6. Thermodynamics and Kinetics

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:

$K_{eq}=e^{-\frac{\Delta E}{RT}}$

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).]

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.

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.

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^-5 sec-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?

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?

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

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)

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

(c) ∆H° = -21.3 kcal/mol ; AS° = -51 cal/mol•K ; T = 373 K

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?

Calculate ∆G°, ∆H°, and ∆S° for the following acid–base reactions. Rationalize the value of ∆H° based on the structure of the conjugate bases. [Assume T = 298 K.]

(a)

Calculate ∆G°, ∆H°, and ∆S° for the following acid–base reactions. Rationalize the value of ∆H° based on the structure of the conjugate bases. [Assume T = 298 K.]

(c)

Write the rate law for the following reaction and identify which molecules are present in the rate-determining step. Draw a possible transition state and propose a mechanism.

Third-order reactions are rare. Why do you think that is?