For the physical adsorption of a gas on a solid surface, the standard enthalpy of adsorption is and — Surface Chemistry Chemistry Question
Question
For the physical adsorption of a gas on a solid surface, the standard enthalpy of adsorption $\Delta H^\circ$ is $-25\text{ kJ mol}^{-1}$ and the standard entropy of adsorption $\Delta S^\circ$ is $-125\text{ J K}^{-1}\text{ mol}^{-1}$. Assuming both values are independent of temperature, calculate the equilibrium temperature (in Kelvin) above which the adsorption becomes non-spontaneous.
💡 Solution & Explanation
For the process to be at equilibrium, the Gibbs free energy change $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 0$. Thus, $T = \Delta H^\circ / \Delta S^\circ$. Plugging in the values: $T = (-25000\text{ J mol}^{-1}) / (-125\text{ J K}^{-1}\text{ mol}^{-1}) = 200\text{ K}$. At temperatures higher than $200\text{ K}$, the $T\Delta S$ term dominates, making $\Delta G > 0$, thereby rendering the adsorption non-spontaneous.