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?