Am I able to get help with this please.

Question

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

A(g)+2B(g)↽−−⇀C(g)+D(g)

were found to be 

Pa=4.95 atm, 

Pb=5.49 atm, 

Pc=4.79 atm, and 

Pd=5.76 atm.

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

Delta Grxn=KJ/mol?

Answer 1

A(g) + 2B(g) <==> C(g) + D(g)

Write the equilibrium expression:

Kp = (C)(D) / (A)(B)²

Solve for Kp:

(4.79)(5.76) / (4.95)(5.49)² = 0.185

Use Kp to solve for ∆G @ 25ºC:

∆G = -RT ln K

∆G = -(8.314 J/Kmol)(298K) ln 0.185

∆G = -2478 J/mol x -1.69

∆G = 4181 J/mol = 4.18 kJ/mol