Q. 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 mol⁻¹) of the protein upon denaturation at 60°C is closest to
A) 11.1
B) 2.0
C) 2000.0
D) 333.0
Explanation
[Solution from source] Equal concentrations mean K = 1, so ΔG° = −RT ln K = 0 and ΔS° = ΔH°/T = 666 kJ mol⁻¹ / 333 K = 2.0 kJ K⁻¹ mol⁻¹. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.