Find Free Energy

Question

At 725 K, ∆G° = -15.7 kJ/mol for the reaction A (g) + 2 B (g) → 3 C (g). If the partial pressures of A, B, and C are 1.21 atm, 3.25 atm, and 0.235 atm respectively, what is the free energy for this reaction?

Answer 1

We were given the following information:

ΔG° = -15.7 kJ*mol^-1

P_A = 1.21 atm

P_B = 3.25 atm

P_C = 0.235 atm

T = 725K

We can use the equation for calculating the ΔG of reaction that is not at standard state conditions:

ΔG_rxn = ΔG° + RTlnQ

So, we just plug in all of our values:

ΔG_rxn = ΔG° + RTlnQ

<=> -15700 J*mol^-1 + (8.314 J*K^-1*mol^-1)(725k)ln[(0.235)³/(1.21)(3.25)²]

<=> -15700 J*mol^-1 - 41545 J*mol^-1

ΔG_rxn = -57.2 kJ*mol^-1

Cheers!