[cvc121]

Hi,

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 P_NO = 0.500 bar, P_O2 = 0.100 bar, P_NO2 = 0.900 bar.

Here is my attempt:
Q = (P_NO2)² / (P_O2)(P_NO)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.

[mjc123]

Looks OK to me, but as I mentioned in another of your posts, be careful about the number of decimal places. You are only given the figure of -72.6 correct to 1 dp, so your answer should be -64.0 rather than -63.98.

PS Why did you insert the last bit completely irrelevantly in a separate post in another forum?