Textbook Question
The following reaction has a value of ΔG° = –2.1 kJ/mol (–0.50 kcal/mol).
CH3Br + H2S ⇌ CH3SH + HBr
b. Starting with a 1 M solution of CH3Br and H2S, calculate the final concentrations of all four species at equilibrium.
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Ch.4 - The Study of Chemical Reactions
Wade9th EditionOrganic ChemistryISBN: 9780135213728
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Table of contents
- Ch.1 - Structure and Bonding107
- Ch. 2 - Acids and Bases; Functional Groups115
- Ch.3 - Structure and Stereochemistry of Alkanes100
- Ch.4 - The Study of Chemical Reactions99
- Ch.5 - Stereochemistry93
- Ch.6 - Alkyl Halides; Nucleophilic Substitution118
- Ch. 7 - Structure and Synthesis of Alkenes; Elimination158
- Ch.8 - Reactions of Alkenes171
- Ch.9 - Alkynes91
- Ch.10 - Structure and Synthesis of Alcohols143
- Ch.11 - Reactions of Alcohols104
- Ch. 12 - Infrared Spectroscopy and Mass Spectrometry22
- Ch. 13 - Nuclear Magnetic Resonance Spectroscopy45
- Ch. 14 - Ethers, Epoxides, and Thioethers72
- Ch. 15 - Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy89
- Ch. 16 - Aromatic Compounds74
- Ch. 17 - Reactions of Aromatic Compounds126
- Ch. 18 - Ketones and Aldehydes193
- Ch. 19 - Amines129
- Ch. 20 - Carboxylic Acids77
- Ch. 21 - Carboxylic Acid Derivatives141
- Ch. 22 - Condensations and Alpha Substitutions of Carbonyl Compounds175
- Ch. 23 - Carbohydrates and Nucleic Acids93
- Ch. 24 - Amino Acids, Peptides, and Proteins54
Chapter 4, Problem 5a
The following reaction has a value of ΔG° = –2.1 kJ/mol (–0.50 kcal/mol).
CH3Br + H2S ⇌ CH3SH + HBr
a. Calculate Keq at room temperature (25 °C) for this reaction as written.
Verified step by step guidance1
Understand the relationship between ΔG (Gibbs free energy change) and the equilibrium constant (Keq). The equation to use is ΔG = -RT ln(Keq), where R is the gas constant and T is the temperature in Kelvin.
Convert the temperature from Celsius to Kelvin. Use the formula T(K) = T(°C) + 273.15. For 25 °C, T = 25 + 273.15 = 298.15 K.
Identify the value of the gas constant R. For this calculation, use R = 8.314 J/(mol·K) (note that ΔG is given in kJ/mol, so ensure consistent units).
Rearrange the equation ΔG = -RT ln(Keq) to solve for Keq. The rearranged formula is Keq = e^(-ΔG / RT).
Substitute the given values into the equation: ΔG = -2.1 kJ/mol (convert to J/mol by multiplying by 1000), R = 8.314 J/(mol·K), and T = 298.15 K. Perform the calculation to find Keq.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Gibbs Free Energy (ΔG)
Gibbs Free Energy (ΔG) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. A negative ΔG indicates that a reaction is spontaneous, meaning it can proceed without external energy input. The magnitude of ΔG also provides insight into the reaction's favorability and equilibrium position.
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Breaking down the different terms of the Gibbs Free Energy equation.
Equilibrium Constant (Keq)
The equilibrium constant (Keq) is a dimensionless value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. It is derived from the law of mass action and provides insight into the extent of a reaction; a larger Keq indicates a greater concentration of products at equilibrium, while a smaller Keq suggests more reactants.
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02:19The relationship between equilibrium constant and pKa.
Relationship between ΔG and Keq
The relationship between Gibbs Free Energy (ΔG) and the equilibrium constant (Keq) is described by the equation ΔG = -RT ln(Keq), where R is the universal gas constant and T is the temperature in Kelvin. This equation allows for the calculation of Keq from ΔG values, indicating how the spontaneity of a reaction correlates with its equilibrium position. A negative ΔG corresponds to a Keq greater than 1, suggesting that products are favored at equilibrium.
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Related Practice
Textbook Question
Use the bond-dissociation enthalpies in Table 4-2 (page 167) to calculate the heats of reaction for the two possible first propagation steps in the chlorination of isobutane. Use this information to draw a reaction-energy diagram like Figure 4-8, comparing the activation energies for formation of the two radicals.
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Textbook Question
Under base-catalyzed conditions, two molecules of acetone can condense to form diacetone alcohol. At room temperature (25 °C), about 5% of the acetone is converted to diacetone alcohol. Determine the value of ΔG° for this reaction.
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Textbook Question
Free-radical chlorination of hexane gives very poor yields of 1-chlorohexane, while cyclohexane can be converted to chlorocyclohexane in good yield.
a. How do you account for this difference?
b. What ratio of reactants (cyclohexane and chlorine) would you use for the synthesis of chlorocyclohexane?
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Textbook Question
The reaction of tert-butyl chloride with methanol(CH3)3C—Cl Tert-butylchloride + CH3—OH methanol —> (CH3)C—OCH3 methyltert-butylether + HCl is found to follow the rate equation rate= Kr[(CH3)3C—Cl] c. What is the kinetic order overall?
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Textbook Question
The dehydrogenation of butane to trans-but-2-ene has ΔH° = +116 kJ/mol (+27.6 kcal/mol) and ΔS° = +117J/kelvin-mol (+28.0 cal/kelvin-mol). a. Compute the value of ΔG° for dehydrogenation at room temperature (25 °C or 298 °K). Is dehydrogenation favored or disfavored?HINT: When you are doing synthesis problems, avoid using these high-temperature industrial methods. They require specialized equipment, and they produce variable mixtures of products.
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