At 25 ∘C , the equilibrium partial pressures for the reaction

Question

3A(g)+4B(g)↽−−⇀2C(g)+3D(g) were found to be 𝑃A=5.56 atm, 𝑃B=5.63 atm, 𝑃C=4.27 atm, and 𝑃D=5.18 atm.

What is the standard change in Gibbs free energy of this reaction at 25 ∘C ?

Answer 1

First, we'll find the equilibrium constant, Kp.

Kp = (C)²(D)³ / (A)³(B)⁴ = (4.27)²(5.18)³ / (5.56)³(5.63)⁴

Kp = 0.0147

∆G = ∆Gº + RT ln K

@ equilibrium, ∆G = 0 so...

∆Gº = -RT ln K

∆Gº = -(8.314 J/Kmol)(298K) ln 0.0147 = -2478 J/mol * -4.22

∆Gº = +10,457 J/mol = +10.5 kJ/mol

(note the positive sign, meaning not spontaneous which agrees with the low value of Kp)