Question

In a study of the temperature-induced reversible denaturation of chymotrypsinogen you determine, using a van’t Hoff plot, that the enthalpy change at 54.5°C is +533 kJ/mol. What is the ΔG° and ΔS° given that the Keq at 54.5°C = 0.27 and R = 8.315 J/mol · K?

Answer 1

Solution-

Given-

Enthalpy change (ΔH) = 533 KJ/mol

Temperature (T) = 54.5 °C + 273 = 327.5 K

Keq = 0.27

R = 8.315 J/mol · K

We know the standard-state free energy of reaction (ΔG⁰)

ΔG⁰ = -RT (ln(Keq) = - 8.315 J/mol · K * 327.5 K ln(0.27)

       = - 2723.16 (-1.309)

ΔG⁰ = 3564.6 J/mol

Answer = ΔG⁰ = 3.56 KJ/mol

Also we know the relation between ΔG⁰ , ΔS⁰ and ΔT⁰

ΔG⁰ = ΔH⁰ - T ΔS⁰

Insert the values

3.56 KJ/mol = 533 KJ/mol - 327.5 K ΔS⁰

327.5K ΔS⁰ = 533 KJ/mol - 3.56 KJ/mol

           ΔS⁰   = 529.44 KJmol^-1/327.5K

   Answer = ΔS⁰ =1.61 KJ/mol.K