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

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?

(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)

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

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?

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.]

(b)

When ethene is treated in a calorimeter with H₂ and a Pt catalyst, the heat of reaction is found to be –137 kJ/mol (–32.7 kcal/mol), and the reaction goes to completion. When the reaction takes place at 1400 K, the equilibrium is found to be evenly balanced, with K_eq =1.  Compute the value of ΔS for this reaction.

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

(a) ∆H° = -15 kcal/mol ; ∆S° = +37 cal/mol•K ; T = 273 K

For each reaction, estimate whether ΔS° for the reaction is positive, negative, or impossible to predict.

(a)