A protein undergoes reversible thermal denaturation from its initial state N to denatured state D

NEET Chemistry PYQ solved by Nitish Sir 🔥 A Chemical Thermodynamics / Biomolecules numerical on protein denaturation, equilibrium constant (K_eq), Gibbs free energy, and standard entropy change (ΔS°). ❓ QUESTION 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⁻¹ mol⁻¹) of the protein upon denaturation at 60 °C is closest to (1) 333.0 (2) 11.1 (3) 2.0 (4) 2000.0 ✅ Correct answer: (3) 2.0 Key idea → At equilibrium, [N] = [D], so K_eq = [D] / [N] = 1. Since ΔG° = -RT ln(K_eq) and ln(1) = 0, we have ΔG° = 0. Using ΔG° = ΔH° - TΔS° = 0, which gives ΔS° = ΔH° / T. Convert T = 60 °C = 333 K. ΔS° = 666 kJ mol⁻¹ / 333 K = 2.0 kJ K⁻¹ mol⁻¹. 👨‍🏫 ABOUT NITISH SIR B.E (Hons) BITS Pilani • AIR-16 GATE 2012 • AIR-732 JEE Main 2008 • 15+ years teaching. 🎯 FREE for re-NEET aspirants → https://neetsambhav.in 🔔 SUBSCRIBE to Chemistry Forum for daily NEET Chemistry PYQs & live doubt sessions. 💬 Telegram: https://t.me/ChemistryForumNitishSir 🌐 Website: https://neetsambhav.in #NEET #NEETChemistry #Thermodynamics #Biomolecules #PhysicalChemistry NEET chemistry, protein denaturation thermodynamics, delta S calculation, entropy change protein, physical chemistry NEET PYQ, thermodynamics NEET, NEET chemistry PYQ solution, Nitish Sir chemistry, NEET Sambhav, Chemistry Forum