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

Step 1: Gather the key equations.
We use \(\Delta G^\circ = -RT\ln K\) and \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\).
Step 2: Find the equilibrium constant.
At 60 degrees Celsius the concentrations of N and D are equal, so \[ K = \frac{[D]}{[N]} = 1. \]
Step 3: Evaluate \(\Delta G^\circ\).
Since \(\ln 1 = 0\), \[ \Delta G^\circ = -RT\ln 1 = 0. \]
Step 4: Apply the Gibbs equation.
With \(\Delta G^\circ = 0\), \[ 0 = \Delta H^\circ - T\Delta S^\circ \implies \Delta S^\circ = \frac{\Delta H^\circ}{T}. \]
Step 5: Convert the temperature.
\(60^\circ C = 333\,K\).
Step 6: Compute \(\Delta S^\circ\).
\[ \Delta S^\circ = \frac{666}{333} = 2.0\,kJ\,K^{-1}\,mol^{-1}. \]
\[ \boxed{2.0} \]