Chemistry problem... Gibbs Free Energy, help please anyone???

Question

Calculate the equilibrium constant (Kp) for the reaction shown below (at 25°C).

2 CO(g) + O2(g) -----> 2 CO2(g)

ΔG°f CO = -137.3 kj/mol

ΔG°f O2 = 0 kj/mol

ΔG°f CO2 = -394.4 kj/mol

I know I need to calculate ΔG° of the reaction first and then use the equation: ΔG°= -R*T*lnK to solve for K... but I can't seem to get the correct answer... I've tried several times, can anyone help please? Or tell me if I am doing it wrong?

10 points for best answer!!!

Answer 1

2 CO(g) + O2(g) -----> 2 CO2(g)

dG = products - reactants

dG = (2)(-394.4) - [(2)(-137.3)]

dG = -788.8 - (-274.6)

dG = -514.2 kJ

=======================

dG = - RTlnK

-514,200 Joules = - (8.314 Joules/mol-K)(298K)lnK

-514,200 Joules = - (2477.57)lnK

207.54 = lnK

e^x both sides

K = 1.36e90

your answer = 1.36e90