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 ( S in kJ K-1 mol-1) of the protein upon denaturation at 60 C is closest to
- 11.1
- 2.0
- 2000.0
- 333.0
Answer
(2) 2.0
Solution
At equilibrium, [N] = [D], so the equilibrium constant K = [D][N] = 1.
Using G = -RT K = -RT (1) = 0.
Since G = H - T S = 0, we have:
S = H T = 666 kJ mol-1(60 + 273) K = 666333 = 2.0 kJ K-1 mol-1. Answer: (2)