Question

Calculate the equilibrium constant at 298 K for the reaction of ammonia with oxygen to form nitrogen and water. The data refer to 298 K.

4NH3(g) + 3O2(g) <> 2N2(g) + 6H2O(l)

Substance NH3(g) O2(g) N2(g) H2O(l)

ΔH°f (kJ/mol) -46 0 0 -285

ΔG°f (kJ/mol) -16 0 0 -237

S°(J/K·mol) 192 205 192 70

I thought Kc is just Molar concentration of Products divide by Molar concentration on Reactants which would be 1^{2 x} 1⁶ divide by 1^{4 x} 1³ which just be 1 wouldnt't it?

Answer 1

ΔG°rxn = ΔG°products - ΔG°reactants

             = (2 x 0 + 6 x -237) - (4 x -16 + 3 x 0)

             = -1422 + 64

             = -1358 kJ / mol

T = 298 K

R = 8.314 x 10^-3 kJ / mol K

ΔG° = - R T ln Keq

-1358 = -8.314 x 10^-3 x 298 x ln Keq

ln Keq = 548 .1

Keq = 1.09 x 10^238

equilibrium constant = Keq = 1.09 x 10^238

note : if products concentrations and reactant concentrations are given at equilibrium condition you can directly use the formula Keq = products / reactants