Question

At 25 °C, the equilibrium partial pressures for the following reaction were found to be PA = 5.11 atm, PB = 5.75 atm, PC = 4.98 atm, and PD = 4.91 atm. 2A(g)+2B(g)=C(g)+2D(g) What is the standard change in Gibbs free energy of this reaction at 25 °C?

Answer 1

G=G⁰ + RTl_nQ

G⁰ at standered conditions means at 25 C, 1 atm pressure is -72.6 kj/moles

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

Q= [C] [D]²/[A]² {B]²

Q=4.98 X[4.91]²/[5.11]².[5.75]²

Q= 0.139

G = G⁰ + RTl_nQ

= -72.6 + 0.008314 x 298 ln 0.139

= -70.12x -0.8569

=60.08 KJ/ mole