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
