Practice question
Question
A protein in denatured state (D) is in equilibrium with native state (N). At 360 K, both N and D states are equally populated. If the standard entropy change for the reaction at this temperature Δ ⁰ = -139 J K ⁻¹ mol ⁻¹ , then the corresponding standard enthalpy change ΔH ᴼ for the reaction in kJ mol ⁻¹ (rounded off to one decimal place) is
Explanation
Mechanistically, molecular bonding involves hybridization and orbital overlap dictate geometry and reactivity. This validates Option NAT because it satisfies bonding and stability criteria, a pattern repeatedly demonstrated in chemistry research.
Discussion
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