At 1120 K, ∆G° = 82.1 kJ/mol for the reaction 3 A (g) + B (g) →2 C (g). If the partial pressures of A, B, and C are 11.5 atm, 8.60 atm, and 0.510 atm respectively

Question

, what is the free energy for this reaction?

__kJ/mol

Answer 1

∆G = ∆G° + RTlnQ, where ∆G is free energy of the reaction, ∆G° is standard free energy, R is a constant equal to 8.314*10^-3 kJ/mol⋅K, T is temperature in K, and Q is the reaction quotient for this particular reaction.

Q is set up like K, so for the reaction 3 A (g) + B (g) → 2 C (g),

Q = (P_C)² / [(P_A)³ (P_B)]= (0.510 atm)² / [(11.5 atm)³(8.60 atm)] = 1.988604 * 10^-5 (unitless)

∆G = ∆G° + RTlnQ

= 82.1 kJ/mol + (8.314*10^-3 kJ/mol⋅K)(1120 K)ln(1.988604 * 10^-5) (K's cancel to give kJ/mol)

= 82.1 kJ/mol - 100.80352 kJ/mol

= -18.7 kJ/mol

This ∆G is negative, which means the direction is spontaneous in the forward direction. This makes sense because the partial pressure of products is so low that the reaction will shift right (proceed forward) to establish equilibrium despite having a positive ∆G°.