Gibbs Free Energy

Question

Given ∆H°= +500 kJ/mol and the equilibrium constant, K = 1 x 10⁻¹⁰ , which of the following is correct concerning the following system at equilibrium at 298 Kelvin? A + B ⇌ C + D

a. ∆G = 0

b. ∆G° = 0

c. ∆G° < 0

Answer 1

Oh, this one is trying to be tricky...

You probably know that ∆G° = ˙∆H° - T∆S°, and that ∆G° = -RT ln K. Although you are given ∆H°, you have no information about ∆S*, so you can't use the first equation. The second equation allows you to calculate ∆G°, which is equal to -(8.314 J/K•mol) (298 K) ln (1 x 10^-10), which is equal to a large positive value (57,000 J/mol, or 57 kJ/mol). Alternatively, you could reason that since K is very small, the reaction will be spontaneous in the reverse direction under standard conditions, which allows you to determine that ∆G° will be positive. Answers B and C are eliminated, which leaves only A.

The easier way to choose the correct answer is to read carefully what the question says... "...the following system at equilibrium..." Since the system is at equilibrium, you should realize that means ∆G = 0 (this is always true at equilibrium). Although ∆G° is not equal to zero, ∆G will be 0 whenever the reaction reaches equilibrium, and the reaction quotient, Q, is equal to K.