The thermodynamics of gas adsorption on a solid surface follows the relation . At , the free energy — Surface Chemistry Chemistry Question
Question
The thermodynamics of gas adsorption on a solid surface follows the relation $\Delta G = \Delta H - T\Delta S$. At $250\text{ K}$, the free energy change $\Delta G$ is measured to be $-15\text{ kJ mol}^{-1}$ and the enthalpy change $\Delta H$ is $-45\text{ kJ mol}^{-1}$. What is the magnitude of the entropy change ($\Delta S$) for this process in $\text{J K}^{-1}\text{ mol}^{-1}$?
💡 Solution & Explanation
Using the Gibbs-Helmholtz equation: $\Delta G = \Delta H - T\Delta S \implies -15\text{ kJ/mol} = -45\text{ kJ/mol} - 250\text{ K} \times \Delta S \implies 30\text{ kJ/mol} = -250\text{ K} \times \Delta S$. Solving for entropy: $\Delta S = -30 / 250\text{ kJ K}^{-1}\text{ mol}^{-1} = -0.120\text{ kJ K}^{-1}\text{ mol}^{-1}$. Converting to Joules, this is $-120\text{ J K}^{-1}\text{ mol}^{-1}$. The magnitude is $120$.