Calculate the equilibrium constant for the following reaction at 25 °C, given that ΔG°f(A) = -105 kJ/mol, ΔG°f(B) = -88 kJ/mol, ΔG°f(C) = -217 kJ/mol, and R = 8.314 J/mol•K 2A + B ⇌ C

Answer 1

We can use the equation ∆G = -RT ln K

First, we find ∆G by taking ∆Gformation of products - ∆Gformation of reactants

∆Gf products = -217 kJ/mol

∆Gf reactants = 2*-105 kJ/mol + (-88 kJ/mol) = -298 kJ/mol

∆G = -217 - (-298) = 81 kJ/mol

∆G = -RT ln K

81 kJ/mol = (-0.008314 kJ/mol-K) (298K) ln K = -2.48 ln K

ln K = - 32.66

K = 6.55x10^-15