Areaction has a standard free-energy change of –10.70 kj mol–1 (–2.557 kcal mol–1). calculate the equilibrium constant for the reaction at 25 °c.

74.9. If ΔG° = -10.70 kJ·mol^(-1) at 25 °C, K = 74.9.

The relationship between ΔG° and K is

ΔG° = -RTlnK

where

R = the gas constant = 8.314 J·K^(-1)mol^(-1)

T is the Kelvin temperature

Thus,

lnK = -ΔG°/(RT)

In this problem,

T = (25 + 273.15) K = 298.15 K

∴ lnK = -[-10 700 J·mol^(-1)]/[8.314 J·K^(-1)mol^(-1) x 298.15 K]

= [10 700 J·mol^(1)]/[2479 J·mol^(-1)] = 4.317

K = e^4.317 = 74.9