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A protein undergoes reversible thermal denaturation from its initial state N to denatured state D according to $N\rightleftharpoons D$. At $60^\circ C$, 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 ($\Delta S^\circ$ in kJ K$^{-1}$ mol$^{-1}$) of the protein upon denaturation at $60^\circ C$ is closest to
Detailed Solution
$N\rightleftharpoons D$, $K=\frac{[D]}{[N]}=1$ since [D] = [N] at equilibrium. $\Delta G^\circ=-2.303RT\log K=0$. $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ \Rightarrow 0=666-333\Delta S^\circ \Rightarrow \Delta S^\circ=\frac{666}{333}=2$ kJ K$^{-1}$ mol$^{-1}$.