Question

question 22

For an equilibrium reaction, ΔG = 0 kJ at 327.0 K. If the standard enthalpy of this reaction is -46.9 kJ, and the standard entropy is 107.0 J/K, what is the equilibrium constant?

Answer 1

ΔHo = -46.9 KJ

ΔSo = 107 J/K

= 0.107 KJ/K

T = 327 K

use:

ΔGo = ΔHo - T*ΔSo

ΔGo = -46.9 - 327.0 * 0.107

ΔGo = -81.889 KJ

Now we have:

T = 327 K

ΔGo = -81.889 KJ/mol

ΔGo = -81889 J/mol

use:

ΔGo = -R*T*ln Kc

-81889 = - 8.314*327.0* ln(Kc)

ln Kc = 30.1209

Kc = 1.206*10^13

Answer: 1.21*10^13