Gibb's Free Energy Question

  • Thread starter cvc121
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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.
 

Answers and Replies

  • #2
20,449
4,357
If you did the arithmetic correctly, this looks correct.
 

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