Q.22 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 ΔS0 = -139 J K-1 mol-1, then the corresponding standard enthalpy change ΔH0 for the reaction in kJ mol-1
(rounded off to one decimal place) is ____
Protein denaturation involves equilibrium between native (N) and denatured (D) states, where thermodynamics dictate stability. At 360 K, equal populations mean ΔG° = 0, allowing direct computation of ΔH° from given ΔS°. The value is -50.0 kJ mol⁻¹.
Thermodynamic Principle
The Gibbs free energy equation is ΔG° = ΔH° – TΔS°. At equilibrium with equal populations, K = [D]/[N] = 1, so ΔG° = -RT ln(1) = 0 J mol⁻¹. Thus, 0 = ΔH° – TΔS°, rearranging gives ΔH° = TΔS°.
Step-by-Step Calculation
Temperature T = 360 K and ΔS° = -139 J K⁻¹ mol⁻¹ = -0.139 kJ K⁻¹ mol⁻¹. ΔH° = 360 × (-0.139) = -50.04 kJ mol⁻¹, rounded to -50.0 kJ mol⁻¹. Negative ΔS° indicates reduced disorder in the forward reaction (likely N → D has positive ΔS°), with negative ΔH° favoring N at lower T.
Reaction Direction Analysis
The problem states “reaction” with ΔS° = -139 J K⁻¹ mol⁻¹, implying N → D where unfolding increases entropy (but given value is for opposite). Standard convention uses ΔG°_unfolding > 0 for stable N; here ΔS°_unfolding would be positive, but signed as given yields ΔH° = -50.0. Matches exam context for numerical answer.
Exam Tips
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Recognize equilibrium implies ΔG° = 0 for direct ΔH° = TΔS°.
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Units: Convert J to kJ by dividing by 1000; round to one decimal.
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Negative values common for folding (enthalpy-driven).
Correct Answer: -50.0
