Consider the equilibrium A(g)=2B(g)+3C(g) at 25 degrees Celsius. When A is loaded into a cylinder at 10 atm and the system is allowed to come to equilibrium, the final pressure is found to be 12.13 atm. What is the standard gibbs free energy of reaction for this reaction.

Answer 1

..........A ==> 2B + 3C
I.........10.....0.....0
C.........-x.....2x...3x
E........10-x....2x...3x

Ptotal = 12.13 = 10-x+3x+2x
Solve for x and find pressures of A, B, C; substitute into Kp expression below and solve for Kp.
Kp = p^3C*p^2B/pA

Then dG = -RT*ln*Kp
T is 25C (from standard temperature in the problem.)