A protein undergoes reversible thermal denaturation from its initial state N to denatured state D according to $N ightleftharpoons D$. At $60^{\circ}C$, the concentrations of both N and D are equal at equilibrium, and the standard enthalpy change of denaturation is $666\ ext{kJ mol}^{-1}$. The standard entropy change $\Delta S^{\circ}$ (in $ ext{kJ K}^{-1}mol^{-1}$) of the protein upon denaturation at $60^{\circ}C$ is closest to

Question

A protein undergoes reversible thermal denaturation from its initial state N to denatured state D according to $N ightleftharpoons D$. At $60^{\circ}C$, the concentrations of both N and D are equal at equilibrium, and the standard enthalpy change of denaturation is $666\ 	ext{kJ mol}^{-1}$. The standard entropy change $\Delta S^{\circ}$ (in $	ext{kJ K}^{-1}mol^{-1}$) of the protein upon denaturation at $60^{\circ}C$ is closest to

Accepted Answer

B. 1. Since the concentrations of N and D are equal at $60^{\circ}C$, the system is at **equilibrium** at this temperature, which means the standard Gibbs free energy change is zero.
$$
\Delta G^{\circ} = 0
$$

2. Using the relation between free energy, enthalpy, and entropy,
$$
\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}
$$

3. Since $\Delta G^{\circ} = 0$, this simplifies to
$$
\Delta H^{\circ} = T\Delta S^{\circ}
$$

4. Converting the given temperature to Kelvin, $T = 60 + 273 = 333\ K$, and rearranging for $\Delta S^{\circ}$,
$$
\Delta S^{\circ} = \frac{\Delta H^{\circ}}{T} = \frac{666}{333}
$$

5. Calculating this ratio,
$$
\Delta S^{\circ} = 2.0\ 	ext{kJ K}^{-1}mol^{-1}
$$

Hence, the correct answer is **B**.