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

Question

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

3A(g)+4B(g)↽−−⇀2C(g)+3D(g)3A(g)+4B(g)↽−−⇀2C(g)+3D(g)

were found to be 𝑃A=4.66, 𝑃B=4.77, 𝑃C=4.24, and 𝑃D=5.65 D=5.65 atm.

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

Answer 1

∆G = -RT ln K

So, first we need to find K and then solve for ∆G

3A + 4B <==> 2C + 3D

4.66...4.77.....4.24...5.65 @ equilibrium

Kp = (C)²(D)³ / (A)³(B)⁴

Kp = (4.24)²(5.65)³ / (4.66)²(4.77)⁴

Kp = 0.288

∆G = -(8.314 x 298) ln 0.288

∆G = -2478 x -1.24

∆G = +3084 J/mol = 3.08 kJ / mol