There are two equations of Gibb's free energy, ΔG°:
ΔG° = ΔH° - TΔS°
ΔG° = -RTlnK
Let's use the first equation. Since ΔS° is negligible, it is equal to 0. Therefore, ΔG° = ΔH° = 7.11 kJ/mol. We use this value to the second equation.
7.11 kJ/mol = -(8.314 J/mol·K)*(1 kJ/1000 J)*(25 + 273 K)*(lnK)
lnK = -2.8697
K = 0.05671
Therefore, the equilibrium constant is equal to 0.05671.