Question

The δg°\' of the reaction is -"8.550" kj ·mol–1. calculate the equilibrium constant for the reaction. (assume a temperature of 25°

c.)

Answer 1

31.5. If ΔG° = -8.550 kJ·mol^(1) at 25 °C, K = 31.5.

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 = -[-8550 J·mol^(-1)]/[8.314 J·K^(-1)mol^(-1) x 298.15 K]

= [8550 J·mol^(1)]/[2479 J·mol^(-1)] = 3.449

K = e^3.449 = 31.5