what is K at 298 K for the reaction, C6F6(g) ⇌ 3C2F2(g)?

Question

Given that the ∆G°_f (kJ/mol) at 298 K for difluoroacetylene (C₂F₂) and hexafluorobenzene (C₆F₆) are 185.3 and 81.3 respectively, 

Answer 1

@Anthony T: Here is my attempt at a solution to this problem. The approach seems correct to me. Any input is appreciated.

C₆F₆ <==> 3C₂F₂

∆Gºrxn = ∑n∆Gºproducts - ∑n∆Gºreactants

∆Gº = 3x185.3 - 81.3 = 555.9 - 81.3 = 474.6 kJ/mol (note that ∆Gº is positive and a large value thus the reaction is not spontaneous)

To find K, we can use ∆Gº = -RT ln K

474.6 kJ/mol = -(0.008314 kJ/molK)(298) ln K

474.6 kJ/mol = -2.48 kJ/mol ln K

ln K = - 191.4 kJ/mol

K = 7.5x10^-84

(This is an incredibly small K which is consistent with the reaction being non spontaneous, but please do check my math to be sure)

Answer 2

The ΔGº for the reaction = 3 x 185.3 j/mole - 81.3 j/mole = 474.6 j/mole.

ΔGº_reaction = -RT ln K Substitute 474.6 j/mole for ΔGº_reaction , R = 8.314 j/mole-deg, T = 298 K and solve for K.

ln K = ΔGº_reaction / -RT = 474.6 j/mole/ (- 8.314 j/mole-deg x 298K)

ln K = - 0.192

K = 0.825

Please check math.