Question

a)Calculate the equilibrium constant at 10 K for a reaction with ΔH^o = 10 kJ and ΔS^o = 100 J/K.

b) Calculate the equilibrium constant at 105 K for the thermodynamic data in the previous question.

Answer 1

Part A

ΔH^o = 10 kJ,   ΔS^o = 100 J/K convert in to joules = 0.1 kJ/K

T = 10K

first find the deltaG using the formula

deltaG = deltaH - T deltaS

delta G = 10 kJ - 10K (0.1 kJ/K)

deltaG = 10kJ - 1 kJ = 9 kJ

now use the relation

ΔGo = - RT ln Keq

R = 0.008314 kJ / mol K

T = 10K

plug in all the values in the above equation

9kJ = - (0.008314 kJ / mol K) (10K) ln Keq

ln Keq = -9 kJ / 0.08314

ln Keq = -108.25

Keq = e^-108.25 = 9.71x 10^-48

i have doubt the given ΔS^o = 100 J/K is it correct or -100 J/K?

Part B

ΔH^o = 10 kJ,   ΔS^o = 100 J/K convert in to joules = 0.1 kJ/K

T = 105K

first find the deltaG using the formula

deltaG = deltaH - T deltaS

delta G = 10 kJ - 105K (0.1 kJ/K)

deltaG = 10kJ - 10.5 kJ = -0.5 kJ

now use the relation

ΔGo = - RT ln Keq

R = 0.008314 kJ / mol K

T = 105K

plug in all the values in the above equation

-0.5kJ = - (0.008314 kJ / mol K) (105K) ln Keq

ln Keq = -0.5 / -0.87297

ln Keq = 0.57275

Keq = e^0.57275 = 1.77