How Do You Calculate ΔG for a Reaction at Non-Standard Conditions?

[cvc121]

Homework Statement

For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 bar for all species. For the reaction 2 NO(gas) + O₂(gas) <---> 2 NO₂(gas), the standard change in Gibbs free energy (ΔG°) = -72.6 kJ. What is the ΔG for this reaction at 298 K when the partial pressures are PNO = 0.500 bar, PO₂ = 0.100 bar, PNO₂ = 0.900 bar.

Homework Equations

Q = (PNO₂)² / (PO₂)(PNO)²
ΔG = ΔG° + RTln(Q)

The Attempt at a Solution

Q = (PNO₂)² / (PO₂)(PNO)²
Q = (0.900 bar)² / (0.100)(0.500)²
Q = 32.4

ΔG = ΔG° + RTln(Q)
ΔG = -72.6 kJ + (0.008314 kJ mol- K-)(298 K)(ln 32.4)
ΔG = -72.6 + 8.6174 = -63.98 kJ

I am not sure about my approach to this question. Can anyone confirm if I am on the right track? Thanks. All help Is very much appreciated.

 

[Chestermiller]

If you did the arithmetic correctly, this looks correct.