A protein undergoes reversible thermal denaturation from its initial state N to denatured state D according to N ⇌ D. At 60 °C, the concentrations of both N and D are equal at equilibrium, and the sta

Question

A protein undergoes reversible thermal denaturation from its initial state N to denatured state D according to N ⇌ D. At 60 °C, the concentrations of both N and D are equal at equilibrium, and the standard enthalpy change of denaturation is 666 kJ mol⁻¹. The standard entropy change (ΔS° in kJ K⁻¹ mo

Accepted Answer

Answer: 2.0 — N ⇌ D K = [D]/[N] K = 1 {as [D] = [N] at equilibrium} Now, ΔG° = −2.303 RT log K ΔG° = 0 {as K = 1} Now, ΔG° = ΔH° − TΔS° 0 = 666 − 333 ΔS° ΔS° = 666/333 ΔS° = 2 kJ K⁻¹ mol⁻¹

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