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Courses/Fullerton_College/Introductory_Biochemistry/17%3A_Nucleic_Acids | The blueprint for the reproduction and the maintenance of each organism is found in the nuclei of its cells, concentrated in elongated, threadlike structures called chromosomes. These complex structures, consisting of DNA and proteins, contain the basic units of heredity, called genes. The number of chromosomes (and genes) varies with each species. Human body cells have 23 pairs of chromosomes having 20,000–40,000 different genes. Sperm and egg cells contain only a single copy of each chromosome; that is, they contain only one member of each chromosome pair. Thus, in sexual reproduction, the entire complement of chromosomes is achieved only when an egg and sperm combine. A new individual receives half its hereditary material from each parent. Calling the unit of heredity a “gene” merely gives it a name. But what really are genes and how is the information they contain expressed? One definition of a gene is that it is a segment of DNA that constitutes the code for a specific polypeptide. If genes are segments of DNA, we need to learn more about the structure and physiological function of DNA. We begin by looking at the small molecules needed to form DNA and RNA (ribonucleic acid)—the nucleotides. 17.1: Prelude to Nucleic Acids In the 1970s, an intense research effort began that eventually led to the production of genetically engineered human insulin—the first genetically engineered product to be approved for medical use. To accomplish this feat, researchers first had to determine how insulin is made in the body and then find a way of causing the same process to occur in nonhuman organisms, such as bacteria or yeast cells. Many aspects of these discoveries are presented in this chapter on nucleic acids. 17.2: Nucleotides Nucleotides are composed of phosphoric acid, a pentose sugar (ribose or deoxyribose), and a nitrogen-containing base (adenine, cytosine, guanine, thymine, or uracil). Ribonucleotides contain ribose, while deoxyribonucleotides contain deoxyribose. 17.3: Nucleic Acid Structure DNA is the nucleic acid that stores genetic information. RNA is the nucleic acid responsible for using the genetic information in DNA to produce proteins. Nucleotides are joined together to form nucleic acids through the phosphate group of one nucleotide connecting in an ester linkage to the OH group on the third carbon atom of the sugar unit of a second nucleotide. Nucleic acid sequences are written starting with the nucleotide having a free phosphate group (the 5′ end). 17.4: Replication and Expression of Genetic Information In DNA replication, each strand of the original DNA serves as a template for the synthesis of a complementary strand. DNA polymerase is the primary enzyme needed for replication. In transcription, a segment of DNA serves as a template for the synthesis of an RNA sequence. RNA polymerase is the primary enzyme needed for transcription. Three types of RNA are formed during transcription: mRNA, rRNA, and tRNA. 17.5: Protein Synthesis and the Genetic Code In translation, the information in mRNA directs the order of amino acids in protein synthesis. A set of three nucleotides (codon) codes for a specific amino acid. 17.6: Mutations and Genetic Diseases The nucleotide sequence in DNA may be modified either spontaneously or from exposure to heat, radiation, or certain chemicals and can lead to mutations. Mutagens are the chemical or physical agents that cause mutations. Genetic diseases are hereditary diseases that occur because of a mutation in a critical gene. 17.7: Viruses Viruses are very small infectious agents that contain either DNA or RNA as their genetic material. The human immunodeficiency virus (HIV) causes acquired immunodeficiency syndrome (AIDS). 17.8: Flow of Genetic Information As the cell’s so-called blueprint, DNA must be copied to pass on to new cells and its integrity safeguarded. The information in the DNA must also be accessed and transcribed to make the RNA instructions that direct the synthesis of proteins. 17.8.1: DNA Replication 17.8.2: DNA Repair 17.8.3: Transcription 17.8.4: Regulation of Transcription 17.8.5: RNA Processing 17.8.6: Translation 17.E: Nucleic Acids (Exercises) Problems and select solutions for the chapter. 17.S: Nucleic Acids (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. Template:HideTOC |
Courses/Indiana_Tech/EWC%3A_CHEM_2300_-_Introductory_Organic_(Budhi)/3%3A_Organic_Compounds_of_Oxygen/3.10%3A_Properties_of_Aldehydes_and_Ketones | Learning Objectives Explain why the boiling points of aldehydes and ketones are higher than those of ethers and alkanes of similar molar masses but lower than those of comparable alcohols. Compare the solubilities in water of aldehydes and ketones of four or fewer carbon atoms with the solubilities of comparable alkanes and alcohols. Name the typical reactions take place with aldehydes and ketones. Describe some of the uses of common aldehydes and ketones. The carbon-to-oxygen double bond is quite polar, more polar than a carbon-to-oxygen single bond. The electronegative oxygen atom has a much greater attraction for the bonding electron pairs than does the carbon atom. The carbon atom has a partial positive charge, and the oxygen atom has a partial negative charge: In aldehydes and ketones, this charge separation leads to dipole-dipole interactions that are great enough to significantly affect the boiling points. Table \(\PageIndex{1}\) shows that the polar single bonds in ethers have little such effect, whereas hydrogen bonding between alcohol molecules is even stronger. Compound Family Molar Mass Type of Intermolecular Forces Boiling Point (°C) CH3CH2CH2CH3 alkane 58 dispersion only –1 CH3OCH2CH3 ether 60 weak dipole 6 CH3CH2CHO aldehyde 58 strong dipole 49 CH3CH2CH2OH alcohol 60 hydrogen bonding 97 Formaldehyde is a gas at room temperature. Acetaldehyde boils at 20°C; in an open vessel, it boils away in a warm room. Most other common aldehydes are liquids at room temperature. Although the lower members of the homologous series have pungent odors, many higher aldehydes have pleasant odors and are used in perfumes and artificial flavorings. As for the ketones, acetone has a pleasant odor, but most of the higher homologs have rather bland odors. The oxygen atom of the carbonyl group engages in hydrogen bonding with a water molecule. The solubility of aldehydes is therefore about the same as that of alcohols and ethers. Formaldehyde, acetaldehyde, and acetone are soluble in water. As the carbon chain increases in length, solubility in water decreases. The borderline of solubility occurs at about four carbon atoms per oxygen atom. All aldehydes and ketones are soluble in organic solvents and, in general, are less dense than water. Oxidation of Aldehydes and Ketones Aldehydes and ketones are much alike in many of their reactions, owing to the presence of the carbonyl functional group in both. They differ greatly, however, in one most important type of reaction: oxidation. Aldehydes are readily oxidized to carboxylic acids, whereas ketones resist oxidation. The aldehydes are, in fact, among the most easily oxidized of organic compounds. They are oxidized by oxygen (\(\ce{O2}\)) in air to carboxylic acids. \[\ce{2RCHO + O_2 -> 2RCOOH} \nonumber \] The ease of oxidation helps chemists identify aldehydes. A sufficiently mild oxidizing agent can distinguish aldehydes not only from ketones but also from alcohols. Tollens’ reagent , for example, is an alkaline solution of silver (\(\ce{Ag^{+}}\)) ion complexed with ammonia (NH 3 ), which keeps the \(\ce{Ag^{+}}\) ion in solution. \[\ce{H_3N—Ag^{+}—NH_3} \nonumber \] When Tollens’ reagent oxidizes an aldehyde, the \(\ce{Ag^{+}}\) ion is reduced to free silver (\(\ce{Ag}\)). \[\underbrace{\ce{RCHO(aq)}}_{\text{an aldehyde}} + \ce{2Ag(NH3)2^{+}(aq) -> RCOO^{-} +} \underbrace{\ce{2Ag(s)}}_{\text{free silver}} + \ce{4NH3(aq) + 2H2O} \nonumber \] Deposited on a clean glass surface, the silver produces a mirror (Figure \(\PageIndex{1}\)). Ordinary ketones do not react with Tollens’ reagent. Although ketones resist oxidation by ordinary laboratory oxidizing agents, they undergo combustion, as do aldehydes. Some Common Carbonyl Compounds Formaldehyde has an irritating odor. Because of its reactivity, it is difficult to handle in the gaseous state. For many uses, it is therefore dissolved in water and sold as a 37% to 40% aqueous solution called formalin . Formaldehyde denatures proteins, rendering them insoluble in water and resistant to bacterial decay. For this reason, formalin is used in embalming solutions and in preserving biological specimens. Aldehydes are the active components in many other familiar substances. Large quantities of formaldehyde are used to make phenol-formaldehyde resins for gluing the wood sheets in plywood and as adhesives in other building materials. Sometimes the formaldehyde escapes from the materials and causes health problems in some people. While some people seem unaffected, others experience coughing, wheezing, eye irritation, and other symptoms. The odor of green leaves is due in part to a carbonyl compound, cis-3-hexenal, which with related compounds is used to impart a “green” herbal odor to shampoos and other products. Acetaldehyde is an extremely volatile, colorless liquid. It is a starting material for the preparation of many other organic compounds. Acetaldehyde is formed as a metabolite in the fermentation of sugars and in the detoxification of alcohol in the liver. Aldehydes are the active components of many other familiar materials (Figure \(\PageIndex{2}\)). Acetone is the simplest and most important ketone. Because it is miscible with water as well as with most organic solvents, its chief use is as an industrial solvent (for example, for paints and lacquers). It is also the chief ingredient in some brands of nail polish remover. To Your Health: Acetone in Blood, Urine, and Breath Acetone is formed in the human body as a by-product of lipid metabolism. Normally, acetone does not accumulate to an appreciable extent because it is oxidized to carbon dioxide and water. The normal concentration of acetone in the human body is less than 1 mg/100 mL of blood. In certain disease states, such as uncontrolled diabetes mellitus, the acetone concentration rises to higher levels. It is then excreted in the urine, where it is easily detected. In severe cases, its odor can be noted on the breath. Ketones are also the active components of other familiar substances, some of which are noted in the accompanying figure. Certain steroid hormones have the ketone functional group as a part of their structure. Two examples are progesterone, a hormone secreted by the ovaries that stimulates the growth of cells in the uterine wall and prepares it for attachment of a fertilized egg, and testosterone, the main male sex hormone. These and other sex hormones affect our development and our lives in fundamental ways. Summary 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. |
Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.MM%3A_Molecular_Models | Objective After completing this section, you should be able to use ball-and-stick molecular models to make models of simple organic compounds (e.g., ethane, ethylene, acetylene, ethanol, formaldehyde, acetone, acetic acid), given their Kekulé structures or molecular formulas. Study Notes You will have noticed that we have given two names for most of the compounds discussed up to this point. In general we shall be using systematic (i.e., IUPAC—International Union of Pure and Applied Chemistry) names throughout the course. However, simple compounds are often known by their common names, which may be more familiar than their IUPAC counterparts. We shall address the subject of nomenclature (naming) in Chapter 3. ethanol formaldehyde (methanal), acetone (propanone), acetic (ethanoic) acid, If your instructor is having you work with molecular models in class, they may use this section for you to practice creating specific structures. Exercises Construct a molecular model of each of the compounds listed below. $\ce{\sf{CH3-CN}}$ $\ce{\sf{CH3-N=C=O}}$ $\ce{\sf{CH3-CH2-O-CH3}}$ Hint: Use the curved sticks to form the multiple bonds and the straight sticks for single bonds. Answers: |
Courses/North_Central_State_College/CHEM_1010%3A_Introductory_Chemistry/08%3A_Chemical_Bonding/8.07%3A_Electronegativity_and_Polarity_-_Why_Oil_and_Water_Dont_Mix | Learning Objectives Explain how polar compounds differ from nonpolar compounds. Determine if a molecule is polar or nonpolar. Given a pair of compounds, predict which would have a higher melting or boiling point. Bond Polarity The ability of an atom in a molecule to attract shared electrons is called electronegativity . When two atoms combine, the difference between their electronegativities is an indication of the type of bond that will form. If the difference between the electronegativities of the two atoms is small, neither atom can take the shared electrons completely away from the other atom, and the bond will be covalent. If the difference between the electronegativities is large, the more electronegative atom will take the bonding electrons completely away from the other atom (electron transfer will occur), and the bond will be ionic. This is why metals (low electronegativities) bonded with nonmetals (high electronegativities) typically produce ionic compounds. A bond may be so polar that an electron actually transfers from one atom to another, forming a true ionic bond. How do we judge the degree of polarity? Scientists have devised a scale called electronegativity , a scale for judging how much atoms of any element attract electrons. Electronegativity is a unitless number; the higher the number, the more an atom attracts electrons. A common scale for electronegativity is shown in Figure \(\PageIndex{1}\). The polarity of a covalent bond can be judged by determining the difference of the electronegativities of the two atoms involved in the covalent bond, as summarized in the following table: Electronegativity Difference Bond Type 0–0.4 pure covalent 0.5–2.0 polar covalent >2.0 likely ionic Nonpolar Covalent Bonds A bond in which the electronegativity difference is less than 1.9 is considered to be mostly covalent in character. However, at this point we need to distinguish between two general types of covalent bonds. A nonpolar covalent bond is a covalent bond in which the bonding electrons are shared equally between the two atoms. In a nonpolar covalent bond, the distribution of electrical charge is balanced between the two atoms. The two chlorine atoms share the pair of electrons in the single covalent bond equally, and the electron density surrounding the \(\ce{Cl_2}\) molecule is symmetrical. Also note that molecules in which the electronegativity difference is very small (<0.5) are also considered nonpolar covalent. An example would be a bond between chlorine and bromine (\(\Delta\)EN \(=3.0 - 2.8 = 0.2\)). Polar Covalent Bonds A bond in which the electronegativity difference between the atoms is between 0.5 and 2.0 is called a polar covalent bond. A polar covalent bond is a covalent bond in which the atoms have an unequal attraction for electrons and so the sharing is unequal. In a polar covalent bond, sometimes simply called a polar bond, the distribution of electrons around the molecule is no longer symmetrical. An easy way to illustrate the uneven electron distribution in a polar covalent bond is to use the Greek letter delta \(\left( \delta \right)\). The atom with the greater electronegativity acquires a partial negative charge, while the atom with the lesser electronegativity acquires a partial positive charge. The delta symbol is used to indicate that the quantity of charge is less than one. A crossed arrow can also be used to indicate the direction of greater electron density. Electronegativity differences in bonding using Pauling scale. Differences in electronegativity classify bonds as covalent, polar covalent, or ionic. Example \(\PageIndex{1}\): Bond Polarity What is the polarity of each bond? C–H O–H Solution Using Figure \(\PageIndex{1}\), we can calculate the difference of the electronegativities of the atoms involved in the bond. For the C–H bond, the difference in the electronegativities is 2.5 − 2.1 = 0.4. Thus we predict that this bond will be non polar covalent. For the O–H bond, the difference in electronegativities is 3.5 − 2.1 = 1.4, so we predict that this bond will be polar covalent. Exercise \(\PageIndex{1}\) What is the polarity of each bond? Rb–F P–Cl Answer a likely ionic Answer b polar covalent Molecular Polarity To determine if a molecule is polar or nonpolar, it is frequently useful to look at Lewis structures. Nonpolar compounds will be symmetric, meaning all of the sides around the central atom are identical—bonded to the same element with no unshared pairs of electrons. Polar molecules are asymmetric, either containing lone pairs of electrons on a central atom or having atoms with different electronegativities bonded. This works pretty well, as long as you can visualize the molecular geometry. That's the hard part. To know how the bonds are oriented in space, you have to have a strong grasp of Lewis structures and VSEPR theory. Assuming that you do, you can look at the structure of each one and decide if it is polar or not, whether or not you know the individual atom's electronegativity . This is because you know that all bonds between dissimilar elements are polar, and in these particular examples, it doesn't matter which direction the dipole moment vectors are pointing (out or in). A polar molecule is a molecule in which one end of the molecule is slightly positive, while the other end is slightly negative. A diatomic molecule that consists of a polar covalent bond, such as \(\ce{HF}\), is a polar molecule. 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. A molecule with two poles is called a dipole (see figure below). Hydrogen fluoride is a dipole. 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. 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 are oriented this way, they cancel out and the overall molecular polarity of \(\ce{CO_2}\) is zero. Water is a bent molecule because of the two lone pairs on the central oxygen atom. The individual dipoles point from the \(\ce{H}\) atoms toward the \(\ce{O}\) atom. Because of the shape, the dipoles do not cancel each other out and the water molecule is polar. In the figure below, the net dipole is shown in blue and 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 \(\PageIndex{14}\)). 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 Example \(\PageIndex{2}\): Label each of the following as polar or nonpolar. Water, H 2 O: Methanol, CH 3 OH: Hydrogen Cyanide, HCN: Oxygen, O 2 : Propane, C 3 H 8 : Solution Water is polar. Any molecule with lone pairs of electrons around the central atom is polar. Methanol is polar. This is not a symmetric molecule. The \(\ce{-OH}\) side is different from the other 3 \(\ce{-H}\) sides. Hydrogen cyanide is polar. The molecule is not symmetric. The nitrogen and hydrogen have different electronegativities, creating an uneven pull on the electrons. Oxygen is nonpolar. The molecule is symmetric. The two oxygen atoms pull on the electrons by exactly the same amount. Propane is nonpolar, because it is symmetric, with \(\ce{H}\) atoms bonded to every side around the central atoms and no unshared pairs of electrons. Exercise \(\PageIndex{2}\) Label each of the following as polar or nonpolar. a. SO 3 b. NH 3 Answer a non polar Answer b polar 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: StackExchange ( thomij ). Marisa Alviar-Agnew ( Sacramento City College ) Henry Agnew (UC Davis) |
Courses/University_of_Wisconsin_Oshkosh/Chem_370%3A_Physical_Chemistry_1_-_Thermodynamics_(Gutow)/zz%3A_Back_Matter/20%3A_Glossary | Words (or words that have the same definition) The definition is case sensitive (Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages] (Optional) Caption for Image (Optional) External or Internal Link (Optional) Source for Definition (Eg. "Genetic, Hereditary, DNA ...") (Eg. "Relating to genes or heredity") NaN The infamous double helix https://bio.libretexts.org/ CC-BY-SA; Delmar Larsen Word(s) Definition Image Caption Link Source Sample Word 1 Sample Definition 1 NaN NaN NaN NaN |
Courses/University_of_Florida/CHM2047%3A_One-Semester_General_Chemistry_(Kleiman)/09%3A_Gases/9.03%3A_Further_Applications_of_the_Ideal-Gas_Equations | Learning Objectives To relate the amount of gas consumed or released in a chemical reaction to the stoichiometry of the reaction. To understand how the ideal gas equation and the stoichiometry of a reaction can be used to calculate the volume of gas produced or consumed in a reaction. With the ideal gas law , we can use the relationship between the amounts of gases (in moles) and their volumes (in liters) to calculate the stoichiometry of reactions involving gases, if the pressure and temperature are known. This is important for several reasons. Many reactions that are carried out in the laboratory involve the formation or reaction of a gas, so chemists must be able to quantitatively treat gaseous products and reactants as readily as they quantitatively treat solids or solutions. Furthermore, many, if not most, industrially important reactions are carried out in the gas phase for practical reasons. Gases mix readily, are easily heated or cooled, and can be transferred from one place to another in a manufacturing facility via simple pumps and plumbing. Gas Densities and Molar Mass The ideal-gas equation can be manipulated to solve a variety of different types of problems. For example, the density, \(\rho\), of a gas, depends on the number of gas molecules in a constant volume. To determine this value, we rearrange the ideal gas equation to \[\dfrac{n}{V}=\dfrac{P}{RT}\label{10.5.1} \] Density of a gas is generally expressed in g/L (mass over volume). Multiplication of the left and right sides of Equation \ref{10.5.1} by the molar mass in g/mol (\(M\)) of the gas gives \[\rho= \dfrac{g}{L}=\dfrac{PM}{RT} \label{10.5.2} \] This allows us to determine the density of a gas when we know the molar mass, or vice versa. The density of a gas INCREASES with increasing pressure and DECREASES with increasing temperature Example \(\PageIndex{1}\) What is the density of nitrogen gas (\(\ce{N_2}\)) at 248.0 Torr and 18º C? Solution Step 1: Write down your given information P = 248.0 Torr V = ? n = ? R = 0.0820574 L•atm•mol -1 K -1 T = 18º C Step 2: Convert as necessary. \[(248 \; \rm{Torr}) \times \dfrac{1 \; \rm{atm}}{760 \; \rm{Torr}} = 0.3263 \; \rm{atm} \nonumber \] \[18\,^oC + 273 = 291 K\nonumber \] Step 3: This one is tricky. We need to manipulate the Ideal Gas Equation to incorporate density into the equation. Write down all known equations: \[PV = nRT \nonumber \] \[\rho=\dfrac{m}{V} \nonumber \] where \(\rho\) is density, \(m\) is mass, and \(V\) is volume. \[m=M \times n \nonumber \] where \(M\) is molar mass and \(n\) is the number of moles. Now take the definition of density (Equation \ref{10.5.1}) \[\rho=\dfrac{m}{V} \nonumber \] Keeping in mind \(m=M \times n\)...replace \((M \times n)\) for \(mass\) within the density formula. \[\begin{align*} \rho &=\dfrac{M \times n}{V} \\[4pt] \dfrac{\rho}{M} &= \dfrac{n}{V} \end{align*} \nonumber \] Now manipulate the Ideal Gas Equation \[ \begin{align*} PV &= nRT \\[4pt] \dfrac{n}{V} &= \dfrac{P}{RT} \end{align*} \nonumber \] \((n/V)\) is in both equations. \[ \begin{align*} \dfrac{n}{V} &= \dfrac{\rho}{M} \\[4pt] &= \dfrac{P}{RT} \end{align*} \nonumber \] Now combine them please. \[\dfrac{\rho}{M} = \dfrac{P}{RT}\nonumber \] Isolate density. \[\rho = \dfrac{PM}{RT} \nonumber \] Step 4: Now plug in the information you have. \[ \begin{align*} \rho &= \dfrac{PM}{RT} \\[4pt] &= \dfrac{(0.3263\; \rm{atm})(2*14.01 \; \rm{g/mol})}{(0.08206\, L\, atm/K mol)(291 \; \rm{K})} \\[4pt] &= 0.3828 \; g/L \end{align*} \nonumber \] An example of varying density for a useful purpose is the hot air balloon, which consists of a bag (called the envelope) that is capable of containing heated air. As the air in the envelope is heated, it becomes less dense than the surrounding cooler air (Equation \(\ref{10.5.2}\)), which is has enough lifting power (due to buoyancy) to cause the balloon to float and rise into the air. Constant heating of the air is required to keep the balloon aloft. As the air in the balloon cools, it contracts, allowing outside cool air to enter, and the density increases. When this is carefully controlled by the pilot, the balloon can land as gently as it rose. Determining Gas Volumes in Chemical Reactions The ideal gas law can be used to calculate volume of gases consumed or produced. The ideal-gas equation frequently is used to interconvert between volumes and molar amounts in chemical equations. Example \(\PageIndex{2A}\) What volume of carbon dioxide gas is produced at STP by the decomposition of 0.150 g \(\ce{CaCO_3}\) via the equation: \[ \ce{CaCO3(s) \rightarrow CaO(s) + CO2(g)} \nonumber \] Solution Begin by converting the mass of calcium carbonate to moles. \[ \dfrac{0.150\;g}{100.1\;g/mol} = 0.00150\; mol \nonumber \] The stoichiometry of the reaction dictates that the number of moles \(\ce{CaCO_3}\) decomposed equals the number of moles \(\ce{CO2}\) produced. Use the ideal-gas equation to convert moles of \(\ce{CO2}\) to a volume. \[ \begin{align*} V &= \dfrac{nRT}{PR} \\[4pt] &= \dfrac{(0.00150\;mol)\left( 0.08206\; \frac{L \cdot atm}{mol \cdot K} \right) ( 273.15\;K)}{1\;atm} \\[4pt] &= 0.0336\;L \; or \; 33.6\;mL \end{align*} \nonumber \] Example \(\PageIndex{2B}\) A 3.00 L container is filled with \(\ce{Ne(g)}\) at 770 mmHg at 27 o C. A \(0.633\;\rm{g}\) sample of \(\ce{CO2}\) vapor is then added. What is the partial pressure of \(\ce{CO2}\) and \(\ce{Ne}\) in atm? What is the total pressure in the container in atm? Solution Step 1: Write down all given information, and convert as necessary. Before: \(P = 770\,mmHg \rigtharrow 1.01 \,atm \nonumber \] \(V = 3.00\,L\) \(n_{\ce{Ne}} = ?\) \(T = 27^o C \rightarrow 300\; K\) Other Unknowns: \(n_{\ce{CO2}}\)= ? \[n_{CO_2} = 0.633\; \rm{g} \;CO_2 \times \dfrac{1 \; \rm{mol}}{44\; \rm{g}} = 0.0144\; \rm{mol} \; CO_2 \nonumber \] Step 2: After writing down all your given information, find the unknown moles of \(\ce{Ne}\). \[ \begin{align*} n_{Ne} &= \dfrac{PV}{RT} \\[4pt] &= \dfrac{(1.01\; \rm{atm})(3.00\; \rm{L})}{(0.08206\;atm\;L/mol\;K)(300\; \rm{K})} \\[4pt] &= 0.123 \; \rm{mol} \end{align*} \nonumber \] Because the pressure of the container before the \(\ce{CO2}\) was added contained only \(\ce{Ne}\), that is your partial pressure of \(Ne\). After converting it to atm, you have already answered part of the question! \[P_{Ne} = 1.01\; \rm{atm} \nonumber \] Step 3: Now that have pressure for \(\ce{Ne}\), you must find the partial pressure for \(CO_2\). Use the ideal gas equation. \[ \dfrac{P_{Ne}\cancel{V}}{n_{Ne}\cancel{RT}} = \dfrac{P_{CO_2}\cancel{V}}{n_{CO_2}\cancel{RT}} \nonumber \] but because both gases share the same Volume (\(V\)) and Temperature (\(T\)) and since the Gas Constant (\(R\)) is constants, all three terms cancel. \[ \begin{align*} \dfrac{P}{n_{Ne}} &= \dfrac{P}{n_{CO_2}} \\[4pt] \dfrac{1.01 \; \rm{atm}}{0.123\; \rm{mol} \;Ne} &= \dfrac{P_{CO_2}}{0.0144\; \rm{mol} \;CO_2} \\[4pt] P_{CO_2} &= 0.118 \; \rm{atm} \end{align*} \nonumber \] This is the partial pressure \(\ce{CO_2}\). Step 4: Now find total pressure. \[\begin{align*} P_{total} &= P_{Ne} + P_{CO_2} \\[4pt] &= 1.01 \; \rm{atm} + 0.118\; \rm{atm} \\[4pt] &= 1.128\; \rm{atm} \\[4pt] &\approx 1.13\; \rm{atm} \; \text{(with appropriate significant figures)} \end{align*} \nonumber \] Example \(\PageIndex{2C}\): Sulfuric Acid Sulfuric acid, the industrial chemical produced in greatest quantity (almost 45 million tons per year in the United States alone), is prepared by the combustion of sulfur in air to give \(\ce{SO2}\), followed by the reaction of \(\ce{SO2}\) with \(\ce{O2}\) in the presence of a catalyst to give \(\ce{SO3}\), which reacts with water to give \(\ce{H2SO4}\). The overall chemical equation is as follows: \[\ce {2S(s) + 3O2(g) + 2H2O(l) \rightarrow 2H2SO4(aq)} \nonumber \] What volume of O 2 (in liters) at 22°C and 745 mmHg pressure is required to produce 1.00 ton (907.18 kg) of H 2 SO 4 ? Given: reaction, temperature, pressure, and mass of one product Asked for: volume of gaseous reactant Strategy: A Calculate the number of moles of H 2 SO 4 in 1.00 ton. From the stoichiometric coefficients in the balanced chemical equation, calculate the number of moles of \(\ce{O2}\) required. B Use the ideal gas law to determine the volume of \(\ce{O2}\) required under the given conditions. Be sure that all quantities are expressed in the appropriate units. Solution: mass of \(\ce{H2SO4}\) → mol e s \(\ce{H2SO4}\) → m oles \(\ce{O2}\) → liters \(\ce{O2}\) A We begin by calculating the number of moles of H 2 SO 4 in 1.00 ton: \[\rm\dfrac{907.18\times10^3\;g\;H_2SO_4}{(2\times1.008+32.06+4\times16.00)\;g/mol}=9250\;mol\;H_2SO_4 \nonumber \] We next calculate the number of moles of \(\ce{O2}\) required: \[\rm9250\;mol\;H_2SO_4\times\dfrac{3mol\; O_2}{2mol\;H_2SO_4}=1.389\times10^4\;mol\;O_2 \nonumber \] B After converting all quantities to the appropriate units, we can use the ideal gas law to calculate the volume of O 2 : \[\begin{align*} V&=\dfrac{nRT}{P} \\[4pt] &=\rm\dfrac{1.389\times10^4\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times(273+22)\;K}{745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}} \\[4pt] &=3.43\times10^5\;L \end{align*} \nonumber \] The answer means that more than 300,000 L of oxygen gas are needed to produce 1 ton of sulfuric acid. These numbers may give you some appreciation for the magnitude of the engineering and plumbing problems faced in industrial chemistry. Exercise \(\PageIndex{2}\) Charles used a balloon containing approximately 31,150 L of \(\ce{H2}\) for his initial flight in 1783. The hydrogen gas was produced by the reaction of metallic iron with dilute hydrochloric acid according to the following balanced chemical equation: \[\ce{ Fe(s) + 2 HCl(aq) \rightarrow H2(g) + FeCl2(aq)} \nonumber \] How much iron (in kilograms) was needed to produce this volume of \(\ce{H2}\) if the temperature were 30°C and the atmospheric pressure was 745 mmHg? Answer 68.6 kg of Fe (approximately 150 lb) Example \(\PageIndex{3}\): Emergency Air bags Sodium azide (\(\ce{NaN_3}\)) decomposes to form sodium metal and nitrogen gas according to the following balanced chemical equation: \[\ce{ 2NaN3 \rightarrow 2Na(s) + 3N2(g)} \nonumber \] This reaction is used to inflate the air bags that cushion passengers during automobile collisions. The reaction is initiated in air bags by an electrical impulse and results in the rapid evolution of gas. If the \(\ce{N_2}\) gas that results from the decomposition of a 5.00 g sample of \(\ce{NaN_3}\) could be collected by displacing water from an inverted flask, what volume of gas would be produced at 21°C and 762 mmHg? Given: reaction, mass of compound, temperature, and pressure Asked for: volume of nitrogen gas produced Strategy: A Calculate the number of moles of \(\ce{N_2}\) gas produced. From the data in Table S3 , determine the partial pressure of \(\ce{N_2}\) gas in the flask. B Use the ideal gas law to find the volume of \(\ce{N_2}\) gas produced. Solution: A Because we know the mass of the reactant and the stoichiometry of the reaction, our first step is to calculate the number of moles of \(\ce{N_2}\) gas produced: \[\rm\dfrac{5.00\;g\;NaN_3}{(22.99+3\times14.01)\;g/mol}\times\dfrac{3mol\;N_2}{2mol\;NaN_3}=0.115\;mol\; N_2 \nonumber \] The pressure given (762 mmHg) is the total pressure in the flask, which is the sum of the pressures due to the N 2 gas and the water vapor present. Table S3 tells us that the vapor pressure of water is 18.65 mmHg at 21°C (294 K), so the partial pressure of the \(\ce{N_2}\) gas in the flask is only \[\begin{align*} \rm(762 − 18.65)\;mmHg \times\dfrac{1\;atm}{760\;mmHg} &= 743.4\; \cancel{mmHg} \times\dfrac{1\;atm}{760\;\cancel{mmHg}} \\[4pt] &= 0.978\; atm. \end{align*} \nonumber \] B Solving the ideal gas law for V and substituting the other quantities (in the appropriate units), we get \[V=\dfrac{nRT}{P}=\rm\dfrac{0.115\;mol\times0.08206\dfrac{atm\cdot L}{mol\cdot K}\times294\;K}{0.978\;atm}=2.84\;L \nonumber \] Exercise\(\PageIndex{3}\) A 1.00 g sample of zinc metal is added to a solution of dilute hydrochloric acid. It dissolves to produce \(\ce{H2}\) gas according to the equation \[\ce{ Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq)}. \nonumber \] The resulting H 2 gas is collected in a water-filled bottle at 30°C and an atmospheric pressure of 760 mmHg. What volume does it occupy? Answer 0.397 L Summary The relationship between the amounts of products and reactants in a chemical reaction can be expressed in units of moles or masses of pure substances, of volumes of solutions, or of volumes of gaseous substances. The ideal gas law can be used to calculate the volume of gaseous products or reactants as needed. In the laboratory, gases produced in a reaction are often collected by the displacement of water from filled vessels; the amount of gas can then be calculated from the volume of water displaced and the atmospheric pressure. A gas collected in such a way is not pure, however, but contains a significant amount of water vapor. The measured pressure must therefore be corrected for the vapor pressure of water, which depends strongly on the temperature. |
Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/13%3A_Spontaneous_Processes_and_Thermodynamic_Equilibrium | Template:HideTOC A spontaneous process is the time-evolution of a system in which it releases free energy and moves to a lower, more thermodynamically stable energy state. 13.1: The Nature of Spontaneous Processes In this section, consider the differences between two types of changes in a system: Those that occur spontaneously and those that occur by force. In doing so, we’ll gain an understanding as to why some systems are naturally inclined to change in one direction under certain conditions and how relatively quickly or slowly that natural change proceeds. We’ll also gain insight into how the spontaneity of a process affects the distribution of energy and matter within the system. 13.2: Entropy and Spontaneity - A Molecular Statistical Interpretation These forms of motion are ways in which the molecule can store energy. The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. A perfectly ordered system with only a single microstate available to it would have an entropy of zero. The only system that meets this criterion is a perfect crystal at a temperature of absolute zero (0 K), in which each component atom, molecule, or ion is fixed in place within a perfect crystal lattice. 13.3: Entropy and Heat - Experimental Basis of the Second Law of Thermodynamics A reversible process is one for which all intermediate states between extremes are equilibrium states; it can change direction at any time. In contrast, an irreversible process occurs in one direction only. The change in entropy of the system or the surroundings is the quantity of heat transferred divided by the temperature. The 2nd law states that in a reversible process, the entropy of the universe is constant and in an irreversible process, the entropy of the universe increases. 13.4: Entropy Changes in Reversible Processes Changes in internal energy, that are not accompanied by a temperature change, might reflect changes in the entropy of the system. Changes in internal energy, that are not accompanied by a temperature change, might reflect changes in the entropy of the system. 13.5: Entropy Changes and Spontaneity In the quest to identify a property that may reliably predict the spontaneity of a process, we have identified a very promising candidate: entropy. Processes that involve an increase in entropy of the system (ΔS > 0) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include the surroundings, we may reach a significant conclusion regarding the relation between this property and spontaneity. 13.6: The Third Law of Thermodynamics This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion (at least classically, quantum mechanics argues for constant motion) means there is but one possible location for each identical atom or molecule comprising the crystal (Ω=1). According to the Boltzmann equation, the entropy of this system is zero. 13.7: The Gibbs Free Energy One of the major goals of chemical thermodynamics is to establish criteria for predicting whether a particular reaction or process will occur spontaneously. We have developed one such criterion, the change in entropy of the universe. This is not particularly useful and a criterion of spontaneity that is based solely on the state functions of a system would be much more convenient and is provided by a new state function: the Gibbs free energy. 13.8: Carnot Cycle, Efficiency, and Entropy The Carnot cycle has the greatest efficiency possible of an engine (although other cycles have the same efficiency) based on the assumption of the absence of incidental wasteful processes such as friction, and the assumption of no conduction of heat between different parts of the engine at different temperatures. 13.E: Spontaneous Processes (Exercises) |
Bookshelves/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/11%3A_Gases/11.10%3A_Gases_in_Chemical_Reactions-_Stoichiometry_Revisited | Earlier in the course we performed stoichiometric calculations with chemical reactions using quantities of moles and mass (typically in grams). These same principles can be applied to chemical reactions involving gases except that we first have to convert volumes of gases into moles. Example Hydrogen gas reacts with oxygen gas to produce water vapor via the following balanced chemical equation: 2H2(g) + O2(g) -> 2H2O(g) If the temperature is 320 K and pressure is 1.34 atm, what volume of oxygen is required to produce 65.0 g of water? Strategy: Since we are given the temperature and pressure, to find the volume of oxygen using the ideal gas law we need to first calculate the moles of oxygen. To find the moles of oxygen required, we can first calculate the moles of water in 65.0g. \[n_{water}=\frac{m_{water}}{mm_{water}}=\frac{\text{65.0 g}}{\text{18.0g/mol}}=\text{3.61 mol of water}\] From the balanced chemical equation, we can see that 1 equivalent of oxygen produces 2 equivalents of water. We can therefore write the following ratio: 1 mol O2 : 2 mol H2O We can now solve for the amount of oxygen: \[(\text{3.61 mol } H_{2}O)\times(\frac{ \text{1 mol } O_{2}}{\text{2 mol } H_{2}O })= \text{1.81 mol } O_{2}\] Finally, now that we know how many moles of oxygen are required, we can calculate the volume of the oxygen using the ideal gas law and the temperature&pressure provided in question: \[PV=nRT\] \[V=\frac{nRT}{P}\] \[{V = \rm \frac{1.81 mol\ \cdot 0.08206 \frac{L atm}{mol K}\cdot 320K}{1.34 atm }}\] \[V = 35.5 \rm L\] |
Courses/Brevard_College/CHE_104%3A_Principles_of_Chemistry_II/07%3A_Acid_and_Base_Equilibria/7.20%3A_Titration_Curves | The \(x\)-\(y\) plot that we know of as a graph was the brainchild of the French mathematician-philosopher Rene Descartes (1596-1650). His studies in mathematics led him to develop what was known as "Cartesian geometry", including the concept of today's graphs. The coordinates are often referred to as Cartesian coordinates. Titration Curves As base is added to acid at the beginning of a titration, the pH rises very slowly. Nearer to the equivalence point, the pH begins to rapidly increase. If the titration is a strong acid with a strong base, the pH at the equivalence point is equal to 7. A bit past the equivalence point, the rate of change of the pH again slows down. A titration curve is a graphical representation of the pH of a solution during a titration. The figure below shows two different examples of a strong acid-strong base titration curve. On the left is a titration in which the base is added to the acid, and so the pH progresses from low to high. On the right is a titration in which the acid is added to the base. In this case, the pH starts out high and decreases during the titration. In both cases, the equivalence point is reached when the moles of acid and base are equal and the pH is 7. This also corresponds to the color change of the indicator. Titration curves can also be generated in the case of a weak acid-strong base titration or a strong acid-weak base titration. The general shape of the titration curve is the same, but the pH at the equivalence point is different. In a weak acid-strong base titration, the pH is greater than 7 at the equivalence point. In a strong acid-weak base titration, the pH is less than 7 at the equivalence point. Summary A titration curve is a graphical representation of the pH of a solution during a titration. In a strong acid-strong base titration, the equivalence point is reached when the moles of acid and base are equal and the pH is 7. In a weak acid-strong base titration, the pH is greater than 7 at the equivalence point. In a strong acid-weak base titration, the pH is less than 7 at the equivalence point. |
Courses/South_Puget_Sound_Community_College/Chem_121%3A_Introduction_to_Chemistry/07%3A_Chapter_6_-_Chemical_Reactions_and_Stoichiometry/7.01%3A_Prelude_to_Chemical_Reactions | The space shuttle—and any other rocket-based system—uses chemical reactions to propel itself into space and maneuver itself when it gets into orbit. The rockets that lift the orbiter are of two different types. The three main engines are powered by reacting liquid hydrogen with liquid oxygen to generate water. Then there are the two solid rocket boosters, which use a solid fuel mixture that contains mainly ammonium perchlorate and powdered aluminum. The chemical reaction between these substances produces aluminum oxide, water, nitrogen gas, and hydrogen chloride. Although the solid rocket boosters each have a significantly lower mass than the liquid oxygen and liquid hydrogen tanks, they provide over 80% of the lift needed to put the shuttle into orbit—all because of chemical reactions. |
Courses/University_of_Toronto/UTSC%3A_First-Year_Chemistry_Textbook_(Winter_2025)/14%3A_Fundamental_Equilibrium_Concepts/14.01%3A_Introduction | Imagine a beach populated with sunbathers and swimmers. As those basking in the sun get too hot, they enter the surf to swim and cool off. As the swimmers tire, they return to the beach to rest. If the rate at which sunbathers enter the surf were to equal the rate at which swimmers return to the sand, then the numbers (though not the identities) of sunbathers and swimmers would remain constant. This scenario illustrates a dynamic phenomenon known as equilibrium , in which opposing processes occur at equal rates. Chemical and physical processes are subject to this phenomenon; these processes are at equilibrium when the forward and reverse reaction rates are equal. Equilibrium systems are pervasive in nature; the various reactions involving carbon dioxide dissolved in blood are examples (see Figure \(\PageIndex{1}\)). This chapter provides a thorough introduction to the essential aspects of chemical equilibria. |
Courses/University_of_North_Carolina_Charlotte/CHEM_2141%3A__Survey_of_Physical_Chemistry/09%3A_Quantum_Basics/9.12%3A_The_Chemical_Bond | The Hydrogen Molecule This four-particle system, two nuclei plus two electrons, is described by the Hamiltonian \[ \hat{H} = -\frac{1}{2} \nabla^2_1 -\frac{1}{2} \nabla^2_2 -\frac{1}{2M_A} \nabla^2_A -\frac{1}{2M_B} \nabla^2_B -\frac{1}{r_{1A}} -\frac{1}{r_{2B}} -\frac{1}{r_{2A}} -\frac{1}{r_{1B}} +\frac{1}{r_{12}} +\frac{1}{R} \label{1}\] in terms of the coordinates shown in Figure \(\PageIndex{1}\). We note first that the masses of the nuclei are much greater than those of the electrons, M proton = 1836 atomic units, compared to m electron = 1 atomic unit. Therefore nuclear kinetic energies will be negligibly small compared to those of the electrons. In accordance with the Born-Oppenheimer approximation, we can first consider the electronic Schrödinger equation \[\hat{H}_{elec} \psi(r_1,r_2,R) = E_{elec}(R) \psi(r_1,r_2,R) \label{2}\] where \[ \hat{H} = -\frac{1}{2} \nabla^2_1 -\frac{1}{2} \nabla^2_2 -\frac{1}{r_{1A}} -\frac{1}{r_{2B}} -\frac{1}{r_{2A}} -\frac{1}{r_{1B}} +\frac{1}{r_{12}} +\frac{1}{R} \label{3}\] The internuclear separation R occurs as a parameter in this equation so that the Schrödinger equation must, in concept, be solved for each value of the internuclear distance R . A typical result for the energy of a diatomic molecule as a function of R is shown in Figure \(\PageIndex{2}\). For a bound state, the energy minimum occurs at for R = R e , known as the equilibrium internuclear distance . The depth of the potential well at R e is called the binding energy or dissociation energy D e . For the H 2 molecule, D e = 4.746 eV and R e =1.400 bohr = 0.7406 Å. Note that as R → 0, E(R) → \(\infty\), since the 1/ R nuclear repulsion will become dominant. The more massive nuclei move much more slowly than the electrons. From the viewpoint of the nuclei, the electrons adjust almost instantaneously to any changes in the internuclear distance. The electronic energy E elec ( R ) therefore plays the role of a potential energy in the Schrödinger equation for nuclear motion \[ \left\{ -\frac{1}{2M_A} \nabla^2_A -\frac{1}{2M_B} \nabla^2_B + V(R)\right\} \chi (r_A,r_B) = E \chi (r_A,r_B) \label{4}\] where \[ V(R) = E_{elec}(R) \label{5}\] from solution of Equation \(\ref{2}\). Solutions of Equation \(\ref{4}\) determine the vibrational and rotational energies of the molecule. These will be considered elsewhere. For the present, we are interested in the obtaining electronic energy from Equation \(\ref{2}\) and \(\ref{3}\). We will thus drop the subscript "elec" on \(\hat{H}\) and E ( R ) for the remainder this Chapter. The first quantum-mechanical account of chemical bonding is due to Heitler and London in 1927, only one year after the Schrödinger equation was proposed. They reasoned that, since the hydrogen molecule H 2 was formed from a combination of hydrogen atoms A and B , a first approximation to its electronic wavefunction might be \[ \psi(r_1,r_2) = \psi_{1s} (r_{1A})\psi_{1s} (r_{2B}) \label{6}\] Using this function into the variational integral \[ \tilde{E}(R) = \frac{\int{ \psi \hat{H} \psi d\tau}}{\int{\psi^2 d\tau}} \label{7}\] the value R e \(\approx\) 1.7 bohr was obtained, indicating that the hydrogen atoms can indeed form a molecule. However, the calculated binding energy D e \(\approx\) 0.25 eV, is much too small to account for the strongly-bound H 2 molecule. Heitler and London proposed that it was necessary to take into account the exchange of electrons, in which the electron labels in Equation \(\ref{6}\) are reversed. The properly symmetrized function \[ \psi(r_1, r_2) = \psi_{1s} (r_{1A})\psi_{1s} (r_{2B}) +\psi_{1s} (r_{1B})\psi_{1s} (r_{2A}) \label{8}\] gave a much more realistic binding energy value of 3.20 eV, with R e = 1.51 bohr. We have already used exchange symmetry (and antisymmetry) in our treatment of the excited states of helium. The variational function (Equation \(\ref{8}\)) was improved (Wang, 1928) by replacing the hydrogen 1 s functions \(e^{-r}\) by \(e^{-\zeta r}\). The optimized value \(\zeta\) = 1.166 gave a binding energy of 3.782 eV. The quantitative breakthrough was the computation of James and Coolidge (1933). Using a 13-parameter function of the form \[ \psi(r_1, r_2) = e^{- \alpha ( \xi_{1}+\xi_{2})} \mbox{ x polynomial in} \{ \xi_{1}, \xi_{2}, \eta_{1}, \eta_{2}, \rho \} , \xi_{i} \equiv \frac{r_{iA} + r_{iB}}{R}, \eta_{i} \equiv \frac{r_{iA}+r_{iB}}{R}, \rho \equiv \frac{r_{12}}{R} \label{9}\] they obtained R e = 1.40 bohr, D e = 4.720 eV. In a sense, this result provided a proof of the validity of quantum mechanics for molecules, in the same sense that Hylleraas' computation on helium was a proof for many-electron atoms. The Valence Bond Theory The basic idea of the Heitler-London model for the hydrogen molecule can be extended to chemical bonds between any two atoms. The orbital function (8) must be associated with the singlet spin function \(\sigma_{0,0}(1,2)\) in order that the overall wavefunction be antisymmetric. This is a quantum-mechanical realization of the concept of an electron-pair bond, first proposed by G. N. Lewis in 1916. It is also now explained why the electron spins must be paired, i.e., antiparallel. It is also permissible to combine an antisymmetric orbital function with a triplet spin function but this will, in most cases, give a repulsive state, as shown by the red curve in Figure \(\PageIndex{2}\). According to valence-bond theory, unpaired orbitals in the valence shells of two adjoining atoms can combine to form a chemical bond if they overlap significantly and are symmetry compatible. A \(\sigma\)-bond is cylindrically symmetrical about the axis joining the atoms. Two s AO's, two p z AO's or an s and a p z can contribute to a \(\sigma\)-bond, as shown in Figure \(\PageIndex{3}\). The z-axis is chosen along the internuclear axis. Two p x or two p y AO's can form a \(\pi\)-bond, which has a nodal plane containing the internuclear axis. Examples of symmetry-incompatible AO's would be an s with a p x or a p x with a p y . In such cases the overlap integral would vanish because of cancelation of positive and negative contributions. Some possible combinations of AO's forming \(\sigma\) and \(\pi\) bonds are shown in Figure \(\PageIndex{3}\). Bonding in the HCl molecule can be attributed to a combination of a hydrogen 1 s with an unpaired 3 p z on chlorine. In Cl 2 , a sigma bond is formed between the 3 p z AO's on each chlorine. As a first approximation, the other doubly-occupied AO's on chlorine-the inner shells and the valence-shell lone pairs-are left undisturbed. The oxygen atom has two unpaired 2 p -electrons, say 2 p x and 2 p y . Each of these can form a \(\sigma\)-bond with a hydrogen 1 s to make a water molecule. It would appear from the geometry of the p -orbitals that the HOH bond angle would be 90°. It is actually around 104.5°. We will resolve this discrepency shortly. The nitrogen atom, with three unpaired 2 p electrons can form three bonds. In NH 3 , each 2 p -orbital forms a \(\sigma\)-bond with a hydrogen 1 s . Again 90° HNH bond angles are predicted, compared with the experimental 107°. The diatomic nitrogen molecule has a triple bond between the two atoms, one \(\sigma\) bond from combining 2 p z AO's and two \(\pi\) bonds from the combinations of 2 p x 's and 2 p y 's, respectively. Hybrid Orbitals and Molecular Geometry To understand the bonding of carbon atoms, we must introduce additional elaborations of valence-bond theory. We can write the valence shell configuration of carbon atom as 2 s 2 2 p x 2 p y , signifying that two of the 2 p orbitals are unpaired. It might appear that carbon would be divalent, and indeed the species CH 2 (carbene or methylene radical) does have a transient existence. But the chemistry of carbon is dominated by tetravalence. Evidently it is a good investment for the atom to promote one of the 2 s electrons to the unoccupied 2 p z orbital. The gain in stability attained by formation of four bonds more than compensates for the small excitation energy. It can thus be understood why the methane molecule CH 4 exists. The molecule has the shape of a regular tetrahedron, which is the result of hybridization , mixing of the s and three p orbitals to form four sp 3 hybrid atomic orbitals. Hybrid orbitals can overlap more strongly with neighboring atoms, thus producing stronger bonds. The result is four C-H \(\sigma\)-bonds, identical except for orientation in space, with 109.5° H-C-H bond angles. Other carbon compounds make use of two alternative hybridization schemes. The s AO can form hybrids with two of the p AO's to give three sp 2 hybrid orbitals, with one p -orbital remaining unhybridized. This accounts for the structure of ethylene (ethene): The C-H and C-C \(\sigma\)-bonds are all trigonal sp 2 hybrids, with 120° bond angles. The two unhybridized p -orbitals form a \(\pi\)-bond, which gives the molecule its rigid planar structure. The two carbon atoms are connected by a double bond, consisting of one \(\sigma\) and one \(\pi\). The third canonical form of sp -hybridization occurs in C-C triple bonds, for example, acetylene (ethyne). Here, two of the p AO's in carbon remain unhybridized and can form two \(\pi\)-bonds, in addition to a \(\sigma\)-bond, with a neighboring carbon: Acetylene H-C\(\equiv\)C-H is a linear molecule since sp -hybrids are oriented 180° apart. The deviations of the bond angles in H 2 O and NH 3 from 90° can be attributed to fractional hybridization. The angle H-O-H in water is 104.5° while H-N-H in ammonia is 107°. It is rationalized that the p -orbitals of the central atom acquire some s -character and increase their angles towards the tetrahedral value of 109.5°. Correspondingly, the lone pair orbitals must also become hybrids. Apparently, for both water and ammonia, a model based on tetrahedral orbitals on the central atoms would be closer to the actual behavior than the original selection of s - and p -orbitals. The hybridization is driven by repulsions between the electron densities of neighboring bonds. Valence Shell Model An elementary, but quite successful, model for determining the shapes of molecules is the valence shell electron repulsion theory (VSEPR), first proposed by Sidgewick and Powell and popularized by Gillespie. The local arrangement of atoms around each multivalent center in the molecule can be represented by AX n - k E k , where X is another atom and E is a lone pair of electrons. The geometry around the central atom is then determined by the arrangement of the n electron pairs (bonding plus nonbonding), which minimizes their mutual repulsion. The following geometric configurations satisfy this condition: n shape 2 linear 5 trigonal bipyramid 3 trigonal planar 6 octahedral 4 tetrahedral 7 pentagonal bipyramid The basic geometry will be distorted if the n surrounding pairs are not identical. The relative strength of repulsion between pairs follows the order E-E > E-X > X-X. In ammonia, for example, which is NH 3 E, the shape will be tetrahedral to a first approximation. But the lone pair E will repel the N-H bonds more than they repel one another. Thus the E-N-H angle will increase from the tetrahedral value of 109.5°, causing the H-N-H angles to decrease slightly. The observed value of 107° is quite well accounted for. In water, OH 2 E 2 , the opening of the E-O-E angle will likewise cause a closing of H-O-H, and again, 104.5° seems like a reasonable value. Valence-bond theory is about 90% successful in explaining much of the descriptive chemistry of ground states. VB theory fails to account for the triplet ground state of O 2 or for the bonding in electron-deffcient molecules such as diborane, B 2 H 6 . It is not very useful in consideration of excited states, hence for spectroscopy. Many of these deficiencies are remedied by molecular orbital theory, which we take up in the next Chapter. |
Courses/Sacramento_City_College/SCC%3A_CHEM_300_-_Beginning_Chemistry/SCC%3A_CHEM_300_-_Beginning_Chemistry_(Faculty)/14%3A_Acids_and_Bases | Acids and bases are common substances found in many every day items, from fruit juices and soft drinks to soap. In this unit we'll exam what the properties are of acids and bases, and learn about the chemical nature of these important compounds. You'll learn what pH is and how to calculate the pH of a solution. 14.1: Sour Patch Kids and International Spy Movies Sour Patch Kids are a soft candy with a coating of invert sugar and sour sugar (a combination of citric acid, tartaric acid and sugar). Its slogan, "Sour. Sweet. Gone.", refers to the sour-to-sweet taste of the candy. 14.2: Acids- Properties and Examples 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. 14.3: Bases- Properties and Examples A base is thought of as a substance which can accept protons or any chemical compound that yields hydroxide ions (OH-) in solution. It is also commonly referred to as any substance that can react with an acid to decrease or neutralize its acidic properties, change the color of indicators (e.g. turn red litmus paper blue), feel slippery to the touch when in solution, taste bitter, react with acids to form salts, and promote certain chemical reactions (e.g. base catalysis). 14.4: Molecular Definitions of Acids and Bases Although the properties of acids and bases had been recognized for a long time, it was Svante Arrhenius in the 1880's who determined that: the properties of acids were due to the presence of hydrogen ions, and the properties of bases were due to the presence of hydroxide ions. 14.5: Reactions of Acids and Bases When an acid and a base are combined, water and a salt are the products. Salts are ionic compounds containing a positive ion other than H+ and a negative ion other than the hydroxide ion, OH-. Double displacement reactions of this type are called neutralization reactions. Salt solutions do not always have a pH of 7, however. Through a process known as hydrolysis, the ions produced when an acid and base combine may react with the water to produce slightly acidic or basic solutions. 14.6: Acid–Base Titration- A Way to Quantify the Amount of Acid or Base in a Solution Acid-base titrations are lab procedures used to determine the concentration of a solution. One of the standard laboratory exercises in General Chemistry is an acid-base titration. During an acid-base titration, an acid with a known concentration (a standard solution) is slowly added to a base with an unknown concentration (or vice versa). A few drops of indicator solution are added to the base. The indicator will signal, by color change, when the base has been neutralized (when [H+] = [OH-]). 14.7: Strong and Weak Acids and Bases Acids are classified as either strong or weak, based on their ionization in water. A strong acid is an acid which is completely ionized in an aqueous solution. A weak acid is an acid that ionizes only slightly in an aqueous solution. Acetic acid (found in vinegar) is a very common weak acid. 14.8: Water- Acid and Base in One Water is an interesting compound in many respects. Here, we will consider its ability to behave as an acid or a base. In some circumstances, a water molecule will accept a proton and thus act as a Brønsted-Lowry base. 14.9: The pH and pOH Scales- Ways to Express Acidity and Basicity pH and pOH are defined as the negative log of hydrogen ion concentration and hydroxide concentration, respectively. Knowledge of ether can be used to calculate either [H+] of [OH-]. pOH is related to pH and can be easily calculated from pH. 14.10: Buffers- Solutions That Resist pH Change A buffer is a solution that resists dramatic changes in pH. Buffers do so by being composed of certain pairs of solutes: either a weak acid plus a salt derived from that weak acid or a weak base plus a salt of that weak base. 14.E: Homework Chapter 14 14.E: Homework Chapter 14 Answers |
Courses/CSU_San_Bernardino/CHEM_2100%3A_General_Chemistry_I_(Mink)/13%3A_Fundamental_Equilibrium_Concepts/13.03%3A_Equilibrium_Constants | Learning Objectives By the end of this section, you will be able to: Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures Relate the magnitude of an equilibrium constant to properties of the chemical system The status of a reversible reaction is conveniently assessed by evaluating its reaction quotient ( Q ) . For a reversible reaction described by \[m A +n B +\rightleftharpoons x C +y D \nonumber \] the reaction quotient is derived directly from the stoichiometry of the balanced equation as \[Q_c=\frac{[ C ]^x[ D ]^y}{[ A ]^m[ B ]^n} \nonumber \] where the subscript c denotes the use of molar concentrations in the expression. If the reactants and products are gaseous, a reaction quotient may be similarly derived using partial pressures: \[Q_p=\frac{P_{ C }^x P_{ D }^y}{P_{ A }{ }^m P_{ B }{ }^n} \nonumber \] Note that the reaction quotient equations above are a simplification of more rigorous expressions that use relative values for concentrations and pressures rather than absolute values. These relative concentration and pressure values are dimensionless (they have no units); consequently, so are the reaction quotients. For purposes of this introductory text, it will suffice to use the simplified equations and to disregard units when computing Q . In most cases, this will introduce only modest errors in calculations involving reaction quotients. Example \(\PageIndex{1}\): Writing Reaction Quotient Expressions Write the concentration-based reaction quotient expression for each of the following reactions: \(\ce{3 O2(g) \rightleftharpoons 2 O3(g)}\) \(\ce{N2(g) + 3 H2(g) \rightleftharpoons 2 NH3(g)}\) \(\ce{4 NH3(g) + 7 O2(g) \rightleftharpoons 4 NO2(g) + 6 H2O(g)}\) Solution \[Q_c=\frac{\left[ \ce{O3} \right]^2}{\left[ \ce{O2} \right]^3} \nonumber \] \[Q_c=\frac{\left[ \ce{NH3} \right]^2}{\left[ \ce{N2} \right]\left[ \ce{H2} \right]^3} \nonumber \] \[Q_c=\frac{\left[ \ce{NO2} \right]^4\left[ \ce{H2O} \right]^6}{\left[ \ce{NH3} \right]^4\left[ \ce{O2} \right]^7} \nonumber \] Exercise \(\PageIndex{1}\) Write the concentration-based reaction quotient expression for each of the following reactions: \(\ce{2SO2(g) + O2(g) \rightleftharpoons 2 SO3(g)}\) \(\ce{C4H8(g) \rightleftharpoons 2 C2H4(g)}\) \(\ce{2 C4H10(g) + 13 O2(g) \rightleftharpoons 8 CO2(g) + 10 H2O (g)}\) Answer \[Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]} \nonumber \] \[Q_c=\frac{\left[ \ce{C2H4} \right]^2}{\left[ \ce{C4H8} \right]} \nonumber \] \[Q_c=\frac{\left[ \ce{CO2} \right]^8\left[ \ce{H2O} \right]^{10}}{\left[ \ce{C4H10} \right]^2\left[ \ce{O2} \right]^{13}} \nonumber \] The numerical value of \(Q\) varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction’s status. To illustrate this point, consider the oxidation of sulfur dioxide: \[\ce{2 SO2(g) + O2(g) \rightleftharpoons 2 SO3(g)} \nonumber \] Two different experimental scenarios are depicted in Figure \(\PageIndex{1}\), one in which this reaction is initiated with a mixture of reactants only, SO 2 and O 2 , and another that begins with only product, SO 3 . For the reaction that begins with a mixture of reactants only, Q is initially equal to zero: \[Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=\frac{0^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=0 \nonumber \] As the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of Q c ), product concentration increases (as does the numerator of Q c ), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of Q c . If the reaction begins with only product present, the value of Q c is initially undefined (immeasurably large, or infinite): \[Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=\frac{\left[ \ce{SO3} \right]^2}{0} \rightarrow \infty \nonumber \] In this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of Q c decrease with time, the reactant concentrations and the denominator of Q c increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium. The constant value of Q exhibited by a system at equilibrium is called the equilibrium con stant, \(K\): \[K \equiv Q \text { at equilibrium } \nonumber \] Comparison of the data plots in Figure \(\PageIndex{1}\) shows that both experimental scenarios resulted in the same value for the equilibrium constant. This is a general observation for all equilibrium systems, known as the law of mass action . Definition: Law of Mass Action At a given temperature, the reaction quotient for a system at equilibrium is constant. Example \(\PageIndex{2}\): Evaluating a Reaction Quotient Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation: \[\ce{2 NO2(g) <=> N2O4(g)} \nonumber \] When 0.10 mol NO 2 is added to a 1.0-L flask at 25 °C, the concentration changes so that at equilibrium, [NO 2 ] = 0.016 M and [N 2 O 4 ] = 0.042 M . What is the value of the reaction quotient before any reaction occurs? What is the value of the equilibrium constant for the reaction? Solution As for all equilibrium calculations in this text, use the simplified equations for \(Q\) and \(K\) and disregard any concentration or pressure units, as noted previously in this section. (a) Before any product is formed \[\left[ \ce{NO2} \right]=\frac{0.10~\text{mol} }{1.0~\text{L} }=0.10~\text{M} \nonumber \] \[[\ce{N2O4}] = 0~\text{M} \nonumber \] Thus \[Q_c=\frac{\left[ \ce{N2O4} \right]}{\left[ \ce{NO2} \right]^2}=\frac{0}{0.10^2}=0 \nonumber \] (b) At equilibrium, \[K_c=Q_c=\frac{\left[ \ce{N2O4} \right]}{\left[ \ce{NO2} \right]^2}=\frac{0.042}{0.016^2}=1.6 \times 10^2. \nonumber \] The equilibrium constant is \(1.6 \times 10^{2}\). Exercise \(\PageIndex{2}\) For the reaction \[\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)} \nonumber \] the equilibrium concentrations are [SO 2 ] = 0.90 M , [O 2 ] = 0.35 M , and [SO 3 ] = 1.1 M . What is the value of the equilibrium constant, K c ? Answer K c = 4.3 By its definition, the magnitude of an equilibrium constant explicitly reflects the composition of a reaction mixture at equilibrium, and it may be interpreted with regard to the extent of the forward reaction. A reaction exhibiting a large K will reach equilibrium when most of the reactant has been converted to product, whereas a small K indicates the reaction achieves equilibrium after very little reactant has been converted. It’s important to keep in mind that the magnitude of K does not indicate how rapidly or slowly equilibrium will be reached. Some equilibria are established so quickly as to be nearly instantaneous, and others so slowly that no perceptible change is observed over the course of days, years, or longer. The equilibrium constant for a reaction can be used to predict the behavior of mixtures containing its reactants and/or products. As demonstrated by the sulfur dioxide oxidation process described above, a chemical reaction will proceed in whatever direction is necessary to achieve equilibrium. Comparing Q to K for an equilibrium system of interest allows prediction of what reaction (forward or reverse), if any, will occur. To further illustrate this important point, consider the reversible reaction shown below: \[\ce{CO(g) + H2O(g) \rightleftharpoons CO2(g) + H2(g)} \quad K_c=0.640 \quad T =800{ }^{\circ} C \nonumber \] The bar charts in Figure \(\PageIndex{2}\) represent changes in reactant and product concentrations for three different reaction mixtures. The reaction quotients for mixtures 1 and 3 are initially lesser than the reaction’s equilibrium constant, so each of these mixtures will experience a net forward reaction to achieve equilibrium. The reaction quotient for mixture 2 is initially greater than the equilibrium constant, so this mixture will proceed in the reverse direction until equilibrium is established. Example \(\PageIndex{3}\): Predicting the Direction of Reaction Given here are the starting concentrations of reactants and products for three experiments involving this reaction: \[\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)} \quad K_c=0.64 \nonumber \] Determine in which direction the reaction proceeds as it goes to equilibrium in each of the three experiments shown. Reactants/Products Experiment 1 Experiment 2 Experiment 3 [CO]i 0.020 M 0.011 M 0.0094 M [H2O]i 0.020 M 0.0011 M 0.0025 M [CO2]i 0.0040 M 0.037 M 0.0015 M [H2]i 0.0040 M 0.046 M 0.0076 M Solution Experiment 1: \[Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.0040)(0.0040)}{(0.020)(0.020)}=0.040 \nonumber \] Q c < K c (0.040 < 0.64) The reaction will proceed in the forward direction. Experiment 2: \[Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.037)(0.046)}{(0.011)(0.0011)}=1.4 \times 10^2 \nonumber \] Q c > K c (140 > 0.64) The reaction will proceed in the reverse direction. Experiment 3: \[Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.0015)(0.0076)}{(0.0094)(0.0025)}=0.48 \nonumber \] Q c < K c (0.48 < 0.64) The reaction will proceed in the forward direction. Exercise \(\PageIndex{3}\) Calculate the reaction quotient and determine the direction in which each of the following reactions will proceed to reach equilibrium. A 1.00-L flask containing 0.0500 mol of NO(g), 0.0155 mol of Cl 2 (g), and 0.500 mol of NOCl: \[\ce{2 NO(g) + Cl2(g) <=> 2 NOCl(g)} \quad K_c=4.6 \times 10^4 \nonumber \] A 5.0-L flask containing 17 g of NH 3 , 14 g of N 2 , and 12 g of H 2 : \[\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)} \quad K_c=0.060 \nonumber \] A 2.00-L flask containing 230 g of SO 3 (g): \[\ce{2 SO3(g) <=> 2 SO2(g) + O2(g)} \quad K_c=0.230 \nonumber \] Answer (a) Q c = 6.45 \times 10^{3} DELMAR, forward. (b) Q c = 0.23, reverse. (c) Q c = 0, forward. Homogeneous Equilibria A homogeneous equilibrium is one in which all reactants and products (and any catalysts, if applicable) are present in the same phase. By this definition, homogeneous equilibria take place in solutions . These solutions are most commonly either liquid or gaseous phases, as shown by the examples below: \[\begin{aligned} \ce{C2H2(aq) + 2 Br2(aq) & \rightleftharpoons C2H2Br4(aq)} & K_c & =\frac{\left[ C_2 H_2 Br_4\right]}{\left[ C_2 H_2\right]\left[ Br_2\right]^2} \\[4pt] \ce{I2(aq) + I^{-}(aq) & \rightleftharpoons I_3^{-}(aq)} & K_c & =\frac{\left[ I_3-\right.}{\left[ I_2\right]\left[ I^{-}\right]} \\[4pt] \ce{HF(aq) + H2O(l) & \rightleftharpoons H3O^{+}(aq) + F^{-}(aq)} & K_c & =\frac{\left[ H_3 O^{+}\right]\left[ F^{-}\right]}{[ HF ]} \\[4pt] \ce{NH3(aq) + H2O(l) & \rightleftharpoons NH4^{+}(aq) + OH^{-}(aq)} & K_c & =\frac{\left[ NH_4^{+}\right]\left[ OH^{-}\right]}{\left[ NH_3\right]} \end{aligned} \nonumber \] These examples all involve aqueous solutions, those in which water functions as the solvent. In the last two examples, water also functions as a reactant, but its concentration is not included in the reaction quotient. The reason for this omission is related to the more rigorous form of the Q (or K ) expression mentioned previously in this chapter, in which relative concentrations for liquids and solids are equal to 1 and needn’t be included . Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species. 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 equilibria below all involve gas-phase solutions: \[\begin{aligned} \ce{C2H6(g) & \rightleftharpoons C2H4(g) + H2(g)} & K_c & =\frac{\left[ C_2 H_4\right]\left[ H_2\right]}{\left[ C_2 H_6\right]} \\[4pt] \ce{3 O2(g) & \rightleftharpoons 2 O3(g)} & K_c & =\frac{\left[ O_3\right]^2}{\left[ O_2\right]^3} \\[4pt] \ce{N2(g) + 3 H2(g) & \rightleftharpoons 2 NH3(g)} & K_c & =\frac{\left[ NH_3\right]^2}{\left[ N_2\right]\left[ H_2\right]^3} \\[4pt] \ce{C3H8(g) + 5 O2(g) & \rightleftharpoons 3 CO2(g) + 4 H2O(g)} & K_c & =\frac{\left[ CO_2\right]^3\left[ H_2 O \right]^4}{\left[ C_3 H_8\right]\left[ O_2\right]^5} \end{aligned} \nonumber \] For gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations ( K c ) or partial pressures ( K p ) of the reactants and products. A relation between these two K values may be simply derived from the ideal gas equation and the definition of molarity: \[\begin{align*} P V &=n R T \\[4pt] P & =\left(\frac{n}{V}\right) R T \\[4pt] & =M R T \end{align*} \nonumber \] where \(P\) is partial pressure, \(V\) is volume, \(n\) is molar amount, \(R\) is the gas constant, \(T\) is temperature, and \(M\) is molar concentration. For the gas-phase reaction \[a A +b B \rightleftharpoons c C +d D \nonumber \] \[\begin{align*} K_P &=\frac{\left(P_C\right)^c\left(P_D\right)^d}{\left(P_A\right)^a\left(P_B\right)^b} \\[4pt] &= \dfrac{([ C ] \times R T)^c([ D ] \times R T)^d}{([ A ] \times R T)^a([ B ] \times R T)^b} \\[4pt] &= \dfrac{[ C ]^c[ D ]^d}{[ A ]^a[ B ]^b} \times \frac{(R T)^{c+d}}{(R T)^{a+b}} \\[4pt] &= K_c(R T)^{(c+d)-(a+b)} \\[4pt] &= K_c(R T)^{\Delta n} \end{align*} \nonumber \] And so, the relationship between K c and K P is \[K_P=K_c(R T)^{\Delta n} \nonumber \] where \(Δn\) is the difference in the molar amounts of product and reactant gases, in this case: \[\Delta n=(c+d)-(a+b) \nonumber \] Example \(\PageIndex{4}\): Calculation of K P Write the equations relating K c to K P for each of the following reactions: \(\ce{C2H6(g) <=> C2H4(g) + H2(g)}\) \(\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}\) \(\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}\) K c is equal to 0.28 for the following reaction at 900 °C: \[\ce{CS2(g) + 4 H2(g) <=> CH4(g) + 2 H2S(g)} \nonumber \] What is K P at this temperature? Solution \(Δn = (2) − (1) = 1\) \[K_P = K_c (RT)^{Δn} = K_c (RT)^1 = K_c (RT) \nonumber \] \(Δn = (2) − (2) = 0\) \[K_P = K_c (RT)^{Δn} = K_c (RT)^0 = K_c \nonumber \] \(Δn = (2) − (1 + 3) = −2\) \[K_P = K_c (RT)^{Δn} = K_c (RT)^{−2} = \dfrac{K_c}{(R T)^2} \nonumber \] \[K_P = K_c (RT)^{Δn} = (0.28)[(0.0821)(1173)]^{−2} = 3.0 \times 10^{−5} \nonumber \] Exercise \(\PageIndex{4}\) Write the equations relating K c to K P for each of the following reactions: \(\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)}\) \(\ce{N2O4(g) <=> 2 NO2(g)}\) \(\ce{C3H8(g) + 5 O2(g) <=> 3 CO2(g) + 4 H2O (g)}\) At 227 °C, the following reaction has K c = 0.0952: \[\ce{CH3OH(g) <=> CO(g) + 2 H2(g)} \nonumber \] What would be the value of K P at this temperature? Answer (a) K P = K c ( RT ) −1 ; (b) K P = K c ( RT ); (c) K P = K c ( RT ); (d) 160 or 1.6 \times 10^{2} DELMAR Heterogeneous Equilibria A heterogeneous equilibrium involves reactants and products in two or more different phases, as illustrated by the following examples: \[\begin{align*} \ce{PbCl2(s) & \rightleftharpoons Pb^{2+}(aq) + 2 Cl^{-}(aq)} & K_c & =\left[ \ce{Pb^{2+}} \right]\left[ \ce{Cl^{-}} \right]^2 \\[4pt] \ce{CaO(s) + CO2(g) & \rightleftharpoons CaCO3(s)} & K_c & =\frac{1}{\left[ \ce{CO_2}\right]} \\[4pt] \ce{C(s) + 2S(g) & \rightleftharpoons CS2(g)} & K_c & =\frac{\left[ CS_2\right]}{\left[ \ce{S^2} \right.} \\[4pt] \ce{Br2(l) & \rightleftharpoons Br2(g)} & K_c & =\left[ \ce{Br_2(g)} \right] \end{align*} \nonumber \] Again, note that concentration terms are only included for gaseous and solute species, as discussed previously. Two of the above examples include terms for gaseous species only in their equilibrium constants, and so K p expressions may also be written: \[\begin{align*} \ce{CaO(s) + CO2(g) & \rightleftharpoons CaCO3(s)} & K_P & =\frac{1}{P_{ \ce{CO2}}} \\[4pt] \ce{C(s) + 2 S(g) & \rightleftharpoons CS2(g)} & K_P & =\frac{P_{ \ce{CS2}}}{\left(P_{ \ce{S} }\right)^2} \end{align*} \nonumber \] Coupled Equilibria The equilibrium systems discussed so far have all been relatively simple, involving just single reversible reactions. Many systems, however, involve two or more coupled equilibrium reactions, those which have in common one or more reactant or product species. Since the law of mass action allows for a straightforward derivation of equilibrium constant expressions from balanced chemical equations, the K value for a system involving coupled equilibria can be related to the K values of the individual reactions. Three basic manipulations are involved in this approach, as described below. 1. Changing the direction of a chemical equation essentially swaps the identities of “reactants” and “products,” and so the equilibrium constant for the reversed equation is simply the reciprocal of that for the forward equation. \[\begin{array}{ll} A \rightleftharpoons B & K_{ c }=\frac{[ B ]}{[ A ]} \\[4pt] B \rightleftharpoons A & K_{ c^{\prime}}=\frac{[ A ]}{[ B ]} \end{array} \nonumber \] \[K_{ c^{\prime}}=\frac{1}{ K_{ c }} \nonumber \] 2. Changing the stoichiometric coefficients in an equation by some factor x results in an exponential change in the equilibrium constant by that same factor: \[\begin{array}{ll} A \rightleftharpoons B & K_{ c }=\frac{[ B ]}{[ A ]} \\[4pt] xA \rightleftharpoons xB & K_{ c }=\frac{[ B ]^{ x }}{[ A ]^{ x }} \end{array} \nonumber \] \[K_{ c^{\prime}}= K_{ c }{ }^{ x } \nonumber \] 3. Adding two or more equilibrium equations together yields an overall equation whose equilibrium constant is the mathematical product of the individual reaction’s K values: \[\begin{array}{ll} A \rightleftharpoons B & K_{ c 1}=\frac{[ B ]}{[ A ]} \\[4pt] B \rightleftharpoons C & K_{ c 2}=\frac{[ C ]}{[ B ]} \end{array} \nonumber \] The net reaction for these coupled equilibria is obtained by summing the two equilibrium equations and canceling any redundancies: \[\begin{aligned} & A + B \rightleftharpoons B + C \\[4pt] & A + B \rightleftharpoons B + C \\[4pt] & A \rightleftharpoons C \end{aligned} \nonumber \] \[K_{ c^{\prime}}=\frac{[ C ]}{[ A ]} \nonumber \] Comparing the equilibrium constant for the net reaction to those for the two coupled equilibrium reactions reveals the following relationship: \[K_{ c 1} K_{ c 2}=\frac{[ B ]}{[ A ]} \times \frac{[ C ]}{[ B ]}=\frac{ \cancel{[ B ]}[ C ]}{[ A ] \cancel{[ B ]}}=\frac{[ C ]}{[ A ]}= K_{ c^{\prime}} \nonumber \] \[K_{ c^{\prime}}= K_{ c 1} K_{ c 2} \nonumber \] Example \(\PageIndex{5}\) demonstrates the use of this strategy in describing coupled equilibrium processes. Example \(\PageIndex{5}\): Equilibrium Constants for Coupled Reactions A mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 °C: \[\ce{2 NH3(g) + 3 I2(g) \rightleftharpoons N2(g) + 6 HI(g)} \nonumber \] Use the information below to calculate Kc for this reaction. \[\begin{align*} \ce{N2(g) + 3 H2(g) \rightleftharpoons 2 NH3(g)} & K_{ c 1}=0.50 \text { at } 400{ }^{\circ} C \\[4pt] \ce{H2(g) + I2(g) \rightleftharpoons 2 HI(g)} & K_{ c 2}=50 \text { at } 400{ }^{\circ} C \end{align*} \nonumber \] Solution The equilibrium equation of interest and its K value may be derived from the equations for the two coupled reactions as follows. Reverse the first coupled reaction equation: \[2 NH_3(g) \rightleftharpoons N_2(g) + 3 H_2(g) \quad K_{ c 1}{ }^{\prime}=\frac{1}{ K_{ c 1}}=\frac{1}{0.50}=2.0 \nonumber \] Multiply the second coupled reaction by 3: \[3 H_2(g) + 3 I_2(g) \rightleftharpoons 6 HI (g) \quad K_{ c 2}{ }^{\prime}= K_{ c 2}^3=50^3=1.2 \times 10^5 \nonumber \] Finally, add the two revised equations: \[\begin{align*} \ce{2 NH3(g) + 3 H2(g) + 3 I2(g) &\rightleftharpoons N2(g) + 3 H2(g) + 6 HI(g)} \\[4pt] \ce{2 NH3(g) + 3 I2(g) &\rightleftharpoons N2(g) + 6HI(g)} \end{align*} \nonumber \] \[K_{ c }= K_{ c 1}, K_{ c 2},=(2.0)\left(1.2 \times 10^5\right)=2.5 \times 10^5 \nonumber \] Exercise \(\PageIndex{5}\) Use the provided information to calculate Kc for the following reaction at 550 °C: \[\begin{align*} \ce{H2(g) + CO2(g) &\rightleftharpoons CO(g) + H2O(g)} && K_{ c }=? \\[4pt] \ce{CoO(s) + CO(g) &\rightleftharpoons Co(s)+ CO2(g)} && K_{ c 1}=490 \\[4pt] \ce{CoO(s) + H2(g) &\rightleftharpoons Co(s)+ H2O(g)} && K_{ c 1}=67 \end{align*} \nonumber \] Answer K c = 0.14 |
Courses/University_of_Arkansas_Little_Rock/Chem_1402%3A_General_Chemistry_1_(Belford)/Homework/08%3A_Bonding_and_Molecular_Structure/8.05_Drawing_Lewis_Dot_Structures | Lewis Dot Structures, Electron and Molecular Geometry Exercise \(\PageIndex{1}\) Draw the best Lewis Dot Structure for each of the following species. Give the name of the electronic arrangement and the name for the molecular geometry for each of the species. BeF 2 , BCl 3 , CCl 4 , PBr 5 , SI 6 , BH 2 – , NI 3 , ClF 4 + , SF 5 – Answer 0 1 2 3 Species Name Lewis Dot Structure Electron Geometry Molecular Geometry BeF2 NaN linear linear BCl3 NaN trigonal planar trigonal planar CCl4 NaN tetrahedral tetrahedral PBr5 NaN trigonal bipyramidal trigonal bipyramidal SI6 NaN octahedral octahedral BH2– NaN trigonal planar bent NI3 NaN tetrahedral trigonal pyramidal ClF4+ NaN trigonal bipyramidal see saw SF5– NaN octahedral square pyramidal |
Courses/Oregon_Institute_of_Technology/OIT%3A_CHE_332_--_Organic_Chemistry_II_(Lund)/10%3A_Nucleophilic_Carbonyl_Addition_Reactions/10.S%3A_Nucleophilic_Carbonyl_Addition_Reactions_(Summary) | Before moving on to the next chapter, you should be confident in your ability to: Recognize aldehyde and ketone groups in organic biomolecules Draw/explain the bonding picture for aldehyde and ketone groups Explain why the carbonyl carbon in an aldehyde or ketone is electrophilic Draw complete curved arrow mechanisms for the following reaction types: formation of a hemiacetal/hemiketal collapse of a hemiacetal/hemiketal to revert to an aldehyde/ketone formation and hydrolysis of an acetal/ketal formation and hydrolysis of an N-glycosidic bond formation and hydrolysis of an imine transimination Explain how the carbocation intermediates in glycosidic bond formation and hydrolysis reactions are stabilized by resonance Explain the stereochemical considerations of a nucleophilic addition to an aldehyde/ketone, especially in the context of glycosidic bond formation. Be able to identify the re and si faces of an aldehyde, ketone, or imine. In addition to these fundamental skills, you should develop your confidence in working with end-of-chapter problems involving more challenging, multi-step biochemical reactions. |
Courses/University_of_British_Columbia/CHEM_100%3A_Foundations_of_Chemistry/02%3A_Measurement_and_Problem_Solving/2.02%3A_Scientific_Notation_-_Writing_Large_and_Small_Numbers | Learning Objectives Express a large number or a small number in scientific notation. Carry out arithmetical operations and express the final answer in scientific notation Chemists often work with numbers that are exceedingly large or small. For example, entering the mass in grams of a hydrogen atom into a calculator would require a display with at least 24 decimal places. A system called scientific notation avoids much of the tedium and awkwardness of manipulating numbers with large or small magnitudes. In scientific notation, these numbers are expressed in the form \[ N \times 10^n \nonumber \] where N is greater than or equal to 1 and less than 10 (1 ≤ N < 10), and n is a positive or negative integer (10 0 = 1). The number 10 is called the base because it is this number that is raised to the power \(n\). Although a base number may have values other than 10, the base number in scientific notation is always 10. A simple way to convert numbers to scientific notation is to move the decimal point as many places to the left or right as needed to give a number from 1 to 10 (N). The magnitude of n is then determined as follows: If the decimal point is moved to the left n places, n is positive. If the decimal point is moved to the right n places, n is negative. Another way to remember this is to recognize that as the number N decreases in magnitude, the exponent increases and vice versa. The application of this rule is illustrated in Example \(\PageIndex{1}\). Example \(\PageIndex{1}\): Expressing Numbers in Scientific Notation Convert each number to scientific notation. 637.8 0.0479 7.86 12,378 0.00032 61.06700 2002.080 0.01020 Solution Unnamed: 0 Explanation Answer a To convert 637.8 to a number from 1 to 10, we move the decimal point two places to the left: 637.8 Because the decimal point was moved two places to the left, n = 2. \(6.378 \times 10^2\) b To convert 0.0479 to a number from 1 to 10, we move the decimal point two places to the right: 0.0479 Because the decimal point was moved two places to the right, n = −2. \(4.79 \times 10^{−2}\) c This is usually expressed simply as 7.86. (Recall that 100 = 1.) \(7.86 \times 10^0\) d Because the decimal point was moved four places to the left, n = 4. \(1.2378 \times 10^4\) e Because the decimal point was moved four places to the right, n = −4. \(3.2 \times 10^{−4}\) f Because the decimal point was moved one place to the left, n = 1. \(6.106700 \times 10^1\) g Because the decimal point was moved three places to the left, n = 3. \(2.002080 \times 10^3\) h Because the decimal point was moved two places to the right, n = -2. \(1.020 \times 10^{−2}\) Addition and Subtraction Before numbers expressed in scientific notation can be added or subtracted, they must be converted to a form in which all the exponents have the same value. The appropriate operation is then carried out on the values of N. Example \(\PageIndex{2}\) illustrates how to do this. Example \(\PageIndex{2}\): Expressing Sums and Differences in Scientific Notation Carry out the appropriate operation and then express the answer in scientific notation. \( (1.36 \times 10^2) + (4.73 \times 10^3) \nonumber\) \((6.923 \times 10^{−3}) − (8.756 \times 10^{−4}) \nonumber\) Solution Unnamed: 0 Explanation Answer a Both exponents must have the same value, so these numbers are converted to either \((1.36 \times 10^2) + (47.3 \times 10^2) = \) \((1.36 + 47.3) \times 10^2 = 48.66 × 10^2\) or \((0.136 \times 10^3) + (4.73 \times 10^3) =\) \((0.136 + 4.73) \times 10^3) = 4.87 \times 10^3\). Choosing either alternative gives the same answer, reported to two decimal places. In converting 48.66 × 102 to scientific notation, \(n\) has become more positive by 1 because the value of \(N\) has decreased. \(4.87 \times 10^3\) b Converting the exponents to the same value gives either \((6.923 \times 10^{-3}) − (0.8756 \times 10^{-3})= \) \((6.923 − 0.8756) \times 10^{−3}\) or \((69.23 \times 10^{-4}) − (8.756 \times 10^{-4}) =\) \( (69.23 − 8.756) \times 10^{−4} = 60.474 \times 10^{−4}\). In converting 60.474 × 10-4 to scientific notation, \(n\) has become more positive by 1 because the value of \(N\) has decreased. \(6.047 \times 10^{−3}\) Multiplication and Division When multiplying numbers expressed in scientific notation, we multiply the values of \(N\) and add together the values of \(n\). Conversely, when dividing, we divide \(N\) in the dividend (the number being divided) by \(N\) in the divisor (the number by which we are dividing) and then subtract n in the divisor from n in the dividend. In contrast to addition and subtraction, the exponents do not have to be the same in multiplication and division. Examples of problems involving multiplication and division are shown in Example \(\PageIndex{3}\). Example \(\PageIndex{3}\): Expressing Products and Quotients in Scientific Notation Perform the appropriate operation and express your answer in scientific notation. \( (6.022 \times 10^{23})(6.42 \times 10^{−2}) \nonumber\) \( \dfrac{ 1.67 \times 10^{-24} }{ 9.12 \times 10 ^{-28} } \nonumber \) \( \dfrac{ (6.63 \times 10^{−34})(6.0 \times 10) }{ 8.52 \times 10^{−2}} \nonumber \) Solution Unnamed: 0 Explanation Unnamed: 2 a In multiplication, we add the exponents: \((6.022 \times 10^{23})(6.42 \times 10^{−2})\)\(= (6.022)(6.42) \times 10^{[23 + (−2)]} = 38.7 \times 10^{21} \nonumber \) In converting \(38.7 \times 10^{21}\) to scientific notation, \(n\) has become more positive by 1 because the value of \(N\) has decreased. \(3.87 \times 10^{22}\) b In division, we subtract the exponents: \({1.67 \times 10^{−24} \over 9.12 \times 10^{−28}} = \) \({1.67 \over 9.12} \times 10^{[−24 − (−28)]} = 0.183 \times 10^4 \) In converting \(0.183 \times 10^4\) to scientific notation, \(n\) has become more negative by 1 because the value of \(N\) has increased. \( 1.83 \times 10^3\) c This problem has both multiplication and division: \( {(6.63 \times 10^{−34})(6.0 \times 10) \over (8.52 \times 10^{−2})} = \) \({39.78 \over 8.52} \times 10^{[−34 + 1 − (−2)]} \nonumber \) \( 4.7\times 10^{-31}\) |
Courses/University_of_San_Diego/Fall_2024_Chem_220_Analytical_Chemistry_David_De_Haan/04%3A_Equilibrium_and_Activity | 4.1: Solving Equilibrium Problems In this section we will learn how to set‐up and solve equilibrium problems. 4.2: Buffer Solutions Adding as little as 0.1 mL of concentrated HCl to a liter of \(\text{H}_2\text{O}\) shifts the pH from 7.0 to 3.0. Adding the same amount of HCl to a liter of a solution that 0.1 M in acetic acid and 0.1 M in sodium acetate, however, results in a negligible change in pH. Why do these two solutions respond so differently to the addition of HCl? A mixture of acetic acid and sodium acetate is one example of an acid–base buffer. 4.3: Activity Effects Careful measurements on the metal–ligand complex \(\text{Fe(SCN)}^{2+}\) suggest its stability, and thus its equilibrium constant, decreases in the presence of inert ions. Understanding why this is so is critical to developing a complete understanding of equilibrium chemistry. |
Courses/UWMilwaukee/CHE_125%3A_GOB_Introductory_Chemistry/02%3A_Measurements_and_Density/2.05%3A_The_Basic_Units_of_Measurement | Learning Objectives State the different measurement systems used in chemistry. Describe how prefixes are used in the metric system and identify how the prefixes milli-, centi-, and kilo- compare to the base unit. How long is a yard? It depends on whom you ask and when you asked the question. Today we have a standard definition of the yard, which you can see marked on every football field. If you move the ball ten yards, you get a first down and it does not matter whether you are playing in Los Angeles, Dallas, or Green Bay. But at one time that yard was arbitrarily defined as the distance from the tip of the king's nose to the end of his outstretched hand. Of course, the problem there is simple: new king, new distance (and then you have to re-mark all of those football fields). SI Base Units All measurements depend on the use of units that are well known and understood. The English system of measurement units (inches, feet, ounces, etc.) are not used in science because of the difficulty in converting from one unit to another. The metric system is used because all metric units are based on multiples of 10, making conversions very simple. The metric system was originally established in France in 1795. The International System of Units is a system of measurement based on the metric system. The acronym SI is commonly used to refer to this system and stands for the French term, Le Système International d'Unités . The SI was adopted by international agreement in 1960 and is composed of seven base units in Table \(\PageIndex{1}\). Quantity SI Base Unit Symbol Length meter \(\text{m}\) Mass kilogram \(\text{kg}\) Temperature kelvin \(\text{K}\) Time second \(\text{s}\) Amount of a Substance mole \(\text{mol}\) Electric Current ampere \(\text{A}\) Luminous Intensity candela \(\text{cd}\) The first units are frequently encountered in chemistry. All other measurement quantities, such as volume, force, and energy, can be derived from these seven base units. Unfortunately, the Metric System is Not Ubiquitous The map below shows the adoption of the SI units in countries around the world. The United States has legally adopted the metric system for measurements, but does not use it in everyday practice. Great Britain and much of Canada use a combination of metric and imperial units. Prefix Multipliers Conversions between metric system units are straightforward because the system is based on powers of ten. For example, meters, centimeters, and millimeters are all metric units of length. There are 10 millimeters in 1 centimeter and 100 centimeters in 1 meter. Metric prefixes are used to distinguish between units of different size. These prefixes all derive from either Latin or Greek terms. For example, mega comes from the Greek word \(\mu \varepsilon \gamma \alpha \varsigma\), meaning "great". Table \(\PageIndex{2}\) lists the most common metric prefixes and their relationship to the central unit that has no prefix. Length is used as an example to demonstrate the relative size of each prefixed unit. Prefix Unit Abbreviation Meaning Example giga \(\text{G}\) 1000000000 1 gigameter \(\left( \text{Gm} \right)=10^9 \: \text{m}\) mega \(\text{M}\) 1000000 1 megameter \(\left( \text{Mm} \right)=10^6 \: \text{m}\) kilo \(\text{k}\) 1000 1 kilometer \(\left( \text{km} \right)=1,000 \: \text{m}\) hecto \(\text{h}\) 100 1 hectometer \(\left( \text{hm} \right)=100 \: \text{m}\) deka \(\text{da}\) 10 1 dekameter \(\left( \text{dam} \right)=10 \: \text{m}\) NaN NaN 1 1 meter \(\left( \text{m} \right)\) deci \(\text{d}\) 1/10 1 decimeter \(\left( \text{dm} \right)=0.1 \: \text{m}\) centi \(\text{c}\) 1/100 1 centimeter \(\left( \text{cm} \right)=0.01 \: \text{m}\) milli \(\text{m}\) 1/1,000 1 millimeter \(\left( \text{mm} \right)=0.001 \: \text{m}\) micro \(\mu\) 1/1,000,000 1 micrometer \(\left( \mu \text{m} \right)=10^{-6} \: \text{m}\) nano \(\text{n}\) 1/1,000,000,000 1 nanometer \(\left( \text{nm} \right)=10^{-9} \: \text{m}\) pico \(\text{p}\) 1/1,000,000,000,000 1 picometer \(\left( \text{pm} \right)=10^{-12} \: \text{m}\) There are a couple of odd little practices with the use of metric abbreviations. Most abbreviations are lowercase. We use "\(\text{m}\)" for meter and not "\(\text{M}\)". However, when it comes to volume, the base unit "liter" is abbreviated as "\(\text{L}\)" and not "\(\text{l}\)". So we would write 3.5 milliliters as \(3.5 \: \text{mL}\). As a practical matter, whenever possible you should express the units in a small and manageable number. If you are measuring the weight of a material that weighs \(6.5 \: \text{kg}\), this is easier than saying it weighs \(6500 \: \text{g}\) or \(0.65 \: \text{dag}\). All three are correct, but the \(\text{kg}\) units in this case make for a small and easily managed number. However, if a specific problem needs grams instead of kilograms, go with the grams for consistency. Example \(\PageIndex{1}\): Unit Abbreviations Give the abbreviation for each unit and define the abbreviation in terms of the base unit. kiloliter microsecond decimeter nanogram Solutions Unnamed: 0 Explanation Answer a The prefix kilo means “1,000 ×,” so 1 kL equals 1,000 L. kL b The prefix micro implies 1/1,000,000th of a unit, so 1 µs equals 0.000001 s. µs c The prefix deci means 1/10th, so 1 dm equals 0.1 m. dm d The prefix nano means 1/1000000000, so a nanogram is equal to 0.000000001 g. ng Exercise \(\PageIndex{1}\) Give the abbreviation for each unit and define the abbreviation in terms of the base unit. kilometer milligram nanosecond centiliter Answer a: km Answer b: mg Answer c: ns Answer d: cL Summary Metric prefixes derive from Latin or Greek terms. The prefixes are used to make the units manageable. The SI system is based on multiples of ten. There are seven basic units in the SI system. Five of these units are commonly used in chemistry. |
Courses/University_of_California_Davis/UCD_Chem_002A/UCD_Chem_2A/Homework/Homework/Homework_3 | 3.1 Write the products of these acid-base reactions. Balance the complete reaction. \(HF_{(aq)} + KOH_{(aq)} \rightarrow\) \(H_2SO_{4\;(aq)} + Ca(OH)_{2\;(aq)} \rightarrow\) 3.2 Balance the following redox reaction in basic conditions with state symbols. \[H_2O_{(l)} + Na_{(s)} \rightarrow Na^+_{(aq)} + H_{2\;(g)}\] 3.3 In a lab experiment, a 20.00 ml \(HNO_2\) solution was titrated with \(0.1234\; M\) sodium hydroxide solution, and the equivalence point was reached when 43.21 ml of this basic solution was added. What is the original molarity of the \(HNO_2\) solution? 3.4 What is the required mL of 1.25 M \(HBr\) necessary to completely neutralize a 175 mL sample of 2.38 M \(Mg(OH)_2\)? 3.5 A rubber duck is initially filled with 100 mL air at \(2 ^oC\) in the bathroom and then set in the sun to be heated. The final temperature increases to \(10 ^oC\) at constant pressure. What is the final condition of the rubber duck? Does it shrink or expand? What is its final volume? 3.6 A commercial balloon is filled with 3000 L helium gas at \(30^oC\). If the pressure is 1 atm, what is the mass of helium gas? Why does the balloon float? 3.7 Uranium hexafluoride (\(UF_6\)) is a compound used in the uranium enrichment process that produces fuel for nuclear reactors and nuclear weapons. It forms solid grey crystals at standard temperature and pressure and boils at 56 °C . Figure: Uranium hexafluoride crystals sealed in an ampoule The density of uranium hexafluoride is 12.6 g/L at 745 torr. What is the temperature in °C under these condition? 3.8 A constant-volume container has 2.00 moles gas sample at 25 °C and the gas sample exerts a pressure of 450 torr. If 2.00 moles more gas is added in the container and the temperature is increased to 50 °C, what is the pressure in atm? |
Courses/Saint_Francis_University/Chem_114%3A_Human_Chemistry_II_(Hargittai)/19%3A_Enzymes_and_Vitamins/19.03%3A_Enzyme_Classification | Learning Objectives Objective 1 Objective 2 Hundreds of enzymes have been purified and studied in an effort to understand how they work so effectively and with such specificity. The resulting knowledge has been used to design drugs that inhibit or activate particular enzymes. An example is the intensive research to improve the treatment of or find a cure for acquired immunodeficiency syndrome (AIDS). AIDS is caused by the human immunodeficiency virus (HIV). Researchers are studying the enzymes produced by this virus and are developing drugs intended to block the action of those enzymes without interfering with enzymes produced by the human body. Several of these drugs have now been approved for use by AIDS patients. Enzyme Nomenclature Most enzymes can be recognized because they have the family name ending – ase . However, the first enzymes to be discovered were named according to their source or method of discovery. The enzyme pepsin , which aids in the hydrolysis of proteins, is found in the digestive juices of the stomach (Greek pepsis , meaning “digestion”). Papain , another enzyme that hydrolyzes protein (in fact, it is used in meat tenderizers), is isolated from papayas. In addition to the family name, more systematic enzyme names will give two sepcific pieces of information: the first part is the substrate upon which the enzyme acts, and the second part is the type of reaction it catalyzes. For example, alcohol dehydrogenase (Figure \(\PageIndex{1}\)) catalyzes the oxidation of an alcohol to an aldehyde. Enzyme Classification As more enzymes were discovered, chemists recognized the need for a more systematic and chemically informative identification scheme. In the current numbering and naming scheme, under the oversight of the Nomenclature Commission of the International Union of Biochemistry, enzymes are arranged into six groups according to the general type of reaction they catalyze (Table \(\PageIndex{1}\)), with subgroups and secondary subgroups that specify the reaction more precisely. Each enzyme is assigned a four-digit number, preceded by the prefix EC—for enzyme classification—that indicates its group, subgroup, and so forth. This is demonstrated in Table \(\PageIndex{2}\) for alcohol dehydrogenase. Main Class Type of Reaction Catalyzed Subclasses Examples Oxidoreductases oxidation-reduction reactions Dehydrogenases catalyze oxidation-reduction reactions involving hydrogen. Alcohol dehydrogenase NaN NaN Oxidases catalyze oxidation by addition of O2 to a substrate. NaN NaN NaN Reductases catalyze reactions in which a substrate is reduced. NaN Transferases transfer reactions of functional groups Transaminases catalyze the transfer of amino group. NaN NaN NaN Kinases catalyze the transfer of a phosphate group. Phosphofructokinase Hydrolases reactions that use water to break a chemical bond Lipases catalyze the hydrolysis of lipids NaN NaN NaN Proteases catalyze the hydrolysis of proteins NaN NaN NaN Amylases catalyze the hydrolysis of carbohydrates NaN NaN NaN Nucleases catalyze the hydrolysis of DNA and RNA NaN Lyases reactions in which functional groups are added or removed without hydrolysis Decarboxylases catalyze the removal of carboxyl groups. NaN NaN NaN Deaminases catalyze the removal of amino groups. NaN NaN NaN Dehydratases catalyze the removal of water. NaN NaN NaN Hydratases catalyze the addition of water. Fumarase Isomerases reactions in which a compound is converted to its isomer Isomerases may catalyze the conversion of an aldose to a ketose. Triose Phosphate Isomerase NaN NaN Mutases catalyze reactions in which a functional group is transferred from one atom in a substrate to another. NaN Ligases reactions in which new bonds are formed between carbon and another atom; energy is required Synthetases catalyze reactions in which two smaller molecules are linked to form a larger one. NaN NaN NaN Carboxylases catalyze the addition of CO2 using ATP Pyruvate Carboxylase Alcohol Dehydrogenase: EC 1.1.1.1 Alcohol Dehydrogenase: EC 1.1.1.1.1 The first digit indicates that this enzyme is an oxidoreductase; that is, an enzyme that catalyzes an oxidation-reduction reaction. The first digit indicates that this enzyme is an oxidoreductase; that is, an enzyme that catalyzes an oxidation-reduction reaction. The second digit indicates that this oxidoreductase catalyzes a reaction involving a primary or secondary alcohol. The second digit indicates that this oxidoreductase catalyzes a reaction involving a primary or secondary alcohol. The third digit indicates that either the coenzyme NAD+ or NADP+ is required for this reaction. The third digit indicates that either the coenzyme NAD+ or NADP+ is required for this reaction. The fourth digit indicates that this was the first enzyme isolated, characterized, and named using this system of nomenclature. The fourth digit indicates that this was the first enzyme isolated, characterized, and named using this system of nomenclature. The systematic name for this enzyme is alcohol:NAD+ oxidoreductase, while the recommended or common name is alcohol dehydrogenase. The systematic name for this enzyme is alcohol:NAD+ oxidoreductase, while the recommended or common name is alcohol dehydrogenase. Reaction catalyzed: NaN Figure \(\PageIndex{1}\): Structure of the alcohol dehydrogenase protein (E.C.1.1.1.1) (EE ISOZYME) complexed wtih nicotinamide adenini dinulceotide (NAD) and zinc (PDB: 1CDO). Summary An enzyme is a biological catalyst, a substance that increases the rate of a chemical reaction without being changed or consumed in the reaction. A systematic process is used to name and classify enzymes. |
Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Introduction_to_Lasers/06_References | http://micro.magnet.fsu.edu/primer/l...sersintro.html D. Sands, Diode Lasers, lOP Publishing, 2005. http://www.bell-labs.com/history/laser/ http://laserstars.org/history nobelprize.org/nobel_prizes/p...ownes-bio.html http://micro.magnet.fsu.edu/optics/t...le/maiman.html J.C. Wright, M.J. Wirth, Anal. Chem. 52, 1980, 988A and 1087 A. W. Demtr ö der , Laser Spectroscopy, Springer, Berlin, 2002 (3 rd Ed). P.W. Milonni, J.H. Eberly Lasers Wiley, NY 1988. http://www.rp-photonics.com/encyclopedia.html M. Gerloch, Orbitals, Terms and States, Wiley, New York, 1986. H.J.R. Dutton, Understanding Optical Communications, IBM Report SG24-5230-00, 1998, http://www.redbooks.ibm.com (History and fiber- optics) http://www1.union.edu/newmanj/Physics100/index.htm High Power Diode Lasers, F. Bachmann, et al. Eds.; Springer: 2007. Additional references are included in the " Applications " section. Author Contact Information Carol Korzeniewski Department of Chemistry & Biochemistry Texas Tech University Lubbock, TX 79409-1061 [email protected] |
Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/11%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.04%3A_Interpretting_IR_Spectra | Guided IR Spectrum Interpretation Now, let’s take a look at the IR spectrum for 1-hexanol. There is a very broad ‘mountain’ centered at about 3400 cm -1 . This signal is characteristic of the O-H stretching mode of alcohols, and is a dead giveaway for the presence of an alcohol group. The breadth of this signal is a consequence of hydrogen bonding between molecules. In the spectrum of octanoic acid we see, as expected, the characteristic carbonyl peak, this time at 1709 cm -1 . We also see a low, broad absorbance band that looks like an alcohol, except that it is displaced slightly to the right (long-wavelength) side of the spectrum, causing it to overlap to some degree with the C-H region. This is the characteristic carboxylic acid O-H single bond stretching absorbance. The spectrum for 1-octene shows two peaks that are characteristic of alkenes: the one at 1642 cm -1 is due to stretching of the carbon-carbon double bond, and the one at 3079 cm-1 is due to stretching of the s bond between the alkene carbons and their attached hydrogens. Alkynes have characteristic IR absorbance peaks in the range of 2100-2250 cm -1 due to stretching of the carbon-carbon triple bond, and terminal alkenes can be identified by their absorbance at about 3300 cm-1, due to stretching of the bond between the sp-hybridized carbon and the terminal hydrogen. It is possible to identify other functional groups such as amines and ethers, but the characteristic peaks for these groups are considerably more subtle and/or variable, and often are overlapped with peaks from the fingerprint region. For this reason, we will limit our discussion here to the most easily recognized functional groups, which are summarized in this table . As you can imagine, obtaining an IR spectrum for a compound will not allow us to figure out the complete structure of even a simple molecule, unless we happen to have a reference spectrum for comparison. In conjunction with other analytical methods, however, IR spectroscopy can prove to be a very valuable tool, given the information it provides about the presence or absence of key functional groups. IR can also be a quick and convenient way for a chemist to check to see if a reaction has proceeded as planned. If we were to run a reaction in which we wished to convert cyclohexanone to cyclohexanol, for example, a quick comparison of the IR spectra of starting compound and product would tell us if we had successfully converted the ketone group to an alcohol. More examples of IR spectra To illustrate the usefulness of infrared absorption spectra, examples for five C 4 H 8 O isomers are presented below their corresponding structural formulas. Try to associate each spectrum with one of the isomers in the row above it. Exercises |
Courses/Lumen_Learning/Book%3A_US_History_I_(AY_Collection)_(Lumen)/15%3A_The_Sectional_Crisis/15.7%3A_Primary_Source_Reading%3A_The_Declaration_of_Sentiments | The Declaration of Sentiments, Seneca Falls Conference, 1848 Elizabeth Cady Stanton and Lucretia Mott, two American activists in the movement to abolish slavery called together the first conference to address Women’s rights and issues in Seneca Falls, New York, in 1848. Part of the reason for doing so had been that Mott had been refused permission to speak at the world anti-slavery convention in London, even though she had been an official delegate. Applying the analysis of human freedom developed in the Abolitionist movement, Stanton and others began the public career of modern feminist analysis The Declaration of the Seneca Falls Convention, using the model of the U.S. Declaration of Independence, forthrightly demanded that the rights of women as right-bearing individuals be acknowledged and respected by society. It was signed by sixty-eight women and thirty-two men. The Declaration of Sentiments When, in the course of human events, it becomes necessary for one portion of the family of man to assume among the people of the earth a position different from that which they have hitherto occupied, but one to which the laws of nature and of nature’s God entitle them, a decent respect to the opinions of mankind requires that they should declare the causes that impel them to such a course. We hold these truths to be self-evident: that all men and women are created equal; that they are endowed by their Creator with certain inalienable rights; that among these are life, liberty, and the pursuit of happiness; that to secure these rights governments are instituted, deriving their just powers from the consent of the governed. Whenever any form of government becomes destructive of these ends, it is the right of those who suffer from it to refuse allegiance to it, and to insist upon the institution of a new government, laying its foundation on such principles, and organizing its powers in such form, as to them shall seem most likely to effect their safety and happiness. Prudence, indeed, will dictate that governments long established should not be changed for light and transient causes; and accordingly all experience hath shown that mankind are more disposed to suffer. while evils are sufferable, than to right themselves by abolishing the forms to which they are accustomed. But when a long train of abuses and usurpations, pursuing invariably the same object, evinces a design to reduce them under absolute despotism, it is their duty to throw off such government, and to provide new guards for their future security. Such has been the patient sufferance of the women under this government, and such is now the necessity which constrains them to demand the equal station to which they are entitled. The history of mankind is a history of repeated injuries and usurpations on the part of man toward woman, having in direct object the establishment of an absolute tyranny over her. To prove this, let facts be submitted to a candid world. The history of mankind is a history of repeated injuries and usurpations on the part of man toward woman, having in direct object the establishment of an absolute tyranny over her. To prove this, let facts be submitted to a candid world. He has never permitted her to exercise her inalienable right to the elective franchise. He has compelled her to submit to laws, in the formation of which she had no voice. He has withheld from her rights which are given to the most ignorant and degraded men–both natives and foreigners. Having deprived her of this first right of a citizen, the elective franchise, thereby leaving her without representation in the halls of legislation, he has oppressed her on all sides. He has made her, if married, in the eye of the law, civilly dead. He has taken from her all right in property, even to the wages she earns. He has made her, morally, an irresponsible being, as she can commit many crimes with impunity, provided they be done in the presence of her husband. In the covenant of marriage, she is compelled to promise obedience to her husband, he becoming, to all intents and purposes, her master–the law giving him power to deprive her of her liberty, and to administer chastisement. He has so framed the laws of divorce, as to what shall be the proper causes, and in case of separation, to whom the guardianship of the children shall be given, as to be wholly regardless of the happiness of women–the law, in all cases, going upon a flase supposition of the supremacy of man, and giving all power into his hands. After depriving her of all rights as a married woman, if single, and the owner of property, he has taxed her to support a government which recognizes her only when her property can be made profitable to it. He has monopolized nearly all the profitable employments, and from those she is permitted to follow, she receives but a scanty remuneration. He closes against her all the avenues to wealth and distinction which he considers most homorable to himself. As a teacher of theology, medicine, or law, she is not known. He has denied her the facilities for obtaining a thorough education, all colleges being closed against her. He allows her in church, as well as state, but a subordinate position, claiming apostolic authority for her exclusion from the ministry, and, with some exceptions, from any public participation in the affairs of the church. He has created a false public sentiment by giving to the world a different code of morals for men and women, by which moral delinquencies which exclude women from society, are not only tolerated, but deemed of little account in man. He has usurped the prerogative of Jehovah himself, claiming it as his right to assign for her a sphere of action, when that belongs to her conscience and to her God. He has endeavored, in every way that he could, to destroy her confidence in her own powers, to lessen her self-respect, and to make her willing to lead a dependent and abject life. Now, in view of this entire disfranchisement of one-half the people of this country, their social and religious degradation–in view of the unjust laws above mentioned, and because women do feel themselves aggrieved, oppressed, and fraudulently deprived of their most sacred rights, we insist that they have immediate admission to all the rights and privileges which belong to them as citizens of the United States. Source : from Elizabeth Cady Stanton, A History of Woman Suffrage , vol. 1 (Rochester, N.Y.: Fowler and Wells, 1889), pages 70-71. Public domain content Internet Modern History Sourcebook. Located at : http://legacy.fordham.edu/halsall/mod/Senecafalls.asp . License : Public Domain: No Known Copyright |
Courses/Grinnell_College/CHM_363%3A_Physical_Chemistry_1_(Grinnell_College)/09%3A_Helmholtz_and_Gibbs_Energies/9.04%3A_Thermodynamic_Functions_have_Natural_Variables | The fundamental thermodynamic equations follow from five primary thermodynamic definitions and describe internal energy, enthalpy, Helmholtz energy, and Gibbs energy in terms of their natural variables. Here they will be presented in their differential forms. Introduction The fundamental thermodynamic equations describe the thermodynamic quantities U, H, G, and A in terms of their natural variables. The term "natural variable" simply denotes a variable that is one of the convenient variables to describe U, H, G, or A. When considered as a whole, the four fundamental equations demonstrate how four important thermodynamic quantities depend on variables that can be controlled and measured experimentally. Thus, they are essentially equations of state, and using the fundamental equations, experimental data can be used to determine sought-after quantities like \(G\) or \(H\). First Law of Thermodynamics The first law of thermodynamics is represented below in its differential form \[ dU = đq+đw \nonumber \] where \(U\) is the internal energy of the system, \(q\) is heat flow of the system, and \(w\) is the work of the system. The "đ" symbol represent inexact differentials and indicates that both \(q\) and \(w\) are path functions. Recall that \(U\) is a state function. The first law states that internal energy changes occur only as a result of heat flow and work done. It is assumed that w refers only to PV work, where \[ w = -\int{pdV} \nonumber \] The fundamental thermodynamic equation for internal energy follows directly from the first law and the principle of Clausius: \[ dU = đq + đw \nonumber \] \[ dS = \dfrac{\delta q_{rev}}{T} \nonumber \] we have \[ dU = TdS + \delta w \nonumber \] Since only \(PV\) work is performed, \[ dU = TdS - pdV \label{DefU} \] The above equation is the fundamental equation for \(U\) with natural variables of entropy \(S\) and volume\(V\). Principle of Clausius The Principle of Clausius states that the entropy change of a system is equal to the ratio of heat flow in a reversible process to the temperature at which the process occurs. Mathematically this is written as \[ dS = \dfrac{\delta q_{rev}}{T} \nonumber \] where \(S\) is the entropy of the system, \(q_{rev}\) is the heat flow of a reversible process, and \(T\) is the temperature in Kelvin. Enthalpy Mathematically, enthalpy is defined as \[ H = U + pV \label{DefEnth} \] where \(H\) is enthalpy of the system, \(p\) is pressure, and \(V\) is volume. The fundamental thermodynamic equation for enthalpy follows directly from it deffinition (Equation \(\ref{DefEnth}\)) and the fundamental equation for internal energy (Equation \(\ref{DefU}\)) : \[ \begin{align*} dH &= dU + d(pV) \\[4pt] &= dU + pdV + VdP \\[4pt] dU &= TdS - pdV \\[4pt] dH &= TdS - pdV + pdV + Vdp \\[4pt] dH &= TdS + Vdp \end{align*} \] The above equation is the fundamental equation for \(H\). The natural variables of enthalpy are \(S\) and \(p\), entropy and pressure. Gibbs Energy The mathematical description of Gibbs energy is as follows \[ \begin{align} G &= U + pV - TS \nonumber \\[4pt] &= H - TS \label{Defgibbs} \end{align}\] where \(G\) is the Gibbs energy of the system. The fundamental thermodynamic equation for Gibbs Energy follows directly from its definition \(\ref{Defgibbs}\) and the fundamental equation for enthalpy \(\ref{DefEnth}\): \[ \begin{align*} dG &= dH - d(TS) \\[4pt] & dH - TdS - SdT \end{align*}\] Since \[ \begin{align} dH &= TdS + VdP \nonumber \\[4pt] dG &= TdS + VdP - TdS - SdT \nonumber \\[4pt] &= VdP - SdT \label{EqGibbs1} \end{align}\] The above equation is the fundamental equation for \(G\). The natural variables of Gibbs energy are \(P\) and \(T\). Helmholtz Energy Mathematically, Helmholtz energy is defined as \[ A = U - TS \label{DefHelm} \] where \(A\) is the Helmholtz energy of the system, which is often written as the symbol \(F\). The fundamental thermodynamic equation for Helmholtz energy follows directly from its definition (Equation \(\ref{DefHelm}\)) and the fundamental equation for internal energy (Equation \(\ref{DefU}\)): \[ \begin{align*} dA &= dU - d(TS) \\[4pt] &= dU - TdS - SdT \end{align*} \] Since \[ \begin{align} dU &= TdS - pdV \nonumber \\[4pt] dA &= TdS - pdV -TdS - SdT \nonumber \\[4pt] &= -pdV - SdT \label{EqHelm1} \end{align} \] Equation \(\ref{EqHelm1}\) is the fundamental equation for \(A\) with natural variables of \(V\) and \(T\). For the definitions to hold, it is assumed that only PV work is done and that only reversible processes are used. These assumptions are required for the first law and the principle of Clausius to remain valid. Also, these equations do not account include n, the number of moles, as a variable. When \(n\) is included, the equations appear different, but the essence of their meaning is captured without including the n-dependence. Chemical Potential The fundamental equations derived above were not dependent on changes in the amounts of species in the system. Below the n-dependent forms are presented 1 ,4 . \[ dU = TdS - PdV + \sum_{i=1}^{N}\mu_idn_i \nonumber \] \[ dH = TdS + VdP + \sum_{i=1}^{N}\mu_idn_i \nonumber \] \[ dG = -SdT + Vdp + \sum_{i=1}^{N}\mu_idn_i \nonumber \] \[ dA = -SdT - PdV + \sum_{i=1}^{N}\mu_idn_i \nonumber \] where μ i is the chemical potential of species i and dn i is the change in number of moles of substance i. Importance/Relevance of Fundamental Equations The differential fundamental equations describe U, H, G, and A in terms of their natural variables. The natural variables become useful in understanding not only how thermodynamic quantities are related to each other, but also in analyzing relationships between measurable quantities (i.e. P, V, T) in order to learn about the thermodynamics of a system. Below is a table summarizing the natural variables for U, H, G, and A: Thermodynamic Quantity Natural Variables U (internal energy) S, V H (enthalpy) S, P G (Gibbs energy) T, P A (Helmholtz energy) T, V Maxwell Relations The fundamental thermodynamic equations are the means by which the Maxwell relations are derived 1 ,4 . The Maxwell Relations can, in turn, be used to group thermodynamic functions and relations into more general "families" 2,3 . As we said dA is an exact differential . Let's write is out in its natural variables (Equation \(\ref{EqHelm1}\)) and take a cross derivative. The dA expression in natural variables is \[dA = \left( \dfrac{\partial A}{\partial V} \right)_T dV + \left( \dfrac{\partial A}{\partial T} \right) _V dT \nonumber \] The partial derivatives of A of first order can already be quite interesting we see e.g. in step 2 that the first partial of A versus V (at T constant) is the negative of the pressure. \[ \left( \dfrac{\partial A}{\partial V} \right)_T = -P \nonumber \] Likewise we find the (isochoric) slope with temperature gives us the negative of the entropy. Thus entropy is one of the first order derivatives of A. \[\left( \dfrac{\partial A}{\partial T} \right) _V = -S \nonumber \] When we apply a cross derivative \[ \left( \dfrac{\partial^2 A}{\partial V \partial T} \right) = \left( \dfrac{\partial (-S)}{\partial V} \right) _T + \left( \dfrac{\partial (-P)}{\partial T} \right) _V \nonumber \] we get what is known as a Maxwell relation : \[ \left( \dfrac{\partial P}{\partial T} \right) _V = \left( \dfrac{\partial S}{\partial V} \right) _T \nonumber \] What does Equation three mean for the heat capacity? answer A similar treatment of dG (Equation \(\ref{EqGibbs1}\) gives: \[\left( \dfrac{\partial G}{\partial T} \right) _P = -S \nonumber \] \[\left( \dfrac{\partial G}{\partial P} \right) _T = V \nonumber \] and another Maxwell relation \[ - \left( \dfrac{\partial S}{\partial P} \right) _T = \left( \dfrac{\partial V}{\partial T} \right) _P \nonumber \] Exercise \(\PageIndex{1}\) If the assumptions made in the derivations above were not made, what would effect would that have? Try to think of examples were these assumptions would be violated. Could the definitions, principles, and laws used to derive the fundamental equations still be used? Why or why not? Answer If it was not assumed that PV-work was the only work done, then the work term in the second law of thermodynamics equation would include other terms (e.g. for electrical work, mechanical work). If reversible processes were not assumed, the Principle of Clausius could not be used. One example of such situations could the movement of charged particles towards a region of like charge (electrical work) or an irreversible process like combustion of hydrocarbons or friction. Exercise \(\PageIndex{2}\) For what kind of system does the number of moles not change? This said, do the fundamental equations without n-dependence apply to a wide range of processes and systems? Answer In general, a closed system of non-reacting components would fit this description. For example, the number of moles would not change for a closed system in which a gas is sealed (to prevent leaks) in a container and allowed to expand/is contracted. Exercise \(\PageIndex{3}\) Derive the Maxwell Relations. Answer See the Maxwell Relations section. Exercise \(\PageIndex{4}\) Derive the expression \[ \left (\dfrac{\partial H}{\partial P} \right)_{T,n} = -T \left(\dfrac{\partial V}{\partial T} \right)_{P,n} +V \nonumber \] Then apply this equation to an ideal gas. Does the result seem reasonable? Answer \((\dfrac{\partial H}{\partial P})_{T,n} = 0 \) for an ideal gas. Since there are no interactions between ideal gas molecules, changing the pressure will not involve the formation or breaking of any intermolecular interactions or bonds. Exercise \(\PageIndex{5}\) Using the definition of Gibbs energy and the conditions observed at phase equilibria, derive the Clapeyron equation. Answer See the third outside link. References DOI: 10.1063/1.1749582 DOI: 10.1063/1.1749549 DOI:10.1103/PhysRev.3.273 A Treatise on Physical Chemistry, 3rd ed.; Taylor, H. S. and Glasstone, S., Eds.; D. Van Nostrand Company: New York, 1942; Vol. 1; p 454-485. |
Courses/Fullerton_College/Introductory_Biochemistry/07%3A_Conformations_and_Stereochemistry/7.04%3A_Solutions_to_7.3_Problems | Exercises on Bond Line Structures, Drawing, and Isomers 1. a. Both compounds are 4-ethyl-2-methylhexane. Therefore, they are identical compounds. b. The first compound is 2,3,5-trimethylheptane, and the second is 3-ethyl-2,5-dimethylhexane. Therefore, these are isomers of one another. c. The first compound is 2,3,3-trimethylpentane, and the second is also 2,3,3-trimethylpentane. Therefore, these are identical. 2. a. Rotating the second structure yields the first structure. Therefore these are identical. b. Non-mirror image and non-superimposable. Therefore, these are diastereomers. c. Mirror images and non-superimposable on it's mirror image. These are enantiomers. 3. 4. Functional groups: carboxylic acid, amine, amide, thiol Amino acids: glutamate/glutamic acid, cysteine, glycine Stereoisomers: two stereocenters, \(2^2 = 4\) stereoisomers |
Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/18%3A_Carboxylic_Acids_and_Their_Derivatives/18.10%3A_Reactions_of_Unsaturated_Carboxylic_Acids_and_Their_Derivatives | Unsaturated carboxylic acids of the type \(\ce{RCH=CH(CH_2)}_n \ce{COOH}\) usually exhibit the properties characteristic of isolated double bonds and isolated carboxyl groups when \(n\) is large and the functional groups are far apart. As expected, exceptional behavior is found most commonly when the groups are sufficiently close together to interact strongly, as in \(\alpha\),\(\beta\)-unsaturated acids, \(\ce{R} \overset{\beta}{\ce{C}} \ce{H=} \overset{\alpha}{\ce{C}} \ce{CO_2H}\). We shall emphasize those properties that are exceptional in the following discussion. Migration of the Double Bond In the presence of strong base, \(\alpha\),\(\beta\)- and \(\beta\),\(\gamma\)-unsaturated carboxylic acids tend to interconvert by migration of the double bond: Ester derivatives, \(\ce{RCH=CH-CH_2COOR'}\), and the corresponding unsaturated aldehydes and ketones, \(\ce{RCH=CH-CH_2COR'}\), are much more prone to this type of rearrangement than are the acids. Hydration and Hydrogen Bromide Addition Like alkenes, the double bonds of \(\alpha\),\(\beta\)-unsaturated acids can be brominated, hydroxylated, hydrated, and hydrobrominated, although the reactions often are relatively slow. In the addition of unsymmetrical reagents the direction of addition is opposite to that observed for alkenes (anti-Markovnikov). Thus propenoic (acrylic) acid adds hydrogen bromide and water to form 3-bromo- and 3-hydroxypropanoic acids: These additions are analogous to the addition of halogens and halogen acids to 1,3-butadiene ( Section 13-2 ). In the first step, a proton is transferred to the carbonyl oxygen. The resulting conjugate acid can be regarded as a resonance hybrid of structures \(16a\)-\(16d\): In the second step, a nucleophile (such as \(\ce{Br}^\ominus\) or a water molecule) attacks an electron-deficient carbon of the hybrid \(16\). Attack at the carboxyl carbon may occur but does not lead to a stable product. Attack of the nucleophile at the \(\beta\) carbon, however, produces the enol form of the \(\beta\)-substituted acid, which then is converted rapidly to the normal carboxylic acid: Lactone Formation When the double bond of an unsaturated acid is farther down the carbon chain than between the alpha and beta positions, the so-called "conjugate addition" is not possible. Nonetheless, the double bond and carboxyl group frequently interact in the presence of acidic catalysts because the carbocation that results from addition of a proton to the double bond has a built-in nucleophile (the carboxyl group), which may attack the cationic center to form a cyclic ester called a lactone . Lactone formation only occurs readily by this mechanism when a five- or six-membered ring can be formed: Five- or six-membered lactones also are formed by internal esterification when either \(\gamma\)- or \(\delta\)-hydroxy acids are heated. Under similar conditions, \(\beta\)-hydroxy acids are dehydrated to \(\alpha\),\(\beta\)-unsaturated acids, whereas \(\alpha\)-hydroxy acids undergo bimolecular esterification to substances with six-membered dilactone rings called lactides : More on the Michael Addition The foregoing examples of addition to the double bonds of unsaturated carboxylic acids all involve activation by an electrophilic species such as \(\ce{H}^\oplus\). Conjugate addition also may occur by nucleophilic attack on acid derivatives, the most important being the base-catalyzed Michael addition ( Section 17-5B ) and 1,4-addition of organometallic compounds ( Section 14-12D ). In all of these reactions a nucleophilic agent, usually a carbanion, attacks the double bond of an \(\alpha\),\(\beta\)-unsaturated compound in which the double bond is conjugated with, and activated by, a strongly electronegative unsaturated group (such as \(\ce{-CN}\), \(\ce{-NO_2}\), etc.). In the Michael addition, the carbanion usually is an enolate salt. The overall reaction is illustrated here by the specific example of the addition of diethyl propanedioate (diethyl malonate) to ethyl 3-phenylpropenoate (ethyl cinnamate): The mechanism of this kind of transformation, with diethyl propanedioate as the addend, is outlined in Equations 18-25 and 18-26. The basic catalyst required for the Michael addition (here symbolized as \(\ce{B:}\)) serves by forming the corresponding anion: A variety of nucleophilic agents can be used; propanedinitrile, 3-oxobutanoate esters, and cyanoethanoate esters all form relatively stable carbanions and function well in Michael addition reactions. Obviously, if the carbanion is too stable, it will have little or no tendency to attack the double bond of the \(\alpha\),\(\beta\)-unsaturated acid derivative. Enamines ( Sections 16-4C and 17-4B ) are excellent addends in many Michael-type reactions. An example is provided by the addition of \(\ce{N}\)-(1-cyclohexenyl)-azacyclopentane to methyl 2-methylpropanoate: |
Courses/Mt._San_Antonio_College/Chem_10_-_Chemistry_for_Allied_Health_Majors_(1st_semester)/02%3A_Matter_and_Energy/2.01%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/Lumen_Learning/Book%3A_Microeconomics-2_(Lumen)/12%3A_Module-_Monopoly/12.17%3A_Outcome-_Revenue_Costs_Profit_and_Losses_in_Monopolies | What you’ll learn to do: calculate and graph a monopoly’s fixed, variable, average, marginal and total costs We know that because a monopolist controls the market for a good or service, they get more say in how much they want to produce and what price to sell it at. In this outcome, you’ll see how they make those decisions. Here are some of the specific things you’ll learn to do in this section: Measure variable and total costs as the area under the average variable and average total cost curves Calculate and graph the firm’s average, marginal and total revenues Measure total revenues as the area under the average revenue curves Determine the profit maximizing output level and price; calculate and graphically illustrate where marginal revenue equals marginal costs Calculate and graphically illustrate profit and losses for a monopolist LEARNING ACTIVITIES The learning activities for this section include the following: Reading: Choosing Output and Price Reading: Illustrating Monopoly Profits Self Check: Revenue, Costs, Profit and Losses in Monopolies Take time to review and reflect on each of these activities in order to improve your performance on the assessment for this section. CC licensed content, Original Authored by : Steven Greenlaw and Lumen Learning. License : CC BY: Attribution |
Courses/can/CHEM_210%3A_General_Chemistry_I_(An_Atoms_Up_Approach)/15%3A_Gases_and_Gas_Laws | 15.1: Gas Pressure - a Result of Collisions Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers. 15.2: The Gas Laws The behavior of gases can be modeled with gas laws. Boyle's law relates a gas's pressure and volume at constant temperature and amount. Charles's law relates a gas's volume and temperature at constant pressure and amount. In gas laws, temperatures must always be expressed in kelvins. 15.3: Other Gas Relationships There are gas laws that relate any two physical properties of a gas. The combined gas law relates pressure, volume, and temperature of a gas. 15.4: Ideal Gases and The Ideal Gas Law 15.5: Dalton's Law of Partial Pressures The pressure exerted by each gas in a gas mixture is independent of the pressure exerted by all other gases present. Consequently, the total pressure exerted by a mixture of gases is the sum of the partial pressures of the components (Dalton’s law of partial pressures). The amount of gas in a mixture may be described by its partial pressure or its mole fraction. In a mixture, the partial pressure of each gas is the product of the total pressure and the mole fraction. 15.5.1: Vapor Pressure 15.6: Ideal Gases and Real Gases We imagine that the results of a large number of experiments are available for our analysis. Our characterization of these results has been that all gases obey the same equations—Boyle’s law, Charles’ law, and the ideal gas equation—and do so exactly. This is an oversimplification. In fact they are always approximations. They are approximately true for all gases under all “reasonable” conditions, but they are not exactly true for any real gas under any condition. 15.7: Gas Stoichiometry The ideal gas law relates the four independent physical properties of a gas at any time. The ideal gas law can be used in stoichiometry problems whose chemical reactions involve gases. Standard temperature and pressure (STP) are a useful set of benchmark conditions to compare other properties of gases. At STP, gases have a volume of 22.4 L per mole. The ideal gas law can be used to determine densities of gases. |
Courses/Oregon_Tech_PortlandMetro_Campus/OT_-_PDX_-_Metro%3A_General_Chemistry_III/00%3A_Front_Matter/02%3A_InfoPage | This text is disseminated via the Open Education Resource (OER) LibreTexts Project ( https://LibreTexts.org ) and like the hundreds of other texts available within this powerful platform, it is freely available for reading, printing and "consuming." Most, but not all, pages in the library have licenses that may allow individuals to make changes, save, and print this book. Carefully consult the applicable license(s) before pursuing such effects. Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new technologies to support learning. The LibreTexts mission is to unite students, faculty and scholars in a cooperative effort to develop an easy-to-use online platform for the construction, customization, and dissemination of OER content to reduce the burdens of unreasonable textbook costs to our students and society. The LibreTexts project is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being optimized by students, faculty, and outside experts to supplant conventional paper-based books. These free textbook alternatives are organized within a central environment that is both vertically (from advance to basic level) and horizontally (across different fields) integrated. The LibreTexts libraries are Powered by NICE CXOne and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. This material is based upon work supported by the National Science Foundation under Grant No. 1246120, 1525057, and 1413739. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation nor the US Department of Education. Have questions or comments? For information about adoptions or adaptions contact [email protected] . More information on our activities can be found via Facebook ( https://facebook.com/Libretexts ), Twitter ( https://twitter.com/libretexts ), or our blog ( http://Blog.Libretexts.org ). This text was compiled on 04/21/2025 |
Courses/Purdue/Chem_26505%3A_Organic_Chemistry_I_(Lipton)/Chapter_2._Functional_Groups_and_Nomenclature/2.03_Alkynes | Alkynes are organic molecules made of the functional group carbon-carbon triple bonds and are written in the empirical formula of \(C_nH_{2n-2}\). They are unsaturated hydrocarbons. Like alkenes have the suffix –ene, alkynes use the ending –yne; this suffix is used when there is only one alkyne in the molecule. Introduction Here are the molecular formulas and names of the first ten carbon straight chain alkynes. 0 1 Name Molecular Formula Ethyne C2H2 Propyne C3H4 1-Butyne C4H6 1-Pentyne C5H8 1-Hexyne C6H10 1-Heptyne C7H12 1-Octyne C8H14 1-Nonyne C9H16 1-Decyne C10H18 The more commonly used name for ethyne is acetylene, which used industrially. Naming Alkynes Like previously mentioned, the IUPAC rules are used for the naming of alkynes. Rule 1 Find the longest carbon chain that includes both carbons of the triple bond. Rule 2 Number the longest chain starting at the end closest to the triple bond. A 1-alkyne is referred to as a terminal alkyne and alkynes at any other position are called internal alkynes. For example: 4-chloro-6-diiodo-7-methyl-2-nonyne Rule 3 After numbering the longest chain with the lowest number assigned to the alkyne, label each of the substituents at its corresponding carbon. While writing out the name of the molecule, arrange the substituents in alphabetical order. If there are more than one of the same substituent use the prefixes di, tri, and tetra for two, three, and four substituents respectively. These prefixes are not taken into account in the alphabetical order. For example: 1-triiodo-4-dimethyl-2-nonyne If there is an alcohol present in the molecule, number the longest chain starting at the end closest to it, and follow the same rules. However, the suffix would be –ynol, because the alcohol group takes priority over the triple bond. 5- methyl-7-octyn-3-ol When there are two triple bonds in the molecule, find the longest carbon chain including both the triple bonds. Number the longest chain starting at the end closest to the triple bond that appears first. The suffix that would be used to name this molecule would be –diyne. For example: 4-methyl-1,5-octadiyne Rule 4 Substituents containing a triple bond are called alkynyl. For example: 1-chloro-1-ethynyl-4-bromocyclohexane Here is a table with a few of the alkynyl substituents: 0 1 Name Molecule Ethynyl -C?CH 2- Propynyl -CH2C?CH 2-Butynyl -CH3C?CH2CH3 Rule 5 A molecule that contains both double and triple bonds is called an alkenyne. The chain can be numbered starting with the end closest to the functional group that appears first. For example: 6-ethyl-3-methyl-1,4-nonenyne Outside links http://en.wikipedia.org/wiki/Alkyne http://www.cem.msu.edu/~reusch/VirtualText/nomen1.htm Reference Vollhardt, Peter, and Neil E. Schore. Organic Chemistry: Structure and Function . 5th Edition. New York: W. H. Freeman & Company, 2007. Problems Name or draw out the following molecules: 1. 4,4-dimethyl-2-pentyne 2. 4-Penten-1-yne 3. 1-ethyl-3-dimethylnonyne 4. Contributors A. Sheth and S. Sujit (UCD) |
Courses/CSU_Chico/CSU_Chico%3A_CHEM_451_-_Biochemistry_I/CHEM_451_Test/07%3A_Binding/7.2%3A_Mathematical_Analyses_of_Binding_Graphs | Learning Objectives derive equations from the equation for fractional saturation Y = ([ML]/[M0] = [L]/(KD+ [L]) equations for 1/Y as a function of 1/[L] (double reciprocal plot), Y/[L] vs Y (bound/free vs bound or the Scatchard Plot) and draw a plot of Y vs log [L] (semilog plot) draw error bars on the plots of Y vs L, 1/Y vs I/L, Y/L vs Y and Y vs log L. explain from these derivative equations and graphs how to calculate determine Kd describe appropriate methods to fit the data from these equations with special attention given to the effect of error bars in the experimental value of Y Template:HideTOC |
Courses/Lumen_Learning/Book%3A_General_College_Chemistry_I_(Lumen)/10%3A_8-_Gases/10.05%3A_Effusion_and_Diffusion_of_Gases | Learning Objectives By the end of this section, you will be able to: Define and explain effusion and diffusion State Graham’s law and use it to compute relevant gas properties If you have ever been in a room when a piping hot pizza was delivered, you have been made aware of the fact that gaseous molecules can quickly spread throughout a room, as evidenced by the pleasant aroma that soon reaches your nose. Although gaseous molecules travel at tremendous speeds (hundreds of meters per second), they collide with other gaseous molecules and travel in many different directions before reaching the desired target. At room temperature, a gaseous molecule will experience billions of collisions per second. The mean free path is the average distance a molecule travels between collisions. The mean free path increases with decreasing pressure; in general, the mean free path for a gaseous molecule will be hundreds of times the diameter of the molecule. In general, we know that when a sample of gas is introduced to one part of a closed container, its molecules very quickly disperse throughout the container; this process by which molecules disperse in space in response to differences in concentration is called diffusion (shown in Figure 1). The gaseous atoms or molecules are, of course, unaware of any concentration gradient, they simply move randomly—regions of higher concentration have more particles than regions of lower concentrations, and so a net movement of species from high to low concentration areas takes place. In a closed environment, diffusion will ultimately result in equal concentrations of gas throughout, as depicted in Figure 1. The gaseous atoms and molecules continue to move, but since their concentrations are the same in both bulbs, the rates of transfer between the bulbs are equal (no net transfer of molecules occurs). Figure 1. (a) Two gases, H 2 and O 2 , are initially separated. (b) When the stopcock is opened, they mix together. The lighter gas, H 2 , passes through the opening faster than O 2 , so just after the stopcock is opened, more H 2 molecules move to the O 2 side than O 2 molecules move to the H 2 side. (c) After a short time, both the slower-moving O 2 molecules and the faster-moving H 2 molecules have distributed themselves evenly on both sides of the vessel. We are often interested in the rate of diffusion , the amount of gas passing through some area per unit time: The diffusion rate depends on several factors: the concentration gradient (the increase or decrease in concentration from one point to another); the amount of surface area available for diffusion; and the distance the gas particles must travel. Note also that the time required for diffusion to occur is inversely proportional to the rate of diffusion, as shown in the rate of diffusion equation. A process involving movement of gaseous species similar to diffusion is effusion , the escape of gas molecules through a tiny hole such as a pinhole in a balloon into a vacuum (Figure 2). Although diffusion and effusion rates both depend on the molar mass of the gas involved, their rates are not equal; however, the ratios of their rates are the same. Figure 2. Diffusion occurs when gas molecules disperse throughout a container. Effusion occurs when a gas passes through an opening that is smaller than the mean free path of the particles, that is, the average distance traveled between collisions. Effectively, this means that only one particle passes through at a time. If a mixture of gases is placed in a container with porous walls, the gases effuse through the small openings in the walls. The lighter gases pass through the small openings more rapidly (at a higher rate) than the heavier ones (Figure 3). In 1832, Thomas Graham studied the rates of effusion of different gases and formulated Graham’s law of effusion : The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles : This means that if two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is inversely proportional to the ratio of the square roots of the masses of their particles: Figure 3. A balloon filled with air (the blue one) remains full overnight. A balloon filled with helium (the green one) partially deflates because the smaller, light helium atoms effuse through small holes in the rubber much more readily than the heavier molecules of nitrogen and oxygen found in air. (credit: modification of work by Mark Ott) Example 1: Applying Graham’s Law to Rates of Effusion Calculate the ratio of the rate of effusion of hydrogen to the rate of effusion of oxygen. [reveal-answer q=”190867″]Show Answer[/reveal-answer] [hidden-answer a=”190867″] From Graham’s law, we can use the molar mass of each gas: Hydrogen effuses four times as rapidly as oxygen. [/hidden-answer] Check Your Learning At a particular pressure and temperature, nitrogen gas effuses at the rate of 79 mL/s. Using the same apparatus at the same temperature and pressure, at what rate will sulfur dioxide effuse? [reveal-answer q=”337487″]Show Answer[/reveal-answer] [hidden-answer a=”337487″]52 mL/s[/hidden-answer] Here’s another example, making the point about how determining times differs from determining rates. Example 2: Effusion Time Calculations It takes 243 s for 4.46 × 10 -5 mol Xe to effuse through a tiny hole. Under the same conditions, how long will it take 4.46 × 10 –5 mol Ne to effuse? [reveal-answer q=”844226″]Show Answer[/reveal-answer] [hidden-answer a=”844226″]It is important to resist the temptation to use the times directly, and to remember how rate relates to time as well as how it relates to mass. Recall the definition of rate of effusion: and combine it with Graham’s law: To get: Noting that amount of A = amount of B , and solving for time for Ne : and substitute values: Finally, solve for the desired quantity: Note that this answer is reasonable: Since Ne is lighter than Xe, the effusion rate for Ne will be larger than that for Xe, which means the time of effusion for Ne will be smaller than that for Xe. [/hidden-answer] Check Your Learning A party balloon filled with helium deflates to of its original volume in 8.0 hours. How long will it take an identical balloon filled with the same number of moles of air (ℳ = 28.2 g/mol) to deflate to of its original volume? [reveal-answer q=”192191″]Show Answer[/reveal-answer] [hidden-answer a=”192191″]32 h[/hidden-answer] Finally, here is one more example showing how to calculate molar mass from effusion rate data. Example 3: Determining Molar Mass Using Graham’s Law An unknown gas effuses 1.66 times more rapidly than CO 2 . What is the molar mass of the unknown gas? Can you make a reasonable guess as to its identity? [reveal-answer q=”422479″]Show Answer[/reveal-answer] [hidden-answer a=”422479″] From Graham’s law, we have: Plug in known data: Solve: The gas could well be CH 4 , the only gas with this molar mass. [/hidden-answer] Check Your Learning Hydrogen gas effuses through a porous container 8.97-times faster than an unknown gas. Estimate the molar mass of the unknown gas. [reveal-answer q=”844180″]Show Answer[/reveal-answer] [hidden-answer a=”844180″]163 g/mol[/hidden-answer] Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment Gaseous diffusion has been used to produce enriched uranium for use in nuclear power plants and weapons. Naturally occurring uranium contains only 0.72% of 235 U, the kind of uranium that is “fissile,” that is, capable of sustaining a nuclear fission chain reaction. Nuclear reactors require fuel that is 2–5% 235 U, and nuclear bombs need even higher concentrations. One way to enrich uranium to the desired levels is to take advantage of Graham’s law. In a gaseous diffusion enrichment plant, uranium hexafluoride (UF 6 , the only uranium compound that is volatile enough to work) is slowly pumped through large cylindrical vessels called diffusers, which contain porous barriers with microscopic openings. The process is one of diffusion because the other side of the barrier is not evacuated. The molecules have a higher average speed and diffuse through the barrier a little faster than the heavier molecules. The gas that has passed through the barrier is slightly enriched in and the residual gas is slightly depleted. The small difference in molecular weights between and only about 0.4% enrichment, is achieved in one diffuser (Figure 4). But by connecting many diffusers in a sequence of stages (called a cascade), the desired level of enrichment can be attained. The large scale separation of gaseous from was first done during the World War II, at the atomic energy installation in Oak Ridge, Tennessee, as part of the Manhattan Project (the development of the first atomic bomb). Although the theory is simple, this required surmounting many daunting technical challenges to make it work in practice. The barrier must have tiny, uniform holes (about 10 –6 cm in diameter) and be porous enough to produce high flow rates. All materials (the barrier, tubing, surface coatings, lubricants, and gaskets) need to be able to contain, but not react with, the highly reactive and corrosive UF 6 . Because gaseous diffusion plants require very large amounts of energy (to compress the gas to the high pressures required and drive it through the diffuser cascade, to remove the heat produced during compression, and so on), it is now being replaced by gas centrifuge technology, which requires far less energy. A current hot political issue is how to deny this technology to Iran, to prevent it from producing enough enriched uranium for them to use to make nuclear weapons. Key Concepts and Summary Gaseous atoms and molecules move freely and randomly through space. Diffusion is the process whereby gaseous atoms and molecules are transferred from regions of relatively high concentration to regions of relatively low concentration. Effusion is a similar process in which gaseous species pass from a container to a vacuum through very small orifices. The rates of effusion of gases are inversely proportional to the square roots of their densities or to the square roots of their atoms/molecules’ masses (Graham’s law). Key Equations Exercises A balloon filled with helium gas is found to take 6 hours to deflate to 50% of its original volume. How long will it take for an identical balloon filled with the same volume of hydrogen gas (instead of helium) to decrease its volume by 50%? Explain why the numbers of molecules are not identical in the left- and right-hand bulbs shown in the center illustration of Figure 1. Starting with the definition of rate of effusion and Graham’s finding relating rate and molar mass, show how to derive the Graham’s law equation, relating the relative rates of effusion for two gases to their molecular masses. Heavy water, D 2 O (molar mass = 20.03 g mol –1 ), can be separated from ordinary water, H 2 O (molar mass = 18.01), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion of H 2 O and D 2 O. Which of the following gases diffuse more slowly than oxygen? F 2 , Ne, N 2 O, C 2 H 2 , NO, Cl 2 , H 2 S During the discussion of gaseous diffusion for enriching uranium, it was claimed that diffuses 0.4% faster than Show the calculation that supports this value. The molar mass of = 235.043930 + 6 × 18.998403 = 349.034348 g/mol, and the molar mass of = 238.050788 + 6 × 18.998403 = 352.041206 g/mol. Calculate the relative rate of diffusion of (molar mass 2.0 g/mol) compared to that of (molar mass 4.0 g/mol) and the relative rate of diffusion of O 2 (molar mass 32 g/mol) compared to that of O 3 (molar mass 48 g/mol). A gas of unknown identity diffuses at a rate of 83.3 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas. When two cotton plugs, one moistened with ammonia and the other with hydrochloric acid, are simultaneously inserted into opposite ends of a glass tube that is 87.0 cm long, a white ring of NH 4 Cl forms where gaseous NH 3 and gaseous HCl first come into contact. (Hint: Calculate the rates of diffusion for both NH 3 and HCl, and find out how much faster NH 3 diffuses than HCl.) At approximately what distance from the ammonia moistened plug does this occur? [reveal-answer q=”693396″]Selected Answers[/reveal-answer] [hidden-answer a=”693396″] 1. Use the rate of effusion equation: 3. Effusion can be defined as the process by which a gas escapes through a pinhole into a vacuum. Graham’s law states that with a mixture of two gases A and B: Both A and B are in the same container at the same temperature, and therefore will have the same kinetic energy: Therefore, 5. Gases with molecular masses greater than that of oxygen (31.9988 g/mol) will diffuse more slowly than O 2 . These gases are F 2 (37.9968 g/mol), N 2 O (44.0128 g/mol ), Cl 2 (70.906 g/mol), and H 2 S (34.082 g/mol). 7. 9. Rate of diffusion for NH 3 is proportional to Rate of diffusion for HCl is proportional to [/hidden-answer] Glossary diffusion: movement of an atom or molecule from a region of relatively high concentration to one of relatively low concentration (discussed in this chapter with regard to gaseous species, but applicable to species in any phase) effusion: transfer of gaseous atoms or molecules from a container to a vacuum through very small openings Graham’s law of effusion: rates of diffusion and effusion of gases are inversely proportional to the square roots of their molecular masses mean free path: average distance a molecule travels between collisions rate of diffusion: amount of gas diffusing through a given area over a given time CC licensed content, Shared previously Chemistry. Provided by : OpenStax College. Located at : http://openstaxcollege.org . License : CC BY: Attribution . License Terms : Download for free at https://openstaxcollege.org/textbooks/chemistry/get |
Courses/University_of_Kansas/CHEM_130%3A_General_Chemistry_I_(Sharpe_Elles)/06%3A_Stoichiometry_of_Chemical_Reactions/6.04%3A_Reaction_Yields | Learning Objectives By the end of this section, you will be able to: Explain the concepts of theoretical yield and limiting reactants/reagents. Derive the theoretical yield for a reaction under specified conditions. Calculate the percent yield for a reaction. The relative amounts of reactants and products represented in a balanced chemical equation are often referred to as stoichiometric amounts . All the exercises of the preceding module involved stoichiometric amounts of reactants. For example, when calculating the amount of product generated from a given amount of reactant, it was assumed that any other reactants required were available in stoichiometric amounts (or greater). In this module, more realistic situations are considered, in which reactants are not present in stoichiometric amounts. Limiting Reactant Consider another food analogy, making grilled cheese sandwiches (Figure \(\PageIndex{1}\)): \[1 \text { slice of cheese }+2 \text { slices of bread } \longrightarrow 1 \text { sandwich } \nonumber \] Stoichiometric amounts of sandwich ingredients for this recipe are bread and cheese slices in a 2:1 ratio. Provided with 28 slices of bread and 11 slices of cheese, one may prepare 11 sandwiches per the provided recipe, using all the provided cheese and having six slices of bread left over. In this scenario, the number of sandwiches prepared has been limited by the number of cheese slices, and the bread slices have been provided in excess . Consider this concept now with regard to a chemical process, the reaction of hydrogen with chlorine to yield hydrogen chloride: \[\ce{H2(g) + Cl2(g) -> 2 HCl (g)} \nonumber \] The balanced equation shows the hydrogen and chlorine react in a 1:1 stoichiometric ratio. If these reactants are provided in any other amounts, one of the reactants will nearly always be entirely consumed, thus limiting the amount of product that may be generated. This substance is the limiting reactant , and the other substance(s) is the excess reactant . Identifying the limiting and excess reactants for a given situation requires computing the molar amounts of each reactant provided and comparing them to the stoichiometric amounts represented in the balanced chemical equation. For example, imagine combining 3 moles of H 2 and 2 moles of Cl 2 . This represents a 3:2 (or 1.5:1) ratio of hydrogen to chlorine present for reaction, which is greater than the stoichiometric ratio of 1:1. Hydrogen, therefore, is present in excess, and chlorine is the limiting reactant. Reaction of all the provided chlorine (2 mol) will consume 2 mol of the 3 mol of hydrogen provided, leaving 1 mol of hydrogen unreacted. An alternative approach to identifying the limiting reactant involves comparing the amount of product expected for the complete reaction of each reactant. Each reactant amount is used to separately calculate the amount of product that would be formed per the reaction’s stoichiometry. The reactant yielding the lesser amount of product is the limiting reactant. For the example in the previous paragraph, complete reaction of the hydrogen would yield \[\text{mol}\,\ce{HCl}\,\text{produced}= 3\, \text{mol}\,\ce{H2} \times \dfrac{2\, \text{mol}\,\ce{HCl}}{1\,\ce{mol}\,\ce{H2}}=6\, \text{mol}\, \ce{HCl} \nonumber \] Complete reaction of the provided chlorine would produce \[\text{mol}\,\ce{HCl}\,\text{produced}=2\,\text{mol}\,\ce{Cl2} \times \frac{2\,\text{mol}\,\ce{HCl}}{1\,\text{mol}\,\ce{Cl2}}=4\,\text{mol}\,\ce{HCl} \nonumber \] The chlorine will be completely consumed once 4 moles of HCl have been produced. Since enough hydrogen was provided to yield 6 moles of HCl, there will be unreacted hydrogen remaining once this reaction is complete. Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant (Figure \(\PageIndex{2}\)). Link to Learning View this interactive simulation illustrating the concepts of limiting and excess reactants. Example \(\PageIndex{1}\): Identifying the Limiting Reactant Silicon nitride is a very hard, high-temperature-resistant ceramic used as a component of turbine blades in jet engines. It is prepared according to the following equation: \[3 Si (s)+2 N_2(g) \longrightarrow Si_3 N_4(s) \nonumber \] Which is the limiting reactant when 2.00 g of Si and 1.50 g of N 2 react? Solution Compute the provided molar amounts of reactants, and then compare these amounts to the balanced equation to identify the limiting reactant. \[mol Si =2.00 g Si \times \frac{1 mol Si }{28.09 g Si }=0.0712 mol Si \nonumber \] \[mol Si =2.00 g Si \times \frac{1 mol Si }{28.09 g Si }=0.0712 mol Si \nonumber \] The provided Si:N 2 molar ratio is: \[\frac{0.0712 mol Si }{0.0535 mol N_2}=\frac{1.33 mol Si }{1 mol N_2} \nonumber \] The stoichiometric Si:N 2 ratio is: \[\frac{3 mol Si }{2 mol N_2}=\frac{1.5 mol Si }{1 mol N_2} \nonumber \] Comparing these ratios shows that Si is provided in a less-than-stoichiometric amount, and so is the limiting reactant. Alternatively, compute the amount of product expected for complete reaction of each of the provided reactants. The 0.0712 moles of silicon would yield \[mol Si_3 N_4 \text { produced }=0.0712 mol Si \times \frac{1 mol Si_3 N_4}{3 mol Si^2}=0.0237 mol Si_3 N_4 \nonumber \] while the 0.0535 moles of nitrogen would produce \[\text { mol Si } N_4 \text { produced }=0.0535 mol N \times \frac{1 mol Si_3 N_4}{2 mol N_2}=0.0268 mol Si_3 N_4 \nonumber \] Since silicon yields the lesser amount of product, it is the limiting reactant. Exercise \(\PageIndex{1}\) Which is the limiting reactant when 5.00 g of H 2 and 10.0 g of O 2 react and form water? Answer O 2 Percent Yield The amount of product that may be produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation, is called the theoretical yield of the reaction. In practice, the amount of product obtained is called the actual yield , and it is often less than the theoretical yield for a number of reasons. Some reactions are inherently inefficient, being accompanied by side reactions that generate other products. Others are, by nature, incomplete (consider the partial reactions of weak acids and bases discussed earlier in this chapter). Some products are difficult to collect without some loss, and so less than perfect recovery will reduce the actual yield. The extent to which a reaction’s theoretical yield is achieved is commonly expressed as its percent yield : \[\text { percent yield }=\frac{\text { actual yield }}{\text { theoretical yield }} \times 100 \% \label{percentyield} \] Actual and theoretical yields may be expressed as masses or molar amounts (or any other appropriate property; e.g., volume, if the product is a gas). As long as both yields are expressed using the same units, these units will cancel when percent yield is calculated. Example \(\PageIndex{2}\): Calculation of Percent Yield Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation: \[\ce{CuSO4(aq) + Zn(s) -> Cu(s) + ZnSO4(aq)} \nonumber \] What is the percent yield? Solution The provided information identifies copper sulfate as the limiting reactant, and so the theoretical yield is found by the approach illustrated in the previous module, as shown here: \[1.274\,\text{g}\,\ce{CuSO4} \times \frac{1\,\text{mol}\,\ce{CuSO4}}{159.62\,\text{g}\,\ce{CuSO4}} \times \frac{1\,\text{mol}\,\ce{Cu} }{1 \,\text{mol}\, \ce{CuSO4}} \times \frac{63.55\,\text{g}\,\ce{Cu}}{1\,\text{mol}\,\ce{Cu}}=0.5072\,\text{g}\,\ce{Cu} \nonumber \] Using this theoretical yield and the provided value for actual yield, the percent yield is calculated (via Equation \ref{percentyield} to be \[\begin{align*} \text { percent yield }& =\left(\frac{\text { actual yield }}{\text { theoretical yield }}\right) \times 100\% \\[4pt] & =\left(\frac{0.392\,\text{g}\,\ce{Cu}}{0.5072\,\text{g}\,\ce{Cu2}}\right) \times 100\% \\[4pt] & =77.3 \% \end{align*} \] Exercise \(\PageIndex{2}\) What is the percent yield of a reaction that produces 12.5 g of the gas Freon CF 2 Cl 2 from 32.9 g of CCl 4 and excess HF? \[\ce{CCl4 + 2 HF -> CF2Cl2 + 2 HCl} \nonumber \] Answer 48.3% How Sciences Interconnect: Green Chemistry and Atom Economy The purposeful design of chemical products and processes that minimize the use of environmentally hazardous substances and the generation of waste is known as green chemistry . Green chemistry is a philosophical approach that is being applied to many areas of science and technology, and its practice is summarized by guidelines known as the “Twelve Principles of Green Chemistry” (see details at this website ). One of the 12 principles is aimed specifically at maximizing the efficiency of processes for synthesizing chemical products. The atom economy of a process is a measure of this efficiency, defined as the percentage by mass of the final product of a synthesis relative to the masses of all the reactants used: \[\text { atom economy }=\frac{\text { mass of product }}{\text { mass of reactants }} \times 100 \% \nonumber \] Though the definition of atom economy at first glance appears very similar to that for percent yield, be aware that this property represents a difference in the theoretical efficiencies of different chemical processes. The percent yield of a given chemical process, on the other hand, evaluates the efficiency of a process by comparing the yield of product actually obtained to the maximum yield predicted by stoichiometry. The synthesis of the common nonprescription pain medication, ibuprofen, nicely illustrates the success of a green chemistry approach (Figure \(\PageIndex{3}\)). First marketed in the early 1960s, ibuprofen was produced using a six-step synthesis that required 514 g of reactants to generate each mole (206 g) of ibuprofen, an atom economy of 40%. In the 1990s, an alternative process was developed by the BHC Company (now BASF Corporation) that requires only three steps and has an atom economy of ~80%, nearly twice that of the original process. The BHC process generates significantly less chemical waste; uses less-hazardous and recyclable materials; and provides significant cost-savings to the manufacturer (and, subsequently, the consumer). In recognition of the positive environmental impact of the BHC process, the company received the Environmental Protection Agency’s Greener Synthetic Pathways Award in 1997. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Calorimetry/Differential_Scanning_Calorimetry | Calorimetry involves the experimental quantification of heat released in a chemical process, either a reaction or a conformational alteration. It can be used to determine parameters such as the Heat of Reaction (\(Δ_{r}H\)), which is the change in enthalpy associated with the process of a chemical reaction. When \(Δ_{r}H\) is a negative value, the process is exothermic and releases heat; when \(Δ_{r}H\) is a positive value, the process is endothermic and requires heat input. Calorimetry uses a closed system, meaning there is a system separated from its surroundings by some boundary, through which heat and energy but not mass are able to flow. Calorimetry may be conducted at either constant pressure or volume and allows one to monitor the change in temperature as a result of the chemical process being investigated. Introduction Differential scanning calorimetry is a specific type of calorimetry including both a sample substance and a reference substance, residing in separate chambers. While the reference chamber contains only a solvent (such as water), the sample chamber contains an equal amount of the same solvent in addition to the substance of interest, of which the Δ r H is being determined. The \(Δ_{r}H\) due to the solvent is constant in both chambers, so any difference between the two can be attributed to the presence of the substance of interest. Each chamber is heated by a separate source in a way that their temperatures are always equal. This is accomplished through the use of thermocouples; the temperature of each chamber is constantly monitored and if a temperature difference is detected, then heat will be added to the cooler chamber to compensate for the difference. The heating rate used to maintain equivalent temperatures is logged as a function with respect to the temperature. For example, if the experimental goal is to determine the \(Δ_{r}H\) of a protein denaturation process, the reference cell could contain 100 mL H 2 O, and the sample cell could contain 1 mg of the protein in addition to the same 100 mL H 2 O. Therefore, the contribution of the solvent (H 2 O) to the heat capacity of each cell would be equal, and the only difference would be the presence of the protein in the sample chamber. Equations The following equation relates the change in temperature to the change in enthalpy: \[ dH = \int^{T_f}_{T_i}nC_p dt \] where \(dH\) is the rate of change in enthalpy, \(C_P\) is the molar heat capacity of the calorimeter, \(dT\) is the rate of change in temperature, \(n\) is the number of moles of material, and \(T_i\) and \(T_f\) are the initial and final temperatures, respectively. This equation can be integrated to yield \[\Delta H = nC_p \Delta T\] where ΔH is the total change in enthalpy and ΔT is the change in temperature. Differential Thermograms The output yielded by differential scanning calorimetry is called a differential thermogram, which plots the required heat flow against temperature. Data analysis is highly dependent on the assumption that both the reference and sample cells are constantly and accurately maintained at equal temperatures. This graph indicates the change in power (electrical heat) as the temperatures of the two cells are gradually increased. A change in specific heat results in a small change in power, and can be either positive or negative depending on the particular process. The advent of an endothermic reaction will cause an increase in power as temperature increases, since additional heat is required to drive the reaction and still maintain the reference temperature. When an exothermic reaction occurs, the opposite effect is observed; power decreases because heat is released by the reaction and less power is required to maintain equivalent temperatures in the chambers. Examples Differential scanning calorimetry can be used to study many different fields including biopolymer energetics where it is used to find the enthalpy of the protein denaturation process. A protein can be changed from its native state, in which it has a specific conformation due to non-covalent intramolecular interactions, to a denatured state where this characteristic structure is altered. Analysis of proteins through DSC can provide both the enthalpy of denaturation and information about the cooperativity of the denaturation process. A sharper peak in the thermogram indicates a higher level of cooperativity, meaning that when one structural association is disturbed, the likelihood of disruption at other points of association will be enhanced. DSC is also used in conjunction with differential thermal analysis. Through the combination of these two techniques, thermal behavior of inorganic compounds can be studied while the melting, boiling and decomposition points of organic compounds and polymers are found. References Skoog, Douglas A. Principles of Instrumental Analysis . Brooks/Cole, 1985. Print. Chang, Raymond. Physical Chemistry for the Biosciences . New York: University Science, 2005. Print. |
Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/02%3A_Elements_Atoms_and_the_Periodic_Table/2.01%3A_The_Elements | Learning Objectives Define a chemical element and give examples of the abundance of different elements. Represent a chemical element with a chemical symbol. An element is a substance that cannot be broken down into simpler chemical substances. There are about 90 naturally occurring elements known on Earth. Using technology, scientists have been able to create nearly 30 additional elements that do not occur in nature. Today, chemistry recognizes 118 elements—some of which were created an atom at a time. Figure \(\PageIndex{1}\) shows some of the chemical elements. Abundance The elements vary widely in abundance. In the universe as a whole, the most common element is hydrogen (about 90% of atoms), followed by helium (most of the remaining 10%). All other elements are present in relatively minuscule amounts, as far as we can detect. Earth’s Crust Earth (overall) Element Percentage Element Percentage oxygen 46.1 iron 34.6 silicon 28.2 oxygen 29.5 aluminum 8.23 silicon 15.2 iron 5.53 magnesium 12.7 calcium 4.15 nickel 2.4 sodium 2.36 sulfur 1.9 magnesium 2.33 all others 3.7 potassium 2.09 NaN NaN titanium 0.565 NaN NaN hydrogen 0.14 NaN NaN phosphorus 0.105 NaN NaN all others 0.174 NaN NaN Source: D. R. Lide, ed. CRC Handbook of Chemistry and Physics, 89th ed. (Boca Raton, FL: CRC Press, 2008–9), 14–17. Source: D. R. Lide, ed. CRC Handbook of Chemistry and Physics, 89th ed. (Boca Raton, FL: CRC Press, 2008–9), 14–17. Source: D. R. Lide, ed. CRC Handbook of Chemistry and Physics, 89th ed. (Boca Raton, FL: CRC Press, 2008–9), 14–17. Source: D. R. Lide, ed. CRC Handbook of Chemistry and Physics, 89th ed. (Boca Raton, FL: CRC Press, 2008–9), 14–17. On the planet Earth, however, the situation is rather different. Oxygen makes up 46.1% of the mass of Earth’s crust (the relatively thin layer of rock forming Earth’s surface), mostly in combination with other elements, while silicon makes up 28.2%. Hydrogen, the most abundant element in the universe, makes up only 0.14% of Earth’s crust. Table \(\PageIndex{1}\) lists the relative abundances of elements on Earth as a whole and in Earth’s crust. Table \(\PageIndex{2}\) lists the relative abundances of elements in the human body. If you compare Table \(\PageIndex{1}\) and Table \(\PageIndex{2}\), you will find disparities between the percentage of each element in the human body and on Earth. Oxygen has the highest percentage in both cases, but carbon, the element with the second highest percentage in the body, is relatively rare on Earth and does not even appear as a separate entry in Table \(\PageIndex{1}\); carbon is part of the 0.174% representing “other” elements. How does the human body concentrate so many apparently rare elements? Element Percentage by Mass oxygen 61 carbon 23 hydrogen 10 nitrogen 2.6 calcium 1.4 phosphorus 1.1 sulfur 0.20 potassium 0.20 sodium 0.14 chlorine 0.12 magnesium 0.027 silicon 0.026 iron 0.006 fluorine 0.0037 zinc 0.0033 all others 0.174 Source: D. R. Lide, ed. CRC Handbook of Chemistry and Physics, 89th ed. (Boca Raton, FL: CRC Press, 2008–9), 7–24. Source: D. R. Lide, ed. CRC Handbook of Chemistry and Physics, 89th ed. (Boca Raton, FL: CRC Press, 2008–9), 7–24. The relative amounts of elements in the body have less to do with their abundances on Earth than with their availability in a form we can assimilate. We obtain oxygen from the air we breathe and the water we drink. We also obtain hydrogen from water. On the other hand, although carbon is present in the atmosphere as carbon dioxide, and about 80% of the atmosphere is nitrogen, we obtain those two elements from the food we eat, not the air we breathe. LOOKING CLOSER: PHOSPHOROUS , THE CHEMICAL BOTTLENECK There is an element that we need more of in our bodies than is proportionately present in Earth’s crust, and this element is not easily accessible. Phosphorus makes up 1.1% of the human body but only 0.105% of Earth’s crust. We need phosphorus for our bones and teeth, and it is a crucial component of all living cells. Unlike carbon, which can be obtained from carbon dioxide through photosynthesis, there is no phosphorus compound present in our surroundings that can serve as a convenient source. Phosphorus, then, is nature’s bottleneck. Its availability limits the amount of life our planet can sustain. Higher forms of life, such as humans, can obtain phosphorus by selecting a proper diet (plenty of protein); but lower forms of life, such as algae, must absorb it from the environment. When phosphorus-containing detergents were introduced in the 1950s, wastewater from normal household activities greatly increased the amount of phosphorus available to algae and other plant life. Lakes receiving this wastewater experienced sudden increases in growth of algae. When the algae died, concentrations of bacteria that ate the dead algae increased. Because of the large bacterial concentrations, the oxygen content of the water dropped, causing fish to die in large numbers. This process, called eutrophication , is considered a negative environmental impact. Today, many detergents are made without phosphorus so the detrimental effects of eutrophication are minimized. You may even see statements to that effect on detergent boxes. It can be sobering to realize how much impact a single element can have on life—or the ease with which human activity can affect the environment. Names and Symbols Each element has a name. Some of these names date from antiquity, while others are quite new. Today, the names for new elements are proposed by their discoverers but must be approved by the International Union of Pure and Applied Chemistry, an international organization that makes recommendations concerning all kinds of chemical terminology. Today, new elements are usually named after famous scientists. The names of the elements can be cumbersome to write in full, especially when combined to form the names of compounds. Therefore, each element name is abbreviated as a one- or two-letter chemical symbol. By convention, the first letter of a chemical symbol is a capital letter, while the second letter (if there is one) is a lowercase letter. The first letter of the symbol is usually the first letter of the element’s name, while the second letter is some other letter from the name. Some elements have symbols that derive from earlier, mostly Latin names, so the symbols may not contain any letters from the English name. Table \(\PageIndex{3}\) lists the names and symbols of some of the most familiar elements. 0 1 2 3 aluminum Al magnesium Mg argon Ar manganese Mn arsenic As mercury Hg* barium Ba neon Ne bismuth Bi nickel Ni boron B nitrogen N bromine Br oxygen O calcium Ca phosphorus P carbon C platinum Pt chlorine Cl potassium K* chromium Cr silicon Si copper Cu* silver Ag* fluorine F sodium Na* gold Au* strontium Sr helium He sulfur S hydrogen H tin Sn* iron Fe tungsten W† iodine I uranium U lead Pb* zinc Zn lithium Li zirconium Zr *The symbol comes from the Latin name of element. *The symbol comes from the Latin name of element. *The symbol comes from the Latin name of element. *The symbol comes from the Latin name of element. †The symbol for tungsten comes from its German name—wolfram. †The symbol for tungsten comes from its German name—wolfram. †The symbol for tungsten comes from its German name—wolfram. †The symbol for tungsten comes from its German name—wolfram. Element names in languages other than English are often close to their Latin names. For example, gold is oro in Spanish and or in French (close to the Latin aurum ), tin is estaño in Spanish (compare to stannum ), lead is plomo in Spanish and plomb in French (compare to plumbum ), silver is argent in French (compare to argentum ), and iron is fer in French and hierro in Spanish (compare to ferrum ). The closeness is even more apparent in pronunciation than in spelling. Example \(\PageIndex{1}\) Write the chemical symbol for each element without consulting Table \(\PageIndex{3}\) "Element Names and Symbols". bromine boron carbon calcium gold Answer a Br Answer b B Answer c C Answer d Ca Answer e Au Exercise \(\PageIndex{1}\) Write the chemical symbol for each element without consulting Table \(\PageIndex{3}\). manganese magnesium neon nitrogen silver Answer a Mn Answer b Mg Answer c Ne Answer d N Answer e Ag Example \(\PageIndex{2}\) What element is represented by each chemical symbol? Na Hg P K I Answer a sodium Answer b mercury Answer c phosphorus Answer d potassium Answer e iodine Exercise \(\PageIndex{2}\) What element is represented by each chemical symbol? Pb Sn U O F Answer a lead Answer b tin Answer c uranium Answer d oxygen Answer e fluorine Concept Review Exercises What is an element? Give some examples of how the abundance of elements varies. Why are chemical symbols so useful? What is the source of the letter(s) for a chemical symbol? Answers An element is the basic chemical building block of matter; it is the simplest chemical substance. Elements vary from being a small percentage to more than 30% of the atoms around us. Chemical symbols are useful to concisely represent the elements present in a substance. The letters usually come from the name of the element. Key Takeaways All matter is composed of elements. Chemical elements are represented by a one- or two-letter symbol. |
Courses/Sacramento_City_College/SCC%3A_Chem_309_-_General_Organic_and_Biochemistry_(Bennett)/Text/16%3A_Proteins_and_Enzymes/16.01%3A_Properties_of_Amino_Acids | Learning Objectives To recognize amino acids and classify them based on the characteristics of their side chains. The proteins in all living species, from bacteria to humans, are constructed from the same set of 20 amino acids, so called because each contains an amino group attached to a carboxylic acid. The amino acids in proteins are α-amino acids, which means the amino group is attached to the α-carbon of the carboxylic acid. Humans can synthesize only about half of the needed amino acids; the remainder must be obtained from the diet and are known as essential amino acids. However, two additional amino acids have been found in limited quantities in proteins: Selenocysteine was discovered in 1986, while pyrrolysine was discovered in 2002. The amino acids are colorless, nonvolatile, crystalline solids, melting and decomposing at temperatures above 200°C. These melting temperatures are more like those of inorganic salts than those of amines or organic acids and indicate that the structures of the amino acids in the solid state and in neutral solution are best represented as having both a negatively charged group and a positively charged group. Such a species is known as a zwitterion. Classification In addition to the amino and carboxyl groups, amino acids have a side chain or R group attached to the α-carbon. Each amino acid has unique characteristics arising from the size, shape, solubility, and ionization properties of its R group. As a result, the side chains of amino acids exert a profound effect on the structure and biological activity of proteins. Although amino acids can be classified in various ways, one common approach is to classify them according to whether the functional group on the side chain at neutral pH is nonpolar, polar but uncharged, negatively charged, or positively charged. The structures and names of the 20 amino acids, their one- and three-letter abbreviations, and some of their distinctive features are given in Table \(\PageIndex{1}\). Common Name Abbreviation Structural Formula (at pH 6) Molar Mass Distinctive Feature Amino acids with a nonpolar R group Amino acids with a nonpolar R group Amino acids with a nonpolar R group Amino acids with a nonpolar R group Amino acids with a nonpolar R group glycine gly (G) NaN 75 the only amino acid lacking a chiral carbon alanine ala (A) NaN 89 — valine val (V) NaN 117 a branched-chain amino acid leucine leu (L) NaN 131 a branched-chain amino acid isoleucine ile (I) NaN 131 an essential amino acid because most animals cannot synthesize branched-chain amino acids phenylalanine phe (F) NaN 165 also classified as an aromatic amino acid tryptophan trp (W) NaN 204 also classified as an aromatic amino acid methionine met (M) NaN 149 side chain functions as a methyl group donor proline pro (P) NaN 115 contains a secondary amine group; referred to as an α-imino acid Amino acids with a polar but neutral R group Amino acids with a polar but neutral R group Amino acids with a polar but neutral R group Amino acids with a polar but neutral R group Amino acids with a polar but neutral R group serine ser (S) NaN 105 found at the active site of many enzymes threonine thr (T) NaN 119 named for its similarity to the sugar threose cysteine cys (C) NaN 121 oxidation of two cysteine molecules yields cystine tyrosine tyr (Y) NaN 181 also classified as an aromatic amino acid asparagine asn (N) NaN 132 the amide of aspartic acid glutamine gln (Q) NaN 146 the amide of glutamic acid Amino acids with a negatively charged R group Amino acids with a negatively charged R group Amino acids with a negatively charged R group Amino acids with a negatively charged R group Amino acids with a negatively charged R group aspartic acid asp (D) NaN 132 carboxyl groups are ionized at physiological pH; also known as aspartate glutamic acid glu (E) NaN 146 carboxyl groups are ionized at physiological pH; also known as glutamate Amino acids with a positively charged R group Amino acids with a positively charged R group Amino acids with a positively charged R group Amino acids with a positively charged R group Amino acids with a positively charged R group histidine his (H) NaN 155 the only amino acid whose R group has a pKa (6.0) near physiological pH lysine lys (K) NaN 147 — arginine arg (R) NaN 175 almost as strong a base as sodium hydroxide The first amino acid to be isolated was asparagine in 1806. It was obtained from protein found in asparagus juice (hence the name). Glycine, the major amino acid found in gelatin, was named for its sweet taste (Greek glykys , meaning “sweet”). In some cases an amino acid found in a protein is actually a derivative of one of the common 20 amino acids (one such derivative is hydroxyproline). The modification occurs after the amino acid has been assembled into a protein. Configuration Notice in Table \(\PageIndex{1}\) that glycine is the only amino acid whose α-carbon is not chiral . Therefore, with the exception of glycine, the amino acids could theoretically exist in either the D- or the L-enantiomeric form and rotate plane-polarized light. As with sugars, chemists used L-glyceraldehyde as the reference compound for the assignment of absolute configuration to amino acids. Its structure closely resembles an amino acid structure except that in the latter, an amino group takes the place of the OH group on the chiral carbon of the L-glyceraldehyde and a carboxylic acid replaces the aldehyde. Modern stereochemistry assignments using the Cahn-Ingold-Prelog priority rules used ubiquitously in chemistry show that all of the naturally occurring chiral amino acids are S except Cys which is R. We learned that all naturally occurring sugars belong to the D series. It is interesting, therefore, that nearly all known plant and animal proteins are composed entirely of L-amino acids. However, certain bacteria contain D-amino acids in their cell walls, and several antibiotics (e.g., actinomycin D and the gramicidins) contain varying amounts of D-leucine, D-phenylalanine, and D-valine. Summary Amino acids can be classified based on the characteristics of their distinctive side chains as nonpolar, polar but uncharged, negatively charged, or positively charged. The amino acids found in proteins are L-amino acids. |
Courses/University_of_Missouri/MU%3A__1330H_(Keller)/21%3A_Nuclear_Chemistry/21.E%3A_Exercises | 21.1: Radioactivity Q21.1.1 Why are many radioactive substances warm to the touch? Why do many radioactive substances glow? Q21.1.2 Describe the differences between nonionizing and ionizing radiation in terms of the intensity of energy emitted and the effect each has on an atom or molecule after collision. Which nuclear decay reactions are more likely to produce ionizing radiation? nonionizing radiation? Q21.1.3 Would you expect nonionizing or ionizing radiation to be more effective at treating cancer? Why? S21.1.3 Ionizing radiation is higher in energy and causes greater tissue damage, so it is more likely to destroy cancerous cells. Q21.1.4 Historically, concrete shelters have been used to protect people from nuclear blasts. Comment on the effectiveness of such shelters. Q21.1.5 Gamma rays are a very high-energy radiation, yet α particles inflict more damage on biological tissue. Why? Q21.1.6 List the three primary sources of naturally occurring radiation. Explain the factors that influence the dose that one receives throughout the year. Which is the largest contributor to overall exposure? Which is the most hazardous? S21.1.6 Three primary naturally occurring radiations are radium, uranium and thorium, each all having long half lives. Inhalation of air, ingestion of food and water,terrestrial radation from the ground and cosmic radiation from space are all factors tat influence the does that a person receives throughout the year. Inhalation of the air is the largest contributor to exposure. Radiation can damage DNA or kill cells. When radiation is exposed to your body, it will collide with atoms and this will change and damage your DNA. Q21.1.7 Because radon is a noble gas, it is inert and generally unreactive. Despite this, exposure to even low concentrations of radon in air is quite dangerous. Describe the physical consequences of exposure to radon gas. Why are people who smoke more susceptible to these effects? Q21.1.8 Most medical imaging uses isotopes that have extremely short half-lives. These isotopes usually undergo only one kind of nuclear decay reaction. Which kind of decay reaction is usually used? Why? Why would a short half-life be preferred in these cases? Beta decay. Alfa decay can be easily stopped by paper, which means it can not be used to see inside people's body. Also, Gamma rays are really dangerous for human, that even a short period of time exploding to it will have negative effect on human body. Thus, Beta decay is the perfect choice. It can be used to see through human's body and stopped by aluminum or some other metals. Since all these radioactive decays are harmful for human body, if the half time of these reactions are short, the time exploding to these reactions will be short too. Q21.1.9 Which would you prefer: one exposure of 100 rem, or 10 exposures of 10 rem each? Explain your rationale. S21.1.9 Ten exposures of 10 rem are less likely to cause major damage. Q21.1.10 A 2.14 kg sample of rock contains 0.0985 g of uranium. How much energy is emitted over 25 yr if 99.27% of the uranium is 238 U, which has a half-life of 4.46 × 10 9 yr, if each decay event is accompanied by the release of 4.039 MeV? If a 180 lb individual absorbs all of the emitted radiation, how much radiation has been absorbed in rads? Q21.1.11 There is a story about a “radioactive boy scout” who attempted to convert thorium-232, which he isolated from about 1000 gas lantern mantles, to uranium-233 by bombarding the thorium with neutrons. The neutrons were generated via bombarding an aluminum target with α particles from the decay of americium-241, which was isolated from 100 smoke detectors. Write balanced nuclear reactions for these processes. The “radioactive boy scout” spent approximately 2 h/day with his experiment for 2 yr. Assuming that the alpha emission of americium has an energy of 5.24 MeV/particle and that the americium-241 was undergoing 3.5 × 10 6 decays/s, what was the exposure of the 60.0 kg scout in rads? The intrepid scientist apparently showed no ill effects from this exposure. Why? 241/95 Am---> 4/2 He + 237/93Np---> 4/2He + 233/91Pa----> 1/0n+ 232/91Th---> 1/1 H + 233/92 U By adding alpha particles to the products side of the reaction, he was able to reduce the mass number by 4 and the atomic number by 2 to get the products he wanted. Bombardment with neutrons and 1 H was required to lower to the mass number to get Th and then raise both the mass number and the atomic number to yield Uranium. 2 hours*365*2= 1460 hours of exposure.*60min/1hr*60s/1min= 5.26*10^6s of exposure 1MeV= (1.6022*10^-13 Joules) * (5.24 MeV/particle)*2 particles= 1.679*10^-12 Joules. (1.679*10^-12 Joules) * (1 amu/ 1.4924*10^-10 Joules)= 2.51*10^-13 amu E=mc^2 E=(2.51*10^-13 amu)(1.66*10^-22kg/amu)(2.9998*10^8m/s)^2= (3.75*10^-18 kgm^2/s)*(3.5*10^6 decays/s)= 1.31*10^-11 joules of exposure per second. The scientist showed no ill effects from this exposure because if we multiple the energy in joules of exposure per second, 1.31*10^-11, by the total amount of seconds of exposure, 5.26*10^6s, we find that he was only exposed to 6.9*10^-5 joules of radiation throughout the span of two years. This is a very small amount of radiation for such a long span of time. In order to plug in the values for this equation, we must convert the given MeV to Joules with the known conversion rate. Similarly, we must convert Joules to amu with another known conversion rate. Then we can plug in the values and multiply by c^2 but we must not forget to multiple the amu by the conversion rate to kg in order to yield Joules. After all of this is done, we multiple the amount of Joules of exposure per second by the total amount of exposure in seconds in order to find out the total amount of exposure over the two year span. 21.2: Patterns of Nuclear Stability Q21.2.1 How do chemical reactions compare with nuclear reactions with respect to mass changes? Does either type of reaction violate the law of conservation of mass? Explain your answers. Q21.2.2 Why is the amount of energy released by a nuclear reaction so much greater than the amount of energy released by a chemical reaction? Q21.2.3 Explain why the mass of an atom is less than the sum of the masses of its component particles. Q21.2.4 The stability of a nucleus can be described using two values. What are they, and how do they differ from each other? Q21.2.5 In the days before true chemistry, ancient scholars (alchemists) attempted to find the philosopher’s stone, a material that would enable them to turn lead into gold. Is the conversion of Pb → Au energetically favorable? Explain why or why not. Q21.2.6 Describe the energy barrier to nuclear fusion reactions and explain how it can be overcome. Q21.2.7 Imagine that the universe is dying, the stars have burned out, and all the elements have undergone fusion or radioactive decay. What would be the most abundant element in this future universe? Why? Q21.2.8 Numerous elements can undergo fission, but only a few can be used as fuels in a reactor. What aspect of nuclear fission allows a nuclear chain reaction to occur? Q21.2.9 How are transmutation reactions and fusion reactions related? Describe the main impediment to fusion reactions and suggest one or two ways to surmount this difficulty. Q21.2.10 Using the information provided in Chapter 33, complete each reaction and calculate the amount of energy released from each in kilojoules. 238 Pa → ? + β − 216 Fr → ? + α 199 Bi → ? + β + Q21.2.22 Using the information provided in Chapter 33, complete each reaction and calculate the amount of energy released from each in kilojoules. 194 Tl → ? + β + 171 Pt → ? + α 214 Pb → ? + β − Q21.2.23 Using the information provided in Chapter 33, complete each reaction and calculate the amount of energy released from each in kilojoules per mole. \(_{91}^{234}\textrm{Pa}\rightarrow \,?+\,_{-1}^0\beta\) \(_{88}^{226}\textrm{Ra}\rightarrow \,?+\,_2^4\alpha\) Q21.2.24 Using the information provided in Chapter 33, complete each reaction and then calculate the amount of energy released from each in kilojoules per mole. \(_{27}^{60}\textrm{Co}\rightarrow\,?+\,_{-1}^0\beta\) (The mass of cobalt-60 is 59.933817 amu.) technicium-94 (mass = 93.909657 amu) undergoing fission to produce chromium-52 and potassium-40 Q21.2.25 Using the information provided in Chapter 33, predict whether each reaction is favorable and the amount of energy released or required in megaelectronvolts and kilojoules per mole. the beta decay of bismuth-208 (mass = 207.979742 amu) the formation of lead-206 by alpha decay Q21.2.26 Using the information provided, predict whether each reaction is favorable and the amount of energy released or required in megaelectronvolts and kilojoules per mole. alpha decay of oxygen-16 alpha decay to produce chromium-52 Q21.2.27 Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for 87 Sr if the measured mass of 87 Sr is 86.908877 amu. the calculated mass the mass defect the nuclear binding energy the nuclear binding energy per nucleon Q21.2.30 The experimentally determined mass of 29 S is 28.996610 amu. Find each of the following. the calculated mass the mass defect the nuclear binding energy the nuclear binding energy per nucleon Q21.2.31 Calculate the amount of energy that is released by the neutron-induced fission of 235 U to give 141 Ba, 92 Kr (mass = 91.926156 amu), and three neutrons. Report your answer in electronvolts per atom and kilojoules per mole. Q21.2.31 Calculate the amount of energy that is released by the neutron-induced fission of 235 U to give 90 Sr, 143 Xe, and three neutrons. Report your answer in electronvolts per atom and kilojoules per mole. Q21.2.33 Calculate the amount of energy released or required by the fusion of helium-4 to produce the unstable beryllium-8 (mass = 8.00530510 amu). Report your answer in kilojoules per mole. Do you expect this to be a spontaneous reaction? Q21.2.34 Calculate the amount of energy released by the fusion of 6 Li and deuterium to give two helium-4 nuclei. Express your answer in electronvolts per atom and kilojoules per mole. Q21.2.35 How much energy is released by the fusion of two deuterium nuclei to give one tritium nucleus and one proton? How does this amount compare with the energy released by the fusion of a deuterium nucleus and a tritium nucleus, which is accompanied by ejection of a neutron? Express your answer in megaelectronvolts and kilojoules per mole. Pound for pound, which is a better choice for a fusion reactor fuel mixture? Numerical Answers 1 \(_{91}^{238}\textrm{Pa}\rightarrow\,_{92}^{238}\textrm{U}+\,_{-1}^0\beta\); −5.540 × 10 −16 kJ \(_{87}^{216}\textrm{Fr}\rightarrow\,_{85}^{212}\textrm{At}+\,_{2}^4\alpha\); −1.470 × 10 −15 kJ \(_{83}^{199}\textrm{Bi}\rightarrow\,_{82}^{199}\textrm{Pb}+\,_{+1}^0\beta\); −5.458 × 10 −16 kJ \(_{91}^{234}\textrm{Pa}\rightarrow\,_{92}^{234}\textrm{U}+\,_{-1}^0\beta\); 2.118 × 10 8 kJ/mol \(_{88}^{226}\textrm{Ra}\rightarrow\,_{86}^{222}\textrm{Rn}+\,_{2}^4\alpha\); 4.700 × 10 8 kJ/mol The beta decay of bismuth-208 to polonium is endothermic (ΔE = 1.400 MeV/atom, 1.352 × 10 8 kJ/mol). The formation of lead-206 by alpha decay of polonium-210 is exothermic (ΔE = −5.405 MeV/atom, −5.218 × 10 8 kJ/mol). 757 MeV/atom, 8.70 MeV/nucleon 53.438245 amu 0.496955 amu 463 MeV/atom 8.74 MeV/nucleon −173 MeV/atom; 1.67 × 10 10 kJ/mol ΔE = + 9.0 × 10 6 kJ/mol beryllium-8; no D–D fusion: ΔE = −4.03 MeV/tritium nucleus formed = −3.89 × 10 8 kJ/mol tritium; D–T fusion: ΔE = −17.6 MeV/tritium nucleus = −1.70 × 10 9 kJ/mol; D–T fusion 21.3: Nuclear Transmutations Q21.3.1 Why do scientists believe that hydrogen is the building block of all other elements? Why do scientists believe that helium-4 is the building block of the heavier elements? Q21.3.2 How does a star produce such enormous amounts of heat and light? How are elements heavier than Ni formed? Q21.3.3 Propose an explanation for the observation that elements with even atomic numbers are more abundant than elements with odd atomic numbers. S21.3.3 The raw material for all elements with Z > 2 is helium (Z = 2), and fusion of helium nuclei will always produce nuclei with an even number of protons. Q21.3.4 During the lifetime of a star, different reactions that form different elements are used to power the fusion furnace that keeps a star “lit.” Explain the different reactions that dominate in the different stages of a star’s life cycle and their effect on the temperature of the star. Q21.3.5 A line in a popular song from the 1960s by Joni Mitchell stated, “We are stardust….” Does this statement have any merit or is it just poetic? Justify your answer. Q21.3.6 If the laws of physics were different and the primary element in the universe were boron-11 (Z = 5), what would be the next four most abundant elements? Propose nuclear reactions for their formation. Q21.3.7 Write a balanced nuclear reaction for the formation of each isotope. 27 Al from two 12 C nuclei 9 Be from two 4 He nuclei Q21.3.8 At the end of a star’s life cycle, it can collapse, resulting in a supernova explosion that leads to the formation of heavy elements by multiple neutron-capture events. Write a balanced nuclear reaction for the formation of each isotope during such an explosion. 106 Pd from nickel-58 selenium-79 from iron-56 Q21.3.9 When a star reaches middle age, helium-4 is converted to short-lived beryllium-8 (mass = 8.00530510 amu), which reacts with another helium-4 to produce carbon-12. How much energy is released in each reaction (in megaelectronvolts)? How many atoms of helium must be “burned” in this process to produce the same amount of energy obtained from the fusion of 1 mol of hydrogen atoms to give deuterium? 21.4: Rates of Radioactive Decay Q21.4.1 What do chemists mean by the half-life of a reaction? Q21.4.2 If a sample of one isotope undergoes more disintegrations per second than the same number of atoms of another isotope, how do their half-lives compare? Q21.4.3 Half-lives for the reaction A + B → C were calculated at three values of [A] 0 , and [B] was the same in all cases. The data are listed in the following table: [A]0 (M) t½ (s) 0.50 420 0.75 280 1.00 210 Does this reaction follow first-order kinetics? On what do you base your answer? S21.4.3 No; the reaction is second order in A because the half-life decreases with increasing reactant concentration according to t 1 /2 = 1/ k [A 0 ]. Q21.4.4 Ethyl-2-nitrobenzoate (NO 2 C 6 H 4 CO 2 C 2 H 5 ) hydrolyzes under basic conditions. A plot of [NO 2 C 6 H 4 CO 2 C 2 H 5 ] versus t was used to calculate t ½, with the following results: [NO2C6H4CO2C2H5] (M/cm3) t½ (s) 0.05 240 0.04 300 0.03 400 Is this a first-order reaction? Explain your reasoning. Q21.4.5 Azomethane (CH 3 N 2 CH 3 ) decomposes at 600 K to C 2 H 6 and N 2 . The decomposition is first order in azomethane. Calculate t ½ from the data in the following table: Time (s) \(P_{\large{\mathrm{CH_3N_2CH_3}}}\) (atm) 0 8.2 × 10−2 2000 3.99 × 10−2 4000 1.94 × 10−2 How long will it take for the decomposition to be 99.9% complete? S21.4.5 t 1 /2 = 1.92 × 10 3 s or 1920 s; 19100 s or 5.32 hrs. Q21.4.6 The first-order decomposition of hydrogen peroxide has a half-life of 10.7 h at 20°C. What is the rate constant (expressed in s −1 ) for this reaction? If you started with a solution that was 7.5 × 10 −3 M H 2 O 2 , what would be the initial rate of decomposition (M/s)? What would be the concentration of H 2 O 2 after 3.3 h? Application Problems Until the 1940s, uranium glazes were popular on ceramic dishware. One brand, Fiestaware, had bright orange glazes that could contain up to 20% uranium by mass. Although this practice is less common today due to the negative association of radiation, it is still possible to buy Depression-era glassware that is quite radioactive. Aqueous solutions in contact with this “hot” glassware can reach uranium concentrations up to 10 ppm by mass. If 1.0 g of uranium undergoes 1.11 × 10 7 decays/s, each to an α particle with an energy of 4.03 MeV, what would be your exposure in rem and rad if you drank 1.0 L of water that had been sitting for an extended time in a Fiestaware pitcher? Assume that the water and contaminants are excreted only after 18 h and that you weigh 70.0 kg. Neutrography is a technique used to take the picture of an object using a beam of neutrons. How does the penetrating power of a neutron compare with alpha, beta, and gamma radiation? Do you expect similar penetration for protons? How would the biological damage of each particle compare with the other types of radiation? (Recall that a neutron’s mass is approximately 2000 times the mass of an electron.) Spent fuel elements in a nuclear reactor contain radioactive fission products in addition to heavy metals. The conversion of nuclear fuel in a reactor is shown here: Neglecting the fission products, write balanced nuclear reactions for the conversion of the original fuel to each product. The first atomic bomb used 235 U as a fissile material, but there were immense difficulties in obtaining sufficient quantities of pure 235 U. A second fissile element, plutonium, was discovered in 1940, and it rapidly became important as a nuclear fuel. This element was produced by irradiating 238 U with neutrons in a nuclear reactor. Complete the series that produced plutonium, all isotopes of which are fissile: \(_{92}^{238}\textrm U+\,_0^1\textrm n\rightarrow\,\textrm U\rightarrow\,\textrm{Np}\rightarrow\,\textrm{Pu}\) Boron neutron capture therapy is a potential treatment for many diseases. As the name implies, when boron-10, one of the naturally occurring isotopes of boron, is bombarded with neutrons, it absorbs a neutron and emits an α particle. Write a balanced nuclear reaction for this reaction. One advantage of this process is that neutrons cause little damage on their own, but when they are absorbed by boron-10, they can cause localized emission of alpha radiation. Comment on the utility of this treatment and its potential difficulties. An airline pilot typically flies approximately 80 h per month. If 75% of that time is spent at an altitude of about 30,000 ft, how much radiation is that pilot receiving in one month? over a 30 yr career? Is the pilot receiving toxic doses of radiation? At a breeder reactor plant, a 72 kg employee accidentally inhaled 2.8 mg of 239 Pu dust. The isotope decays by alpha decay and has a half-life of 24,100 yr. The energy of the emitted α particles is 5.2 MeV, and the dust stays in the employee’s body for 18 h. How many plutonium atoms are inhaled? What is the energy absorbed by the body? What is the physical dose in rads? What is the dose in rems? Will the dosage be fatal? For many years, the standard source for radiation therapy in the treatment of cancer was radioactive 60 Co, which undergoes beta decay to 60 Ni and emits two γ rays, each with an energy of 1.2 MeV. Show the sequence of nuclear reactions. If the half-life for beta decay is 5.27 yr, how many 60 Co nuclei are present in a typical source undergoing 6000 dps that is used by hospitals? The mass of 60 Co is 59.93 amu. It is possible to use radioactive materials as heat sources to produce electricity. These radioisotope thermoelectric generators (RTGs) have been used in spacecraft and many other applications. Certain Cold War–era Russian-made RTGs used a 5.0 kg strontium-90 source. One mole of strontium-90 releases β particles with an energy of 0.545 MeV and undergoes 2.7 × 10 13 decays/s. How many watts of power are available from this RTG (1 watt = 1 J/s)? Potassium consists of three isotopes (potassium-39, potassium-40, and potassium-41). Potassium-40 is the least abundant, and it is radioactive, decaying to argon-40, a stable, nonradioactive isotope, by the emission of a β particle with a half-life of precisely 1.25 × 10 9 yr. Thus the ratio of potassium-40 to argon-40 in any potassium-40–containing material can be used to date the sample. In 1952, fragments of an early hominid, Meganthropus, were discovered near Modjokerto in Java. The bone fragments were lying on volcanic rock that was believed to be the same age as the bones. Potassium–argon dating on samples of the volcanic material showed that the argon-40-to-potassium-40 molar ratio was 0.00281:1. How old were the rock fragments? Could the bones also be the same age? Could radiocarbon dating have been used to date the fragments? Answers 6.6 × 10 −3 rad; 0.13 rem \(_{92}^{235}\textrm U+\,_0^1\textrm n\rightarrow\,_{92}^{236}\textrm U+\,\gamma\) \(_{92}^{238}\textrm U+\,_0^1\textrm n\rightarrow\,_{92}^{239}\textrm U+\,\gamma\) \(_{92}^{239}\textrm U\rightarrow\,_{93}^{239}\textrm{Np}+\,_{-1}^0\beta\) \(_{93}^{239}\textrm{Np}\rightarrow\,_{94}^{239}\textrm{Pu}+\,_{-1}^0\beta\) \(_{94}^{239}\textrm{Pu}\rightarrow\,_{95}^{239}\textrm{Am}+\,_{-1}^0\beta\) \(_{95}^{239}\textrm{Am}\rightarrow\,_{96}^{239}\textrm{Cm}+\,_{-1}^0\beta\) 7.1 × 10 18 atoms of Pu 0.35 J 0.49 rad 9.8 rem; this dose is unlikely to be fatal. 130 W |
Courses/Oregon_Institute_of_Technology/OIT_(Lund)%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)/10%3A_Nucleophilic_Carbonyl_Addition_Reactions | 10.0: Prelude to Nucleophilic Carbonyl Addition Reactions Acetals are derived from aldehydes. The reactions that occur at the carbonyl carbon of aldehydes and ketones is absolutely central to the chemistry of carbohydrates such as starch and cellulose, and it is this chemistry that is the subject of the chapter we are about to begin. 10.1: Nucleophilic Additions to Aldehydes and Ketones - An Overview Recall from chapter 1 that the ketone functional group is made up of a carbonyl bonded to two carbons, while in an aldehyde one (or both) of the neighboring atoms is a hydrogen. 10.2: Hemiacetals, Hemiketals, and Hydrates One of the most important examples of a nucleophilic addition reaction in biochemistry, and in carbohydrate chemistry in particular, is the addition of an alcohol to a ketone or aldehyde. When an alcohol adds to an aldehyde, the result is called a hemiacetal; when an alcohol adds to a ketone the resulting product is a hemiketal. 10.3: Acetals and Ketals Hemiacetals and hemiketals can react with a second alcohol nucleophile to form an acetal or ketal. The second alcohol may be the same as the first (ie. if R2=R3 in the scheme below), or different. 10.4: N-glycosidic Bonds We have just seen that when a second alcohol attacks a hemiacetal or hemiketal, the result is an acetal or ketal, with the glycosidic bonds in carbohydrates providing a biochemical example. But if a hemiacetal is attacked not by a second alcohol but by an amine, what results is a kind of ‘mixed acetal’ in which the anomeric carbon is bonded to one oxygen and one nitrogen. 10.5: Imines The electrophilic carbon atom of aldehydes and ketones can be the target of nucleophilic attack by amines as well as alcohols. The end result of attack by an amine nucleophile is a functional group in which the C=O double bond is replaced by a C=N double bond, and is known as an imine. 10.6: A Look Ahead - Addition of Carbon and Hydride Nucleophiles to Carbonyls We have seen in this chapter a number of reactions in which oxygen and nitrogen nucleophiles add to carbonyl groups. Other nucleophiles are possible in carbonyl addition mechanisms: in chapters 12 and 13, for example, we will examine in detail some enzyme-catalyzed reactions where the attacking nucleophile is a resonance stabilized carbanion (usually an enolate ion). 10.E: Nucleophilic Carbonyl Addition Reactions (Exercises) 10.S: Nucleophilic Carbonyl Addition Reactions (Summary) |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/09%3A_The_Second_Law_-_Entropy_and_Spontaneous_Change/9.24%3A_The_Free_Energy_Changes_for_A_Spontaneous_Process_at_Constant_T | Now let us consider the change in the Helmholtz free energy when a system undergoes a spontaneous change while in thermal contact with surroundings whose temperature remains constant at \(\hat{T}\). We begin by considering an arbitrarily small increment of change in a process in which the temperature of the system remains constant at \(T=\hat{T}\). The change in the Helmholtz free energy for this process is \({\left(dA\right)}_T=dE-TdS\). Substituting \(dE=dq^{spon}+dw^{spon}\) gives \[{\left(dA\right)}_T=dq^{spon}+dw^{spon}-TdS \nonumber \] (spontaneous process, constant \(T\)) Rearranging, we have \({\left(dA\right)}_T-dw^{spon}+TdS=dq^{spon}\). Using the inequality \(dq^{spon}<\hat{T}dS\), we have \[{\left(dA\right)}_T-dw^{spon}+TdS<\hat{T}dS \nonumber \] When we stipulate that \(T=\hat{T}=\mathrm{constant}\), this becomes \[{\left(dA\right)}_T<dw^{spon} \nonumber \] (spontaneous process, constant \(T\)) where \(dw^{spon}\) is all of the work of any kind done on the system during a small increment of the spontaneous process. If we introduce the still further requirement that the volume is constant, we have \(dw^{spon}_{PV}=0\) and \(dw^{spon}=dw^{spon}_{NPV}\). Then \[{\left(dA\right)}_{TV}<dw^{spon}_{npv} \nonumber \] (spontaneous process, constant \(T\) and \(V\)) and if only pressure–volume work is possible, \[{\left(dA\right)}_{TV}<0 \nonumber \] (spontaneous process, constant \(T\) and \(V\), only \(PV\) work) From our earlier discussion of reversible processes, we have the parallel relationships \[{\left(dA\right)}_T=dw^{rev}_{net} \nonumber \] (reversible isothermal process) \[{\left(dA\right)}_{TV}=dw^{rev}_{NPV} \nonumber \] (reversible process at constant \(T\) and \(V\)) \[{\left(dA\right)}_{TV}=0 \nonumber \] (reversible process at constant \(T\) and \(V\), only \(PV\) work) Similarly, under these conditions, the change in the Gibbs free energy for a spontaneous isothermal process is \[\begin{aligned} \left(dG\right)_T & =dH-TdS \\ ~& =dE+d\left(PV\right)-TdS \\ ~ & =dq^{spon}+dw^{spon}_{PV}+dw^{spon}_{NPV}+d\left(PV\right)-TdS \end{aligned} \nonumber \] Rearranging, we have \[{\left(dG\right)}_T-dw^{spon}_{PV}-dw^{spon}_{NPV}-d\left(PV\right)+TdS ={dq}^{spon}<\hat{T}dS \nonumber \] and since \(T=\hat{T}=\mathrm{constant}\), \[{\left(dG\right)}_T<dw^{spon}_{pv}+dw^{spon}_{npv}+d\left(pv\right) \nonumber \] (spontaneous process, constant \(T\)) As we did when considering the enthalpy change for a spontaneous process, we introduce the additional constraints that the system is subjected to a constant applied pressure, \(P_{applied}\), and that \({P=P}_{applied}\) throughout the process. The irreversible pressure–volume work done by the surroundings on the system becomes \(dw^{spon}_{PV}={-P}_{applied}dV\), and the change in the pressure volume product becomes \({d\left(PV\right)=P}_{applied}dV\). The Gibbs free energy inequality becomes \[{\left(dG\right)}_{TP}<dw^{spon}_{npv} \nonumber \] (spontaneous process, constant \(P_{applied}\) and \(T\)) If only pressure–volume work is possible, this becomes \[{\left(dG\right)}_{TP}<0 \nonumber \] (spontaneous process, constant \(P_{applied}\) and \(T\), only \(PV\) work) From our earlier discussion of reversible processes, we have the parallel relationships \[{\left(dG\right)}_{TP}=dw^{rev}_{NPV} \nonumber \] (reversible process, constant \(P\) and \(T\)) \[{\left(dG\right)}_{TP}=0 \nonumber \] (reversible process, constant \(P\) and \(T\), only \(PV\) work) Since each of these differential-expression criteria applies to every incremental part of a reversible change that falls within its scope, we have the following criteria for finite spontaneous changes when the temperature of the system is constant: \[{\left(\Delta A\right)}_T<w^{spon} \nonumber \] (spontaneous process, constant \(T\)) \[{\left(\Delta A\right)}_{TV}<w^{spon}_{npv} \nonumber \] (spontaneous process, constant \(T\) and \(V\)) \[{\left(\Delta A\right)}_{TV}<0 \nonumber \] (spontaneous process, constant \(T\) and \(V\), only \(PV\) work) \[{\left(\Delta G\right)}_{TP}<w^{spon}_{npv} \nonumber \] (spontaneous process, constant \(P_{applied}\) and \(T\)) \[{\left(\Delta G\right)}_{TP}=0 \nonumber \] (spontaneous process, constant \(P_{applied}\) and \(T\), only \(PV\) work) While the development we have just made assumes that the system temperature is strictly constant, the validity of these finite-change inequalities is not restricted to the condition of strictly constant system temperature. We can derive these finite-change inequalities by essentially the same argument from less restrictive conditions. Let us consider a spontaneous process in which a system goes from state B to state C while in contact with surroundings whose temperature remains constant at \(\hat{T}\). We suppose that in both state B and state C the system temperature is equal to the surroundings temperature; that is, \(T_B=T_C=\hat{T}=\mathrm{constant}\). However, at any intermediate point in the process, the system can have any temperature whatsoever. In states B and C, the Helmholtz free energies are \(A_B=E_B-\hat{T}S_B\) and \(A_C=E_C-\hat{T}S_C\). The change in the Helmholtz free energy is \(\left(A_C-A_B\right)=\left(E_C{-E}_B\right)-\hat{T}\left(S_C-S_B\right)\) or \({\left(\Delta A\right)}_{\hat{T}}=\Delta E-\hat{T}\Delta S=q^{spon}+w^{spon}-\hat{T}\Delta S\). Rearranging, and using \(q^{spon}<\hat{T}\Delta S\), we have \({\left(\Delta A\right)}_{\hat{T}}-w^{spon}+\hat{T}\Delta S=q^{spon}<\hat{T}\Delta S\), so that \[{\left(\Delta A\right)}_{\hat{T}}<w^{spon} \nonumber \] (spontaneous process, constant \(\hat{T}\)) If we require further that the system volume remain constant, there is no pressure–volume work, and we have \[{\left(\Delta A\right)}_{V\hat{T}}<w^{spon}_{npv} \nonumber \] (spontaneous process, constant \(\hat{T}\) and \(V\)) If only pressure–volume work is possible, \(w^{spon}_{NPV}=0\), and \[{\left(\Delta A\right)}_{V\hat{T}}<0 \nonumber \] (spontaneous process, constant \(\hat{T}\) and \(V\), only \(PV\) work) Under the same temperature assumptions, and assuming that \(P_B=P_C=P_{applied}=\mathrm{constant}\), the Gibbs free energies are \(G_B=E_B+P_{applied}V_B-\hat{T}S_B\) and \(G_c=E_C+P_{applied}V_C-\hat{T}S_C\). So that \(\left(G_C-G_B\right)=\left(E_C{-E}_B\right)+P_{applied}\left(V_C-V_B\right)-\hat{T}\left(S_C-S_B\right)\) or \[ \begin{aligned} {\left(\Delta G\right)}_{P\hat{T}} & =\Delta E+P_{applied}\Delta V-\hat{T}\Delta S \\ ~ & =q^{spon}+w^{spon}_{PV}+w^{spon}_{NPV}+P_{applied}\Delta V-\hat{T}\Delta S \end{aligned} \nonumber \] The pressure–volume work is \(w^{spon}_{PV}=-P_{applied}\Delta V\). Cancelling and rearranging, we have \[{\left(\Delta G\right)}_{P\hat{T}}-w^{spon}_{NPV}+\hat{T}\Delta S=q^{spon}<\hat{T}\Delta S \nonumber \] and \[{\left(\Delta G\right)}_{P\hat{T}}<w^{spon}_{npv} \nonumber \] (spontaneous process, constant \(\hat{T}\) and \(P\)) If only pressure–volume work is possible, \[{\left(\Delta G\right)}_{P\hat{T}}<0 \nonumber \] (spontaneous process, constant \(\hat{T}\) and \(P\), only \(PV\) work) We find \({\left(\Delta G\right)}_{P\hat{T}}<w^{spon}_{npv}\)> for any spontaneous process that occurs at constant pressure, while the system is in contact with surroundings at the constant temperature \(\hat{T}\), and in which the initial and final system temperatures are equal to \(\hat{T}\). These are the most common conditions for carrying out a chemical reaction. Consider the situation after we mix non-volatile reactants in an open vessel in a constant-temperature bath. We suppose that the initial temperature of the mixture is the same as that of the bath. The atmosphere applies a constant pressure to the system. The reaction is an irreversible process. It proceeds spontaneously until its equilibrium position is reached. Until equilibrium is reached, the reaction cannot be reversed by an arbitrarily small change in the applied pressure or the temperature of the surroundings. \({\left(\Delta G\right)}_{P\hat{T}}<w^{spon}_{npv}\) and \({\left(\Delta G\right)}_{P\hat{T}}<0\) are criteria for spontaneous change that apply to this situation whatever the temperature of the system might be during any intermediate part of the process. |
Courses/Widener_University/CHEM_105%3A_Introduction_to_General_Organic_and_Biological_Chemistry_Fall_22/02%3A_Measurements_and_Problem_Solving/2.02%3A_Scientific_Notation_-_Writing_Large_and_Small_Numbers | Learning Objectives Express a large number or a small number in scientific notation. Carry out arithmetical operations and express the final answer in scientific notation Chemists often work with numbers that are exceedingly large or small. For example, entering the mass in grams of a hydrogen atom into a calculator would require a display with at least 24 decimal places. A system called scientific notation avoids much of the tedium and awkwardness of manipulating numbers with large or small magnitudes. In scientific notation, these numbers are expressed in the form \[ N \times 10^n \nonumber \] where N is greater than or equal to 1 and less than 10 (1 ≤ N < 10), and n is a positive or negative integer (10 0 = 1). The number 10 is called the base because it is this number that is raised to the power \(n\). Although a base number may have values other than 10, the base number in scientific notation is always 10. A simple way to convert numbers to scientific notation is to move the decimal point as many places to the left or right as needed to give a number from 1 to 10 (N). The magnitude of n is then determined as follows: If the decimal point is moved to the left n places, n is positive. If the decimal point is moved to the right n places, n is negative. Another way to remember this is to recognize that as the number N decreases in magnitude, the exponent increases and vice versa. The application of this rule is illustrated in Example \(\PageIndex{1}\). Example \(\PageIndex{1}\): Expressing Numbers in Scientific Notation Convert each number to scientific notation. 637.8 0.0479 7.86 12,378 0.00032 61.06700 2002.080 0.01020 Solution Unnamed: 0 Explanation Answer a To convert 637.8 to a number from 1 to 10, we move the decimal point two places to the left: 637.8 Because the decimal point was moved two places to the left, n = 2. \(6.378 \times 10^2\) b To convert 0.0479 to a number from 1 to 10, we move the decimal point two places to the right: 0.0479 Because the decimal point was moved two places to the right, n = −2. \(4.79 \times 10^{−2}\) c This is usually expressed simply as 7.86. (Recall that 100 = 1.) \(7.86 \times 10^0\) d Because the decimal point was moved four places to the left, n = 4. \(1.2378 \times 10^4\) e Because the decimal point was moved four places to the right, n = −4. \(3.2 \times 10^{−4}\) f Because the decimal point was moved one place to the left, n = 1. \(6.106700 \times 10^1\) g Because the decimal point was moved three places to the left, n = 3. \(2.002080 \times 10^3\) h Because the decimal point was moved two places to the right, n = -2. \(1.020 \times 10^{−2}\) Addition and Subtraction Before numbers expressed in scientific notation can be added or subtracted, they must be converted to a form in which all the exponents have the same value. The appropriate operation is then carried out on the values of N. Example \(\PageIndex{2}\) illustrates how to do this. Example \(\PageIndex{2}\): Expressing Sums and Differences in Scientific Notation Carry out the appropriate operation and then express the answer in scientific notation. \( (1.36 \times 10^2) + (4.73 \times 10^3) \nonumber\) \((6.923 \times 10^{−3}) − (8.756 \times 10^{−4}) \nonumber\) Solution Unnamed: 0 Explanation Answer a Both exponents must have the same value, so these numbers are converted to either \((1.36 \times 10^2) + (47.3 \times 10^2) = (1.36 + 47.3) \times 10^2 = 48.66 × 10^2\) or \((0.136 \times 10^3) + (4.73 \times 10^3) = (0.136 + 4.73) \times 10^3) = 4.87 \times 10^3\). Choosing either alternative gives the same answer, reported to two decimal places. In converting 48.66 × 102 to scientific notation, \(n\) has become more positive by 1 because the value of \(N\) has decreased. \(4.87 \times 10^3\) b Converting the exponents to the same value gives either \((6.923 \times 10^{-3}) − (0.8756 \times 10^{-3}) = (6.923 − 0.8756) \times 10^{−3}\) or \((69.23 \times 10^{-4}) − (8.756 \times 10^{-4}) = (69.23 − 8.756) \times 10^{−4} = 60.474 \times 10^{−4}\). In converting 60.474 × 10-4 to scientific notation, \(n\) has become more positive by 1 because the value of \(N\) has decreased. \(6.047 \times 10^{−3}\) Multiplication and Division When multiplying numbers expressed in scientific notation, we multiply the values of \(N\) and add together the values of \(n\). Conversely, when dividing, we divide \(N\) in the dividend (the number being divided) by \(N\) in the divisor (the number by which we are dividing) and then subtract n in the divisor from n in the dividend. In contrast to addition and subtraction, the exponents do not have to be the same in multiplication and division. Examples of problems involving multiplication and division are shown in Example \(\PageIndex{3}\). Example \(\PageIndex{3}\): Expressing Products and Quotients in Scientific Notation Perform the appropriate operation and express your answer in scientific notation. \( (6.022 \times 10^{23})(6.42 \times 10^{−2}) \nonumber\) \( \dfrac{ 1.67 \times 10^{-24} }{ 9.12 \times 10 ^{-28} } \nonumber \) \( \dfrac{ (6.63 \times 10^{−34})(6.0 \times 10) }{ 8.52 \times 10^{−2}} \nonumber \) Solution Unnamed: 0 Explanation Answer a In multiplication, we add the exponents: \[(6.022 \times 10^{23})(6.42 \times 10^{−2})= (6.022)(6.42) \times 10^{[23 + (−2)]} = 38.7 \times 10^{21} \nonumber \] In converting \(38.7 \times 10^{21}\) to scientific notation, \(n\) has become more positive by 1 because the value of \(N\) has decreased. NaN \(3.87 \times 10^{22}\) b In division, we subtract the exponents: \[{1.67 \times 10^{−24} \over 9.12 \times 10^{−28}} = {1.67 \over 9.12} \times 10^{[−24 − (−28)]} = 0.183 \times 10^4 \nonumber \] In converting \(0.183 \times 10^4\) to scientific notation, \(n\) has become more negative by 1 because the value of \(N\) has increased. \( 1.83 \times 10^3\) c This problem has both multiplication and division: \[ {(6.63 \times 10^{−34})(6.0 \times 10) \over (8.52 \times 10^{−2})} = {39.78 \over 8.52} \times 10^{[−34 + 1 − (−2)]} \nonumber \] \( 4.7\times 10^{-31}\) |
Ancillary_Materials/Laboratory_Experiments/Wet_Lab_Experiments/General_Chemistry_Labs/Chemistry_I_Laboratory_Manual_(Garber-Morales)/00%3A_Front_Matter/02%3A_InfoPage | This text is disseminated via the Open Education Resource (OER) LibreTexts Project ( https://LibreTexts.org ) and like the hundreds of other texts available within this powerful platform, it is freely available for reading, printing and "consuming." Most, but not all, pages in the library have licenses that may allow individuals to make changes, save, and print this book. Carefully consult the applicable license(s) before pursuing such effects. Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new technologies to support learning. The LibreTexts mission is to unite students, faculty and scholars in a cooperative effort to develop an easy-to-use online platform for the construction, customization, and dissemination of OER content to reduce the burdens of unreasonable textbook costs to our students and society. The LibreTexts project is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being optimized by students, faculty, and outside experts to supplant conventional paper-based books. These free textbook alternatives are organized within a central environment that is both vertically (from advance to basic level) and horizontally (across different fields) integrated. The LibreTexts libraries are Powered by NICE CXOne and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. This material is based upon work supported by the National Science Foundation under Grant No. 1246120, 1525057, and 1413739. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation nor the US Department of Education. Have questions or comments? For information about adoptions or adaptions contact [email protected] . More information on our activities can be found via Facebook ( https://facebook.com/Libretexts ), Twitter ( https://twitter.com/libretexts ), or our blog ( http://Blog.Libretexts.org ). This text was compiled on 04/21/2025 |
Courses/Los_Angeles_Trade_Technical_College/LATTC_Hybrid_Chem_51/00%3A_Front_Matter/02%3A_InfoPage | This text is disseminated via the Open Education Resource (OER) LibreTexts Project ( https://LibreTexts.org ) and like the hundreds of other texts available within this powerful platform, it is freely available for reading, printing and "consuming." Most, but not all, pages in the library have licenses that may allow individuals to make changes, save, and print this book. Carefully consult the applicable license(s) before pursuing such effects. Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new technologies to support learning. The LibreTexts mission is to unite students, faculty and scholars in a cooperative effort to develop an easy-to-use online platform for the construction, customization, and dissemination of OER content to reduce the burdens of unreasonable textbook costs to our students and society. The LibreTexts project is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being optimized by students, faculty, and outside experts to supplant conventional paper-based books. These free textbook alternatives are organized within a central environment that is both vertically (from advance to basic level) and horizontally (across different fields) integrated. The LibreTexts libraries are Powered by NICE CXOne and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. This material is based upon work supported by the National Science Foundation under Grant No. 1246120, 1525057, and 1413739. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation nor the US Department of Education. Have questions or comments? For information about adoptions or adaptions contact [email protected] . More information on our activities can be found via Facebook ( https://facebook.com/Libretexts ), Twitter ( https://twitter.com/libretexts ), or our blog ( http://Blog.Libretexts.org ). This text was compiled on 04/21/2025 |
Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/21%3A_Carboxylic_Acids/21.06%3A_Condensation_of_Acids_with_Alcohols-_The_Fischer_Esterification | Carboxylic acids can react with alcohols to form esters in a process called Fischer esterification. An acid catalyst is required and the alcohol is also used as the reaction solvent. The oxygen atoms are color-coded in the reaction below to help understand the reaction mechanism. For example, butanoic acid reacts with methanol to synthsize methylbutanoate. It is important to note that any proton source can be used as the catalyst. Sulfuric acid is shown in the example below. Mechanism 1) Protonation of the carbonyl by the acid. The carbonyl is now activated toward nucleophilic reactions. 2) Nucleophilic reaction at the carbonyl 3) Proton transfer 4) Water leaves 5) Deprotonation Isotopic Labeling Evidence to support the Fischer esterfication mechanism comes from isotopic labeling experiments with oxygen-18. If the reaction is carried out with oxygen-18 labeled alcohol, the isotope is found exclusively in the ester and not the water generated. Exercise 6. Draw the bond-line structures for the products of the following reactions. Answer 6. |
Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/05%3A_Cooperativity/19%3A_Self-Assembly | Cooperative self-assembly refers to the the spontaneous formation of sophisticated structures from many molecular units. Generally, we think of this as involving many molecules (cooperative units), although single- and bi-molecular problems can be wrapped into this description, as in the helix–coil transition. Examples include: Peptides and proteins Protein folding, binding, and association Amyloid fibrilization Assembly of multi-protein complexes Viral capsid self-assembly Nucleic acids DNA hybridization, DNA origami Folding and association of RNA structures: pseudoknots, ribozym es Lipids Bilayer structures Micelle formation Although molecular structures also assemble with the input of energy, the emphasis here in on spontaneous self-assembly in the absence of external input. 19.1: Micelle Formation In particular, we will focus on micellar structures formed from a single species of amphiphilic molecule in aqueous solution. These are typically lipids or surfactants that have a charged or polar head group linked to one or more long hydrocarbon chains. 19.2: Classical Nucleation Theory 19.3: Why Are Micelles Uniform in Size? 19.4: Shape of Self-Assembled Amphiphiles |
Courses/can/CHEM_210_General_Chemistry_I_(Puenzo)/09%3A_Periodic_Properties_of_the_Elements/9.03%3A_Electron_Configurations-_How_Electrons_Occupy_Orbitals | Learning Objectives Derive the predicted ground-state electron configurations of atoms Identify and explain exceptions to predicted electron configurations for atoms and ions Relate electron configurations to element classifications in the periodic table Having introduced the basics of atomic structure and quantum mechanics, we can use our understanding of quantum numbers to determine how atomic orbitals relate to one another. This allows us to determine which orbitals are occupied by electrons in each atom. The specific arrangement of electrons in orbitals of an atom determines many of the chemical properties of that atom. Orbital Energies and Atomic Structure The energy of atomic orbitals increases as the principal quantum number, \(n\), increases. In any atom with two or more electrons, the repulsion between the electrons makes energies of subshells with different values of \(l\) differ so that the energy of the orbitals increases within a shell in the order s < p < d < f. Figure \(\PageIndex{1}\) depicts how these two trends in increasing energy relate. The 1 s orbital at the bottom of the diagram is the orbital with electrons of lowest energy. The energy increases as we move up to the 2 s and then 2 p , 3 s , and 3 p orbitals, showing that the increasing n value has more influence on energy than the increasing l value for small atoms. However, this pattern does not hold for larger atoms. The 3 d orbital is higher in energy than the 4 s orbital. Such overlaps continue to occur frequently as we move up the chart. Electrons in successive atoms on the periodic table tend to fill low-energy orbitals first. Thus, many students find it confusing that, for example, the 5 p orbitals fill immediately after the 4 d , and immediately before the 6 s . The filling order is based on observed experimental results, and has been confirmed by theoretical calculations. As the principal quantum number, n , increases, the size of the orbital increases and the electrons spend more time farther from the nucleus. Thus, the attraction to the nucleus is weaker and the energy associated with the orbital is higher (less stabilized). But this is not the only effect we have to take into account. Within each shell, as the value of l increases, the electrons are less penetrating (meaning there is less electron density found close to the nucleus), in the order s > p > d > f . Electrons that are closer to the nucleus slightly repel electrons that are farther out, offsetting the more dominant electron–nucleus attractions slightly (recall that all electrons have −1 charges, but nuclei have + Z charges). This phenomenon is called shielding and will be discussed in more detail in the next section. Electrons in orbitals that experience more shielding are less stabilized and thus higher in energy. For small orbitals (1 s through 3 p ), the increase in energy due to n is more significant than the increase due to l ; however, for larger orbitals the two trends are comparable and cannot be simply predicted. We will discuss methods for remembering the observed order. The arrangement of electrons in the orbitals of an atom is called the electron configuration of the atom. We describe an electron configuration with a symbol that contains three pieces of information ( Figure \(\PageIndex{2}\)): The number of the principal quantum shell, n , The letter that designates the orbital type (the subshell, l ), and A superscript number that designates the number of electrons in that particular subshell. For example, the notation 2 p 4 (read "two–p–four") indicates four electrons in a p subshell ( l = 1) with a principal quantum number ( n ) of 2. The notation 3 d 8 (read "three–d–eight") indicates eight electrons in the d subshell (i.e., l = 2) of the principal shell for which n = 3. The Aufbau Principle To determine the electron configuration for any particular atom, we can “build” the structures in the order of atomic numbers. Beginning with hydrogen, and continuing across the periods of the periodic table, we add one proton at a time to the nucleus and one electron to the proper subshell until we have described the electron configurations of all the elements. This procedure is called the Aufbau principle , from the German word Aufbau (“to build up”). Each added electron occupies the subshell of lowest energy available (in the order shown in Figure \(\PageIndex{3}\)), subject to the limitations imposed by the allowed quantum numbers according to the Pauli exclusion principle. Electrons enter higher-energy subshells only after lower-energy subshells have been filled to capacity. Figure \(\PageIndex{3}\) illustrates the traditional way to remember the filling order for atomic orbitals. Since the arrangement of the periodic table is based on the electron configurations, Figure \(\PageIndex{4}\) provides an alternative method for determining the electron configuration. The filling order simply begins at hydrogen and includes each subshell as you proceed in increasing Z order. For example, after filling the 3 p block up to Ar, we see the orbital will be 4s (K, Ca), followed by the 3 d orbitals. We will now construct the ground-state electron configuration and orbital diagram for a selection of atoms in the first and second periods of the periodic table. Orbital diagrams are pictorial representations of the electron configuration, showing the individual orbitals and the pairing arrangement of electrons. We start with a single hydrogen atom (atomic number 1), which consists of one proton and one electron. Referring to either Figure \(\PageIndex{3}\) or \(\PageIndex{4}\), we would expect to find the electron in the 1 s orbital. By convention, the \(m_s=+\dfrac{1}{2}\) value is usually filled first. The electron configuration and the orbital diagram are: Following hydrogen is the noble gas helium, which has an atomic number of 2. The helium atom contains two protons and two electrons. The first electron has the same four quantum numbers as the hydrogen atom electron ( n = 1, l = 0, m l = 0, \(m_s=+\dfrac{1}{2}\)). The second electron also goes into the 1 s orbital and fills that orbital. The second electron has the same n , l , and m l quantum numbers, but must have the opposite spin quantum number, \(m_s=−\dfrac{1}{2}\). This is in accord with the Pauli exclusion principle: No two electrons in the same atom can have the same set of four quantum numbers. For orbital diagrams, this means two arrows go in each box (representing two electrons in each orbital) and the arrows must point in opposite directions (representing paired spins). The electron configuration and orbital diagram of helium are: The n = 1 shell is completely filled in a helium atom. The next atom is the alkali metal lithium with an atomic number of 3. The first two electrons in lithium fill the 1 s orbital and have the same sets of four quantum numbers as the two electrons in helium. The remaining electron must occupy the orbital of next lowest energy, the 2 s orbital (Figure \(\PageIndex{3}\) or \(\PageIndex{4}\)). Thus, the electron configuration and orbital diagram of lithium are: An atom of the alkaline earth metal beryllium, with an atomic number of 4, contains four protons in the nucleus and four electrons surrounding the nucleus. The fourth electron fills the remaining space in the 2 s orbital. An atom of boron (atomic number 5) contains five electrons. The n = 1 shell is filled with two electrons and three electrons will occupy the n = 2 shell. Because any s subshell can contain only two electrons, the fifth electron must occupy the next energy level, which will be a 2 p orbital. There are three degenerate 2 p orbitals ( m l = −1, 0, +1) and the electron can occupy any one of these p orbitals. When drawing orbital diagrams, we include empty boxes to depict any empty orbitals in the same subshell that we are filling. Carbon (atomic number 6) has six electrons. Four of them fill the 1 s and 2 s orbitals. The remaining two electrons occupy the 2 p subshell. We now have a choice of filling one of the 2 p orbitals and pairing the electrons or of leaving the electrons unpaired in two different, but degenerate, p orbitals. The orbitals are filled as described by Hund’s rule : the lowest-energy configuration for an atom with electrons within a set of degenerate orbitals is that having the maximum number of unpaired electrons. Thus, the two electrons in the carbon 2 p orbitals have identical n , l , and m s quantum numbers and differ in their m l quantum number (in accord with the Pauli exclusion principle). The electron configuration and orbital diagram for carbon are: Nitrogen (atomic number 7) fills the 1 s and 2 s subshells and has one electron in each of the three 2 p orbitals, in accordance with Hund’s rule. These three electrons have unpaired spins. Oxygen (atomic number 8) has a pair of electrons in any one of the 2 p orbitals (the electrons have opposite spins) and a single electron in each of the other two. Fluorine (atomic number 9) has only one 2 p orbital containing an unpaired electron. All of the electrons in the noble gas neon (atomic number 10) are paired, and all of the orbitals in the n = 1 and the n = 2 shells are filled. The electron configurations and orbital diagrams of these four elements are: The alkali metal sodium (atomic number 11) has one more electron than the neon atom. This electron must go into the lowest-energy subshell available, the 3 s orbital, giving a 1 s 2 2 s 2 2 p 6 3 s 1 configuration. The electrons occupying the outermost shell orbital(s) (highest value of n ) are called valence electrons, and those occupying the inner shell orbitals are called core electrons ( Figure \PageIndex5\PageIndex5). Since the core electron shells correspond to noble gas electron configurations, we can abbreviate electron configurations by writing the noble gas that matches the core electron configuration, along with the valence electrons in a condensed format. For our sodium example, the symbol [Ne] represents core electrons, (1 s 2 2 s 2 2 p 6 ) and our abbreviated or condensed configuration is [Ne]3 s 1 . Similarly, the abbreviated configuration of lithium can be represented as [He]2 s 1 , where [He] represents the configuration of the helium atom, which is identical to that of the filled inner shell of lithium. Writing the configurations in this way emphasizes the similarity of the configurations of lithium and sodium. Both atoms, which are in the alkali metal family, have only one electron in a valence s subshell outside a filled set of inner shells. \[\ce{Li:[He]}\,2s^1\\ \ce{Na:[Ne]}\,3s^1 \nonumber \] The alkaline earth metal magnesium (atomic number 12), with its 12 electrons in a [Ne]3 s 2 configuration, is analogous to its family member beryllium, [He]2 s 2 . Both atoms have a filled s subshell outside their filled inner shells. Aluminum (atomic number 13), with 13 electrons and the electron configuration [Ne]3 s 2 3 p 1 , is analogous to its family member boron, [He]2 s 2 2 p 1 . The electron configurations of silicon (14 electrons), phosphorus (15 electrons), sulfur (16 electrons), chlorine (17 electrons), and argon (18 electrons) are analogous in the electron configurations of their outer shells to their corresponding family members carbon, nitrogen, oxygen, fluorine, and neon, respectively, except that the principal quantum number of the outer shell of the heavier elements has increased by one to n = 3. Figure \(\PageIndex{6}\) shows the lowest energy, or ground-state, electron configuration for these elements as well as that for atoms of each of the known elements. When we come to the next element in the periodic table, the alkali metal potassium (atomic number 19), we might expect that we would begin to add electrons to the 3 d subshell. However, all available chemical and physical evidence indicates that potassium is like lithium and sodium, and that the next electron is not added to the 3 d level but is, instead, added to the 4 s level (Figure \(\PageIndex{3}\) or \(\PageIndex{4}\)). As discussed previously, the 3 d orbital with no radial nodes is higher in energy because it is less penetrating and more shielded from the nucleus than the 4 s , which has three radial nodes. Thus, potassium has an electron configuration of [Ar]4 s 1 . Hence, potassium corresponds to Li and Na in its valence shell configuration. The next electron is added to complete the 4 s subshell and calcium has an electron configuration of [Ar]4 s 2 . This gives calcium an outer-shell electron configuration corresponding to that of beryllium and magnesium. Beginning with the transition metal scandium (atomic number 21), additional electrons are added successively to the 3 d subshell. This subshell is filled to its capacity with 10 electrons (remember that for l = 2 [ d orbitals], there are 2 l + 1 = 5 values of m l , meaning that there are five d orbitals that have a combined capacity of 10 electrons). The 4 p subshell fills next. Note that for three series of elements, scandium (Sc) through copper (Cu), yttrium (Y) through silver (Ag), and lutetium (Lu) through gold (Au), a total of 10 d electrons are successively added to the ( n – 1) shell next to the n shell to bring that ( n – 1) shell from 8 to 18 electrons. For two series, lanthanum (La) through lutetium (Lu) and actinium (Ac) through lawrencium (Lr), 14 f electrons ( l = 3, 2 l + 1 = 7 m l values; thus, seven orbitals with a combined capacity of 14 electrons) are successively added to the ( n – 2) shell to bring that shell from 18 electrons to a total of 32 electrons. Example \(\PageIndex{1}\): Quantum Numbers and Electron Configurations What is the electron configuration and orbital diagram for a phosphorus atom? What are the four quantum numbers for the last electron added? Solution The atomic number of phosphorus is 15. Thus, a phosphorus atom contains 15 electrons. The order of filling of the energy levels is 1 s , 2 s , 2 p , 3 s , 3 p , 4 s , . . . The 15 electrons of the phosphorus atom will fill up to the 3 p orbital, which will contain three electrons: The last electron added is a 3 p electron. Therefore, n = 3 and, for a p -type orbital, l = 1. The m l value could be –1, 0, or +1. The three p orbitals are degenerate, so any of these m l values is correct. For unpaired electrons, convention assigns the value of \(+\dfrac{1}{2}\) for the spin quantum number; thus, \(m_s=+\dfrac{1}{2}\). Exercise \(\PageIndex{1}\) Identify the atoms from the electron configurations given: [Ar]4 s 2 3 d 5 [Kr]5 s 2 4 d 10 5 p 6 Answer a Mn Answer b Xe The periodic table can be a powerful tool in predicting the electron configuration of an element. However, we do find exceptions to the order of filling of orbitals that are shown in Figure \(\PageIndex{3}\) or \(\PageIndex{4}\). For instance, the electron configurations of the transition metals chromium (Cr; atomic number 24) and copper (Cu; atomic number 29), among others, are not those we would expect. In general, such exceptions involve subshells with very similar energy, and small effects can lead to changes in the order of filling. In the case of Cr and Cu, we find that half-filled and completely filled subshells apparently represent conditions of preferred stability. This stability is such that an electron shifts from the 4 s into the 3 d orbital to gain the extra stability of a half-filled 3 d subshell (in Cr) or a filled 3 d subshell (in Cu). Other exceptions also occur. For example, niobium (Nb, atomic number 41) is predicted to have the electron configuration [Kr]5 s 2 4 d 3 . Experimentally, we observe that its ground-state electron configuration is actually [Kr]5 s 1 4 d 4 . We can rationalize this observation by saying that the electron–electron repulsions experienced by pairing the electrons in the 5 s orbital are larger than the gap in energy between the 5 s and 4 d orbitals. There is no simple method to predict the exceptions for atoms where the magnitude of the repulsions between electrons is greater than the small differences in energy between subshells. Electron Configurations and the Periodic Table As described earlier, the periodic table arranges atoms based on increasing atomic number so that elements with the same chemical properties recur periodically. When their electron configurations are added to the table (Figure \(\PageIndex{6}\)), we also see a periodic recurrence of similar electron configurations in the outer shells of these elements. Because they are in the outer shells of an atom, valence electrons play the most important role in chemical reactions. The outer electrons have the highest energy of the electrons in an atom and are more easily lost or shared than the core electrons. Valence electrons are also the determining factor in some physical properties of the elements. Elements in any one group (or column) have the same number of valence electrons; the alkali metals lithium and sodium each have only one valence electron, the alkaline earth metals beryllium and magnesium each have two, and the halogens fluorine and chlorine each have seven valence electrons. The similarity in chemical properties among elements of the same group occurs because they have the same number of valence electrons. It is the loss, gain, or sharing of valence electrons that defines how elements react. It is important to remember that the periodic table was developed on the basis of the chemical behavior of the elements, well before any idea of their atomic structure was available. Now we can understand why the periodic table has the arrangement it has—the arrangement puts elements whose atoms have the same number of valence electrons in the same group. This arrangement is emphasized in Figure \(\PageIndex{6}\), which shows in periodic-table form the electron configuration of the last subshell to be filled by the Aufbau principle. The colored sections of Figure \(\PageIndex{6}\) show the three categories of elements classified by the orbitals being filled: main group, transition, and inner transition elements. These classifications determine which orbitals are counted in the valence shell , or highest energy level orbitals of an atom. Main group elements (sometimes called representative elements ) are those in which the last electron added enters an s or a p orbital in the outermost shell, shown in blue and red in Figure \(\PageIndex{6}\). This category includes all the nonmetallic elements, as well as many metals and the intermediate semimetallic elements. The valence electrons for main group elements are those with the highest n level. For example, gallium (Ga, atomic number 31) has the electron configuration [Ar] 4 s 2 3 d 10 4 p 1 , which contains three valence electrons (underlined). The completely filled d orbitals count as core, not valence, electrons. Transition elements or transition metals . These are metallic elements in which the last electron added enters a d orbital. The valence electrons (those added after the last noble gas configuration) in these elements include the ns and ( n – 1) d electrons. The official IUPAC definition of transition elements specifies those with partially filled d orbitals. Thus, the elements with completely filled orbitals (Zn, Cd, Hg, as well as Cu, Ag, and Au in Figure \(\PageIndex{6}\)) are not technically transition elements. However, the term is frequently used to refer to the entire d block (colored yellow in Figure \(\PageIndex{6}\)), and we will adopt this usage in this textbook. Inner transition elements are metallic elements in which the last electron added occupies an f orbital. They are shown in green in Figure \(\PageIndex{6}\). The valence shells of the inner transition elements consist of the ( n – 2) f, the ( n – 1) d , and the ns subshells. There are two inner transition series: The lanthanide series: lanthanide (La) through lutetium (Lu) The actinide series: actinide (Ac) through lawrencium (Lr) Lanthanum and actinium, because of their similarities to the other members of the series, are included and used to name the series, even though they are transition metals with no f electrons. Electron Configurations of Ions We have seen that ions are formed when atoms gain or lose electrons. A cation (positively charged ion) forms when one or more electrons are removed from a parent atom. For main group elements, the electrons that were added last are the first electrons removed. For transition metals and inner transition metals, however, electrons in the s orbital are easier to remove than the d or f electrons, and so the highest ns electrons are lost, and then the ( n – 1) d or ( n – 2) f electrons are removed. An anion (negatively charged ion) forms when one or more electrons are added to a parent atom. The added electrons fill in the order predicted by the Aufbau principle. Example \(\PageIndex{2}\): Predicting Electron Configurations of Ions What is the electron configuration and orbital diagram of: Na + P 3– Al 2 + Fe 2 + Sm 3 + Solution First, write out the electron configuration for each parent atom. We have chosen to show the full, unabbreviated configurations to provide more practice for students who want it, but listing the core-abbreviated electron configurations is also acceptable. Next, determine whether an electron is gained or lost. Remember electrons are negatively charged, so ions with a positive charge have lost an electron. For main group elements, the last orbital gains or loses the electron. For transition metals, the last s orbital loses an electron before the d orbitals. Na: 1 s 2 2 s 2 2 p 6 3 s 1 . Sodium cation loses one electron, so Na + : 1 s 2 2 s 2 2 p 6 3 s 1 = Na + : 1 s 2 2 s 2 2 p 6 . P: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 3 . Phosphorus trianion gains three electrons, so P 3− : 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 . Al: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 1 . Aluminum dication loses two electrons Al 2 + : 1 s 2 2 s 2 2 p 6 3 s 2 3 p 1 = Al 2 + : 1 s 2 2 s 2 2 p 6 3 s 1 . Fe: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 6 . Iron(II) loses two electrons and, since it is a transition metal, they are removed from the 4 s orbital Fe 2 + : 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 6 = 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 6 . Sm: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 6 5 s 2 4 d 10 5 p 6 6 s 2 4 f 6 . Samarium trication loses three electrons. The first two will be lost from the 6 s orbital, and the final one is removed from the 4 f orbital. Sm 3 + : 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 6 5 s 2 4 d 10 5 p 6 6 s 2 4 f 6 = 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 6 5 s 2 4 d 10 5 p 6 4 f 5 . Exercise \(\PageIndex{2}\) Which ion with a +2 charge has the electron configuration 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 4 d 5 ? Which ion with a +3 charge has this configuration? Answer a Tc 2 + Answer b Ru 3 + Summary The relative energy of the subshells determine the order in which atomic orbitals are filled (1 s , 2 s , 2 p , 3 s , 3 p , 4 s , 3 d , 4 p , and so on). Electron configurations and orbital diagrams can be determined by applying the Pauli exclusion principle (no two electrons can have the same set of four quantum numbers) and Hund’s rule (whenever possible, electrons retain unpaired spins in degenerate orbitals). Electrons in the outermost orbitals, called valence electrons, are responsible for most of the chemical behavior of elements. In the periodic table, elements with analogous valence electron configurations usually occur within the same group. There are some exceptions to the predicted filling order, particularly when half-filled or completely filled orbitals can be formed. The periodic table can be divided into three categories based on the orbital in which the last electron to be added is placed: main group elements ( s and p orbitals), transition elements ( d orbitals), and inner transition elements ( f orbitals). Glossary Aufbau principle procedure in which the electron configuration of the elements is determined by “building” them in order of atomic numbers, adding one proton to the nucleus and one electron to the proper subshell at a time core electron electron in an atom that occupies the orbitals of the inner shells electron configuration electronic structure of an atom in its ground state given as a listing of the orbitals occupied by the electrons Hund’s rule every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin orbital diagram pictorial representation of the electron configuration showing each orbital as a box and each electron as an arrow valence electrons electrons in the outermost or valence shell (highest value of n ) of a ground-state atom; determine how an element reacts valence shell outermost shell of electrons in a ground-state atom; for main group elements, the orbitals with the highest n level ( s and p subshells) are in the valence shell, while for transition metals, the highest energy s and d subshells make up the valence shell and for inner transition elements, the highest s , d, and f subshells are included |
Courses/Pasadena_City_College/CHEM_001A%3A_General_Chemistry_and_Chemical_Analysis/15%3A_Chemical_Kinetics/15.07%3A_Catalysis | Learning Objectives To understand how catalysts increase the reaction rate and the selectivity of chemical reactions. Catalysts are substances that increase the reaction rate of a chemical reaction without being consumed in the process. A catalyst, therefore, does not appear in the overall stoichiometry of the reaction it catalyzes, but it must appear in at least one of the elementary reactions in the mechanism for the catalyzed reaction. The catalyzed pathway has a lower E a , but the net change in energy that results from the reaction (the difference between the energy of the reactants and the energy of the products) is not affected by the presence of a catalyst ( Figur e \(\PageIndex{1}\)). Nevertheless, because of its lower E a , the reaction rate of a catalyzed reaction is faster than the reaction rate of the uncatalyzed reaction at the same temperature. Because a catalyst decreases the height of the energy barrier, its presence increases the reaction rates of both the forward and the reverse reactions by the same amount. In this section, we will examine the three major classes of catalysts: heterogeneous catalysts, homogeneous catalysts, and enzymes. A catalyst affects E a , not Δ E . Heterogeneous Catalysis In heterogeneous catalysis , the catalyst is in a different phase from the reactants. At least one of the reactants interacts with the solid surface in a physical process called adsorption in such a way that a chemical bond in the reactant becomes weak and then breaks. Poisons are substances that bind irreversibly to catalysts, preventing reactants from adsorbing and thus reducing or destroying the catalyst’s efficiency. An example of heterogeneous catalysis is the interaction of hydrogen gas with the surface of a metal, such as Ni, Pd, or Pt. As shown in part (a) in Figure \(\PageIndex{2}\) , the hydrogen–hydrogen bonds break and produce individual adsorbed hydrogen atoms on the surface of the metal. Because the adsorbed atoms can move around on the surface, two hydrogen atoms can collide and form a molecule of hydrogen gas that can then leave the surface in the reverse process, called desorption. Adsorbed H atoms on a metal surface are substantially more reactive than a hydrogen molecule. Because the relatively strong H–H bond (dissociation energy = 432 kJ/mol) has already been broken, the energy barrier for most reactions of H 2 is substantially lower on the catalyst surface. Figure \(\PageIndex{2}\) shows a process called hydrogenation , in which hydrogen atoms are added to the double bond of an alkene, such as ethylene, to give a product that contains C–C single bonds, in this case ethane. Hydrogenation is used in the food industry to convert vegetable oils, which consist of long chains of alkenes, to more commercially valuable solid derivatives that contain alkyl chains. Hydrogenation of some of the double bonds in polyunsaturated vegetable oils, for example, produces margarine, a product with a melting point, texture, and other physical properties similar to those of butter. Several important examples of industrial heterogeneous catalytic reactions are in Table \(\PageIndex{1}\). Although the mechanisms of these reactions are considerably more complex than the simple hydrogenation reaction described here, they all involve adsorption of the reactants onto a solid catalytic surface, chemical reaction of the adsorbed species (sometimes via a number of intermediate species), and finally desorption of the products from the surface. Commercial Process Catalyst Initial Reaction Final Commercial Product contact process V2O5 or Pt 2SO2 + O2 → 2SO3 H2SO4 Haber process Fe, K2O, Al2O3 N2 + 3H2 → 2NH3 NH3 Ostwald process Pt and Rh 4NH3 + 5O2 → 4NO + 6H2O HNO3 water–gas shift reaction Fe, Cr2O3, or Cu CO + H2O → CO2 + H2 H2 for NH3, CH3OH, and other fuels steam reforming Ni CH4 + H2O → CO + 3H2 H2 methanol synthesis ZnO and Cr2O3 CO + 2H2 → CH3OH CH3OH Sohio process bismuth phosphomolybdate \(\mathrm{CH}_2\textrm{=CHCH}_3+\mathrm{NH_3}+\mathrm{\frac{3}{2}O_2}\rightarrow\mathrm{CH_2}\textrm{=CHCN}+\mathrm{3H_2O}\) \(\underset{\textrm{acrylonitrile}}{\mathrm{CH_2}\textrm{=CHCN}}\) catalytic hydrogenation Ni, Pd, or Pt RCH=CHR′ + H2 → RCH2—CH2R′ partially hydrogenated oils for margarine, and so forth Homogeneous Catalysis In homogeneous catalysis , the catalyst is in the same phase as the reactant(s). The number of collisions between reactants and catalyst is at a maximum because the catalyst is uniformly dispersed throughout the reaction mixture. Many homogeneous catalysts in industry are transition metal compounds ( Table \(\PageIndex{2}\)), but recovering these expensive catalysts from solution has been a major challenge. As an added barrier to their widespread commercial use, many homogeneous catalysts can be used only at relatively low temperatures, and even then they tend to decompose slowly in solution. Despite these problems, a number of commercially viable processes have been developed in recent years. High-density polyethylene and polypropylene are produced by homogeneous catalysis. Commercial Process Catalyst Reactants Final Product Union Carbide [Rh(CO)2I2]− CO + CH3OH CH3CO2H hydroperoxide process Mo(VI) complexes CH3CH=CH2 + R–O–O–H NaN hydroformylation Rh/PR3 complexes RCH=CH2 + CO + H2 RCH2CH2CHO adiponitrile process Ni/PR3complexes 2HCN + CH2=CHCH=CH2 NCCH2CH2CH2CH2CN used to synthesize nylon olefin polymerization (RC5H5)2ZrCl2 CH2=CH2 –(CH2CH2–)n: high-density polyethylene Enzymes Enzymes, catalysts that occur naturally in living organisms, are almost all protein molecules with typical molecular masses of 20,000–100,000 amu. Some are homogeneous catalysts that react in aqueous solution within a cellular compartment of an organism. Others are heterogeneous catalysts embedded within the membranes that separate cells and cellular compartments from their surroundings. The reactant in an enzyme-catalyzed reaction is called a substrate . Because enzymes can increase reaction rates by enormous factors (up to 10 17 times the uncatalyzed rate) and tend to be very specific, typically producing only a single product in quantitative yield, they are the focus of active research. At the same time, enzymes are usually expensive to obtain, they often cease functioning at temperatures greater than 37 °C, have limited stability in solution, and have such high specificity that they are confined to turning one particular set of reactants into one particular product. This means that separate processes using different enzymes must be developed for chemically similar reactions, which is time-consuming and expensive. Thus far, enzymes have found only limited industrial applications, although they are used as ingredients in laundry detergents, contact lens cleaners, and meat tenderizers. The enzymes in these applications tend to be proteases, which are able to cleave the amide bonds that hold amino acids together in proteins. Meat tenderizers, for example, contain a protease called papain, which is isolated from papaya juice. It cleaves some of the long, fibrous protein molecules that make inexpensive cuts of beef tough, producing a piece of meat that is more tender. Some insects, like the bombadier beetle, carry an enzyme capable of catalyzing the decomposition of hydrogen peroxide to water ( Figure \(\PageIndex{3}\)). Enzyme inhibitors cause a decrease in the reaction rate of an enzyme-catalyzed reaction by binding to a specific portion of an enzyme and thus slowing or preventing a reaction from occurring. Irreversible inhibitors are therefore the equivalent of poisons in heterogeneous catalysis. One of the oldest and most widely used commercial enzyme inhibitors is aspirin, which selectively inhibits one of the enzymes involved in the synthesis of molecules that trigger inflammation. The design and synthesis of related molecules that are more effective, more selective, and less toxic than aspirin are important objectives of biomedical research. Summary Catalysts participate in a chemical reaction and increase its rate. They do not appear in the reaction’s net equation and are not consumed during the reaction. Catalysts allow a reaction to proceed via a pathway that has a lower activation energy than the uncatalyzed reaction. In heterogeneous catalysis, catalysts provide a surface to which reactants bind in a process of adsorption. In homogeneous catalysis, catalysts are in the same phase as the reactants. Enzymes are biological catalysts that produce large increases in reaction rates and tend to be specific for certain reactants and products. The reactant in an enzyme-catalyzed reaction is called a substrate. Enzyme inhibitors cause a decrease in the reaction rate of an enzyme-catalyzed reaction. |
Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Case_Studies%3A_Proteins | Angiotnesin Peptide Enkephalines Membrane Transport Membrane transport is essential for cellular life. As cells proceed through their life cycle, a vast amount of exchange is necessary to maintain function. Transport may involve the incorporation of biological molecules and the discharge of waste products that are necessary for normal function. 1 Membrane transport refers to the movement of particles (solute) across or through a membranous barrier. 2 These membranous barriers, in the case of the cell for example, consist of a phospholipid bilayer. The phospholipids orient themselves in such a way so that the hydrophilic (polar) heads are nearest the extracellular and intracellular mediums, and the hydrophobic (non-polar) tails align between the two hydrophilic head groups. Membrane transport is dependent upon the permeability of the membrane, transmembrane solute concentration, and the size and charge of the solute. 2 Solute particles can traverse the membrane via three mechanisms: passive, facilitated, and active transport. 1 Some of these transport mechanisms require the input of energy and use of a transmembrane protein, whereas other mechanisms do not incorporate secondary molecules. 3 Permanent Hair Wave The formation of disulfide bonds has a direct application in producing curls in hair by the permanent wave process. Hair keratin consists of many protein alpha-helices. Three alpha-helices are interwoven into a left-handed coil called a protofibril. Eleven protofibrils are bonded and coiled together to make a microfibril. Hundreds of these microfibrils are cemented into an irregular bundle called a macrofibril. These in turn are mixed with dead and living cells to make a complete strand of hair. Sickle Cell Anemia The incorrect amino acid sequence in a protein may lead to fatal consequences. For example, the inherited disease, sickle cell anemia, results from a single incorrect amino acid at the 6th position of the beta - protein chain out of 146. Hemoglobin consists of four protein chains - two beta and two alpha. Template:HideTOC |
Bookshelves/General_Chemistry/Chemistry_2e_(OpenStax)/10%3A_Liquids_and_Solids/10.05%3A_The_Solid_State_of_Matter | Learning Objectives By the end of this section, you will be able to: Define and describe the bonding and properties of ionic, molecular, metallic, and covalent network crystalline solids Describe the main types of crystalline solids: ionic solids, metallic solids, covalent network solids, and molecular solids Explain the ways in which crystal defects can occur in a solid When most liquids are cooled, they eventually freeze and form crystalline solids , solids in which the atoms, ions, or molecules are arranged in a definite repeating pattern. It is also possible for a liquid to freeze before its molecules become arranged in an orderly pattern. The resulting materials are called amorphous solids or noncrystalline solids (or, sometimes, glasses). The particles of such solids lack an ordered internal structure and are randomly arranged (Figure \(\PageIndex{1}\)). Metals and ionic compounds typically form ordered, crystalline solids. Substances that consist of large molecules, or a mixture of molecules whose movements are more restricted, often form amorphous solids. For examples, candle waxes are amorphous solids composed of large hydrocarbon molecules. Some substances, such as silicon dioxide (shown in Figure \(\PageIndex{2}\)), can form either crystalline or amorphous solids, depending on the conditions under which it is produced. Also, amorphous solids may undergo a transition to the crystalline state under appropriate conditions. Crystalline solids are generally classified according to the nature of the forces that hold its particles together. These forces are primarily responsible for the physical properties exhibited by the bulk solids. The following sections provide descriptions of the major types of crystalline solids: ionic, metallic, covalent network, and molecular. Ionic Solids Ionic solids , such as sodium chloride and nickel oxide, are composed of positive and negative ions that are held together by electrostatic attractions, which can be quite strong (Figure \(\PageIndex{3}\)). Many ionic crystals also have high melting points. This is due to the very strong attractions between the ions—in ionic compounds, the attractions between full charges are (much) larger than those between the partial charges in polar molecular compounds. This will be looked at in more detail in a later discussion of lattice energies. Although they are hard, they also tend to be brittle, and they shatter rather than bend. Ionic solids do not conduct electricity; however, they do conduct when molten or dissolved because their ions are free to move. Many simple compounds formed by the reaction of a metallic element with a nonmetallic element are ionic. Metallic Solids Metallic solids such as crystals of copper, aluminum, and iron are formed by metal atoms Figure \(\PageIndex{4}\). The structure of metallic crystals is often described as a uniform distribution of atomic nuclei within a “sea” of delocalized electrons. The atoms within such a metallic solid are held together by a unique force known as metallic bonding that gives rise to many useful and varied bulk properties. All exhibit high thermal and electrical conductivity, metallic luster, and malleability. Many are very hard and quite strong. Because of their malleability (the ability to deform under pressure or hammering), they do not shatter and, therefore, make useful construction materials. The melting points of the metals vary widely. Mercury is a liquid at room temperature, and the alkali metals melt below 200 °C. Several post-transition metals also have low melting points, whereas the transition metals melt at temperatures above 1000 °C. These differences reflect differences in strengths of metallic bonding among the metals. Covalent Network Solids Covalent network solids include crystals of diamond, silicon, some other nonmetals, and some covalent compounds such as silicon dioxide (sand) and silicon carbide (carborundum, the abrasive on sandpaper). Many minerals have networks of covalent bonds. The atoms in these solids are held together by a network of covalent bonds, as shown in Figure \(\PageIndex{5}\). To break or to melt a covalent network solid, covalent bonds must be broken. Because covalent bonds are relatively strong, covalent network solids are typically characterized by hardness, strength, and high melting points. For example, diamond is one of the hardest substances known and melts above 3500 °C. Molecular Solids Molecular solids , such as ice, sucrose (table sugar), and iodine, as shown in Figure \(\PageIndex{6}\): , are composed of neutral molecules. The strengths of the attractive forces between the units present in different crystals vary widely, as indicated by the melting points of the crystals. Small symmetrical molecules (nonpolar molecules), such as H 2 , N 2 , O 2 , and F 2 , have weak attractive forces and form molecular solids with very low melting points (below −200 °C). Substances consisting of larger, nonpolar molecules have larger attractive forces and melt at higher temperatures. Molecular solids composed of molecules with permanent dipole moments (polar molecules) melt at still higher temperatures. Examples include ice (melting point, 0 °C) and table sugar (melting point, 185 °C). Properties of Solids A crystalline solid, like those listed in Table \(\PageIndex{1}\), has a precise melting temperature because each atom or molecule of the same type is held in place with the same forces or energy. Thus, the attractions between the units that make up the crystal all have the same strength and all require the same amount of energy to be broken. The gradual softening of an amorphous material differs dramatically from the distinct melting of a crystalline solid. This results from the structural nonequivalence of the molecules in the amorphous solid. Some forces are weaker than others, and when an amorphous material is heated, the weakest intermolecular attractions break first. As the temperature is increased further, the stronger attractions are broken. Thus amorphous materials soften over a range of temperatures. Type of Solid Type of Particles Type of Attractions Properties Examples ionic ions ionic bonds hard, brittle, conducts electricity as a liquid but not as a solid, high to very high melting points NaCl, Al2O3 metallic atoms of electropositive elements metallic bonds shiny, malleable, ductile, conducts heat and electricity well, variable hardness and melting temperature Cu, Fe, Ti, Pb, U covalent network atoms of electronegative elements covalent bonds very hard, not conductive, very high melting points C (diamond), SiO2, SiC molecular molecules (or atoms) IMFs variable hardness, variable brittleness, not conductive, low melting points H2O, CO2, I2, C12H22O11 How Sciences Interconnect: Graphene: Material of the Future Carbon is an essential element in our world. The unique properties of carbon atoms allow the existence of carbon-based life forms such as ourselves. Carbon forms a huge variety of substances that we use on a daily basis, including those shown in Figure \(\PageIndex{7}\). You may be familiar with diamond and graphite, the two most common allotropes of carbon. (Allotropes are different structural forms of the same element.) Diamond is one of the hardest-known substances, whereas graphite is soft enough to be used as pencil lead. These very different properties stem from the different arrangements of the carbon atoms in the different allotropes. You may be less familiar with a recently discovered form of carbon: graphene. Graphene was first isolated in 2004 by using tape to peel off thinner and thinner layers from graphite. It is essentially a single sheet (one atom thick) of graphite. Graphene, illustrated in Figure \(\PageIndex{8}\): , is not only strong and lightweight, but it is also an excellent conductor of electricity and heat. These properties may prove very useful in a wide range of applications, such as vastly improved computer chips and circuits, better batteries and solar cells, and stronger and lighter structural materials. The 2010 Nobel Prize in Physics was awarded to Andre Geim and Konstantin Novoselov for their pioneering work with graphene. Crystal Defects In a crystalline solid, the atoms, ions, or molecules are arranged in a definite repeating pattern, but occasional defects may occur in the pattern. Several types of defects are known, as illustrated in Figure \(\PageIndex{9}\). Vacancies are defects that occur when positions that should contain atoms or ions are vacant. Less commonly, some atoms or ions in a crystal may occupy positions, called interstitial sites , located between the regular positions for atoms. Other distortions are found in impure crystals, as, for example, when the cations, anions, or molecules of the impurity are too large to fit into the regular positions without distorting the structure. Trace amounts of impurities are sometimes added to a crystal (a process known as doping) in order to create defects in the structure that yield desirable changes in its properties. For example, silicon crystals are doped with varying amounts of different elements to yield suitable electrical properties for their use in the manufacture of semiconductors and computer chips. |
Courses/Smith_College/Organic_Chemistry_(LibreTexts)/23%3A_Carbonyl_Condensation_Reactions | Learning Objectives When you have completed Chapter 23, you should be able to fulfill all of the detailed objectives listed under each individual section. design multi‑step syntheses in which the reactions introduced in this unit are used in conjunction with any of the reactions described in previous units. solve road‑map problems that require a knowledge of carbonyl condensation reactions. define, and use in context, any of the key terms introduced. In this chapter, we consider the fourth and final general type of reaction that carbonyl compounds undergo—the carbonyl condensation reaction. Carbonyl condensation reactions take place between two carbonyl‑containing reactants, one of which must possess an alpha‑hydrogen atom. The first step of the reaction involves the removal of an alpha‑hydrogen atom by a base. In the second step, the enolate anion that results from this removal attacks the carbonyl‑carbon of the second reacting molecule. In the final step of the reaction, a proton is transferred to the tetrahedral intermediate formed in the second step, although in some cases the product that results may subsequently be dehydrated. 23.1: Chapter Objectives 23.2: Carbonyl Condensations - The Aldol Reaction 23.3: Carbonyl Condensations versus Alpha Substitutions 23.4: Dehydration of Aldol Products - Synthesis of Enones 23.5: Using Aldol Reactions in Synthesis 23.6: Mixed Aldol Reactions 23.7: Intramolecular Aldol Reactions Molecules which contain two carbonyl functionalities have the possibility of forming a ring through an intramolecular aldol reaction. The term “Intramolecular” means “within the same molecule.” In this case, it means that the enolate donor and the electrophilic acceptor of an aldol reaction are contained in the same molecule such as dialdehydes, keto aldehydes, or diketones. In these cases, the small distance between the donor and acceptor leads to faster reaction rates. 23.8: The Claisen Condensation Reaction 23.9: Mixed Claisen Condensations Claisen condensations between different ester reactants are called Crossed Claisen reactions. Crossed Claisen reactions in which both reactants can serve as Claisen donors and Claisen acceptors generally give complex mixtures which are difficult to separate. 23.10: Intramolecular Claisen Condensations - The Dieckmann Cyclization 23.11: Conjugate Carbonyl Additions - The Michael Reaction Certain nucleophiles undergo conjugate addition with the alkene of an α, β-unsaturated carbonyl compounds rather than undergo direct nucleophilic addition with the carbonyl. During conjugate addition, a nucleophile adds to the electrophilic β-alkene carbon to from a C-Nuc bond. If the starting materials contains an ester the corresponding alkoxide is used as the base in the reaction. Otherwise a hydroxide base, such as sodium or potassium hydroxide, is commonly used. 23.12: Carbonyl Condensations with Enamines - The Stork Reaction 23.13: The Robinson Annulation Reaction 23.14: Some Biological Carbonyl Condensation Reactions 23.S: Carbonyl Condensation Reactions (Summary) |
Courses/Brevard_College/LNC_216_CHE/01%3A_Introduction_to_Chemistry/1.02%3A_Using_the_Scientific_Method | Learning Objectives To identify the components of the scientific method Classify measurements as being quantitative or qualitative. Evaluate science in the media. The Scientific Method Scientists search for answers to questions and solutions to problems by using a procedure called the scientific method. This procedure consists of making observations, formulating hypotheses, and designing experiments, which in turn lead to additional observations, hypotheses, and experiments in repeated cycles (Figure \(\PageIndex{1}\)). Step 1: Make observations Observations can be qualitative or quantitative. Qualitative observations describe properties or occurrences in ways that do not rely on numbers. Examples of qualitative observations include the following: the outside air temperature is cooler during the winter season, table salt is a crystalline solid, sulfur crystals are yellow, and dissolving a penny in dilute nitric acid forms a blue solution and a brown gas. Quantitative observations are measurements, which by definition consist of both a number and a unit. Examples of quantitative observations include the following: the melting point of crystalline sulfur is 115.21° Celsius, and 35.9 grams of table salt—whose chemical name is sodium chloride—dissolves in 100 grams of water at 20° Celsius. For the question of the dinosaurs’ extinction, the initial observation was quantitative: iridium concentrations in sediments dating to 66 million years ago were 20–160 times higher than normal. Step 2: Formulate a hypothesis After deciding to learn more about an observation or a set of observations, scientists generally begin an investigation by forming a hypothesis, a tentative explanation for the observation(s). The hypothesis may not be correct, but it puts the scientist’s understanding of the system being studied into a form that can be tested. For example, the observation that we experience alternating periods of light and darkness corresponding to observed movements of the sun, moon, clouds, and shadows is consistent with either of two hypotheses: Earth rotates on its axis every 24 hours, alternately exposing one side to the sun, or the sun revolves around Earth every 24 hours. Suitable experiments can be designed to choose between these two alternatives. For the disappearance of the dinosaurs, the hypothesis was that the impact of a large extraterrestrial object caused their extinction. Unfortunately (or perhaps, fortunately), this hypothesis does not lend itself to direct testing by any obvious experiment, but scientists can collect additional data that either support or refute it. Step 3: Design and perform experiments After a hypothesis has been formed, scientists conduct experiments to test its validity. Experiments are systematic observations or measurements, preferably made under controlled conditions—that is, under conditions in which a single variable changes. Step 4: Accept or modify the hypothesis A properly designed and executed experiment enables a scientist to determine whether the original hypothesis is valid. In which case he can proceed to step 5. In other cases, experiments often demonstrate that the hypothesis is incorrect or that it must be modified thus requiring further experimentation. Step 5: Development into law and/or theory More experimental data are then collected and analyzed, at which point a scientist may begin to think that the results are sufficiently reproducible (i.e., dependable) to merit being summarized in law, a verbal or mathematical description of a phenomenon that allows for general predictions. A law simply says what happens; it does not address the question of why. One example of a law, the law of definite proportions, which was discovered by the French scientist Joseph Proust (1754–1826), states that a chemical substance always contains the same proportions of elements by mass. Thus, sodium chloride (table salt) always contains the same proportion by mass of sodium to chlorine, in this case, 39.34% sodium and 60.66% chlorine by mass, and sucrose (table sugar) is always 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. Whereas a law states only what happens, a theory attempts to explain why nature behaves as it does. Laws are unlikely to change greatly over time unless a major experimental error is discovered. In contrast, a theory, by definition, is incomplete and imperfect, evolving with time to explain new facts as they are discovered. Because scientists can enter the cycle shown in Figure \(\PageIndex{1}\) at any point, the actual application of the scientific method to different topics can take many different forms. For example, a scientist may start with a hypothesis formed by reading about work done by others in the field, rather than by making direct observations. A Real-World Application of the Scientific Method In 2007, my husband and I journeyed to China to adopt our daughter. Upon arrival in Beijing, I became violently ill. Due to her visa paperwork, my husband, daughter, and I were required to stay in China for two weeks. Unfortunately, I was ill the entire time. Once the two-week period was up, the three of us flew back to the United States where I continued to be sick. For the next year, I remained ill and lost a total of 30 pounds. The picture below shows me holding my daughter eight months after we returned home from China. I would like you to attempt to perform the scientific method on my situation described above. List the steps of the scientific method along with some plausible explanations. * Please have this ready to discuss in class .* Exercise \(\PageIndex{1}\) Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation. Ice always floats on liquid water. Birds evolved from dinosaurs. According to Albert Einstein, mass X speed of light = energy When 10 g of ice was added to 100 mL of water at 25°C, the temperature of the water decreased to 15.5°C after the ice melted. The ingredients of Ivory soap were analyzed to see whether it really is 99.44% pure, as advertised. Solution This is a general statement of a relationship between the properties of liquid and solid water, so it is a law. This is an educated guess regarding the origin of birds, so it is a hypothesis. This is a theory that explains an explanation of events and can be disproven at any time. The temperature is measured before and after a change is made in a system, so these are quantitative observations. This is an analysis designed to test a hypothesis (in this case, the manufacturer’s claim of purity), so it is an experiment. Exercise \(\PageIndex{2}\) Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation. Measured amounts of acid were added to a Rolaids tablet to see whether it really “consumes 47 times its weight in excess stomach acid.” Heat always flows from hot objects to cooler ones, not in the opposite direction. The universe was formed by a massive explosion that propelled matter into a vacuum. Michael Jordan is the greatest pure shooter ever to play professional basketball. Limestone is relatively insoluble in water but dissolves readily in dilute acid with the evolution of a gas. Gas mixtures that contain more than 4% hydrogen in air are potentially explosive. Answer a experiment Answer b law Answer c theory Answer d hypothesis Answer e qualitative observation Answer f quantitative observation Evaluating Science in the Media |
Courses/Georgia_Southern_University/CHEM_1152%3A_Survey_of_Chemistry_II_(Osborne)/10%3A_Metabolism/10.01%3A_Prelude_to_Metabolism | ↵ The discovery of the link between insulin and diabetes led to a period of intense research aimed at understanding exactly how insulin works in the body to regulate glucose levels. Hormones in general act by binding to some protein, known as the hormone’s receptor, thus initiating a series of events that lead to a desired outcome. In the early 1970s, the insulin receptor was purified, and researchers began to study what happens after insulin binds to its receptor and how those events are linked to the uptake and metabolism of glucose in cells. The insulin receptor is located in the cell membrane and consists of four polypeptide chains: two identical chains called α chains and two identical chains called β chains. The α chains, positioned on the outer surface of the membrane, consist of 735 amino acids each and contain the binding site for insulin. The β chains are integral membrane proteins, each composed of 620 amino acids. The binding of insulin to its receptor stimulates the β chains to catalyze the addition of phosphate groups to the specific side chains of tyrosine (referred to as phosphorylation) in the β chains and other cell proteins, leading to the activation of reactions that metabolize glucose. In this chapter we will look at the pathway that breaks down glucose—in response to activation by insulin—for the purpose of providing energy for the cell. Life requires energy. Animals, for example, require heat energy to maintain body temperature, mechanical energy to move their limbs, and chemical energy to synthesize the compounds needed by their cells. Living cells remain organized and functioning properly only through a continual supply of energy. But only specific forms of energy can be used. Supplying a plant with energy by holding it in a flame will not prolong its life. On the other hand, a green plant is able to absorb radiant energy from the sun, the most abundant source of energy for life on the earth. Plants use this energy first to form glucose and then to make other carbohydrates, as well as lipids and proteins. Unlike plants, animals cannot directly use the sun’s energy to synthesize new compounds. They must eat plants or other animals to get carbohydrates, fats, and proteins and the chemical energy stored in them. Once digested and transported to the cells, the nutrient molecules can be used in either of two ways: as building blocks for making new cell parts or repairing old ones or “burned” for energy. The thousands of coordinated chemical reactions that keep cells alive are referred to collectively as metabolism . In general, metabolic reactions are divided into two classes: the breaking down of molecules to obtain energy is catabolism , and the building of new molecules needed by living systems is anabolism . Definition: Metabolite Any chemical compound that participates in a metabolic reaction is a metabolite . Most of the energy required by animals is generated from lipids and carbohydrates. These fuels must be oxidized, or “burned,” for the energy to be released. The oxidation process ultimately converts the lipid or carbohydrate to carbon dioxide (CO 2 ) and water (H 2 O). Carbohydrate : \[\ce{C6H12O6 + 6O2 → 6CO2 + 6H2O + 670 kcal} \nonumber \] Lipid : \[\ce{C16H32O2 + 23O2 → 16CO2 + 16H2O + 2,385 kcal} \nonumber \] These two equations summarize the biological combustion of a carbohydrate and a lipid by the cell through respiration. Respiration is the collective name for all metabolic processes in which gaseous oxygen is used to oxidize organic matter to carbon dioxide, water, and energy. Like the combustion of the common fuels we burn in our homes and cars (wood, coal, gasoline), respiration uses oxygen from the air to break down complex organic substances to carbon dioxide and water. But the energy released in the burning of wood is manifested entirely in the form of heat, and excess heat energy is not only useless but also injurious to the living cell. Living organisms instead conserve much of the energy respiration releases by channeling it into a series of stepwise reactions that produce adenosine triphosphate (ATP) or other compounds that ultimately lead to the synthesis of ATP . The remainder of the energy is released as heat and manifested as body temperature. |
Courses/Saint_Marys_College_Notre_Dame_IN/CHEM_118_(Under_Construction)/CHEM_118_Textbook/03%3A_Compounds/3.11%3A_Shapes_and_Properties-_Polar_and_Nonpolar_Molecules | Learning Objective Determine if a molecule is polar or nonpolar. Molecular Polarity To determine if a molecule is polar or nonpolar, it is frequently useful to look at Lewis structures. Nonpolar compounds will be symmetric, meaning all of the sides around the central atom are identical - bonded to the same element with no unshared pairs of electrons. Notice that a tetrahedral molecule such as \(\ce{CCl_4}\) is nonpolar Figure (\(\PageIndex{1}\). Another non polar molecule shown below is boron trifluoride, BF 3 . BF 3 is a trigonal planar molecule and all three peripheral atoms are the same. Polar molecules are asymmetric, either containing lone pairs of electrons on a central atom or having atoms with different electronegativities bonded. This works pretty well - as long as you can visualize the molecular geometry. That's the hard part. To know how the bonds are oriented in space, you have to have a strong grasp of Lewis structures and VSEPR theory. Assuming you do, you can look at the structure of each one and decide if it is polar or not - whether or not you know the individual atom electronegativity . This is because you know that all bonds between dissimilar elements are polar, and in these particular examples, it doesn't matter which direction the dipole moment vectors are pointing (out or in). A polar molecule is a molecule in which one end of the molecule is slightly positive, while the other end is slightly negative. A diatomic molecule that consists of a polar covalent bond, such as \(\ce{HF}\), is a polar molecule. As mentioned in section 4.7, because the electrons in the bond are nearer to the F atom, this side of the molecule takes on a partial negative charge, which is represented by δ− (δ is the lowercase Greek letter delta). The other side of the molecule, the H atom, adopts a partial positive charge, which is represented by δ+. 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. A molecule with two poles is called a dipole (see figure below). Hydrogen fluoride is a dipole. 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. 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 are oriented this way, they cancel out and the overall molecular polarity of \(\ce{CO_2}\) is zero. Water is a bent molecule because of the two lone pairs on the central oxygen atom. The individual dipoles point from the \(\ce{H}\) atoms toward the \(\ce{O}\) atom. Because of the shape, the dipoles do not cancel each other out and the water molecule is polar. In the figure below, the net dipole is shown in blue and points upward. Three other polar molecules are shown below with the arrows pointing to the more electron dense atoms. Just like the water molecule, none of the bond moments cancel out. 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 does 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. Example \(\PageIndex{1}\): Label each of the following as polar or nonpolar. Water, H 2 O: Methanol, CH 3 OH: Hydrogen Cyanide, HCN: Oxygen, O 2 : Propane, C 3 H 8 : Solution Water is polar. Any molecule with lone pairs of electrons around the central atom is polar. Methanol is polar. This is not a symmetric molecule. The \(\ce{-OH}\) side is different from the other 3 \(\ce{-H}\) sides. Hydrogen cyanide is polar. The molecule is not symmetric. The nitrogen and hydrogen have different electronegativities, creating an uneven pull on the electrons. Oxygen is nonpolar. The molecule is symmetric. The two oxygen atoms pull on the electrons by exactly the same amount. Propane is nonpolar, because it is symmetric, with \(\ce{H}\) atoms bonded to every side around the central atoms and no unshared pairs of electrons. Exercise \(\PageIndex{1}\) Label each of the following as polar or nonpolar. a. SO 3 b. NH 3 Answer a non polar Answer b polar Summary Non polar molecules are symmetric with no unshared electrons. Polar molecules are asymmetric, either containing lone pairs of electrons on a central atom or having atoms with different electronegativities bonded. |
Courses/University_of_Kansas/KU%3A_CHEM_110_GOB_Chemistry_(Sharpe_Elles)/03%3A_Nuclear_Chemistry/3.3%3A_Units_of_Radioactivity | Skills to Develop To express amounts of radioactivity in a variety of units. Previously, we used mass to indicate the amount of radioactive substance present. 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. Perhaps the direct way of reporting radioactivity is the number of radioactive decays per second. One decay per second is called one becquerel (Bq) . Even in a small mass of radioactive material, however, there are many thousands of decays or disintegrations per second. The unit curie (Ci) , now defined as 3.7 × 10 10 decays per second, was originally defined as the number of decays per second in 1 g of radium. Many radioactive samples have activities that are on the order of microcuries (µCi) or more. Both the becquerel and curie can be used in place of grams to describe quantities of radioactive material. As an example, the amount of americium in an average smoke detector has an activity of 0.9 µCi. The unit becquerel is named after Henri Becquerel, who discovered radioactivity in 1896. The unit curie is named after Polish scientist Marie Curie, who performed some of the initial investigations into radioactive phenomena and discovered the elements, polonium (Po) and radium (Ra) in the early 1900s. Example \(\PageIndex{1}\) A sample of radium has an activity of 16.0 mCi (millicuries). If the half-life of radium is 1,600 y, how long before the sample’s activity is 1.0 mCi? SOLUTION The following table shows the activity of the radium sample over multiple half-lives: Time in Years Activity 0 16.0 mCi 1600 8.0 mCi 3200 4.0 mCi 4800 2.0 mCi 6400 1.0 mCi Over a period of 4 half-lives, the activity of the radium will be halved four times, at which point its activity will be 1.0 mCi. Thus, it takes 4 half-lives, or 4 × 1,600 y = 6,400 y, for the activity to decrease to 1.0 mCi. Exercise \(\PageIndex{1}\) A sample of radon has an activity of 60,000 Bq. If the half-life of radon is 15 h, how long before the sample’s activity is 3,750 Bq? Other measures of radioactivity are based on the effects it has on living tissue. Radioactivity can transfer energy to tissues in two ways: through the kinetic energy of the particles hitting the tissue and through the electromagnetic energy of the gamma rays being absorbed by the tissue. Either way, the transferred energy—like thermal energy from boiling water—can damage the tissue. The rad (an acronym for radiation absorbed dose) is a unit equivalent to a gram of tissue absorbing 0.01 J: 1 rad = 0.01 J/g Another unit of radiation absorption is the gray (Gy): 1 Gy = 100 rad The rad is more common. To get an idea of the amount of energy this represents, consider that the absorption of 1 rad by 70,000 g of H 2 O (approximately the same mass as a 150 lb person) would increase its temperature by only 0.002°C. This may not seem like a lot, but it is enough energy to break about 1 × 10 21 molecular C–C bonds in a person’s body. That amount of damage would not be desirable. Predicting the effects of radiation is complicated by the fact that various tissues are affected differently by different types of emissions. To quantify these effects, the unit rem (an acronym for roentgen equivalent, man) is defined as rem = rad × factor where factor is a number greater than or equal to 1 that takes into account the type of radioactive emission and sometimes the type of tissue being exposed. For beta particles, the factor equals 1. For alpha particles striking most tissues, the factor is 10, but for eye tissue, the factor is 30. Most radioactive emissions that people are exposed to are on the order of a few dozen millirems (mrem) or less; a medical X ray is about 20 mrem. A sievert (Sv) is a related unit and is defined as 100 rem. What is a person’s annual exposure to radioactivity and radiation? Table \(\PageIndex{1}\) lists the sources and annual amounts of radiation exposure. It may surprise you to learn that fully 82% of the radioactivity and radiation exposure we receive is from natural sources—sources we cannot avoid. Fully 10% of the exposure comes from our own bodies—largely from 14 C and 40 K. Source Amount (mrem) radon gas 200.00 medical sources 53.00 radioactive atoms in the body naturally 39.00 terrestrial sources 28.00 cosmic sources 28.00 consumer products 10.00 nuclear energy 0.05 Total 358.00 Flying from New York City to San Francisco adds 5 mrem to your overall radiation exposure because the plane flies above much of the atmosphere, which protects us from most cosmic radiation. The actual effects of radioactivity and radiation exposure on a person’s health depend on the type of radioactivity, the length of exposure, and the tissues exposed. Table \(\PageIndex{2}\) lists the potential threats to health at various amounts of exposure over short periods of time (hours or days). Exposure (rem) Effect 1 (over a full year) no detectable effect ∼20 increased risk of some cancers ∼100 damage to bone marrow and other tissues; possible internal bleeding; decrease in white blood cell count 200–300 visible “burns” on skin, nausea, vomiting, and fatigue >300 loss of white blood cells; hair loss ∼600 death One of the simplest ways of detecting radioactivity is by using a piece of photographic film embedded in a badge or a pen. On a regular basis, the film is developed and checked for exposure. A comparison of the exposure level of the film with a set of standard exposures indicates the amount of radiation a person was exposed to. Figure \(\PageIndex{1}\) : Detecting Radioactivity. A Geiger counter is a common instrument used to detect radioactivity. Another means of detecting radioactivity is an electrical device called a Geiger counter (Figure \(\PageIndex{1}\)). It contains a gas-filled chamber with a thin membrane on one end that allows radiation emitted from radioactive nuclei to enter the chamber and knock electrons off atoms of gas (usually argon). The presence of electrons and positively charged ions causes a small current, which is detected by the Geiger counter and converted to a signal on a meter or, commonly, an audio circuit to produce an audible “click.” Key Takeaway Radioactivity can be expressed in a variety of units, including rems, rads, and curies. |
Bookshelves/Inorganic_Chemistry/Inorganic_Chemistry_(Saito) | Inorganic chemistry is of fundamental importance not only as a basic science but also as one of the most useful sources for modern technologies. Elementary substances and solid-state inorganic compounds are widely used in the core of information, communication, automotive, aviation and space industries as well as in traditional ones. Inorganic compounds are also indispensable in the frontier chemistry of organic synthesis using metal complexes, homogeneous catalysis, bioinorganic functions, etc. One of the reasons for the rapid progress of inorganic chemistry is the development of the structural determination of compounds by X-ray and other analytical instruments. It has now become possible to account for the structure-function relationships to a considerable extent by the accumulation of structural data on inorganic compounds. It is no exaggeration to say that a revolution of inorganic chemistry is occurring. We look forward to the further development of inorganic chemistry in near future. This text book describes important compounds systematically along the periodic table, and readers are expected to learn typical ones both in the molecular and solid states. The necessary theories to explain these properties of compounds come from physical chemistry and basic concepts for learning inorganic chemistry are presented in the first three chapters. Front Matter 1: Elements and Periodicity 2: Bonding and Structure 3: Reactions 4: Chemistry of Nonmetallic Elements 5: Chemistry of Main-Group Metals 6: Chemistry of Transition Metals 7: Lanthanoids and Actinoids 8: Reaction and Physical Properties 9: Solution of problems Back Matter Thumbnail: The ball-and-stick model of diisobutylaluminium hydride, showing aluminium as pink, carbon as black, and hydrogen as white. (Public Domain; Benjah-bmm27 via Wikipedia ). |
Courses/Barry_University/CHE360%3A_Inorganic_Chemistry/02%3A_Atomic_Structure/2.04%3A_The_Schrodinger_equation_particle_in_a_box_and_atomic_wavefunctions/2.4.01%3A_Particle_in_a_Box | The particle in the box is a model that can illustrate how a wave equation works. Although it does not represent a real situation, we can limit our model to just one dimension (the x-dimension, for instance) such that the Schrödinger equation becomes significantly simplified. Despite being unrealistic, this simplification is quite useful for gaining an understanding of the Schrödinger equation. The model of a particle in a box The particle in the box is a hypothetical situation with a particle trapped in a one-dimensional “box”. Let’s not get hung-up on the fact that the common object called a “box” is typically an object with three dimensions instead of just one dimension. And let’s also not get hung up on the word “particle”. This is a particle that has properties of a wave…so it is unlike the macroscopic particle that you’re probably imagining. This “box” is more like a line, or an x-axis; it is just a one-dimensional space in which a particle-wave is trapped. The particle-wave can only exist inside the walls (where \(0<x<a\)), along the x axis in the figure shown above. In terms of energy, the potential energy is zero (\(V=0\)) because the particle is in an energetically-favorable position here. On the other hand, outside the box, the particle cannot exist and the potential energy is infinitely large (\(V=\infty\)) outside the walls (where \(x<0\) or \(x>a\)). This means that it is infinitely unfavorable for the particle-wave to exist outside the box, and so it never does. The particle-wave is trapped between the walls, along the 1-dimensional \(x\) axis, and there are no forces acting on the particle-wave inside this “box”. The Schrödinger wave equation for a particle in a box The particle in a box model lets us consider a simple version of the Schr ö dinger equation. Before we simplify, let's take another look at the full Hamiltonian for a particle-wave in three dimensions (see equation 2.2.2 ) and the simplest form of the Schrödinger equation ( see equation 2.2.1 ). Both of these equations are described in the previous section and are written below for convenience. The Schrödinger Equation (from equation 2.2.1): \(\hat{H}\psi = E\psi\) The Hamiltonian in three dimensions (from equation 2.2.2): \[\hat{H}=\dfrac{-h{^2}}{8\pi{^2}m_e}\left(\dfrac{\partial{^2}}{\partial{x^2}}+\dfrac{\partial{^2}}{\partial{y^2}}+\dfrac{\partial{^2}}{\partial{z^2}}\right)-\dfrac{Ze^2}{4\pi{}\epsilon_0{r}}\nonumber \] In the particle in a box model, there is only one dimension, \(x\). Because the \(y\) and \(z\) values are zero, we can drop \(y\) and \(z\) out of our Hamiltonian equation. And, since \(V=0\) inside the box, we can drop the whole part of the Hamiltonian equation that describes the potential energy (\(\frac{-Ze^2}{4\pi{}\epsilon_0{r}}\)). This leaves us with a much simplified Hamiltonian operator, which we can then use to write the Schrödinger wave equation for a particle moving in one dimension. One last thing we'll do is use a more general value \(m\) for the mass of the particle, rather than \(m_e\) that specifically represents the mass of an electron. The Schrödinger equation for a particle-wave in a one-dimensional box is: \[\dfrac{-h{^2}}{8\pi{^2}m}\left(\dfrac{\partial{^2}\psi(x)}{\partial{x^2}}\right) = E\psi \label{1DSchr} \] Exercise \(\PageIndex{1}\) Follow the steps that were described in the last paragraph to find Equation \ref{1DSchr} from equation 2.2.1 and 2.2.2. In other words, derive the Schr ö dinger Equation for the particle in a box given the three-dimensional Hamiltonian and Schr ö dinger Equations. Answer Here is the Hamitlonian Equation for in three dimensions. \(\hat{H}=\textcolor{red}{\dfrac{-h{^2}}{8\pi{^2}m}\left(\dfrac{\partial{^2}}{\partial{x^2}}+\dfrac{\partial{^2}}{\partial{y^2}}+\dfrac{\partial{^2}}{\partial{z^2}}\right)}-\textcolor{blue}{\dfrac{Ze^2}{4\pi{}\epsilon_0{r}}}\) There are two parts to this equation: the kinetic energy contribution and the potential energy contribution. While the first part of this expression, \(\textcolor{red}{\dfrac{-h{^2}}{8\pi{^2}m}\left(\dfrac{\partial{^2}}{\partial{x^2}}+\dfrac{\partial{^2}}{\partial{y^2}}+\dfrac{\partial{^2}}{\partial{z^2}}\right)}\), is the kinetic energy contribution, the second part of the expression, \(\textcolor{blue}{-\dfrac{Ze^2}{4\pi{}\epsilon_0{r}}}\), is the potential energy (\(V\)) contribution. (1) Simplify the expression for kinetic energy Your particle-wave is moving in only one dimension. We can arbitrarily choose the dimension \(x\). This means that \(x\) is a variable and will have some value from \(0\) to \(a\). On the other hand, the particle does not move in the other two dimensions, \(y\) and \(z\). You can assign the position of the particle along \(x\) and \(y\) axes as zero or you can just understand that they do not exist in any significant capacity in our 1-D space. Either way, you can just drop them out of the kinetic energy part of the Hamiltonian, as so: When \(y=z=0\) and the particle-wave only exists/moves in the \(x\) dimension, \(\textcolor{red}{\dfrac{-h{^2}}{8\pi{^2}m}\left(\dfrac{\partial{^2}}{\partial{x^2}}+\dfrac{\partial{^2}}{\partial{y^2}}+\dfrac{\partial{^2}}{\partial{z^2}}\right)}\) = \(\textcolor{red}{\dfrac{-h{^2}}{8\pi{^2}m}\left(\dfrac{\partial{^2}}{\partial{x^2}}\right)}\) And the overall Hamiltonian is simplified: \(\hat{H}=\textcolor{red}{\dfrac{-h{^2}}{8\pi{^2}m}\left(\dfrac{\partial{^2}}{\partial{x^2}}\right)}-\textcolor{blue}{\dfrac{Ze^2}{4\pi{}\epsilon_0{r}}}\) (2) Simplify the expression for potential energy (\(V\)) . The particle's potential energy inside the box is zero (\(V=0\)). Therefore, \(\textcolor{blue}{V=-\dfrac{Ze^2}{4\pi{}\epsilon_0{r}}}=0\). We can simply drop the potential energy portion from the Hamiltonian because its value is zero. The simplified Hamiltonian is now: \(\hat{H}=\textcolor{red}{\dfrac{-h^2}{8\pi^2m}\left(\dfrac{\partial{^2}}{\partial{x^2}}\right)}\) (3) Substitute the simplified Hamiltonian into the Schr ö dinger Equation. The Schr ö dinger is: \(\textcolor{red}{\hat{H}}\psi = E\psi\) When the simplified Hamiltonian is substituted into this equation, the result is the 1-D Schr ö dinger Equation \ref{1DSchr}. \(\textcolor{red}{\left(\dfrac{-h{^2}}{8\pi{^2}m}\left(\dfrac{\partial{^2}\psi(x)}{\partial{x^2}}\right)\right)} = E\psi\) The wavefunction of a one-dimensional wave Although we've simplified the Schrödinger Equation by considering the particle in the box, Equation \ref{1DSchr} may still look mysterious to you. But it's simpler than you might realize. Let's unpack it! Here we'll go through the steps of deriving the 1-dimensional wavefunction for the particle in a box. We won't try to derive a three-dimensional wavefunction for a "real" electron, but when you understand how to find the 1-dimensional wavefunction, and what it tells us, then you can conceptually extend this to understand what the wavefunction tells us about electrons in three dimensions. Let's dissect Equation \ref{1DSchr}. Here it is again for reference: \[\textcolor{green}{\dfrac{-h{^2}}{8\pi{^2}m_e}}\left(\dfrac{\partial{^2}\psi(x)}{\partial{x^2}}\right) = \textcolor{green}{E}\psi\nonumber \] The expression \(\textcolor{green}{\frac{-h^2}{8\pi^2m_e}}\) is just a negative constant. This negative constant is multiplied by the second derivative of the wavefunction in the expression above. And \(\textcolor{green}{E}\) is a constant multiplied by that same function. We can rearrange this equation as follows: \[\dfrac{\partial{^2}\psi(x)}{\partial{x^2}} = \textcolor{green}{\dfrac{-8\pi{^2}m_eE}{h{^2}}}\psi = -\textcolor{green}{C}\psi \label{Schrx} \] where \(C\) is just the constant; \(\textcolor{green}{C}=\textcolor{green}{\dfrac{8\pi{^2}m_eE}{h{^2}}}\) To an expert in the mathematical field of Differential Equations, this expression \ref{1DSchr} follows a familiar pattern that makes it easy to translate. If you're not an expert in differential equations, that is OK; you'll just have to bear with this while things get a little "hand-wavy" and try Exercise \(\PageIndex{2}\) to prove it to yourself that the following "hand waviness" is true. An expert in differential equations could tell you that when the second derivative of a function is equal to a negative constant times that same function, the function can be written in terms of \(sin\) and \(cos\), as shown in Equation \ref{wf1} (that was the hand-wavy part...see, not so bad): \[\psi=A\sin(rx) + B\cos(sx) \label{wf1} \] \(A, B, r,\) and \(s\) are constants in Equation \ref{wf1}. This expression is useful because it is in a form that we could plot with a graphing program (or calculator). However, to do so we must know the constants \(A, B, r,\) and \(s\). What are \(r\) and \(s\)? If we substitute expression \ref{wf1} into Equation \ref{1DSchr} and solve for r and s, we find the expressions below. \[r=s=\sqrt{2m_eE}\left(\frac{2\pi}{h}\right) \label{rs} \] This shows us that r and s are equal and constant; we know the values for \(\pi, h,\) and \(m_e\), and we can solve for the constant \(E\). Exercise \(\PageIndex{2}\): Prove that Equation \ref{wf1} is true Show that Equation \ref{wf1} is true in the case that Equation \ref{Schrx} ( \(\frac{\partial{^2}\psi(x)}{\partial{x^2}}=-C\psi \)) is true. Also consider Equation \ref{rs} (\(r,s\) are constants and \(r=s\)) and explain why the constant \(r\) must equal \(s\). To do this problem, you'll also need to know the basic differentiation rules of \(\sin\) and \(\cos\) functions, given below for convenience. Differentiation Rules for \(sin\) and \(cos\) : \(\frac{d}{dx}\big(\sin(rx)\big)=r\cos(rx)\), where r is a constant. \(\frac{d}{dx}\big(\cos(rx)\big)=-r\sin(rx)\), where r is a constant. Answer You can show that \ref{wf1} is true by substituting \ref{wf1} back into Equation \ref{Schrx}; in other words take the second derivative of the function, \(\psi=A\sin(rx) + B\cos(sx)\) to show that \(\frac{\partial{^2}\psi(x)}{\partial{x^2}}=-C\psi \). \(C, r,\) and \(s\) are arbitrary constants, and \(r=s\). The actual value of these constants is irrelevant for this problem. Here is one way to approach the problem: (1) Substitute \(\textcolor{red}{\psi=A\sin(rx) + B\cos(sx)}\) into \(\textcolor{blue}{\frac{\partial{^2}\psi(x)}{\partial{x^2}}=-C\psi} \). \(\textcolor{blue}{\frac{\partial{^2}}{\partial{x^2}}}\left(\textcolor{red}{A\sin(rx) + B\cos(sx)}\right)= \textcolor{blue}{-C}\textcolor{red}{\psi} \) (2) Take the first and second derivative of \(\textcolor{red}{A\sin(rx) + B\cos(sx)}\). The result of taking the first derivative... \(\textcolor{blue}{\frac{\partial{}}{\partial{x}}}\left(\textcolor{green}{r}\textcolor{red}{A\cos(rx) - \textcolor{green}{s}B\sin(sx)}\right)= \textcolor{blue}{-C}\textcolor{red}{\psi} \) ...and the second derivative... \(\textcolor{green}{-r^2}\textcolor{red}{A\sin(rx)} - \textcolor{green}{s^2}\textcolor{red}{B\cos(sx)} = \textcolor{blue}{-C}\textcolor{red}{\psi}\) This is where we see that the only way \(\psi=A\sin(rx) + B\cos(sx)\) is if we can factor the values \(r^2\) and \(s^2\) out of the left side. We have a hint that \(r=s\) from the discussion above and Equation \ref{rs}. Now we see that this is a necessary condition for the expression \(\textcolor{green}{-r^2}\textcolor{red}{A\sin(rx)} - \textcolor{green}{s^2}\textcolor{red}{B\cos(sx)} = \textcolor{blue}{-C}\textcolor{red}{\psi}\) to be true. (3) Simplify If \(r=s\), then \(r^2=s^2\) and we can replace s with r in the equation above, as so... \(\textcolor{green}{-r^2}\textcolor{red}{A\sin(rx)} - \textcolor{green}{r^2}\textcolor{red}{B\cos(sx)} = \textcolor{blue}{-C}\textcolor{red}{\psi}\) Now, we can factor the constant, \(-r^2\) out of the left side of this expression, \(\textcolor{green}{-r^2}\left(\textcolor{red}{A\sin(rx) + B\cos(sx)}\right) = \textcolor{blue}{-C}\textcolor{red}{\psi}\) Now we see that \(\psi=A\sin(rx) + B\cos(sx)\) when \(-C=-r^2\). In other words, we have shown that \(\psi=A\sin(rx) + B\cos(sx)\) is true. Why is this a useful exercise? Although it is difficult to explain the derivation of the expression \(\psi=A\sin(rx) + B\cos(sx)\) without differential equations and only knowing that \(\frac{\partial{^2}\psi(x)}{\partial{x^2}}=-C\psi \), to show this is true is as simple as solving this exercise. This exercise is here to help you review your calculus and to show you that the "hand waving" in the text above is not magic, but it is coming from math that you have seen in your calculus courses. Math is crucial for explaining the nature of the universe! Eat your vegetables and practice your math. Exercise \ref{rs} *Complete Exercise \(\PageIndex{2}\) before attempting this one. Derive the expression for r below (from Equation \ref{rs}) using the 1-D wavefunction (\ref{wf1}: \(\psi=A\sin(rx) + B\cos(sx) \)) and the 1-D Schrödinger equation (\ref{1DSchr}: \(\left(\frac{-h{^2}}{8\pi{^2}m_e}\left(\frac{\partial{^2}\psi(x)}{\partial{x^2}}\right) = E\psi\right)\)). \(r=\dfrac{2\pi}{h}\sqrt{2m_eE}\) Answer The steps below could be carried out in a different sequence: (1) Rearrange Let's move all constants to the right side of \ref{1DSchr} so that we arrive at the expression that is shown in part of Equation \ref{Schrx} . \(\dfrac{\partial{^2}\psi(x)}{\partial{x^2}} = \textcolor{green}{\dfrac{-8\pi{^2}m_eE}{h{^2}}}\psi\) (2) Take the second derivative of \(\psi\) , which we found in Exercise \(\PageIndex{2}\). \(\dfrac{\partial{^2}\psi(x)}{\partial{x^2}} =\textcolor{green}{-r^2}\left(\textcolor{red}{A\sin(rx) + B\cos(sx)}\right) = \textcolor{green}{\dfrac{-8\pi{^2}m_eE}{h{^2}}}\textcolor{red}{\psi}\) And since \(\textcolor{red}{\psi=A\sin(rx) + B\cos(sx)}\), we can simplify this by dividing both sides by \(\psi\) (or by \(A\sin(rx) + B\cos(sx) \)). **recall that \(s=r\). So, you could have substituted \(r\) for \(s\) in the expression above, and you'd still be on the right track. \(\textcolor{green}{-r^2}= \textcolor{green}{\dfrac{-8\pi{^2}m_eE}{h{^2}}}\) (3) Solve for r (*we already showed \(r=s\) in the Exercise \(\PageIndex{2}\)). \(\textcolor{green}{r} = \sqrt{\dfrac{8\pi{^2}m_eE}{h{^2}}} = \dfrac{2\pi}{h}\sqrt{2m_eE}\) What are \(A\) and \(B\)? We can find the possible values of \(A\) and \(B\) by looking at the possible extremes for the value of \(x\). In our model in Figure \(\PageIndex{1}\), we can see that x can have values from \(0\rightarrow a\). The two extremes, where \(x=0\) and \(x=a\), lie at the walls of the box. The electron cannot exist beyond these values because it is trapped inside the walls, so its wavefunction must be zero at \(x=0\) and \(x=a\). The case where \(x=0\) can help us find the value of \(B\). We just said that the wavefunction must be zero where \(x=0\). This means: \[\psi_{x=0} = A\sin(rx) + B\cos(sx) = A\sin(0) + B\cos(0) = 0 \nonumber \] Because \(A\sin(0)=0\), the expression can be simplified to: \[\psi_{x=0} = B\cos(0) = B(1) = 0 \nonumber \] Now we can see that there is only one possible value for the constant, \(B\), that would allow this expression to be true, \(B=0\). And since \(B\) is constant then it will always be zero despite the value of \(x\). This simplifies our 1-D wavefunction: \[\psi=A\sin(rx) \label{wf2} \] The case where \(x=a\) can help us find the value for \(A\). Since the wavefunction must be zero where \(x=a\): \[\psi_{x=a} = A\sin(rx) = A\sin(ra) = 0 \nonumber \] In this case, \(A\) cannot be zero because if both A and B are zero, the wavefunction is zero at all values of \(x\), and so our wavefunction would not exist inside the box. So, let's assume A is not zero. Then, it must be the case that \[\sin(ra)=0 \nonumber \] and then because of the way that \(\sin\) functions work, the quantity \(ra\) must be an integer value of \(\pi\) if \(\sin(ra)=0\). \[ra = \pm n\pi \nonumber \] where \(n\ = 1, 2, 3...\) any non-zero integer. We can ignore negative values here since both positive and negative values of \(r\) will give the same value for \(\sin(ra)\). We can then solve for r: \[r=\frac{n\pi}{a} \label{r} \] and substitute the value \(r=\dfrac{n\pi}{a}\) into Equation \ref{wf2} to get: \[\psi = A\sin\left(\frac{n\pi{x}}{a}\right) \label{wf3} \] Recall that the square of the wavefunction (\(\psi^2\)) gives the probability of finding the electron anywhere in space. In our model, the probability of finding the particle-wave inside the box is unity (it is 1). Mathematically, this is the normalizing requirement (expressed as \(\int \psi_A \psi_A^* d\tau=1\)), which can be solved to find the value of A. \[A = \sqrt{\frac{2}{a}} \nonumber \] Now, we can substitute the value of \(A\) into expression \ref{wf3}, and we have a wavefunction that can be visualized using any graphing program/calculator! \[\psi=\sqrt{\frac{2}{a}}\left(\sin\left(\frac{n\pi{x}}{a}\right)\right) \label{wf4} \] What is the energy (E) of a particle in the box? So, now you can see how a 1-D wavefunction can be solved. But how do we find the E of the electron from this? Well, in order to not distract you earlier, we skipped over a very simple way to do this. The E falls right out of the equations you just saw. We can set the two expressions we found above for the constant \(r\), (equations \ref{rs} and \ref{r}) equal to one another and then solve for E: \[r=\dfrac{n\pi}{a}=\sqrt{2mE}\left(\dfrac{2\pi}{h}\right) \nonumber \] \[E=\frac{n^2h^2}{8ma^2} \label{E} \] Expression \ref{E} can be used to calculate the energy of a particle in a one-dimensional box of length \(a\), given its integer energy level, \(n\). Here we can see that the energy is quantized because \(n\) is an integer (\(n=1,2,3...\)). In other words, \(n\) is a quantum number . Making sense of the particle in a box: plotting the 1-D \(\psi\) and \(\psi^2\). No matter whether you're dealing with a 1-dimensional or 3-dimensional wave equation, the wavefunction itself (\(\psi\)) describes the particle's wave properties. This \(\psi\) doesn't have actual physical meaning, so it's hard to imagine what it "looks like" other than just plotting its function. However, the probability of finding the particle-wave at any specific position along the x-axis between \(x=0\) and \(x=a\) is more physically meaningful. The probability of finding the particle is proportional to the square of the wave function, which is represented by either \(\psi^2\) or, sometimes, \(\psi\psi^*\). The plots of the functions for \(\psi\) and \(\psi^2\) for the first three possible values of \(n\) are shown below in Figure \(\PageIndex{2}\). You could create plots similar to these simply by plotting the function shown in Equation \ref{wf4}. To plot, you just need to assign a value of \(n\), and it is convenient to assign the length of the box as \(a=1\). The plot generated would have the general ratio of \(\frac{x}{a}\) on the x axis, and thus would be relevant for any length box. The graphs above represent "solutions" to the wave function. For example, the solution \(n=1\) gives the plot shown in Figure \(\PageIndex{2}\) B. The solution of \(n=2\) is plotted in Figure \(\PageIndex{2}\) C, and so on. These values for \(n\) are some of the possible solutions to the 1-dimensional \(\psi\), and they yield descriptions of the wave behavior and probability of finding a particle in 1-dimensional space. How does this apply to atoms? The 1-dimensional particle in a box does not represent a real situation; but rather, it is a simple model that we can use to understand a more complex system, like a electron orbiting the nucleus in three dimensions. It's useful to recognize the analogies here that might represent something familiar in a real situation. \(x=0\) is analogous to the nucleus in an atom. Here in the particle in a box model, \(\psi=0\) and there is zero probability of finding an electron. When this is extended to an atom, this position is analogous to the nucleus (at the origin of a coordinate system where \(x=y=z=0\). A 3-dimensional \(\psi\) is also zero at the nucleus and the electron cannot exist there; also, \(\psi\) approaches zero as it approaches this position in both the particle in a box model and in a more realistic case of an electron in an atom. \(x=a\) is analogous to a boundary surface far from the nucleus. In the particle in a box model, the 1-dimensional \(\psi\) is zero at \(x=a\). This is analogous to an electron's \(\psi\) approaching zero as it gets further from the nucleus. There will be a more in depth description of boundary surface in the next section (2.2.2) . You already know this as the outermost surface of an electron orbital. One minor but notable difference between the 1-dimensional particle in a box and the case of a real atom is that the \(\psi\) in three dimensions approaches but never quite reaches zero as distance from the nucleus increases, while \(\psi=0\) for \(x=a\) in the more simple case of a 1-dimensional particle in a box. A change in sign of the \(\psi\) (and where both \(psi=0\) and \(\psi^2=0\)) is a node. In both the 1-dimensional particle in a box model and in the more realistic 3-dimensional case, a node is found where the \(\psi\) changes sign. At this point, \(\psi=\psi^2=0\); in other words there is zero probability of finding the particle or electron at these points. It is easy to spot these points in the plots above because the wave function crosses the x-axis and the \(\psi^2\) meets zero. Exercise \(\PageIndex{4}\) Identify the points on plots B, C, and D in Figure \(\PageIndex{2}\) that are nodes. How many nodes are there for \(n=1, n=2,\) and \(n=3\)? Answer The nodes are annotated with red circles on the figure below. Panel B (\(n=1\)) has zero nodes. There are two points where \(\psi=0\) in the case of \(n=1\), but these points are at the walls of the box and they are not nodes. The walls of the box are analogous to the nucleus and the boundary surface of an electron orbital. Panel C (\(n=2\)) has one node where the \(\psi\) crosses zero. Panel D (\(n=3\)) shows two nodes. Exercise \(\PageIndex{5}\) Use a graphing program to plot the 1-dimensional \(\psi\) and \(\psi^2\) for \(n=1,2,3,4\). How many nodes are there for \(n=4\)? Is this expected? Predict how many nodes we should expect for \(n=5\). You can use any plotting program to do this, and if you aren't familiar with any, try this one: Desmos Answer You can use any plotting program to do this. This is the function you should plot (it is Equation \ref{wf4}): \(\psi=\sqrt{\frac{2}{a}}\left(\sin\left(\frac{n\pi{x}}{a}\right)\right)\) For n=1: Assign a value of \(n=1\), as stated in the problem. In the text above, it also states that it is convenient to assign a value of \(a=1\). Assigning these values results in the following function: \(\psi=\sqrt{\frac{2}{1}}\left(\sin\left(\frac{1\pi{x}}{1}\right)\right)\) and simplifying leads to \(\psi=\sqrt{2}\left(\sin\left(\pi{x}\right)\right)\) For n=2,3,4 you would repeat the process above. You will get the following functions: For n=2: \(\psi=\sqrt{2}\left(\sin\left(2\pi{x}\right)\right)\) For n=3: \(\psi=\sqrt{2}\left(\sin\left(3\pi{x}\right)\right)\) For n=4: \(\psi=\sqrt{2}\left(\sin\left(4\pi{x}\right)\right)\) In a program like Desmos, you need to input the function correctly. It's useful to know the code for at least one graphing program. For Desmos, and most others, you can find help online. For example, the correct input for Desmos for the n=4 case is sqrt(2)*(sin(4*pi*x)). But, you have to type it in because copy/paste doesn't work. You'll also want to display your graph only from \(x=0\) to \(x=1\) by hitting the settings button (a little wrench in the upper right corner) and changing the x scale. Go ahead and change the y scale too, to \(y=-2\) to \(y=2\). If you do this, you should get something that looks like this: To graph the square of each wavefunction, you'd just square the functions that we just plotted: In both the case of \(\psi\) and \(\psi^2\), the \(n=4\) function shows three nodes (see the bolded purple line, where nodes are tiny grey spots on the axis). This follows a pattern that you might have noticed with the cases of \(n=1,2,\) and \(3\) where the number of nodes is \(n-1\) (nodes = \(n-1\)). We should expect three nodes for \(n=4\) based on this pattern, and four nodes for \(n=5\). In the next section, we will extend these ideas qualitatively to three dimensions. While there is only one quantum number in one dimension, there are three quantum numbers in three dimensions that (when combined) give discrete descriptions of an electron in three-dimensional space (plus a fourth quantum number that explains other properties of the electron). |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/18%3A_Quantum_Mechanics_and_Molecular_Energy_Levels/18.07%3A_Particle_Spins_and_Statistics-_Bose-Einstein_and_Fermi-Dirac_Statistics | Our goal is to develop the theory of statistical thermodynamics from Boltzmann statistics. In this chapter, we explore the rudiments of quantum mechanics in order to become familiar with the idea that we can describe a series of discrete energy levels for any given molecule. For our purposes, that is all we need. We should note, however, that we are not developing the full story about the relationship between quantum mechanics and statistical thermodynamics. The spin of a particle is an important quantum mechanical property. It turns out that quantum mechanical solutions depend on the spin of the particle being described. Particles with integral spins behave differently from particles with half-integral spins. When we treat the statistical distribution of these particles, we need to treat particles with integral spins differently from particles with half-integral spins. Particles with integral spins are said to obey Bose-Einstein statistics ; particles with half-integral spins obey Fermi-Dirac statistics . Fortunately, both of these treatments converge to the Boltzmann distribution if the number of quantum states available to the particles is much larger than the number of particles. For macroscopic systems at ordinary temperatures, this is the case. In Chapters 19 and 20 , we introduce the ideas underlying the theory of statistical mechanics. In Chapter 21 , we derive the Boltzmann distribution from a set of assumptions that does not correspond to either the Bose-Einstein or the Fermi-Dirac requirement. In Chapter 25 , we derive the Bose-Einstein and Fermi-Dirac distributions and show how they become equivalent to the Boltzmann distribution for most systems of interest in chemistry. |
Courses/Saint_Marys_College_Notre_Dame_IN/CHEM_122%3A_Principles_of_Chemistry_II_(Under_construction)/1%3A_Review_from_CHEM_121/1.14%3A_Liquids_and_Intermolecular_Forces_(Exercises) | 11.1: A Molecular Comparison of Gases, Liquids, and Solids Q11.1.1 A liquid, unlike a gas, is virtually incompressible . Explain what this means using macroscopic and microscopic descriptions. What general physical properties do liquids share with solids? What properties do liquids share with gases? Q11.1.2 Using a kinetic molecular approach, discuss the differences and similarities between liquids and gases with regard to thermal expansion. fluidity. diffusion. Q11.1.3 How must the ideal gas law be altered to apply the kinetic molecular theory of gases to liquids? Explain. Q11.1.4 Why are the root mean square speeds of molecules in liquids less than the root mean square speeds of molecules in gases? 11.2: Intermolecular Forces Conceptual Problems Q11.2.1 What is the main difference between intramolecular interactions and intermolecular interactions? Which is typically stronger? How are changes of state affected by these different kinds of interactions? Q11.2.2 Describe the three major kinds of intermolecular interactions discussed in this chapter and their major features. The hydrogen bond is actually an example of one of the other two types of interaction. Identify the kind of interaction that includes hydrogen bonds and explain why hydrogen bonds fall into this category. Q11.2.3 Which are stronger—dipole–dipole interactions or London dispersion forces? Which are likely to be more important in a molecule with heavy atoms? Explain your answers. Q11.2.4 Explain why hydrogen bonds are unusually strong compared to other dipole–dipole interactions. How does the strength of hydrogen bonds compare with the strength of covalent bonds? Q11.2.5 Liquid water is essential for life as we know it, but based on its molecular mass, water should be a gas under standard conditions. Why is water a liquid rather than a gas under standard conditions? Q11.2.6 Describe the effect of polarity, molecular mass, and hydrogen bonding on the melting point and boiling point of a substance. Q11.2.7 Why are intermolecular interactions more important for liquids and solids than for gases? Under what conditions must these interactions be considered for gases? Q11.2.8 Using acetic acid as an example, illustrate both attractive and repulsive intermolecular interactions. How does the boiling point of a substance depend on the magnitude of the repulsive intermolecular interactions? Q11.2.9 In group 17, elemental fluorine and chlorine are gases, whereas bromine is a liquid and iodine is a solid. Why? Q11.2.10 The boiling points of the anhydrous hydrogen halides are as follows: HF, 19°C; HCl, −85°C; HBr, −67°C; and HI, −34°C. Explain any trends in the data, as well as any deviations from that trend. Q11.2.11 Identify the most important intermolecular interaction in each of the following. SO 2 HF CO 2 CCl 4 CH 2 Cl 2 Q11.2.12 Identify the most important intermolecular interaction in each of the following. LiF I 2 ICl NH 3 NH 2 Cl Q11.2.13 Would you expect London dispersion forces to be more important for Xe or Ne? Why? (The atomic radius of Ne is 38 pm, whereas that of Xe is 108 pm.) Q11.2.14 Arrange Kr, Cl 2 , H 2 , N 2 , Ne, and O 2 in order of increasing polarizability. Explain your reasoning. Q11.2.15 Both water and methanol have anomalously high boiling points due to hydrogen bonding, but the boiling point of water is greater than that of methanol despite its lower molecular mass. Why? Draw the structures of these two compounds, including any lone pairs, and indicate potential hydrogen bonds. Q11.2.16 The structures of ethanol, ethylene glycol, and glycerin are as follows: Arrange these compounds in order of increasing boiling point. Explain your rationale. Q11.2.17 Do you expect the boiling point of H 2 S to be higher or lower than that of H 2 O? Justify your answer. Q11.2.18 Ammonia (NH 3 ), methylamine (CH 3 NH 2 ), and ethylamine (CH 3 CH 2 NH 2 ) are gases at room temperature, while propylamine (CH 3 CH 2 CH 2 NH 2 ) is a liquid at room temperature. Explain these observations. Q11.2.19 Why is it not advisable to freeze a sealed glass bottle that is completely filled with water? Use both macroscopic and microscopic models to explain your answer. Is a similar consideration required for a bottle containing pure ethanol? Why or why not? Q11.2.20 Which compound in the following pairs will have the higher boiling point? Explain your reasoning. NH 3 or PH 3 ethylene glycol (HOCH 2 CH 2 OH) or ethanol 2,2-dimethylpropanol [CH 3 C(CH 3 ) 2 CH 2 OH] or n -butanol (CH 3 CH 2 CH 2 CH 2 OH) Q11.2.21 Some recipes call for vigorous boiling, while others call for gentle simmering. What is the difference in the temperature of the cooking liquid between boiling and simmering? What is the difference in energy input? Q11.2.22 Use the melting of a metal such as lead to explain the process of melting in terms of what is happening at the molecular level. As a piece of lead melts, the temperature of the metal remains constant, even though energy is being added continuously. Why? Q11.2.23 How does the O–H distance in a hydrogen bond in liquid water compare with the O–H distance in the covalent O–H bond in the H 2 O molecule? What effect does this have on the structure and density of ice? Q11.2.24 Explain why the hydrogen bonds in liquid HF are stronger than the corresponding intermolecular H⋅⋅⋅I interactions in liquid HI. In which substance are the individual hydrogen bonds stronger: HF or H 2 O? Explain your reasoning. For which substance will hydrogen bonding have the greater effect on the boiling point: HF or H 2 O? Explain your reasoning. Answers Water is a liquid under standard conditions because of its unique ability to form four strong hydrogen bonds per molecule. As the atomic mass of the halogens increases, so does the number of electrons and the average distance of those electrons from the nucleus. Larger atoms with more electrons are more easily polarized than smaller atoms, and the increase in polarizability with atomic number increases the strength of London dispersion forces. These intermolecular interactions are strong enough to favor the condensed states for bromine and iodine under normal conditions of temperature and pressure. The V-shaped SO 2 molecule has a large dipole moment due to the polar S=O bonds, so dipole–dipole interactions will be most important. The H–F bond is highly polar, and the fluorine atom has three lone pairs of electrons to act as hydrogen bond acceptors; hydrogen bonding will be most important. Although the C=O bonds are polar, this linear molecule has no net dipole moment; hence, London dispersion forces are most important. This is a symmetrical molecule that has no net dipole moment, and the Cl atoms are relatively polarizable; thus, London dispersion forces will dominate. This molecule has a small dipole moment, as well as polarizable Cl atoms. In such a case, dipole–dipole interactions and London dispersion forces are often comparable in magnitude. Water has two polar O–H bonds with H atoms that can act as hydrogen bond donors, plus two lone pairs of electrons that can act as hydrogen bond acceptors, giving a net of four hydrogen bonds per H 2 O molecule. Although methanol also has two lone pairs of electrons on oxygen that can act as hydrogen bond acceptors, it only has one O–H bond with an H atom that can act as a hydrogen bond donor. Consequently, methanol can only form two hydrogen bonds per molecule on average, versus four for water. Hydrogen bonding therefore has a much greater effect on the boiling point of water. Vigorous boiling causes more water molecule to escape into the vapor phase, but does not affect the temperature of the liquid. Vigorous boiling requires a higher energy input than does gentle simmering. 11.3: Some Properties of Liquids Conceptual Problems Why is a water droplet round? How is the environment of molecules on the surface of a liquid droplet different from that of molecules in the interior of the droplet? How is this difference related to the concept of surface tension? Explain the role of intermolecular and intramolecular forces in surface tension. A mosquito is able to walk across water without sinking, but if a few drops of detergent are added to the water, the insect will sink. Why? Explain how soaps or surfactants decrease the surface tension of a liquid. How does the meniscus of an aqueous solution in a capillary change if a surfactant is added? Illustrate your answer with a diagram. Of CH 2 Cl 2 , hexane, and ethanol, which has the lowest viscosity? Which has the highest surface tension? Explain your reasoning in each case. At 25°C, cyclohexanol has a surface tension of 32.92 mN/m 2 , whereas the surface tension of cyclohexanone, which is very similar chemically, is only 25.45 mN/m 2 . Why is the surface tension of cyclohexanone so much less than that of cyclohexanol? What is the relationship between surface tension and temperature? viscosity and temperature? Explain your answers in terms of a microscopic picture. What two opposing forces are responsible for capillary action? How do these forces determine the shape of the meniscus? Which of the following liquids will have a concave meniscus in a glass capillary? Explain your reasoning. pentane diethylene glycol (HOCH 2 CH 2 OCH 2 CH 2 OH) carbon tetrachloride How does viscosity depend on molecular shape? What molecular features make liquids highly viscous? Conceptual Answers Adding a soap or a surfactant to water disrupts the attractive intermolecular interactions between water molecules, thereby decreasing the surface tension. Because water is a polar molecule, one would expect that a soap or a surfactant would also disrupt the attractive interactions responsible for adhesion of water to the surface of a glass capillary. As shown in the sketch, this would decrease the height of the water column inside the capillary, as well as making the meniscus less concave. As the structures indicate, cyclohexanol is a polar substance that can engage in hydrogen bonding, much like methanol or ethanol; consequently, it is expected to have a higher surface tension due to stronger intermolecular interactions. Cohesive forces are the intermolecular forces that hold the molecules of the liquid together, while adhesive forces are the attractive forces between the molecules of the liquid and the walls of the capillary. If the adhesive forces are stronger than the cohesive forces, the liquid is pulled up into the capillary and the meniscus is concave. Conversely, if the cohesive forces are stronger than the adhesive forces, the level of the liquid inside the capillary will be lower than the level outside the capillary, and the meniscus will be convex. Viscous substances often consist of molecules that are much longer than they are wide and whose structures are often rather flexible. As a result, the molecules tend to become tangled with one another (much like overcooked spaghetti), which decreases the rate at which they can move through the liquid. Numerical Problems The viscosities of five liquids at 25°C are given in the following table. Explain the observed trends in viscosity. Compound Molecular Formula Viscosity (mPa•s) benzene C6H6 0.604 aniline C6H5NH2 3.847 1,2-dichloroethane C2H4Cl2 0.779 heptane C7H16 0.357 1-heptanol C7H15OH 5.810 The following table gives values for the viscosity, boiling point, and surface tension of four substances. Examine these data carefully to see whether the data for each compound are internally consistent and point out any obvious errors or inconsistencies. Explain your reasoning. Compound Viscosity (mPa•s at 20°C) Boiling Point (°C) Surface Tension (dyn/cm at 25°C) A 0.41 61 27.16 B 0.55 65 22.55 C 0.92 105 36.76 D 0.59 110 28.53 Surface tension data (in dyn/cm) for propanoic acid (C 3 H 6 O 2 ), and 2-propanol (C 3 H 8 O), as a function of temperature, are given in the following table. Plot the data for each compound and explain the differences between the two graphs. Based on these data, which molecule is more polar? Compound 25°C 50°C 75°C propanoic acid 26.20 23.72 21.23 2-propanol 20.93 18.96 16.98 Numerical Answer 3. The plots of surface tension versus temperature for propionic acid and isopropanol have essentially the same slope, but at all temperatures the surface tension of propionic acid is about 30% greater than for isopropanol. Because surface tension is a measure of the cohesive forces in a liquid, these data suggest that the cohesive forces for propionic acid are significantly greater than for isopropanol. Both substances consist of polar molecules with similar molecular masses, and the most important intermolecular interactions are likely to be dipole–dipole interactions. Consequently, these data suggest that propionic acid is more polar than isopropanol. 11.4: Phase Changes Conceptual Problems In extremely cold climates, snow can disappear with no evidence of its melting. How can this happen? What change(s) in state are taking place? Would you expect this phenomenon to be more common at high or low altitudes? Explain your answer. Why do car manufacturers recommend that an automobile should not be left standing in subzero temperatures if its radiator contains only water? Car manufacturers also warn car owners that they should check the fluid level in a radiator only when the engine is cool. What is the basis for this warning? What is likely to happen if it is ignored? Use Hess’s law and a thermochemical cycle to show that, for any solid, the enthalpy of sublimation is equal to the sum of the enthalpy of fusion of the solid and the enthalpy of vaporization of the resulting liquid. Three distinct processes occur when an ice cube at −10°C is used to cool a glass of water at 20°C. What are they? Which causes the greatest temperature change in the water? When frost forms on a piece of glass, crystals of ice are deposited from water vapor in the air. How is this process related to sublimation? Describe the energy changes that take place as the water vapor is converted to frost. What phase changes are involved in each process? Which processes are exothermic, and which are endothermic? ice melting distillation condensation forming on a window the use of dry ice to create a cloud for a theatrical production What phase changes are involved in each process? Which processes are exothermic, and which are endothermic? evaporation of methanol crystallization liquefaction of natural gas the use of naphthalene crystals to repel moths Why do substances with high enthalpies of fusion tend to have high melting points? Why is the enthalpy of vaporization of a compound invariably much larger than its enthalpy of fusion? What is the opposite of fusion, sublimation, and condensation? Describe the phase change in each pair of opposing processes and state whether each phase change is exothermic or endothermic. Draw a typical heating curve (temperature versus amount of heat added at a constant rate) for conversion of a solid to a liquid and then to a gas. What causes some regions of the plot to have a positive slope? What is happening in the regions of the plot where the curve is horizontal, meaning that the temperature does not change even though heat is being added? If you know the mass of a sample of a substance, how could you use a heating curve to calculate the specific heat of the substance, as well as the change in enthalpy associated with a phase change? Draw the heating curve for a liquid that has been superheated. How does this differ from a normal heating curve for a liquid? Draw the cooling curve for a liquid that has been supercooled. How does this differ from a normal cooling curve for a liquid? Conceptual Answers When snow disappears without melting, it must be subliming directly from the solid state to the vapor state. The rate at which this will occur depends solely on the partial pressure of water, not on the total pressure due to other gases. Consequently, altitude (and changes in atmospheric pressure) will not affect the rate of sublimation directly. 3 The general equations and enthalpy changes for the changes of state involved in converting a solid to a gas are: \[ \text{solid} \rightarrow \text{liquid}: \Delta H_{fus} \] \[ \text{liquid} \rightarrow \text{gas}: \Delta H_{vap} \] \[ \text{solid} \rightarrow \text{gas}: \Delta{H_{sub}}= \Delta{H_{fus}} + \Delta{H_{vap}}\] The relationship between these enthalpy changes is shown schematically in the thermochemical cycle below: The formation of frost on a surface is an example of deposition, which is the reverse of sublimation. The change in enthalpy for deposition is equal in magnitude, but opposite in sign, to Δ H sub , which is a positive number: Δ H sub = Δ H fus + Δ H vap . liquid + heat → vapor: endothermic liquid → solid + heat: exothermic gas → liquid + heat: exothermic solid + heat → vapor: endothermic The enthalpy of vaporization is larger than the enthalpy of fusion because vaporization requires the addition of enough energy to disrupt all intermolecular interactions and create a gas in which the molecules move essentially independently. In contrast, fusion requires much less energy, because the intermolecular interactions in a liquid and a solid are similar in magnitude in all condensed phases. Fusion requires only enough energy to overcome the intermolecular interactions that lock molecules in place in a lattice, thereby allowing them to move more freely. The portions of the curve with a positive slope correspond to heating a single phase, while the horizontal portions of the curve correspond to phase changes. During a phase change, the temperature of the system does not change, because the added heat is melting the solid at its melting point or evaporating the liquid at its boiling point. A superheated liquid exists temporarily as liquid with a temperature above the normal boiling point of the liquid. When a supercooled liquid boils, the temperature drops as the liquid is converted to vapor. Conversely, a supercooled liquid exists temporarily as a liquid with a temperature lower than the normal melting point of the solid. As shown below, when a supercooled liquid crystallizes, the temperature increases as the liquid is converted to a solid. Numerical Problems The density of oxygen at 1 atm and various temperatures is given in the following table. Plot the data and use your graph to predict the normal boiling point of oxygen. 0 1 2 3 4 5 6 7 T (K) 60.0 70.0 80.0 90.0 100.000 120.000 140.000 d (mol/L) 40.1 38.6 37.2 35.6 0.123 0.102 0.087 The density of propane at 1 atm and various temperatures is given in the following table. Plot the data and use your graph to predict the normal boiling point of propane. 0 1 2 3 4 5 6 7 8 T (K) 100.0 125.0 150.0 175.0 200.0 225.0 250.000 275.000 d (mol/L) 16.3 15.7 15.0 14.4 13.8 13.2 0.049 0.044 Draw the cooling curve for a sample of the vapor of a compound that has a melting point of 34°C and a boiling point of 77°C as it is cooled from 100°C to 0°C. Propionic acid has a melting point of −20.8°C and a boiling point of 141°C. Draw a heating curve showing the temperature versus time as heat is added at a constant rate to show the behavior of a sample of propionic acid as it is heated from −50°C to its boiling point. What happens above 141°C? A 0.542 g sample of I 2 requires 96.1 J of energy to be converted to vapor. What is the enthalpy of sublimation of I 2 ? A 2.0 L sample of gas at 210°C and 0.762 atm condenses to give 1.20 mL of liquid, and 476 J of heat is released during the process. What is the enthalpy of vaporization of the compound? One fuel used for jet engines and rockets is aluminum borohydride [Al(BH 4 ) 3 ], a liquid that readily reacts with water to produce hydrogen. The liquid has a boiling point of 44.5°C. How much energy is needed to vaporize 1.0 kg of aluminum borohydride at 20°C, given a Δ H vap of 30 kJ/mol and a molar heat capacity ( C p ) of 194.6 J/(mol•K)? How much energy is released when freezing 100.0 g of dimethyl disulfide (C 2 H 6 S 2 ) initially at 20°C? Use the following information: melting point = −84.7°C, Δ H fus = 9.19 kJ/mol, C p = 118.1 J/(mol•K). The following four problems use the following information (the subscript p indicates measurements taken at constant pressure): Δ H fus (H 2 O) = 6.01 kJ/mol, Δ H vap (H 2 O) = 40.66 kJ/mol, C p(s) (crystalline H 2 O) = 38.02 J/(mol•K), C p(l) (liquid H 2 O) = 75.35 J/(mol•K), and C p(g) (H 2 O gas) = 33.60 J/(mol•K). How much heat is released in the conversion of 1.00 L of steam at 21.9 atm and 200°C to ice at −6.0°C and 1 atm? How much heat must be applied to convert a 1.00 g piece of ice at −10°C to steam at 120°C? How many grams of boiling water must be added to a glass with 25.0 g of ice at −3°C to obtain a liquid with a temperature of 45°C? How many grams of ice at −5.0°C must be added to 150.0 g of water at 22°C to give a final temperature of 15°C? Numerical Answers The transition from a liquid to a gaseous phase is accompanied by a drastic decrease in density. According to the data in the table and the plot, the boiling point of liquid oxygen is between 90 and 100 K (actually 90.2 K). 45.0 kJ/mol 488 kJ 32.6 kJ 57 g 11.5: Vapor Pressure Conceptual Problems What is the relationship between the boiling point, vapor pressure, and temperature of a substance and atmospheric pressure? What is the difference between a volatile liquid and a nonvolatile liquid? Suppose that two liquid substances have the same molecular mass, but one is volatile and the other is nonvolatile. What differences in the molecular structures of the two substances could account for the differences in volatility? An “old wives’ tale” states that applying ethanol to the wrists of a child with a very high fever will help to reduce the fever because blood vessels in the wrists are close to the skin. Is there a scientific basis for this recommendation? Would water be as effective as ethanol? Why is the air over a strip of grass significantly cooler than the air over a sandy beach only a few feet away? If gasoline is allowed to sit in an open container, it often feels much colder than the surrounding air. Explain this observation. Describe the flow of heat into or out of the system, as well as any transfer of mass that occurs. Would the temperature of a sealed can of gasoline be higher, lower, or the same as that of the open can? Explain your answer. What is the relationship between the vapor pressure of a liquid and its temperature? the surface area of the liquid? the pressure of other gases on the liquid? its viscosity? At 25°C, benzene has a vapor pressure of 12.5 kPa, whereas the vapor pressure of acetic acid is 2.1 kPa. Which is more volatile? Based on the intermolecular interactions in the two liquids, explain why acetic acid has the lower vapor pressure. Numerical Problems Acetylene (C 2 H 2 ), which is used for industrial welding, is transported in pressurized cylinders. Its vapor pressure at various temperatures is given in the following table. Plot the data and use your graph to estimate the vapor pressure of acetylene at 293 K. Then use your graph to determine the value of Δ H vap for acetylene. How much energy is required to vaporize 2.00 g of acetylene at 250 K? 0 1 2 3 4 5 6 7 T (K) 145.0 155.0 175.0 200 225 250 300 P (mmHg) 1.3 7.8 32.2 190 579 1370 5093 The following table gives the vapor pressure of water at various temperatures. Plot the data and use your graph to estimate the vapor pressure of water at 25°C and at 75°C. What is the vapor pressure of water at 110°C? Use these data to determine the value of Δ H vap for water. 0 1 2 3 4 5 6 7 T (°C) 0.0 10.0 30.0 50.0 60 80 100 P (mmHg) 4.6 9.2 31.8 92.6 150 355 760 The Δ H vap of carbon tetrachloride is 29.8 kJ/mol, and its normal boiling point is 76.8°C. What is its boiling point at 0.100 atm? The normal boiling point of sodium is 883°C. If Δ H vap is 97.4 kJ/mol, what is the vapor pressure (in millimeters of mercury) of liquid sodium at 300°C? An unknown liquid has a vapor pressure of 0.860 atm at 63.7°C and a vapor pressure of 0.330 atm at 35.1°C. Use the data in Table 11.6 in Section 11.5 to identify the liquid. An unknown liquid has a boiling point of 75.8°C at 0.910 atm and a boiling point of 57.2°C at 0.430 atm. Use the data in Table 11.6 in Section 11.5 to identify the liquid. If the vapor pressure of a liquid is 0.850 atm at 20°C and 0.897 atm at 25°C, what is the normal boiling point of the liquid? If the vapor pressure of a liquid is 0.799 atm at 99.0°C and 0.842 atm at 111°C, what is the normal boiling point of the liquid? The vapor pressure of liquid SO 2 is 33.4 torr at −63.4°C and 100.0 torr at −47.7 K. What is the Δ H vap of SO 2 ? What is its vapor pressure at −24.5 K? At what temperature is the vapor pressure equal to 220 torr? The vapor pressure of CO 2 at various temperatures is given in the following table: 0 1 2 3 4 T (°C) −120 −110 −100 −90 P (torr) 9.81 34.63 104.81 279.5 What is Δ H vap over this temperature range? What is the vapor pressure of CO 2 at −70°C? At what temperature does CO 2 have a vapor pressure of 310 torr? Numerical Answers vapor pressure at 273 K is 3050 mmHg; Δ H vap = 18.7 kJ/mol, 1.44 kJ 12.5°C Δ H vap = 28.9 kJ/mol, n -hexane Δ H vap = 7.81 kJ/mol, 36°C 11.6: Phase Diagrams Conceptual Problems A phase diagram is a graphic representation of the stable phase of a substance at any combination of temperature and pressure. What do the lines separating different regions in a phase diagram indicate? What information does the slope of a line in a phase diagram convey about the physical properties of the phases it separates? Can a phase diagram have more than one point where three lines intersect? If the slope of the line corresponding to the solid/liquid boundary in the phase diagram of water were positive rather than negative, what would be the effect on aquatic life during periods of subzero temperatures? Explain your answer. Conceptual Answer The lines in a phase diagram represent boundaries between different phases; at any combination of temperature and pressure that lies on a line, two phases are in equilibrium. It is physically impossible for more than three phases to coexist at any combination of temperature and pressure, but in principle there can be more than one triple point in a phase diagram. The slope of the line separating two phases depends upon their relative densities. For example, if the solid–liquid line slopes up and to the right , the liquid is less dense than the solid, while if it slopes up and to the left , the liquid is denser than the solid. Numerical Problems Naphthalene (C 10 H 8 ) is the key ingredient in mothballs. It has normal melting and boiling points of 81°C and 218°C, respectively. The triple point of naphthalene is 80°C at 1000 Pa. Use these data to construct a phase diagram for naphthalene and label all the regions of your diagram. Argon is an inert gas used in welding. It has normal boiling and freezing points of 87.3 K and 83.8 K, respectively. The triple point of argon is 83.8 K at 0.68 atm. Use these data to construct a phase diagram for argon and label all the regions of your diagram. 11.7: Structure of Solids Conceptual Problems 1. Compare the solid and liquid states in terms of a. rigidity of structure. b. long-range order. c. short-range order. 2. How do amorphous solids differ from crystalline solids in each characteristic? Which of the two types of solid is most similar to a liquid? a. rigidity of structure b. long-range order c. short-range order 3. Why is the arrangement of the constituent atoms or molecules more important in determining the properties of a solid than a liquid or a gas? 4. Why are the structures of solids usually described in terms of the positions of the constituent atoms rather than their motion? 5. What physical characteristics distinguish a crystalline solid from an amorphous solid? Describe at least two ways to determine experimentally whether a material is crystalline or amorphous. 6. Explain why each characteristic would or would not favor the formation of an amorphous solid. a. slow cooling of pure molten material b. impurities in the liquid from which the solid is formed c. weak intermolecular attractive forces 7. A student obtained a solid product in a laboratory synthesis. To verify the identity of the solid, she measured its melting point and found that the material melted over a 12°C range. After it had cooled, she measured the melting point of the same sample again and found that this time the solid had a sharp melting point at the temperature that is characteristic of the desired product. Why were the two melting points different? What was responsible for the change in the melting point? Conceptual Answers II 3. The arrangement of the atoms or molecules is more important in determining the properties of a solid because of the greater persistent long-range order of solids. Gases and liquids cannot readily be described by the spatial arrangement of their components because rapid molecular motion and rearrangement defines many of the properties of liquids and gases. 7. The initial solid contained the desired compound in an amorphous state, as indicated by the wide temperature range over which melting occurred. Slow cooling of the liquid caused it to crystallize, as evidenced by the sharp second melting point observed at the expected temperature. Conceptual Problems II Why is it valid to represent the structure of a crystalline solid by the structure of its unit cell? What are the most important constraints in selecting a unit cell? All unit cell structures have six sides. Can crystals of a solid have more than six sides? Explain your answer. Explain how the intensive properties of a material are reflected in the unit cell. Are all the properties of a bulk material the same as those of its unit cell? Explain your answer. The experimentally measured density of a bulk material is slightly higher than expected based on the structure of the pure material. Propose two explanations for this observation. The experimentally determined density of a material is lower than expected based on the arrangement of the atoms in the unit cell, the formula mass, and the size of the atoms. What conclusion(s) can you draw about the material? Only one element (polonium) crystallizes with a simple cubic unit cell. Why is polonium the only example of an element with this structure? What is meant by the term coordination number in the structure of a solid? How does the coordination number depend on the structure of the metal? Arrange the three types of cubic unit cells in order of increasing packing efficiency. What is the difference in packing efficiency between the hcp structure and the ccp structure? The structures of many metals depend on pressure and temperature. Which structure—bcc or hcp—would be more likely in a given metal at very high pressures? Explain your reasoning. A metal has two crystalline phases. The transition temperature, the temperature at which one phase is converted to the other, is 95°C at 1 atm and 135°C at 1000 atm. Sketch a phase diagram for this substance. The metal is known to have either a ccp structure or a simple cubic structure. Label the regions in your diagram appropriately and justify your selection for the structure of each phase. Numerical Problems II Metallic rhodium has an fcc unit cell. How many atoms of rhodium does each unit cell contain? Chromium has a structure with two atoms per unit cell. Is the structure of this metal simple cubic, bcc, fcc, or hcp? The density of nickel is 8.908 g/cm 3 . If the metallic radius of nickel is 125 pm, what is the structure of metallic nickel? The density of tungsten is 19.3 g/cm 3 . If the metallic radius of tungsten is 139 pm, what is the structure of metallic tungsten? An element has a density of 10.25 g/cm 3 and a metallic radius of 136.3 pm. The metal crystallizes in a bcc lattice. Identify the element. A 21.64 g sample of a nonreactive metal is placed in a flask containing 12.00 mL of water; the final volume is 13.81 mL. If the length of the edge of the unit cell is 387 pm and the metallic radius is 137 pm, determine the packing arrangement and identify the element. A sample of an alkali metal that has a bcc unit cell is found to have a mass of 1.000 g and a volume of 1.0298 cm 3 . When the metal reacts with excess water, the reaction produces 539.29 mL of hydrogen gas at 0.980 atm and 23°C. Identify the metal, determine the unit cell dimensions, and give the approximate size of the atom in picometers. A sample of an alkaline earth metal that has a bcc unit cell is found to have a mass 5.000 g and a volume of 1.392 cm 3 . Complete reaction with chlorine gas requires 848.3 mL of chlorine gas at 1.050 atm and 25°C. Identify the metal, determine the unit cell dimensions, and give the approximate size of the atom in picometers. Lithium crystallizes in a bcc structure with an edge length of 3.509 Å. Calculate its density. What is the approximate metallic radius of lithium in picometers? Vanadium is used in the manufacture of rust-resistant vanadium steel. It forms bcc crystals with a density of 6.11 g/cm 3 at 18.7°C. What is the length of the edge of the unit cell? What is the approximate metallic radius of the vanadium in picometers? A simple cubic cell contains one metal atom with a metallic radius of 100 pm. Determine the volume of the atom(s) contained in one unit cell [the volume of a sphere = (\({4 \over 3} \))πr 3 ]. What is the length of one edge of the unit cell? (Hint: there is no empty space between atoms.) Calculate the volume of the unit cell. Determine the packing efficiency for this structure. Use the steps in Problem 11 to calculate the packing efficiency for a bcc unit cell with a metallic radius of 1.00 Å. Numerical Answers 1. four 3. fcc 5. molybdenum 7. sodium, unit cell edge = 428 pm, r = 185 pm 9. d = 0.5335 g/cm 3 , r =151.9 pm 11.8: Bonding in Solids Conceptual Problems 1. Four vials labeled A–D contain sucrose, zinc, quartz, and sodium chloride, although not necessarily in that order. The following table summarizes the results of the series of analyses you have performed on the contents: Unnamed: 0 A B C D Melting Point high high high low Thermal Conductivity poor poor good poor Electrical Conductivity in Solid State moderate poor high poor Electrical Conductivity in Liquid State high poor high poor Hardness hard hard soft soft Luster none none high none Match each vial with its contents. 2. Do ionic solids generally have higher or lower melting points than molecular solids? Why? Do ionic solids generally have higher or lower melting points than covalent solids? Explain your reasoning. 3. The strength of London dispersion forces in molecular solids tends to increase with molecular mass, causing a smooth increase in melting points. Some molecular solids, however, have significantly lower melting points than predicted by their molecular masses. Why? 4. Suppose you want to synthesize a solid that is both heat resistant and a good electrical conductor. What specific types of bonding and molecular interactions would you want in your starting materials? 5. Explain the differences between an interstitial alloy and a substitutional alloy. Given an alloy in which the identity of one metallic element is known, how could you determine whether it is a substitutional alloy or an interstitial alloy? 6. How are intermetallic compounds different from interstitial alloys or substitutional alloys? Conceptual Answers 1. a. NaCl, ionic solid b. quartz, covalent solid c. zinc, metal d. sucrose, molecular solid 5. In a substitutional alloy, the impurity atoms are similar in size and chemical properties to the atoms of the host lattice; consequently, they simply replace some of the metal atoms in the normal lattice and do not greatly perturb the structure and physical properties. In an interstitial alloy, the impurity atoms are generally much smaller, have very different chemical properties, and occupy holes between the larger metal atoms. Because interstitial impurities form covalent bonds to the metal atoms in the host lattice, they tend to have a large effect on the mechanical properties of the metal, making it harder, less ductile, and more brittle. Comparing the mechanical properties of an alloy with those of the parent metal could be used to decide whether the alloy were a substitutional or interstitial alloy. Numerical Problems 1. Will the melting point of lanthanum(III) oxide be higher or lower than that of ferrous bromide? The relevant ionic radii are as follows: La 3 + , 104 pm; O 2− , 132 pm; Fe 2 + , 83 pm; and Br − , 196 pm. Explain your reasoning. 2. Draw a graph showing the relationship between the electrical conductivity of metallic silver and temperature. 3. Which has the higher melting point? Explain your reasoning in each case. a. Os or Hf b. SnO 2 or ZrO 2 c. Al 2 O 3 or SiO 2 4. Draw a graph showing the relationship between the electrical conductivity of a typical semiconductor and temperature. Numerical Answer 3. a. Osmium has a higher melting point, due to more valence electrons for metallic bonding. b. Zirconium oxide has a higher melting point, because it has greater ionic character. c. Aluminum oxide has a higher melting point, again because it has greater ionic character. |
Courses/University_of_South_Carolina__Upstate/CHEM_U109%3A_Chemistry_of_Living_Things_-_Mueller/08%3A_Solids_Liquids_and_Gases/8.1%3A_Intermolecular_Interactions | Skills to Develop 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 liquid 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. 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 interactions between the particles of substances. (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 Substances Substances with the highest melting and boiling points have covalent network bonding. This substances are held together by covalent bonds. In these substances, all the atoms in a sample are covalently bonded to other atoms. In effect, the entire sample is essentially one large molecule so the term intermolecular (or between molecules) is often not used for these interactions (since there are not multiple, discreet molecules). 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 \(\PageIndex{1}\)), which is a form of pure carbon. At temperatures over 3,500°C, diamond finally vaporizes into gas-phase atoms. Figure \(\PageIndex{1}\) : Diamond. Diamond, a form of pure carbon, has covalent network bonding. It takes a very high temperature—over 3,500°C—for diamond to leave the solid state. Source: Photo © Thinkstock 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. While not ions are not technically molecules, ionic interactions between particles are usually classified as a 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 \(\PageIndex{2}\)) is an example of a substance whose particles experience ionic interactions (Table \(\PageIndex{1}\)). Figure \(\PageIndex{2}\): Sodium Chloride. Solid NaCl is held together by ionic interactions. Source: Photo © Thinkstock Covalent Compounds (also called Molecular Compounds) Many substances that experience covalent bonding exist as discrete molecules. Very strong covalent bonds within the molecule hold the atoms together to make each molecule. It is not, however, covalent bonds that are responsible for bringing the separate molecules together to make liquids and solids. Instead, there are three types of intermolecular interactions that exist between molecules that hold liquids and solids together. All thee types result from the attractions between partial opposite charges and are weaker than covalent or ionic bonds. In many molecules, the electrons that are shared in a covalent bond are not shared equally among the two atoms in the bond. Typically, one of the atoms attracts the electrons more strongly than the other, leading to an unequal sharing of electrons in the bond. This idea is illustrated in Figure \(\PageIndex{3}\), which shows a diagram of the covalent bond in hydrogen fluoride (HF). The fluorine atom attracts the electrons in the bond more than the hydrogen atom does. An arrow is often used to show in which direction the electrons are pulled. The result is an unequal distribution of electrons in the bond, favoring the fluorine side of the covalent bond. Because of this unequal distribution, the fluorine side of the covalent bond actually takes on a partial negative charge (indicated by the δ− in Figure \(\PageIndex{3}\)), while the hydrogen side of the bond, being electron deficient, takes on a partial positive charge (indicated by the δ+ in Figure \(\PageIndex{3}\)). A 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. Figure \(\PageIndex{3}\): Polar Covalent Bonds. The electrons in the HF molecule are not equally shared by the two atoms in the bond. Because the fluorine atom has nine protons in its nucleus, it attracts the negatively charged electrons in the bond more than the hydrogen atom does with its one proton in its nucleus. Thus, electrons are more strongly attracted to the fluorine atom, leading to an imbalance in the electron distribution between the atoms. The fluorine side of the bond picks up a partial overall negative charge (represented by the δ− in the diagram), while the hydrogen side of the bond has an overall partial positive charge (represented by the δ+ in the diagram). Such a bond is called a polar covalent bond . 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. These types of intermolecular interactions are called a dipole-dipole forces. Many molecules with polar covalent bonds experience dipole-dipole forces. The covalent bonds in some molecules are oriented in space symmetrically so that the bonds in the molecules cancel each other out. The individual bonds are polar, but the overall molecule is not polar; rather, the molecule is nonpolar due to its overall symmetry. Such molecules experience little or no dipole-dipole interactions. Carbon dioxide (CO 2 ) and carbon tetrachloride (CCl 4 ) are examples of such molecules (Figure \(\PageIndex{4}\)). Figure \(\PageIndex{4}\): Nonpolar Molecules. Although the individual bonds in both CO 2 and CCl 4 are polar, their effects cancel out because of the symmetrical distribution of the bonds in each molecule. As a result, such molecules experience little or no dipole-dipole interaction. The H–F, O–H, and N–H bonds are strongly polar; in molecules that have these bonds, particularly strong dipole-dipole interactions (as strong as 10% of a true covalent bond) can occur. Because of this strong interaction, the term hydrogen bonding is used to describe this dipole-dipole interaction. The physical properties of water, which has two O–H bonds, are strongly affected by the presence of hydrogen bonding forces of attraction between water molecules. Figure \(\PageIndex{5}\) shows how molecules experiencing hydrogen bonding can interact. Hydrogen-bonding forces between water molecules (illustrated below as green dashes) are different than the covalent O-H bonds within a water molecule (illustrated as solid orange lines below). Hydrogen-bonding forces between molecules are stronger than dipole-dipole forces between molecules, but not as strong as covalent bonds within molecules. Figure \(\PageIndex{5}\) : Hydrogen-Bonding forces between Water Molecules (dashed green lines). The presence of hydrogen bonding in molecules like water can have a large impact on the physical properties of a substance. 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. Because dispersion forces are caused by the instantaneous distribution of electrons in a molecule, larger molecules with a large number of electrons can experience substantial dispersion forces. 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. The phase that a substance adopts depends on the type and magnitude of the intermolecular interactions the particles of a substance experience. If the intermolecular interactions are relatively strong, then a large amount of energy—in terms of temperature—is necessary for a substance to change phases. If the intermolecular interactions are weak, a low temperature is all that is necessary to move a substance out of the solid phase. Example \(\PageIndex{1}\): Intermolecular Forces What intermolecular interactions 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 fully charged ions, so the intermolecular interactions in potassium chloride are ionic bonds. Because ionic bonds 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 forces. It is also polar so has dipole-dipole forces, but the H-bonding forces are more significant. 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. Concept Review Exercise What types of interactions might cause a gas to stick together as liquid or solid in: A. a covalent network substance B. an ionic compound C. a molecular compound? Answer A. covalent bonds (sharing of electrons) B. ionic bonds (attraction between full opposite charges of ions) C. All molecules will have London-dispersion forces, polar molecules will have dipole-dipole forces, and molecules with H directly bonded to N, O, or F will have hydrogen-bonding forces. (All three are attractions of partial opposite charges between molecules.) Key Takeaways A phase is a form of matter that has the same physical properties throughout. Atoms and molecules interact with each other through various forces: ionic and covalent bonds, dipole-dipole interactions, hydrogen bonding, and dispersion forces. Exercises List the three common phases in the order you are likely to find them—from lowest temperature to highest temperature. List these intermolecular interactions from weakest to strongest: London forces, hydrogen-bonding forces, dipole-dipole forces and ionic interactions. What type of intermolecular interaction is predominate in each substance? ammonia (NH 3 ) sodium sulfate (Na 2 SO 4 ) decane (C 10 H 22 ) Explain how a molecule like carbon dioxide (CO 2 ) can have polar covalent bonds but be nonpolar overall. What types of intermolecular forces would exist in a sample of sulfur dichloride (SCl 2 )? What are some of the physical properties of substances that experience covalent network bonding? What are some of the physical properties of substances that experience only dispersion forces? 8. Heat is required to boil water (endothermic). What types of interactions are being overcome when water is boiled? Answers solid, liquid, and gas London forces < dipole-dipole forces < hydrogen-bonding forces < ionic interactions a. hydrogen-bonding forces predominate (but NH 3 also has dipole-dipole and London-dispersion); b. ionic interactions; c. dipsersion forces (C-C and C-H bonds are not polar) In linear carbon dioxide, the two covalent bonds are oriented on opposite sides (symmetrically) such that the polarity of each bond cancels out. Sulfur dichloride has polar bonds, and because of the two lone pairs on the central sulfur atom, would have a bent geometry. The molecule is not symmetrical so one side is more negative and the other more positive. Like all molecules, it would have London-dispersion forces and, because it is polar, also dipole-dipole forces. very hard, high melting point low melting point and low boiling point London-dispersion, dipole-dipole, and hydrogen-bonding forces of attraction are being overcome to separate the liquid water molecules into gaseous water molecules. (Covalent bonds bind oxygen to the two hydrogens in each water, but these are not broken when water is boiled.) |
Courses/Oregon_Institute_of_Technology/OIT%3A_CHE_333_-_Organic_Chemistry_III_(Lund)/Appendix_I%3A_Index_of_enzymatic_reactions_by_pathway/Nucleotide_catabolism | Sections/problems listed with an asterisk (*) do not discuss the exact reaction indicated, but do discuss a closely related reaction. Cytidine ( to uridine ) Cytidine deaminase (EC 3.5.4.5) P11.4 Uridine Uridine phosphorylase (EC 2.4.2.3) Section 9.2 *, Section 11.5 ; P9.1 Dihydropyrimidine dehydrogenase (EC 1.3.1.2) Section 16.5D Dihydropyrimidase (EC 3.5.2.2) Section 12.5 *; P12.11a beta -ureidopropionase (EC 3.5.1.6) P13.11 Thymidine Thymidine phosphorylase (EC 2.4.2.4) Section 9.2 * Dihydropyrimidine dehydrogenase (EC 1.3.1.2) Section 16.5D Dihydropyrimidase (EC 3.5.2.2) Section 12.5 *; P12.11a beta -ureidopropionase ( EC 3.5.1.6 ) P13.11 Adenosine ( to uric acid via xanthine ) Adenosine deaminase ( EC 3.5.4.4 ) P11.5 Purine nucleoside phosphorylase (EC 2.4.2.1) Section 9.2 * Xanthine oxidase (EC 1.17.1.4; EC 1.17.3.2) not discussed Guanosine ( to uric acid via xanthine ) Purine nucleoside phosphorylase ( EC 2.4.2.1 ) Section 9.2 * Guanine deaminase ( EC 3.5.4.3 ) P11.5 * Xanthine oxidase ( EC 1.17.1.4 ; EC 1.17.3.2 ) not discussed Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) |
Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/20%3A_Amines/20.04%3A_Synthesis_of_Amines | Reductive Amination of Aldehydes and Ketones (Carbonyls) Aldehydes and ketones can be converted into 1 o , 2 o and 3 o amines using reductive amination. The reaction takes place in two parts. The first step is the nucleophiic addition of the carbonyl group to form an imine. The second step is the reduction of the imine to an amine using an reducing agent. A reducing agent commonly used for this reaction is sodium cyanoborohydride (NaBH 3 CN). The nitrogen gains a bond to carbon during this reaction sequence. When carbonyls react with ammonia, a primary amine is produced. The reaction pattern continues for each amine classification. For example, pyrrolidine reacts with 2-butanone to produce the imine, which can be reduced by LiAlH 4 , sodium cyanoborohydride (NaBH 3 CN), or H 2 with an active metal catalyst to produce a tertiary amine. Amide Reduction to 1°, 2° or 3° Amines using LiAlH 4 There is a direct correlation between the structure of the amide and the structure of the amine produced. Primary amides are reduced to primary amines. Secondary amides are reduced to secondary amines. Tertiary amides are reduced to tertiary amines. Lithium aluminum hydride is a stronger reducing agent than sodium borohydride, which is not strong enough for this reaction. Exercise 8. Add the missing reactants/products to the following reactions. Answer 8. |
Courses/Widener_University/CHEM_105%3A_Introduction_to_General_Organic_and_Biological_Chemistry_Fall_22/06%3A_Covalent_Bonding_and_Simple_Molecular_Compounds/6.05%3A_Characteristics_of_Covalent_Bonds | Learning Objectives Recognize bond characteristics of covalent compounds: bond length and bond polarity. Use electronegativity values to predict bond polarity. Bond Length We previously stated that the covalent bond in the hydrogen molecule (H 2 ) has a certain length (about 7.4 × 10 −11 m). Other covalent bonds also have known bond lengths, which are dependent on both the identities of the atoms in the bond and whether the bonds are single, double, or triple bonds. Table \(\PageIndex{1}\) lists the approximate bond lengths for some single covalent bonds. The exact bond length may vary depending on the identity of the molecule but will be close to the value given in the table. Bond Length (× \(10^{−12}\, m\)) H–H 74 H–C 110 H–N 100 H–O 97 H–I 161 C–C 154 C–N 147 C–O 143 N–N 145 O–O 145 Table \(\PageIndex{2}\) compares the lengths of single covalent bonds with those of double and triple bonds between the same atoms. Without exception, as the number of covalent bonds between two atoms increases, the bond length decreases. With more electrons between the two nuclei, the nuclei can get closer together before the internuclear repulsion is strong enough to balance the attraction. Bond Length (× \(10^{−12}\, m\)) C–C 154 C=C 134 C≡C 120 C–N 147 C=N 128 C≡N 116 C–O 143 C=O 120 C≡O 113 N–N 145 N=N 123 N≡N 110 O–O 145 O=O 121 Electronegativity and Bond Polarity Although we defined covalent bonding as electron sharing, the electrons in a covalent bond are not always shared equally by the two bonded atoms. Unless the bond connects two atoms of the same element, as in H 2 , there will always be one atom that attracts the electrons in the bond more strongly than the other atom does, as in HCl, shown in Figure \(\PageIndex{1}\). A covalent bond that has an equal sharing of electrons (Figure \(\PageIndex{1a}\)) is called a nonpolar covalent bond . A covalent bond that has an unequal sharing of electrons, as in Figure \(\PageIndex{1b}\), is called a polar covalent bond . The distribution of electron density in a polar bond is uneven. It is greater around the atom that attracts the electrons more than the other. For example, the electrons in the H–Cl bond of a hydrogen chloride molecule spend more time near the chlorine atom than near the hydrogen atom. Note that the shaded area around Cl in Figure \(\PageIndex{1b}\) is much larger than it is around H. This imbalance in electron density results in a buildup of partial negative charge (designated as δ−) on one side of the bond (Cl) and a partial positive charge (designated δ+) on the other side of the bond (H). This is seen in Figure \(\PageIndex{2a}\). The separation of charge in a polar covalent bond results in an electric dipole (two poles), represented by the arrow in Figure \(\PageIndex{2b}\). The direction of the arrow is pointed toward the δ− end while the + tail of the arrow indicates the δ+ end of the bond. Any covalent bond between atoms of different elements is a polar bond, but the degree of polarity varies widely. Some bonds between different elements are only minimally polar, while others are strongly polar. Ionic bonds can be considered the ultimate in polarity, with electrons being transferred rather than shared. To judge the relative polarity of a covalent bond, chemists use electronegativity , which is a relative measure of how strongly an atom attracts electrons when it forms a covalent bond. There are various numerical scales for rating electronegativity. Figure \(\PageIndex{3}\) shows one of the most popular—the Pauling scale. Looking Closer: Linus Pauling Arguably the most influential chemist of the 20th century, Linus Pauling (1901–94) is the only person to have won two individual (that is, unshared) Nobel Prizes. In the 1930s, Pauling used new mathematical theories to enunciate some fundamental principles of the chemical bond. His 1939 book The Nature of the Chemical Bond is one of the most significant books ever published in chemistry. By 1935, Pauling’s interest turned to biological molecules, and he was awarded the 1954 Nobel Prize in Chemistry for his work on protein structure. (He was very close to discovering the double helix structure of DNA when James Watson and James Crick announced their own discovery of its structure in 1953.) He was later awarded the 1962 Nobel Peace Prize for his efforts to ban the testing of nuclear weapons. In his later years, Pauling became convinced that large doses of vitamin C would prevent disease, including the common cold. Most clinical research failed to show a connection, but Pauling continued to take large doses daily. He died in 1994, having spent a lifetime establishing a scientific legacy that few will ever equal. The polarity of a covalent bond can be judged by determining the difference in the electronegativities of the two atoms making the bond. The greater the difference in electronegativities, the greater the imbalance of electron sharing in the bond. Although there are no hard and fast rules, the general rule is if the difference in electronegativities is less than about 0.4, the bond is considered nonpolar; if the difference is greater than 0.4, the bond is considered polar. If the difference in electronegativities is large enough ( generally greater than about 1.8), the resulting compound is considered ionic rather than covalent . An electronegativity difference of zero, of course, indicates a nonpolar covalent bond. Example \(\PageIndex{1}\) Describe the electronegativity difference between each pair of atoms and the resulting polarity (or bond type). C and H H and H Na and Cl O and H Solution Carbon has an electronegativity of 2.5, while the value for hydrogen is 2.1. The difference is 0.4, which is rather small. The C–H bond is therefore considered nonpolar. Both hydrogen atoms have the same electronegativity value—2.1. The difference is zero, so the bond is nonpolar. Sodium’s electronegativity is 0.9, while chlorine’s is 3.0. The difference is 2.1, which is rather high, and so sodium and chlorine form an ionic compound. With 2.1 for hydrogen and 3.5 for oxygen, the electronegativity difference is 1.4. We would expect a very polar bond. The sharing of electrons between O and H is unequal with the electrons more strongly drawn towards O. Exercise \(\PageIndex{1}\) Describe the electronegativity (EN) difference between each pair of atoms and the resulting polarity (or bond type). C and O K and Br N and N Cs and F Answer a: The EN difference is 1.0 , hence polar. The sharing of electrons between C and O is unequal with the electrons more strongly drawn towards O. Answer b: The EN difference is greater than 1.8, hence ionic. Answer c: Identical atoms have zero EN difference, hence nonpolar. Answer d: The EN difference is greater than 1.8, hence ionic. Bond Polarity and Molecular 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 . A polar molecule is a molecule in which 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 is called a dipole . A simplified way to depict polar molecules like HF is pictured below (see figure below). When placed between oppositely charged plates, polar molecules orient themselves so that their positive ends are closer to the negative plate and their negative ends are closer to the positive plate (see Figure 4.4.6 below). Experimental techniques involving electric fields can be used to determine if a certain substance is composed of polar molecules and to measure the degree of polarity. 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. (Figure \(\PageIndex{7}\)) is a comparison between carbon dioxide and water. Carbon dioxide (CO 2 ) is a linear molecule. The oxygen atoms are more electronegative than the carbon atom, so there are two individual dipoles pointing outward from the C atom to each O atom. However, since the dipoles are of equal strength and are oriented directly opposite each other, they cancel each other out, and the overall molecular polarity of CO 2 is zero. CO 2 is a nonpolar molecule. In H 2 O, the orientation of the two O–H bonds is bent : one end of the molecule has a partial positive charge, and the other end has a partial negative charge. In short, the molecule itself is polar . The polarity of water has an enormous impact on its physical and chemical properties. For example, the boiling point of water (100°C) is high for such a small molecule due to the fact that polar molecules attract each other strongly. On the other hand, the nonpolar carbon dioxide becomes a gas at −77°C, almost 200° lower than the temperature at which water boils. More details will be presented in the next section. Key Takeaways Covalent bonds between different atoms have different bond lengths. Covalent bonds can be polar or nonpolar, depending on the electronegativity difference between the atoms involved. |
Courses/American_River_College/Chemistry_305_(S21_Zarzana)/Map%3A_Introductory_Chemistry_(Tro)/07%3A_Chemical_Reactions/7.08%3A_AcidBase_and_Gas_Evolution_Reactions | Learning Objectives Identify when a reaction will evolve a gas. Neutralization Reactions Acids and bases react chemically with each other to form salts . A salt is a general chemical term for any ionic compound formed from an acid and a base. In reactions where the acid is a hydrogen-ion-containing compound and the base is a hydroxide-ion-containing compound, water is also a product. The general reaction is as follows: \[\text{acid + base} → \text{water + salt}\] The reaction of acid and base to make water and a salt is called neutralization . Like any chemical equation, a neutralization chemical equation must be properly balanced. For example, the neutralization reaction between sodium hydroxide and hydrochloric acid is as follows: \[\ce{NaOH (aq) + HCl (aq) \rightarrow NaCl (aq) + H_2O (ℓ)} \label{Eq2}\] with coefficients all understood to be one. The neutralization reaction between sodium hydroxide and sulfuric acid is as follows: \[\ce{2NaOH (aq) + H_2SO_4 (aq) \rightarrow Na_2SO_4(aq) + 2H_2O (ℓ)} \label{Eq3}\] Example \(\PageIndex{1}\): Neutralizing Nitric Acid Nitric acid (HNO 3 (aq)) can be neutralized by calcium hydroxide (Ca(OH) 2 (aq)). Write a balanced chemical equation for the reaction between these two compounds and identify the salt that it produces. Solution Steps Explanation Equation Write the unbalanced equation. This is a double displacement reaction, so the cations and anions swap to create new products. Ca(OH)2(aq) + HNO3(aq) → Ca(NO3)2(aq) + H2O(ℓ) Balance the equation. Because there are two OH− ions in the formula for Ca(OH)2, we need two moles of HNO3 to provide H+ ions Ca(OH)2(aq) + 2HNO3(aq) → Ca(NO3)2(aq) + 2H2O(ℓ) Additional step: Identify the salt. NaN The salt formed is calcium nitrate. Exercise \(\PageIndex{1}\) Hydrocyanic acid (\(\ce{HCN(aq)}\)) can be neutralized by potassium hydroxide (\(\ce{KOH(aq)}\)). Write a balanced chemical equation for the reaction between these two compounds and identify the salt that it produces. Answer \[\ce{KOH (aq) + HCN(aq) → KCN (aq) + H2O(ℓ)} \nonumber\] Gas Evolving Reactions A gas evolution reaction is a chemical process that produces a gas, such as oxygen or carbon dioxide. In the following examples, an acid reacts with a carbonate, producing salt, carbon dioxide, and water, respectively. For example, nitric acid reacts with sodium carbonate to form sodium nitrate, carbon dioxide, and water (Table \(\PageIndex{1}\)): \[\ce{2HNO3(aq)+Na2CO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)}\] Sulfuric acid reacts with calcium carbonate to form calcium sulfate, carbon dioxide, and water: \[\ce{H2SO4(aq) + CaCO3(aq) → CaSO4(aq) + CO2(g)+H2O(l)}\] Hydrochloric acid reacts with calcium carbonate to form calcium chloride, carbon dioxide, and water: \[\ce{2HCl(aq) + CaCO3(aq) → CaCl2(aq) + CO2(g) + H2O(l)}\] Figure \(\PageIndex{1}\) demonstrates this type of reaction: In this reaction setup, lime water, a dilute calcium hydroxide ( \(Ca(OH)_2\) ) solution, is poured into one of the test tubes and sealed with a stopper. A small amount of hydrochloric acid is carefully poured into the remaining test tube. A small amount of sodium carbonate is added to the acid, and the tube is sealed with a rubber stopper. The two tubes are connected. As a result of the acid-carbonate reaction, carbon dioxide is produced and the lime water turns milky. Reactant Type Intermediate Product Gas Evolved Example sulfide none \(\ce{H2S}\) \(\ce{2HCl(aq) + K2S \rightarrow H2S (g) + 2KCl (aq)}\) carbonates and bicarbonates \(\ce{H2CO3}\) \(\ce{CO2}\) \(\ce{2HCl(aq) + K2CO2 \rightarrow H2O (l) + CO2(g) + 2KCl (aq)}\) sulfites and bisulfites \(\ce{H2SO3}\) \(\ce{SO2}\) \(\ce{2HCl(aq) + K2SO2 \rightarrow H2O (l) + SO2(g) + 2KCl (aq)}\) ammonia \(\ce{NH4OH}\) \(\ce{NH3}\) \(\ce{NH4Cl(aq) + KOH \rightarrow H2O (l) + NH3(g) + 2KCl (aq)}\) The gas-evolving experiment lime water is illustrated in the following video: Video \(\PageIndex{1}\): Carbon Dioxide (\(CO_2\)) & Limewater (Chemical Reaction). As the reaction proceeds, the limewater on the turns from clear to milky; this is due to the \(CO_2(g)\) reacting with the aqueous calcium hydroxide to form calcium carbonate, which is only slightly soluble in water. When this experiment is repeated with nitric or sulfuric acid instead of \(HCl\), it yields the same results: the clear limewater turns milky, indicating the production of carbon dioxide. Another method to chemically generate gas is the oxidation of metals in acidic solutions. This reaction will yield a metal salt and hydrogen gas. \[\ce{2HCl (aq) + Zn(s) \rightarrow ZnCl_2 (aq) + H_2 (g)}\] Here, hydrochloric acid oxidizes zinc to produce an aqueous metal salt and hydrogen gas bubbles. Recall that oxidation refers to a loss of electrons, and reduction refers to the gain of electrons. In the above redox reaction, neutral zinc is oxidized to \(Zn^{2+}\), and the acid, \(H^+\), is reduced to \(H_2(g)\). The oxidation of metals by strong acids is another common example of a gas evolution reaction. Contributors & Affiliations 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: Boundless (www.boundless.com) Wikipedia (CC-BY-SA-3.0) Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ). Marisa Alviar-Agnew ( Sacramento City College ) Henry Agnew (UC Davis) |
Courses/Los_Angeles_Trade_Technical_College/Chem_51/zz%3A_Back_Matter/20.04%3A_Naming_Alkanes | Learning Objectives To identify and name simple (straight-chain) alkanes given formulas and write formulas for straight-chain alkanes given their names. To name branched chain alkanes by the IUPAC system and write formulas for branched chain alkanes given IUPAC names We begin our study of organic chemistry with the hydrocarbons , the simplest organic compounds, which are composed of carbon and hydrogen atoms only. As we noted, there are several different kinds of hydrocarbons. They are distinguished by the types of bonding between carbon atoms and the properties that result from that bonding. Hydrocarbons with only carbon-to-carbon single bonds (C–C) and existing as a continuous chain of carbon atoms also bonded to hydrogen atoms are called alkanes (or saturated hydrocarbons) . Saturated , in this case, means that each carbon atom is bonded to four other atoms (hydrogen or carbon)—the most possible; there are no double or triple bonds in the molecules. The word saturated has the same meaning for hydrocarbons as it does for the dietary fats and oils: the molecule has no carbon-to-carbon double bonds (C=C). We previously introduced the three simplest alkanes—methane (CH 4 ), ethane (C 2 H 6 ), and propane (C 3 H 8 ) and they are shown again in Figure \(\PageIndex{1}\). The flat representations shown do not accurately portray bond angles or molecular geometry. Methane has a tetrahedral shape that chemists often portray with wedges indicating bonds coming out toward you and dashed lines indicating bonds that go back away from you. An ordinary solid line indicates a bond in the plane of the page. Recall that the VSEPR theory correctly predicts a tetrahedral shape for the methane molecule (Figure \(\PageIndex{2}\)). Methane (CH 4 ), ethane (C 2 H 6 ), and propane (C 3 H 8 ) are the beginning of a series of compounds in which any two members in a sequence differ by one carbon atom and two hydrogen atoms—namely, a CH 2 unit. The first 10 members of this series are given in Table \(\PageIndex{1}\). Name Molecular Formula (CnH2n + 2) Condensed Structural Formula Number of Possible Isomers methane CH4 CH4 — ethane C2H6 CH3CH3 — propane C3H8 CH3CH2CH3 — butane C4H10 CH3CH2CH2CH3 2 pentane C5H12 CH3CH2CH2CH2CH3 3 hexane C6H14 CH3CH2CH2CH2CH2CH3 5 heptane C7H16 CH3CH2CH2CH2CH2CH2CH3 9 octane C8H18 CH3CH2CH2CH2CH2CH2CH2CH3 18 nonane C9H20 CH3CH2CH2CH2CH2CH2CH2CH2CH3 35 decane C10H22 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3 75 Consider the series in Figure \(\PageIndex{3}\). The sequence starts with C 3 H 8 , and a CH 2 unit is added in each step moving up the series. Any family of compounds in which adjacent members differ from each other by a definite factor (here a CH 2 group) is called a homologous series . The members of such a series, called homologs , have properties that vary in a regular and predictable manner. The principle of homology gives organization to organic chemistry in much the same way that the periodic table gives organization to inorganic chemistry. Instead of a bewildering array of individual carbon compounds, we can study a few members of a homologous series and from them deduce some of the properties of other compounds in the series. The principle of homology allows us to write a general formula for alkanes: C n H 2 n + 2 . Using this formula, we can write a molecular formula for any alkane with a given number of carbon atoms. For example, an alkane with eight carbon atoms has the molecular formula C 8 H (2 × 8) + 2 = C 8 H 18 . Branched Chain Alkanes As noted in previously, the number of isomers increases rapidly as the number of carbon atoms increases. There are 3 pentanes, 5 hexanes, 9 heptanes, and 18 octanes. It would be difficult to assign unique individual names that we could remember. A systematic way of naming hydrocarbons and other organic compounds has been devised by the International Union of Pure and Applied Chemistry (IUPAC). These rules, used worldwide, are known as the IUPAC System of Nomenclature . (Some of the names we used earlier, such as isobutane, isopentane, and neopentane, do not follow these rules and are called common names .) A stem name (Table \(\PageIndex{1}\)) indicates the number of carbon atoms in the longest continuous chain (LCC). Atoms or groups attached to this carbon chain, called substituents , are then named, with their positions indicated by numbers. For now, we will consider only those substituents called alkyl groups. Stem Number meth- 1 eth- 2 prop- 3 but- 4 pent- 5 hex- 6 hept- 7 oct- 8 non- 9 dec- 10 An alkyl group is a group of atoms that results when one hydrogen atom is removed from an alkane. The group is named by replacing the -ane suffix of the parent hydrocarbon with -yl . For example, the -CH 3 group derived from methane (CH 4 ) results from subtracting one hydrogen atom and is called a methyl group . The alkyl groups we will use most frequently are listed in Table \(\PageIndex{2}\). Alkyl groups are not independent molecules; they are parts of molecules that we consider as a unit to name compounds systematically. Parent Alkane Parent Alkane.1 Alkyl Group Alkyl Group.1 Condensed Structural Formula methane NaN methyl NaN CH3– ethane NaN ethyl NaN CH3CH2– propane NaN propyl NaN CH3CH2CH2– NaN NaN isopropyl NaN (CH3)2CH– butane NaN butyl NaN CH3CH2CH2CH2– NaN NaN sec-butyl NaN CH3CH2(CH3)CH- NaN NaN isobutyl NaN -CH2CH(CH3)2 NaN NaN tert-butyl NaN -C(CH3)3 Simplified IUPAC rules for naming alkanes are as follows (demonstrated in Example \(\PageIndex{1}\)). 1. Name alkanes according to the LCC (longest continuous chain) of carbon atoms in the molecule (rather than the total number of carbon atoms). This LCC, considered the parent chain, determines the base name, to which we add the suffix - ane to indicate that the molecule is an alkane. 2. If the hydrocarbon is branched, number the carbon atoms of the LCC. Numbers are assigned in the direction that gives the lowest numbers to the carbon atoms with attached substituents. Hyphens are used to separate numbers from the names of substituents; commas separate numbers from each other. (The LCC need not be written in a straight line; for example, the LCC in the following has five carbon atoms.) 3. Place the names of the substituent groups in alphabetical order before the name of the parent compound. If the same alkyl group appears more than once, the numbers of all the carbon atoms to which it is attached are expressed. If the same group appears more than once on the same carbon atom, the number of that carbon atom is repeated as many times as the group appears. Moreover, the number of identical groups is indicated by the Greek prefixes di -, tri -, tetra -, and so on. These prefixes are not considered in determining the alphabetical order of the substituents. For example, ethyl is listed before dimethyl; the di- is simply ignored. The last alkyl group named is prefixed to the name of the parent alkane to form one word. When these rules are followed, every unique compound receives its own exclusive name. The rules enable us to not only name a compound from a given structure but also draw a structure from a given name. The best way to learn how to use the IUPAC system is to put it to work, not just memorize the rules. It’s easier than it looks. Example \(\PageIndex{1}\) Name each compound. SOLUTION The LCC has five carbon atoms, and so the parent compound is pentane (rule 1). There is a methyl group (rule 2) attached to the second carbon atom of the pentane chain. The name is therefore 2-methylpentane. The LCC has six carbon atoms, so the parent compound is hexane (rule 1). Methyl groups (rule 2) are attached to the second and fifth carbon atoms. The name is 2,5-dimethylhexane. The LCC has eight carbon atoms, so the parent compound is octane (rule 1). There are methyl and ethyl groups (rule 2), both attached to the fourth carbon atom (counting from the right gives this carbon atom a lower number; rule 3). The correct name is thus 4-ethyl-4-methyloctane. Exercise \(\PageIndex{1}\) Name each compound. Example \(\PageIndex{2}\) Draw the structure for each compound. 2,3-dimethylbutane 4-ethyl-2-methylheptane SOLUTION In drawing structures, always start with the parent chain. The parent chain is butane, indicating four carbon atoms in the LCC. Then add the groups at their proper positions. You can number the parent chain from either direction as long as you are consistent; just don’t change directions before the structure is done. The name indicates two methyl (CH 3 ) groups, one on the second carbon atom and one on the third. Finally, fill in all the hydrogen atoms, keeping in mind that each carbon atom must have four bonds. The parent chain is heptane in this case, indicating seven carbon atoms in the LCC. –C–C–C–C–C–C–C– Adding the groups at their proper positions gives Filling in all the hydrogen atoms gives the following condensed structural formulas: Note that the bonds (dashes) can be shown or not; sometimes they are needed for spacing. Exercise \(\PageIndex{2}\) Draw the structure for each compound. 4-ethyloctane 3-ethyl-2-methylpentane 3,3,5-trimethylheptane Key Takeaway Alkanes have both common names and systematic names, specified by IUPAC. |
Bookshelves/Analytical_Chemistry/Instrumental_Analysis_(LibreTexts)/11%3A_Atomic_Mass_Spectrometry | 11.1: General Features of Atomic Mass Spectrometry In mass spectrometry we convert the analyte into ions and then separate these ions based on the ratio of their masses to their charges. In this section we give careful attention to what we mean by mass, by charge, and by mass-to-charge ratio. We also give brief consideration to how we generate and measure ions, topics covered in greater detail in subsequent sections. 11.2: Mass Spectrometers A mass spectrometer has three essential needs: a means for producing ions, in this case (mostly) singly charged atoms; a means for separating these ions in space or in time by their mass-to-charge ratios; and a means for counting the number of ions for each mass-to-charge ratio. 11.3: Inductively Coupled Plasma Mass Spectrometer An inductively coupled plasma in ICP is formed by ionizing a flowing stream of argon gas, producing argon ions and electrons. The sample is introduced into the plasma where the high operating temperature of 6000–8000 K is sufficient to atomize and ionize the sample. In ICP-MS we use the plasma as a source of ions that we can send to a mass spectrometer for analysis. 11.4: Other Forms of Atomic Mass Spectrometry Although ICP-MS is the most widely used method of atomic mass spectrometry, there are other forms of atomic mass spectrometry, three of which we highlight here. |
Courses/University_of_Illinois_Springfield/CHE_124%3A_General_Chemistry_for_the_Health_Professions_(Morsch_and_Andrews)/03%3A_Ionic_Bonding_and_Simple_Ionic_Compounds/3.S%3A_Ionic_Bonding_and_Simple_Ionic_Compounds_(Summary) | To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms and ask yourself how they relate to the topics in the chapter. Atoms combine into compounds by forming chemical bonds . A survey of stable atoms and molecules leads to the octet rule , which says that stable atoms tend to have eight electrons in their outermost, or valence, shell. One way atoms obtain eight electrons in the valence shell is for some atoms to lose electrons while other atoms gain them. When this happens, the atoms take on an electrical charge. Charged atoms are called ions . Ions having opposite charges attract each other. This attraction is called ionic bonding , and the compounds formed are called ionic compounds . Positively charged ions are called cations , while negatively charged ions are called anions . The formation of both cations and anions can be illustrated using electron configurations. Because elements in a column of the periodic table have the same valence shell electron configuration, atoms in the same column of the periodic table tend to form ions having the same charge. Electron dot diagrams , or Lewis diagrams , can also be used to illustrate the formation of cations and anions. Ionic compounds are represented in writing by a chemical formula , which gives the lowest ratio of cations and anions present in the compound. In a formula, the symbol of the cation is written first, followed by the symbol of the anion. Formula unit is considered the basic unit of an ionic compound because ionic compounds do not exist as discrete units. Instead, they exist as crystals , three-dimensional arrays of ions, with cations surrounded by anions and anions surrounded by cations. Chemical formulas for ionic compounds are determined by balancing the positive charge from the cation(s) with the negative charge from the anion(s). A subscript to the right of the ion indicates that more than one of that ion is present in the chemical formula. Some ions are groups of atoms bonded together and having an overall electrical charge. These are called polyatomic ions . Writing formulas with polyatomic ions follows the same rules as with monatomic ions, except that when more than one polyatomic ion is present in a chemical formula, the polyatomic ion is enclosed in parentheses and the subscript is outside the right parenthesis. Ionic compounds typically form between metals and nonmetals or between polyatomic ions. Names of ionic compounds are derived from the names of the ions, with the name of the cation coming first, followed by the name of the anion. If an element can form cations of different charges, there are two alternate systems for indicating the compound’s name. In the Stock system , a roman numeral in parentheses indicates the charge on the cation. An example is the name for FeCl 2 , which is iron(II) chloride. In the common system, the suffixes - ous and - ic are used to stand for the lower and higher possible charge of the cation, respectively. These suffixes are attached to a stem representing the element (which frequently comes from the Latin form of the element name). An example is the common name for FeCl 2 , which is ferrous chloride. The formula mass of an ionic compound is the sum of the masses of each individual atom in the formula. Care must be taken when calculating formula masses for formulas containing multiple polyatomic ions because the subscript outside the parentheses refers to all the atoms in the polyatomic ion. |
Courses/BethuneCookman_University/B-CU%3ACH-331_Physical_Chemistry_I/CH-331_Text/CH-331_Text/06%3A_The_Hydrogen_Atom/6.04%3A_Hydrogen_Atomic_Orbitals_Depend_upon_Three_Quantum_Numbers | Recognize how the hydrogen atomic orbitals vary as a function of the three primary quantum numbers The solutions to the hydrogen atom Schrödinger equation discussed previously are functions that are products of a spherical harmonic function and a radial function . \[ \psi _{n, l, m_l } (r, \theta , \varphi) = \underbrace{R_{n,l} (r)}_{radial} \underbrace{ Y^{m_l}_l (\theta , \varphi)}_{angular} \label {6.1.14} \] The wavefunctions for the hydrogen atom depend upon the three variables \(r\), \(\theta\), and \(\varphi \) and the three quantum numbers \(n\), \(l\), and \(m_l\). The variables give the position of the electron relative to the proton in spherical coordinates. The absolute square of the wavefunction, \(| \psi (r, \theta , \varphi )|^2\), evaluated at \(r\), \(\theta \), and \(\varphi\) gives the probability density of finding the electron inside a differential volume \(d \tau\), centered at the position specified by \(r\), \(\theta \), and \(\varphi\). Evaluate the following integrals \( \langle \psi (r, \theta, \varphi )| \psi (r, \theta , \varphi ) \rangle \nonumber\) \( \langle \psi (r, \theta, \varphi )| \psi (r', \theta' , \varphi' ) \rangle \nonumber\) Answer a. This integral is equal to one since \(\psi(r, \theta, \varphi)\) are normalized eigenstates. b. However, we can explicitly evaluate this integral for any arbitrary pair of eigenstates \[\begin{align*} \langle\psi(r,\theta,\varphi)|\psi(r',\theta',\varphi')\rangle & = \int\limits_{all space}\psi^*(r,\theta,\varphi)\psi(r',\theta',\varphi')d\tau \\[4pt] &=\int\limits_{0}^{\infty} dr \int\limits_{0}^{\pi}d\theta\int\limits_{0}^{2\pi}d\varphi(r^2\sin(\theta))\overbrace{\psi*(r,\theta,\varphi)}^{R_{n,l}(r)Y_{l}^{m_l}(\theta,\varphi)}\overbrace{\psi(r,\theta,\varphi)}^{R_{n',l'}(r)Y_{l'}^{m'_l}(\theta,\varphi)} \\[4pt] &=\int \limits_{0}^{\infty} dr \int\limits_{0}^{\pi}d\theta \int \limits_{0}^{2\pi} d\varphi(r^2\sin(\theta))[R_{n,l}(r)Y_{l}^{m_l}(\theta,\varphi)][R_{n',l'}(r)Y_{l'}^{m'_l}(\theta,\varphi)] \\[4pt] &=\left[\int\limits_{0}^{\infty}r^2[R_{n,l}(r)R_{n',l'}(r)]dr\right]\left[\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\sin(\theta)[Y_{l}^{m_l}(\theta,\varphi)Y_{l'}^{m'_l}(\theta,\varphi)]d\theta d\varphi \right] \\[4pt] &=\langle R_{n,l}(r)|R_{n',l'}(r)\rangle\langle Y_{l}^{m_l}(\theta,\varphi)|Y_{l'}^{m'_l}(\theta,\varphi)\rangle \\[4pt] &=(\delta_{nn'}\delta_{ll'})(\delta_{ll'}\delta_{mm'}) =\delta_{nn'}\delta_{ll'}\delta_{mm'} \end{align*} \] While part a demonstrates normality of the eigenstates, part b demonstrates the orthogonality of the eigenstate (and normality too). The quantum numbers \(n\), \(l\), and \(m_l\) have names: \(n\) is called the principal quantum number , \(l\) is called the angular momentum quantum number , and \(m_l\) is called the magnetic quantum number because the energy in a magnetic field depends upon \(m_l\). Often \(l\) is called the azimuthal quantum number because it is a consequence of the \(\theta\)-equation, which involves the azimuthal angle \(\Theta \), referring to the angle to the zenith. The three quantum numbers have specific values that are dictated by the physical constraints or boundary conditions imposed upon the Schrödinger equation: n must be an integer greater than 0, \(l\) can have the values 0 to \(n‑1\), and \(m_l\) can have \(2l + 1\) values ranging from \(-l\) to \(+l\) in unit or integer steps. The values of the quantum number \(l\) usually are coded by a letter: s means 0, p means 1, d means 2, f means 3; the next codes continue alphabetically (e.g., g means \(l = 4\)). The quantum numbers specify the quantization of physical quantities. The discrete energies of different states of the hydrogen atom are given by n, the magnitude of the angular momentum is given by \(l\), and one component of the angular momentum (usually chosen by chemists to be the z‑component) is given by \(m_l\). The total number of orbitals with a particular value of \(n\) is \(n^2\). Consider several values for \(n\), and show that the number of orbitals for each \(n\) is \(n^2\). Construct a table summarizing the allowed values for the quantum numbers \(n\), \(l\), and \(m_l\) for energy levels 1 through 7 of hydrogen. The notation 3d specifies the quantum numbers for an electron in the hydrogen atom. What are the values for \(n\) and \(l\) ? What are the values for the energy and angular momentum? What are the possible values for the magnetic quantum number? What are the possible orientations for the angular momentum vector? Radial Part of the Wavefunction The asymptotic behavior (i.e., far away from the nucleus) to the radial part of the wavefunction is \[ R_\text{asymptotic} (r) \sim \exp \left(-\dfrac {r}{n} a_0 \right) \label {6.1.15} \] where \(n\) will turn out to be a quantum number and \(a_0\) is the Bohr radius (~52.9 pm). Note that this function decreases exponentially with distance, in a manner similar to the decaying exponential portion of the harmonic oscillator wavefunctions, but with a different distance dependence, \(r\) vs. \(r^2\). What happens to the magnitude of \(R_\text{asymptotic}(r)\) as the distance \(r\) from the proton approaches infinity? Sketch a graph of the function, \(R_\text{asymptotic}(r)\). Why might this behavior be expected for an electron in a hydrogen atom? Answer \[R(r)=e^{-\frac{c}{n} a_{0}} \nonumber \] As \(r\) approaches infinity, the exponential decay goes to zero, this is to be expected as the likelihood of an electron being found at an infinite distance away is almost zero too. The polynomials produced by the truncation of the power series are related to the associated Laguerre polynomials, \(L_n , _l(r)\), where the set of \(c_i\) are constant coefficients. \[L_{n, l} (r) = \sum _{r=0}^{n-l-1} c_i r^i \label {6.1.16} \] These polynomials are identified by two indices or quantum numbers, \(n\) and \(l\). Physically acceptable solutions require that \(n\) must be greater than or equal to \(l +1\). The smallest value for \(l\) is zero, so the smallest value for \(n\) is 1. The angular momentum quantum number affects the solution to the radial equation because it appears in the radial differential equation, (Equation \(\ref{6.1.14}\)). The \(R(r)\) functions that solve the radial differential Equation \(\ref{6.1.14}\), are products of the associated Laguerre polynomials and the exponential factor, multiplied by a normalization factor \((N_{n,l})\) and \(\left (\dfrac {r}{a_0} \right ) ^l\). \[R (r) = N_{n,l} \left ( \dfrac {r}{a_0} \right ) ^l L_{n,l} (r) e^{-\frac {r}{n {a_0}}} \label {6.1.17} \] The decreasing exponential term overpowers the increasing polynomial term so that the overall wavefunction exhibits the desired approach to zero at large values of \(r\). The first six radial functions are provided in Table 6.4.1 . Note that the functions in the table exhibit a dependence on \(Z\), the atomic number of the nucleus. As discussed later in this chapter, other one electron systems have electronic states analogous to those for the hydrogen atom, and inclusion of the charge on the nucleus allows the same wavefunctions to be used for all one-electron systems. For hydrogen, \(Z = 1\). Unnamed: 0 \(n\) \(l\) \(R_{n,l} (\rho)\) \(R_{10}\) 1 0 \(2 \left (\dfrac {Z}{a_0} \right ) ^{3/2} e^{-\rho}\) \(R_{20}\) 2 0 \( \dfrac {1}{2 \sqrt {2}}\left (\dfrac {Z}{a_0} \right ) ^{3/2} (2 - \rho) e^{-\rho/2}\) \(R_{21}\) 2 1 \( \dfrac {1}{2 \sqrt {6}}\left (\dfrac {Z}{a_0} \right ) ^{3/2} \rho e^{-\rho/2}\) \(R_{30}\) 3 0 \( \dfrac {2}{81 \sqrt {3}}\left (\dfrac {Z}{a_0} \right ) ^{3/2} (27 - 18 \rho + 2\rho ^2) e^{-\rho/3}\) \(R_{31}\) 3 1 \( \dfrac {1}{81 \sqrt {6}}\left (\dfrac {Z}{a_0} \right ) ^{3/2} (6 \rho + \rho ^2) e^{-\rho/3}\) \(R_{32}\) 3 2 \( \dfrac {1}{81 \sqrt {30}}\left (\dfrac {Z}{a_0} \right ) ^{3/2} \rho ^2 e^{-\rho/3}\) The constraint that \(n\) be greater than or equal to \(l +1\) also turns out to quantize the energy, producing the same quantized expression for hydrogen atom energy levels that was obtained from the Bohr model of the hydrogen atom. \[ E_n = - \dfrac {\mu e^4}{8 \epsilon ^2_0 h^2 n^2} \nonumber \] It is interesting to compare the results obtained by solving the Schrödinger equation with Bohr’s model of the hydrogen atom. There are several ways in which the Schrödinger and Bohr models differ. First, and perhaps most strikingly, the Schrödinger model does not produce well-defined orbits for the electron. The wavefunctions only give us the probability for the electron to be at various directions and distances from the proton. Second, the quantization of angular momentum is different from that proposed by Bohr. Bohr proposed that the angular momentum is quantized in integer units of \(\hbar\), while the Schrödinger model leads to an angular momentum of \( \sqrt{(l (l +1)} \hbar\). Third, the quantum numbers appear naturally during solution of the Schrödinger equation while Bohr had to postulate the existence of quantized energy states. Although more complex, the Schrödinger model leads to a better correspondence between theory and experiment over a range of applications that was not possible for the Bohr model. Explain how the Schrödinger equation leads to the conclusion that the angular momentum of the hydrogen atom can be zero, and explain how the existence of such states with zero angular momentum contradicts Bohr's idea that the electron is orbiting around the proton in the hydrogen atom. |
Courses/University_of_Wisconsin_Oshkosh/Chem_370%3A_Physical_Chemistry_1_-_Thermodynamics_(Gutow)/01%3A_Thermodynamics/1.10%3A_Extended_Explanations_-_Phase_Equilibria/1.10.03%3A_Gibbs_Energies_and_Phase_Diagrams | First Order Transitions The following plot shows the Gibbs energy as a function of temperature, including phase changes from solid to liquid (melting) and liquid to gas (boiling). Gibbs energy (\(\bar{G}\)) as a function of temperature (\(T\)). Although the \(G\) curve is continuous, its first order derivatives (\(-S\)) is discontinuous at the phase changes. This is why this transition it called a first order transition . We could say that: \(G\) is continuous but has a kink The first order derivatives (\(H\),\(S\),..) are discontinuous (have a jump) The second order derivatives (\(C_P\), ..) have a singularity (go to ∞) Second Order Transitions More subtle transitions where \(G\) is continuous, \(H\) and \(S\) are also continuous but have a kink and the discontinuity is only found in the second order derivatives (such as \(C_P\)) also exist. They are called second order transitions . In such a case: \(G\) is continuous and has no kink The first order derivatives (\(H\),\(S\),..) are continuous (but have a kink) The second order derivatives (\(C_P\), ..) are discontinuous (have a jump) Transition Order Function 1st Order 2nd Order 0 G,A kink smooth 1 H,S,V,.. jump kink 2 CP,CV,α,κ sing. ∞ jump This classification goes back to Ehrenfest. Obviously it based on the question: what order derivative is the first to go discontinuous? Of course we could extend this principle and define third order transitions but there are reasons to be doubtful that such things exist. Another problem is that it is assumed that the order must be integer: 1,2, etc. Is it possible to have a transition of intermediate non-integer order, say 1.3? Although derivatives of fractional order are beyond the scope of the chemistry curriculum the mathematics does exist (Liouville). Schematic comparison of \(G\), \(S\) and \(C_P\) for 1st and 2nd order transitions The Gibbs free energy is a particularly important function in the study of phases and phase transitions. The behavior of \(G(N, P,T)\), particularly as a function of \(P\) and \(T\), can signify a phase transition and can tell us some of the thermodynamic properties of different phases. Consider, first, the behavior of \(G\) vs. \(T\) between the solid and liquid phases of benzene: We immediately notice several things. First, although the free energy is continuous across the phase transition, its first derivative, \(\partial G/\partial T\) is not: The slope of \(G(T)\) in the solid region is different from the slope in the liquid region. When the first derivative of the free energy with respect to one of its dependent thermodynamic variables is discontinuous across a phase transition, this is an example of what is called a first order phase transition . The solid-liquid-gas phase transition of most substances is first order. When the free energy exhibits continuous first derivatives but discontinuous second derivatives, the phase transition is called second order . Examples of this type of phase transition are the order-disorder transition in paramagnetic materials. Now, recall that \[S = -\dfrac{\partial G}{\partial T} \label{13.1}\] Consider the slopes in the solid and liquid parts of the graph: \[\dfrac{\partial G^\text{(solid)}}{\partial T} = -S^\text{(solid)}, \: \: \: \: \: \: \: \dfrac{\partial G^\text{(liquid)}}{\partial T} = -S^\text{(liquid)} \label{13.2}\] However, since \[\dfrac{\partial G^\text{(liquid)}}{\partial T} < \dfrac{\partial G^\text{(solid)}}{\partial T} \label{13.3}\] ( note that the slopes are all negative, and the slope of the liquid line is more negative than that of the solid line ), it follows that \(-S^\text{(liquid)} < -S^\text{(solid)}\) or \(S^\text{(liquid)} > S^\text{(solid)}\). This is what we might expect considering that the liquid phase is higher in entropy than the solid phase. The same argument can be made with regards to the gaseous phase. Similarly, if we consider the dependence of \(G\) on pressure, we obtain a curve like that shown in the figure below: As noted previously, here again, we see that the first derivative of \(\bar{G} (P)\) is discontinuous, signifying a first-order phase transition. Recalling that the average molar volume is \[\bar{V} = \dfrac{\partial \bar{G}}{\partial P} \label{13.4}\] From the graph, we see that the slopes obey \[\bar{V}^\text{(gas)} \gg \bar{V}^\text{(liquid)} > \bar{V}^\text{(solid)} \label{13.5}\] as one might expect for a normal substance like benzene at a temperature above its triple point. Because the temperature is above the triple point, the free energy follows a continuous path (even though it is not everywhere differentiable) from gas to liquid to solid. On the other hand, for water, we see something a bit different, namely, that \[\bar{V}^\text{(gas)} \gg \bar{V}^\text{(solid)} > \bar{V}^\text{(liquid)} \label{13.6}\] at a temperature below the triple point. This, again, indicates, the unusual property of water that its solid phase is less dense than its liquid phase in the coexistence region. Interestingly, if we look at how the plot of \(G(P)\) changes with \(T\), we obtain a plot like that shown below: Below the triple point, it is easy to see from the benzene phase diagram that the system proceeds directly from solid to gas. There is a liquid curve on this plot that is completely disconnected from the gas-solid curve, suggesting that, below the triple point, the liquid state can exist metastably if at all. AT the triple point, the solid can transition into the liquid or gas phases depending on the value of the free energy. Near the critical temperature, we see the liquid-gas transition line, while the solid line is disconnected. Above the critical temperature, the system exists as a supercritical fluid, which is shown on the lower line, and this line now shows derivative discontinuity. Conjugate Variables As discussed before there are many other forms of work possible, such as electrical work, magnetic work or elastic work. These they are commonly incorporated in the formalism of thermodynamics by adding other terms, e.g: \[dG = -SdT + VdP + ℰde + MdH + FdL + γdA \nonumber \] ℰde stands for the electromotoric force ℰ and de the amount of charge transported against it. MdH stand for magnetization and (change in) magnetic field. F stands for the elastic force of e.g. a rubber band dL for the length it is stretched γ stands for the surface tension (e.g. of a soap bubble), A for its surafce area. The terms always appear in a pair of what is known as conjugate variables. That is even clearer if we write out the state function rather than its differential form: \[G = U + PV -TS + ℰe + MH + FL + γA + ... \nonumber \] The PV term can also be generalized -and needs to be so- for a viscous fluid to a stress-strain conjugate pair. It then involves a stress tensor. We will soon encounter another conjugate pair: μdn that deals with changes in composition (n) and the thermodynamic potential μ. |
Courses/Madera_Community_College/Concepts_of_Physical_Science/04%3A_Fluid_Mechanics_and_Waves/4.06%3A_Wave_Mechanics/4.6.02%3A_Simple_Harmonic_Motion-_A_Special_Periodic_Motion | Learning Objectives Describe a simple harmonic oscillator. Explain the link between simple harmonic motion and waves. The oscillations of a system in which the net force can be described by Hooke’s law are of special importance, because they are very common. They are also the simplest oscillatory systems. Simple Harmonic Motion (SHM) is the name given to oscillatory motion for a system where the net force can be described by Hooke’s law, and such a system is called a simple harmonic oscillator . If the net force can be described by Hooke’s law and there is no damping (by friction or other non-conservative forces), then a simple harmonic oscillator will oscillate with equal displacement on either side of the equilibrium position, as shown for an object on a spring in Figure \(\PageIndex{1}\). The maximum displacement from equilibrium is called the amplitude \(X\). The units for amplitude and displacement are the same, but depend on the type of oscillation. For the object on the spring, the units of amplitude and displacement are meters; whereas for sound oscillations, they have units of pressure (and other types of oscillations have yet other units). Because amplitude is the maximum displacement, it is related to the energy in the oscillation. TAKE-HOME EXPERIMENT: SHM AND THE MARBLE Find a bowl or basin that is shaped like a hemisphere on the inside. Place a marble inside the bowl and tilt the bowl periodically so the marble rolls from the bottom of the bowl to equally high points on the sides of the bowl. Get a feel for the force required to maintain this periodic motion. What is the restoring force and what role does the force you apply play in the simple harmonic motion (SHM) of the marble? What is so significant about simple harmonic motion? One special thing is that the period \(T\) and frequency \(f\) of a simple harmonic oscillator are independent of amplitude. The string of a guitar, for example, will oscillate with the same frequency whether plucked gently or hard. Because the period is constant, a simple harmonic oscillator can be used as a clock. The Link between Simple Harmonic Motion and Waves If a time-exposure photograph of the bouncing car were taken as it drove by, the headlight would make a wavelike streak, as shown in Figure \(\PageIndex{2}\). Similarly, Figure \(\PageIndex{3}\) shows an object bouncing on a spring as it leaves a wavelike trace of its position on a moving strip of paper. Both waves are sine functions. All simple harmonic motion is intimately related to sine and cosine waves. The displacement \(x(t)\), the velocity \(v(t)\), and the acceleration \(a(t)\) of simple harmonic motion sketches out characteristic shapes (called sinusoidal functions) as a function of time. You just need to scale them horizontally and vertically, so that they have correct period and amplitude, respectively. Figure \(\PageIndex{4}\) shows the simple harmonic motion of an object on a spring and presents graphs of \(x(t)\),\(v(t)\), and \(a(t)\) versus time. The most important point here is that these relationships are valid for all simple harmonic motion. They are very useful in visualizing waves associated with simple harmonic motion, including visualizing how waves add with one another. Exercise \(\PageIndex{1}\) Suppose you pluck a banjo string. You hear a single note that starts out loud and slowly quiets over time. Describe what happens to the sound waves in terms of period, frequency and amplitude as the sound decreases in volume. Answer Frequency and period remain essentially unchanged. Only amplitude decreases as volume decreases. Exercise \(\PageIndex{2}\) A babysitter is pushing a child on a swing. At the point where the swing reaches \(x\), where would the corresponding point on a wave of this motion be located? Answer \(x\) is the maximum deformation, which corresponds to the amplitude of the wave. The point on the wave would either be at the very top or the very bottom of the curve. Section Summary Simple harmonic motion is oscillatory motion for a system that can be described only by Hooke’s law. Such a system is also called a simple harmonic oscillator. Displacement, velocity, and acceleration in simple harmonic motion as a function of time are represented by sinusoidal functions in time, that is functions that are similar to \(\sin (t)\) and \(\cos (t)\). Glossary amplitude the maximum displacement from the equilibrium position of an object oscillating around the equilibrium position simple harmonic motion the oscillatory motion in a system where the net force can be described by Hooke’s law simple harmonic oscillator a device that implements Hooke’s law, such as a mass that is attached to a spring, with the other end of the spring being connected to a rigid support such as a wall |
Courses/Ripon_College/CHM_321%3A_Inorganic_Chemistry/01%3A_Atomic_Structure_Molecular_Structure_and_Bonding/1.06%3A_Molecular_Orbital_Theory/1.6.03%3A_Homonuclear_Diatomic_Molecules/1.6.3.04%3A_Photoelectron_Spectroscopy | A photoelectron spectrum can show the relative energies of occupied molecular orbitals by ionization. The ionization energy is a direct measure of the energy required to just remove the electron concerned from its initial level to the vacuum level (free electron). Photoelectron spectroscopy measures the relative energies of the ground and excited positive ion states that are obtained by removal of single electrons from the neutral molecule. \[A + \text{photon} \rightarrow A^+ + e^- \nonumber \] The information obtained from photoelectron spectroscopy is typically discussed in terms of the electronic structure and bonding in the ground states of neutral molecules, with ionization of electrons occurring from bonding molecular orbitals, lone pairs, antibonding molecular orbitals, or atomic cores. These descriptions reflect the relationship of ionization energies to the molecular orbital model of electronic structure. Ionization energies are directly related to the energies of molecular orbitals (by Koopmans' theorem ). Example: Photoelectron spectrum of dihydrogen The molecular orbital description of dihydrogen involves two \(1s\) atomic orbitals generating two molecular orbitals: a bonding \(\sigma_g\) and an antibonding \(\sigma_u^*\). The two electrons occupy the \(\sigma_g\) bonding orbital, leaving the molecule with a bond order of one (Figure \(\PageIndex{1}\)). The PES spectrum of dihydrogen (Figure \(\PageIndex{1}\)) has a single band that corresponds to the ionization of one electron from the \(\sigma_g\). The multiple peaks are due to electrons ejecting from a range of stimulated vibrational energy levels. When extensive vibrational structure is resolved in a PES molecular orbital, then the removal of an electron from that molecular orbital induces a significant change in the bonding (in this case an increase in the bond length due to decrease in bond order). Example: Photoelectron spectrum of dinitrogen Diatomic nitrogen is more complex than hydrogen since multiple molecular orbitals are occupied. Five molecular orbitals are occupied; two of them are degenerate. Three bands in the photoelectron spectrum correspond to ionization of an electron in \(\sigma_g(2p)\), \(\pi_u(2p)\) and \(\sigma_u^*(2s)\) molecular orbitals. Ionization of the fourth type of orbital, \(\sigma_g(2s)\), does not appear in Figure \(\PageIndex{2}\) because it is either off scale or because the incident light \(h\nu\) used did not have sufficient energy to ionize electrons in that deeply stabilized molecular orbital. Note that extensive vibrational structure for the \(\pi_u(2p)\) band indicates that the removal of an electron from this molecular orbital causes a significant change in the bonding. From the photoelectron spectrum of dinitrogen, we can see that the electrons in the \(\sigma_g(2p)\) orbital can be ionized using less energy than required to ionize electrons in the \(\pi_u(2p)\) orbital. This is evidence for \(\sigma_g(2p)\) existing at a higher energy than the \(\pi_u(2p)\) orbitals. |
Courses/Lebanon_Valley_College/CHM_312%3A_Physical_Chemistry_II_(Lebanon_Valley_College)/07%3A_Chemical_Equilibrium/7.02%3A_Reaction_Quotient_and_Equilibrium_Constant | Let’s consider a prototypical reaction at constant \(T,P\): \[ a\mathrm{A} + b\mathrm{B} \rightarrow c\mathrm{C} + d\mathrm{D} \label{10.1.1} \] The Gibbs free energy of the reaction is defined as: \[ \Delta_{\text{rxn}} G = G_{\text{products}} - G_{\text{reactants}} = G^{\text{C}} + G^{\text{D}} - G^{\text{A}}-G^{\text{B}}, \label{10.1.2} \] and replacing the absolute Gibbs free energies with the chemical potentials \(\mu_i\), we obtain: \[ \Delta_{\text{rxn}} G = c \mu_{\text{C}} + d \mu_{\text{D}} - a \mu_{\text{A}}- b\mu_{\text{B}}. \label{10.1.3} \] Assuming the reaction is happening in the gas phase, we can then use Equation 9.4.6 to replace the chemical potentials with their value in the reaction mixture, as: \[\begin{equation} \begin{aligned} \mkern-60mu \Delta_{\text{rxn}} G =& \; c (\mu_{\text{C}}^{-\kern-6pt{\ominus}\kern-6pt-}+RT \ln P_{\text{C}}) + d (\mu_{\text{D}}^{-\kern-6pt{\ominus}\kern-6pt-}+RT \ln P_{\text{D}}) - a (\mu_{\text{A}}^{-\kern-6pt{\ominus}\kern-6pt-}+RT \ln P_{\text{A}}) - b (\mu_{\text{B}}^{-\kern-6pt{\ominus}\kern-6pt-}+RT \ln P_{\text{B}}) \\[4pt] =& \; \underbrace{c \mu_{\text{C}}^{-\kern-6pt{\ominus}\kern-6pt-}+ d \mu_{\text{D}}^{-\kern-6pt{\ominus}\kern-6pt-}- a \mu_{\text{A}}^{-\kern-6pt{\ominus}\kern-6pt-}- b\mu_{\text{B}}^{-\kern-6pt{\ominus}\kern-6pt-}}_{\Delta_{\text{rxn}} G^{-\kern-6pt{\ominus}\kern-6pt-}} +RT \ln \dfrac{P_{\text{C}}^c \cdot P_{\text{D}}^d}{P_{\text{A}}^a \cdot P_{\text{B}}^b}. \end{aligned} \end{equation} \label{10.1.4} \] We can define a new quantity called the reaction quotient as a function of the partial pressures of each substance:\(^1)\) \[ Q_P = \dfrac{P_{\text{C}}^c \cdot P_{\text{D}}^d}{P_{\text{A}}^a \cdot P_{\text{B}}^b}, \label{10.1.5} \] and we can then simply rewrite Equation \ref{10.1.4} using Equation \ref{10.1.5} as: \[ \Delta_{\text{rxn}} G = \Delta_{\text{rxn}} G^{-\kern-6pt{\ominus}\kern-6pt-}+ RT \ln Q_P. \label{10.1.6} \] This equation tells us that the sign of \(\Delta_{\text{rxn}} G\) is influenced by the reaction quotient \(Q_P\). For a spontaneous reaction at the beginning, the partial pressures of the reactants are much higher than the partial pressures of the products, therefore \(Q_P \ll 1\) and \(\Delta_{\text{rxn}} G < 0\), as we expect. As the reaction proceeds, the partial pressures of the products will increase, while the partial pressures of the reactants will decrease. Consequently, both \(Q_P\) and \(\Delta_{\text{rxn}} G\) will increase. The reaction will completely stop when \(\Delta_{\text{rxn}} G = 0\), which is the chemical equilibrium point. At the reaction equilibrium: \[ \Delta_{\text{rxn}} G = 0 = \Delta_{\text{rxn}} G^{-\kern-6pt{\ominus}\kern-6pt-}+ RT \ln K_P, \label{10.1.7} \] where we have defined a new quantity called equilibrium constant , as the value the reaction quotient assumes when the reaction reaches equilibrium, and we have denoted it with the symbol \(K_P\).\(^2\) From Equation \ref{10.1.7} we can derive the following fundamental equation on the standard Gibbs free energy of reaction: \[ \Delta_{\text{rxn}} G^{-\kern-6pt{\ominus}\kern-6pt-}= - RT \ln K_P. \label{10.1.8} \] To extend the concept of \(K_P\) beyond the four species in the prototypical reaction (10.1), we can use the product of a series symbol \(\left( \prod_i \right)\), and write: \[ K_P=\prod_i P_{i,\text{eq}}^{\nu_i}, \label{10.1.9} \] where \(P_{i,\text{eq}}\) are the partial pressure of each species at equilibrium. Eq. (10.1.9) is in principle valid for ideal gases only. However, reaction involving ideal gases are pretty rare. As such, we can further extend the concept of equilibrium constant and write: \[ K_{\text{eq}} =\prod_i a_{i,\text{eq}}^{\nu_i}, \label{10.1.10} \] where we have replaced the partial pressure at equilibrium, \(P_{i,\text{eq}}\), with a new concept introduced initially by Gilbert Newton Lewis (1875–1946),\(^3\) that he termed activity , and represented by the letter \(a\). For ideal gases, it is clear that \(a_i=P_i/P^{-\kern-6pt{\ominus}\kern-6pt-}\). For non-ideal gases, the activity is equal to the fugacity \(a_i=f_i/P^{-\kern-6pt{\ominus}\kern-6pt-}\), a concept that we will investigate in the next chapter. For pure liquids and solids, the activity is simply \(a_i=1\). For diluted solutions, the activity is equal to a measured concentration (such as, for example, the mole fraction \(x_i\) in the liquid phase, and \(y_i\) in the gas phase, or the molar concentration \([i]/[i]^{-\kern-6pt{\ominus}\kern-6pt-}\) with \([i]^{-\kern-6pt{\ominus}\kern-6pt-}= 1\;\text[mol/L]\)). Finally for concentrated solutions, the activity is related to the measured concentration via an activity coefficient. We will return to the concept of activity in chapter 14, when we will specifically deal with solutions. For now, it is interesting to use the activity to write the definition of the following two constants: \[ K_y =\prod_i \left( y_{i,\text{eq}} \right)^{\nu_i} \qquad \qquad \qquad \qquad K_C =\left( \prod_i [i]_{\text{eq}}/[i]^{-\kern-6pt{\ominus}\kern-6pt-}\right)^{\nu_i}, \label{10.1.11} \] which can then be related with \(K_P\) for a mixture of ideal gases using: \[ P_i = y_i P \qquad \qquad \qquad P_i=\dfrac{n_i}{V}RT=[i]RT, \label{10.1.12} \] which then results in: \[ K_P = K_y\cdot \left(\dfrac{P}{P^{-\kern-6pt{\ominus}\kern-6pt-}}\right)^{\Delta \nu} \qquad \qquad K_P = K_C \left( \dfrac{[i]^{-\kern-6pt{\ominus}\kern-6pt-}RT}{P^{-\kern-6pt{\ominus}\kern-6pt-}} \right)^{\Delta \nu}, \label{10.1.13} \] with \(\Delta \nu =\sum_i \nu_i\). Using the general equilibrium constant, \(K_{\text{eq}}\), we can also rewrite the fundamental equation on \(\Delta_{\text{rxn}} G^{-\kern-6pt{\ominus}\kern-6pt-}\) that we derived in Equation \ref{10.1.8} to be applicable at most conditions, as: \[ \Delta_{\text{rxn}} G^{-\kern-6pt{\ominus}\kern-6pt-}= - RT \ln K_{\text{eq}}, \label{10.1.14} \] and since \(\Delta_{\text{rxn}} G^{-\kern-6pt{\ominus}\kern-6pt-}\) depends on \(T,P\) and \(\{n_i\}\), it is useful to explore how \(K_{\text{eq}}\) depends on those variables as well. Notice that since we used Equation 9.4.5 to derive the reaction quotient, the partial pressures inside it are always dimensionless since they are divided by \(P^{-\kern-6pt{\ominus}\kern-6pt-}\).︎ The subscript \(P\) refers to the fact that the equilibrium constant is measured in terms of partial pressures.︎ Gilber Lewis is the same scientist that invented the concept of Lewis Structures. |
Courses/San_Diego_Miramar_College/Chem_103%3A_Fundamentals_of_GOB_Chemistry_(Garces)/07%3A_Stoichiometry/7.10%3A_Chemical_Formulas_as_Conversion_Factors | Learning Objectives Use chemical formulas as conversion factors. Figure \(\PageIndex{1}\) shows that we need 2 hydrogen atoms and 1 oxygen atom to make one water molecule. If we want to make two water molecules, we will need 4 hydrogen atoms and 2 oxygen atoms. If we want to make five molecules of water, we need 10 hydrogen atoms and 5 oxygen atoms. The ratio of atoms we will need to make any number of water molecules is the same: 2 hydrogen atoms to 1 oxygen atom. Using formulas to indicate how many atoms of each element we have in a substance, we can relate the number of moles of molecules to the number of moles of atoms. For example, in 1 mol of water (H 2 O) we can construct the relationships given in (Table \(\PageIndex{1}\)). 1 Molecule of \(H_2O\) Has 1 Mol of \(H_2O\) Has Molecular Relationships 2 H atoms 2 mol of H atoms \(\mathrm{\dfrac{2\: mol\: H\: atoms}{1\: mol\: H_2O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: H_2O\: molecules}{2\: mol\: H\: atoms}}\) 1 O atom 1 mol of O atoms \(\mathrm{\dfrac{1\: mol\: O\: atoms}{1\: mol\: H_2O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: H_2O\: molecules}{1\: mol\: O\: atoms}}\) The Mole is big A mole represents a very large number! The number 602,214,129,000,000,000,000,000 looks about twice as long as a trillion, which means it’s about a trillion trillion. (CC BY-SA NC; what if? [what-if.xkcd.com] ). A trillion trillion kilograms is how much a planet weighs. If 1 mol of quarters were stacked in a column, it could stretch back and forth between Earth and the sun 6.8 billion times. 1 Molecule of \(C_2H_6O\) Has 1 Mol of \(C_2H_6O\) Has Molecular and Mass Relationships 2 C atoms 2 mol of C atoms \(\mathrm{\dfrac{2\: mol\: C\: atoms}{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{2\: mol\: C\: atoms}}\) 6 H atoms 6 mol of H atoms \(\mathrm{\dfrac{6\: mol\: H\: atoms}{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{6\: mol\: H\: atoms}}\) 1 O atom 1 mol of O atoms \(\mathrm{\dfrac{1\: mol\: O\: atoms}{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{1\: mol\: O\: atoms}}\) 2 (12.01 amu) C 24.02 amu C 2 (12.01 g) C 24.02 g C \(\mathrm{\dfrac{24.02\: g\: C\: }{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{24.02\: g\: C\: }}\) 6 (1.008 amu) H 6.048 amu H 6 (1.008 g) H 6.048 g H \(\mathrm{\dfrac{6.048\: g\: H\: }{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{6.048\: g\: H\: }}\) 1 (16.00 amu) O 16.00 amu O 1 (16.00 g) O 16.00 g O \(\mathrm{\dfrac{16.00\: g\: O\: }{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{16.00\: g\: O\: }}\) The following example illustrates how we can use the relationships in Table \(\PageIndex{2}\) as conversion factors. Example \(\PageIndex{1}\): Ethanol If a sample consists of 2.5 mol of ethanol (C 2 H 6 O), how many moles of carbon atoms does it have? Solution Steps for Problem Solving If a sample consists of 2.5 mol of ethanol (C2H6O), how many moles of carbon atoms does it have? Identify the "given" information and what the problem is asking you to "find." Given: 2.5 mol C2H6O Find: mol C atoms List other known quantities. 1 mol C2H6O = 2 mol C Prepare a concept map and use the proper conversion factor. NaN Cancel units and calculate. Note how the unit mol C2H6O molecules cancels algebraically. \(\mathrm{2.5\: \cancel{mol\: C_2H_6O\: molecules}\times\dfrac{2\: mol\: C\: atoms}{1\: \cancel{mol\: C_2H_6O\: molecules}}=5.0\: mol\: C\: atoms}\) Think about your result. There are twice as many C atoms in one C2H6O molecule, so the final amount should be double. Exercise \(\PageIndex{1}\) If a sample contains 6.75 mol of Na 2 SO 4 , how many moles of sodium atoms, sulfur atoms, and oxygen atoms does it have? Answer 13.5 mol Na atoms, 6.75 mol S atoms, and 27.0 mol O atoms The fact that 1 mol equals 6.022 × 10 23 items can also be used as a conversion factor. Example \(\PageIndex{2}\): Oxygen Mass Determine the mass of Oxygen in 75.0g of C 2 H 6 O. Solution Steps for Problem Solving Determine the mass of Oxygen in 75.0g of C2H6O Identify the "given" information and what the problem is asking you to "find." Given: 75.0g C2H6O Find: g O List other known quantities. 1 mol O = 16.0g O 1 mol C2H6O = 1 mol O 1 mol C2H6O = 46.07g C2H6O Prepare a concept map and use the proper conversion factor. The conversion factors are 1 mol C2H6O over 46.07 g C2H6O, 1 mol O over 1 mol C2H6O, and 16.00 g O over 1 mole O. Cancel units and calculate. \(\require{cancel}\mathrm{75.0\: \cancel{g\: C_2H_6O}\times\dfrac{1\: \cancel{mol\: C_2H_6O}}{46.07\:\cancel{g\: C_2H_6O}}\times\dfrac{1\: \cancel{mol\:O}}{1\: \cancel{mol\:C_2H_6O}}\times\dfrac{16.00\: g\: O}{1\: \cancel{mol\:O}}=26.0\: g\: O}\) Think about your result. NaN Exercise \(\PageIndex{2}\) How many molecules are present in 16.02 mol of C 4 H 10 ? How many C atoms are in 16.02 mol? How many moles of each type of atom are in 2.58 mol of Na 2 SO 4 ? Answer a: 9.647 x 10 24 C 4 H 10 molecules and 3.859 x 10 25 C atoms Answer b: 5.16 mol Na atoms, 2.58 mol S atoms, and 10.3 mol O atoms Summary In any given formula, the ratio of the number of moles of molecules (or formula units) to the number of moles of atoms can be used as a conversion factor. |
Courses/Georgian_College/Chemistry_-_Academic_and_Career_Preparation/02%3A_Trends_and_Bonding/2.03%3A_Periodic_Trends-_Ionization_Energy | The incredible green lights in this cold northern sky consist of charged particles known as ions. Their swirling pattern is caused by the pull of Earth’s magnetic field. Called the northern lights, this phenomenon of nature shows that ions respond to a magnetic field. Do you know what ions are? Read on to find out. Atoms Are Neutral The northern lights aren’t caused by atoms, because atoms are not charged particles. An atom always has the same number of electrons as protons. Electrons have an electric charge of -1 and protons have an electric charge of +1. Therefore, the charges of an atom’s electrons and protons “cancel out.” This explains why atoms are neutral in electric charge. Q: What would happen to an atom’s charge if it were to gain extra electrons? A: If an atom were to gain extra electrons, it would have more electrons than protons. This would give it a negative charge, so it would no longer be neutral. Atoms to Ions Atoms cannot only gain extra electrons. They can also lose electrons. In either case, they become ions . Ions are atoms that have a positive or negative charge because they have unequal numbers of protons and electrons. If atoms lose electrons, they become positive ions, or cations. If atoms gain electrons, they become negative ions, or anions. Consider the example of fluorine (see figure below). A fluorine atom has nine protons and nine electrons, so it is electrically neutral. If a fluorine atom gains an electron, it becomes a fluoride ion with an electric charge of -1. Names and Symbols Like fluoride, other negative ions usually have names ending in –ide . Positive ions, on the other hand, are just given the element name followed by the word ion . For example, when a sodium atom loses an electron, it becomes a positive sodium ion. The charge of an ion is indicated by a plus (+) or minus sign (-), which is written to the right of and just above the ion’s chemical symbol. For example, the fluoride ion is represented by the symbol F - , and the sodium ion is represented by the symbol Na + . If the charge is greater than one, a number is used to indicate it. For example, iron (Fe) may lose two electrons to form an ion with a charge of plus two. This ion would be represented by the symbol Fe 2 + . This and some other common ions are listed with their symbols in the table below. Cations Unnamed: 1 Anions Unnamed: 3 Name of Ion Chemical Symbol Name of Ion Chemical Symbol Calcium ion Ca2+ Chloride Cl- Hydrogen ion H+ Fluoride F- Iron(II) ion Fe2+ Bromide Br- Iron(III) ion Fe3+ Oxide O2- Q: How does the iron(III) ion differ from the iron(II) ion? A: The iron(III) ion has a charge of +3, so it has one less electron than the iron(II) ion, which has a charge of +2. Q: What is the charge of an oxide ion? How does its number of electrons compare to its number of protons? A: An oxide ion has a charge of -2. It has two more electrons than protons. How Ions Form The process in which an atom becomes an ion is called ionization. It may occur when atoms are exposed to high levels of radiation. The radiation may give their outer electrons enough energy to escape from the attraction of the positive nucleus. However, most ions form when atoms transfer electrons to or from other atoms or molecules. For example, sodium atoms may transfer electrons to chlorine atoms. This forms positive sodium ions (Na + ) and negative chloride ions (Cl - ). Q: Why do you think atoms lose electrons to, or gain electrons from, other atoms? A: Atoms form ions by losing or gaining electrons because it makes them more stable and this state takes less energy to maintain. The most stable state for an atom is to have its outermost energy level filled with the maximum possible number of electrons. In the case of metals such as lithium, with just one electron in the outermost energy level, a more stable state can be achieved by losing that one outer electron. In the case of nonmetals such as fluorine, which has seven electrons in the outermost energy level, a more stable state can be achieved by gaining one electron and filling up the outer energy level. Properties of Ions Ions are highly reactive, especially as gases. They usually react with ions of opposite charge to form neutral compounds. For example, positive sodium ions and negative chloride ions react to form the neutral compound sodium chloride, commonly known as table salt. This occurs because oppositely charged ions attract each other. Ions with the same charge, on the other hand, repel each other. Ions are also deflected by a magnetic field, as you saw in the opening image of the northern lights. Summary Atoms have equal numbers of positive protons and negative electrons, so they are neutral in electric charge. Atoms can gain or lose electrons and become ions, which are atoms that have a positive or negative charge because they have unequal numbers of protons and electrons. The process in which an atom becomes an ion is called ionization. It may occur when atoms are exposed to high levels of radiation or when atoms transfer electrons to or from other atoms. Ions are reactive, attracted or repulsed by other charged particles, and deflected by a magnetic field. Review Why are atoms neutral in electric charge? Define ion. Compare and contrast cations and anions, and give an example of each. Describe how ions form. List properties of ions. The model in the illustration below represents an atom of lithium (Li). If the lithium atom becomes an ion, which type of ion will it be, a cation or an anion? What will be the electric charge of this ion? What will the ion be named? What symbol will be used to represent it? |
Courses/Nassau_Community_College/Organic_Chemistry_I_and_II/08%3A_Reactions_of_Alkenes/8.07%3A_Stereochemistry_of_Reactions_-_Hydration_of_Achiral_Alkenes | Learning Objective discern the stereochemical differences between the EAR of chiral and achiral alkenes Stereochemistry and the Subtle Details Organic reactions in the laboratory or in living systems can produce chiral centres. Consider reaction of 1-pentene with water (acid catalyzed). Markovnikov regiochemistry occurs and the OH adds to the second carbon. However, both R and S products occur giving a racemic (50/50) mixture of 2‑butanol. How does this occur? The proton addition to 1‑pentene results in a planar carbocation intermediate. A molecule of water is then equally likely to react from the top or the bottom of this cation to produce either (S)‑2‑pentanol or (R)‑2‑pentanol, respectively, as shown in the mechanism below. |
Courses/University_of_Arkansas_Little_Rock/Chem_1402%3A_General_Chemistry_1_(Belford)/Text/1.B%3A_Review_of_the_Tools_of_Quantitative_Chemistry/1B.1%3A_Units_of_Measurement | Learning Objectives Recognize different ways to quantity matter (measured, exact, ande defined numbers) Recognize English Units and Metric Units and SI units Understand the evolution of the definitions of units over time and their uses in society Convert between different units Quantification of Matter Scientists use three basic types of numbers when quantifying matter; counted numbers, measured numbers and defined numbers. Counted Numbers: A counted number is an exact number in the sense that you have an entity, and you count the number of entities. So an exact number has two parts, the entity, and the number of entities. If you have 16 rocks in your hand, you have sixteen rocks, not more, not less. If you break one of the rocks into two, you now have 17 rocks in your hand, not more, not less. These are exact numbers. Counted numbers are exact and there is no uncertainty about their value. Measured Numbers: Now the mass of the rocks is a measured number and requires a third aspect to describe, which is the unit, defined by the scale used to measure it. So a measured number has 3 parts; magnitude, unit and entity. Sixteen one pound rocks weighs less than one 500 pound rock, although the number 16 is more than the number one, and this is because different units are being used to describe the mass. If you break each of the sixteen one pound rocks in half, you have 32 rocks, but they still weigh the same. Measured numbers have uncertainty that is indicated by their number of significant digits ( section 1B.2.2 ). Defined Numbers: A defined number has a value inherent in its definition, and like a counted number is an exact number, but it may, or may not have uncertainty. Defined numbers are often used in measurements, for example, twelve inches is defined to be one foot and there is no uncertainty in that equivalence. 12inches ≡ 1foot But not all defined numbers are integers, like the number \(\pi\), which is defined to be ratio of the circumference to the diameter of a circle. \(\pi\)≡\(\frac{circumference}{diameter}\) \(\pi\) is an exact number in the sense that it is exactly the ratio of the circumference to the diameter of a circle, but it is an irrational number and like a measured number, when we write down the value of \(\pi\) we give it a precision (3.14, 3.142, 3.1416, 3.14159,...), which defines how precisely we express the number. Understanding defined numbers is of importance, and as we shall learn, as of May 20, 2019, the basic units of the metric system became based on defined numbers ( 2019 SI unit definitions ). Metrology Metrology is the science of measurement. There are two fundamental tenets to metrology, traceability and uncertainty. Traceability is the ability to relate a measurement to a standard through calibrations, where each calibration has an element of uncertainty, which relates to how precisely the measurement (calibration) was made (see Section 1B.2 ). Inherent in a standard are the definition upon which it is based, and the "method" used to put this definition into practice, which is often known by the French term "mise en pratique". In modern metrology the "mise en pratique" is a set of instructions that allows implementation of the standard into the practice of measurement, and results in the traceability and uncertainty of a measurement to the standard. The earliest standards were related to physical objects, and from these, tools like a ruler could be created that allowed the measurement of other objects in the unit of the standard. Early units included body parts, like the foot, which varied from person to person, and so a standard foot was required. A multitude of units could describe the same measurement (see Figure 1B1.1 for different ways to measure length). 0 1 NaN NaN Figure \(\PageIndex{1}\): On the left are some of the units to describe length, some of which were based on human body parts (foot) or actions (pace). Figure \(\PageIndex{1}\): On the left are some of the units to describe length, some of which were based on human body parts (foot) or actions (pace). Today science uses the metric system. In fact most countries in the world use the metric system as the normal units of measurement. The customary units of measurement in the US are the foot, inch and pound, even though most of the rest of the world uses the metric system, as does the scientific community. What may be surprising for most students is that a year after the civil war in 1866 the U.S. congress passed the metric act, which made the metric system legal for purpose of commerce. Metric Units Early Definitions The Metric unit system evolved out of the French Revolutionary War, and was originally based on a unit of length, the meter. The earliest definition was the meridional definition, defining the meter as one ten-millionth of the distance from the north pole to the equator along the meridian going through Paris. This was done by surveying the distance from a belfry in Dunkirk France to Montjuic castle in Barcelona Spain, which was assumed to be one tenth the length of the Paris meridian. While the meridional metre survey was being conducted a series of prototype platinum bars were developed, and in 1799 the one that was closest to the meridional meter was chosen to be the standard meter (Wikipedia: History of the Metre ). It should be noted that the original metric system was based entirely on this length, where the unit of mass was the gram, which was defined as the mass of a cube of water at 4 o C that was one hundredth of a meter on each side (1 cm 3 ). As this unit was very small compared to items that would be used in commerce, a prototype artifact of pure platinum was created that weighed 1000g, which became the first standard kilogram. This lasted until 1875 and the Treaty of the Meter. Property Unit Symbol length meter m mass gram g volume liter L Note, one of the advantages of the metric system is the decimalization of the magnitude of units. This allowed easy conversion from small to large scale measurements. To go from centimeters to kilometers you simply multiple by 10,000, which is much easier to do than to convert the similar English units of inches to miles (see section 1B.3.5 Unit Conversions). Treaty of the Meter On May 20, 1875 the US was one of 17 nations to sign the "Treaty of the Meter" (the Metric Convention), which established the "Bureau International des Poids et Mesures" ( BIPM ), that was tasked to develop an international unified system of weights and measures. Through the BIPM's ongoing " Conférence Générale des Poids et Mesures " ( CGPM) , the standards continue to be revised and redefined to the present time. In addition to defining the standards the BIPM is also responsible for producing the " mise en pratique," which are published online . As a result of the metric convention 31 prototype meter standards of platinum/iridium were built, and "National Prototype standard 27" was the U.S. reference standard for length until 1983. The mass standard was now an iridium/platinum alloy kept in a vault in Sèvres France, under conditions specified by the Treaty of the Meter. Over time the metric unit system for measurement evolved into the SI or International System of Units. International System of Units In 1921 during the 6 th meeting of the CGPM the scope and responsibilities of the BIPM was extended into other fields of physics (like temperature and time). In 1960 during during the 11 th meeting of the CGPM, the International System of Units was established, and the system now officially became known as the SI system for its French spelling, Système international d'unités . The SI units are commonly called the metric units, they are the units used in scientific publications and as we shall see, are continually being revised and redefined through meetings of the CGPM. There are seven SI base units that underpin all SI measurements. Through proper combinations they can describe other measured phenomena, like volume and energy, whose units of liter and joule are classified as derived SI units ( section 1B.3.3.2 ). SI Base Units The seven SI base units are listed in Table \(\PageIndex{2}\). These can be combined to form derived units that describe any measurable physical quantity. Thus by defining these seven SI base units, any physical measurement can be reproduced. In this class we will use the first five units extensively; the meter, kilogram, second, kelvin and mole (and general chemistry 2 will also use the ampere). 0 1 2 length meter m mass kilogram kg time second s temperature (absolute) kelvin K amount of substance mole mol electric current ampere A luminous intensity candela cd Note a kilogram is 1,000 g (SI prefixes section 1B.3.3.3 ), which is roughly the volume of one liter of water (with the liter being a derived unit for volume). The kilogram is the only SI base unit with an SI prefix. A few special points about some of the SI base units are worth noting The base unit of mass is unique in that an SI Prefixe is built into it (k for 1,000); i.e., the base SI unit is not the gram, but the Kilogram . This historically occurred because a gram is a real small quantity, and did not represent the magnitude of mass used in normal daily activities or commerce. Although the kilogram has an SI prefix in the base unit, you do not use Si prefixes on the kg, but on the gram (10 9 g is a gigagram, not a megakilogram). The base unit of time is the second, and time is not expressed as a decimalized metric number. Numerous attempts to make it so have never garnered any success; we are still stuck with the 24:60:60 system that we inherited from ancient times. The ancient Egyptians of around 1500 BC invented the 12-hour day, and the 60:60 part is a remnant of the base-60 system that the Sumerians used for their astronomical calculations around 100 BCE. Of special interest to Chemistry is the mole , the base unit for expressing the quantity of matter . It is now defined as exactly 6.02214076 x 10 23 entities of anything. 1960-2019 Definitions The following SI base unit definitions are adapted from NIST Special Publication 330 as obtained from the website, NIST Reference on Constants, Units and Uncertainty . 0 1 meter The distance of length traveled by light in a vacuum over the time interval of 1/299,792,458 of a second. (Note the speed of light in a vacuum is 299,792,458m/sec kilogram The mass the international iridium/platinum prototype kept in Paris France. second The duration of 9,192,631,770 periods of radiation corresponding to transitions on the cesium 133 atom (atomic clock) kelvin 1/273.16 of the thermodynamic temperature of the triple point of water mole the number of atoms in 12 grams of carbon-12 (the symbol is mol) ampere constant current maintained by two straight parallel conductors of infinite length in a vacuum that are 1 meter apart and produce between them a force of 2x10-7N/m candela the luminous intensity of a source of monochromatic radiation in a direction with a frequency of 540x1012 hertz in that direction with a flux of 1/683 watt per steradian. The definitions in Table \(\PageIndex{3}\) are based on both artifacts and constants/phenomena. For example, the meter was originally based on an artifact in France, but in 1983 was changed to being defined by the speed of light (c) in a vacuum. In 1967 the second became defined by the cesium-133 atomic clock. In 1971 the mole was defined by the number of atoms in 12 grams of carbon-12, which required a sample of carbon-12. In 1954 the kelvin was defined by the triple point of water (a concept covered in general chemistry 2, section 12.7.3 ), which requires a sample of water (that has variable isotopic composition, a topic that will be on the first exam, section 2.3 ). But throughout this time, the definition of the kilogram never changed, it was still based on the standard kilogram in France. There are some problems with the above definitions, the most obvious of which is the artifact-based definition of the kilogram. At the time of the creation of the original standard there were also "witness" standards made of identical mass, many of which became national prototypes. Over the years they would on occasion have their masses verified, and were shown to be changing (fig. 1B.1.6). The problem is, if a witness standard is getting "heavier", no one really know if it is really heavier, or if the prototype was getting lighter. 2019 Definitions On May 20, 2019 the definitions of the SI Base units were changed, and instead of defining 7 base units, seven constants that are considered to be invariants of nature are defined as exact numbers with no uncertainty. These constants are defined numbers , and this is a fundamental change as what is being defined are constants of nature and not the base units. The base unit is then obtained by applying the constant to the "mise en pratique" . To do this, scientists first had to be able to measure these constants very accurately, and once done, they defined the constant based on the measurement as an exact (defined) number. Then to apply it they sort of reverse engineered the process of measuring the constant, with the constant now being treated as a defined (exact) number, and could measure the value of a sample based on the defined constant. This sounds sort of complicated, but the process is actually easy to understand if we look at a constant we are familiar with, like density, the ratio of the mass of a substance to its volume. For an incompressible substance like water the density can be considered a constant, and if you define the density of water to be 1 g/ml as an exact definition, and have a device to measure volume, you can then use that to determine the mass. So the mass is determined by the defined value of the density as 1 g/mL (an exact number by definition). Of course the density of water as a constant is not an invariant of nature, and changes as temperature, pressure and other factors change. But the principle is the same. Many of these constants will be encountered as we proceed through the material of this class and although students need to know the SI base units, they do not need to be concerned with deriving the SI base units from the 7 defined constants. It is hoped that by the time this Chapter is finished students can understand the logic of the new definitions, and use units in their calculations. The key to this is that the constants must have the units of the base unit that they describe (density has units of mass over volume, so could be used to describe units of mass or volume). When you look at the card in Figure 1B.1.7 you see units like J/Hz for Planck's constant. Those are actually SI derived units ( section 1B.3.2 ), and Planck's constant can be described in units of \(\frac{kg\cdot m^{2}}{s}\), so it could be used to define the SI base units of Kilogram, meter and second. In sections 1B.3 Mathematics in Chemistry and section 1B.4 Dimensional Analysis, students will learn how to use units in calculations, and these type of math skills are what is required to mathematically relate the fundamental constants to the SI base units. Optional: A Deeper Look To get a better understanding of the change in definitions it may be prudent to look at the respective values of the new and old definitions. This information is obtained from The International System of Units , 8 th ed. Section 2.1.1, BIPM (old definitions) and Appendix 3, Resolution 1 of 26 th CGPM (new definitions). The kilogram is probably the unit whose definition has undergone the most radical change as it was the only previous definition that depended on one specific artifact, and now involves the use a a device called the Watt balance. The Watt balance essentially balances gravitational forces with electromagnetic to determine Planck's constant. Once Planck's constant was determined, new "secondary standard" kilograms can be created by sort of reverse engineering the use of the device, but even this is very complicated, as the Watt balance operates in a vacuum and cryogenic temperatures, and the secondary standard needs to be in atmospheric conditions where normal measurements occur. The article " A LEGO Watt balance: An apparatus to determine a mass based on the new SI ", gives a pretty good walk though on how the Watt balance is used. Students who wish to learn more about the new SI definitions are encouraged to look at the resources provided through NIST's SI Redefintion website. The American Scientist article "Weighing the Kilogram" by Paul Karol is also an easy to read article going over many of the issues associated with defining the kilogram and the mole. Note: All students need to know what the SI Base units are, and if they see the definitions of the base units on an exam, they need to determine if they relate to the pre or post May 2019 revisions, as many textbooks and exam banks will still be using the old ones. SI Derived Units In the previous section we stated that the SI base units could be combined to measure different types of physical phenomena. Figure 1B.1.8 shows common units of measure that can be derived from the base units. There are two types, those that do not have special SI names and symbols, like area (m 2 ), and those that do, like pressure, work,... So for example, the SI unit of energy is the Joule, which is a "derived unit" and can be described in terms of the base units of mass, length and time: \[\underbrace{1J=1\frac{kg\cdot m^{2}}{s^{2}}}_{\text{SI Derived Unit of Energy in terms of SI Base Units}} \] Table represents common SI units: Quantity Unit Explanation Work N Newton = kg m s-2 Pressure Pa Pascal = N m-2 Energy J Joule = N.m Charge C Coulomb = A.s Electric Potential V Volt = J/C Power Watt 1 watt = 1 J/s Volume L 1L = 1dm3 Volume mL 1mL = 1cm3 Molarity M mol/L Note, there are many non-SI units like the calorie, atm, foot, etc., which can also be described in terms of the SI base units. Conversions between units is covered in section 1B.3.5 . SI Prefixes SI prefixs are commonly used to represent very large and small numbers. You need to memorize many of these and know how to convert between them. Multiplier Name Abbreviation Abbreviation.1 Name.1 Multiplier.1 10+30 quetta* Q q quecto* 10-30 10+27 ronna* R r ronto* 10-27 10+24 yotta Y y yocto 10-24 10+21 zetta Z z zepto 10-21 10+18 exa E a atto 10-18 10+15 peta P f femto 10-15 10+12 tera T p pico 10-12 10+9 giga G n nano 10-9 10+6 mega M m micro 10-6 10+3 kilo k m milli 10-3 10+2 hecto h c centi 10-2 10+1 deca da d deci 10-1 *quetta, ronna, ronto & quecto were added in 2022 ( https://www.bipm.org/en/cgpm-2022/resolution-3 ). NOTE: The Angstrom is a common unit of length Å = 10 - 10 m). It is not an SI prefix, but is a common unit of length that is used when describing bond lengths and distances between atoms (and ions) in a crystal. Use of SI prefixes For real large or small numbers, the convention is to place a number with between 1 and 999 in front of the SI prefix, and use the appropriate prefix to show the value. So a memory stick with 1,200,000 bytes would be written as 1.2 Mbyte, not as 1,200 kbytes. What is the ENG key on scientific calculators? The ENG key takes large or small numbers and expresses them as integer multiples of 10 3 or 10 -3 , effectively converting them to values that can be expressed with SI prefixes. If you type 1,200,000 into your calculator and press ENG, it gives 1.2 x 10 6 , so 1,200,000 bytes of memory on a memory stick would be 1.2 MBytes (you do not say 1,200 kBytes). If you try 0.12 and press ENG, it gives 120 x 10 -3 , so 0.12g is 120 mg. So the ENG function quickly converts numbers to values that can easily be expressed with SI prefixes. section 1B.3 we will go over mathematical operations using SI prefixes, and the conversions of units. Vocabulary Metric - decimal based unit of measure originally based on the meter, but evolved into SI. Metrology - Science of Measurement SI Base Unit - seven base units that all physical measurements can be derived from. SI Derived Unit - a unit of measurment created by combining SI Base Units. SI Prefixes - used to describe large or small values of SI units, usually of multiples of 10 3 . SI Unit - System international, unit of measure started in 1960 and evolving through the "Bureau International des Poids et Mesures" ( BIPM ), which was established by the Metric Convention of 1875. It is the established unit of measure for the practice of science. Test Yourself Query \(\PageIndex{1}\) Flash cards that go from SI prefix name to value. NOTE: If you already know an answer you can just click "I got it right" and move on to the next. Everytime you reload these should rearrange. Query \(\PageIndex{1}\) Flash cards that go from SI prefix value to name. NOTE: If you already know an answer you can just click "I got it right" and move on to the next. Everytime you reload these should rearrange. Query \(\PageIndex{1}\) Flash cards that go from SI prefix symbol to value. NOTE: If you already know an answer you can just click "I got it right" and move on to the next. Everytime you reload these should rearrange. Query \(\PageIndex{1}\) |
Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/30%3A_Natural_Products_and_Biosynthesis/30.05%3A_Biosynthesis | Fatty Acids The idea that ethanoic acid (acetic acid) is a possible common starting material for the biosynthesis of many organic compounds was first proposed by Collie (1893) on purely structural grounds. He recognized a structural connection between a linear chain of recurring \(\ce{CH_3CO}\) units (a polyketomethylene chain, \(\ce{CH_3COCH_2COCH_2COCH_2CO}-\)) and certain cyclic natural products. In the example given below, orsellinic acid is represented as if it were derived from a chain of four \(\ce{CH_3CO}\) units by a condensation-cyclization reaction: Experimental verification of Collie's hypothesis came many years later when isotopic hydrogen and carbon (\(\ce{^2H}\), \(\ce{^3H}\), \(\ce{^{13}C}\), and \(\ce{^{14}C}\)) became available. Tracer studies showed that long-chain fatty acids are made by plants and animals from \(\ce{CH_3CO}\) units by successively linking together the carbonyl group of one to the methyl group of another (K. Bloch and F. Lynen, Nobel Prize, 1964). If ethanoic acid supplied to the organism is labeled at the carboxyl group with \(\ce{^{14}C} \: \left( \overset{*}{\ce{C}} \right)\), the fatty acid has the label at alternate carbons: \[\ce{CH_3} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{O_2H}\] However, if the carbon of the methyl group is labeled, the product comes out labeled at the other set of alternate carbons: \[\overset{*}{\ce{C}} \ce{H_3CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CO_2H}\] Ethanoic acid is activated for biosynthesis by combination with the thiol, coenzyme A (\textbf{CoA} \ce{SH}\), Figure 18-7) to give the thioester, ethanoyl (acetyl) coenzyme A \(\left( \ce{CH_3COS} \textbf{CoA} \right)\). You may recall that the metabolic degradation of fats also involves this coenzyme ( Section 18-8F ) and it is tempting to assume that fatty acid biosynthesis is simply the reverse of fatty acid metabolism to \(\ce{CH_3COS} \textbf{CoA}\). However, this is not quite the case. In fact, it is a general observation in biochemistry that primary metabolites are synthesized by different routes from those by which they are metabolized (for example, compare the pathways of carbon in photosynthesis and metabolism of carbohydrates in Sectionss 20-9 and 20-10 ). A brief description of the main events in fatty-acid biosynthesis follows, and all of these steps must be understood to be under control of appropriate enzymes and their coenzymes even though they are omitted here. The \(\ce{CH_3CO}\) group of ethanoyl coenzyme A is first transferred to a protein having a free thiol \(\left( \ce{SH} \right)\) group to make another thioester, represented here as \(\ce{CH_3COS}-\textbf{ACP}\), where ACP stands for A cyl- C arrier- P rotein. The growing carbon chain remains bound to this protein throughout the synthesis: \[\ce{CH_3COS} \textbf{CoA} + \ce{HS}-\textbf{ACP} \rightarrow \ce{CH_3COS}-\textbf{ACP} + \textbf{CoA} \ce{SH}\] Carboxylation of \(\ce{CH_3COS}-\textbf{ACP}\) yields a propanedioyl thioester, \(6\), which then undergoes a Claisen condensation with a second mole of \(\ce{CH_3COS}-\textbf{ACP}\) accompanied by decarboxylation to yield a 3-oxobutanoyl thioester, \(7\): Reduction of the ketone group of the thioester (by \(\ce{NADPH}\)) leads to a thiol ester of a four-carbon carboxylic acid. Repetitive condensations with thioester \(6\) followed by reduction eventually lead to fatty acids. Each repetition increases the chain length by two carbons: The preceding scheme is representative of fatty acid biosynthesis in plants, animals, and bacteria. The major difference is that plant and bacterial fatty acids usually contain more double bonds (or even triple bonds) than do animal fatty acids. Biosynthesis of Aromatic Rings Collie's hypothesis that aromatic compounds are made biologically from ethanoic acid was greatly expanded by A. J. Birch to include an extraordinary number of diverse compounds. The generic name "acetogenin" has been suggested as a convenient classification for ethanoate (acetate)-derived natural products, but the name "polyketides" also is used. Naturally occurring aromatic compounds and quinones are largely made in this way. An example is 2-hydroxy-6-methylbenzoic acid formed as a metabolite of the mold Penicillium urticae ; using \(\ce{^{14}C}\)-carboxyl-labeled ethanoic acid, the label has been shown to be at the positions indicated below: Terpene Biosynthesis The biosynthesis of terpenes clearly follows a somewhat different course from fatty acids in that branched-chain compounds are formed. One way that this can come about is for 2-oxobutanoyl coenzyme A to undergo an aldol addition at the keto carbonyl group with the ethanoyl coenzyme A to give the 3-methyl-3-hydroxypentanedioic acid derivative, \(8\): The next step is reduction of one of the carboxyl groups of \(8\) to give mevalonic acid: This substance has been shown by tracer studies to be an efficient precursor of terpenes and steroids. Mevalonic acid has six carbon atoms, whereas the isoprene unit has only five. Therefore, if mevalonic acid is the precursor of isoprene units, it must lose one carbon atom at some stage. Synthesis of mevalonic acid labeled at the carboxyl group with \(\ce{^{14}C}\), and use of this material as a starting material for production of cholesterol, gives unlabeled cholesterol. Therefore, the carboxyl carbon is the one that is lost: Formation of the "biological isoprene unit" from mevalonic acid has been shown to proceed by stepwise phosphorylation of both alcohol groups, then elimination and decarboxylation to yield 3-methyl-3-butenyl pyrophosphate, \(9\) (often called \(\Delta^3\)-isopentenyl pyrophosphate): The coupling of the five-carbon units, \(9\), to give isoprenoid compounds has been suggested to proceed by the following steps. First, isomerization of the double bond is effected by an enzyme \(\left( \ce{E} \right)\) carrying an \(\ce{SH}\) group: The ester, \(10\), then becomes connected to the double bond of a molecule of \(9\), probably in an enzyme-induced carbocation type of polymerization ( Section 10-8B ): The product of the combination of two units of the pyrophosphate, \(9\), through this sequence is geranyl pyrophosphate if, as shown, the proton is lost to give a trans double bond. Formation of a cis double bond would give neryl pyrophosphate ( Section 30-3B ). Continuation of the head-to-tail addition of five-carbon units to geranyl (or neryl) pyrophosphate can proceed in the same way to farnesyl pyrophosphate and so to gutta-percha (or natural rubber). At some stage, a new process must be involved because, although many isoprenoid compounds are head-to-tail type polymers of isoprene, others, such as squalene, lycopene, and \(\beta\)- and \(\gamma\)-carotene (Table 30-1), are formed differently. Squalene, for example, has a structure formed from head-to-head reductive coupling of two farnesyl pyrophosphates: Since squalene can be produced from farnesyl pyrophosphate with \(\ce{NADPH}\) and a suitable enzyme system, the general features of the above scheme for terpene biosynthesis are well supported by experiment. In summary, the sequence from ethanoate to squalene has been traced as \[\text{ethanoyl coenzyme A} \rightarrow \text{mevalonic acid} \rightarrow \text{isopentenyl pyrophosphate} \rightarrow \text{farnesyl pyrophosphate} \rightarrow \text{squalene}\] Cholesterol Biosynthesis Isotopic labeling experiments show that cholesterol is derived from ethanoate by way of squalene and lanosterol. The evidence for this is that homogenized liver tissue is able to convert labeled squalene to labeled lanosterol and thence to labeled cholesterol. The conversion of squalene to lanosterol is particularly interesting because, although squalene is divisible into isoprene units, lanosterol is not - a methyl being required at \(\ce{C_8}\) and not \(\ce{C_{13}}\): As a result, some kind of rearrangement must be required to get from squalene to lanosterol. The nature of this rearrangement becomes clearer if we write the squalene formula so as to take the shape of lanosterol: When squalene is written in this form, we see that it is beautifully constructed for cyclization to lanosterol. The key intermediate that initiates the cyclization is the 2,3-epoxide of squalene. Enzymatic cleavage of the epoxide ring is followed by cyclization and then manifold hydride \(\left( \ce{H} \colon \right)\) and methide \(\left( \ce{CH_3} \colon \right)\) shifts to give lanosterol: The evidence is strong that the biosynthesis of lanosterol actually proceeds by a route of this type. With squalene made from either methyl- or carboxyl-labeled ethanoate, all the carbons of lanosterol and cholesterol are labeled just as predicted from the mechanism. Furthermore, ingenious double-labeling experiments have shown that the methyl at \(\ce{C_{13}}\) of lanosterol is the one that was originally located at \(\ce{C_{14}}\), whereas the one at \(\ce{C_{14}}\) is the one that came from \(\ce{C_8}\). The conversion of lanosterol to cholesterol involves removal of the three methyl groups at the 4,4- and 14-positions, shift of the double bond at the \(B\)/\(C\) junction between \(\ce{C_5}\) and \(\ce{C_6}\), and reduction of the \(\ce{C_{24}}\)-\(\ce{C_{25}}\) double bond. The methyl groups are indicated by tracer experiments to be eliminated by oxidation to carbon dioxide. The biosynthetic connection between ethanoyl coenzyme A and the complex natural products briefly discussed is summarized in Figure 30-1. |
Bookshelves/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/01%3A_Atoms/1.09%3A_Atomic_Mass-_The_Average_Mass_of_an_Elements_Atoms | Learning Objectives to know the meaning of isotopes and atomic masses. There are 21 elements with only one isotope, so all their atoms have identical masses. All other elements have two or more isotopes, so their atoms have at least two different masses. However, all elements obey the law of definite proportions when they combine with other elements, so they behave as if they had just one kind of atom with a definite mass. To solve this dilemma, we define the atomic mass as the weighted average mass of all naturally occurring isotopes of the element. A atomic mass is defined as \[\text{Atomic mass} = \left(\dfrac{\%\text{ abundance isotope 1}}{100}\right)\times \left(\text{mass of isotope 1}\right) + \left(\dfrac{\%\text{ abundance isotope 2}}{100}\right)\times \left(\text{mass of isotope 2}\right)~ ~ ~ + ~ ~ ... \label{amass}\] Similar terms would be added for all the isotopes that would be found in a bulk sample from nature. GPAs The weighted average is analogous to the method used to calculate grade point averages in most colleges: \[\text{GPA} = \left(\dfrac{\text{Credit Hours Course 1}}{\text{total credit hours}}\right)\times \left(\text{Grade in Course 1}\right) + \left(\dfrac{\text{Credit Hours Course 2}}{\text{total credit hours}}\right)\times \left(\text{Grade in Course 2}\right)~ + ~... \nonumber\] The periodic table lists the atomic masses of all the elements. Comparing these values with those given for some of the isotopes reveals that the atomic masses given in the periodic table never correspond exactly to those of any of the isotopes Figure \(\PageIndex{1}\). Because most elements exist as mixtures of several stable isotopes, the atomic mass of an element is defined as the weighted average of the masses of the isotopes. For example, naturally occurring carbon is largely a mixture of two isotopes: 98.89% 12 C (mass = 12 amu by definition) and 1.11% 13 C (mass = 13.003355 amu). The percent abundance of 14 C is so low that it can be ignored in this calculation. The average atomic mass of carbon is then calculated as follows: \[ \rm(0.9889 \times 12 \;amu) + (0.0111 \times 13.003355 \;amu) = 12.01 \;amu \label{Eq5} \] Carbon is predominantly 12 C, so its average atomic mass should be close to 12 amu, which is in agreement with this calculation. The value of 12.01 is shown under the symbol for C in the periodic table, although without the abbreviation amu, which is customarily omitted. Thus the tabulated atomic mass of carbon or any other element is the weighted average of the masses of the naturally occurring isotopes. Example \(\PageIndex{1}\): Lead Naturally occurring lead is found to consist of four isotopes: 1.40% \({}_{\text{82}}^{\text{204}}\text{Pb}\) whose isotopic mass is 203.973. 24.10% \({}_{\text{82}}^{\text{206}}\text{Pb}\) whose isotopic mass is 205.974. 22.10% \({}_{\text{82}}^{\text{207}}\text{Pb}\) whose isotopic mass is 206.976. 52.40% \({}_{\text{82}}^{\text{208}}\text{Pb}\) whose isotopic mass is 207.977. Calculate the atomic mass of an average naturally occurring sample of lead. Solution This is a direct application of Equation \ref{amass} and is best calculated term by term. Suppose that you had 1 mol lead. This would contain 1.40% (\(\dfrac{1.40}{100}\) × 1 mol) \({}_{\text{82}}^{\text{204}}\text{Pb}\) whose molar mass is 203.973 g mol –1 . The mass of 204 82 Pb would be \[\begin{align*} \text{m}_{\text{204}} &=n_{\text{204}}\times \text{ }M_{\text{204}} \\[4pt] &=\left( \frac{\text{1}\text{.40}}{\text{100}}\times \text{ 1 mol} \right)\text{ (203}\text{.973 g mol}^{\text{-1}}\text{)} \\[4pt] &=\text{2}\text{0.86 g} \end{align*}\] Similarly for the other isotopes \[\begin{align*}\text{m}_{\text{206}}&=n_{\text{206}}\times \text{ }M_{\text{206}}\\[4pt] &=\left( \frac{\text{24}\text{.10}}{\text{100}}\times \text{ 1 mol} \right)\text{ (205}\text{.974 g mol}^{\text{-1}}\text{)}\\[4pt] &=\text{49}\text{0.64 g} \\[6pt]\text{m}_{\text{207}}&=n_{\text{207}}\times \text{ }M_{\text{207}}\\[4pt] &=\left( \frac{\text{22}\text{.10}}{\text{100}}\times \text{ 1 mol} \right)\text{ (206}\text{.976 g mol}^{\text{-1}}\text{)}\\[4pt] &=\text{45}\text{0.74 g} \\[6pt] \text{m}_{\text{208}}&=n_{\text{208}}\times \text{ }M_{\text{208}}\\[4pt] &=\left( \frac{\text{52}\text{.40}}{\text{100}}\times \text{ 1 mol} \right)\text{ (207}\text{.977 g mol}^{\text{-1}}\text{)}\\[4pt] &=\text{108}\text{0.98 g} \end{align*}\] Upon summing all four results, the mass of 1 mol of the mixture of isotopes is to be found \[2.86\, g + 49.64\, g + 45.74\, g + 108.98\, g = 207.22\, g\nonumber\] The mass of an average lead atom, and thus lead's atomic mass, is 207.2 g/mol. This should be confirmed by consulting the Periodic Table of the Elements . Exercise \(\PageIndex{1}\): Boron Boron has two naturally occurring isotopes. In a sample of boron, \(20\%\) of the atoms are \(\ce{B}-10\), which is an isotope of boron with 5 neutrons and mass of \(10 \: \text{amu}\). The other \(80\%\) of the atoms are \(\ce{B}-11\), which is an isotope of boron with 6 neutrons and a mass of \(11 \: \text{amu}\). What is the atomic mass of boron? Answer The mass of an average boron atom, and thus boron's atomic mass, is \(10.8 \: \text{amu}\). This should be confirmed by consulting the Periodic Table of the Elements . But which Natural Abundance should be used? An important corollary to the existence of isotopes should be emphasized at this point. When highly accurate results are obtained, atomic weights may vary slightly depending on where a sample of an element was obtained. For this reason, the Commission on Isotopic Abundance and Atomic Weights of IUPAC (IUPAC/CIAAWhas redefined the atomic masses of 10 elements having two or more isotopes. The percentages of different isotopes often depends on the source of the element. For example, oxygen in Antarctic precipitation has an atomic weight of 15.99903, but oxygen in marine \(\ce{N2O}\) has an atomic mass of 15.9997. "Fractionation" of the isotopes results from slightly different rates of chemical and physical processes caused by small differences in their masses. The difference can be more dramatic when an isotope is derived from nuclear reactors. Mass Spectrometry: Measuring the Mass of Atoms and Molecules Although the masses of the electron, the proton, and the neutron are known to a high degree of precision, the mass of any given atom is not simply the sum of the masses of its electrons, protons, and neutrons. For example, the ratio of the masses of 1 H (hydrogen) and 2 H (deuterium) is actually 0.500384, rather than 0.49979 as predicted from the numbers of neutrons and protons present. Although the difference in mass is small, it is extremely important because it is the source of the huge amounts of energy released in nuclear reactions. Because atoms are much too small to measure individually and do not have charges, there is no convenient way to accurately measure absolute atomic masses. Scientists can measure relative atomic masses very accurately, however, using an instrument called a mass spectrometer. The technique is conceptually similar to the one Thomson used to determine the mass-to-charge ratio of the electron. First, electrons are removed from or added to atoms or molecules, thus producing charged particles called ions. When an electric field is applied, the ions are accelerated into a separate chamber where they are deflected from their initial trajectory by a magnetic field, like the electrons in Thomson’s experiment. The extent of the deflection depends on the mass-to-charge ratio of the ion. By measuring the relative deflection of ions that have the same charge, scientists can determine their relative masses ( Figure \(\PageIndex{2}\) ). Thus it is not possible to calculate absolute atomic masses accurately by simply adding together the masses of the electrons, the protons, and the neutrons, and absolute atomic masses cannot be measured, but relative masses can be measured very accurately. It is actually rather common in chemistry to encounter a quantity whose magnitude can be measured only relative to some other quantity, rather than absolutely. We will encounter many other examples later in this text. In such cases, chemists usually define a standard by arbitrarily assigning a numerical value to one of the quantities, which allows them to calculate numerical values for the rest. The arbitrary standard that has been established for describing atomic mass is the atomic mass unit (amu or u), defined as one-twelfth of the mass of one atom of 12 C. Because the masses of all other atoms are calculated relative to the 12 C standard, 12 C is the only atom whose exact atomic mass is equal to the mass number. Experiments have shown that 1 amu = 1.66 × 10 −24 g. Mass spectrometric experiments give a value of 0.167842 for the ratio of the mass of 2 H to the mass of 12 C, so the absolute mass of 2 H is \[\rm{\text{mass of }^2H \over \text{mass of }^{12}C} \times \text{mass of }^{12}C = 0.167842 \times 12 \;amu = 2.104104\; amu \label{Eq4}\] The masses of the other elements are determined in a similar way. Example \(\PageIndex{2}\): Bromine Naturally occurring bromine consists of the two isotopes listed in the following table: Isotope Exact Mass (amu) Percent Abundance (%) 79Br 78.9183 50.69 81Br 80.9163 49.31 Calculate the atomic mass of bromine. Given : exact mass and percent abundance Asked for : atomic mass Strategy : Convert the percent abundances to decimal form to obtain the mass fraction of each isotope. Multiply the exact mass of each isotope by its corresponding mass fraction (percent abundance ÷ 100) to obtain its weighted mass. Add together the weighted masses to obtain the atomic mass of the element. Check to make sure that your answer makes sense. Solution : A The atomic mass is the weighted average of the masses of the isotopes (Equation \ref{amass}. In general, we can write Bromine has only two isotopes. Converting the percent abundances to mass fractions gives \[\ce{^{79}Br}: {50.69 \over 100} = 0.5069 \nonumber\] \[\ce{^{81}Br}: {49.31 \over 100} = 0.4931 \nonumber\] B Multiplying the exact mass of each isotope by the corresponding mass fraction gives the isotope’s weighted mass: \(\ce{^{79}Br}: 79.9183 \;amu \times 0.5069 = 40.00\; amu\) \(\ce{^{81}Br}: 80.9163 \;amu \times 0.4931 = 39.90 \;amu\) C The sum of the weighted masses is the atomic mass of bromine is 40.00 amu + 39.90 amu = 79.90 amu D This value is about halfway between the masses of the two isotopes, which is expected because the percent abundance of each is approximately 50%. Exercise \(\PageIndex{2}\) Magnesium has the three isotopes listed in the following table: Isotope Exact Mass (amu) Percent Abundance (%) 24Mg 23.98504 78.70 25Mg 24.98584 10.13 26Mg 25.98259 11.17 Use these data to calculate the atomic mass of magnesium. Answer 24.31 amu Summary The mass of an atom is a weighted average that is largely determined by the number of its protons and neutrons, whereas the number of protons and electrons determines its charge. Each atom of an element contains the same number of protons, known as the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the atomic mass unit (amu), which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. The atomic mass of an element is the weighted average of the masses of the naturally occurring isotopes. When one or more electrons are added to or removed from an atom or molecule, a charged particle called an ion is produced, whose charge is indicated by a superscript after the symbol. |
Courses/Pasadena_City_College/Chem_2A_(Ku)_Textbook/07%3A_Intermolecular_Forces_and_Functional_Groups/7.02%3A_Functional_Groups_and_Intermolecular_Forces/7.2.02%3A_Functional_Groups_w__Oxygen_and_Sulfur/7.2.2.02%3A_Properties_of_Aldehydes_and_Ketones | Learning Objectives Explain why the boiling points of aldehydes and ketones are higher than those of ethers and alkanes of similar molar masses but lower than those of comparable alcohols. Compare the solubilities in water of aldehydes and ketones of four or fewer carbon atoms with the solubilities of comparable alkanes and alcohols. Name the typical reactions take place with aldehydes and ketones. Describe some of the uses of common aldehydes and ketones. The carbon-to-oxygen double bond is quite polar, more polar than a carbon-to-oxygen single bond. The electronegative oxygen atom has a much greater attraction for the bonding electron pairs than does the carbon atom. The carbon atom has a partial positive charge, and the oxygen atom has a partial negative charge: In aldehydes and ketones, this charge separation leads to dipole-dipole interactions that are great enough to significantly affect the boiling points. Table \(\PageIndex{1}\) shows that the polar single bonds in ethers have little such effect, whereas hydrogen bonding between alcohol molecules is even stronger. Compound Family Molar Mass Type of Intermolecular Forces Boiling Point (°C) CH3CH2CH2CH3 alkane 58 dispersion only –1 CH3OCH2CH3 ether 60 weak dipole 6 CH3CH2CHO aldehyde 58 strong dipole 49 CH3CH2CH2OH alcohol 60 hydrogen bonding 97 Formaldehyde is a gas at room temperature. Acetaldehyde boils at 20°C; in an open vessel, it boils away in a warm room. Most other common aldehydes are liquids at room temperature. Although the lower members of the homologous series have pungent odors, many higher aldehydes have pleasant odors and are used in perfumes and artificial flavorings. As for the ketones, acetone has a pleasant odor, but most of the higher homologs have rather bland odors. The oxygen atom of the carbonyl group engages in hydrogen bonding with a water molecule. The solubility of aldehydes is therefore about the same as that of alcohols and ethers. Formaldehyde, acetaldehyde, and acetone are soluble in water. As the carbon chain increases in length, solubility in water decreases. The borderline of solubility occurs at about four carbon atoms per oxygen atom. All aldehydes and ketones are soluble in organic solvents and, in general, are less dense than water. Oxidation of Aldehydes and Ketones Aldehydes and ketones are much alike in many of their reactions, owing to the presence of the carbonyl functional group in both. They differ greatly, however, in one most important type of reaction: oxidation. Aldehydes are readily oxidized to carboxylic acids, whereas ketones resist oxidation. The aldehydes are, in fact, among the most easily oxidized of organic compounds. They are oxidized by oxygen (\(\ce{O2}\)) in air to carboxylic acids. \[\ce{2RCHO + O_2 -> 2RCOOH} \nonumber \] The ease of oxidation helps chemists identify aldehydes. A sufficiently mild oxidizing agent can distinguish aldehydes not only from ketones but also from alcohols. Tollens’ reagent , for example, is an alkaline solution of silver (\(\ce{Ag^{+}}\)) ion complexed with ammonia (NH 3 ), which keeps the \(\ce{Ag^{+}}\) ion in solution. \[\ce{H_3N—Ag^{+}—NH_3} \nonumber \] When Tollens’ reagent oxidizes an aldehyde, the \(\ce{Ag^{+}}\) ion is reduced to free silver (\(\ce{Ag}\)). \[\underbrace{\ce{RCHO(aq)}}_{\text{an aldehyde}} + \ce{2Ag(NH3)2^{+}(aq) -> RCOO^{-} +} \underbrace{\ce{2Ag(s)}}_{\text{free silver}} + \ce{4NH3(aq) + 2H2O} \nonumber \] Deposited on a clean glass surface, the silver produces a mirror (Figure \(\PageIndex{1}\)). Ordinary ketones do not react with Tollens’ reagent. Although ketones resist oxidation by ordinary laboratory oxidizing agents, they undergo combustion, as do aldehydes. Some Common Carbonyl Compounds Formaldehyde has an irritating odor. Because of its reactivity, it is difficult to handle in the gaseous state. For many uses, it is therefore dissolved in water and sold as a 37% to 40% aqueous solution called formalin . Formaldehyde denatures proteins, rendering them insoluble in water and resistant to bacterial decay. For this reason, formalin is used in embalming solutions and in preserving biological specimens. Aldehydes are the active components in many other familiar substances. Large quantities of formaldehyde are used to make phenol-formaldehyde resins for gluing the wood sheets in plywood and as adhesives in other building materials. Sometimes the formaldehyde escapes from the materials and causes health problems in some people. While some people seem unaffected, others experience coughing, wheezing, eye irritation, and other symptoms. The odor of green leaves is due in part to a carbonyl compound, cis-3-hexenal, which with related compounds is used to impart a “green” herbal odor to shampoos and other products. Acetaldehyde is an extremely volatile, colorless liquid. It is a starting material for the preparation of many other organic compounds. Acetaldehyde is formed as a metabolite in the fermentation of sugars and in the detoxification of alcohol in the liver. Aldehydes are the active components of many other familiar materials (Figure \(\PageIndex{2}\)). Acetone is the simplest and most important ketone. Because it is miscible with water as well as with most organic solvents, its chief use is as an industrial solvent (for example, for paints and lacquers). It is also the chief ingredient in some brands of nail polish remover. To Your Health: Acetone in Blood, Urine, and Breath Acetone is formed in the human body as a by-product of lipid metabolism. Normally, acetone does not accumulate to an appreciable extent because it is oxidized to carbon dioxide and water. The normal concentration of acetone in the human body is less than 1 mg/100 mL of blood. In certain disease states, such as uncontrolled diabetes mellitus, the acetone concentration rises to higher levels. It is then excreted in the urine, where it is easily detected. In severe cases, its odor can be noted on the breath. Ketones are also the active components of other familiar substances, some of which are noted in the accompanying figure. Certain steroid hormones have the ketone functional group as a part of their structure. Two examples are progesterone, a hormone secreted by the ovaries that stimulates the growth of cells in the uterine wall and prepares it for attachment of a fertilized egg, and testosterone, the main male sex hormone. These and other sex hormones affect our development and our lives in fundamental ways. Summary 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. |
Courses/University_of_WisconsinStevens_Point/CHEM_101%3A_Basic_Chemistry_(D'Acchioli)/04%3A_Atomic_Structure/4.03%3A_Three_Types_of_Radioactivity | Learning Objective Differentiate between alpha particles, beta particles and gamma rays. Ernest Rutherford , (30 August 1871 – 19 October 1937), was a New Zealand who came to be known as the father of nuclear physics. While working in Thomson’s laboratory shortly after Thomson discovered the electron, Rutherford showed that compounds of uranium and other elements emitted at least two distinct types of radiation. One was readily absorbed by matter and seemed to consist of particles that had a positive charge and were massive compared to electrons. Because it was the first kind of radiation to be discovered, Rutherford called these substances alpha (α) particles . Rutherford also showed that the particles in the second type of radiation, beta (β) particles , had the same charge and mass-to-charge ratio as Thomson’s electrons; they are now known to be high-speed electrons. In 1908 was awarded the "for his investigations into the disintegration of the elements, and the chemistry of radioactive substances", for which he was the first Canadian and Oceanian Nobel laureate. A third type of radiation, gamma (γ) rays , was discovered somewhat later and found to be similar to the lower-energy form of radiation called x-rays, now used to produce images of bones and teeth. Table \(\PageIndex{1}\): Names, Symbols, Representations, and Descriptions of the most common types of radioactivity. Summary An alpha particle is a high energy helium nucleus. A beta particle is a high energy electron. A gamma ray is a very high energy electromagnetic radiation |
Courses/SUNY_Oneonta/Chem_221%3A_Organic_Chemistry_I_(Bennett)/1%3ALecture_Textbook/10%3A_Electrophilic_Reactions/10.05%3A_Electrophilic_Substitution | Until now, have already been introduced to electrophilic addition and electrophilic isomerization - now, let's move to the third variation on the electrophilic theme, that of electrophilic substitution. In an electrophilic substitution reaction, a pair of \(\pi\)-bonded electrons first attacks an electrophile - usually a carbocation species - and a proton is then abstracted from an adjacent carbon to reestablish the double bond, either in the original position or with isomerization. Electrophilic substitution mechanism: Electrophilic substitution reactions in isoprenoid biosynthesis Electrophilic substitution steps are very important in the biosynthetic pathways of isoprenoid compounds. In an early, chain-elongating reaction (EC 2.5.1.1) of the pathways of many isoprenoids, building blocks IPP and DMAPP combine to form a 10-carbon isoprenoid product called geranyl diphosphate (GPP): In a preliminary step (step a below), the diphosphate group on DMAPP departs to form an allylic carbocation. In step 1, the \(\pi \) electrons in IPP then attack the electrophilic carbocation from step a, resulting in a new carbon-carbon bond and a tertiary carbocation intermediate. Proton abstraction (step 2) leads to re-establishment of a double bond one carbon over from where it started out in IPP. Exercise \(\PageIndex{1}\) DMAPP is much more prone to spontaneous hydrolysis than IPP when they are dissolved in water. Explain why. Exercise \(\PageIndex{2}\) Farnesyl diphosphate ( FPP ) is synthesized by adding another five-carbon building block to geranyl diphosphate . What is this building block - IPP or DMAPP ? Draw a mechanism for the formation of FPP . Exercise \(\PageIndex{3}\) Propose a likely mechanism for the following transformation, which is the first stage in a somewhat complex reaction in the synthesis of an isoprenoid compound in plants. ( Science 1997 , 277 , 1815) Exercise \(\PageIndex{4}\) The electrophilic carbon in an electrophilic substitution reaction is often a carbocation , but it can also be the methyl group on S-adenosylmethionine (SAM - see section 8.8A). Propose a likely mechanism for this methylation reaction. ( Biochemistry 2012 , 51 , 3003) Electrophilic aromatic substitution Until now, we have been focusing mostly on electrophilic reactions of alkenes. Recall from section 2.2 that \(\pi \) bonds in aromatic rings are substantially less reactive than those in alkenes. Aromatic systems, however, do in fact undergo electrophilic substitution reactions given a powerful electrophile such as a carbocation, and if the carbocation intermediate that forms can be sufficiently stabilized. Electrophilic aromatic substitution (Friedel-Crafts alkylation) mechanism Organic chemists often refer to electrophilic aromatic substitution reactions with carbocation electrophiles as Friedel-Crafts alkylation reactions . Exercise \(\PageIndex{5}\) Aromatic rings generally do not undergo electrophilic addition reactions. Why not? The Friedel-Crafts reaction below is part of the biosynthesis of vitamin K and related biomolecules. Loss of diphosphate creates a powerful carbocation electrophile (step a) which attracts the \(\pi \) electrons of the aromatic ring to form a carbocation intermediate with a new carbon-carbon bond (step 1). Substitution is completed by proton abstraction (step 2) which re-establishes the aromatic sextet. An important point must be made here: because aromatic \(\pi \) bonds are substantially less reactive than alkene p bonds, the electrophilic must be VERY electrophilic - usually a carbocation. In addition, the carbocation intermediate that results from attack by aromatic \(\pi \) electrons is generally stabilized by resonance with lone pair electrons on a nearby oxygen or nitrogen (look at the resonance forms of the positively-charged intermediate that forms as the result of step 1 in the above figure). Remember that stabilizing the intermediate formed in a rate-limiting step has the effect of lowering the activation energy for the step, and thus accelerating the reaction. Organic chemists use the term ring activation to refer to the rate-accelerating effect of electron-donating heteroatoms in electrophilic aromatic substitution reactions. Aromatic rings lacking any activating oxygen or nitrogen atoms are less reactive towards electrophilic substitution. An example of the ring-activating effect of the nitrogen atom on an aromatic ring can be found in the following Friedel-Crafts reaction (EC 2.5.1.34), which should be familiar from the introduction to this chapter: Recall that this is a key early step in the biosynthetic pathway for the ergot alkaloids which are hypothesized to have been the root cause of the 'bewitchment' of several young girls in 17th century Salem, Massachusetts. (J. Am. Chem. Soc. 1992,114, 7354). Exercise \(\PageIndex{6}\) Draw a likely mechanism for the biosynthesis of dimethylallyl tryptophan , including a resonance structure showing how the carbocation intermediate in the rate determining step is stabilized by lone pair electrons on the ring nitrogen (in other words, show how the nitrogen serves to activate the ring). Friedel-Crafts reactions, in addition to being important biochemical transformations, are commonly carried out in the laboratory. It is instructive to consider a few examples to see how the same principles of structure and reactivity apply to both biochemical and laboratory reactions. Below is an example of a laboratory Friedel-Crafts alkylation reaction: Recall that a powerful electrophile - such as a carbocation - is required for an electrophilic aromatic substitution to occur. The 2-chloropropane reactant is electrophilic, but not electrophilic enough to react with benzene. Here's where the aluminum trichloride catalyst comes in: it reacts as a Lewis acid with the alkyl chloride to generate a secondary carbocation: The carbocation thus generated is sufficiently electrophilic to react with the aromatic \(\pi \) electrons, in a manner that should be familiar from the biochemical examples discussed above: You may have noticed, however, that one element from the biochemical Friedel-Crafts reactions is missing here: there is no activating group to stabilize the ring carbocation intermediate. Indeed, the presence of an activating group - for example, the oxygen atom of a methoxy substituent - greatly increases the rate of a Friedel-Crafts alkylation. Note in the example shown above that two products are formed: one is an ortho-disubstituted benzene and one is para-disubstituted. Note also that no meta-disubstituted product is formed. This phenomenon is referred to as the ortho-para directing effect, and you are led towards an explanation in the exercise below. Exercise \(\PageIndex{7}\) Draw the lowest-energy resonance contributors of the carbocation intermediates leading to formation of the ortho and para products in the reaction above. Use resonance structures to illustrate how the methoxy substituent is a ring-activating group. Draw the hypothetical carbocation intermediate in a reaction leading to formation of a meta-disubstituted product. Is this carbocation stabilized by the methoxy oxygen? Can you see why no meta product forms? Exercise \(\PageIndex{8}\) a) Just as there are ring-activating groups in electrophilic aromatic substitutions, there are also ring-deactivating groups. For each of the substituted benzene reactants below, draw the carbocation intermediate leading to the ortho substitution product and decide whether the substituent is ring-activating or ring-deactivating in a Friedel-Crafts reaction with 2-chloropropane and AlCl3 (in other words, which compounds would react faster than benzene, and which would react slower?) Explain how the ring-deactivating effect works. (challenging!) Ring-deactivating substituents are usually also meta-directing. Use one of your carbocation intermediate drawings from part (a) of this exercise, and the concept od resonance, to explain this observation. (answer part (b) first) Look again at the vitamin K biosynthesis reaction, and discuss the ring activating/directing effects of the two substituents on the substrate. Contributors Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) |
Courses/Prince_Georges_Community_College/CHEM_1010%3A_General_Chemistry_(Miller)/02%3A_Atoms_and_Elements/2.06%3A_The_Average_Mass_of_an_Elements_Atoms | Learning Objectives to know the meaning of isotopes and atomic masses. Atomic and Molecular Weights The subscripts in chemical formulas, and the coefficients in chemical equations represent exact quantities. \(\ce{H_2O}\), for example, indicates that a water molecule comprises exactly two atoms of hydrogen and one atom of oxygen. The following equation: \[ \ce{C3H8(g) + 5O2(g) \rightarrow 3CO2(g) + 4H2O(l)} \label{Eq1} \] not only tells us that propane reacts with oxygen to produce carbon dioxide and water, but that 1 molecule of propane reacts with 5 molecules of oxygen to produce 3 molecules of carbon dioxide and 4 molecules of water. Since counting individual atoms or molecules is a little difficult, quantitative aspects of chemistry rely on knowing the masses of the compounds involved. Atoms of different elements have different masses. Early work on the separation of water into its constituent elements (hydrogen and oxygen) indicated that 100 grams of water contained 11.1 grams of hydrogen and 88.9 grams of oxygen: \[\text{100 grams Water} \rightarrow \text{11.1 grams Hydrogen} + \text{88.9 grams Oxygen} \label{Eq2} \] Later, scientists discovered that water was composed of two atoms of hydrogen for each atom of oxygen. Therefore, in the above analysis, in the 11.1 grams of hydrogen there were twice as many atoms as in the 88.9 grams of oxygen. Therefore, an oxygen atom must weigh about 16 times as much as a hydrogen atom: \[ \dfrac{\dfrac{88.9\;g\;Oxygen}{1\; atom}}{\dfrac{111\;g\;Hydrogen}{2\;atoms}} = 16 \label{Eq3} \] Hydrogen, the lightest element, was assigned a relative mass of '1', and the other elements were assigned 'atomic masses' relative to this value for hydrogen. Thus, oxygen was assigned an atomic mass of 16. We now know that a hydrogen atom has a mass of 1.6735 x 10 -24 grams, and that the oxygen atom has a mass of 2.6561 X 10 -23 grams. As we saw earlier, it is convenient to use a reference unit when dealing with such small numbers: the atomic mass unit . The atomic mass unit ( amu ) was not standardized against hydrogen, but rather, against the 12 C isotope of carbon ( amu = 12 ). Thus, the mass of the hydrogen atom ( 1 H) is 1.0080 amu , and the mass of an oxygen atom ( 16 O) is 15.995 amu . Once the masses of atoms were determined, the amu could be assigned an actual value: 1 amu = 1.66054 x 10 -24 grams conversely: 1 gram = 6.02214 x 10 23 amu Mass Numbers and Atomic Mass of Elements: Mass Numbers and Atomic Mass of Elements, YouTube(opens in new window) [youtu.be] Average Atomic Mass Although the masses of the electron, the proton, and the neutron are known to a high degree of precision ( Table 2.3.1 ), the mass of any given atom is not simply the sum of the masses of its electrons, protons, and neutrons. For example, the ratio of the masses of 1 H (hydrogen) and 2 H (deuterium) is actually 0.500384, rather than 0.49979 as predicted from the numbers of neutrons and protons present. Although the difference in mass is small, it is extremely important because it is the source of the huge amounts of energy released in nuclear reactions. Because atoms are much too small to measure individually and do not have charges, there is no convenient way to accurately measure absolute atomic masses. Scientists can measure relative atomic masses very accurately, however, using an instrument called a mass spectrometer. The technique is conceptually similar to the one Thomson used to determine the mass-to-charge ratio of the electron. First, electrons are removed from or added to atoms or molecules, thus producing charged particles called ions. When an electric field is applied, the ions are accelerated into a separate chamber where they are deflected from their initial trajectory by a magnetic field, like the electrons in Thomson’s experiment. The extent of the deflection depends on the mass-to-charge ratio of the ion. By measuring the relative deflection of ions that have the same charge, scientists can determine their relative masses ( Figure \(\PageIndex{1}\) ). Thus it is not possible to calculate absolute atomic masses accurately by simply adding together the masses of the electrons, the protons, and the neutrons, and absolute atomic masses cannot be measured, but relative masses can be measured very accurately. It is actually rather common in chemistry to encounter a quantity whose magnitude can be measured only relative to some other quantity, rather than absolutely. We will encounter many other examples later in this text. In such cases, chemists usually define a standard by arbitrarily assigning a numerical value to one of the quantities, which allows them to calculate numerical values for the rest. The arbitrary standard that has been established for describing atomic mass is the atomic mass unit (amu or u), defined as one-twelfth of the mass of one atom of 12 C. Because the masses of all other atoms are calculated relative to the 12 C standard, 12 C is the only atom listed in Table 2.3.2 whose exact atomic mass is equal to the mass number. Experiments have shown that 1 amu = 1.66 × 10 −24 g. Mass spectrometric experiments give a value of 0.167842 for the ratio of the mass of 2 H to the mass of 12 C, so the absolute mass of 2 H is \[\rm{\text{mass of }^2H \over \text{mass of }^{12}C} \times \text{mass of }^{12}C = 0.167842 \times 12 \;amu = 2.104104\; amu \label{Eq4} \] The masses of the other elements are determined in a similar way. The periodic table lists the atomic masses of all the elements. Comparing these values with those given for some of the isotopes in Table 2.3.2 reveals that the atomic masses given in the periodic table never correspond exactly to those of any of the isotopes. Because most elements exist as mixtures of several stable isotopes, the atomic mass of an element is defined as the weighted average of the masses of the isotopes. For example, naturally occurring carbon is largely a mixture of two isotopes: 98.89% 12 C (mass = 12 amu by definition) and 1.11% 13 C (mass = 13.003355 amu). The percent abundance of 14 C is so low that it can be ignored in this calculation. The average atomic mass of carbon is then calculated as follows: \[ \rm(0.9889 \times 12 \;amu) + (0.0111 \times 13.003355 \;amu) = 12.01 \;amu \label{Eq5} \] Carbon is predominantly 12 C, so its average atomic mass should be close to 12 amu, which is in agreement with this calculation. The value of 12.01 is shown under the symbol for C in the periodic table, although without the abbreviation amu, which is customarily omitted. Thus the tabulated atomic mass of carbon or any other element is the weighted average of the masses of the naturally occurring isotopes. Example \(\PageIndex{1}\): Bromine Naturally occurring bromine consists of the two isotopes listed in the following table: Isotope Exact Mass (amu) Percent Abundance (%) 79Br 78.9183 50.69 81Br 80.9163 49.31 Calculate the atomic mass of bromine. Given : exact mass and percent abundance Asked for : atomic mass Strategy : Convert the percent abundances to decimal form to obtain the mass fraction of each isotope. Multiply the exact mass of each isotope by its corresponding mass fraction (percent abundance ÷ 100) to obtain its weighted mass. Add together the weighted masses to obtain the atomic mass of the element. Check to make sure that your answer makes sense. Solution : A The atomic mass is the weighted average of the masses of the isotopes. In general, we can write atomic mass of element = [(mass of isotope 1 in amu) (mass fraction of isotope 1)] + [(mass of isotope 2) (mass fraction of isotope 2)] + … Bromine has only two isotopes. Converting the percent abundances to mass fractions gives \[\ce{^{79}Br}: {50.69 \over 100} = 0.5069 \nonumber \] \[\ce{^{81}Br}: {49.31 \over 100} = 0.4931 \nonumber \] B Multiplying the exact mass of each isotope by the corresponding mass fraction gives the isotope’s weighted mass: \(\ce{^{79}Br}: 79.9183 \;amu \times 0.5069 = 40.00\; amu\) \(\ce{^{81}Br}: 80.9163 \;amu \times 0.4931 = 39.90 \;amu\) C The sum of the weighted masses is the atomic mass of bromine is 40.00 amu + 39.90 amu = 79.90 amu D This value is about halfway between the masses of the two isotopes, which is expected because the percent abundance of each is approximately 50%. Exercise \(\PageIndex{1}\) Magnesium has the three isotopes listed in the following table: Isotope Exact Mass (amu) Percent Abundance (%) 24Mg 23.98504 78.70 25Mg 24.98584 10.13 26Mg 25.98259 11.17 Use these data to calculate the atomic mass of magnesium. Answer 24.31 amu Finding the Averaged Atomic Weight of an Element: Finding the Averaged Atomic Weight of an Element(opens in new window) [youtu.be] Summary The mass of an atom is a weighted average that is largely determined by the number of its protons and neutrons, whereas the number of protons and electrons determines its charge. Each atom of an element contains the same number of protons, known as the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the atomic mass unit (amu), which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. The atomic mass of an element is the weighted average of the masses of the naturally occurring isotopes. When one or more electrons are added to or removed from an atom or molecule, a charged particle called an ion is produced, whose charge is indicated by a superscript after the symbol. |
Courses/University_of_California_Davis/Chem_110B%3A_Physical_Chemistry_II/Text/16%3A_The_Properties_of_Gases | 16.1: All Dilute Gases Behave Ideally For sufficiently low pressures (hence, low densities), each gas approaches ideal-gas behavior. 16.2: van der Waals and Redlich-Kwong Equations The van der Waals Equation of State is an equation relating the density of gases and liquids to the pressure, volume, and temperature conditions. The Redlich–Kwong equation of state is an empirical, algebraic equation that relates temperature, pressure, and volume of gases. It is generally more accurate than the van der Waals equation and the ideal gas equation at temperatures above the critical temperature. 16.3: A Cubic Equation of State Cubic equations of state are called such because they can be rewritten as a cubic function of molar volume. The Van der Waals equation of state is the most well known of cubic EOS, but many others have been developed. 16.4: The Law of Corresponding States The theorem of corresponding states (or principle of corresponding states) indicates that all fluids, when compared at the same reduced temperature and reduced pressure, have approximately the same compressibility factor and all deviate from ideal gas behavior to about the same degree. 16.5: The Second Virial Coefficient Because the perfect gas law is an imperfect description of a real gas, we can combine the perfect gas law and the compressibility factors of real gases to develop an Equation to describe the isotherms of a real gas. This Equation is known as the virial Equation of state, which expresses the deviation from ideality in terms of a power series in the density. The second virial coefficient describes the contribution of the pair-wise potential to the pressure of the gas. 16.6: The Repulsive Term in the Lennard-Jones Potential The Lennard-Jones intermolecular pair potential takes its name from Sir John Edward Lennard-Jones and consists of two 'parts'; a steep repulsive term, and smoother attractive term, representing the London dispersion forces. Apart from being an important model in itself, the Lennard-Jones potential frequently forms one of 'building blocks' of many force fields. 16.7: Van der Waals Constants in Terms of Molecular Parameters 16.E: The Properties of Gases (Exercises) These are homework exercises to accompany Chapter 16 of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap. |
Courses/Saint_Marys_College_Notre_Dame_IN/CHEM_342%3A_Bio-inorganic_Chemistry/Assignments/Worksheets/4.1%3A_Introduction_to_Ligand_Field_Theory_and_Crystal_Field_Theory | This worksheet is confidential because it is based on Copyrighted material that was published in the Journal of American Chemical Society. If you are a student in this course, please access this worksheet by clicking here (Worksheet 4.1) . You have access to this document only if you are a Saint Mary's College student. If you are not a student in this course, but would like to see this content, the original version is available in the Journal of Chemical Education . 1 Attribution 1. Worksheet 4.1 was adapted from Day 1 of A Guided Inquiry Activity for Teaching Ligand Field Theory : J. Chem. Educ. 2015, 92, 8, 1369-1372. Publication Date:June 17, 2015. DOI: 10.1021/acs.jchemed.5b00019 . |
Courses/Sacramento_City_College/SCC%3A_CHEM_300_-_Beginning_Chemistry/SCC%3A_CHEM_300_-_Beginning_Chemistry_(Alviar-Agnew)/12%3A_Liquids_Solids_and_Intermolecular_Forces/12.08%3A_Water_-_A_Remarkable_Molecule | Learning Objectives Interpret the unique properties of water in terms of a phase diagram. 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? Phase Diagram for Water Water is a unique substance in many ways. One of these special properties is the fact that solid water (ice) is less dense than liquid water just above the freezing point. The phase diagram for water is shown in the figure below. Notice one key difference between the general phase diagram and the phase diagram for water. In water's diagram, the slope of the line between the solid and liquid states is negative rather than positive. The reason for this is that water is an unusual substance, as its solid state is less dense than the liquid state. Ice floats in liquid water. Therefore, a pressure change has the opposite effect on those two phases. If ice is relatively near its melting point, it can be changed into liquid water by the application of pressure. The water molecules are actually closer together in the liquid phase than they are in the solid phase. Refer again to water's phase diagram (figure above). Notice point \(E\), labeled the critical point . What does that mean? At \(373.99^\text{o} \text{C}\), particles of water in the gas phase are moving very, very rapidly. At any temperature higher than that, the gas phase cannot be made to liquefy, no matter how much pressure is applied to the gas. The critical pressure \(\left( P_\text{C} \right)\) is the pressure that must be applied to the gas at the critical temperature in order to turn it into a liquid. For water, the critical pressure is very high, \(217.75 \: \text{atm}\). The critical point is the intersection point of the critical temperature and the critical pressure. Summary Solid water is less dense than liquid water just above the freezing point. The critical temperature \(\left( T_\text{C} \right)\) of a substance is the highest temperature at which the substance can possibly exist as a liquid. The critical pressure \(\left( P_\text{C} \right)\) is the pressure that must be applied to the gas at the critical temperature in order to turn it into a liquid. The critical point is the intersection point of the critical temperature and the critical pressure. |
Courses/Oregon_Institute_of_Technology/OIT%3A_CHE_332_--_Organic_Chemistry_II_(Lund)/9%3A_Phosphate_Transfer_Reactions/9.3%3A_ATP_The_Principal_Phosphate_Group_Donor | Thus far we have been very general in our discussion of phosphate transfer reactions, referring only to generic 'donor' and 'acceptor' species. It's time to get more specific. The most important donor of phosphate groups in the cell is a molecule called adenosine triphosphate, commonly known by its abbreviation ATP. that there are essentially three parts to the ATP molecule: an adenine nucleoside 'base', a five-carbon sugar (ribose), and triphosphate. The three phosphates are designated by Greek letters a, b, and g, with the a phosphate being the one closest to the ribose. Adenosine diphosphate (ADP) and adenosine monophosphate (AMP) are also important players in the reactions of this chapter. ATP is a big molecule, but the bond-breaking and bond-forming events we will be studying in this chapter all happen in the phosphate part of the molecule. You will see structural drawings of ATP, ADP, and AMP abbreviated in many different ways in this text and throughout the biochemical literature, depending on what is being illustrated. For example, the three structures below are all abbreviated depictions of ATP: The following exercise will give you some practice in recognizing different abbreviations for ATP and other biological molecules that contain phosphate groups. Below are a number of representations, labeled A-S, of molecules that contain phosphate groups. Different abbreviations are used. Arrange A-S into groups of drawings that depict the same species (for example, group together all of the abbreviations which depict ATP). You are probably familiar with the physiological role of ATP from your biology classes - it is commonly called 'the energy currency of the cell'. What this means is that ATP stores energy we get from the oxidation of fuel molecules such as carbohydrates or fats. The energy in ATP is stored in the two high-energy phosphate anhydride linkages. When one or both of these phosphate anhydride links are broken as a phosphate group is transferred to an acceptor, a substantial amount of energy is released. The negative charges on the phosphate groups are separated, eliminating some of electrostatic repulsion that existed in ATP. One way to picture this is as a coil springing open, releasing potential energy. In addition, cleavage of a phosphate anhydride bond means that surrounding water molecules are able to form more stabilizing hydrogen-bonding interactions with the products than was possible with the starting materials, again making the reaction more 'downhill', or exergonic. It is important to understand that while the phosphate anhydride bonds in ATP are thermodynamically unstable (they contain a great deal of chemical energy), they are at the same time kinetically stable: ATP-cleaving reactions are exothermic, but also have a high energy barrier, making them very slow unless catalyzed by an enzyme. In other words, the release of the energy contained in ATP is highly energetic but also subject to tight control by the interaction of highly evolved enzymes in our metabolic pathways. ATP is a versatile phosphate group donor: depending on the site of nucleophilic attack (at the \(\alpha \), \(\beta \), or \(\gamma \) phosphorus), different phosphate transfer outcomes are possible. Below are the three most common patterns seen in the central metabolic pathways. A 'squigly' line in each figure indicates the \(P-O\) bond being broken. We will study specific examples of each of these in the coming sections. Attack at the \(\gamma \)-phosphate: Attack at the \(\beta \)-phosphate: Attack at the \(\alpha \)-phosphate: The common thread running through all of the ATP-dependent reactions we will see in this section is the idea that the phosphate acceptor molecule is undergoing a thermodynamically 'uphill' transformation to become a more reactive species. The energy for this uphill transformation comes from breaking a high-energy phosphate anhydride bond in ATP. That is why ATP is often referred to as 'energy currency': the energy in its anhydride bonds is used to 'pay for' a thermodynamically uphill chemical step. Propose a fourth hypothetical phosphate transfer reaction between ATP and the generic acceptor molecule in the figure above, in which inorganic phosphate (P i) is a by-product. Why is this hypothetical phosphate transfer reaction less energetically favorable compared to all of the possible ATP-cleaving reactions shown in the figure above? |
Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20%3A_Energy_Metabolism/20.02%3A_Stage_I_of_Catabolism | Learning Objectives To describe how carbohydrates, fats, and proteins are broken down during digestion. We have said that animals obtain chemical energy from the food—carbohydrates, fats, and proteins—they eat through reactions defined collectively as catabolism . We can think of catabolism as occurring in three stages (Figure \(\PageIndex{1}\)). In stage I, carbohydrates, fats, and proteins are broken down into their individual monomer units: carbohydrates into simple sugars, fats into fatty acids and glycerol, and proteins into amino acids. One part of stage I of catabolism is the breakdown of food molecules by hydrolysis reactions into the individual monomer units—which occurs in the mouth, stomach, and small intestine—and is referred to as digestion. In stage II , these monomer units (or building blocks) are further broken down through different reaction pathways, one of which produces ATP , to form a common end product that can then be used in stage III to produce even more ATP. In this chapter, we will look at each stage of catabolism—as an overview and in detail. The conversion of food into cellular energy (as ATP) occurs in three stages. Digestion of Carbohydrates Carbohydrate digestion begins in the mouth (Figure \(\PageIndex{2}\)) where salivary α-amylase attacks the α-glycosidic linkages in starch, the main carbohydrate ingested by humans. Cleavage of the glycosidic linkages produces a mixture of dextrins, maltose, and glucose. The α-amylase mixed into the food remains active as the food passes through the esophagus, but it is rapidly inactivated in the acidic environment of the stomach. The primary site of carbohydrate digestion is the small intestine. The secretion of α-amylase in the small intestine converts any remaining starch molecules, as well as the dextrins, to maltose. Maltose is then cleaved into two glucose molecules by maltase. Disaccharides such as sucrose and lactose are not digested until they reach the small intestine, where they are acted on by sucrase and lactase, respectively. The major products of the complete hydrolysis of disaccharides and polysaccharides are three monosaccharide units: glucose, fructose, and galactose. These are absorbed through the wall of the small intestine into the bloodstream. Digestion of Proteins Protein digestion begins in the stomach (Figure \(\PageIndex{3}\)), where the action of gastric juice hydrolyzes about 10% of the peptide bonds. Gastric juice is a mixture of water (more than 99%), inorganic ions, hydrochloric acid, and various enzymes and other proteins. The pain of a gastric ulcer is at least partially due to irritation of the ulcerated tissue by acidic gastric juice. The hydrochloric acid (HCl) in gastric juice is secreted by glands in the stomach lining. The pH of freshly secreted gastric juice is about 1.0, but the contents of the stomach may raise the pH to between 1.5 and 2.5. HCl helps to denature food proteins; that is, it unfolds the protein molecules to expose their chains to more efficient enzyme action. The principal digestive component of gastric juice is pepsinogen, an inactive enzyme produced in cells located in the stomach wall. When food enters the stomach after a period of fasting, pepsinogen is converted to its active form—pepsin—in a series of steps initiated by the drop in pH. Pepsin catalyzes the hydrolysis of peptide linkages within protein molecules. It has a fairly broad specificity but acts preferentially on linkages involving the aromatic amino acids tryptophan, tyrosine, and phenylalanine, as well as methionine and leucine. Protein digestion is completed in the small intestine. Pancreatic juice, carried from the pancreas via the pancreatic duct, contains inactive enzymes such as trypsinogen and chymotrypsinogen. They are activated in the small intestine as follows (Figure \(\PageIndex{4}\)): The intestinal mucosal cells secrete the proteolytic enzyme enteropeptidase, which converts trypsinogen to trypsin; trypsin then activates chymotrypsinogen to chymotrypsin (and also completes the activation of trypsinogen). Both of these active enzymes catalyze the hydrolysis of peptide bonds in protein chains. Chymotrypsin preferentially attacks peptide bonds involving the carboxyl groups of the aromatic amino acids (phenylalanine, tryptophan, and tyrosine). Trypsin attacks peptide bonds involving the carboxyl groups of the basic amino acids (lysine and arginine). Pancreatic juice also contains procarboxypeptidase, which is cleaved by trypsin to carboxypeptidase. The latter is an enzyme that catalyzes the hydrolysis of peptide linkages at the free carboxyl end of the peptide chain, resulting in the stepwise liberation of free amino acids from the carboxyl end of the polypeptide. Aminopeptidases in the intestinal juice remove amino acids from the N-terminal end of peptides and proteins possessing a free amino group. Figure \(\PageIndex{5}\) illustrates the specificity of these protein-digesting enzymes. The amino acids that are released by protein digestion are absorbed across the intestinal wall into the circulatory system, where they can be used for protein synthesis. This diagram illustrates where in a peptide the different peptidases we have discussed would catalyze hydrolysis the peptide bonds. Digestion of Lipids Lipid digestion begins in the upper portion of the small intestine (Figure \(\PageIndex{6}\)). A hormone secreted in this region stimulates the gallbladder to discharge bile into the duodenum. The principal constituents of bile are the bile salts, which emulsify large, water-insoluble lipid droplets, disrupting some of the hydrophobic interactions holding the lipid molecules together and suspending the resulting smaller globules (micelles) in the aqueous digestive medium. These changes greatly increase the surface area of the lipid particles, allowing for more intimate contact with the lipases and thus rapid digestion of the fats. Another hormone promotes the secretion of pancreatic juice, which contains these enzymes. The lipases in pancreatic juice catalyze the digestion of triglycerides first to diglycerides and then to 2‑monoglycerides and fatty acids: The monoglycerides and fatty acids cross the intestinal lining into the bloodstream, where they are resynthesized into triglycerides and transported as lipoprotein complexes known as chylomicrons. Phospholipids and cholesteryl esters undergo similar hydrolysis in the small intestine, and their component molecules are also absorbed through the intestinal lining. The further metabolism of monosaccharides, fatty acids, and amino acids released in stage I of catabolism occurs in stages II and III of catabolism. Summary During digestion, carbohydrates are broken down into monosaccharides, proteins are broken down into amino acids, and triglycerides are broken down into glycerol and fatty acids. Most of the digestion reactions occur in the small intestine. |
Courses/University_of_Kansas/CHEM_130%3A_General_Chemistry_I_(Sharpe_Elles)/17%3A_Appendices/17.13%3A_Answer_Key/17.13.10%3A_Chapter_11 | 1. A solution can vary in composition, while a compound cannot vary in composition. Solutions are homogeneous at the molecular level, while other mixtures are heterogeneous. 3. (a) The process is endothermic as the solution is consuming heat. (b) Attraction between the K + and N O 3 − N O 3 − ions is stronger than between the ions and water molecules (the ion-ion interactions have a lower, more negative energy). Therefore, the dissolution process increases the energy of the molecular interactions, and it consumes the thermal energy of the solution to make up for the difference. (c) No, an ideal solution is formed with no appreciable heat release or consumption. 5. (a) ion-dipole forces; (b) dipole-dipole forces; (c) dispersion forces; (d) dispersion forces; (e) hydrogen bonding 7. Heat is released when the total intermolecular forces (IMFs) between the solute and solvent molecules are stronger than the total IMFs in the pure solute and in the pure solvent: Breaking weaker IMFs and forming stronger IMFs releases heat. Heat is absorbed when the total IMFs in the solution are weaker than the total of those in the pure solute and in the pure solvent: Breaking stronger IMFs and forming weaker IMFs absorbs heat. 9. Crystals of NaCl dissolve in water, a polar liquid with a very large dipole moment, and the individual ions become strongly solvated. Hexane is a nonpolar liquid with a dipole moment of zero and, therefore, does not significantly interact with the ions of the NaCl crystals. 11. (a) Fe(NO 3 ) 3 is a strong electrolyte, thus it should completely dissociate into Fe 3 + and NO 3 − NO 3 − ions. Therefore, (z) best represents the solution. (b) Fe ( NO 3 ) 3 ( s ) ⟶ Fe 3 + ( a q ) + 3 NO 3 − ( a q ) Fe ( NO 3 ) 3 ( s ) ⟶ Fe 3 + ( a q ) + 3 NO 3 − ( a q ) 13. (a) high conductivity (solute is an ionic compound that will dissociate when dissolved); (b) high conductivity (solute is a strong acid and will ionize completely when dissolved); (c) nonconductive (solute is a covalent compound, neither acid nor base, unreactive towards water); (d) low conductivity (solute is a weak base and will partially ionize when dissolved) 15. (a) ion-dipole; (b) hydrogen bonds; (c) dispersion forces; (d) dipole-dipole attractions; (e) dispersion forces 17. The solubility of solids usually decreases upon cooling a solution, while the solubility of gases usually decreases upon heating. 19. 40% 21. 2.8 g 23. 2.9 atm 25. 102 L HCl 27. The strength of the bonds between like molecules is stronger than the strength between unlike molecules. Therefore, some regions will exist in which the water molecules will exclude oil molecules and other regions will exist in which oil molecules will exclude water molecules, forming a heterogeneous region. 29. Both form homogeneous solutions; their boiling point elevations are the same, as are their lowering of vapor pressures. Osmotic pressure and the lowering of the freezing point are also the same for both solutions. 31. (a) Find number of moles of HNO 3 and H 2 O in 100 g of the solution. Find the mole fractions for the components. (b) The mole fraction of HNO 3 is 0.378. The mole fraction of H 2 O is 0.622. 33. (a) X Na 2 CO 3 = 0.0119 ; X Na 2 CO 3 = 0.0119 ; X H 2 O = 0.988 ; X H 2 O = 0.988 ; (b) X NH 4 NO 3 = 0.0928 ; X NH 4 NO 3 = 0.0928 ; X H 2 O = 0.907 ; X H 2 O = 0.907 ; (c) X Cl 2 = 0.192 ; X Cl 2 = 0.192 ; X CH 2 CI 2 = 0.808 ; X CH 2 CI 2 = 0.808 ; (d) X C 5 H 9 N = 0.00426 ; X C 5 H 9 N = 0.00426 ; X CHCl 3 = 0.997 X CHCl 3 = 0.997 35. In a 1 M solution, the mole is contained in exactly 1 L of solution. In a 1 m solution, the mole is contained in exactly 1 kg of solvent. 37. (a) Determine the molar mass of HNO 3 . Determine the number of moles of acid in the solution. From the number of moles and the mass of solvent, determine the molality. (b) 33.7 m 39. (a) 6.70 × × 10 −1 m ; (b) 5.67 m ; (c) 2.8 m ; (d) 0.0358 m 41. 1.08 m 43. (a) Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the boiling point of water and the boiling point of the solution; determine the new boiling point. (b) 100.5 °C 45. (a) Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the freezing temperature of water and the freezing temperature of the solution; determine the new freezing temperature. (b) −1.8 °C 47. (a) Determine the molar mass of Ca(NO 3 ) 2 ; determine the number of moles of Ca(NO 3 ) 2 in the solution; determine the number of moles of ions in the solution; determine the molarity of ions, then the osmotic pressure. (b) 2.67 atm 49. (a) Determine the molal concentration from the change in boiling point and K b ; determine the moles of solute in the solution from the molal concentration and mass of solvent; determine the molar mass from the number of moles and the mass of solute. (b) 2.1 × × 10 2 g mol −1 51. No. Pure benzene freezes at 5.5 °C, and so the observed freezing point of this solution is depressed by Δ T f = 5.5 − 0.4 = 5.1 °C. The value computed, assuming no ionization of HCl, is Δ T f = (1.0 m)(5.14 °C/ m ) = 5.1 °C. Agreement of these values supports the assumption that HCl is not ionized. 53. 144 g mol −1 55. 0.870 °C 57. S 8 59. 1.39 × × 10 4 g mol −1 61. 54 g 63. 100.26 °C 65. (a) X CH 3 OH = 0.590 ; X CH 3 OH = 0.590 ; X C 2 H 5 OH = 0.410 ; X C 2 H 5 OH = 0.410 ; (b) Vapor pressures are: CH 3 OH: 55 torr; C 2 H 5 OH: 18 torr; (c) CH 3 OH: 0.75; C 2 H 5 OH: 0.25 67. The ions and compounds present in the water in the beef lower the freezing point of the beef below −1 °C. 69. Δ bp = K b m = ( 1.20 ° C / m ) ( 9.41 g × 1 mol Hg Cl 2 271.496 g 0.03275 kg ) = 1.27 ° C Δ bp = K b m = ( 1.20 ° C / m ) ( 9.41 g × 1 mol Hg Cl 2 271.496 g 0.03275 kg ) = 1.27 ° C The observed change equals the theoretical change; therefore, no dissociation occurs. 71. Colloidal System Dispersed Phase Dispersion Medium starch dispersion starch water smoke solid particles air fog water air pearl water calcium carbonate (CaCO3) whipped cream air cream floating soap air soap jelly fruit juice pectin gel milk butterfat water ruby chromium(III) oxide (Cr2O3) aluminum oxide (Al2O3) 73. Colloidal dispersions consist of particles that are much bigger than the solutes of typical solutions. Colloidal particles are either very large molecules or aggregates of smaller species that usually are big enough to scatter light. Colloids are homogeneous on a macroscopic (visual) scale, while solutions are homogeneous on a microscopic (molecular) scale. 75. If they are placed in an electrolytic cell, dispersed particles will move toward the electrode that carries a charge opposite to their own charge. At this electrode, the charged particles will be neutralized and will coagulate as a precipitate. |
Courses/University_of_South_Carolina__Upstate/CHEM_U109%3A_Chemistry_of_Living_Things_-_Mueller/11%3A_Organic_Chemistry/11.01_Organic_Chemistry | Skills to Develop To recognize the composition and properties typical of organic and inorganic compounds. Scientists of the 18th and early 19th centuries studied compounds obtained from plants and animals and labeled them organic because they were isolated from “organized” (living) systems. Compounds isolated from nonliving systems, such as rocks and ores, the atmosphere, and the oceans, were labeled inorganic . For many years, scientists thought organic compounds could be made by only living organisms because they possessed a vital force found only in living systems. The vital force theory began to decline in 1828, when the German chemist Friedrich Wöhler synthesized urea from inorganic starting materials. He reacted silver cyanate (AgOCN) and ammonium chloride (NH 4 Cl), expecting to get ammonium cyanate (NH 4 OCN). What he expected is described by the following equation. \[AgOCN + NH_4Cl \rightarrow AgCl + NH_4OCN \label{Eq1}\] Instead, he found the product to be urea (NH 2 CONH 2 ), a well-known organic material readily isolated from urine. This result led to a series of experiments in which a wide variety of organic compounds were made from inorganic starting materials. The vital force theory gradually went away as chemists learned that they could make many organic compounds in the laboratory. Today organic chemistry is the study of the chemistry of the carbon compounds, and inorganic chemistry is the study of the chemistry of all other elements. It may seem strange that we divide chemistry into two branches—one that considers compounds of only one element and one that covers the 100-plus remaining elements. However, this division seems more reasonable when we consider that of tens of millions of compounds that have been characterized, the overwhelming majority are carbon compounds. The word organic has different meanings. Organic fertilizer, such as cow manure, is organic in the original sense; it is derived from living organisms. Organic foods generally are foods grown without synthetic pesticides or fertilizers. Organic chemistry is the chemistry of compounds of carbon. Carbon is unique among the other elements in that its atoms can form stable covalent bonds with each other and with atoms of other elements in a multitude of variations. The resulting molecules can contain from one to millions of carbon atoms. We previously surveyed organic chemistry by dividing its compounds into families based on functional groups. We begin with the simplest members of a family and then move on to molecules that are organic in the original sense—that is, they are made by and found in living organisms. These complex molecules (all containing carbon) determine the forms and functions of living systems and are the subject of biochemistry. Organic compounds, like inorganic compounds, obey all the natural laws. Often there is no clear distinction in the chemical or physical properties among organic and inorganic molecules. Nevertheless, it is useful to compare typical members of each class, as in Table \(\PageIndex{1}\). Organic Hexane Unnamed: 2 Inorganic NaCl low melting points −95°C NaN high melting points 801°C low boiling points 69°C NaN high boiling points 1,413°C low solubility in water; high solubility in nonpolar solvents insoluble in water; soluble in gasoline NaN greater solubility in water; low solubility in nonpolar solvents soluble in water; insoluble in gasoline flammable highly flammable NaN nonflammable nonflammable aqueous solutions do not conduct electricity nonconductive NaN aqueous solutions conduct electricity conductive in aqueous solution exhibit covalent bonding covalent bonds NaN exhibit ionic bonding ionic bonds Keep in mind, however, that there are exceptions to every category in this table. To further illustrate typical differences among organic and inorganic compounds, Table \(\PageIndex{1}\) also lists properties of the inorganic compound sodium chloride (common table salt, NaCl) and the organic compound hexane (C 6 H 14 ), a solvent that is used to extract soybean oil from soybeans (among other uses). Many compounds can be classified as organic or inorganic by the presence or absence of certain typical properties, as illustrated in Table \(\PageIndex{1}\). |
Courses/Duke_University/Textbook%3A_Modern_Applications_of_Chemistry_(Cox)/02%3A_Nuclear_Chemistry | Nuclear reactions differ from other chemical processes in one critical way: in a nuclear reaction, the identities of the elements change. In addition, nuclear reactions are often accompanied by the release of enormous amounts of energy, as much as a billion times more than the energy released by chemical reactions. Moreover, the yields and rates of a nuclear reaction are generally unaffected by changes in temperature, pressure, or the presence of a catalyst. 2.1: Components of the Nucleus Subatomic particles of the nucleus (protons and neutrons) are called nucleons. A nuclide is an atom with a particular number of protons and neutrons. An unstable nucleus that decays spontaneously is radioactive, and its emissions are collectively called radioactivity. Isotopes that emit radiation are called radioisotopes. Each nucleon is attracted to other nucleons by the strong nuclear force. 2.2: Nuclear Reactions Protons and neutrons are called nucleons and a nuclide is an atom with a specific number nucleons. Unstable nuclei decay spontaneously are radioactive and its emissions are called radioactivity. Nuclei are bound by the strong nuclear force. Stable nuclei generally have even numbers of protons and neutrons with a ratio of at least 1. Nuclei that contain magic numbers of protons and neutrons are often especially stable including superheavy elements, with atomic numbers near 126. 2.3: Nuclear Radiation Nuclei can undergo reactions that change their number of protons, number of neutrons, or energy state. Many different particles can be involved and the most common are protons, neutrons, positrons, alpha (α) particles, beta (β) particles (high-energy electrons), and gamma (γ) rays (which compose high-energy electromagnetic radiation). As with chemical reactions, nuclear reactions are always balanced. When a nuclear reaction occurs, the total mass (number) and the total charge remain unchanged. 2.4: Rates of Radioactive Decay Unstable nuclei undergo spontaneous radioactive decay. The most common types of radioactivity are α decay, β decay, γ emission, positron emission, and electron capture. Nuclear reactions also often involve γ rays, and some nuclei decay by electron capture. Each of these modes of decay leads to the formation of a new stable nuclei sometimes via multiple decays before ending in a stable isotope. All nuclear decay processes follow first-order kinetics and each radioisotope has its own half-life. 2.5: Stability of the Atomic Nucleus The energy changes in nuclear reactions are enormous compared with those of even the most energetic chemical reactions. In fact, the energy changes in a typical nuclear reaction are so large that they result in a measurable change of mass. In this section, we describe the relationship between mass and energy in nuclear reactions and show how the seemingly small changes in mass that accompany nuclear reactions result in the release of enormous amounts of energy. 2.6: The Origin of the Elements The relative abundances of the elements in the known universe vary by more than 12 orders of magnitude. In this section, we explain why 1H and 2He together account for at least 99% of all the atoms in the known universe. We also describe the nuclear reactions that take place in stars, which transform one nucleus into another and create all the naturally occurring elements. 2.7: Transmutation of the Elements 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 experimentally. The conversion of one element to another is the process of transmutation. 2.8: Nuclear Fission Many heavier elements with smaller binding energies per nucleon can decompose into more stable elements that have intermediate mass numbers and larger binding energies per nucleon. Sometimes neutrons are also produced. This decomposition is called fission, the breaking of a large nucleus into smaller pieces. The breaking is rather random with the formation of a large number of different products. Fission usually does not occur naturally, but is induced by bombardment with neutrons. 2.9: Nuclear Fusion The process of converting very light nuclei into heavier nuclei is also accompanied by the conversion of mass into large amounts of energy, a process called fusion. The principal source of energy in the sun is a net fusion reaction in which four hydrogen nuclei fuse and produce one helium nucleus and two positrons. This is a net reaction of a more complicated series of events. 2.10: Nuclear Chemistry (Exercises) These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. |
Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(LibreTexts)/10%3A_Chemical_Bonding/10.04%3A_Covalent_Lewis_Structures-_Electrons_Shared | Learning Objectives Define covalent bond . Illustrate covalent bond formation with Lewis electron dot diagrams. Ionic bonding typically occurs when it is easy for one atom to lose one or more electrons and another atom to gain one or more electrons. However, some atoms won’t give up or gain electrons easily. Yet they still participate in compound formation. How? There is another mechanism for obtaining a complete valence shell: sharing electrons. When electrons are shared between two atoms, they make a bond called a covalent bond . Let us illustrate a covalent bond by using H atoms, with the understanding that H atoms need only two electrons to fill the 1 s subshell. Each H atom starts with a single electron in its valence shell: \[\mathbf{H\, \cdot }\; \; \; \; \; \mathbf{\cdot \: H} \nonumber \] The two H atoms can share their electrons: \[\mathbf{H}\: \mathbf{: H} \nonumber \] We can use circles to show that each H atom has two electrons around the nucleus, completely filling each atom’s valence shell: Because each H atom has a filled valence shell, this bond is stable, and we have made a diatomic hydrogen molecule. (This explains why hydrogen is one of the diatomic elements.) For simplicity’s sake, it is not unusual to represent the covalent bond with a dash, instead of with two dots: H–H Because two atoms are sharing one pair of electrons, this covalent bond is called a single bond . As another example, consider fluorine. F atoms have seven electrons in their valence shell: These two atoms can do the same thing that the H atoms did; they share their unpaired electrons to make a covalent bond. Note that each F atom has a complete octet around it now: We can also write this using a dash to represent the shared electron pair: There are two different types of electrons in the fluorine diatomic molecule. The bonding electron pair makes the covalent bond. Each F atom has three other pairs of electrons that do not participate in the bonding; they are called lone pair electrons . Each F atom has one bonding pair and three lone pairs of electrons. Covalent bonds can be made between different elements as well. One example is HF . Each atom starts out with an odd number of electrons in its valence shell: The two atoms can share their unpaired electrons to make a covalent bond: We note that the H atom has a full valence shell with two electrons, while the F atom has a complete octet of electrons. Example \(\PageIndex{1}\): Use Lewis electron dot diagrams to illustrate the covalent bond formation in HBr. Solution HBr is very similar to HF, except that it has Br instead of F. The atoms are as follows: The two atoms can share their unpaired electron: Exercise \(\PageIndex{1}\) Use Lewis electron dot diagrams to illustrate the covalent bond formation in Cl 2 . Answer When working with covalent structures, it sometimes looks like you have leftover electrons. You apply the rules you learned so far, and there are still some electrons that remain unattached. You can't just leave them there. So where do you put them? Multiple Covalent Bonds Some molecules are not able to satisfy the octet rule by making only single covalent bonds between the atoms. Consider the compound ethene, which has a molecular formula of \(\ce{C_2H_4}\). The carbon atoms are bonded together, with each carbon also bonded to two hydrogen atoms. two \(\ce{C}\) atoms \(= 2 \times 4 = 8\) valence electrons four \(\ce{H}\) atoms \(= 4 \times 1 = 4\) valence electrons total of 12 valence electrons in the molecule If the Lewis electron dot structure was drawn with a single bond between the carbon atoms and with the octet rule followed, it would look like this: This Lewis structure is incorrect because it contains a total of 14 electrons. However, the Lewis structure can be changed by eliminating the lone pairs on the carbon atoms and having to share two pairs instead of only one pair. A double covalent bond is a covalent bond formed by atoms that share two pairs of electrons. The double covalent bond that occurs between the two carbon atoms in ethane can also be represented by a structural formula and with a molecular model as shown in the figure below. A triple covalent bond is a covalent bond formed by atoms that share three pairs of electrons. The element nitrogen is a gas that composes the majority of Earth's atmosphere. A nitrogen atom has five valence electrons, which can be shown as one pair and three single electrons. When combining with another nitrogen atom to form a diatomic molecule, the three single electrons on each atom combine to form three shared pairs of electrons. Each nitrogen atom follows the octet rule with one lone pair of electrons, and six electrons that are shared between the atoms. Summary Covalent bonds are formed when atoms share electrons. Lewis electron dot diagrams can be drawn to illustrate covalent bond formation. Double bonds or triple bonds between atoms may be necessary to properly illustrate the bonding in some molecules. Contributions & Attributions Anonymous by request |
Courses/Kutztown_University_of_Pennsylvania/CHM_320%3A_Advanced_Inorganic_Chemistry_textbook/02%3A_Atomic_Structure/2.02%3A_The_Schrodinger_equation_particle_in_a_box_and_atomic_wavefunctions | Considering the failures of the Bohr model, Erwin Schr ö dinger and Werner Heisenberg proposed a major change in the paradigm regarding the electron. In several breakthrough papers (1925-1927) they attributed wave properties to electrons and they each received Nobel prizes for developing the theories of Wave Mechanics (or the "New Quantum Mechanics"). This approach treated electrons as having "dual" nature: possessing properties of both waves and particles. Schrödinger's Equation describes the behavior of the electron (in a hydrogen atom) in three dimensions. It is a mathematical equation that defines the electron’s position, mass, total energy, and potential energy. The simplest form of the Schrödinger Equation is as follows: \[\hat{H}\psi = E\psi \nonumber \] where \(\hat{H}\) is the Hamiltonian operator , \(E\) is the energy of the electron , and \(\psi\) is the wavefunction . The Hamiltonian, \(\hat{H}\) The Hamiltonian operator is like a set of instructions that tells us what to do with the function that follows it. A Hamiltonian operator is a function over three-dimensional space that corresponds to the sum of kinetic energies and potential energies of the particles in a system, one electron and its nucleus in this case. The Hamiltonian operator for a one-electron system is: \[\hat{H}=\dfrac{-h{^2}}{8\pi{^2}m_e}\left(\dfrac{\partial{^2}}{\partial{x^2}}+\dfrac{\partial{^2}}{\partial{y^2}}+\dfrac{\partial{^2}}{\partial{z^2}}\right)-\dfrac{Ze^2}{4\pi{}\epsilon_0{r}}, \nonumber \] where \(h\) is Planck's constant, \(m_e\) is the mass of the electron, \(e\) is the charge of the electron, \(r\) is the distance from the nucleus (\(r=\sqrt{x^2+y^2+z^2}\)), \(Z\) is the charge of the nucleus, and \(4\pi{}\epsilon_0\) is the permittivity of a vacuum. Kinetic Energy The first part of the Hamiltonian written above, \(\dfrac{-h{^2}}{8\pi{^2}m_e}\left(\dfrac{\partial{^2}}{\partial{x^2}}+\dfrac{\partial{^2}}{\partial{y^2}}+\dfrac{\partial{^2}}{\partial{z^2}}\right)\) describes the kinetic energy of the electron. This is the energy due to motion of the electron. Potential energy The second part written above, \(\dfrac{-Ze^2}{4\pi{}\epsilon_0{r}}\), describes the potential energy of the electron, and is commonly written as \(V(r)\) or \(V(x,y,z)\). \[V(x,y,x) = \dfrac{-Ze^2}{4\pi{}\epsilon_0{r}} = \dfrac{-Ze^2}{4\pi{}\epsilon_0{\sqrt{x^2+y^2+z^2}}} \nonumber \] The potential energy depends on the attractive electrostatic force between the electron and the nucleus. You might notice that this attraction is essentially the same as the electrostatic force defined by Coulomb's law. And, just as in Coulomb's law, when two opposite charges are attracted to one another, the potential energy of the force is negative. Thus, when an electron is close to the nucleus, the potential energy is a large negative number corresponding to a strong attractive force. When an electron is farther from the nucleus, the potential energy is still negative but with a smaller magnitude, corresponding to a weaker attractive force. If the electron is very far from the nucleus (\(r = \infty\)) then the attractive force, and the potential energy, is zero. The Wavefunction, \(\psi\) In simple terms, the wavefunction (\(\psi\)) of an electron describes the electron's position in space, relative to the nucleus. The square of the \(\psi\) describes an atomic orbital. We can't define the position too exactly because we would violate the Heisenberg Uncertainty principle, but we can define its wave. A simple example of a \(\psi\) is described in the next section: Particle in a Box. Here, we will describe the \(\psi\) in general terms. Generally, in a one-electron atom, the electron \(\psi\) is defined by the wave's distance from the nucleus and its angle with respect to the x, y, and z axes of the atom's Cartesian coordinates (the nucleus is at the origin). The general form of the (\(\psi\)) for an electron in a hydrogen atom can be written as follows: \[\psi_{n,l,m_l} = R_{n,l}(r) + Y_{l,m_l}(\theta,\phi) \nonumber \] Quantum Numbers define \(\psi\) The (\(\psi\)) is defined by three of the quantum numbers: \(n\), \(l\), and \(m_l\). These quantum numbers will be discussed more in a later section (2.2.2) The radial variation, \(R\), depends on the electron's distance from the nucleus. The quantum numbers \(n\) (energy level) and \(l\) (orbital type) define \(R\). Since \(n\) must be an integer, there are only certain allowed values for the solution to the wavefunction. The angular variation, \(Y\), depends on the angle with respect to the x, y, and z coordinates, and depends on the quantum numbers \(l\) (the orbital type) and \(m_l\) (the angular momentum, or the specific orbital). For example \(p_x\) lies along the \(x\) axis, while \(p_y\) points in a different direction in space. For review, a list of the quantum numbers, their values, and meanings are in the table below. SYMBOL NAME VALUES MEANING \(n\) principal \(1,2,3...\)(any integer) energy level, shell \(l\) angular momentum \(0 \rightarrow n-1\) subshell, \(0=s, 1=p, 2=d, 3=f...\) this is the angular dependence of the orbital, shape of the orbital *letters have historical meaning, sharp, principle, diffuse, fundamental \(m_l\) magnetic \(+l \rightarrow -l\) orientation of angular momentum in space, orbital \(m_s\) spin \(+\frac{1}{2}, -\frac{1}{2}\) the imaginary property we call "spin", up or down Some important considerations and limitations Although it might seem like there could be any value of x, y, and z for the Hamiltonian, these values are limited by the allowed positions of electrons according to \(\psi\), which is limited by integer values of \(n\). In other words the allowed solutions are quantized . However, there are an infinite number of values for \(n\) from \(n=1\rightarrow\infty\), so there are also infinite solutions to the Schr ö dinger equation. The \(\psi\) describes the wave properties of an electron. The probability of finding the electron somewhere in space is the square of the wavefunction (\(\psi^2\) or \(\psi \psi^*\)). In other words, \(\psi^2\) describes the shape and size of an electron's orbital (the shapes you already know). There are some requirements for a physically realistic and meaningful solution for \(\psi\), and thus \(\psi^2\). There is only one possible value for \(\psi\) for any set of the three quantum numbers \(n, l, m_l\). The \(\psi\) approaches zero as \(r\) approaches infinity, and so \(\psi^2\) also approaches zero as \(r\rightarrow\infty\). The wavefunction must be normalized. In other words, the total probability of finding the electron in all of space must be 1. \[\int_{\text {all space}} \psi_{A} \psi_{A}^{*} d \tau=1 \nonumber \] Any two orbitals must not occupy the same space. In other words, any two orbitals in an atom are orthogonal. If \(\psi_{A}\) and \(\psi_{B}\) are wavefunctions for different orbitals in the same atom, \[\int_{\text {all space}} \psi_{A} \psi_{B}^{*} d \tau=0 \nonumber \] The probability of finding the electron anywhere in infinite space must be defined. This means that the wave functions and their first derivatives must be continuous (i.e. not change abruptly from one point to the next). Awesome Sources for further reading +Plus : https://plus.maths.org/content/schrodinger-1 |
Courses/Madera_Community_College/Concepts_of_Physical_Science/02%3A_Matter_and_Motion/2.08%3A_Motion_Graphs | Learning Objectives Interpret motion graphs and the relationship between different types of motion graphs. Predict motion graphs. In the previous sections of this text we looked at several example problems in which motion graphs were used to model the solution to those problems. In this section, we will revisit some of those problems, and focus specifically on the trends observed within the motion graphs created for those problems. Finally, we will also hint at some forthcoming material related to a common type of motion as we consider some motion graphs we haven’t seen yet. These ideas are explored in three subsections which follow. |
Bookshelves/Organic_Chemistry/Organic_Chemistry_III_(Morsch_et_al.)/26%3A_Amino_Acids_Peptides_and_Proteins/26.01%3A_Introduction | Objectives After completing this section, you should be able to give examples of the various biological roles played by proteins. identify amino acids as being the building blocks from which all proteins are made. show, in a general way, how the joining together of a number of amino acids through the formation of peptide bonds results in the formation of proteins. Key Terms Make certain that you can define, and use in context, the key terms below. amino acid enzyme peptide bond protein Study Notes The “peptide bond” or “peptide linkage” that is formed between the amino group of one amino acid and the carboxyl group of a second amino acid is identical to the C$\ce{-}$N bond present in amides (see Section 21.7). We shall review the nature of such bonds in Section 26.4. Proteins are polymers of amino acids , linked by amide groups known as peptide bond s. An amino acid can be thought of as having two components: a 'backbone', or 'main chain', composed of an ammonium group, an 'alpha-carbon', and a carboxylate, and a variable 'side chain' (in green below) bonded to the alpha-carbon. There are twenty different side chains in naturally occurring amino acids , and it is the identity of the side chain that determines the identity of the amino acid: for example, if the side chain is a -CH 3 group, the amino acid is alanine, and if the side chain is a -CH 2 OH group, the amino acid is serine. Many amino acid side chains contain a functional group (the side chain of serine, for example, contains a primary alcohol), while others, like alanine, lack a functional group, and contain only a simple alkane. The two 'hooks' on an amino acid monomer are the amine and carboxylate groups. Proteins (polymers of ~50 amino acids or more) and peptides (shorter polymers) are formed when the amino group of one amino acid monomer reacts with the carboxylate carbon of another amino acid to form an amide linkage, which in protein terminology is a peptide bond . Which amino acids are linked, and in what order - the protein sequence - is what distinguishes one protein from another, and is coded for by an organism's DNA. Protein sequences are written in the amino terminal (N-terminal) to carboxylate terminal (C-terminal) direction, with either three-letter or single-letter abbreviations for the amino acids (see amino acid table ). Below is a four amino acid peptide with the sequence "cysteine - histidine - glutamate - methionine". Using the single-letter code, the sequence is abbreviated CHEM. When an amino acid is incorporated into a protein it loses a molecule of water and what remains is called a residue of the original amino acid. Thus we might refer to the 'glutamate residue' at position 3 of the CHEM peptide above. Once a protein polymer is constructed, it in many cases folds up very specifically into a three-dimensional structure, which often includes one or more 'binding pockets' in which other molecules can be bound. It is this shape of this folded structure, and the precise arrangement of the functional groups within the structure (especially in the area of the binding pocket) that determines the function of the protein. Enzymes are proteins which catalyze biochemical reactions. One or more reacting molecules - often called substrates - become bound in the active site pocket of an enzyme, where the actual reaction takes place. Receptors are proteins that bind specifically to one or more molecules - referred to as ligands - to initiate a biochemical process. For example, we saw in the introduction to this chapter that the TrpVI receptor in mammalian tissues binds capsaicin (from hot chili peppers) in its binding pocket and initiates a heat/pain signal which is sent to the brain. Shown below is an image of the glycolytic enzyme fructose-1,6-bisphosphate aldolase (in grey), with the substrate molecule bound inside the active site pocket. (x-ray crystallographic data are from Protein Science 1999 , 8, 291 ; pdb code 4ALD. Image produced with JMol First Glance ) Intro to nucleic acids ⇒ Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) |
Courses/Fresno_City_College/Introductory_Chemistry_Atoms_First_for_FCC/01%3A__Basics_of_Measurement/1.03%3A_Conversions/1.3.02%3A_Solving_Multi-step_Conversion_Problems | Multiple Conversions Sometimes you will have to perform more than one conversion to obtain the desired unit. For example, suppose you want to convert 54.7 km into millimeters. We will set up a series of conversion factors so that each conversion factor produces the next unit in the sequence. W e first convert the given amount in km to the base unit, which is meters. We know that 1,000 m =1 km. Then we convert meters to mm, remembering that \(1\; \rm{mm}\) = \( 10^{-3}\; \rm{m}\). Concept Map Calculation \[ \begin{align*} 54.7 \; \cancel{\rm{km}} \times \dfrac{1,000 \; \cancel{\rm{m}}}{1\; \cancel{\rm{km}}} \times \dfrac{1\; \cancel{\rm{mm}}}{\cancel{10^{-3} \rm{m}}} & = 54,700,000 \; \rm{mm} \\ &= 5.47 \times 10^7\; \rm{mm} \end{align*} \nonumber \] In each step, the previous unit is canceled and the next unit in the sequence is produced, each successive unit canceling out until only the unit needed in the answer is left. Example \(\PageIndex{1}\): Unit Conversion Convert 58.2 ms to megaseconds in one multi-step calculation. Solution Steps for Problem Solving Unit Conversion Identify the "given" information and what the problem is asking you to "find." Given: 58.2 ms Find: Ms List other known quantities \(1 ms = 10^{-3} s \) \(1 Ms = 10^6s \) Prepare a concept map. Convert milliseconds to seconds to microseconds: use conversion factors 0.001 second per millisecond and 1 microsecond per 1 million seconds Calculate. \[ \begin{align} 58.2 \; \cancel{\rm{ms}} \times \dfrac{10^{-3} \cancel{\rm{s}}}{1\; \cancel{\rm{ms}}} \times \dfrac{1\; \rm{Ms}}{1,000,000\; \cancel{ \rm{s}}} & =0.0000000582\; \rm{Ms} \nonumber\\ &= 5.82 \times 10^{-8}\; \rm{Ms}\nonumber \end{align}\nonumber \] Neither conversion factor affects the number of significant figures in the final answer. Example \(\PageIndex{2}\): Unit Conversion How many seconds are in a day? Solution Steps for Problem Solving Unit Conversion Identify the "given" information and what the problem is asking you to "find." Given: 1 day Find: s List other known quantities. 1 day = 24 hours 1 hour = 60 minutes 1 minute = 60 seconds Prepare a concept map. Convert day to hour to minute to second: use conversion factors 24 hours per day, 60 minutes per hour, and 60 seconds per minute Calculate. \[1 \: \text{d} \times \dfrac{24 \: \text{hr}}{1 \: \text{d}}\times \dfrac{60 \: \text{min}}{1 \: \text{hr}} \times \dfrac{60 \: \text{s}}{1 \: \text{min}} = 86,400 \: \text{s} \nonumber \] Exercise \(\PageIndex{1}\) Perform each conversion in one multi-step calculation. 43.007 ng to kg 1005 in to ft 12 mi to km Answer a \(4.3007 \times 10^{-11} kg \) Answer b \(83.75\, ft\) Answer c \(19\, km\) Career Focus: Pharmacist A pharmacist dispenses drugs that have been prescribed by a doctor. Although that may sound straightforward, pharmacists in the United States must hold a doctorate in pharmacy and be licensed by the state in which they work. Most pharmacy programs require four years of education in a specialty pharmacy school. Pharmacists must know a lot of chemistry and biology so they can understand the effects that drugs (which are chemicals, after all) have on the body. Pharmacists can advise physicians on the selection, dosage, interactions, and side effects of drugs. They can also advise patients on the proper use of their medications, including when and how to take specific drugs properly. Pharmacists can be found in drugstores, hospitals, and other medical facilities. Curiously, an outdated name for pharmacist is chemist , which was used when pharmacists formerly did a lot of drug preparation, or compounding . In modern times, pharmacists rarely compound their own drugs, but their knowledge of the sciences, including chemistry, helps them provide valuable services in support of everyone’s health. Summary In multi-step conversion problems, the previous unit is canceled for each step and the next unit in the sequence is produced, each successive unit canceling out until only the unit needed in the answer is left. |
Courses/University_of_California_Davis/Chem_205%3A_Symmetry_Spectroscopy_and_Structure/02%3A_Electronic_Spectroscopy/2.08%3A_Symmetry_and_Formaldehyde | These five symmetry elements are tabulated in Table 2.8.1 with their corresponding operators. Symbol Elements Description Symbol Operator Symbol \(E\) identity \(\hat{E}\) no change \(C_n\) \(n\)-fold axis of rotation \(\hat{C}_n\) Rotation by \(360°/n\) leaves the molecule unchanged \(\sigma\) plane of symmetry \(\hat{\sigma}\) Reflection in the plane leaves the molecule unchanged \(i\) center of symmetry. \(\hat{i}\) Inversion through the center of symmetry leaves the molecule unchanged. \(S_n\) \(n\)-fold improper rotation \(\hat{S}_n\) The rotary reflection operation consists of rotating through an angle \(360°/n\) about the axis, followed by reflecting in a plane perpendicular to the axis. Every molecule has a point group associated with it, which are assigned by a set for rules (explained by Group theory ). The character tables takes the point group and represents all of the symmetry that the molecule has. A symmetry operation is a permutation of atoms such that the molecule is transformed into a state indistinguishable from the starting state. Character Tables A character table is a two dimensional chart associated with a point group that contains the irreducible representations of each point group along with their corresponding matrix characters. It also contains the Mulliken symbols used to describe the dimensions of the irreducible representations, and the functions for symmetry symbols for the Cartesian coordinates as well as rotations about the Cartesian coordinates. A character table can be separated into 6 different parts, namely: The Point Group The Symmetry Operation The Mulliken Symbols The Characters for the Irreducible Representations The Functions for Symmetry Symbols for Cartesian Coordinates and Rotations The Function for Symmetry Symbols for Square and Binary Products In many applications of group theory, we only need to know the characters of the representative matrices. Luckily, when each basis function transforms as a 1D irreducible representation (which is true in many cases of interest) there is a simple shortcut to determining the characters. All we have to do is to look at the way the individual basis functions transform under each symmetry operation. Characters For a given operation, step through the basis functions as follows: Add \(1\) to the character if the basis function is unchanged by the symmetry operation (i.e. the basis function is mapped onto itself); Add \(-1\) to the character if the basis function changes sign under the symmetry operation (i.e the basis function is mapped onto minus itself); Add \(0\) to the character if the basis function moves when the symmetry operation is applied (i.e the basis function is mapped onto something different from itself). Formaldehyde The MO’s form a basis for irreducible representation of the \(C_{2v}\) point-group of \(\ce{H2CO}\). Conventionally, the z-axis is along the \(\ce{C=O}\) bond and the x-axis is \(┴\) to place of the molecule. The symmetry operations for \(C_{2v}\) are \(E\), \(C_2\), \(σ_v(xy)\) and \(σ_{v’}(yx)\). For : \(π\) and \(π^{*}\) \[\hat{E} \pi=(+1) \pi\nonumber \] \[\hat{C}_2 \pi=(-1) \pi\nonumber \] \[\hat{\sigma}_{v} \pi=(+1) \pi\nonumber \] \[\hat{\sigma}_{v'} \pi=(-1) \pi\nonumber \] So \(π\) and \(π^{*}\) transform as the \(B_1\) irreducible representation \(σ\) and \(σ^{*}\) \[\hat{E} \sigma=(+1) \sigma\nonumber \] \[\hat{C}_2 \sigma=(+1) \sigma\nonumber \] \[\hat{\sigma}_{v} \sigma=(+1) \sigma\nonumber \] \[\hat{\sigma}_{v'} \sigma=(+1) \sigma\nonumber \] So \(σ\) and \(σ^{*}\) transform as the \(A_1\) irreducible representation n a \(n_a\) also transform as \(B_1\) it is a core electron in this orbital n b \[\hat{E} n_b=(+1) n_b\nonumber \] \[\hat{C}_2 n_b=(-1) n_b\nonumber \] \[\hat{\sigma}_{v} n_b=(-1) n_b\nonumber \] \[\hat{\sigma}_{v'} n_b=(+1) n_b\nonumber \] So \(n_b\) transforms as the \(B_2\) irreducible representation Having ascertained the symmetry species of the MO’s: \(A_1(σ, σ^{*}, n_a)\); \(B_1(π, π^*)\); \(B_2(n)b)\), we can ask about the symmetry of the state produced by a configuration. Direct Products of Representations Here are listed some helpful general rules for the product of two irreducible representations. For specific combinations not listed here, one can work out the product by multiplying the characters of each irreducible representation and solving the linear combination of the irreducible representations from the point group that generates that product. Often this process is simple, especially when one or both of the irreducible representations are non-degenerate (in most cases A or B ). Representation of an Electronic State The representation of a specific quantum electronic state can be evaluated by the direct product of the representations of the molecular orbitals of the occupied electrons. Direct products can be extracted from the tables above. For example, the non-degenerate representations below \[A \times A = A \nonumber \] \[B \times B = A \nonumber \] \[A \times B = B \nonumber \] Remember that the ground state valence electronic configuration of formaldehyde is: \[n_{a}^{2} \sigma^{2} \pi^{2} n_{b}^{2}\left(\pi^{*}\right)^{0}\left(\sigma^{*}\right)^0 \nonumber \] so the representation of the ground-state is \[\Gamma = (B_1 \times B_1) (A_1 \times A_1) (B_1 \times B_1) (B_2 \times B_2) \label{pre} \] This can be simplified using the direct product tables \[\Gamma = (A_1) (A_1) (A_1) (A_1) \nonumber \] so the ground state has a representation of \(A_1\). We can make a general "rule" from this. This is a “closed shell” configuration and corresponds to a state with all molecular orbitals doubly occupied or empty and must be a singlet state! Since there are no odd electrons in the orbitals in the ground state, the configuration has a \(^1A_1\) symmetry (totally symmetry). Rule: Simpler Calculation The symmetry representation of an electronic state is the direct product of the symmetry representations of each of the odd electrons orbitals. Since doubly filled orbitals do not contribute since their products are always totally symmetry: \[B_1 \times B_1=A_1\nonumber \] \[A_2 \times A_{2}=A_1\nonumber \] etc. There are no odd electrons in the orbitals in the ground state. Possible excited state symmetry includes: \[n_{b} \rightarrow \pi^{*} \text { or }^{1}\left(n_{b}, \pi^{*}\right)\nonumber \] and pay attention to only the unpair electrons give the following representation Notation We use lower case representations for designating 1e- orbitals. We use capital case representation for designated are electronic states \[\Gamma=b_{2} \times b_{1}=A_{2}\nonumber \] The electronic state symmetry is thus \(^1A_2\). Conventionally, the \(n_{b} \rightarrow \pi^{*}\) transition, in terms of states, is described as \[{ }^{1} A_{2} \leftarrow{ }^{1} A_{1}\nonumber \] or \[{ }^{1} A_{2}\left(n, \pi^{*}\right) \leftarrow{ }^{1} A_{1}\nonumber \] Higher energy state is conventionally places on the left hand side and the arrow points in the direction of the transition. Example \(\PageIndex{1}\) The \({ }^{1}\left(\pi, \pi^{*}\right)\) state has the symmetry of \[\mathrm{b}_{1} \times \mathrm{b}_{1}=\mathrm{a}_{1}\nonumber \] thus this transition is designated \[{ }^{1} A_{1}\left(\pi, \pi^{*}\right) \leftarrow{ }^{1} A_{1}\nonumber \] and this transition moves electron density from the O to the C, this is because \[\pi=a\left(2 p_{x}^{C}\right)+b\left(2 p_{x}^{o}\right)\nonumber \] \(b > a\) since \(O\) is more electronegative \[\pi^{*}=b^{\prime}\left(2 p_{x}^{C}\right)+a^{\prime}\left(p_{x}^{O}\right)\nonumber \] \(b’ > a’\) since for orthogonality with \(π\). Now, the \({ }^{1} A_{2} \leftarrow{ }^{1} A_{1}\) transition also moves electron density from the O to the C. (For a similar argument). References https://chem.libretexts.org/Bookshel...aracter_Tables http://www1.udel.edu/pchem/C444/Lect.../Lecture31.pdf |
Courses/Brevard_College/CHE_201%3A_Organic_Chemistry_I/01%3A_Introduction_to_Organic_Chemistry/1.03%3A_Covalent_Bonds | Learning Objectives To describe how a covalent bond forms. To apply the octet rule to covalent compounds You have already seen examples of substances that contain covalent bonds. One substance mentioned previously was water (\(\ce{H2O}\)). You can tell from its formula that it is not an ionic compound; it is not composed of a metal and a nonmetal. Consequently, its properties are different from those of ionic compounds. Electron Sharing Previously, we discussed ionic bonding where electrons can be transferred from one atom to another so that both atoms have an energy-stable outer electron shell. Because most filled electron shells have eight electrons in them, chemists called this tendency the octet rule. However, there is another way an atom can achieve a full valence shell: atoms can share electrons. This concept can be illustrated by using two hydrogen atoms, each of which has a single electron in its valence shell. (For small atoms such as hydrogen atoms, the valence shell will be the first shell, which holds only two electrons.) We can represent the two individual hydrogen atoms as follows: In contrast, when two hydrogen atoms get close enough together to share their electrons, they can be represented as follows: By sharing their valence electrons, both hydrogen atoms now have two electrons in their respective valence shells. Because each valence shell is now filled, this arrangement is more stable than when the two atoms are separate. The sharing of electrons between atoms is called a covalent bond, and the two electrons that join atoms in a covalent bond are called a bonding pair of electrons. A discrete group of atoms connected by covalent bonds is called a molecule—the smallest part of a compound that retains the chemical identity of that compound. Chemists frequently use Lewis diagrams to represent covalent bonding in molecular substances. For example, the Lewis diagrams of two separate hydrogen atoms are as follows: The Lewis diagram of two hydrogen atoms sharing electrons looks like this: This depiction of molecules is simplified further by using a dash to represent a covalent bond. The hydrogen molecule is then represented as follows: Remember that the dash, also referred to as a single bond, represents a pair of electrons. The bond in a hydrogen molecule, measured as the distance between the two nuclei, is about 7.4 × 10 −11 m, or 74 picometers (pm; 1 pm = 1 × 10 −12 m). This particular bond length represents a balance between several forces: the attractions between oppositely charged electrons and nuclei, the repulsion between two negatively charged electrons, and the repulsion between two positively charged nuclei. If the nuclei were closer together, they would repel each other more strongly; if the nuclei were farther apart, there would be less attraction between the positive and negative particles. Fluorine is another element whose atoms bond together in pairs to form diatomic (two-atom) molecules. Two separate fluorine atoms have the following electron dot diagrams: Each fluorine atom contributes one valence electron, making a single bond and giving each atom a complete valence shell, which fulfills the octet rule: The circles show that each fluorine atom has eight electrons around it. As with hydrogen, we can represent the fluorine molecule with a dash in place of the bonding electrons: Each fluorine atom has six electrons, or three pairs of electrons, that are not participating in the covalent bond. Rather than being shared, they are considered to belong to a single atom. These are called nonbonding pairs (or lone pairs) of electrons. Covalent Bonds between Different Atoms Now that we have looked at electron sharing between atoms of the same element, let us look at covalent bond formation between atoms of different elements. Consider a molecule composed of one hydrogen atom and one fluorine atom: Each atom needs one additional electron to complete its valence shell. By each contributing one electron, they make the following molecule: In this molecule, the hydrogen atom does not have nonbonding electrons, while the fluorine atom has six nonbonding electrons (three lone electron pairs). The circles show how the valence electron shells are filled for both atoms. Example \(\PageIndex{1}\) Use Lewis diagrams to indicate the formation of the following: Cl 2 HBr Solution a. When two chlorine atoms form a chlorine molecule, they share one pair of electrons. In Cl 2 molecule, each chlorine atom is surrounded by an octet number of electrons. The Lewis diagram for a Cl 2 molecule is similar to the one for F 2 (shown above). b. When a hydrogen atom and a bromine atom form HBr, they share one pair of electrons. In the HBr molecule, H achieves a full valence of two electrons ( duet ) while Br achieves an octet . The Lewis diagram for HBr is similar to that for HF shown above. Exercise \(\PageIndex{1}\) Draw the Lewis diagram for each compound. a molecule composed of one chlorine atom and one fluorine atom a molecule composed of one hydrogen atom and one iodine atom Answer a: Answer b: Covalent Bonds in Larger Molecules The formation of a water molecule from two hydrogen atoms and an oxygen atom can be illustrated using Lewis dot symbols (shown below). The structure on the right is the Lewis electron structure , or Lewis structure , for \(\ce{H2O}\). With two bonding pairs and two lone pairs, the oxygen atom has now completed its octet . Moreover, by sharing a bonding pair with oxygen, each hydrogen atom now has a full valence shell of two electrons. Chemists usually indicate a bonding pair by a single line, as shown (below). Other large molecules are constructed in a similar fashion, with some atoms participating in more than one covalent bond. For example, methane (\(\ce{CH4}\)), the central carbon atom bonded to four hydrogen atoms, can be represented using either of the Lewis structures below. Again, sharing electrons between C and H atoms results in C achieving and octet while H achieving a duet number of electrons. How Many Covalent Bonds Are Formed? 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). In the Lewis structure, the number of bonds formed by an element in a neutral compound is the same as the number of unpaired electrons it must share with other atoms to complete its octet of electrons. For example, each atom of a group 4A (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 CH 4 (methane). Group 5A (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 6A (16) obtain an octet by forming two covalent bonds . Fluorine and the other halogens in group 7A (17) have seven valence electrons and can obtain an octet by forming one covalent bond . Typically, the atoms of group 4A form 4 covalent bonds; group 5A form 3 bonds; group 6A form 2 bonds; and group 7A form one bond. The number of electrons required to obtain an octet determines the number of covalent bonds an atom can form. This is summarized in the table below. In each case, the sum of the number of bonds and the number of lone pairs is 4, which is equivalent to eight (octet) electrons. Atom (Group number) Number of Bonds Number of Lone Pairs Carbon (Group 14 or 4A) 4 0 Nitrogen (Group 15 or 5A) 3 1 Oxygen (Group 16 or 6A) 2 2 Fluorine (Group 17 or 7A) 1 3 Because hydrogen only needs two electrons to fill its valence shell, it follows the duet rule. It is an exception to the octet rule. Hydrogen only needs to form one bond. This is the reason why H is always a terminal atom and never a central atom. Figure \(\PageIndex{1}\) shows the number of covalent bonds various atoms typically form. 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. Example \(\PageIndex{2}\) Examine the Lewis structure of OF 2 below. Count the number of bonds formed by each element. Based on the element's location in the periodic table, does it correspond to the expected number of bonds shown in Table 4.1? Does the Lewis structure below follow the octet rule? Solution Yes. F (group 7A) forms one bond and O (group 6A) forms 2 bonds. Each atom is surrounded by 8 electrons. This structure satisfies the octet rule. Examine the Lewis structure of NCl 3 below. Count the number of bonds formed by each element. Based on the element's location in the periodic table, does it correspond to the expected number of bonds shown in Table 4.1? Does the Lewis structure below follow the octet rule? Answer Both Cl and N form the expected number of bonds. Cl (group 7A) has one bond and 3 lone pairs. The central atom N (group 5A) has 3 bonds and one lone pair. Yes, the Lewis structure of NCl 3 follows the octet rule. Key Takeaways A covalent bond is formed between two atoms by sharing electrons. The number of bonds an element forms in a covalent compound is determined by the number of electrons it needs to reach octet. Hydrogen is an exception to the octet rule. H forms only one bond because it needs only two electrons. |
Courses/University_of_Illinois_UrbanaChampaign/Chem_2363A_Fundamental_Organic_Chemistry_I_(Chan)/09%3A_The_Chemistry_of_Alkyl_Halides/9.11%3A_Determining_SN2%2C_SN%C2%AC1%2C_E2_or_E1 | The functional group of alkyl halides is a carbon-halogen bond, the common halogens being fluorine, chlorine, bromine and iodine. With the exception of iodine, these halogens have electronegativities significantly greater than carbon. Consequently, this functional group is polarized so that the carbon is electrophilic and the halogen is nucleophilic, as shown in the drawing below. Two characteristics other than electronegativity also have an important influence on the chemical behavior of these compounds. The first of these is covalent bond strength . The strongest of the carbon-halogen covalent bonds is that to fluorine. Remarkably, this is the strongest common single bond to carbon, being roughly 30 kcal/mole stronger than a carbon-carbon bond and about 15 kcal/mole stronger than a carbon-hydrogen bond. Because of this, alkyl fluorides and fluorocarbons in general are chemically and thermodynamically quite stable , and do not share any of the reactivity patterns shown by the other alkyl halides. The carbon-chlorine covalent bond is slightly weaker than a carbon-carbon bond, and the bonds to the other halogens are weaker still, the bond to iodine being about 33% weaker. The second factor to be considered is the relative stability of the corresponding halide anions , which is likely the form in which these electronegative atoms will be replaced. This stability may be estimated from the relative acidities of the H-X acids, assuming that the strongest acid releases the most stable conjugate base (halide anion). With the exception of HF (pK a = 3.2), all the hydrohalic acids are very strong, small differences being in the direction HCl < HBr < HI. The characteristics noted above lead us to anticipate certain types of reactions that are likely to occur with alkyl halides. The following table summarizes the expected outcome of alkyl halide reactions with nucleophiles. It is assumed that the alkyl halides have one or more beta-hydrogens, making elimination possible; and that low dielectric solvents (e.g. acetone, ethanol, tetrahydrofuran & ethyl acetate) are used. When a high dielectric solvent would significantly influence the reaction this is noted in red. Note that halogens bonded to sp 2 or sp hybridized carbon atoms do not normally undergo substitution or elimination reactions with nucleophilic reagents . 0 1 2 3 Nucleophile Ionic Nucleophiles ( Weak Bases: I–, Br–, SCN–, N3–, CH3CO2– , RS–, CN– etc. ) pKa's from -9 to 10 (left to right) Anionic Nucleophiles ( Strong Bases: HO–, RO– ) pKa's > 15 Neutral Nucleophiles ( H2O, ROH, RSH, R3N ) pKa's ranging from -2 to 11 Alkyl Group Ionic Nucleophiles ( Weak Bases: I–, Br–, SCN–, N3–, CH3CO2– , RS–, CN– etc. ) pKa's from -9 to 10 (left to right) Anionic Nucleophiles ( Strong Bases: HO–, RO– ) pKa's > 15 Neutral Nucleophiles ( H2O, ROH, RSH, R3N ) pKa's ranging from -2 to 11 Primary RCH2– Rapid SN2 substitution. The rate may be reduced by substitution of β-carbons, as in the case of neopentyl. Rapid SN2 substitution. E2 elimination may also occur. e.g. ClCH2CH2Cl + KOH → CH2=CHCl SN2 substitution. (N ≈ S >>O) Secondary R2CH– SN2 substitution and / or E2 elimination (depending on the basicity of the nucleophile). Bases weaker than acetate (pKa = 4.8) give less elimination. The rate of substitution may be reduced by branching at the β-carbons, and this will increase elimination. E2 elimination will dominate. SN2 substitution. (N ≈ S >>O) In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be formed slowly. Tertiary R3C– E2 elimination will dominate with most nucleophiles (even if they are weak bases). No SN2 substitution due to steric hindrance. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be expected. E2 elimination will dominate. No SN2 substitution will occur. In high dielectric ionizing solvents SN1 and E1 products may be formed. E2 elimination with nitrogen nucleophiles (they are bases). No SN2 substitution. In high dielectric ionizing solvents SN1 and E1 products may be formed. Allyl H2C=CHCH2– Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides. E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides. Benzyl C6H5CH2– Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides (note there are no β hydrogens). E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides. Contributors William Reusch, Professor Emeritus ( Michigan State U. ), Virtual Textbook of Organic Chemistry Additional Resources MasterOrganicChemistry Trapped In SN1/SN2/E1/E2 Hell? Some Resources Deciding SN1/SN2/E1/E2 (1) – The Substrate Deciding SN1/SN2/E1/E2 (2) – The Nucleophile/Base Deciding SN1/SN2/E1/E2 (3) – The Solvent Deciding SN1/SN2/E1/E2 (4) – The Temperature Wrapup – The Quick N’ Dirty Guide To SN1/SN2/E1/E2 Khan Academy Nucleophilicity and Basicity Primary and tertiary alkyl halides Secondary Alkyl Halides Leah4Sci Alkyl Halide Carbon Chain Analysis for SN1 SN2 E1 E2 Reactions Slide Presentations SN1 SN2 E1 E2 comparison slides Web Pages Elimination vs Substitution S N 1, S N 2, E1, E2 Summary Summary of S N 1, S N 2, E1 and E2 * Comparison of S N 1, S N 2, E1, E2 Substituation vs elimination considerations Relationship between Sn1 and E1 Substitution vs Elimination summary Good summation of SN2/SN1/E1/E2 reaction properties Good handouts of substitution and elimination Good summary of SN1, Sn2, E1, E2 SN1 SN2 E1 E2 comparison Videos S N 1, S N 2, E1, E2 Role of solvent in S N 1, S N 2, E1, E2 LONG video on SN1 SN2 E1 E2 Practice Problems Answers to elimination questions |
Courses/University_of_Pittsburgh_at_Bradford/CHEM_0089_-_Concepts_of_Chemistry/10%3A_Equilibria/10.21%3A_Solubility_Product_Constant_(left(_K_textsp_right)) | At one time, a major analytical technique was gravimetric analysis. Gravimetric analysis involves an ion being precipitated out of solution, purified, and weighed to determine the amount of that ion in the original material. As an example, measurement of \(\ce{Ca^{2+}}\) involved dissolving the sample in water, precipitating the calcium as calcium oxalate, purifying the precipitate, drying it, and weighing the final product. Although this approach can be very accurate (atomic weights for many elements were determined this way), the process is slow, tedious, and prone to a number of errors in technique. Newer methods are now available that measure minute amounts of calcium ions in solution without the long, involved gravimetric approach. Solubility Product Constant Ionic compounds have widely differing solubilities. Sodium chloride has a solubility of about \(360 \: \text{g}\) per liter of water at \(25^\text{o} \text{C}\). Salts of alkali metals tend to be quite soluble. On the other end of the spectrum, the solubility of zinc hydroxide is only \(4.2 \times 10^{-4} \text{g/L}\) of water at the same temperature. Many ionic compounds containing hydroxide are relatively insoluble. Most ionic compounds that are considered to be insoluble will still dissolve to a small extent in water. These "mostly insoluble" compounds are considered to be strong electrolytes, because the portion of the compound that dissolves also dissociates. As an example, silver chloride dissociates to a small extent into silver ions and chloride ions upon being added to water. \[\ce{AgCl} \left( s \right) \rightleftharpoons \ce{Ag^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)\nonumber \] The process is written as an equilibrium because the dissociation occurs only to a small extent. Therefore, an equilibrium expression can be written for the process. Keep in mind that the solid silver chloride does not have a variable concentration, and so is not included in the expression. \[K_\text{sp} = \left[ \ce{Ag^+} \right] \left[ \ce{Cl^-} \right]\nonumber \] This equilibrium constant is called the solubility product constant \(\left( K_\text{sp} \right)\) and is equal to the mathematical product of the ions each raised to the power of the coefficient of the ion in the dissociation equation. The stoichiometry of the formula of the ionic compound dictates the form of the \(K_\text{sp}\) expression. For example, the formula of calcium phosphate is \(\ce{Ca_3(PO_4)_2}\). The dissociation equation and \(K_\text{sp}\) expression are shown below: \[\ce{Ca_3(PO_4)_2} \left( s \right) \rightleftharpoons 3 \ce{Ca^{2+}} \left( aq \right) + 2 \ce{PO_4^{3-}} \left( aq \right) \: \: \: K_\text{sp} = \left[ \ce{Ca^{2+}} \right]^3 \left[ \ce{PO_4^{3-}} \right]^2\nonumber \] The table below lists solubility product constants for some common nearly insoluble ionic compounds. Table \(\PageIndex{1}\): Solubility Product Constants \(\left( 25^\text{o} \text{C} \right)\) Table \(\PageIndex{1}\): Solubility Product Constants \(\left( 25^\text{o} \text{C} \right)\).1 Table \(\PageIndex{1}\): Solubility Product Constants \(\left( 25^\text{o} \text{C} \right)\).2 Table \(\PageIndex{1}\): Solubility Product Constants \(\left( 25^\text{o} \text{C} \right)\).3 Compound \(K_\text{sp} \) Compound \(K_\text{sp}\) \(\ce{AgBr}\) \(5.0 \times 10^{-13}\) \(\ce{CuS}\) \(8.0 \times 10^{-37}\) \(\ce{AgCl}\) \(1.8 \times 10^{-10}\) \(\ce{Fe(OH)_2}\) \(7.9 \times 10^{-16}\) \(\ce{Al(OH)_3}\) \(3.0 \times 10^{-34}\) \(\ce{Mg(OH)_2}\) \(7.1 \times 10^{-12}\) \(\ce{BaCO_3}\) \(5.0 \times 10^{-9}\) \(\ce{PbCl_2}\) \(1.7 \times 10^{-5}\) \(\ce{BaSO_4}\) \(1.1 \times 10^{-10}\) \(\ce{PbCO_3}\) \(7.4 \times 10^{-14}\) \(\ce{CaCO_3}\) \(4.5 \times 10^{-9}\) \(\ce{PbI_2}\) \(7.1 \times 10^{-9}\) \(\ce{Ca(OH)_2}\) \(6.5 \times 10^{-6}\) \(\ce{PbSO_4}\) \(6.3 \times 10^{-7}\) \(\ce{Ca_3(PO_4)_2}\) \(1.2 \times 10^{-26}\) \(\ce{Zn(OH)_2}\) \(3.0 \times 10^{-16}\) \(\ce{CaSO_4}\) \(2.4 \times 10^{-5}\) \(\ce{ZnS}\) \(3.0 \times 10^{-23}\) Summary The solubility product constant is equal to the mathematical product of ions each raised to the power of their coefficients in a dissociation equation. Calculations using solubility product constants are illustrated. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/20%3A_Entropy_and_The_Second_Law_of_Thermodynamics/20.04%3A_The_Second_Law_of_Thermodynamics | Spontaneity of an isolated system An isolated system is a little more than just adiabatic. In the latter heat cannot get in or out. In an isolated system nothing gets in or out, neither heat nor mass nor even any radiation, such as light. The isolated system is like a little universe all to itself. Let us consider a zero law process. We have two identical blocks of metal, say aluminum. They are each at thermal equilibrium, but at different temperatures. They are brought into contact with each other but isolated from the rest of the universe. Zeroth law : Heat will flow from hot to cold First law : There is no change in total energy so: \[dU_A= -dU_B \nonumber \] There is also no work so: \[dU_A = \delta q_A + 0 \nonumber \] Because U is a state function this makes \(q\) a state function as well, otherwise this equality does not hold. As there is only one term on the right there is only one path (along q). So we could write: \[dU_A = dq_A \nonumber \] This implies that we do not need to worry about reversible and irreversible paths as there is only one path. Since: \[dS = \dfrac{\delta q_{rev}}{T} \nonumber \] In this particular case: \[TdS = \delta q_{rev} = dU \nonumber \] Thus we get: \[ dS = \dfrac{dU_A}{T_A} + \dfrac{dU_B}{T_B} = \dfrac{dU_A}{T_A} - \dfrac{dU_A}{T_B} = dU_A \left( \dfrac{1}{T_A} - \dfrac{1}{T_B} \right) \nonumber \] Clearly as long as the two temperatures are not the same \(dS\) is not zero and entropy is not conserved. Instead it is increasing . Over time, the temperatures will become the same (if the blocks are identical, the final temperature is the average of \(T_A\) and \(T_B\)) and the entropy will reach a maximum. For our two identical blocks of metal (with same heat capacity, \(C_V\)), we can, in fact, derive that the entropy change: \[\Delta S = C_V \ln \left[\dfrac{T_A^2 + T_B^2}{4T_AT_B}\right]. \nonumber \] This is indeed a positive quantity. In general, we can say for an isolated system: \[dS \ge 0 \nonumber \] Thus if we are dealing with a spontaneous (and isolated) process \(dS >0\) and entropy is being produced . This gives us a criterion for spontaneity . Entropy exchange of an open system In an isolated system \(dS\) represents the produced entropy \(dS_{prod}\) and this is a good criterion for spontaneity. Of course the requirement that the system is isolated is very restrictive and makes the criterion as good as useless... What happens in a system that can exchange heat with the rest of the universe? We do have entropy changes in that case, but part of them may have nothing to do with production, because we also have to consider the heat that is exchanged. \[dS = dS_{prod} + dS_{exchange} \nonumber \] If the process is reversible (that is completely non-spontaneous) we are dealing with \(\delta q_{rev}\) so that \(dS_{exchange} = \delta q_{rev} /T\), but that is also what \(dS_{tot}\) is equal to (by definition). This leaves no room for entropy production. So we have: Isolated: \(dS = dS_{prod} + 0\) Reversible \(dS = 0 + \delta q_{rev}/T\) Notice that this demonstrates that for non-isolated systems entropy change is not a good criterion for spontaneity at all... In the case the heat exchange is irreversible part of the entropy is entropy production by the system: Irreversible: \(dS = dS_{prod} + \delta q_{irrev}/T\) \(dS > \delta q_{irrev}/T\) in this case. Generalizing the isolated, irreversible and reversible cases we may say: \[dS \ge \dfrac{δq}{T} \nonumber \] This is the Clausius inequality . |
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