Sapling Week 5/6 #19

[Zoe FC 1C]

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

For the reaction

N2(g)+3H2(g)↽−−⇀2NH3(g)
the standard change in Gibbs free energy is ΔG∘=−69.0 kJ/mol. What is ΔG for this reaction at 298 K when the partial pressures are P(N2)=0.500 bar, P(H2)=0.200 bar, and P(NH3)=0.600 bar?

[SainehaMaddineni_3I]

You would be using this equation: ΔG = ΔG° + RT*ln(Q)
Since you know the partial pressures, you can plug those into the K expression to get Q --> Q = (Pnh3)^2 / ((Pn2)*(Ph2)^3) = (0.600)^2 / (0.500 * (0.200^3))=90
ΔG° is already given: -69.0 kJ/mol
R = 8.314 J/mol*K but you want it to be in kJ/mol*K --> R = 0.008314 kJ/mol*K

Plug all these values into the equation:
ΔG = -69.0 kJ/mol + (0.008314 kJ/mol*K)*(298 K)*ln(90) = -57.85 kJ/mol

[Deivy Gonzalez 3H]

I thought I was using the same equation. I must have been plugging in something incorrectly but after seeing this I was able to get the correct answer. Thank you.