A reaction 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