A protein undergoes reversible thermal denaturation Nleftharpoons D. At 60 ^°C the concentrations of N and D are equal at equilibrium, and the standard enthalpy change of denaturation is 666 kJ mol⁻¹. The standard entropy change Δ S^° (in kJ K⁻¹mol⁻¹) of the protein upon denaturation at 60 ^°C is closest to:

Solution & Explanation

Concept: at equilibrium ΔG° = −RT ln K, and ΔG° = ΔH° − TΔS°. At 60 °C the equilibrium N ⇌ D has [N] = [D], so K = [D]/[N] = 1. Then ΔG° = −RT ln(1) = 0, since ln 1 = 0. Setting ΔG° = ΔH° − TΔS° = 0 gives ΔS° = ΔH°/T. Convert temperature: T = 60 °C = 333 K (this is the key step — using 60 instead of 333 is the trap). ΔS° = 666 kJ mol⁻¹ / 333 K = 2.0 kJ K⁻¹ mol⁻¹. The large positive ΔS° fits denaturation, where the folded protein unfolds into a more disordered (higher-entropy) state. Hence the answer is A — ΔS° ≈ 2.0 kJ K⁻¹ mol⁻¹.