Answer to Question #126472 in General Chemistry for Roman Kishinevsky

Question

For the reaction

A(aq) ---> B(aq)

the change in the standard free enthalpy is 2.74 kJ at 25 oC and 5.32 kJ at 45 oC. Calculate the value of this reaction's equilibrium constant at 75 oC.

Answer 1

ΔG = −RTlnK

K = e^−ΔG/RT

K(25) = e^{2.74/(0.008314×298)} = 3.02

K(45) = e^{5.32/(0.008314×318)} = 7.48

The Van't Hoff equation

ln(K₂/K₁) = (−ΔH/R)(1/T₂ − 1/T₁)

Enthalpy:

ΔH = −(ln(K₂/K₁)/(1/T₂ − 1/T₁))×R = 35.73 kJ/mol

The entropy change for the reaction at 298 K:

ΔG = ΔH − T×ΔS

ΔS = −(ΔG −ΔH)/T = 0.11 kJ/mol×K

Assume that both enthalpy and entropy changes do not change with temperature , and so we can get the Gibbs free energy value at 75 ^oC:

ΔG = ΔH − T×ΔS = 35.73 − (348×0.11) = −2.55 kJ

The equilibrium constant for this reaction at 75 ^oC:

K = e^−ΔG/RT = e^{2.55/(0.008314×348)} = 2.41

Answer: 2.41