Question

A protein undergoes reversible thermal denaturation from its initial state N to denatured state D according to \(N \rightleftharpoons D\). At \(60^\circ C\), 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 \((\Delta S^\circ)\) in \(kJ\ K^{-1}\ mol^{-1}\) of the protein upon denaturation at \(60^\circ C\) is closest to

• A. \(11.1\)
• B. \(2.0\)
• C. \(2000.0\)
• D. \(333.0\)

Hint:

Whenever equilibrium concentrations of reactants and products are equal, \[ K=1 \] and therefore \[ \Delta G^\circ=0. \] This shortcut frequently appears in thermodynamics and biochemistry problems.

Answer 1

The Correct Option is B

Concept: At equilibrium, the Gibbs free energy change is related to the equilibrium constant by \[ \Delta G^\circ = -RT\ln K \] Also, \[ \Delta G^\circ=\Delta H^\circ-T\Delta S^\circ \] Combining these equations allows determination of entropy change when enthalpy change and equilibrium information are known.

Step 1: Determine equilibrium constant. The problem states that at equilibrium the concentrations of N and D are equal. Therefore, \[ [N]=[D] \] Hence, \[ K=\frac{[D]}{[N]}=1 \]

Step 2: Calculate Gibbs free energy change. Using \[ \Delta G^\circ=-RT\ln K \] Since \[ K=1 \] and \[ \ln1=0 \] Therefore, \[ \Delta G^\circ=0 \]

Step 3: Apply Gibbs-Helmholtz relation. \[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \] Substituting \(\Delta G^\circ=0\), \[ 0 = \Delta H^\circ - T\Delta S^\circ \] \[ T\Delta S^\circ = \Delta H^\circ \] \[ \Delta S^\circ = \frac{\Delta H^\circ}{T} \]

Step 4: Substitute the given values. \[ \Delta H^\circ = 666\ kJ\ mol^{-1} \] Temperature: \[ 60^\circ C = 333\ K \] Therefore, \[ \Delta S^\circ = \frac{666}{333} \] \[ \Delta S^\circ = 2.0\ kJ\ K^{-1}\ mol^{-1} \]

Step 5: Final answer. \[ \boxed{\Delta S^\circ=2.0\ kJ\ K^{-1}\ mol^{-1}} \] Hence option (B) is correct.