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

Gibbs Free Energy

Organic Chemistry

6. Thermodynamics and Kinetics

Gibbs Free Energy

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8:09 minutes

Problem 22

Textbook Question

a. For a reaction with ∆H° = -12 kcal/mol and ∆S° = 0.01 kcal mol^-1 K^-1, calculate the ∆G° and the equilibrium constant at:
1. 30 °C and 2. 150 °C.
b. How does ∆G° change as T increases?
c. How does K_eq change as T increases?

Verified step by step guidance

Step 1: Recall the Gibbs free energy equation: ΔG° = ΔH° - TΔS°. Here, ΔH° is the enthalpy change, ΔS° is the entropy change, and T is the temperature in Kelvin. Convert the given temperatures (30 °C and 150 °C) to Kelvin using the formula T(K) = T(°C) + 273.

Step 2: For part (a), calculate ΔG° at each temperature using the Gibbs free energy equation. Substitute the values of ΔH° (-12 kcal/mol), ΔS° (0.01 kcal/mol·K), and the respective temperatures in Kelvin into the equation. Perform the subtraction to find ΔG° for each temperature.

Step 3: To calculate the equilibrium constant (Keq), use the relationship between ΔG° and Keq: ΔG° = -RT ln(Keq), where R is the gas constant (0.001987 kcal/mol·K). Rearrange the equation to solve for Keq: Keq = e^(-ΔG°/RT). Substitute the values of ΔG°, R, and T for each temperature to find Keq.

Step 4: For part (b), analyze how ΔG° changes as T increases. Since ΔG° = ΔH° - TΔS°, observe that the term TΔS° becomes larger as T increases. If ΔS° is positive, the magnitude of ΔG° decreases (becomes less negative or more positive) as T increases.

Step 5: For part (c), analyze how Keq changes as T increases. Since Keq = e^(-ΔG°/RT), if ΔG° becomes less negative or more positive as T increases, the exponent (-ΔG°/RT) becomes smaller, leading to a decrease in Keq. Conversely, if ΔG° becomes more negative, Keq increases.

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Key Concepts

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

Gibbs Free Energy (∆G°)

Gibbs Free Energy (∆G°) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. It is calculated using the equation ∆G° = ∆H° - T∆S°, where ∆H° is the change in enthalpy, T is the temperature in Kelvin, and ∆S° is the change in entropy. A negative ∆G° indicates a spontaneous reaction, while a positive value suggests non-spontaneity.

Equilibrium Constant (Keq)

The equilibrium constant (Keq) quantifies the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction. It is related to Gibbs Free Energy by the equation ∆G° = -RT ln(Keq), where R is the universal gas constant and T is the temperature in Kelvin. Changes in temperature can affect Keq, as they influence the position of equilibrium and the favorability of the reaction.

Temperature Dependence of ∆G° and Keq

The temperature dependence of Gibbs Free Energy (∆G°) and the equilibrium constant (Keq) is significant in thermodynamics. As temperature increases, the term T∆S° becomes more influential in the Gibbs Free Energy equation, potentially leading to a decrease in ∆G° if ∆S° is positive. Consequently, an increase in temperature can shift the equilibrium constant (Keq), favoring the formation of products or reactants depending on the sign of ∆H° and ∆S°.

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Related Practice

Textbook Question

Looking ahead In Chapter 5, we explain that the equilibrium constant (K_eq) for a reaction can be calculated based on the difference in energy between reactants and products, according to the following equation:

$K_{e q} = e^{- \frac{Δ E}{R T}}$

Using this equation, calculate the equilibrium constant for the 'reaction' shown. [For the rest of the book, if not otherwise specified, assume room temperature (298K).]

Textbook Question

The following reaction has a value of ΔG° = –2.1 kJ/mol (–0.50 kcal/mol).

CH₃Br + H₂S ⇌ CH₃SH + HBr

a. Calculate K_eq at room temperature (25 °C) for this reaction as written.

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.

Textbook Question

a. Which reaction has a greater equilibrium constant: one with a rate constant of 1 × 10^-3 sec^-1 for the forward reaction and a rate constant of 1 × 10^-5sec-1 for the reverse reaction, or one with a rate constant of 1 × 10^-2 sec^-1 for the forward reaction and a rate constant of 1 × 10^-3 sec^-1 for the reverse reaction?

b. If both reactions start with a reactant concentration of 1.0 M, which reaction will form the most product when the reactions have reached equilibrium?

Textbook Question

(b) Mechanistically, the reaction occurs as shown below. Why is this reaction favored? Based on the stability of the anions, estimate K_eq.

Textbook Question

For the following equilibrium processes and the corresponding ∆G°, indicate whether you expect the equilibrium constant to be greater than, equal to, or less than 1. Justify your expectation in words.

(a)

Textbook Question

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

Textbook Question

A certain process has ∆H° = 11.7 kcal/mol and AS° = +33cal/mol•K . That is, this reaction has an unfavorable enthalpy but a favorable entropy term. At what temperature will the process be neither favored nor disfavored?

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a. For a reaction with ∆H° = -12 kcal/mol and ∆S° = 0.01 kcal mol-1 K-1, calculate the ∆G° and the equilibrium constant at: 1. 30 °C and 2. 150 °C. b. How does ∆G° change as T increases? c. How does Keq change as T increases?

Key Concepts

Gibbs Free Energy (∆G°)

Equilibrium Constant (Keq)

Temperature Dependence of ∆G° and Keq

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a. For a reaction with ∆H° = -12 kcal/mol and ∆S° = 0.01 kcal mol^-1 K^-1, calculate the ∆G° and the equilibrium constant at:
1. 30 °C and 2. 150 °C.
b. How does ∆G° change as T increases?
c. How does K_eq change as T increases?