For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction

Question

Anybody willing to explain this to me?

 

For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction

 

C₂H₆(g) + H₂(g) <----> 2CH₄

 

the standard change in Gibbs free energy is ΔG° = -72.6 kJ/mol. What is ΔG for this reaction at 298 K when the partial pressures are

 

P C₂H₄= 0.500atm, P H₂=0.350 atm, P CH₄= 0.600

Answer 1

Q = (P_CH4)²/(P_C2H4)(P_H2)

∆G = ∆Gº + RT ln Q

Q = (0.600)²/(0.500)(0.350) = 2.057

Plug in values being careful of units. Note that ∆Gº is given in kJ/mole and R = 8.31 J/mole-K (not kJ).