The standard entropy change (in cal/K-mol) for the reaction: , if the value of and at 300 K. — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

First calculate standard free energy: $\Delta G^\circ = -RT \ln K_c = - (2\text{ cal/K-mol}) \times (300\text{ K}) \times \ln(e^{-10}) = -600 \times (-10) = 6000\text{ cal/mol} = 6.0\text{ kcal/mol}$. From the relation $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$, rearranging gives $\Delta S^\circ = \frac{\Delta H^\circ - \Delta G^\circ}{T} = \frac{7500 - 6000}{300} = \frac{1500}{300} = 5\text{ cal/K-mol}$.