\times 10^{-5}\) mole potassium acetate \(1\times 10^{-4}\) mole ammonium nitrate \(1\times 10^{-2}\) mole sodium fluoride The K sp for strontium chromate is \(3.6 \times 10^{-5}\) and K sp for barium chromate is \(1.2\times 10^{-10}\). What concentration of potassium chromate will precipitate the maximum amount of either the barium or the strontium chromate from an equimolar 0.10 M solution of barium and strontium ions without precipitating the other? Iron(II) hydroxide is only sparingly soluble in water at 25 °C; its K sp is equal to \(7.9\times 10^{-16}\). Calculate the solubility of iron(II) hydroxide in a solution of pH 6.0. Solutions Answer 1.08e-13 Consider... x = molar solubility; (2x) 2 (0.0010 + 3x) 3 = K sp ; x = ? Answer 0.0030 M Consider... Evaluate \(\ce{[Na+]}\), \(\ce{[K+]}\), \(\ce{[H+]}\) and \(\ce{[OH- ]}\) for fun! Answer d. Consider... Calculate the \(\ce{[Ag+] [Cl- ]}\) for A, B, and C, and compare their values with K sp . Answer e. Consider... A lot of calculations to figure out, but it's a common ion problem. Answer 3.6e-4 Consider... This problem requires some thinking. Answer 7.9 M Consider... \(\ce{[OH- ]}\) = 10 (-14+6) = 1e-8. x * (1e-8 + x) 2 = K sp ; x = ? 17.2: Buffer Solutions 17.3: Acid-Base Indicators Q17.3.1 There are numerous natural indicators present in plants. The dye in red cabbage, the purple color of grapes, even the color of some flowers are some examples. What is the cause for some fruits to change color when they ripen? S17.3.1 \(\ce{[H+]}\) of the juice changes. The changes in pH or \(\ce{[H+]}\) cause the dye to change color if their conjugate acid-base pairs have different colors. There may be other reasons too. Do colors indicate how good or bad they taste? Q17.3.2 Choose the true statement: All weak acids are indicators. All weak bases are indicators. Weak acids and bases are indicators. All indicators are weak acids. An acid-base conjugate pair has different colors. Any indicator changes color when the pH of its solution is 7. S17.3.2 d. Color change is a requirement for indicators. Q17.3.3 Do all indicators change color at pH 7 (y/n)? S17.3.3 No! Phenolphthalein changes color at pH ~9. Bromothymol blue has a p K n value of 7.1. At pH 7, its color changes from yellow to blue. Some indicators change color at pH other than 7. 17.4: Neutralization Reactions and Titration Curves 17.5: Solutions of Salts of Polyprotic Acids 17.6: Acid-Base Equilibrium Calculations: A Summary |
Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/03%3A__The_Vocabulary_of_Analytical_Chemistry/3.09%3A_Additional_Resources | The International Union of Pure and Applied Chemistry (IUPAC) maintains a web-based compendium of analytical terminology. You can find it at the following web site. old.iupac.org/publications/an...al_compendium/ The following papers provide alternative schemes for classifying analytical methods. Booksh, K. S.; Kowalski, B. R. “Theory of Analytical Chemistry,” Anal. Chem. 1994 , 66 , 782A– 791A. Phillips, J. B. “Classification of Analytical Methods,” Anal. Chem. 1981 , 53 , 1463A–1470A. Valcárcel, M.; Luque de Castro, M. D. “A Hierarchical Approach to Analytical Chemistry,” Trends Anal. Chem. 1995 , 14 , 242–250. Valcárcel, M.; Simonet, B. M. “Types of Analytical Information and Their Mutual Relationships,” Trends Anal. Chem. 1995 , 14 , 490–495. Further details on criteria for evaluating analytical methods are found in the following series of papers. Wilson, A. L. “The Performance-Characteristics of Analytical Methods”, Part I- Talanta , 1970 , 17 , 21–29; Part II- Talanta , 1970 , 17 , 31–44; Part III- Talanta , 1973 , 20 , 725–732; Part IV- Talanta , 1974 , 21 , 1109–1121. For a point/counterpoint debate on the meaning of sensitivity consult the following two papers and two letters of response. Ekins, R.; Edwards, P. “On the Meaning of ‘Sensitivity’,” Clin. Chem. 1997 , 43 , 1824–1831. Ekins, R.; Edwards, P. “On the Meaning of ‘Sensitivity:’ A Rejoinder,” Clin. Chem. 1998 , 44 , 1773–1776. Pardue, H. L. “The Inseparable Triangle: Analytical Sensitivity, Measurement Uncertainty, and Quantitative Resolution,” Clin. Chem. 1997 , 43 , 1831–1837. Pardue, H. L. “Reply to ‘On the Meaning of ‘Sensitivity:’ A Rejoinder’,” Clin. Chem. 1998 , 44 , 1776–1778. Several texts provide analytical procedures for specific analytes in well-defined matrices. Basset, J.; Denney, R. C.; Jeffery, G. H.; Mendham, J. Vogel’s Textbook of Quantitative Inorganic Analysis , 4th Edition; Longman: London, 1981. Csuros, M. Environmental Sampling and Analysis for Technicians , Lewis: Boca Raton, 1994. Keith, L. H. (ed) Compilation of EPA’s Sampling and Analysis Methods , Lewis: Boca Raton, 1996 Rump, H. H.; Krist, H. Laboratory Methods for the Examination of Water, Wastewater and Soil , VCH Publishers: NY, 1988. Standard Methods for the Analysis of Waters and Wastewaters , 21st Edition, American Public Health Association: Washington, D. C.; 2005. For a review of the importance of analytical methodology in today’s regulatory environment, consult the following text. Miller, J. M.; Crowther, J. B. (eds) Analytical Chemistry in a GMP Environment , John Wiley & Sons: New York, 2000. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/03%3A_First_Law_of_Thermodynamics/3.04%3A_Calorimetry | As chemists, we are concerned with chemical changes and reactions. The thermodynamics of chemical reactions can be very important in terms of controlling the production of desired products and preventing safety hazards such as explosions. As such, measuring and understanding the thermochemistry of chemical reactions is not only useful, but essential! Calorimetry The techniques of calorimetry can be used to measure q for a chemical reaction directly. The enthalpy change for a chemical reaction is of significant interest to chemists. An exothermic reaction will release heat (\(q_{reaction} < 0\), \(q_{surroundings} > 0\)) causing the temperature of the surrounding to increase. Conversely, an endothermic reaction (\(q_{reaction} > 0\), \(q_{surroundings} < 0\)) will draw heat from the surroundings, causing the temperature of the surrounding to drop. Measuring the temperature change in the surroundings allows for the determination of how much heat was released or absorbed in the reaction. Bomb Calorimetry Bomb calorimetry is used predominantly to measure the heat evolved in combustion reactions, but can be used for a wide variety of reactions. A typical bomb calorimetry set up is shown here. The reaction is contained in a heavy metallic container (the bomb) forcing the reaction to occur at constant volume. As such, the heat evolved (or absorbed) by the reaction is equal to the change in internal energy ( D U rxn ). The bomb is then submerged in a reproducible quantity of water, the temperature of which is monitored with a high-precision thermometer. For combustion reactions, the bomb will be loaded with a small sample of the compound to be combusted, and then the bomb is filled with a high pressure (typically about 10 atm) of O 2 . The reaction is initiated by supplying heat using a short piece of resistive wire carrying an electrical current. The calorimeter must be calibrated by carrying out a reaction for which \(\Delta U_{rxn}\) is well known, so that the resulting temperature change can be related to the amount of heat released or absorbed. A commonly used reaction is the combustion of benzoic acid. This makes a good choice since benzoic acid reacts reliably and reproducibly under normal bomb calorimetry conditions. The “water equivalent” of the calorimeter can then be calculated from the temperature change using the following relationship: \[W = \dfrac{n\Delta U_c +e_{wrire}+e_{other}}{\Delta T} \nonumber \] where n is the number of moles of benzoic acid used, \(\Delta U_c\) is the internal energy of combustion for benzoic acid (3225.7 kJ mol -1 at 25 o C), \(e_{wire}\) accounts for the energy released in the combustion of the fuse wire, e other account for any other corrections (such as heat released due to the combustion of residual nitrogen in the bomb), and D T is the measured temperature change in the surrounding water bath. Once the “water equivalent” is determined for a calorimeter, the temperature change can be used to find \(\Delta U_c\) for an unknown compound from the temperature change created upon combustion of a known quantity of the substance. \[ \Delta U_c = \dfrac{W \Delta T - e_{wire} - e_{other}}{n_{sample}} \nonumber \] The experiment above is known as “isothermal bomb calorimetry” as the entire assembly sits in a constant temperature laboratory. Another approach is to employ “adiabatic bomb calorimetry” in which the assembly sits inside of a water jacket, the temperature of which is controlled to match the temperature of the water inside the insulated container. By matching this temperature, there is no thermal gradient, and thus no heat leaks into or out of the assembly during an experiment (and hence the experiment is effectively “adiabatic”). Finding \(\Delta U_c\) The enthalpy of combustion can be calculated from the internal energy change if the balanced chemical reaction is known. Recall from the definition of enthalpy \[\Delta H = \Delta U + \Delta (pV) \nonumber \] and if the gas-phase reactants and products can be treated as ideal gases (\(pV = nRT\)) \[\Delta H = \Delta U + RT \Delta n_{gas} \nonumber \] at constant temperature. For the combustion of benzoic acid at 25 o C \[\ce{C6H5COOH (s) + 15/2 O_2(g) -> 7 CO2(g) + 3 H2O(l)} \nonumber \] it can be seen that \(\Delta n_{gas}\) is -0.5 mol of gas for every mole of benzoic acid reacted. Example \(\PageIndex{1}\): Combustion of Naphthalene A student burned a 0.7842 g sample of benzoic acid (\(\ce{C7H6O2}\)) in a bomb calorimeter initially at 25.0 o C and saw a temperature increase of 2.02 o C. She then burned a 0.5348 g sample of naphthalene (\(\ce{C10H8}\)) (again from an initial temperature of 25 o C) and saw a temperature increase of 2.24 o C. From this data, calculate \(\Delta H_c\) for naphthalene (assuming e wire and e other are unimportant.) Solution First, the water equivalent: \[W = \dfrac{\left[ (0.7841\,g) \left(\frac{1\,mol}{122.124 \, g} \right)\right] (3225.7 \,kJ/mol)}{2.02 \,°C} = 10.254 \, kJ/°C \nonumber \] Then \(\Delta U_c\) for the sample: \[\Delta U_c = \dfrac{(10.254\, kJ/\,°C)(2.24\,°C )}{(0.5308 \,g)\left(\frac{1\,mol}{128.174 \, g} \right) } = 5546.4 \, kJ/°C \nonumber \] \(\Delta H_c\) is then given by \[\Delta H_c = \Delta U_c + RT \Delta n_{gas} \nonumber \] The reaction for the combustion of naphthalene at 25 o C is: \[\ce{ C10H8(s) + 12 O2(g) -> 10 CO2(g) + 4 H2O(l)} \nonumber \] with \(\Delta n_{gas} = -2\). So \[ \Delta H_c = 5546.4 \,kJ/mol + \left( \dfrac{8.314}{1000} kJ/(mol \, K) \right) (298 \,L) (-2) = 5541\, kJ/mol \nonumber \] The literature value (Balcan, Arzik, & Altunata, 1996) is 5150.09 kJ/mol. So that’s not too far off! |
Courses/University_of_WisconsinStevens_Point/CHEM_101%3A_Basic_Chemistry_(D'Acchioli)/08%3A_Acids_and_Bases/8.01%3A_Acids_and_Bases_-_Experimental_Definitions | Learning Objectives Examine properties of acids and bases. Many people enjoy drinking coffee. A cup first thing in the morning helps start the day. But keeping the coffee maker clean can be a problem. Lime deposits build up after a while and slow down the brewing process. The best cure for this is to put vinegar (dilute acetic acid) in the pot and run it through the brewing cycle. The vinegar dissolves the deposits and cleans the maker, which will speed up the brewing process back to its original rate. Just be sure to run water through the brewing process after the vinegar, or you will get some really horrible coffee. Perhaps you have eaten too much pizza and felt very uncomfortable hours later. This feeling is due to excess stomach acid being produced. The discomfort can be dealt with by taking an antacid. The base in the antacid will react with the \(\ce{HCl}\) in the stomach and neutralize it, taking care of that unpleasant feeling. Figure \(\PageIndex{1}\) provides some examples of acids and bases. Acids Acids are very common in some of the foods that we eat. Citrus fruits such as oranges and lemons contain citric acid and ascorbic acid, which is better known as vitamin C. Carbonated sodas contain phosphoric acid. Vinegar contains acetic acid. Your own stomach utilizes hydrochloric acid to digest food. Acids are a distinct class of compounds because of the properties of their aqueous solutions as outlined below: Aqueous solutions of acids are electrolytes, meaning that they conduct electrical current. Some acids are strong electrolytes because they ionize completely in water, yielding a great many ions. Other acids are weak electrolytes that exist primarily in a non-ionized form when dissolved in water. Acids have a sour taste. Lemons, vinegar, and sour candies all contain acids. Acids change the color of certain acid-base indicators. Two common indicators are litmus and phenolphthalein. Blue litmus turns red in the presence of an acid, while phenolphthalein turns colorless. Acids are corrosive to certain metals. Acids react with bases to produce a salt compound and water. When equal moles of an acid and a base are combined, the acid is neutralized by the base. The products of this reaction are an ionic compound, which is labeled as a salt, and water. Common examples: Lemons, oranges, vinegar, urine, sulfuric acid, hydrochloric acid Bases Bases have properties that mostly contrast with those of acids. Aqueous solutions of bases are also electrolytes. Bases can be either strong or weak, just as acids can. Bases often have a bitter taste and are found in foods less frequently than acids. Many bases, like soaps, are slippery to the touch. Bases also change the color of indicators. Litmus turns blue in the presence of a base while phenolphthalein turns pink. Bases do not react with metals in the way that acids do. However, they can be corrosive to the skin. Bases react with acids to produce a salt and water. Common Examples: Soap, toothpaste, bleach, cleaning agents, limewater, ammonia water, sodium hydroxide. Warning Tasting chemicals and touching them are NOT good lab practices and should be avoided - in other words, don't do this at home. Summary A brief summary of properties of acids and bases are given. Properties of acids and bases mostly contrast each other. |
Courses/Lansing_Community_College/LCC%3A_Chem_151_-_General_Chemistry_I/Text/07%3A_Electronic_Structure_of_Atoms/7.10%3A_Electronic_Structure_of_Atoms_(Exercises) | 6.1: The Wave Nature of Light Conceptual Problems What are the characteristics of a wave? What is the relationship between electromagnetic radiation and wave energy? At constant wavelength, what effect does increasing the frequency of a wave have on its speed? its amplitude? List the following forms of electromagnetic radiation in order of increasing wavelength: x-rays, radio waves, infrared waves, microwaves, ultraviolet waves, visible waves, and gamma rays. List them in order of increasing frequency. Which has the highest energy? A large industry is centered on developing skin-care products, such as suntan lotions and cosmetics, that cannot be penetrated by ultraviolet radiation. How does the wavelength of visible light compare with the wavelength of ultraviolet light? How does the energy of visible light compare with the energy of ultraviolet light? Why is this industry focused on blocking ultraviolet light rather than visible light? Numerical Problems The human eye is sensitive to what fraction of the electromagnetic spectrum, assuming a typical spectral range of 10 4 to 10 20 Hz? If we came from the planet Krypton and had x-ray vision (i.e., if our eyes were sensitive to x-rays in addition to visible light), how would this fraction be changed? What is the frequency in megahertz corresponding to each wavelength? 755 m 6.73 nm 1.77 × 10 3 km 9.88 Å 3.7 × 10 −10 m What is the frequency in megahertz corresponding to each wavelength? 5.8 × 10 −7 m 2.3 Å 8.6 × 10 7 m 6.2 mm 3.7 nm Line spectra are also observed for molecular species. Given the following characteristic wavelengths for each species, identify the spectral region (ultraviolet, visible, etc.) in which the following line spectra will occur. Given 1.00 mol of each compound and the wavelength of absorbed or emitted light, how much energy does this correspond to? NH 3 , 1.0 × 10 −2 m CH 3 CH 2 OH, 9.0 μm Mo atom, 7.1 Å What is the speed of a wave in meters per second that has a wavelength of 1250 m and a frequency of 2.36 × 10 5 s −1 ? A wave travels at 3.70 m/s with a frequency of 4.599 × 10 7 Hz and an amplitude of 1.0 m. What is its wavelength in nanometers? An AM radio station broadcasts with a wavelength of 248.0 m. What is the broadcast frequency of the station in kilohertz? An AM station has a broadcast range of 92.6 MHz. What is the corresponding wavelength range in meters for this reception? An FM radio station broadcasts with a wavelength of 3.21 m. What is the broadcast frequency of the station in megahertz? An FM radio typically has a broadcast range of 82–112 MHz. What is the corresponding wavelength range in meters for this reception? A microwave oven operates at a frequency of approximately 2450 MHz. What is the corresponding wavelength? Water, with its polar molecules, absorbs electromagnetic radiation primarily in the infrared portion of the spectrum. Given this fact, why are microwave ovens used for cooking food? 6.2: Quantized Energy and Photons Conceptual Problems Describe the relationship between the energy of a photon and its frequency. How was the ultraviolet catastrophe explained? If electromagnetic radiation with a continuous range of frequencies above the threshold frequency of a metal is allowed to strike a metal surface, is the kinetic energy of the ejected electrons continuous or quantized? Explain your answer. The vibrational energy of a plucked guitar string is said to be quantized. What do we mean by this? Are the sounds emitted from the 88 keys on a piano also quantized? Which of the following exhibit quantized behavior: a human voice, the speed of a car, a harp, the colors of light, automobile tire sizes, waves from a speedboat? Conceptual Answers The energy of a photon is directly proportional to the frequency of the electromagnetic radiation. Quantized: harp, tire size, speedboat waves; continuous: human voice, colors of light, car speed. Numerical Problems What is the energy of a photon of light with each wavelength? To which region of the electromagnetic spectrum does each wavelength belong? 4.33 × 10 5 m 0.065 nm 786 pm How much energy is contained in each of the following? To which region of the electromagnetic spectrum does each wavelength belong? 250 photons with a wavelength of 3.0 m 4.2 × 10 6 photons with a wavelength of 92 μm 1.78 × 10 22 photons with a wavelength of 2.1 Å A 6.023 x 10 23 photons are found to have an energy of 225 kJ. What is the wavelength of the radiation? Use the data in Table 2.1.1 to calculate how much more energetic a single gamma-ray photon is than a radio-wave photon. How many photons from a radio source operating at a frequency of 8 × 10 5 Hz would be required to provide the same amount of energy as a single gamma-ray photon with a frequency of 3 × 10 19 Hz? Use the data in Table 2.1.1 to calculate how much more energetic a single x-ray photon is than a photon of ultraviolet light. A radio station has a transmitter that broadcasts at a frequency of 100.7 MHz with a power output of 50 kW. Given that 1 W = 1 J/s, how many photons are emitted by the transmitter each second? Numerical Answers 4.59 × 10 −31 J/photon, radio 3.1 × 10 −15 J/photon, gamma ray 2.53 × 10 −16 J/photon, gamma ray 532 nm 6.3: Line Spectra and the Bohr Model Conceptual Problems Is the spectrum of the light emitted by isolated atoms of an element discrete or continuous? How do these spectra differ from those obtained by heating a bulk sample of a solid element? Explain your answers. Explain why each element has a characteristic emission and absorption spectra. If spectral emissions had been found to be continuous rather than discrete, what would have been the implications for Bohr’s model of the atom? Explain the differences between a ground state and an excited state. Describe what happens in the spectrum of a species when an electron moves from a ground state to an excited state. What happens in the spectrum when the electron falls from an excited state to a ground state? What phenomenon causes a neon sign to have a characteristic color? If the emission spectrum of an element is constant, why do some neon signs have more than one color? How is light from a laser different from the light emitted by a light source such as a light bulb? Describe how a laser produces light. Numerical Problems Using a Bohr model and the transition from n = 2 to n = 3 in an atom with a single electron, describe the mathematical relationship between an emission spectrum and an absorption spectrum. What is the energy of this transition? What does the sign of the energy value represent in this case? What range of light is associated with this transition? If a hydrogen atom is excited from an n = 1 state to an n = 3 state, how much energy does this correspond to? Is this an absorption or an emission? What is the wavelength of the photon involved in this process? To what region of the electromagnetic spectrum does this correspond? The hydrogen atom emits a photon with a 486 nm wavelength, corresponding to an electron decaying from the n = 4 level to which level? What is the color of the emission? An electron in a hydrogen atom can decay from the n = 3 level to n = 2 level. What is the color of the emitted light? What is the energy of this transition? Calculate the wavelength and energy of the photon that gives rise to the third line in order of increasing energy in the Lyman series in the emission spectrum of hydrogen. In what region of the spectrum does this wavelength occur? Describe qualitatively what the absorption spectrum looks like. The wavelength of one of the lines in the Lyman series of hydrogen is 121 nm. In what region of the spectrum does this occur? To which electronic transition does this correspond? The emission spectrum of helium is shown. Estimate what change in energy (Δ E ) gives rise to each line? Removing an electron from solid potassium requires 222 kJ/mol. Would you expect to observe a photoelectric effect for potassium using a photon of blue light (λ = 485 nm)? What is the longest wavelength of energy capable of ejecting an electron from potassium? What is the corresponding color of light of this wavelength? The binding energy of an electron is the energy needed to remove an electron from its lowest energy state. According to Bohr’s postulates, calculate the binding energy of an electron in a hydrogen atom. There are 6.02 x 10 23 atoms in 1g of hydrogen atoms What wavelength in nanometers is required to remove such an electron from one hydrogen atom? As a radio astronomer, you have observed spectral lines for hydrogen corresponding to a state with n = 320, and you would like to produce these lines in the laboratory. Is this feasible? Why or why not? Numerical Answers 656 nm; red light n = 2, blue-green light 97.2 nm, 2.04 × 10 −18 J/photon, ultraviolet light, absorption spectrum is a single dark line at a wavelength of 97.2 nm Violet: 390 nm, 307 kJ/mol photons; Blue-purple: 440 nm, 272 kJ/mol photons; Blue-green: 500 nm, 239 kJ/mol photons; Orange: 580 nm, 206 kJ/mol photons; Red: 650 nm, 184 kJ/mol photons 1313 kJ/mol, λ ≤ 91.1 nm 6.4: The Wave Behavior of Matter Conceptual Problems Explain what is meant by each term and illustrate with a sketch: standing wave fundamental overtone node How does Einstein’s theory of relativity illustrate the wave–particle duality of light? What properties of light can be explained by a wave model? What properties can be explained by a particle model? In the modern theory of the electronic structure of the atom, which of de Broglie’s ideas have been retained? Which proved to be incorrect? According to Bohr, what is the relationship between an atomic orbit and the energy of an electron in that orbit? Is Bohr’s model of the atom consistent with Heisenberg’s uncertainty principle? Explain your answer. The development of ideas frequently builds on the work of predecessors. Complete the following chart by filling in the names of those responsible for each theory shown. Conceptual Answer Numerical Problems How much heat is generated by shining a carbon dioxide laser with a wavelength of 1.065 μm on a 68.95 kg sample of water if 1.000 mol of photons is absorbed and converted to heat? Is this enough heat to raise the temperature of the water by 4°C? Show the mathematical relationship between energy and mass and between wavelength and mass. What is the effect of doubling the mass of an object on its energy? mass of an object on its wavelength? frequency on its mass? What is the de Broglie wavelength of a 39 g bullet traveling at 1020 m/s ± 10 m/s? What is the minimum uncertainty in the bullet’s position? What is the de Broglie wavelength of a 6800 tn aircraft carrier traveling at 18 ± 0.1 knots (1 knot = 1.15 mi/h)? What is the minimum uncertainty in its position? Calculate the mass of a particle if it is traveling at 2.2 × 10 6 m/s and has a frequency of 6.67 × 10 7 Hz. If the uncertainty in the velocity is known to be 0.1%, what is the minimum uncertainty in the position of the particle? Determine the wavelength of a 2800 lb automobile traveling at 80 mi/h ± 3%. How does this compare with the diameter of the nucleus of an atom? You are standing 3 in. from the edge of the highway. What is the minimum uncertainty in the position of the automobile in inches? Numerical Answers E = 112.3 kJ, Δ T = 0.3893°C, over ten times more light is needed for a 4.0°C increase in temperature 1.7 × 10 −35 m, uncertainty in position is ≥ 1.4 × 10 −34 m 9.1 × 10 −39 kg, uncertainty in position ≥ 2.6 m 6.5: Quantum Mechanics and Atomic Orbitals Conceptual Problems Why does an electron in an orbital with n = 1 in a hydrogen atom have lower energy than a free electron ( n = ∞)? What four variables are required to fully describe the position of any object in space? In quantum mechanics, one of these variables is not explicitly considered. Which one and why? Chemists generally refer to the square of the wave function rather than to the wave function itself. Why? Orbital energies of species with only one electron are defined by only one quantum number. Which one? In such a species, is the energy of an orbital with n = 2 greater than, less than, or equal to the energy of an orbital with n = 4? Justify your answer. In each pair of subshells for a hydrogen atom, which has the higher energy? Give the principal and the azimuthal quantum number for each pair. 1 s , 2 p 2 p , 2 s 2 s , 3 s 3 d , 4 s What is the relationship between the energy of an orbital and its average radius? If an electron made a transition from an orbital with an average radius of 846.4 pm to an orbital with an average radius of 476.1 pm, would an emission spectrum or an absorption spectrum be produced? Why? In making a transition from an orbital with a principal quantum number of 4 to an orbital with a principal quantum number of 7, does the electron of a hydrogen atom emit or absorb a photon of energy? What would be the energy of the photon? To what region of the electromagnetic spectrum does this energy correspond? What quantum number defines each of the following? the overall shape of an orbital the orientation of an electron with respect to a magnetic field the orientation of an orbital in space the average energy and distance of an electron from the nucleus In an attempt to explain the properties of the elements, Niels Bohr initially proposed electronic structures for several elements with orbits holding a certain number of electrons, some of which are in the following table: Element Number of Electrons Electrons in orbits with n = Unnamed: 0_level_1 Unnamed: 1_level_1 4 3 2 1 H 1 NaN NaN NaN 1 He 2 NaN NaN NaN 2 Ne 10 NaN NaN 8.0 2 Ar 18 NaN 8.0 8.0 2 Li 3 NaN NaN 1.0 2 Na 11 NaN 1.0 8.0 2 K 19 1.0 8.0 8.0 2 Be 4 NaN NaN 2.0 2 Draw the electron configuration of each atom based only on the information given in the table. What are the differences between Bohr’s initially proposed structures and those accepted today? Using Bohr’s model, what are the implications for the reactivity of each element? Give the actual electron configuration of each element in the table. Numerical Problems How many subshells are possible for n = 3? What are they? How many subshells are possible for n = 5? What are they? What value of l corresponds to a d subshell? How many orbitals are in this subshell? What value of l corresponds to an f subshell? How many orbitals are in this subshell? State the number of orbitals and electrons that can occupy each subshell. 2 s 3 p 4 d 6 f State the number of orbitals and electrons that can occupy each subshell. 1 s 4 p 5 d 4 f How many orbitals and subshells are found within the principal shell n = 6? How do these orbital energies compare with those for n = 4? How many nodes would you expect a 4 p orbital to have? A 5 s orbital? A p orbital is found to have one node in addition to the nodal plane that bisects the lobes. What would you predict to be the value of n ? If an s orbital has two nodes, what is the value of n ? Numerical Answers Three subshells, with l = 0 ( s ), l = 1 ( p ), and l = 2 ( d ). A d subshell has l = 2 and contains 5 orbitals. 2 electrons; 1 orbital 6 electrons; 3 orbitals 10 electrons; 5 orbitals 14 electrons; 7 orbitals A principal shell with n = 6 contains six subshells, with l = 0, 1, 2, 3, 4, and 5, respectively. These subshells contain 1, 3, 5, 7, 9, and 11 orbitals, respectively, for a total of 36 orbitals. The energies of the orbitals with n = 6 are higher than those of the corresponding orbitals with the same value of l for n = 4. 6.6: Representation of Orbitals 6.7: Many-Electron Atoms Conceptual Problems A set of four quantum numbers specifies each wave function. What information is given by each quantum number? What does the specified wave function describe? List two pieces of evidence to support the statement that electrons have a spin. The periodic table is divided into blocks. Identify each block and explain the principle behind the divisions. Which quantum number distinguishes the horizontal rows? Identify the element with each ground state electron configuration. [He]2 s 2 2 p 3 [Ar]4 s 2 3 d 1 [Kr]5 s 2 4 d 10 5 p 3 [Xe]6 s 2 4 f 6 Identify the element with each ground state electron configuration. [He]2 s 2 2 p 1 [Ar]4 s 2 3 d 8 [Kr]5 s 2 4 d 10 5 p 4 [Xe]6 s 2 Propose an explanation as to why the noble gases are inert. Numerical Problems How many magnetic quantum numbers are possible for a 4 p subshell? A 3 d subshell? How many orbitals are in these subshells? How many magnetic quantum numbers are possible for a 6 s subshell? A 4 f subshell? How many orbitals does each subshell contain? If l = 2 and m l = 2, give all the allowed combinations of the four quantum numbers ( n , l , m l , m s ) for electrons in the corresponding 3 d subshell. Give all the allowed combinations of the four quantum numbers ( n , l , m l , m s ) for electrons in a 4 d subshell. How many electrons can the 4 d orbital accommodate? How would this differ from a situation in which there were only three quantum numbers ( n , l , m )? Given the following sets of quantum numbers ( n , l , m l , m s ), identify each principal shell and subshell. 1, 0, 0, ½ 2, 1, 0, ½ 3, 2, 0, ½ 4, 3, 3, ½ Is each set of quantum numbers allowed? Explain your answers. n = 2; l = 1; m l = 2; m s = +½ n = 3, l = 0; m l = −1; m s = −½ n = 2; l = 2; m l = 1; m s = +½ n = 3; l = 2; m l = 2; m s = +½ List the set of quantum numbers for each electron in the valence shell of each element. beryllium xenon lithium fluorine List the set of quantum numbers for each electron in the valence shell of each element. carbon magnesium bromine sulfur Sketch the shape of the periodic table if there were three possible values of m s for each electron (+½, −½, and 0); assume that the Pauli principle is still valid. Predict the shape of the periodic table if eight electrons could occupy the p subshell. If the electron could only have spin +½, what would the periodic table look like? If three electrons could occupy each s orbital, what would be the electron configuration of each species? sodium titanium fluorine calcium If Hund’s rule were not followed and maximum pairing occurred, how many unpaired electrons would each species have? How do these numbers compare with the number found using Hund’s rule? phosphorus iodine manganese Write the electron configuration for each element in the ground state. aluminum calcium sulfur tin nickel tungsten neodymium americium Write the electron configuration for each element in the ground state. boron rubidium bromine germanium vanadium palladium bismuth europium Give the complete electron configuration for each element. magnesium potassium titanium selenium iodine uranium germanium Give the complete electron configuration for each element. tin copper fluorine hydrogen thorium yttrium bismuth Write the valence electron configuration for each element: samarium praseodymium boron cobalt Using the Pauli exclusion principle and Hund’s rule, draw valence orbital diagrams for each element. barium neodymium iodine Using the Pauli exclusion principle and Hund’s rule, draw valence orbital diagrams for each element. chlorine silicon scandium How many unpaired electrons does each species contain? lead cesium copper silicon selenium How many unpaired electrons does each species contain? helium oxygen bismuth silver boron For each element, give the complete electron configuration, draw the valence electron configuration, and give the number of unpaired electrons present. lithium magnesium silicon cesium lead Use an orbital diagram to illustrate the aufbau principle, the Pauli exclusion principle, and Hund’s rule for each element. carbon sulfur Numerical Answers For a 4 p subshell, n = 4 and l = 1. The allowed values of the magnetic quantum number, ml , are therefore +1, 0, −1, corresponding to three 4 p orbitals. For a 3 d subshell, n = 3 and l = 2. The allowed values of the magnetic quantum number, ml , are therefore +2, +1, 0, −1, −2, corresponding to five 3 d orbitals. 6.8: Electron Configurations 6.9: Electron Configurations and the Periodic Table |
Courses/University_of_Pittsburgh_at_Bradford/CHEM_0106%3A_Chemistry_of_the_Environment/02%3A_Chemical_Elements/2.19%3A_Hydrogen_and_Alkali_Metals | Can you guess what kind of reaction is taking place in this picture? Some chemistry students just enjoy learning about the science, while others are intrigued by the violent reactions that sometimes can occur. Many chemistry classes have been enlivened by the demonstration of how reactive sodium is with water. In some instances, the demonstration has gone off safely. Unfortunately, in other situations students and instructors have incurred serious injury due to their failure to observe proper safety precautions. One value of the periodic table is the ability to make predictions about the behavior of individual elements. By knowing which group an element is in, we can determine the number of reactive electrons and say something about how that element will behave. Hydrogen and Alkali Metals The periodic table is arranged on the basis of atomic numbers (number of protons in the nucleus). One of the valuable consequences of this arrangement is that we can learn a lot about the electron distribution in these atoms. The colors in the table below indicate the different groupings of atoms based on the location and number of electrons in the atom. If we look at Group I (red column), we see that it is labeled alkali metals. Also note the green H above the alkali metals. All of these elements have a similar configuration of outer-shell electrons (see table below). In each case, there is one electron in the outer orbital and that is an s-orbital electron. Hydrogen is not an alkali metal itself, but has some similar properties due to its simple one proton (located in the nucleus), one electron arrangement. The lone electron exists in a s-orbital around the nucleus. For lithium, there are two 1s electrons in an inner orbit and one 2s electron in the outer orbit. The same pattern holds for sodium and potassium. Element Symbol Electron Configuration hydrogen H \(\1s^1\) lithium Li \(\left[ \ce{He} \right] 2s^1\) sodium Na \(\left[ \ce{Ne} \right] 3s^1\) potassium K \(\left[ \ce{Ar} \right] 4s^1\) rubidium Rb \(\left[ \ce{Kr} \right] 5s^1\) cesium Cs \(\left[ \ce{Xe} \right] 6s^1\) francium Fr \(\left[ \ce{Rn} \right] 7s^1\) Even an atom with a very complex electron composition such as cesium still has the single \(s\) electron in its outer orbital (see figure below). This one electron is very easily removed during chemical reactions. The group I elements react rapidly with oxygen to produce metal oxides. They are very soft metals, which become liquid just above room temperature. Li reacts with water to produce hydrogen gas. Sodium also reacts the same way, just more rapidly. Potassium reacts rapidly with water producing hydrogen gas and heat which ignites the hydrogen gas. Rubidium and cesium react yet more vigorously and explode on contact with water. Summary Group I (alkali metals and hydrogen) elements all have one electron in their outer shell. This electron is in a \(s\) orbital. The Group I metals are all very reactive with water. Review What group are the alkali metals and hydrogen in? What is the outer shell electron configuration in this group? How reactive are the alkali metals with oxygen? How reactive are these metals with water? |
Courses/Roosevelt_University/General_Organic_and_Biochemistry_with_Problems_Case_Studies_and_Activities/09%3A_Acids_and_Bases/9.01%3A_What_are_Acids_and_Bases | Learning Objectives By the end of this section, you will be able to: Identify acids, bases, and conjugate acid-base pairs according to the Brønsted-Lowry definition Write equations for acid and base ionization reactions Use the ion-product constant for water to calculate hydronium and hydroxide ion concentrations Describe the acid-base behavior of amphiprotic substances The acid-base reaction class has been studied for quite some time. In 1680, Robert Boyle reported traits of acid solutions that included their ability to dissolve many substances, to change the colors of certain natural dyes, and to lose these traits after coming in contact with alkali (base) solutions. In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be \(\ce{CO2}\)), and interact with alkalis to form neutral substances. In 1815, Humphry Davy contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Svante Arrhenius defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions. Johannes Brønsted and Thomas Lowry proposed a more general description in 1923 in which acids and bases were defined in terms of the transfer of hydrogen ions, H + . (Note that these hydrogen ions are often referred to simply as protons , since that subatomic particle is the only component of cations derived from the most abundant hydrogen isotope, 1 H.) A compound that donates a proton to another compound is called a Brønsted-Lowry acid , and a compound that accepts a proton is called a Brønsted-Lowry base . An acid-base reaction is, thus, the transfer of a proton from a donor (acid) to an acceptor (base). The concept of conjugate pairs is useful in describing Brønsted-Lowry acid-base reactions (and other reversible reactions, as well). When an acid donates \(\ce{H^{+}}\), the species that remains is called the conjugate base of the acid because it reacts as a proton acceptor in the reverse reaction. Likewise, when a base accepts \(\ce{H^{+}}\), it is converted to its conjugate acid . The reaction between water and ammonia illustrates this idea. In the forward direction, water acts as an acid by donating a proton to ammonia and subsequently becoming a hydroxide ion, \(\ce{OH^{−}}\), the conjugate base of water. The ammonia acts as a base in accepting this proton, becoming an ammonium ion, \(\ce{NH4^{+}}\), the conjugate acid of ammonia. In the reverse direction, a hydroxide ion acts as a base in accepting a proton from ammonium ion, which acts as an acid. The reaction between a Brønsted-Lowry acid and water is called acid ionization . For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions: Base ionization of a species occurs when it accepts protons from water molecules. In the example below, pyridine molecules, \(\ce{C5NH5}\), undergo base ionization when dissolved in water, yielding hydroxide and pyridinium ions: The preceding ionization reactions suggest that water may function as both a base (as in its reaction with hydrogen fluoride) and an acid (as in its reaction with ammonia). Species capable of either donating or accepting protons are called amphiprotic , or more generally, amphoteric , a term that may be used for acids and bases per definitions other than the Brønsted-Lowry one. The equations below show the two possible acid-base reactions for two amphiprotic species, bicarbonate ion and water: \[\ce{HCO3^{–}(aq) + H2O(l) <=> CO3^{2–}(aq) + H3O^{+} (aq)} \nonumber \] \[\ce{HCO3^{-}(aq) + H2O(l) <=> H2CO3(aq) + OH^{-} (aq)} \nonumber \] The first equation represents the reaction of bicarbonate as an acid with water as a base, whereas the second represents reaction of bicarbonate as a base with water as an acid. When bicarbonate is added to water, both these equilibria are established simultaneously and the composition of the resulting solution may be determined through appropriate equilibrium calculations, as described later in this chapter. In the liquid state, molecules of an amphiprotic substance can react with one another as illustrated for water in the equations below: The process in which like molecules react to yield ions is called autoionization . Liquid water undergoes autoionization to a very slight extent; at 25 °C, approximately two out of every billion water molecules are ionized. The extent of the water autoionization process is reflected in the value of its equilibrium constant, the ion-product constant for water, \(K_w\): \[\ce{2H2O(l) <=> H3O^{+}(aq) + OH^{-}(aq)} \label{autoionization} \] with \[K_w=[\ce{H3O^{+}}][\ce{OH^{-}}] \label{Kw} \] The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 °C, \(K_w\) has a value of \(1.0 \times 10^{−14}\). The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at 100 °C, the value for \(K_w\) is about \(5.6 \times 10^{−13}\), roughly 50 times larger than the value at 25 °C. Example \(\PageIndex{1}\): Ion Concentrations in Pure Water What are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25 °C? Solution The autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, [H 3 O + ] = [OH − ] = x . At 25 °C: \[K_{ w }=\left[ H_3 O^{+}\right]\left[ OH^{-}\right]=(x)(x)=x^2=1.0 \times 10^{-14} \nonumber \] So: \[x=\left[ H_3 O^{+}\right]=\left[ OH^{-}\right]=\sqrt{1.0 \times 10^{-14}}=1.0 \times 10^{-7} M \nonumber \] The hydronium ion concentration and the hydroxide ion concentration are the same, \(1.0 \times 10^{−7}\, M\). Exercise \(\PageIndex{1}\) The ion product of water at 80 °C is 2.4 \times 10^{−13}. What are the concentrations of hydronium and hydroxide ions in pure water at 80 °C? Answer \[[\ce{H3O^{+}}] = [\ce{OH^{-}}] = 4.9 \times 10^{−7}\, M \nonumber \] Example \(\PageIndex{2}\): The Inverse Relation between [H 3 O + ] and [OH − ] A solution of an acid in water has a hydronium ion concentration of \(2.0 \times 10^{−6} M\). What is the concentration of hydroxide ion at 25 °C? Solution Use the value of the ion-product constant for water at 25 °C (Equations \ref{autoionization} and \ref{Kw}) to calculate the missing equilibrium concentration. \[\ce{2H2O(l) <=> H3O^{+}(aq) + OH^{-}(aq)} \nonumber \] with \(K_w=[\ce{H3O^{+}}][\ce{OH^{-}}] \nonumber\). Rearrangement of the K w expression shows that [OH − ] is inversely proportional to [H 3 O + ]: \[K_{ w }=\left[\ce{H3O^{+}}\right]\left[\ce{OH^{-}}\right]=\left(2.0 \times 10^{-6}\right)\left(5.0 \times 10^{-9}\right)=1.0 \times 10^{-14} \nonumber \] Compared with pure water, a solution of acid exhibits a higher concentration o f hydronium ions (due to ionization of the acid) and a proportionally lower concentration of hydroxide ions. This may be explained via Le Chatelier’s principle as a left shift in the water autoionization equilibrium resulting from the stress of increased hydronium ion concentration. Substituting the ion concentrations into the K w expression confirms this calculation, resulting in the expected value: \[K_w=[\ce{H3O^{+}}][\ce{OH^{-}}]=(2.0 \times 10^{−6})(5.0 \times 10^{−9})=1.0 \times 10^{−14} \nonumber \] Exercise \(\PageIndex{2}\) What is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M at 25 °C? Answer \[\ce{[H3O+] = 1 \times 10^{−11}\, M} \nonumber \] Example \(\PageIndex{3}\): Representing the Acid-Base Behavior of an Amphoteric Substance Write separate equations representing the reaction of \(\ce{HSO3^{-}}\) as an acid with \(\ce{OH^{-}}\) as a base with \(\ce{HI}\) Solution \(\ce{HSO3^{-}(aq) + OH^{-}(aq) <=> SO3^{2-}(aq)+H2O(l)}\) \(\ce{HSO3^{-}(aq) + HI(aq) <=> H2SO3(aq) + I^{-}(aq)}\) Exercise \(\PageIndex{3}\) Write separate equations representing the reaction of \(\ce{H2PO4^{-}}\) as a base with \(\ce{HBr}\) as an acid with \(\ce{OH^{-}}\) Answer \(\ce{H2PO4^{-}(aq) + HBr(aq) <=> H3PO4(aq) + Br^{-}(aq)}\) \(\ce{H2PO4^{-}(aq) + OH^{-}(aq) <=> HPO4^{2-}(aq) + H2O(l)}\) |
Bookshelves/Organic_Chemistry/Catalytic_Asymmetric_Synthesis_(Punniyamurthy)/10%3A_Organocatalysis/10.01%3A_Chiral_Proline_Based_Reactions | Enantioselective organocatalysis has emerged as a powerful synthetic method complementary to the metal- and enzyme-catalyzed reactions. The low toxicity associated with organocatalysis and operational simplicity makes it an attractive method to synthesize complex structures. Among the organocatalysts, small molecules like chiral proline, chiral thiourea, chiral TADDOL and chiral alkaloids have special reactivity in the asymmetric synthesis. Chiral proline is termed as the simplest bifunctional organocatalysts (Scheme \(\PageIndex{1}\)). This amino acid is called as “simplest enzyme” due to its ability to catalyze reactions with high stereoselectivity. L-Proline is a small molecule, non-toxic, inexpensive, readily available in both enantiomeric forms having bifunctional acid-base sites (Scheme \(\PageIndex{2}\)). The reaction may proceed through either iminium catalysis, or enamine catalysis or bifunctional acid–base catalysis. In the early 1970, the first L-proline-catalyzed aldol cyclization was appeared (Scheme \(\PageIndex{3}\)). After nearly 25 years, the expected transition state for the reaction has been illustrated (Scheme \(\PageIndex{4}\)). Intermolecular Aldol Reaction The enantioselective aldol reaction is one of the most powerful methods for the construction of chiral polyol. The first intermolecular direct enantioselective aldol reaction catalyzed by L-proline appeared employing acetone and 4-nitrobenzaldehyde as the substrates (Scheme \(\PageIndex{5}\)). This result sparked high interest from several groups in further investigating proline-catalyzed direct asymmetric aldol reactions. Subsequently, modified chiral proline derived catalysts L1-3 has been developed to enhance the selectivity of the reaction. For the mechanism, reaction of pyrrolidine with the carbonyl donor can give enamine a that could proceed reaction with the re -face of the aldehydes to give the iminium ion b (Scheme \(\PageIndex{6}\)). The latter can undergo hydrolysis to afford chiral β-hydroxyketone. The proposed transition state illustrates that enamine attack occurs on the re -face of the aldehyde d and e . This facial selectivity of attack by the enamine is dictated by minimizing steric interactions between the aldehyde substituent and the enamine substituent. The attack of the enamine on the si -face of the aldehyde leads to the unfavorable transition state c . Mannich Reaction Parallel to the aldol reaction, enantioselective Mannich reaction of aldehyde, acetone and p -anisidine as the substrates has been explored with 50% yield and 94% ee (Scheme \(\PageIndex{7}\)). The mechanism is analogous to that of the aldol reactions (Scheme \(\PageIndex{8}\)). The reaction of proline with aldehyde or ketone can give enamine that could undergo reaction with the imine to form new stereocenters as iminium product. The latter on hydrolysis can give the target Mannich product. The reaction of ( E )-aldimine with the enamine on its si -face can give the syn product. Because of the re -face is blocked by steric interactions between the aromatic ring of the p -methoxyphenyl group and the ring of proline. The proline-catalyzed Mannich reactions of N -PMP-protected α-imino ethyl glyoxylate with a variety of ketones afford functionalized α-amino acids (Scheme 9). These reactions can generate two adjacent stereogenic centers simultaneously upon C-C bond formation with complete syn -stereocontrol and can be performed in a gram scale with operational simplicity. The proline-catalyzed reaction of N -PMP-protected α-imino ethyl glyoxylate with aliphatic aldehydes provides a general method for synthesis of β-amino and α-amino acid derivatives (Scheme \(\PageIndex{10}\)). The diastereoselectivity depends on the bulkiness of the substituents of the aldehyde donor. In most of cases high syn stereoselectivity can be achieved. The synthesis of chiral quaternary amino acid derivatives can be accomplished using proline based catalysis (Scheme \(\PageIndex{11}\)). The nitrogen is tethered to the α -aryl amine in order to increase the reactivity through ring strain and the products are obtained with high enantioselectivity. ( S )-Proline-catalyzed Mannich-type reaction of aldehydes with α-imino ethyl glyoxylate affords syn -products, while the reaction utilizing (3 R , 5 R )-5-methyl-3-pyrrolidinecarboxylic acid gives anti -selective product (Scheme \(\PageIndex{12}\)). In addition , (R )-3-pyrrolidinecarboxylic acid catalyzes the Mannich-type reactions of ketones with α-imino ethyl glyoxylate to give anti -products, while (S)- proline based reactions give syn -products (Scheme \(\PageIndex{13}\)). Thus, the position of the carboxylic acid group on the pyrrolidine ring directs the stereoselection of the catalyzed reaction providing either syn - or anti -Mannich products. Michael Reaction In 2001, the first example for a direct asymmetric Michael reaction employing an enamine-activated donor appeared. The proline-catalyzed reaction of acetone and cyclopentanone with benzalmalonate and nitrostyrene affords the Michael product with low enantiomeric excess. However, the use of chiral diamine improves the ee significantly with both nitrostyrene and alkylidene malonates as acceptors and ketone donors (Scheme \(\PageIndex{14}\)). Possible stereochemical result has been accounted by assuming acyclic transition states A and B . These Michael reactions constituted the first direct catalytic asymmetric reactions of any type s involving aldehyde donors and encouraged the development of aldehyde-based reactions with a range of electrophiles (Scheme \(\PageIndex{15}\)). The iminium-enamine activation mode can be envisaged to explain the domino oxa-Michael–Michael reaction occurring between 3-methylbut-2-enal and ( E )-2-(2-nitrovinyl)-benzene-1,4-diol upon catalysis with chiral diphenyl prolinol silyl ether, which afford the corresponding enantiopure oxa-Michael–Michael cycloadduct in 76% yield and 99% ee (Scheme \(\PageIndex{16}\)). The latter can be further implicated in a Michael–aldol sequence through the reaction with crotonaldehyde to afford corresponding hexahydro-6H-benzo-chromene in 74% yield. These two domino reactions have constituted the key steps of the first asymmetric total synthesis of the natural biologically active product (+)-conicol. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/17%3A_Electrochemistry/17.09%3A_Electrochemistry_and_Conductivity | From the considerations we have discussed, it is evident that any electrolytic cell involves a flow of electrons in an external circuit and a flow of ions within the materials comprising the cell. The function of the current collectors is to transfer electrons back and forth between the external circuit and the cell reagents. The measurement of solution conductivity is a useful technique for determining the concentrations and mobilities of ions in solution. Since conductivity measurements involve the passage of electrical current through a liquid medium, the process must involve electrode reactions as well as motion of ions through the liquid. Normally, the electrode reactions are of little concern in conductivity measurements. The applied potential is made large enough to ensure that some electrode reaction occurs. When the liquid medium is water, the electrode reactions are usually the reduction of water at the cathode and its oxidation at the anode. The conductivity attributable to a given ionic species is approximately proportional to its concentration. In the absence of dissolved ions, little current is passed. For aqueous solutions, this just restates the familiar observation that pure water is a poor electrical conductor. When few ions are present, it is not possible to move charge through the cell quickly enough to support a significant current in the external circuit. |
Courses/Smith_College/CHM_222_Chemistry_II%3A_Organic_Chemistry_(2025)/13%3A_Stereochemistry_at_Tetrahedral_Centers/13.01%3A_Chapter_Objectives_and_Introduction | The opposite of chiral is achiral . Achiral objects are superimposable with their mirror images. If the molecules are superimposable, they are identical to each other. For example, two pieces of paper are achiral. In contrast, chiral objects, like our hands, are non-superimposable mirror images of each other. Try to line up your left hand perfectly with your right hand, so that the palms are both facing in the same directions. Spend about a minute doing this. Do you see that they cannot line up exactly? The same thing applies to some molecules. A chiral molecule has a mirror image that cannot line up with it perfectly - the mirror images are non-superimposable. This pair of non-superimposable mirror image molecules are called enantiomers . But why are chiral molecules so interesting? Just like your left hand will not fit properly in your right glove, one of the enantiomers of a molecule may not work the same way in your body, as the other. It turns out that many of the biological molecules such as our DNA, amino acids and sugars, are chiral molecules. |
Courses/Saint_Marys_College_Notre_Dame_IN/CHEM_431%3A_Inorganic_Chemistry_(Haas)/CHEM_431_Readings/11%3A_Ligand_Field_Theory_(LFT)_and_Crystal_Field_Theory_(CFT)_of_Octahedral_Complexes/11.03%3A_Ligand_Field_Stabilisation_Energy_(LFSE) | There is a variation on how to think about d orbital splitting diagrams that can be useful in deciding how the d electrons are configured in transition metal complexes. We can use the relative energy levels of the d orbitals in a given complex to calculate whether the overall energy would be higher or lower in a high-spin vs. a low-spin case, for example. The calculation provides us with a value that is called the ligand field stabilisation energy (LFSE). Although we have been thinking of bonding in transition metal complexes in terms of molecular orbital ideas, ligand field stabilisation energy actually has its roots in a separate approach called crystal field theory. Origin Story: LFSE originally comes from CFT Crystal field theory (CFT) was independently developed around 1930 by German physicist Hans Bethe and American physicist John Hasbrouck van Vleck ; the two later became professors of physics at Cornell and Harvard, respectively. CFT actually pre-dates the molecular orbital approach that we have been using so far, but it reaches similar conclusions about transition metal electron configurations. Both scientists were interested in the magnetic properties of metals and metal salts. They knew these properties were related to unpaired electrons. Compounds with unpaired electrons (paramagnetic) are attracted by magnetic fields, whereas compounds having only paired electrons (diamagnetic) are not. They were interested in the factors that influenced the d electron configuration of transition metal salts, because how the d electrons filled could result in different numbers of unpaired electrons. Those differences influenced how strongly the compounds interacted with magnetic fields. Crystal field theory does not consider any bonding interactions in transition metal compounds. It focuses only on the repulsion between the electrons on an anion and the electrons on a metal cation. Of course, all of these electrons have negative charges. As an anion approaches a metal, the physicists reasoned, both sets of electrons would experience repulsive forces that would increase their energy level. If the metal were surrounded by a spherical field of electrons (physicists like to assume things are spherical because it makes the math easier), then all of the electrons on the metal would be raised equally high in energy by these repulsive forces. But in a metal salt, electrons are not approaching the metal from all directions. The geometry comes in because of the fact that the cations and anions pack together in specific arrays. For example, if a metal cation is sitting in an octahedral hole, electrons would only approach from six directions: at each end of the x, y, and z axes. In that case, only the electrons in d orbitals that were aligned along the x, y and z axes would be raised significantly by repulsion with the approaching electrons on the anions. The electrons in the off-axis orbitals would be lower in energy than they would have been in a perfectly spherical field. Scientists later adapted this idea to octahedral coordination complexes, in which the metal sits amidst six ligands rather than getting packed among six anions. That result leads to a picture that is pretty similar to what we get from molecular orbital theory, at least as far as the d orbitals are concerned. In fact, the true mathematical approach to molecular orbital theory does take these electron-electron repulsions into account, but it also factors in attractions between the electrons of the ligands and the nucleus of the metal (and vice versa). The biggest difference is that molecular orbital theory includes what happens to the kinetic energy of the ligand lone pairs as they become shared with the metal (it goes down; that's a big part of why the bonds form). The calculation of LFSE You don't really have to appreciate where this approach came from to be able to see how it is commonly used. If we take the five d orbitals after they have been placed in an octahedral environment, we see that they are at two different energy levels. The average energy level is not, as it might first appear, halfway between these levels . That's because there are three lower energy level orbitals and two higher energy level orbitals. The average is two-fifths of the way up from the bottom three, or three-fifths of the way down from the top two . That average energy level of the five d orbitals is called the " barycenter "; it is assigned a relative energy of zero. The difference between the top level and the bottom level is the field splitting; for an octahedral complex, this field splitting is given the symbol Δ o ; here, Δ stands for the energy difference and o stands for octahedral. If the barycenter is at an energy level of zero, then the lower orbitals are below zero. How far? They are two-fifths of the total Δ o below the barycenter. Any electrons in those orbitals are -0.4Δ o below 0 in energy. Electrons in the upper orbitals are +0.6Δ o above zero in energy. Example \(d^4\): Suppose there are four electrons. The first three electrons sit in each of the three lower orbitals, the ones we sometimes label the t 2g orbitals. The fourth will either go into the upper level, which we sometimes call the e g level, or else pair up with another electron at the lower level. We can now calculate the energy difference between these two possible cases. We can calculate what is called the ligand field stabilisation energy, LFSE (sometimes called crystal field stabilisation energy, or CFSE). It's just the sum of the energies of each of the electrons. \[LFSE = [(0.6 \times number \: of \: e_{g} \: electrons) - (0.4 \times number \: of \: t_{2g} \: electrons)] \Delta_{o}\] or if that's too much jargon, \(LFSE = [(0.6 \times \: \# upper \: e^{-}) - (0.4 \times \: \# lower \: e^{-})] \Delta_{o}\) For the \(d^4\) high-spin case: \[LFSE = [(0.6 \times 1) - (0.4 \times 3)] \Delta_{o} = [0.6-1.2] \Delta_{o} = -0.6 \Delta_{o}\] For the \(d^4 low-spin case: \[LFSE = [(0.6 \times 0) - (0.4 \times 4)] \Delta_{o} = -1.6 \Delta_{o}\] So far, it certainly seems like the low-spin case is at lower energy and thus more stable. Remember, though, that low-spin configuration requires that we place two electrons together into the same orbital. Pairing electrons in the same orbital is going to cost some energy. But how much? This repulsion between a pair of electrons in one orbital is called the pairing energy (PE). For each pair of electrons that occupy the same orbital, that energy must be added to take that repulsion into account. As a result, a calculation of the overall stabilization energy (SE) includes both the LFSE and the PE. \[SE = LFSE + PE\] If there were two sets of paired electrons, we would add 2PE; if there were three sets of paired electrons, we would add 3PE, and so on. Overall, deciding quantitatively whether a complex will be high spin or low spin is the most useful application of an LFSE calculation. However, you do have to know the values of the parameters (the field splitting and the pairing energy) for a definitive decision. Exercise \(\PageIndex{1}\) Use SE calculations to determine stabilisation in both high spin and low spin cases. Just leave your answer expressed in terms of Δ o and PE. a. Fe +2 b. Co +2 c. Co +3 d. Mn +2 e. Ti +3 Answer a) Fe +2 d 6 low spin SE = [-0.4(6)+ 0.6(0)] Δo + 3PE = [-2.4] Δo + 3PE d 6 high spin SE = [-0.4(4)+ 0.6(2)] Δo + 1PE = [-.4] Δo + 1PE Answer b) Co +2 d 7 low spin SE = [-0.4(6)+ 0.6(1)] Δo + 3PE = [-1.8] Δo + 3PE d 7 high spin SE = [-0.4(5)+ 0.6(2)] Δo + 2PE = [-.8] Δo + 2PE Answer c) Co +3 d 6 so this looks the same as Fe +2 Answer d) Mn +2 d 5 low spin SE = [-0.4(5)+ 0.6(0)] Δo + 2PE = [-2.0] Δo + 2PE d 5 high spin SE = [-0.4(3)+ 0.6(2)] Δo + 0PE = [0] Δo Answer e) Ti +3 d 1 so there is no possibility of low spin or high spin SE depends on the Magnitude of \(\Delta_o\) and total PE Right now, let's think about Δ o . How much energy is it? How big is the gap between the d orbitals? And how do we know? Well, we can easily measure this gap using a simple spectrophotometer. A spectrophotometer measures how much light is absorbed by a sample. Furthermore, it measures what specific colors of light are absorbed by the sample. As it happens, the absorption of ultraviolet and visible light by a material is associated with electrons in that material becoming promoted to a higher energy level. Shine a light on a transition metal complex, and an electron may jump the gap. So all we have to do is shine different colors of monochromatic light on the complex and see whether it absorbs one of the colors. This experiment is quite reproducible. A specific compound will always absorb the same colors of light. That's why specific materials have specific colors; the color we see represents the light that is not absorbed by the material. Different colors of light have different amounts of energy. Blue light has higher energy than orange light, which has higher energy than red light. Max Planck and Albert Einstein worked out that the energy of a photon (a "particle" or "packet" of light) is directly proportional to the frequency at which the photon oscillates or vibrates. Blue light has a higher frequency than red light, so it has higher energy. Planck and Einstein expressed this idea with an equation, E = h ν in which the Greek letter ν, pronounced "noo," stands for the frequency (some people just use f for frequency) and h stands for a sort of conversion factor called Planck's constant. Blue light has too much energy to promote an electron to the next energy level in this case. It would send the electron way past the next level, and quantum mechanics does not allow that sort of thing. Red light does not have enough energy for the electron to get there. Orange light, in this case, is just right. It has exactly the right amount of energy to jump the gap. So, if we know the frequency of orange light, we know how much energy there is in an orange photon, and we know how big the field splitting is between the d orbitals. Historically, it turns out that people have most often described visible light in terms of wavelength rather than frequencies. The wavelength is just the distance from one "peak" to the next as the wave of the photon rolls along. The higher the frequency, the closer these peaks are together, and the shorter the wavelength. Blue light has a shorter wavelength, around 400 nm, than red light, around 700 nm. A nanometer (nm) is 10 -9 meters (m); a meter is around a yard. Orange light has a wavelength of around 600 nm, or 600 x 10 -9 m, or 6 x 10 -7 m. For reasons we won't get into, spectroscopists in the past (people who measure the interaction of light and matter) sometimes preferred to work in centimeters, cm; there are 100 cm in 1 m, so orange light has a wavelength of 6 x 10 -5 cm. Now, because they knew there was an inverse relationship between wavelength and energy (the longer the wavelength, the lower the energy; the shorter the wavelength, the higher the energy), they simply took the reciprocal of the wavelength in centimeters to get a number in cm -1 , which they called wavenumbers. They used this as a unit of energy. An orange photon has an energy of 1/0.00006 cm = 16,000 cm -1 . So the gap, the field splitting, Δ o in this complex is 16,000 cm -1 . Let's go back to our comparison between the high-spin and low-spin case. For high spin: \[LFSE = [(0.6 \times 1) - (0.4 \times 3)] \Delta_{o} = [0.6 -1.2] \Delta_{o} = -0.6 \Delta_{o} = -0.6 \times 16000 cm^{-1} = -9600 cm^{-1}\] For the low-spin case: \[LFSE = [(0.6 \times 0) -(0.4 \times 4)] \Delta_{o} = -1.6 \Delta_{o} = -1.6 \times 16000 cm^{-1} = -25600 cm^{-1}\] These LFSE calculations show that the low-spin case is lower in energy, by 14,000 cm -1 . However, we still need to include the pairing energy. Like the field splitting, the pairing energy varies from one complex to another. 20,000 cm -1 is a ballpark estimate of a typical pairing energy. If we use this average value for PE in the example we were discussing above for the high-spin case: \[SE = LFSE + PE = -9600 + 0 cm^{-1} = -9600 cm^{-1}\] For the low-spin case: \[SE = LFSE + PE = -25600 + 20000 cm^{-1} = -5600 cm^{-1}\] Taking the pairing energy into account in this case suggests that the high-spin case is favored. It cost less energy to jump the gap and put an electron in a high-lying e g orbital than it did to pair electrons in a low-lying t 2g orbital. Reminder: Factors that influence \(\Delta_o\) Let's take a look at some real examples of field splitting values to get an idea of how large they are, and what factors they depend on. The metal influences \(\Delta_o\) For example, we can look at the charge on the metal ion: 0 1 2 3 4 5 6 Table of Field Splitting Values, Δo, in Hexaquo Complexes of Differing Chargesa Table of Field Splitting Values, Δo, in Hexaquo Complexes of Differing Chargesa Table of Field Splitting Values, Δo, in Hexaquo Complexes of Differing Chargesa Table of Field Splitting Values, Δo, in Hexaquo Complexes of Differing Chargesa Table of Field Splitting Values, Δo, in Hexaquo Complexes of Differing Chargesa Table of Field Splitting Values, Δo, in Hexaquo Complexes of Differing Chargesa Table of Field Splitting Values, Δo, in Hexaquo Complexes of Differing Chargesa Metal Complex [Mn(OH2)6]2+ [Mn(OH2)6]3+ [Fe(OH2)6]2+ [Fe(OH2)6]3+ [Co(OH2)6]2+ [Co(OH2)6]3+ Δo (cm-1) 7800 21100 10400 13800 9300 18300 a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. NaN NaN In this case, it looks like increased charge on the metal ion results in an increased field splitting. The same thing happens in all three examples: manganese, iron, and cobalt. As the metal ion becomes more charged, attraction of the d electrons toward the nucleus increases. The lower-lying t 2g level is more strongly attracted to the nucleus because it is a little closer; consequently, it drops a little further than the e g level, and the gap gets bigger. Charge on the metal ion is one of two key factors that influences the size of the field splitting. The other factor is the period or row in the periodic table. 0 1 2 3 Table of Field Splitting Values, Δo, in Hexammine Complexes of Group 9a,b Table of Field Splitting Values, Δo, in Hexammine Complexes of Group 9a,b Table of Field Splitting Values, Δo, in Hexammine Complexes of Group 9a,b Table of Field Splitting Values, Δo, in Hexammine Complexes of Group 9a,b Metal Complex [Co(NH3)6]3+ [Rh(NH3)6]3+ [Ir(NH3)6]3+ Δo (cm-1) 21500 33100 41100 a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. b) Miessler, G. L.; Tarr, D. A. Inorganic Chemistry, 4th Ed. Pearson, 2010. a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. b) Miessler, G. L.; Tarr, D. A. Inorganic Chemistry, 4th Ed. Pearson, 2010. a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. b) Miessler, G. L.; Tarr, D. A. Inorganic Chemistry, 4th Ed. Pearson, 2010. a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. b) Miessler, G. L.; Tarr, D. A. Inorganic Chemistry, 4th Ed. Pearson, 2010. Moving from one row to the next in group 9 of the periodic table, we see that the field splitting increases by about 10,000 cm -1 each row. In general: Δ o increases with charge on the metal Δ o increases with the period in the periodic table The ligand identity influences \(\Delta_o\) The metal ion is not the only factor that affects the field splitting. The ligands also play an important role, as seen in the table below. 0 1 2 3 4 5 6 7 Table of Field Splitting Values, Δo, in Assorted Complexesa Table of Field Splitting Values, Δo, in Assorted Complexesa Table of Field Splitting Values, Δo, in Assorted Complexesa Table of Field Splitting Values, Δo, in Assorted Complexesa Table of Field Splitting Values, Δo, in Assorted Complexesa Table of Field Splitting Values, Δo, in Assorted Complexesa Table of Field Splitting Values, Δo, in Assorted Complexesa Table of Field Splitting Values, Δo, in Assorted Complexesa Metal Complex [CrCl6]3- [CrF6]3- [Fe(OH2)6]2+ [Fe(CN)6]4- [Co(OH2)6]3+ [Co(NH3)6]3+ [Co(CN)6]3- Δo (cm-1) 13300 15300 10400 33200 18300 23000 33700 a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. NaN If we sort these ligands into three broad categories, we can form a trend for these effects. It may be easiest to start with the three entries on the right, involving cobalt. The oxygen donor in water has two lone pairs, whereas the nitrogen donor in ammonia has only one. The extra lone pair allows water to act as a π (pi) donor; the lone pair can donate to an orbital on the metal to form a pi bond. In this case, it looks like the π donor (water) results in a smaller Δ o than the normal σ (sigma) donor (ammonia). The cyanide ligand has a donor atom that is also participating in a π bond within the ligand; there is a triple bond between the C and N of the cyanide. That means there is also an associated antibonding orbital, π*. That antibonding orbital raises the possibility of back-bonding from the metal. The metal can donate into the π* orbital to make a pi bond. Cyanide is a π-acceptor. Its field splitting is much larger than the sigma donor. Those conclusions can be confirmed by looking at the entries for iron in the middle of the table. Cyanide is a π-acceptor whereas water is a π-donor. The field splitting in the aquo complex should be much smaller than the field splitting in the cyano complex, and it is. The two entries for chromium on the left both show halides, which are π-donors. They illustrate a factor that can be used to predict field strength between two ligands from the same group, such as two π-donors or two π-acceptors. In general, the more basic the ligand, the greater the field splitting. Fluoride ion is more basic than chloride ion (because chloride is a more stable anion than fluoride) so it results in a slightly greater value of Δ o . The Pairing Energy In order to thoroughly estimate the stabilisation energy, we also need reliable values for the pairing energy. Pairing energies can be calculated for free metal ions; correction factors can be applied to arrive at a corresponding value in a complex. In general, the values in coordination complexes are somewhat lower than the values in free metal ions. In the table below, we have made a general estimate that the value in the complex is about 20% lower than in the free metal ion. 0 1 2 3 4 5 6 Table of Pairing Energies, PE, in Free Ions, with Estimated PE in Complexes Table of Pairing Energies, PE, in Free Ions, with Estimated PE in Complexes Table of Pairing Energies, PE, in Free Ions, with Estimated PE in Complexes Table of Pairing Energies, PE, in Free Ions, with Estimated PE in Complexes Table of Pairing Energies, PE, in Free Ions, with Estimated PE in Complexes Table of Pairing Energies, PE, in Free Ions, with Estimated PE in Complexes Table of Pairing Energies, PE, in Free Ions, with Estimated PE in Complexes Metal Ion Mn2+ Mn3+ Fe2+ Fe3+ Co3+ Ru3+ PE, free ion (cm-1) 28000 25500 17700 30100 21100 ~15,000b PE, complex (est., cm-1) 20400 22900 14200 24100 16900 ~12,000b a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. b) Estimate. a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. b) Estimate. a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. b) Estimate. a) Holleman, A. F.; Wiberg, E.; Wiberg, N. Inorganic Chemistry, 34th Ed. Academic Press: Berlin, 2001. b) Estimate. NaN NaN NaN If we compare the pairing energies (PE) of the 2+ ions to those of the 3+ ions, we see that PE is always lower in the more highly charged ion. That's the opposite of the trend in Δ o . Pairing energy is largely dependent on the size of the ion. The smaller the ion, the smaller the orbital, and the more repulsion between two electrons in the same orbital. Because a Mn 3+ ion is smaller than a Mn 2+ ion (Coulomb's Law says there is greater attraction for the electrons in the former case, so the ion shrinks), a pair of electrons on Mn 3+ is closer together than a pair of electrons on Mn 2+ . In general, pairing energies also get smaller upon moving down a column in the periodic table. Second row transition metals are larger than first-row transition metals, so the pairing energy is smaller in the second row than in the first. Third row transition metals are about the same size as first row metals, so their pairing energies are similar to those of the second row. That's because of a phenomenon called "the lanthanide contraction": third row transition metals contain an extra set of protons in their nuclei because of the f block elements before them; consequently they are a little smaller than might be expected. Exercise \(\PageIndex{2}\) Compare pairing energies to field splitting values to determine whether the following complexes would be high-spin or low spin. a) Mn(OH 2 ) 6 2+ b) Fe(CN) 6 4- c) Co(NH 3 ) 6 3+ Answer a) Δ o < PE, so high-spin. Answer b) Δ o > PE, so low-spin. Answer c) Δ o > PE, so low-spin. Attribution Chris P Schaller, Ph.D. , (College of Saint Benedict / Saint John's University) Curated or created by Kathryn Haas |
Courses/Portland_Community_College/CH104%3A_Allied_Health_Chemisty_I_(2nd_Edition)/03%3A_Dimensional_Anlaysis_and_Density | 3.1: Problem Solving and Unit Conversions Converting from one unit to another is a particularly important skill in science and engineering. Dimensional analysis (also called the unit factor method) is a technique for making these conversions correctly. Dimensional analysis will be used in later chapters to perform other chemistry calculations besides just unit conversions. 3.2: Multi-Step Conversion Problems Sometimes you will have to perform more than one conversion to obtain the desired unit. 3.3: Units Raised to a Power Conversion factors for area and volume can also be produced by the dimensional analysis method. Just remember that if a quantity is raised to a power both the number and the unit must be raised to that same power. 3.4: Units in the Numerator and the Denominator Some complex units are composed of a unit in the numerator and a unit in the denominator. These include units of speed and mileage. Dimensional analysis can be used to make conversions with these units as well. 3.5: Density Density is a physical property found by dividing the mass of an object by its volume. Regardless of the sample size, density is always constant. 3.6: Temperature Three different scales are commonly used to measure temperature: Fahrenheit (expressed as °F), Celsius (°C), and Kelvin (K). |
Courses/UWMilwaukee/CHE_125%3A_GOB_Introductory_Chemistry/09%3A_Solutions/9.S%3A_Solutions_(Summary) | To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter. A solution is a homogeneous mixture. The major component is the solvent , while the minor component is the solute . Solutions can have any phase; for example, an alloy is a solid solution. Solutes are soluble or insoluble , meaning they dissolve or do not dissolve in a particular solvent. The terms miscible and immiscible , instead of soluble and insoluble, are used for liquid solutes and solvents. The statement like dissolves like is a useful guide to predicting whether a solute will dissolve in a given solvent. The amount of solute in a solution is represented by the concentration of the solution. The maximum amount of solute that will dissolve in a given amount of solvent is called the solubility of the solute. Such solutions are saturated . Solutions that have less than the maximum amount are unsaturated . Most solutions are unsaturated, and there are various ways of stating their concentrations. Mass/mass percent , volume/volume percent , and mass/volume percent indicate the percentage of the overall solution that is solute. Parts per million (ppm) and parts per billion (ppb) are used to describe very small concentrations of a solute. Molarity , defined as the number of moles of solute per liter of solution, is a common concentration unit in the chemistry laboratory. Equivalents express concentrations in terms of moles of charge on ions. When a solution is diluted, we use the fact that the amount of solute remains constant to be able to determine the volume or concentration of the final diluted solution. Dissolving occurs by solvation , the process in which particles of a solvent surround the individual particles of a solute, separating them to make a solution. For water solutions, the word hydration is used. If the solute is molecular, it dissolves into individual molecules. If the solute is ionic, the individual ions separate from each other, forming a solution that conducts electricity. Such solutions are called electrolytes . If the dissociation of ions is complete, the solution is a strong electrolyte . If the dissociation is only partial, the solution is a weak electrolyte . Solutions of molecules do not conduct electricity and are called nonelectrolytes . Solutions have properties that differ from those of the pure solvent. Some of these are colligative properties, which are due to the number of solute particles dissolved, not the chemical identity of the solute. Colligative properties include vapor pressure depression , boiling point elevation , freezing point depression , and osmotic pressure . Osmotic pressure is particularly important in biological systems. It is caused by osmosis , the passage of solvents through certain membranes like cell walls. The osmolarity of a solution is the product of a solution’s molarity and the number of particles a solute separates into when it dissolves. Osmosis can be reversed by the application of pressure; this reverse osmosis is used to make fresh water from saltwater in some parts of the world. Because of osmosis, red blood cells placed in hypotonic or hypertonic solutions lose function through either hemolysis or crenation. If they are placed in isotonic solutions, however, the cells are unaffected because osmotic pressure is equal on either side of the cell membrane. |
Courses/University_of_Kansas/CHEM_110%3A_Introductory_Chemistry_(Sharpe_Elles)_SP25/06%3A_Quantities_in_Chemical_Reactions/6.02%3A_Gram-Mole_Conversions | Learning Objectives To convert between mass units and mole units. As we just discussed, molar mass is defined as the mass (in grams) of 1 mole of substance (or Avogadro's number of molecules or formula units). The simplest type of manipulation using molar mass as a conversion factor is a mole-gram conversion (or its reverse, a gram-mole conversion). We also established that 1 mol of Al has a mass of 26.98 g (Example \(\PageIndex{1}\)). Stated mathematically, 1 mol Al = 26.98 g Al We can divide both sides of this expression by either side to get one of two possible conversion factors: \[\mathrm{\dfrac{1\: mol\: Al}{26.98\: g\: Al}\quad and \quad \dfrac{26.98\: g\: Al}{1\: mol\: Al}} \nonumber\] The first conversion factor can be used to convert from mass to moles, and the second converts from moles to mass. Both can be used to solve problems that would be hard to do “by eye.” Example \(\PageIndex{1}\) What is the mass of 3.987 mol of Al? Solution The first step in a conversion problem is to decide what conversion factor to use. Because we are starting with mole units, we want a conversion factor that will cancel the mole unit and introduce the unit for mass in the numerator. Therefore, we should use the \(\mathrm{\dfrac{26.98\: g\: Al}{1\: mol\: Al}}\) conversion factor. We start with the given quantity and multiply by the conversion factor: \(\mathrm{3.987\: mol\: Al\times\dfrac{26.98\: g\: Al}{1\: mol\: Al}}\) Note that the mol units cancel algebraically. (The quantity 3.987 mol is understood to be in the numerator of a fraction that has 1 in the unwritten denominator.) Canceling and solving gives \(\mathrm{3.987\: mol\: Al\times \dfrac{26.98\: g\: Al}{1\: mol\: Al}=107.6\: g\: Al}\) Our final answer is expressed to four significant figures. Exercise \(\PageIndex{1}\) How many moles are present in 100.0 g of Al? (Hint: you will have to use the other conversion factor we obtained for aluminum.) Answer \(\mathrm{100.0\: g\: Al\times \dfrac{1\: mol\: Al}{26.98\: g\: Al}=3.706\: mol\: Al}\) Conversions like this are possible for any substance, as long as the proper atomic mass, formula mass, or molar mass is known (or can be determined) and expressed in grams per mole. Figure \(\PageIndex{1}\) is a chart for determining what conversion factor is needed, and Figure \(\PageIndex{2}\) is a flow diagram for the steps needed to perform a conversion. Example \(\PageIndex{2}\) A biochemist needs 0.00655 mol of bilirubin (C 33 H 36 N 4 O 6 ) for an experiment. How many grams of bilirubin will that be? Solution To convert from moles to mass, we need the molar mass of bilirubin, which we can determine from its chemical formula: 0 1 2 33 C molar mass: 33 × 12.01 g = 396.33 g 36 H molar mass: 36 × 1.01 g = 36.36 g 4 N molar mass: 4 × 14.01 g = 56.04 g 6 O molar mass: 6 × 16.00 g = 96.00 g Total: NaN 584.73 g The molar mass of bilirubin is 584.73 g. Using the relationship 1 mol bilirubin = 584.73 g bilirubin we can construct the appropriate conversion factor for determining how many grams there are in 0.00655 mol. Following the steps from Figure \(\PageIndex{2}\): \(\mathrm{0.00655\: mol\: bilirubin \times \dfrac{584.73\: g\: bilirubin}{mol\: bilirubin}=3.83\: g\: bilirubin}\) The mol bilirubin unit cancels. The biochemist needs 3.83 g of bilirubin. Exercise \(\PageIndex{2}\) A chemist needs 457.8 g of KMnO 4 to make a solution. How many moles of KMnO 4 is that? Answer \(\mathrm{457.8\: g\: KMnO_4\times \dfrac{1\: mol\: KMnO_4}{158.04\: g\: KMnO_4}=2.897\: mol\: KMnO_4}\) To Your Health: Minerals For our bodies to function properly, we need to ingest certain substances from our diets. Among our dietary needs are minerals, the noncarbon elements our body uses for a variety of functions, such developing bone or ensuring proper nerve transmission. The US Department of Agriculture has established some recommendations for the RDIs of various minerals. The accompanying table lists the RDIs for minerals, both in mass and moles, assuming a 2,000-calorie daily diet. Mineral Male (age 19–30 y) Male (age 19–30 y).1 Female (age 19–30 y) Female (age 19–30 y).1 Ca 1,000 mg 0.025 mol 1,000 mg 0.025 mol Cr 35 µg 6.7 × 10−7 mol 25 µg 4.8 × 10−7 mol Cu 900 µg 1.4 × 10−5 mol 900 µg 1.4 × 10−5 mol F 4 mg 2.1 × 10−4 mol 3 mg 1.5 × 10−4 mol I 150 µg 1.2 × 10−6 mol 150 µg 1.2 × 10−6 mol Fe 8 mg 1.4 × 10−4 mol 18 mg 3.2 × 10−4 mol K 3,500 mg 9.0 × 10−2 mol 3,500 mg 9.0 × 10−2 mol Mg 400 mg 1.6 × 10−2 mol 310 mg 1.3 × 10−2 mol Mn 2.3 mg 4.2 × 10−5 mol 1.8 mg 3.3 × 10−5 mol Mo 45 mg 4.7 × 10−7 mol 45 mg 4.7 × 10−7 mol Na 2,400 mg 1.0 × 10−1 mol 2,400 mg 1.0 × 10−1 mol P 700 mg 2.3 × 10−2 mol 700 mg 2.3 × 10−2 mol Se 55 µg 7.0 × 10−7 mol 55 µg 7.0 × 10−7 mol Zn 11 mg 1.7 × 10−4 mol 8 mg 1.2 × 10−4 mol Table \(\PageIndex{1}\) illustrates several things. First, the needs of men and women for some minerals are different. The extreme case is for iron; women need over twice as much as men do. In all other cases where there is a different RDI, men need more than women. Second, the amounts of the various minerals needed on a daily basis vary widely—both on a mass scale and a molar scale. The average person needs 0.1 mol of Na a day, which is about 2.5 g. On the other hand, a person needs only about 25–35 µg of Cr per day, which is under one millionth of a mole. As small as this amount is, a deficiency of chromium in the diet can lead to diabetes-like symptoms or neurological problems, especially in the extremities (hands and feet). For some minerals, the body does not require much to keep itself operating properly. Although a properly balanced diet will provide all the necessary minerals, some people take dietary supplements. However, too much of a good thing, even minerals, is not good. Exposure to too much chromium, for example, causes a skin irritation, and certain forms of chromium are known to cause cancer (as presented in the movie Erin Brockovich ). |
Bookshelves/General_Chemistry/Chemistry_2e_(OpenStax)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.06%3A_Exercises | 1. Write the electron configurations for each of the following elements: Sc Ti Cr Fe Ru 2. Write the electron configurations for each of the following elements and its ions: Ti Ti 2 + Ti 3 + Ti 4 + 3. Write the electron configurations for each of the following elements and its 3+ ions: La Sm Lu 4. Why are the lanthanoid elements not found in nature in their elemental forms? 5. Which of the following elements is most likely to be used to prepare La by the reduction of La 2 O 3 : Al, C, or Fe? Why? 6. Which of the following is the strongest oxidizing agent: or 7. Which of the following elements is most likely to form an oxide with the formula MO 3 : Zr, Nb, or Mo? 8. The following reactions all occur in a blast furnace. Which of these are redox reactions? (f) 9. Why is the formation of slag useful during the smelting of iron? 10. Would you expect an aqueous manganese(VII) oxide solution to have a pH greater or less than 7.0? Justify your answer. 11. Iron(II) can be oxidized to iron(III) by dichromate ion, which is reduced to chromium(III) in acid solution. A 2.5000-g sample of iron ore is dissolved and the iron converted into iron(II). Exactly 19.17 mL of 0.0100 M Na 2 Cr 2 O 7 is required in the titration. What percentage of the ore sample was iron? 12. How many cubic feet of air at a pressure of 760 torr and 0 °C is required per ton of Fe 2 O 3 to convert that Fe 2 O 3 into iron in a blast furnace? For this exercise, assume air is 19% oxygen by volume. 13. Find the potentials of the following electrochemical cell: Cd | Cd 2+ , M = 0.10 ‖ Ni 2+ , M = 0.50 | Ni 14. A 2.5624-g sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g. What was the percent by mass of chloride ion in the original compound? What is the identity of the salt? 15. The standard reduction potential for the reaction is about 1.8 V. The reduction potential for the reaction is +0.1 V. Calculate the cell potentials to show whether the complex ions, [Co(H 2 O) 6 ] 2+ and/or [Co(NH 3 ) 6 ] 2+ , can be oxidized to the corresponding cobalt(III) complex by oxygen. 16. Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.) (f) 17. Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.) (f) 18. Describe the electrolytic process for refining copper. 19. Predict the products of the following reactions and balance the equations. (f) FeCl 3 is added to an aqueous solution of NaOH. 20. What is the gas produced when iron(II) sulfide is treated with a nonoxidizing acid? 21. Predict the products of each of the following reactions and then balance the chemical equations. (f) FeCO 3 is added to a solution of HClO 4 . (g) Fe is heated in air. 22. Balance the following equations by oxidation-reduction methods; note that three elements change oxidation state. 23. Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility. 24. Predict which will be more stable, [CrO 4 ] 2− or [WO 4 ] 2− , and explain. 25. Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula M 3 O 4 are examples of mixed valence compounds in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format MO·M 2 O 3 , to permit estimation of the metal’s two oxidation states.) (i) Cu 2 O 26. Indicate the coordination number for the central metal atom in each of the following coordination compounds: (f) [Fe(en) 2 (CN) 2 ] + (en = ethylenediamine, C 2 H 8 N 2 ) 27. Give the coordination numbers and write the formulas for each of the following, including all isomers where appropriate: (f) hexaamminecobalt(III) hexacyanochromate(III) (g) dibromobis(ethylenediamine) cobalt(III) nitrate 28. Give the coordination number for each metal ion in the following compounds: [Co(CO 3 ) 3 ] 3− (note that CO 3 2− is bidentate in this complex) [Cu(NH 3 ) 4 ] 2+ [Co(NH 3 ) 4 Br 2 ] 2 (SO 4 ) 3 [Pt(NH 3 ) 4 ][PtCl 4 ] [Cr(en) 3 ](NO 3 ) 3 [Pd(NH 3 ) 2 Br 2 ] (square planar) K 3 [Cu(Cl) 5 ] [Zn(NH 3 ) 2 Cl 2 ] 29. Sketch the structures of the following complexes. Indicate any cis , trans , and optical isomers. (f) [Co(C 2 O 4 ) 2 Cl 2 ] 3− (note that is the bidentate oxalate ion, 30. Draw diagrams for any cis , trans , and optical isomers that could exist for the following (en is ethylenediamine): [Co(en) 2 (NO 2 )Cl] + [Co(en) 2 Cl 2 ] + [Pt(NH 3 ) 2 Cl 4 ] [Cr(en) 3 ] 3+ [Pt(NH 3 ) 2 Cl 2 ] 31. Name each of the compounds or ions given in Exercise 19.28, including the oxidation state of the metal. 32. Name each of the compounds or ions given in Exercise 19.30. 33. Specify whether the following complexes have isomers. tetrahedral [Ni(CO) 2 (Cl) 2 ] trigonal bipyramidal [Mn(CO) 4 NO] [Pt(en) 2 Cl 2 ]Cl 2 34. Predict whether the carbonate ligand will coordinate to a metal center as a monodentate, bidentate, or tridentate ligand. 35. Draw the geometric, linkage, and ionization isomers for [CoCl 5 CN][CN]. 36. Determine the number of unpaired electrons expected for [Fe(NO 2 ) 6 ] 3− and for [FeF 6 ] 3− in terms of crystal field theory. 37. Draw the crystal field diagrams for [Fe(NO 2 ) 6 ] 4− and [FeF 6 ] 3− . State whether each complex is high spin or low spin, paramagnetic or diamagnetic, and compare Δ oct to P for each complex. 38. Give the oxidation state of the metal, number of d electrons, and the number of unpaired electrons predicted for [Co(NH 3 ) 6 ]Cl 3 . 39. The solid anhydrous solid CoCl 2 is blue in color. Because it readily absorbs water from the air, it is used as a humidity indicator to monitor if equipment (such as a cell phone) has been exposed to excessive levels of moisture. Predict what product is formed by this reaction, and how many unpaired electrons this complex will have. 40. Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain. 41. How many unpaired electrons are present in each of the following? [CoF 6 ] 3− (high spin) [Mn(CN) 6 ] 3− (low spin) [Mn(CN) 6 ] 4− (low spin) [MnCl 6 ] 4− (high spin) [RhCl 6 ] 3− (low spin) 42. Explain how the diphosphate ion, [O 3 P−O−PO 3 ] 4− , can function as a water softener that prevents the precipitation of Fe 2+ as an insoluble iron salt. 43. For complexes of the same metal ion with no change in oxidation number, the stability increases as the number of electrons in the t 2 g orbitals increases. Which complex in each of the following pairs of complexes is more stable? [Fe(H 2 O) 6 ] 2+ or [Fe(CN) 6 ] 4− [Co(NH 3 ) 6 ] 3+ or [CoF 6 ] 3− [Mn(CN) 6 ] 4− or [MnCl 6 ] 4− 44. Trimethylphosphine, P(CH 3 ) 3 , can act as a ligand by donating the lone pair of electrons on the phosphorus atom. If trimethylphosphine is added to a solution of nickel(II) chloride in acetone, a blue compound that has a molecular mass of approximately 270 g and contains 21.5% Ni, 26.0% Cl, and 52.5% P(CH 3 ) 3 can be isolated. This blue compound does not have any isomeric forms. What are the geometry and molecular formula of the blue compound? 45. Would you expect the complex [Co(en) 3 ]Cl 3 to have any unpaired electrons? Any isomers? 46. Would you expect the Mg 3 [Cr(CN) 6 ] 2 to be diamagnetic or paramagnetic? Explain your reasoning. 47. Would you expect salts of the gold(I) ion, Au + , to be colored? Explain. 48. [CuCl 4 ] 2− is green. [Cu(H 2 O) 6 ] 2+ is blue. Which absorbs higher-energy photons? Which is predicted to have a larger crystal field splitting? |
Courses/Victor_Valley_College/CHEM100_Victor_Valley_College/16%3A_Nucleic_Acids/16.E%3A_Nucleic_Acids_(Exercises) | 19.1: Nucleotides Concept Review Exercises Identify the three molecules needed to form the nucleotides in each nucleic acid. DNA RNA Classify each compound as a pentose sugar, a purine, or a pyrimidine. adenine guanine deoxyribose thymine ribose cytosine Answers nitrogenous base (adenine, guanine, cytosine, and thymine), 2-deoxyribose, and H 3 PO 4 nitrogenous base (adenine, guanine, cytosine, and uracil), ribose, and H 3 PO 4 purine purine pentose sugar pyrimidine pentose sugar pyrimidine Exercises What is the sugar unit in each nucleic acid? RNA DNA Identify the major nitrogenous bases in each nucleic acid. DNA RNA For each structure, circle the sugar unit and identify the nucleotide as a ribonucleotide or a deoxyribonucleotide. For each structure, circle the sugar unit and identify the nucleotide as a ribonucleotide or a deoxyribonucleotide. For each structure, circle the nitrogenous base and identify it as a purine or pyrimidine. For each structure, circle the nitrogenous base and identify it as a purine or pyrimidine. Answers ribose deoxyribose 19.2: Nucleic Acid Structure Concept Review Exercises Name the two kinds of nucleic acids. Which type of nucleic acid stores genetic information in the cell? What are complementary bases? Why is it structurally important that a purine base always pair with a pyrimidine base in the DNA double helix? Answers deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) DNA the specific base pairings in the DNA double helix in which guanine is paired with cytosine and adenine is paired with thymine The width of the DNA double helix is kept at a constant width, rather than narrowing (if two pyrimidines were across from each other) or widening (if two purines were across from each other). Exercises For this short RNA segment, identify the 5′ end and the 3′ end of the molecule. circle the atoms that comprise the backbone of the nucleic acid chain. write the nucleotide sequence of this RNA segment. For this short DNA segment, identify the 5′ end and the 3′ end of the molecule. circle the atoms that comprise the backbone of the nucleic acid chain. write the nucleotide sequence of this DNA segment. Which nitrogenous base in DNA pairs with each nitrogenous base? cytosine adenine guanine thymine Which nitrogenous base in RNA pairs with each nitrogenous base? cytosine adenine guanine thymine How many hydrogen bonds can form between the two strands in the short DNA segment shown below? 5′ ATGCGACTA 3′ 3′ TACGCTGAT 5′ How many hydrogen bonds can form between the two strands in the short DNA segment shown below? 5′ CGATGAGCC 3′ 3′ GCTACTCGG 5′ Answers c. ACU guanine thymine cytosine adenine 22 (2 between each AT base pair and 3 between each GC base pair) 19.3: Replication and Expression of Genetic Information Concept Review Exercises In DNA replication, a parent DNA molecule produces two daughter molecules. What is the fate of each strand of the parent DNA double helix? What is the role of DNA in transcription? What is produced in transcription? Which type of RNA contains the codon? Which type of RNA contains the anticodon? Answers Each strand of the parent DNA double helix remains associated with the newly synthesized DNA strand. DNA serves as a template for the synthesis of an RNA strand (the product of transcription). codon: mRNA; anticodon: tRNA Exercises Describe how replication and transcription are similar. Describe how replication and transcription differ. A portion of the coding strand for a given gene has the sequence 5′‑ATGAGCGACTTTGCGGGATTA‑3′. What is the sequence of complementary template strand? What is the sequence of the mRNA that would be produced during transcription from this segment of DNA? A portion of the coding strand for a given gene has the sequence 5′‑ATGGCAATCCTCAAACGCTGT‑3′. What is the sequence of complementary template strand? What is the sequence of the mRNA that would be produced during transcription from this segment of DNA? Answers Both processes require a template from which a complementary strand is synthesized. 3. 3′‑TACTCGCTGAAACGCCCTAAT‑5′ 5′‑AUGAGCGACUUUGCGGGAUUA‑3′ 19.4: Protein Synthesis and the Genetic Code Concept Review Exercises What are the roles of mRNA and tRNA in protein synthesis? What is the initiation codon? What are the termination codons and how are they recognized? Answers mRNA provides the code that determines the order of amino acids in the protein; tRNA transports the amino acids to the ribosome to incorporate into the growing protein chain. AUG UAA, UAG, and UGA; they are recognized by special proteins called release factors, which signal the end of the translation process. Exercises Write the anticodon on tRNA that would pair with each mRNA codon. 5′‑UUU‑3′ 5′‑CAU‑3′ 5′‑AGC‑3′ 5′‑CCG‑3′ Write the codon on mRNA that would pair with each tRNA anticodon. 5′‑UUG‑3′ 5′‑GAA‑3′ 5′‑UCC‑3′ 5′‑CAC‑3′ The peptide hormone oxytocin contains 9 amino acid units. What is the minimum number of nucleotides needed to code for this peptide? Myoglobin, a protein that stores oxygen in muscle cells, has been purified from a number of organisms. The protein from a sperm whale is composed of 153 amino acid units. What is the minimum number of nucleotides that must be present in the mRNA that codes for this protein? Use Figure \(\PageIndex{3}\) to identify the amino acids carried by each tRNA molecule in Exercise 1. Use Figure \(\PageIndex{3}\) to identify the amino acids carried by each tRNA molecule in Exercise 2. Use Figure \(\PageIndex{3}\) to determine the amino acid sequence produced from this mRNA sequence: 5′‑AUGAGCGACUUUGCGGGAUUA‑3′. Use Figure \(\PageIndex{3}\) to determine the amino acid sequence produced from this mRNA sequence: 5′‑AUGGCAAUCCUCAAACGCUGU‑3′ Answers 3′‑AAA‑5′ 3′‑GUA‑5′ 3′‑UCG‑5′ 3′‑GGC‑5′ 27 nucleotides (3 nucleotides/codon) 1a: phenyalanine; 1b: histidine; 1c: serine; 1d: proline met-ser-asp-phe-ala-gly-leu 19.5: Mutations and Genetic Diseases Concept Review Exercises What effect can UV radiation have on DNA? Is UV radiation an example of a physical mutagen or a chemical mutagen? What causes PKU? How is PKU detected and treated? Answers It can lead to the formation of a covalent bond between two adjacent thymines on a DNA strand, producing a thymine dimer. physical mutagen the absence of the enzyme phenylalanine hydroxylase PKU is diagnosed by assaying a sample of blood or urine for phenylalanine or one of its metabolites; treatment calls for an individual to be placed on a diet containing little or no phenylalanine. Exercises A portion of the coding strand of a gene was found to have the sequence 5′‑ATGAGCGACTTTCGCCCATTA‑3′. A mutation occurred in the gene, making the sequence 5′‑ATGAGCGACCTTCGCCCATTA‑3′. Identify the mutation as a substitution, an insertion, or a deletion. What effect would the mutation have on the amino acid sequence of the protein obtained from this mutated gene (use Figure 19.14)? A portion of the coding strand of a gene was found to have the sequence 5′‑ATGGCAATCCTCAAACGCTGT‑3′. A mutation occurred in the gene, making the sequence 5′‑ATGGCAATCCTCAACGCTGT‑3′. Identify the mutation as a substitution, an insertion, or a deletion. What effect would the mutation have on the amino acid sequence of the protein obtained from this mutated gene (use Figure 19.14)? What is a mutagen? Give two examples of mutagens. For each genetic disease, indicate which enzyme is lacking or defective and the characteristic symptoms of the disease. PKU Tay-Sachs disease Answers substitution Phenylalanine (UUU) would be replaced with leucine (CUU). a chemical or physical agent that can cause a mutation UV radiation and gamma radiation (answers will vary) 19.6: Viruses Questions Describe the general structure of a virus. How does a DNA virus differ from an RNA virus? Why is HIV known as a retrovirus? Describe how a DNA virus invades and destroys a cell. Describe how an RNA virus invades and destroys a cell. How does this differ from a DNA virus? What HIV enzyme does AZT inhibit? What HIV enzyme does raltegravir inhibit? Answers A virus consists of a central core of nucleic acid enclosed in a protective shell of proteins. There may be lipid or carbohydrate molecules on the surface. A DNA virus has DNA as its genetic material, while an RNA virus has RNA as its genetic material. In a cell, a retrovirus synthesizes a DNA copy of its RNA genetic material. The DNA virus enters a host cell and induces the cell to replicate the viral DNA and produce viral proteins. These proteins and DNA assemble into new viruses that are released by the host cell, which may die in the process. - reverse transcriptase - Additional Exercises For this nucleic acid segment, classify this segment as RNA or DNA and justify your choice. determine the sequence of this segment, labeling the 5′ and 3′ ends. For this nucleic acid segment, classify this segment as RNA or DNA and justify your choice. determine the sequence of this segment, labeling the 5′ and 3′ ends. One of the key pieces of information that Watson and Crick used in determining the secondary structure of DNA came from experiments done by E. Chargaff, in which he studied the nucleotide composition of DNA from many different species. Chargaff noted that the molar quantity of A was always approximately equal to the molar quantity of T, and the molar quantity of C was always approximately equal to the molar quantity of G. How were Chargaff’s results explained by the structural model of DNA proposed by Watson and Crick? Suppose Chargaff (see Exercise 3) had used RNA instead of DNA. Would his results have been the same; that is, would the molar quantity of A approximately equal the molar quantity of T? Explain. In the DNA segment 5′‑ATGAGGCATGAGACG‑3′ (coding strand) 3′‑TACTCCGTACTCTGC‑5′ (template strand) What products would be formed from the segment’s replication? Write the mRNA sequence that would be obtained from the segment’s transcription. What is the amino acid sequence of the peptide produced from the mRNA in Exercise 5b? In the DNA segment 5′‑ATGACGGTTTACTAAGCC‑3′ (coding strand) 3′‑TACTGCCAAATGATTCGG‑5′ (template strand) What products would be formed from the segment’s replication? Write the mRNA sequence that would be obtained from the segment’s transcription. What is the amino acid sequence of the peptide produced from the mRNA in Exercise 6b? A hypothetical protein has a molar mass of 23,300 Da. Assume that the average molar mass of an amino acid is 120. How many amino acids are present in this hypothetical protein? What is the minimum number of codons present in the mRNA that codes for this protein? What is the minimum number of nucleotides needed to code for this protein? Bradykinin is a potent peptide hormone composed of nine amino acids that lowers blood pressure. The amino acid sequence for bradykinin is arg-pro-pro-gly-phe-ser-pro-phe-arg. Postulate a base sequence in the mRNA that would direct the synthesis of this hormone. Include an initiation codon and a termination codon. What is the nucleotide sequence of the DNA that codes for this mRNA? A particular DNA coding segment is ACGTTA G CCCCAGCT. Write the sequence of nucleotides in the corresponding mRNA. Determine the amino acid sequence formed from the mRNA in Exercise 9a during translation. What amino acid sequence results from each of the following mutations? replacement of the underlined guanine by adenine insertion of thymine immediately after the underlined guanine deletion of the underlined guanine A particular DNA coding segment is TAC G ACGTAAC A AGC. Write the sequence of nucleotides in the corresponding mRNA. Determine the amino acid sequence formed from the mRNA in Exercise 10a during translation. What amino acid sequence results from each of the following mutations? replacement of the underlined guanine by adenine replacement of the underlined adenine by thymine Two possible point mutations are the substitution of lysine for leucine or the substitution of serine for threonine. Which is likely to be more serious and why? Two possible point mutations are the substitution of valine for leucine or the substitution of glutamic acid for histidine. Which is likely to be more serious and why? Answers RNA; the sugar is ribose, rather than deoxyribose 5′‑GUA‑3′ In the DNA structure, because guanine (G) is always paired with cytosine (C) and adenine (A) is always paired with thymine (T), you would expect to have equal amounts of each. Each strand would be replicated, resulting in two double-stranded segments. 5′‑AUGAGGCAUGAGACG‑3′ met-arg-his-glu-thr 194 194 582 5′‑ACGUUAGCCCCAGCU‑3′ thr-leu-ala-pro-ala thr-leu-thr-pro-ala thr-leu-val-pro-ser thr-leu-pro-gin substitution of lysine for leucine because you are changing from an amino acid with a nonpolar side chain to one that has a positively charged side chain; both serine and threonine, on the other hand, have polar side chains containing the OH group. |
Bookshelves/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/07%3A_Review_Section/7.02%3A_Review-_Photons | 43 Review: Photons D1.3 Blackbody Radiation Have you noticed that some LED light bulbs are a different color from others? Some are “warm” and others are “cool” (see Figure 6). Candlelight is “warmer” (yellower) than fluorescent lights: the latter have a blue cast that can make people’s faces look sickly. These observations are related to blackbody radiation —the continuous spectrum of light emitted by matter as the matter is heated to higher temperatures. Figure 7 shows the spectra , graphs of intensity versus wavelength, for blackbody radiation emitted by matter at several different temperatures. Here, each spectrum depends only on temperature. The wavelength of the maximum in each curve, \[\lambda_{max} \nonumber \] shifts to shorter wavelengths as the temperature increases. This corresponds to a heated metal first becoming red hot, then brighter and white hot as the temperature increases. Exercise 3: Interpreting Blackbody Spectra Additional Practice 1 Query \(\PageIndex{2}\) Additional Practice 2 Query \(\PageIndex{3}\) D1.4 Planck’s Quantum Theory In the late 1800s, physicists derived mathematical expressions for the blackbody curves using well-accepted concepts of mechanics and electromagnetism. They assumed that as the temperature increased the energies of atoms in a metal object would increase as the atoms vibrated more vigorously; these vibrations were assumed to create electromagnetic waves—the blackbody radiation. But no theory based on these ideas was able to predict the shapes of the curves shown in Figure 7. Even worse, the theory predicted that the intensity would become infinitely large for very short wavelengths, an absurd result. In 1900, Max Planck introduced a revolutionary idea, from which he was able to derive a theoretical expression for blackbody spectra that fit the experimental observations within experimental error. Instead of assuming that the vibrating atoms could have a continuous set of energy values, Planck restricted the vibrational energies to discrete values—that is, he assumed that there must be some minimum quantity of energy that could be transferred between vibrating atoms. That quantity of energy is proportional to the frequency of vibration and is called a quantum . \[\text{E}_{\text{quantum}} = h\nu;\;\;\;\;\;\text{because}\;c = \lambda\nu,\;\nu = \dfrac{c}{\lambda},\;\text{E}_{\text{quantum}} = \dfrac{hc}{\lambda} \nonumber \] The proportionality constant “ h ” is now known as Planck’s constant . Its value is very small, 6.626 × 10 −34 J·s (joule-seconds). According to Planck’s theory, electromagnetic radiation occurs in small, indivisible quantities (quanta), just as matter consists of small, chemically indivisible quantities (atoms). Exercise 4: Radiation and Quanta Additional Practice Query \(\PageIndex{5}\) Although Planck had developed a theory of blackbody radiation that worked, he was not satisfied with the assumption of quantized energies for the vibrating atoms. But not for long—a few years later Albert Einstein used the idea of quantization of electromagnetic radiation to explain another puzzling phenomenon: the photoelectric effect, which is a topic for the next section. D2.1 The Photoelectric Effect In your course notebook, make a heading for Photoelectric Effect. After the heading write down what you recall about the photoelectric effect from courses you have already taken. If you remember having a question about the photoelectric effect or if there is anything you remember being puzzled about, write that down as well. We will ask you to refer back to what you have written when you complete this section. Planck’s quantum theory was able to predict accurately the distribution of wavelengths emitted by a blackbody at various temperatures. However, Planck found it difficult to justify his assumption that vibration energies had to be multiples of a minimum energy—a quantum. When Albert Einstein used Planck’s quantum hypothesis to explain a different phenomenon, the photoelectric effect, the validity of quantum theory became clearer. Before doing any calculation or looking at the hint, write in your class notebook an explanation of how you plan to work out the problem. Do all the steps in the calculation in your notebook . Once you have arrived at an answer, then submit your results below and click the “Check” button to see if it is correct. If one or more parts of your answer is incorrect, go over your work in your notebook carefully and check for errors. “Retry” with your new answer. Look at the hint (click on it to expand for view) only after you have attempted to answer the question at least twice. Query \(\PageIndex{6}\) When electromagnetic radiation shines on a metal, such as sodium, electrons can be emitted and an electric current (a flow of electrons) can occur. This is called the photoelectric effect . The effect is complicated: for some wavelengths no electrons are emitted, but at other wavelengths electrons are emitted. Watch this photoelectric effect animation , where sodium is already selected, to answer the questions below. Write down your observations as you watch the animation, and then answer the questions in your course notebook. For which colors of visible light are electrons emitted by sodium? Determine the maximum wavelength at which electrons are emitted. At a wavelength where electrons are emitted, describe the shape of a graph of number of electrons emitted vs. light intensity. Determine the shape of a graph of electron energy vs. light frequency. How does this graph differ from the electrons vs. intensity graph? Based on your observations, draw a rough graph with labeled axes to show how the number of electrons emitted varies with wavelength of light. How would the graph change if the intensity of light increased? If you would like to experiment with the simulation further, download and save the simulation program , go to the location where you saved it, and double-click on the file name (photoelectric_en.jar) to install and run it. You need to have Java installed for this to work. Query \(\PageIndex{7}\) Additional Practice Query \(\PageIndex{8}\) In your course notebook, write a brief summary of the results you obtained from experimenting with the photoelectric effect simulation. Also write a summary of how your experimental results can be interpreted based on the idea that light consists of quanta. Compare what you wrote with the summary below [also available at this link ] and, if necessary, revise what you wrote. Query \(\PageIndex{9}\) Exercise 3: Photoelectric Effect Additional Practice Query \(\PageIndex{11}\) Refer to what you wrote (including things that puzzled you) when you made the heading Photoelectric Effect in your class notebook. Revise what you wrote based on what you have just learned. Write a summary so that it will be a good study aid when you review for an exam. If you still have questions, ask them on Piazza. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/An_Introduction_to_the_Electronic_Structure_of_Atoms_and_Molecules_(Bader)/06%3A_The_Chemical_Bond/6.06%3A_Literature_References | Literature references 0 1 1.0 The electrostatic method used in this book for the interpretation of chemical binding is based on the Hellmann-Feynman theorem. The theorem was proposed independently by both H. Hellmann and R. P. Feynman. Feynman's account of the theorem anticipates many of the applications to chemistry including the electrostatic interpretation of van der Waals forces. R. P. Feynman, Phys. Rev. 56, 340 (1939). 2.0 The wave functions used in the calculation of the density distributions for H2 were determined by G. Das and A. C. Wahl, J. Chem. Phys. 44, 87 (1966). These wave functions include configuration interaction and hence provide suitable descriptions for the H2 systems for large values of the internuclear separation. The wave functions for He2 are from N. R. Kestner, J. Chem. Phys. 48, 252 (1968). |
Courses/Whitworth_University/Science_of_Food_(Russel)/02%3A_Bonding_in_Molecules/2.06%3A_Intermolecular_Interactions | Learning Objectives Define phase . Identify the types of interactions between molecules. A phase is a certain form of matter that includes a specific set of physical properties. That is, the atoms, the molecules, or the ions that make up the phase do so in a consistent manner throughout the phase. Science recognizes three stable phases: the solid phase , in which individual particles can be thought of as in contact and held in place; the l iquid phase , in which individual particles are in contact but moving with respect to each other; and the gas phase , in which individual particles are separated from each other by relatively large distances (see Figure 8.1.1). Not all substances will readily exhibit all phases. For example, carbon dioxide does not exhibit a liquid phase unless the pressure is greater than about six times normal atmospheric pressure. Other substances, especially complex organic molecules, may decompose at higher temperatures, rather than becoming a liquid or a gas. For many substances, there are different arrangements the particles can take in the solid phase, depending on temperature and pressure. Which phase a substance adopts depends on the pressure and the temperature it experiences. Of these two conditions, temperature variations are more obviously related to the phase of a substance. When it is very cold, H 2 O exists in the solid form as ice. When it is warmer, the liquid phase of H 2 O is present. At even higher temperatures, H 2 O boils and becomes steam. Pressure changes can also affect the presence of a particular phase (as we indicated for carbon dioxide), but its effects are less obvious most of the time. We will mostly focus on the temperature effects on phases, mentioning pressure effects only when they are important. Most chemical substances follow the same pattern of phases when going from a low temperature to a high temperature: the solid phase, then the liquid phase, and then the gas phase. However, the temperatures at which these phases are present differ for all substances and can be rather extreme. Table \(\PageIndex{1}\) shows the temperature ranges for solid, liquid, and gas phases for three substances. As you can see, there is extreme variability in the temperature ranges. Substance Solid Phase Below Liquid Phase Above Gas Phase Above hydrogen (H2) −259°C −259°C −253°C water (H2O) 0°C 0°C 100°C sodium chloride (NaCl) 801°C 801°C 1413°C The melting point of a substance is the temperature that separates a solid and a liquid. The boiling point of a substance is the temperature that separates a liquid and a gas. The melting point of a substance is the temperature that separates a solid and a liquid. The boiling point of a substance is the temperature that separates a liquid and a gas. The melting point of a substance is the temperature that separates a solid and a liquid. The boiling point of a substance is the temperature that separates a liquid and a gas. The melting point of a substance is the temperature that separates a solid and a liquid. The boiling point of a substance is the temperature that separates a liquid and a gas. What accounts for this variability? Why do some substances become liquids at very low temperatures, while others require very high temperatures before they become liquids? It all depends on the strength of the intermolecular forces (IMF) between the particles of substances and the kinetic energies (KE) of its molecules. (Although ionic compounds are not composed of discrete molecules, we will still use the term intermolecular to include interactions between the ions in such compounds.) Substances that experience strong intermolecular interactions require higher temperatures to become liquids and, finally, gases. Substances that experience weak intermolecular interactions do not need much energy (as measured by temperature) to become liquids and gases and will exhibit these phases at lower temperatures. Covalent Network Materials Substances with the highest melting and boiling points have covalent network bonding. This type of intermolecular interaction is actually a covalent bond. In these substances, all the atoms in a sample are covalently bonded to one another; in effect, the entire sample is essentially one giant molecule . Many of these substances are solid over a large temperature range because it takes a lot of energy to disrupt all the covalent bonds at once. One example of a substance that shows covalent network bonding is diamond (Figure \(\PageIndex{2}\)). Diamond is composed entirely of carbon atoms, each bonded to four other carbon atoms in a tetrahedral geometry. Melting a covalent network solid is not accomplished by overcoming the relatively weak intermolecular forces. Rather, all of the covalent bonds must be broken, a process that requires extremely high temperatures. Diamond, in fact, does not melt at all. Instead, it vaporizes to a gas at temperatures above 3,500°C. Diamond is extremely hard and is one of the few materials that can cut glass. Ionic Compounds The strongest force between any two particles is the ionic bond , in which two ions of opposing charge are attracted to each other. Thus, ionic interactions between particles are another type of intermolecular interaction. Substances that contain ionic interactions are relatively strongly held together, so these substances typically have high melting and boiling points. Sodium chloride (Figure \(\PageIndex{3}\)) is an example of a substance whose particles experience ionic interactions (Table \(\PageIndex{1}\)). These attractive forces are sometimes referred to as ion-ion interactions. Covalent Molecular Compounds There are two different covalent structures: molecular and network. Covalent network compounds like SiO 2 (quartz) have structures of atoms in a network like diamond described earlier. In this section, we are dealing with the molecular type that contains individual molecules. The bonding between atoms in the individual molecule is covalent but the attractive forces between the molecules are called intermolecular forces (IMF) . In contrast to intramolecular forces (see Figure 8.1.4), such as the covalent bonds that hold atoms together in molecules and polyatomic ions, intermolecular forces hold molecules together in a liquid or solid . Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances without breaking covalent bonds . In this section, we will discuss the three types of IMF in molecular compounds: dipole-dipole, hydrogen bonding and London dispersion forces. Dipole-dipole Intermolecular Forces As discussed in Section 4.4, covalent bond that has an unequal sharing of electrons is called a polar covalent bond. (A covalent bond that has an equal sharing of electrons, as in a covalent bond with the same atom on each side, is called a nonpolar covalent bond.) A molecule with a net unequal distribution of electrons in its covalent bonds is a polar molecule. HF is an example of a polar molecule (see Figure 8.1.5). The charge separation in a polar covalent bond is not as extreme as is found in ionic compounds, but there is a related result: oppositely charged ends of different molecules will attract each other. This type of intermolecular interaction is called a dipole-dipole interaction . Many molecules with polar covalent bonds experience dipole-dipole interactions. The covalent bonds in some molecules are oriented in space in such a way that the bonds in the molecules cancel each other out. The individual bonds are polar, but due to molecular symmetry, the overall molecule is not polar; rather, the molecule is nonpolar . Such molecules experience little or no dipole-dipole interactions. Carbon dioxide (CO 2 ) and carbon tetrachloride (CCl 4 ) are examples of such molecules (Figure \(\PageIndex{6}\)). Recall from the Sections 4.4 and 4.5, on chemical bonding and molecular geometry that polar molecules have a partial positive charge on one side and a partial negative charge on the other side of the molecule—a separation of charge called a dipole . Consider a polar molecule such as hydrogen chloride, HCl. In the HCl molecule, the more electronegative Cl atom bears the partial negative charge, whereas the less electronegative H atom bears the partial positive charge. An attractive force between HCl molecules results from the attraction between the positive end of one HCl molecule and the negative end of another. This attractive force is called a dipole-dipole attraction —the electrostatic force between the partially positive end of one polar molecule and the partially negative end of another, as illustrated in Figure \(\PageIndex{7}\). The effect of a dipole-dipole attraction is apparent when we compare the properties of HCl molecules to nonpolar F 2 molecules. Both HCl and F 2 consist of the same number of atoms and have approximately the same molecular mass. At a temperature of 150 K, molecules of both substances would have the same average kinetic energy. However, the dipole-dipole attractions between HCl molecules are sufficient to cause them to “stick together” to form a liquid, whereas the relatively weaker dispersion forces between nonpolar F 2 molecules are not, and so this substance is gaseous at this temperature. The higher normal boiling point of HCl (188 K) compared to F 2 (85 K) is a reflection of the greater strength of dipole-dipole attractions between HCl molecules, compared to the attractions between nonpolar F 2 molecules. We will often use values such as boiling or freezing points as indicators of the relative strengths of IMFs of attraction present within different substances. Example \(\PageIndex{1}\) Predict which will have the higher boiling point: N 2 or CO. Explain your reasoning. Solution CO and N 2 are both diatomic molecules with masses of about 28 amu, so they experience similar London dispersion forces. Because CO is a polar molecule, it experiences dipole-dipole attractions. Because N 2 is nonpolar, its molecules cannot exhibit dipole-dipole attractions. The dipole-dipole attractions between CO molecules are comparably stronger than the dispersion forces between nonpolar N 2 molecules, so CO is expected to have the higher boiling point. Predict which will have the higher boiling point: \(\ce{ICl}\) or \(\ce{Br2}\). Explain your reasoning. Answer ICl. ICl and Br 2 have similar masses (~160 amu) and therefore experience similar London dispersion forces. ICl is polar and thus also exhibits dipole-dipole attractions; Br 2 is nonpolar and does not. The relatively stronger dipole-dipole attractions require more energy to overcome, so ICl will have the higher boiling point. Hydrogen Bonding Intermolecular Forces Molecules with hydrogen atoms bonded to electronegative atoms such as O, N, and F tend to exhibit unusually strong intermolecular interactions due to a particularly strong type of dipole-dipole attraction called hydrogen bonding . The very large difference in electronegativity between the H atom (2.1) and the atom to which it is bonded (4.0 for an F atom, 3.5 for an O atom, or 3.0 for a N atom), combined with the very small size of a H atom and the relatively small sizes of F, O, or N atoms, leads to highly concentrated partial charges with these atoms. Because the hydrogen atom does not have any electrons other than the ones in the covalent bond, its positively charged nucleus is almost completely exposed, allowing strong attractions to other nearby lone pairs of electrons. Examples of hydrogen bonds include HF⋯HF, H 2 O⋯HOH, and H 3 N⋯HNH 2 , in which the hydrogen bonds are denoted by dots . Figure \(\PageIndex{8}\) illustrates hydrogen bonding between water molecules. The physical properties of water, which has two O–H bonds, are strongly affected by the presence of hydrogen bonding between water molecules. Most molecular compounds that have a mass similar to water are gases at room temperature. However, because of the strong hydrogen bonds, water molecules are able to stay condensed in the liquid state. A hydrogen bond is an intermolecular attractive force in which a hydrogen atom , that is covalently bonded to a small, highly electronegative atom, is attracted to a lone pair of electrons on an atom in a neighboring molecule . Figure \(\PageIndex{9}\) shows how methanol (CH 3 OH) molecules experience hydrogen bonding. Methanol contains both a hydrogen atom attached to O; methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor (lone pair). The hydrogen-bonded structure of methanol is as follows: Despite use of the word “bond,” keep in mind that hydrogen bonds are intermolecular attractive forces, not intramolecular attractive forces (covalent bonds). Hydrogen bonds are much weaker than covalent bonds, but are generally much stronger than other dipole-dipole attractions and dispersion forces. Effect of Hydrogen Bonding on Boiling Points Consider the compounds dimethylether (CH 3 OCH 3 ), ethanol (CH 3 CH 2 OH), and propane (CH 3 CH 2 CH 3 ). Their boiling points, not necessarily in order, are −42.1 °C, −24.8 °C, and 78.4 °C. Match each compound with its boiling point. Explain your reasoning. Solution The shapes of CH 3 OCH 3 , CH 3 CH 2 OH, and CH 3 CH 2 CH 3 are similar, as are their molar masses (46 g/mol, 46 g/mol, and 44 g/mol, respectively), so they will exhibit similar dispersion forces. Since CH 3 CH 2 CH 3 is nonpolar, it may exhibit only dispersion forces. Because CH 3 OCH 3 is polar, it will also experience dipole-dipole attractions. Finally, CH 3 CH 2 OH has an −OH group, and so it will experience the uniquely strong dipole-dipole attraction known as hydrogen bonding. So the ordering in terms of strength of IMFs, and thus boiling points, is CH 3 CH 2 CH 3 < CH 3 OCH 3 < CH 3 CH 2 OH. The boiling point of propane is −42.1 °C, the boiling point of dimethylether is −24.8 °C, and the boiling point of ethanol is 78.5 °C. Exercise \(\PageIndex{2}\) Ethane (CH 3 CH 3 ) has a melting point of −183 °C and a boiling point of −89 °C. Predict the melting and boiling points for methylamine (CH 3 NH 2 ). Explain your reasoning. Answer The melting point and boiling point for methylamine are predicted to be significantly greater than those of ethane. CH 3 CH 3 and CH 3 NH 2 are similar in size and mass, but methylamine possesses an −NH group and therefore may exhibit hydrogen bonding. This greatly increases its IMFs, and therefore its melting and boiling points. It is difficult to predict values, but the known values are a melting point of −93 °C and a boiling point of −6 °C. Hydrogen Bonding and DNA Deoxyribonucleic acid (DNA) is found in every living organism and contains the genetic information that determines the organism’s characteristics, provides the blueprint for making the proteins necessary for life, and serves as a template to pass this information on to the organism’s offspring. A DNA molecule consists of two (anti-)parallel chains of repeating nucleotides, which form its well-known double helical structure, as shown in Figure \(\PageIndex{10}\). Each nucleotide contains a (deoxyribose) sugar bound to a phosphate group on one side, and one of four nitrogenous bases on the other. Two of the bases, cytosine (C) and thymine (T), are single-ringed structures known as pyrimidines. The other two, adenine (A) and guanine (G), are double-ringed structures called purines. These bases form complementary base pairs consisting of one purine and one pyrimidine, with adenine pairing with thymine, and cytosine with guanine. Each base pair is held together by hydrogen bonding. A and T share two hydrogen bonds, C and G share three, and both pairings have a similar shape and structure Figure \(\PageIndex{11}\) The cumulative effect of millions of hydrogen bonds effectively holds the two strands of DNA together. Importantly, the two strands of DNA can relatively easily “unzip” down the middle since hydrogen bonds are relatively weak compared to the covalent bonds that hold the atoms of the individual DNA molecules together. This allows both strands to function as a template for replication. London Dispersion Forces Finally, there are forces between all molecules that are caused by electrons being in different places in a molecule at any one time, which sets up a temporary separation of charge that disappears almost as soon as it appears. These are very weak intermolecular interactions and are called dispersion forces (or London forces) . (An alternate name is London dispersion forces.) Molecules that experience no other type of intermolecular interaction will at least experience dispersion forces. Substances that experience only dispersion forces are typically soft in the solid phase and have relatively low melting points. Examples include waxes , which are long hydrocarbon chains that are solids at room temperature because the molecules have so many electrons. The resulting dispersion forces between these molecules make them assume the solid phase at normal temperatures. Dispersion forces that develop between atoms in different molecules can attract the two molecules to each other. The forces are relatively weak, however, and become significant only when the molecules are very close. Larger and heavier atoms and molecules exhibit stronger dispersion forces than do smaller and lighter atoms and molecules. F 2 and Cl 2 are gases at room temperature (reflecting weaker attractive forces); Br 2 is a liquid, and I 2 is a solid (reflecting stronger attractive forces). Trends in observed melting and boiling points for the halogens clearly demonstrate this effect, as seen in Table 8.1.2. Halogen Molar Mass Atomic Radius Melting Point Boiling Point fluorine, F2 38 g/mol 72 pm 53 K 85 K chlorine, Cl2 71 g/mol 99 pm 172 K 238 K bromine, Br2 160 g/mol 114 pm 266 K 332 K iodine, I2 254 g/mol 133 pm 387 K 457 K astatine, At2 420 g/mol 150 pm 575 K 610 K The increase in melting and boiling points with increasing atomic/molecular size may be rationalized by considering how the strength of dispersion forces is affected by the electronic structure of the atoms or molecules in the substance. In a larger atom, the valence electrons are, on average, farther from the nuclei than in a smaller atom. Thus, they are less tightly held and can more easily form the temporary dipoles that produce the attraction. The measure of how easy or difficult it is for another electrostatic charge (for example, a nearby ion or polar molecule) to distort a molecule’s charge distribution (its electron cloud) is known as polarizability . A molecule that has a charge cloud that is easily distorted is said to be very polarizable and will have large dispersion forces; one with a charge cloud that is difficult to distort is not very polarizable and will have small dispersion forces. Example \(\PageIndex{3}\) Order the following compounds of a group 14 element and hydrogen from lowest to highest boiling point: CH 4 , SiH 4 , GeH 4 , and SnH 4 . Explain your reasoning. Solution Applying the skills acquired in the chapter on chemical bonding and molecular geometry, all of these compounds are predicted to be nonpolar, so they may experience only dispersion forces: the smaller the molecule, the less polarizable and the weaker the dispersion forces; the larger the molecule, the larger the dispersion forces. The molar masses of CH 4 , SiH 4 , GeH 4 , and SnH 4 are approximately 16 g/mol, 32 g/mol, 77 g/mol, and 123 g/mol, respectively. Therefore, CH 4 is expected to have the lowest boiling point and SnH 4 the highest boiling point. The ordering from lowest to highest boiling point is expected to be CH 4 < SiH 4 < GeH 4 < SnH 4 A graph of the actual boiling points of these compounds versus the period of the group 14 elementsshows this prediction to be correct: Order the following hydrocarbons from lowest to highest boiling point: C 2 H 6 , C 3 H 8 , and C 4 H 10 . Answer All of these compounds are nonpolar and only have London dispersion forces: the larger the molecule, the larger the dispersion forces and the higher the boiling point. The ordering from lowest to highest boiling point is therefore C 2 H 6 < C 3 H 8 < C 4 H 10 . Applications: Geckos and Intermolecular Forces Geckos have an amazing ability to adhere to most surfaces. They can quickly run up smooth walls and across ceilings that have no toe-holds, and they do this without having suction cups or a sticky substance on their toes. And while a gecko can lift its feet easily as it walks along a surface, if you attempt to pick it up, it sticks to the surface. How are geckos (as well as spiders and some other insects) able to do this? Although this phenomenon has been investigated for hundreds of years, scientists only recently uncovered the details of the process that allows geckos’ feet to behave this way. Geckos’ toes are covered with hundreds of thousands of tiny hairs known as setae , with each seta, in turn, branching into hundreds of tiny, flat, triangular tips called spatulae . The huge numbers of spatulae on its setae provide a gecko, shown in Figure 8.1.12, with a large total surface area for sticking to a surface. In 2000, Kellar Autumn, who leads a multi-institutional gecko research team, found that geckos adhered equally well to both polar silicon dioxide and nonpolar gallium arsenide. This proved that geckos stick to surfaces because of dispersion forces—weak intermolecular attractions arising from temporary, synchronized charge distributions between adjacent molecules. Although dispersion forces are very weak, the total attraction over millions of spatulae is large enough to support many times the gecko’s weight. In 2014, two scientists developed a model to explain how geckos can rapidly transition from “sticky” to “non-sticky.” Alex Greaney and Congcong Hu at Oregon State University described how geckos can achieve this by changing the angle between their spatulae and the surface. Geckos’ feet, which are normally nonsticky, become sticky when a small shear force is applied. By curling and uncurling their toes, geckos can alternate between sticking and unsticking from a surface, and thus easily move across it. Further investigations may eventually lead to the development of better adhesives and other applications. Boiling Points and Bonding Types In order for a substance to enter the gas phase, its particles must completely overcome the intermolecular forces holding them together. Therefore, a comparison of boiling points is essentially equivalent to comparing the strengths of the attractive intermolecular forces exhibited by the individual molecules. For small molecular compounds, London dispersion forces are the weakest intermolecular forces. Dipole-dipole forces are somewhat stronger, and hydrogen bonding is a particularly strong form of dipole-dipole interaction. However, when the mass of a nonpolar molecule is sufficiently large, its dispersion forces can be stronger than the dipole-dipole forces in a lighter polar molecule. Thus, nonpolar Cl 2 has a higher boiling point than polar HCl. Substance Strongest Intermolecular Force Boiling Point \(\left( ^\text{o} \text{C} \right)\) \(\ce{H_2}\) dispersion -253 \(\ce{Ne}\) dispersion -246 \(\ce{O_2}\) dispersion -183 \(\ce{Cl_2}\) dispersion -34 \(\ce{HCl}\) dipole-dipole -85 \(\ce{HBr}\) dipole-dipole -66 \(\ce{H_2S}\) dipole-dipole -61 \(\ce{NH_3}\) hydrogen bonding -33 \(\ce{HF}\) hydrogen bonding 20 \(\ce{H_2O}\) hydrogen bonding 100 Example \(\PageIndex{4}\): Intermolecular Forces What intermolecular forces besides dispersion forces, if any, exist in each substance? Are any of these substances solids at room temperature? potassium chloride (KCl) ethanol (C 2 H 5 OH) bromine (Br 2 ) Solution Potassium chloride is composed of ions, so the intermolecular interaction in potassium chloride is ionic forces. Because ionic interactions are strong, it might be expected that potassium chloride is a solid at room temperature. Ethanol has a hydrogen atom attached to an oxygen atom, so it would experience hydrogen bonding. If the hydrogen bonding is strong enough, ethanol might be a solid at room temperature, but it is difficult to know for certain. (Ethanol is actually a liquid at room temperature.) Elemental bromine has two bromine atoms covalently bonded to each other. Because the atoms on either side of the covalent bond are the same, the electrons in the covalent bond are shared equally, and the bond is a nonpolar covalent bond. Thus, diatomic bromine does not have any intermolecular forces other than dispersion forces. It is unlikely to be a solid at room temperature unless the dispersion forces are strong enough. Bromine is a liquid at room temperature. Exercise \(\PageIndex{4}\) What intermolecular forces besides dispersion forces, if any, exist in each substance? Are any of these substances solids at room temperature? methylamine (CH 3 NH 2 ) calcium sulfate (CaSO 4 ) carbon monoxide (CO) Answer a. dipole-dipole, hydrogen bonding b. ionic forces (solid at room temperature) c. dipole-dipole Key Takeaways A phase is a form of matter that has the same physical properties throughout. Molecules interact with each other through various forces: dipole-dipole interactions, hydrogen bonding, and dispersion forces. Dipole-dipole attractions result from the electrostatic attraction of the partial negative end of one dipolar molecule for the partial positive end of another. Hydrogen bonds are a special type of dipole-dipole attraction that results when hydrogen is bonded to one of the three most electronegative elements: F, O, or N. The temporary dipole that results from the motion of the electrons in an atom can induce a dipole in an adjacent atom and give rise to the London dispersion force. London forces increase with increasing molecular size. |
Ancillary_Materials/Laboratory_Experiments/Wet_Lab_Experiments/MIT_Labs/Lab_9%3A_Esterfication/9.1%3A_Purpose_of_the_Experiment | I. Purpose of the Experiment This is an integrated experiment, which combines techniques from Organic, Thermodynamics, Physical, and Analytical chemistry. It has been designed to introduce the student to: Organic Synthesis: Synthesizing an ester from an acid catalyzed reaction of a carboxylic acid and alcohol Fundamental Chemical Engineering Separation Process principles on refluxing Two phase solvent extraction using a Separatory Funnel Distillation at atmospheric pressure Precise handling and measurement techniques for volatile organic liquids Correct handling and operation of a Mass Spectrometer & NMR Spectrometer Thermodynamics, solution chemistry, intermolecular bonding, and organic nomenclature |
Courses/Bellarmine_University/BU%3A_Chem_104_(Christianson)/Phase_1%3A_The_Phases_of_Matter/1%3A_Gases/1.3%3A_The_Gas_Laws | ↵ Learning Objectives Identify the mathematical relationships between the various properties of gases Use the ideal gas law, and related gas laws, to compute the values of various gas properties under specified conditions During the seventeenth and especially eighteenth centuries, driven both by a desire to understand nature and a quest to make balloons in which they could fly (Figure \(\PageIndex{1}\)), a number of scientists established the relationships between the macroscopic physical properties of gases, that is, pressure, volume, temperature, and amount of gas. Although their measurements were not precise by today’s standards, they were able to determine the mathematical relationships between pairs of these variables (e.g., pressure and temperature, pressure and volume) that hold for an ideal gas—a hypothetical construct that real gases approximate under certain conditions. Eventually, these individual laws were combined into a single equation—the ideal gas law —that relates gas quantities for gases and is quite accurate for low pressures and moderate temperatures. We will consider the key developments in individual relationships (for pedagogical reasons not quite in historical order), then put them together in the ideal gas law. Pressure and Temperature: Amontons’s Law Imagine filling a rigid container attached to a pressure gauge with gas and then sealing the container so that no gas may escape. If the container is cooled, the gas inside likewise gets colder and its pressure is observed to decrease. Since the container is rigid and tightly sealed, both the volume and number of moles of gas remain constant. If we heat the sphere, the gas inside gets hotter (Figure \(\PageIndex{2}\)) and the pressure increases. This relationship between temperature and pressure is observed for any sample of gas confined to a constant volume. An example of experimental pressure-temperature data is shown for a sample of air under these conditions in Figure \(\PageIndex{3}\). We find that temperature and pressure are linearly related, and if the temperature is on the kelvin scale, then P and T are directly proportional (again, when volume and moles of gas are held constant ); if the temperature on the kelvin scale increases by a certain factor, the gas pressure increases by the same factor. Guillaume Amontons was the first to empirically establish the relationship between the pressure and the temperature of a gas (~1700), and Joseph Louis Gay-Lussac determined the relationship more precisely (~1800). Because of this, the P - T relationship for gases is known as either Amontons’s law or Gay-Lussac’s law . Under either name, it states that the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is held constant . Mathematically, this can be written: \[P∝T\ce{\:or\:}P=\ce{constant}×T\ce{\:or\:}P=k×T \nonumber \] where ∝ means “is proportional to,” and k is a proportionality constant that depends on the identity, amount, and volume of the gas. For a confined, constant volume of gas, the ratio \(\dfrac{P}{T}\) is therefore constant (i.e., \(\dfrac{P}{T}=k\)). If the gas is initially in “Condition 1” (with P = P 1 and T = T 1 ), and then changes to “Condition 2” (with P = P 2 and T = T 2 ), we have that \(\dfrac{P_1}{T_1}=k\) and \(\dfrac{P_2}{T_2}=k\), which reduces to \(\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\). This equation is useful for pressure-temperature calculations for a confined gas at constant volume. Note that temperatures must be on the kelvin scale for any gas law calculations (0 on the kelvin scale and the lowest possible temperature is called absolute zero ). (Also note that there are at least three ways we can describe how the pressure of a gas changes as its temperature changes: We can use a table of values, a graph, or a mathematical equation.) Example \(\PageIndex{1}\): Predicting Change in Pressure with Temperature A can of hair spray is used until it is empty except for the propellant, isobutane gas. On the can is the warning “Store only at temperatures below 120 °F (48.8 °C). Do not incinerate.” Why? The gas in the can is initially at 24 °C and 360 kPa, and the can has a volume of 350 mL. If the can is left in a car that reaches 50 °C on a hot day, what is the new pressure in the can? Solution The can contains an amount of isobutane gas at a constant volume, so if the temperature is increased by heating, the pressure will increase proportionately. High temperature could lead to high pressure, causing the can to burst. (Also, isobutane is combustible, so incineration could cause the can to explode.) We are looking for a pressure change due to a temperature change at constant volume, so we will use Amontons’s/Gay-Lussac’s law. Taking P 1 and T 1 as the initial values, T 2 as the temperature where the pressure is unknown and P 2 as the unknown pressure, and converting °C to K, we have: \(\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\textrm{ which means that }\dfrac{360\:\ce{kPa}}{297\:\ce K}=\dfrac{P_2}{323\:\ce K}\) Rearranging and solving gives: \(P_2=\mathrm{\dfrac{360\:kPa×323\cancel{K}}{297\cancel{K}}=390\:kPa}\) Exercise \(\PageIndex{1}\) A sample of nitrogen, N 2 , occupies 45.0 mL at 27 °C and 600 torr. What pressure will it have if cooled to –73 °C while the volume remains constant? Answer 400 torr Volume and Temperature: Charles’s Law If we fill a balloon with air and seal it, the balloon contains a specific amount of air at atmospheric pressure, let’s say 1 atm. If we put the balloon in a refrigerator, the gas inside gets cold and the balloon shrinks (although both the amount of gas and its pressure remain constant). If we make the balloon very cold, it will shrink a great deal, and it expands again when it warms up. These examples of the effect of temperature on the volume of a given amount of a confined gas at constant pressure are true in general: The volume increases as the temperature increases, and decreases as the temperature decreases. Volume-temperature data for a 1-mole sample of methane gas at 1 atm are listed and graphed in Figure \(\PageIndex{2}\). The relationship between the volume and temperature of a given amount of gas at constant pressure is known as Charles’s law in recognition of the French scientist and balloon flight pioneer Jacques Alexandre César Charles. Charles’s law states that the volume of a given amount of gas is directly proportional to its temperature on the kelvin scale when the pressure is held constant . Mathematically, this can be written as: \[VαT\ce{\:or\:}V=\ce{constant}·T\ce{\:or\:}V=k·T \nonumber \] with k being a proportionality constant that depends on the amount and pressure of the gas. For a confined, constant pressure gas sample, \(\dfrac{V}{T}\) is constant (i.e., the ratio = k ), and as seen with the P - T relationship, this leads to another form of Charles’s law: \(\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\). Example \(\PageIndex{2}\): P redicting Change in Volume with Temperature A sample of carbon dioxide, CO 2 , occupies 0.300 L at 10 °C and 750 torr. What volume will the gas have at 30 °C and 750 torr? Solution Because we are looking for the volume change caused by a temperature change at constant pressure, this is a job for Charles’s law. Taking V 1 and T 1 as the initial values, T 2 as the temperature at which the volume is unknown and V 2 as the unknown volume, and converting °C into K we have: \(\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\textrm{ which means that }\dfrac{0.300\:\ce L}{283\:\ce K}=\dfrac{V_2}{303\:\ce K}\) Rearranging and solving gives: \(V_2=\mathrm{\dfrac{0.300\:L×303\cancel{K}}{283\cancel{K}}=0.321\:L}\) This answer supports our expectation from Charles’s law, namely, that raising the gas temperature (from 283 K to 303 K) at a constant pressure will yield an increase in its volume (from 0.300 L to 0.321 L). Exercise \(\PageIndex{2}\) A sample of oxygen, O 2 , occupies 32.2 mL at 30 °C and 452 torr. What volume will it occupy at –70 °C and the same pressure? Answer 21.6 mL Example \(\PageIndex{3}\): Measuring Temperature with a Volume Change Temperature is sometimes measured with a gas thermometer by observing the change in the volume of the gas as the temperature changes at constant pressure. The hydrogen in a particular hydrogen gas thermometer has a volume of 150.0 cm 3 when immersed in a mixture of ice and water (0.00 °C). When immersed in boiling liquid ammonia, the volume of the hydrogen, at the same pressure, is 131.7 cm 3 . Find the temperature of boiling ammonia on the kelvin and Celsius scales. Solution A volume change caused by a temperature change at constant pressure means we should use Charles’s law. Taking V 1 and T 1 as the initial values, T 2 as the temperature at which the volume is unknown and V 2 as the unknown volume, and converting °C into K we have: \[\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\textrm{ which means that }\mathrm{\dfrac{150.0\:cm^3}{273.15\:K}}=\dfrac{131.7\:\ce{cm}^3}{T_2} \nonumber \] Rearrangement gives \(T_2=\mathrm{\dfrac{131.7\cancel{cm}^3×273.15\:K}{150.0\:cm^3}=239.8\:K}\) Subtracting 273.15 from 239.8 K, we find that the temperature of the boiling ammonia on the Celsius scale is –33.4 °C. Exercise \(\PageIndex{3}\) What is the volume of a sample of ethane at 467 K and 1.1 atm if it occupies 405 mL at 298 K and 1.1 atm? Answer 635 mL Volume and Pressure: Boyle’s Law If we partially fill an airtight syringe with air, the syringe contains a specific amount of air at constant temperature, say 25 °C. If we slowly push in the plunger while keeping temperature constant, the gas in the syringe is compressed into a smaller volume and its pressure increases; if we pull out the plunger, the volume increases and the pressure decreases. This example of the effect of volume on the pressure of a given amount of a confined gas is true in general. Decreasing the volume of a contained gas will increase its pressure, and increasing its volume will decrease its pressure. In fact, if the volume increases by a certain factor, the pressure decreases by the same factor, and vice versa. Volume-pressure data for an air sample at room temperature are graphed in Figure \(\PageIndex{5}\). Unlike the P-T and V-T relationships, pressure and volume are not directly proportional to each other. Instead, \(P\) and \(V\) exhibit inverse proportionality: Increasing the pressure results in a decrease of the volume of the gas. Mathematically this can be written: \[P \propto \dfrac{1}{V} \nonumber \] or \[P=k⋅ \dfrac{1}{V} \nonumber \] or \[PV=k \nonumber \] or \[P_1V_1=P_2V_2 \nonumber \] with \(k\) being a constant. Graphically, this relationship is shown by the straight line that results when plotting the inverse of the pressure \(\left(\dfrac{1}{P}\right)\) versus the volume ( V ), or the inverse of volume \(\left(\dfrac{1}{V}\right)\) versus the pressure (\(P\)). Graphs with curved lines are difficult to read accurately at low or high values of the variables, and they are more difficult to use in fitting theoretical equations and parameters to experimental data. For those reasons, scientists often try to find a way to “linearize” their data. If we plot P versus V , we obtain a hyperbola (Figure \(\PageIndex{6}\)). The relationship between the volume and pressure of a given amount of gas at constant temperature was first published by the English natural philosopher Robert Boyle over 300 years ago. It is summarized in the statement now known as Boyle’s law : The volume of a given amount of gas held at constant temperature is inversely proportional to the pressure under which it is measured. Example \(\PageIndex{4}\): Volume of a Gas Sample The sample of gas has a volume of 15.0 mL at a pressure of 13.0 psi. Determine the pressure of the gas at a volume of 7.5 mL, using: the P - V graph in Figure \(\PageIndex{6a}\) the \(\dfrac{1}{P}\) vs. V graph in Figure \(\PageIndex{6b}\) the Boyle’s law equation Comment on the likely accuracy of each method. Solution Estimating from the P - V graph gives a value for P somewhere around 27 psi. Estimating from the \(\dfrac{1}{P}\) versus V graph give a value of about 26 psi. From Boyle’s law, we know that the product of pressure and volume ( PV ) for a given sample of gas at a constant temperature is always equal to the same value. Therefore we have P 1 V 1 = k and P 2 V 2 = k which means that P 1 V 1 = P 2 V 2 . Using P 1 and V 1 as the known values 13.0 psi and 15.0 mL, P 2 as the pressure at which the volume is unknown, and V 2 as the unknown volume, we have: \[P_1V_1=P_2V_2\mathrm{\:or\:13.0\:psi×15.0\:mL}=P_2×7.5\:\ce{mL} \nonumber \] Solving: \[P_2=\mathrm{\dfrac{13.0\:psi×15.0\cancel{mL}}{7.5\cancel{mL}}=26\:psi} \nonumber \] It was more difficult to estimate well from the P - V graph, so (a) is likely more inaccurate than (b) or (c). The calculation will be as accurate as the equation and measurements allow. Exercise \(\PageIndex{4}\) The sample of gas has a volume of 30.0 mL at a pressure of 6.5 psi. Determine the volume of the gas at a pressure of 11.0 psi, using: the P - V graph in Figure \(\PageIndex{6a}\) the \(\dfrac{1}{P}\) vs. V graph in Figure \(\PageIndex{6b}\) the Boyle’s law equation Comment on the likely accuracy of each method. Answer a about 17–18 mL Answer b ~18 mL Answer c 17.7 mL; it was more difficult to estimate well from the P - V graph, so (a) is likely more inaccurate than (b); the calculation will be as accurate as the equation and measurements allow Breathing and Boyle’s Law What do you do about 20 times per minute for your whole life, without break, and often without even being aware of it? The answer, of course, is respiration, or breathing. How does it work? It turns out that the gas laws apply here. Your lungs take in gas that your body needs (oxygen) and get rid of waste gas (carbon dioxide). Lungs are made of spongy, stretchy tissue that expands and contracts while you breathe. When you inhale, your diaphragm and intercostal muscles (the muscles between your ribs) contract, expanding your chest cavity and making your lung volume larger. The increase in volume leads to a decrease in pressure (Boyle’s law). This causes air to flow into the lungs (from high pressure to low pressure). When you exhale, the process reverses: Your diaphragm and rib muscles relax, your chest cavity contracts, and your lung volume decreases, causing the pressure to increase (Boyle’s law again), and air flows out of the lungs (from high pressure to low pressure). You then breathe in and out again, and again, repeating this Boyle’s law cycle for the rest of your life (Figure \(\PageIndex{7}\)). Moles of Gas and Volume: Avogadro’s Law The Italian scientist Amedeo Avogadro advanced a hypothesis in 1811 to account for the behavior of gases, stating that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain the same number of molecules. Over time, this relationship was supported by many experimental observations as expressed by Avogadro’s law : For a confined gas, the volume (V) and number of moles (n) are directly proportional if the pressure and temperature both remain constant . In equation form, this is written as: \[V∝n\textrm{ or }V=k×n\textrm{ or }\dfrac{V_1}{n_1}=\dfrac{V_2}{n_2} \nonumber \] Mathematical relationships can also be determined for the other variable pairs, such as P versus n , and n versus T . Visit this interactive PhET simulation to investigate the relationships between pressure, volume, temperature, and amount of gas. Use the simulation to examine the effect of changing one parameter on another while holding the other parameters constant (as described in the preceding sections on the various gas laws). The Ideal Gas Law To this point, four separate laws have been discussed that relate pressure, volume, temperature, and the number of moles of the gas: Boyle’s law: PV = constant at constant T and n Amontons’s law: \(\dfrac{P}{T}\) = constant at constant V and n Charles’s law: \(\dfrac{V}{T}\) = constant at constant P and n Avogadro’s law: \(\dfrac{V}{n}\) = constant at constant P and T Combining these four laws yields the ideal gas law , a relation between the pressure, volume, temperature, and number of moles of a gas: \[PV=nRT \nonumber \] where P is the pressure of a gas, V is its volume, n is the number of moles of the gas, T is its temperature on the kelvin scale, and R is a constant called the ideal gas constant or the universal gas constant. The units used to express pressure, volume, and temperature will determine the proper form of the gas constant as required by dimensional analysis, the most commonly encountered values being 0.08206 L atm mol –1 K –1 and 8.3145 kPa L mol –1 K –1 . The ideal gas law is easy to remember and apply in solving problems , as long as you use the proper values and units for the gas constant, R. Gases whose properties of P , V , and T are accurately described by the ideal gas law (or the other gas laws) are said to exhibit ideal behavior or to approximate the traits of an ideal gas . An ideal gas is a hypothetical construct that may be used along with kinetic molecular theory to effectively explain the gas laws as will be described in a later module of this chapter. Although all the calculations presented in this module assume ideal behavior, this assumption is only reasonable for gases under conditions of relatively low pressure and high temperature. In the final module of this chapter, a modified gas law will be introduced that accounts for the non-ideal behavior observed for many gases at relatively high pressures and low temperatures. The ideal gas equation contains five terms, the gas constant R and the variable properties P , V , n , and T . Specifying any four of these terms will permit use of the ideal gas law to calculate the fifth term as demonstrated in the following example exercises. Example \(\PageIndex{5}\): Using the Ideal Gas Law Methane, CH 4 , is being considered for use as an alternative automotive fuel to replace gasoline. One gallon of gasoline could be replaced by 655 g of CH 4 . What is the volume of this much methane at 25 °C and 745 torr? Solution Exercise \(\PageIndex{5}\) Calculate the pressure in bar of 2520 moles of hydrogen gas stored at 27 °C in the 180-L storage tank of a modern hydrogen-powered car. Answer 350 bar If the number of moles of an ideal gas are kept constant under two different sets of conditions, a useful mathematical relationship called the combined gas law is obtained: \(\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\) using units of atm, L, and K. Both sets of conditions are equal to the product of n × R (where n = the number of moles of the gas and R is the ideal gas law constant). Example \(\PageIndex{6}\): Using the Combined Gas Law When filled with air, a typical scuba tank with a volume of 13.2 L has a pressure of 153 atm (Figure \(\PageIndex{8}\)). If the water temperature is 27 °C, how many liters of air will such a tank provide to a diver’s lungs at a depth of approximately 70 feet in the ocean where the pressure is 3.13 atm? Letting 1 represent the air in the scuba tank and 2 represent the air in the lungs, and noting that body temperature (the temperature the air will be in the lungs) is 37 °C, we have: \[\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}⟶\mathrm{\dfrac{(153\:atm)(13.2\:L)}{(300\:K)}=\dfrac{(3.13\:atm)(\mathit{V}_2)}{(310\:K)}} \nonumber \] Solving for V 2 : \[V_2=\mathrm{\dfrac{(153\cancel{atm})(13.2\:L)(310\cancel{K})}{(300\cancel{K})(3.13\cancel{atm})}=667\:L} \nonumber \] (Note: Be advised that this particular example is one in which the assumption of ideal gas behavior is not very reasonable, since it involves gases at relatively high pressures and low temperatures. Despite this limitation, the calculated volume can be viewed as a good “ballpark” estimate.) Exercise \(\PageIndex{6}\) A sample of ammonia is found to occupy 0.250 L under laboratory conditions of 27 °C and 0.850 atm. Find the volume of this sample at 0 °C and 1.00 atm. Answer 0.193 L The Interdependence between Ocean Depth and Pressure in Scuba Diving Whether scuba diving at the Great Barrier Reef in Australia (shown in Figure \(\PageIndex{9}\)) or in the Caribbean, divers must understand how pressure affects a number of issues related to their comfort and safety. Pressure increases with ocean depth, and the pressure changes most rapidly as divers reach the surface. The pressure a diver experiences is the sum of all pressures above the diver (from the water and the air). Most pressure measurements are given in units of atmospheres, expressed as “atmospheres absolute” or ATA in the diving community: Every 33 feet of salt water represents 1 ATA of pressure in addition to 1 ATA of pressure from the atmosphere at sea level. As a diver descends, the increase in pressure causes the body’s air pockets in the ears and lungs to compress; on the ascent, the decrease in pressure causes these air pockets to expand, potentially rupturing eardrums or bursting the lungs. Divers must therefore undergo equalization by adding air to body airspaces on the descent by breathing normally and adding air to the mask by breathing out of the nose or adding air to the ears and sinuses by equalization techniques; the corollary is also true on ascent, divers must release air from the body to maintain equalization. Buoyancy, or the ability to control whether a diver sinks or floats, is controlled by the buoyancy compensator (BCD). If a diver is ascending, the air in his BCD expands because of lower pressure according to Boyle’s law (decreasing the pressure of gases increases the volume). The expanding air increases the buoyancy of the diver, and she or he begins to ascend. The diver must vent air from the BCD or risk an uncontrolled ascent that could rupture the lungs. In descending, the increased pressure causes the air in the BCD to compress and the diver sinks much more quickly; the diver must add air to the BCD or risk an uncontrolled descent, facing much higher pressures near the ocean floor. The pressure also impacts how long a diver can stay underwater before ascending. The deeper a diver dives, the more compressed the air that is breathed because of increased pressure: If a diver dives 33 feet, the pressure is 2 ATA and the air would be compressed to one-half of its original volume. The diver uses up available air twice as fast as at the surface. Standard Conditions of Temperature and Pressure We have seen that the volume of a given quantity of gas and the number of molecules (moles) in a given volume of gas vary with changes in pressure and temperature. Chemists sometimes make comparisons against a standard temperature and pressure (STP) for reporting properties of gases: 273.15 K (0.00 °C) and 1 atm (101.325 kPa). At STP , an ideal gas has a volume of about 22.4 L—this is referred to as the standard molar volume (Figure \(\PageIndex{10}\)). Summary The behavior of gases can be described by several laws based on experimental observations of their properties. The pressure of a given amount of gas is directly proportional to its absolute temperature, provided that the volume does not change (Amontons’s law). The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure (Charles’s law). The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle’s law). Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules (Avogadro’s law). The equations describing these laws are special cases of the ideal gas law, PV = nRT , where P is the pressure of the gas, V is its volume, n is the number of moles of the gas, T is its kelvin temperature, and R is the ideal (universal) gas constant. Key Equations PV = nRT Summary absolute zero temperature at which the volume of a gas would be zero according to Charles’s law. Amontons’s law (also, Gay-Lussac’s law) pressure of a given number of moles of gas is directly proportional to its kelvin temperature when the volume is held constant Avogadro’s law volume of a gas at constant temperature and pressure is proportional to the number of gas molecules Boyle’s law volume of a given number of moles of gas held at constant temperature is inversely proportional to the pressure under which it is measured Charles’s law volume of a given number of moles of gas is directly proportional to its kelvin temperature when the pressure is held constant ideal gas hypothetical gas whose physical properties are perfectly described by the gas laws ideal gas constant ( R ) constant derived from the ideal gas equation R = 0.08226 L atm mol –1 K –1 or 8.314 L kPa mol –1 K –1 ideal gas law relation between the pressure, volume, amount, and temperature of a gas under conditions derived by combination of the simple gas laws standard conditions of temperature and pressure (STP) 273.15 K (0 °C) and 1 atm (101.325 kPa) standard molar volume volume of 1 mole of gas at STP, approximately 22.4 L for gases behaving ideally |
Bookshelves/General_Chemistry/Chem1_(Lower)/09%3A_Chemical_Bonding_and_Molecular_Structure/9.01%3A_Three_Views_of_Chemical_Bonding | 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. Classical models By classical, we mean models that do not take into account the quantum behavior of small particles, notably the electron. These models generally assume that electrons and ions behave as point charges which attract and repel according to the laws of electrostatics. Although this completely ignores what has been learned about the nature of the electron since the development of quantum theory in the 1920's, these classical models have not only proven extremely useful, but the major ones also serve as the basis for the chemist's general classification of compounds into "covalent" and "ionic" categories. Electrostatic (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. Shared-Electrons (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. Quantum Models Quantum models of bonding take into account the fact that a particle as light a the electron cannot really be said to be in any single location. The best we can do is define a region of space in which the probability of finding the electron has some arbitrary value which will always be less than unity. The shape of this volume of space is called an orbital and is defined by a mathematical function that relates the probability to the (x,y,z) coordinates of the molecule. Like other models of bonding, the quantum models attempt to show how more electrons can be simultaneously close to more nuclei. Instead of doing so through purely geometrical arguments, they attempt this by predicting the nature of the orbitals which the valence electrons occupy in joined atoms. The hybrid orbital model : This was developed by Linus Pauling in 1931 and was the first quantum-based model of bonding. It is based on the premise that if the atomic s, p, and d orbitals occupied by the valence electrons of adjacent atoms are combined in a suitable way, the hybrid orbitals that result will have the character and directional properties that are consistent with the bonding pattern in the molecule. The rules for bringing about these combinations turn out to be remarkably simple, so once they were worked out it became possible to use this model to predict the bonding behavior in a wide variety of molecules. The hybrid orbital model is most usefully applied to the p-block elements the first two rows of the periodic table, and is especially important in organic chemistry. The molecular orbital model : This model takes a more fundamental approach by regarding a molecule as a collection of valence electrons and positive cores. Just as the nature of atomic orbitals derives from the spherical symmetry of the atom, so will the properties of these new molecular orbitals be controlled by the interaction of the valence electrons with the multiple positive centers of these atomic cores. These new orbitals, unlike those of the hybrid model, are delocalized ; that is, they do not "belong" to any one atom but extend over the entire region of space that encompasses the bonded atoms. The available (valence) electrons then fill these orbitals from the lowest to the highest, very much as in the Aufbau principle that you learned for working out atomic electron configurations. For small molecules (which are the only ones we will consider here), there are simple rules that govern the way that atomic orbitals transform themselves into molecular orbitals as the separate atoms are brought together. The real power of molecular orbital theory, however, comes from its mathematical formation which lends itself to detailed predictions of bond energies and other properties. The electron-tunneling model : A common theme uniting all of the models we have discussed is that bonding depends on the fall in potential energy that occurs when opposite charges are brought together. In the case of covalent bonds, the shared electron pair acts as a kind of "electron glue" between the joined nuclei. In 1962, however, it was shown that this assumption is not strictly correct, and that instead of being concentrated in the space between the nuclei, the electron orbitals become even more concentrated around the bonded nuclei. At the same time however, they are free to "move" between the two nuclei by a process known as tunneling . This refers to a well-known quantum mechanical effect that allows electrons (or other particles small enough to exhibit wavelike properties) to pass ("tunnel") through a barrier separating two closely adjacent regions of low potential energy. One result of this is that the effective volume of space available to the electron is increased, and according to the uncertainty principle this will reduce the kinetic energy of the electron. The electron-tunneling model According to this model, the bonding electrons act as a kind of fluid that concentrates in the region of each nucleus (lowering the potential energy) and at the same time is able to freely flow between them (reducing the kinetic energy). A summary of the concept, showing its application to a simple molecule, is shown on the next page. Despite its conceptual simplicity and full acknowledgment of the laws of quantum mechanics, this model is not widely known and is rarely taught. Chemical bonding occurs when one or more electrons can be simultaneously close to two nuclei. But how can this be arranged? The conventional picture of the shared electron bond places the bonding electrons in the region between the two nuclei. This makes a nice picture, but it is not consistent with the principle that opposite charges attract. This would imply that the electrons would be "happiest" (at the lowest potential energy) when they are very close to a nucleus, not half a bond-length away from two of them! This plot shows how the potential energy of an electron in the hydrogen atom varies with its distance from the nucleus. Notice how the energy falls without limit as the electron approaches the nucleus, represented here as a proton, \(H^+\). If potential energy were the only consideration, the electron would fall right into the nucleus where its potential energy would be minus infinity. When an electron is added to the proton to make a neutral hydrogen atom, it tries to get as close to the nucleus as possible. The Heisenberg uncertainty principle requires the total energy of the electron energy to increase as the volume of space it occupies diminishes. As the electron gets closer to the nucleus, the nuclear charge confines the electron to such a tiny volume of space that its energy rises, allowing it to "float" slightly away from the nucleus without ever falling into it. The shaded region above shows the range of energies and distances from the nucleus the electron is able to assume within the 1 s orbital. The electron can thus be regarded as a fluid that occupies a vessel whose walls conform to the red potential energy curves shown above. Note that as the potential energy falls, the kinetic energy increases, but only half as fast (this is called the virial theorem ). Thus close to the nucleus, the kinetic energy is large and so is the electron's effective velocity. The top of the shaded area defines the work required to raise its potential energy to zero, thus removing it from the atom; this corresponds, of course, to the ionization energy. The Tunneling Effect A quantum particle can be described by a waveform which is the plot of a mathematical function related to the probability of finding the particle at a given location at any time. If the particle is confined to a box, it turns out that the wave does not fall to zero at the walls of the box, but has a finite probability of being found outside it. This means that a quantum particle is able to penetrate, or "tunnel through" its confining boundaries. This remarkable property is called the tunnel effect . In terms of the electron fluid model introduced above, the fluid is able to "leak out" of the atom if another low-energy location can be found nearby. Electron tunneling in the simplest molecule Suppose we now bring a bare proton up close to a hydrogen atom. Each nucleus has its own potential well, but only that of the hydrogen atom is filled, as indicated by the shading in the leftmost potential well. But the electron fluid is able to tunnel through the potential energy barrier separating the two wells; like any liquid, it will seek a common level in the two sides of the container as shown below. The electron is now "simultaneously close to two nuclei" while never being in between them. Bear in mind that this would be physically impossible for a real liquid composed of real molecules; this is purely a quantum effect that is restricted to a low-mass particle such as the electron. Because the same amount of electron fluid is now shared between the two wells, its level in both is lower. The difference between what it is now and what is was before corresponds to the bond energy of the hydrogen molecule ion. The dihydrogen molecule Now let's make a molecule of dihydrogen . We start with two hydrogen atoms, each with one electron. But there is a problem here: both potential energy wells are already filled with electron fluid; there is no room for any more without pushing the energy way up. But quantum theory again comes to the rescue! If the two electrons have opposite spins, the two fluids are able to interpenetrate each other, very much as two gases are able to occupy the same container. This is depicted by the double shading in the diagram below. When the two hydrogen atoms are within tunneling distance, half of the electron fluid (really the probability of finding the electron) from each well flows into the other well. Because the two fluids are able to interpenetrate, the level is not much different from what it was in the H 2 + ion, but the greater density of the electron-fluid between the two nuclei makes H 2 a strongly bound molecule. So why does dihelium not exist? If we tried to join two helium atoms in this way, we would be in trouble. The electron well of He already contains two electrons of opposite spin. There is no room for more electron fluid (without raising the energy), and thus no way the electrons in either He atom can be simultaneously close to two nuclei. Ionic Bonding Even before G.N.Lewis developed his theory of the shared electron pair bond, it was believed that bonding in many solid salts could be satisfactorily explained on the basis of simple electrostatic forces between the positive and negative ions which are assumed to be the basic molecular units of these compounds. Lewis himself continued to recognize this distinction, which has continued to be a part of the tradition of chemistry; the shared electron pair bond is known as the covalent bond, while the other type is the ionic or electrovalent bond. 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? Bonding in ionic solids 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. Is there such as thing as a "purely" ionic bond? When two elements form an ionic compound, is an electron really lost by one atom and transferred to the other one? In order to deal with this question, consider the data on the ionic solid LiF. The average radius of the neutral Li atom is about 2.52Å. Now if this Li atom reacts with an atom of F to form LiF, what is the average distance between the Li nucleus and the electron it has "lost" to the fluorine atom? The answer is 1.56Å; the electron is now closer to the lithium nucleus than it was in neutral lithium! So the answer to the above question is both yes and no: yes, the electron that was now in the 2 s orbital of Li is now within the grasp of a fluorine 2 p orbital, but no, the electron is now even closer to the Li nucleus than before, so how can it be "lost"? The one thing that is inarguably true about LiF is that there are more electrons closer to positive nuclei than there are in the separated Li and F atoms. But this is just the condition that gives rise to all forms of chemical bonding: Chemical bonds form when electrons can be simultaneously near two or more nuclei It is obvious that the electron-pair bond brings about this situation, and this is the reason for the stability of the covalent bond. What is not so obvious (until you look at the numbers such as were quoted for LiF above) is that the "ionic" bond results in the same condition; even in the most highly ionic compounds, both electrons are close to both nuclei, and the resulting mutual attractions bind the nuclei together. This being the case, is there really any fundamental difference between the ionic and covalent bond? The answer, according to modern chemical thinking is probably "no"; in fact, there is some question as to whether it is realistic to consider that these solids consist of "ions" in the usual sense. The preferred picture that seems to be emerging is one in which the electron orbitals of adjacent atom pairs are simply skewed so as to place more electron density around the "negative" element than around the "positive" one. This being said, it must be reiterated that the ionic model of bonding is a useful one for many purposes, and there is nothing wrong with using the term "ionic bond" to describe the interactions between the atoms in "ionic solids" such as LiF and NaCl. Polar covalence If there is no such thing as a "completely ionic" bond, can we have one that is completely covalent? The answer is yes, if the two nuclei have equal electron attracting powers. This situation is guaranteed to be the case with homonuclear diatomic molecules— molecules consisting of two identical atoms. Thus in Cl 2 , O 2 , and H 2 , electron sharing between the two identical atoms must be exactly even; in such molecules, the center of positive charge corresponds exactly to the center of negative charge: halfway between the two nuclei. Electronegativity This term was introduced earlier in the course to denote the relative electron-attracting power of an atom. The electronegativity is not the same as the electron affinity; the latter measures the amount of energy released when an electron from an external source "falls into" a vacancy within the outermost orbital of the atom to yield an isolated negative ion. The products of bond formation, in contrast, are not ions and they are not isolated; the two nuclei are now drawn closely together by attraction to the region of high electron density between them. Any shift of electron density toward one atom takes place at the energetic expense of stealing it from the other atom. It is important to understand that electronegativity is not a measurable property of an atom in the sense that ionization energies and electron affinity are; electronegativity is a property that an atom displays when it is bonded to another. Any measurement one does make must necessarily depend on the properties of both of the atoms. By convention, electronegativities are measured on a scale on which the highest value, 4.0, is arbitrarily assigned to fluorine. A number of electronegativity scales have been proposed, each based on slightly different criteria. The most well known of these is due to Linus Pauling, and is based on a study of bond energies in a variety of compounds. The periodic trends in electronegativity are about what one would expect; the higher the nuclear charge and the smaller the atom, the more strongly attractive will it be to an outer-shell electron of an atom within binding distance. The division between the metallic and nonmetallic elements is largely that between those that have Pauling electronegativies greater than about 1.7, and those that have smaller electronegativities. The greater the electronegativity difference between two elements A and B , the more polar will be their molecule AB . It is important to point out, however, that the pairs having the greatest electronegativity differences, the alkali halides, are solids order ordinary conditions and exist as molecules only in the rarefied conditions of the gas phase. Even these ionic solids possess a certain amount of covalent character, so, as discussed above, there is no such thing as a "purely ionic" bond. It has become more common to place binary compounds on a scale something like that shown here, in which the degree of shading is a rough indication of the number of compounds at any point on the covalent-ionic scale. Covalent or ionic: a false dichotomy The covalent-ionic continuum described above is certainly an improvement over the old covalent - versus - ionic dichotomy that existed only in the textbook and classroom, but it is still only a one-dimensional view of a multidimensional world, and thus a view that hides more than it reveals. The main thing missing is any allowance for the type of bonding that occurs between more pairs of elements than any other: metallic bonding. Intermetallic compounds are rarely even mentioned in introductory courses, but since most of the elements are metals, there are a lot of them, and many play an important role in metallurgy. In metallic bonding, the valence electrons lose their association with individual atoms; they form what amounts to a mobile "electron fluid" that fills the space between the crystal lattice positions occupied by the atoms, (now essentially positive ions.) The more readily this electron delocalization occurs, the more "metallic" the element. Thus instead of the one-dimension chart shown above, we can construct a triangular diagram whose corners represent the three extremes of "pure" covalent, ionic, and metallic bonding. We can take this a step farther by taking into account collection of weaker binding effects known generally as van der Waals forces. Contrary to what is often implied in introductory textbooks, these are the major binding forces in most of the common salts that are not alkali halides; these include NaOH, CaCl 2 , MgSO 4 . They are also significant in solids such as CuCl 2 and solid SO 3 in which infinite covalently-bound chains are held together by ion-induced dipole and similar forces. The only way to represent this four-dimensional bonding-type space in two dimensions is to draw a projection of a tetrahedron, each of its four corners representing the "pure" case of one type of bonding. Note that some of the entries on this diagram (ice, CH 4 , and the two parts of NH 4 ClO 4 ) are covalently bound units, and their placement refers to the binding between these units. Thus the H 2 O molecules in ice are held together mainly by hydrogen bonding, which is a van der Waals force, with only a small covalent contribution. Note: the triangular and tetrahedral diagrams above were adapted from those in the excellent article by William B. Jensen, "Logic, history and the chemistry textbook", Part II, J. Chemical Education 1998: 817-828. |
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