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Chemistry · PYQ NEET UG 2026
हिन्दीA protein undergoes reversible thermal denaturation from its initial state \( \mathbf{N} \) to denatured state \( \mathbf{D} \) according to \( \mathbf{N \rightleftharpoons D} \). At \( 60\;^\circ\text{C} \), the concentrations of both \( \mathbf{N} \) and \( \mathbf{D} \) are equal at equilibrium, and the standard enthalpy change of denaturation is \( 666\text{ kJ mol}^{-1} \). The standard entropy change (\( \Delta S^\circ \) in \( \text{kJ K}^{-1}\text{mol}^{-1} \)) of the protein upon denaturation at \( 60\;^\circ\text{C} \) is closest to
Correct answer · Option 1
\( 2.0 \)
Explanation
Since [N] = [D] at equilibrium, \( K_{eq} = 1 \), so \( \Delta G^\circ = 0 \). From \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 0 \), we get \( \Delta S^\circ = \frac{\Delta H^\circ}{T} = \frac{666 \text{ kJ mol}^{-1}}{333 \text{ K}} = 2.0 \text{ kJ K}^{-1}\text{mol}^{-1} \).
