For a given reaction at , the standard enthalpy change is and the standard entropy change is . Based — Chemical Equilibrium Chemistry Question
Question
For a given reaction at $298\text{ K}$, the standard enthalpy change is $\Delta H^\circ = -54.07\text{ kJ mol}^{-1}$ and the standard entropy change is $\Delta S^\circ = 10\text{ J K}^{-1}\text{mol}^{-1}$. Based on this information (using $2.303 \times R \times 298 \approx 5705\text{ J mol}^{-1}$), which of the following statements is/are correct?
💡 Solution & Explanation
Using the Gibbs free energy equation: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -54070\text{ J/mol} - (298\text{ K})(10\text{ J/K}\cdot\text{mol}) = -57050\text{ J mol}^{-1}$ (A is correct). $\log_{10} K = -\Delta G^\circ / (2.303 RT) = 57050 / 5705 = 10$ (B is correct). Because $\Delta G^\circ$ is highly negative, the reaction is spontaneous (C is false). Since $K = 10^{10} \gg 1$, the equilibrium heavily favors the products (D is correct).