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RE-NEET2026Chemistry-Gibbs Free Energy & Equilibrium

RE-NEET 2026 Chemistry Entropy and Enthalpy MCQ Question

Type: MCQ-numerical-Medium-Class 11

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⁻¹mol⁻¹) of the protein upon denaturation at 60 °C is closest to

A

2000.0

B

333.0

C

11.1

D

2.0

Correct Answer

Option D

Detailed Explanation

To find the standard entropy change (ΔS°) for the denaturation of the protein, we can use the Gibbs free energy equation at equilibrium:

ΔG°=ΔH°−TΔS°\Delta G° = \Delta H° - T\Delta S°

At equilibrium, ΔG° = 0. Given that ΔH° = 666 kJ mol⁻¹ and T = 60 °C = 333 K, we can rearrange the equation:

0=666−333ΔS°0 = 666 - 333 \Delta S°

Solving for ΔS° gives:

ΔS°=666333=2.0 kJ K−1 mol−1\Delta S° = \frac{666}{333} = 2.0 \text{ kJ K}^{-1} \text{ mol}^{-1}

Thus, the correct answer is D. The other options are incorrect because they do not satisfy the relationship derived from the Gibbs free energy equation.

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