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

Step-by-step solution

For the reversible thermal denaturation N D , the equilibrium constant is given by K = [ D ] [ N ] . At 60^ C, the concentrations of N and D are equal, which means K = 1 . The standard Gibbs free energy change is G^ = -RT K . Since K = 1 , G^ = 0 . Using the thermodynamic relation G^ = H^ - T S^ , we get: H^ = T S^ Given H^ = 666 kJ mol ⁻¹ and T = 60 + 273 = 333 K. S^ = H^ T = 666 333 = 2.0 kJ K ⁻¹ mol ⁻¹ Answer: 2.0