Table of contents

1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

6. Thermodynamics and Kinetics

Entropy

Organic Chemistry

6. Thermodynamics and Kinetics

Entropy

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4:03 minutes

Problem 61

Textbook Question

a. What is the equilibrium constant for a reaction that is carried out at 25 °C (298 K) with ∆H° = 20 kcal/mol and ∆S° = 5.0 × 10^-2 kcal mol^-1 K^-1?
b. What is the equilibrium constant for the same reaction carried out at 125 °C?

Verified step by step guidance

Step 1: Recall the relationship between the Gibbs free energy change (ΔG°) and the equilibrium constant (K). The equation is ΔG° = -RT ln(K), where R is the gas constant (1.987 cal mol⁻¹ K⁻¹), T is the temperature in Kelvin, and K is the equilibrium constant.

Step 2: Use the thermodynamic relationship ΔG° = ΔH° - TΔS° to calculate ΔG° at each temperature. For part (a), substitute ΔH° = 20 kcal/mol, ΔS° = 5.0 × 10⁻² kcal mol⁻¹ K⁻¹, and T = 298 K into the equation. For part (b), use T = 125 °C, which must first be converted to Kelvin (T = 125 + 273 = 398 K).

Step 3: Convert ΔH° and ΔS° into consistent units if necessary. Since R is given in cal mol⁻¹ K⁻¹, convert ΔH° and ΔS° from kcal to cal by multiplying by 1000 (1 kcal = 1000 cal). This ensures all units are consistent for calculations.

Step 4: Calculate ΔG° for each temperature using the equation ΔG° = ΔH° - TΔS°. Substitute the values for ΔH°, T, and ΔS° into the equation for both temperatures (298 K and 398 K).

Step 5: Rearrange the equation ΔG° = -RT ln(K) to solve for K. Use the calculated ΔG° values for each temperature and substitute them into the equation ln(K) = -ΔG° / (RT). Finally, exponentiate both sides to solve for K: K = e^(-ΔG° / (RT)).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Gibbs Free Energy

Gibbs Free Energy (G) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. The change in Gibbs Free Energy (∆G) is related to the equilibrium constant (K) of a reaction through the equation ∆G = -RT ln(K), where R is the universal gas constant and T is the temperature in Kelvin. A negative ∆G indicates a spontaneous reaction, while a positive ∆G suggests non-spontaneity.

Equilibrium Constant (K)

The equilibrium constant (K) 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 standard Gibbs Free Energy change (∆G°) of the reaction, and it can be calculated using the equation K = e^(-∆G°/RT). The value of K provides insight into the position of equilibrium, indicating whether products or reactants are favored.

Van 't Hoff Equation

The Van 't Hoff equation relates the change in the equilibrium constant (K) of a reaction to the change in temperature (T) and the enthalpy change (∆H°) of the reaction. It is expressed as ln(K2/K1) = -∆H°/R(1/T2 - 1/T1). This equation allows for the calculation of the equilibrium constant at different temperatures, providing a way to understand how temperature influences the position of equilibrium in a chemical reaction.

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Related Practice

Textbook Question

In Chapter 13, we discuss the ring-opening reactions of epoxides, such as the one shown here.

(b) Predict the sign of ∆S°.

Textbook Question

Would you expect ∆S to be greater than, less than, or equal to zero in the following reactions?

Textbook Question

For a reaction carried out at 25 °C with an equilibrium constant of 1 × 10^-3, to increase the equilibrium constant by a factor of 10:

a. how much must ∆G° change?

b. how much must ∆H° change if ∆S° = 0 kcal mol^-1K^-1?

c. how much must ∆S° change if ∆H° = 0 kcal mol^-1?

Textbook Question

From the following rate constants, determined at five temperatures, calculate the experimental energy of activation and ∆G^‡, ∆H^‡, and ∆S^‡ for the reaction at 30 °C:

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Textbook Question

(ii) Which side of the reaction is favored by entropy? (iii) If ∆S° = 0 for these reactions, calculate ∆G° (Assume T = 298 K) [BDE for O―H = 110 kcal /mol.]

(b)

Textbook Question

For the following values of ∆H° , ∆S°, and T, tell whether the process would be favored.

(b) ∆H° = +7.34 kcal/mol ; ∆S° = +43 cal/mol•K ; T = 325 K

Textbook Question

Considering the process described in Assessment 5.13, will it be favored or disfavored at a temperature higher than the one you calculated? How about at a temperature below what you calculated?

Textbook Question

For the following values of ∆H° , ∆S°, and T, tell whether the process would be favored.

(d) ∆H° = -8.3 kcal/mol ; ∆S° = -12 cal/mol•K ; T = 298 K

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