Question

At 298 K for reaction \(X \rightleftharpoons Y\), if \(\Delta H^\circ=28.4\;kJ\) and equilibrium constant is \(10^{-7}\), then the standard entropy change for the reaction is:

• A. \(+17.5\)
• B. \(+38.5\)
• C. \(-17.5\)
• D. \(-38.5\)

Hint:

Remember: \[ \Delta G^\circ=-RT\ln K \] A very small equilibrium constant gives a large positive value of \(\Delta G^\circ\).

Answer 1

The Correct Option is C

Concept: \[ \Delta G^\circ = \Delta H^\circ-T\Delta S^\circ \] and \[ \Delta G^\circ=-RT\ln K \]

Step 1: Calculate \(\Delta G^\circ\). \[ K=10^{-7} \] \[ \ln K = -7(2.303) = -16.121 \] \[ \Delta G^\circ = -(8.3)(298)(-16.121) \] \[ \Delta G^\circ \approx39900J \] \[ =39.9kJ \]

Step 2: Use Gibbs equation. \[ 39.9 = 28.4 - 298\Delta S^\circ \] \[ 298\Delta S^\circ = -11.5 \] \[ \Delta S^\circ = -0.0385kJ\,K^{-1} \] \[ = -38.5J\,K^{-1} \] \[ \boxed{-38.5J\,K^{-1}} \]