Question

At 25 °C, the equilibrium partial pressures for the following reaction were found to be PA = 6.30 atm, PB = 7.20 atm, PC = 6.40 atm, and PD = 9.10 atm.
2A(g) + 2B(g)-->C(g) + 3D(g)

What is the standard change in Gibbs free energy of this reaction at 25 °C.

Answer 1

Go = -RT*lnK
lnK = p^2*A*p^2*B/(pC* p^3*D

Answer 2

Go = -RT*lnK
K = p^2*A*p^2*B/(pC* p^3*D

Answer 3

K=(pC^1*pD^3)/(pA^2*pB^2)

then go deltaG=-RT*ln(K)