In a chemisorption process, of a gas adsorbs on a metallic surface releasing of heat. If the entropy — Surface Chemistry Chemistry Question
Question
In a chemisorption process, $0.5\text{ moles}$ of a gas adsorbs on a metallic surface releasing $60\text{ kJ}$ of heat. If the entropy change $\Delta S$ for the adsorption of $1\text{ mole}$ of this gas is $-150\text{ J K}^{-1}\text{ mol}^{-1}$, calculate the change in Gibbs free energy ($\Delta G$) in $\text{kJ mol}^{-1}$ for the adsorption of $1\text{ mole}$ of the gas at $300\text{ K}$. (Report the magnitude only)
💡 Solution & Explanation
Heat released for the adsorption of $0.5\text{ mol} = 60\text{ kJ}$, so for $1\text{ mole}$, $\Delta H = -120\text{ kJ mol}^{-1}$. The entropy change is $\Delta S = -150\text{ J K}^{-1}\text{ mol}^{-1} = -0.15\text{ kJ K}^{-1}\text{ mol}^{-1}$. Using the temperature $T = 300\text{ K}$, we find $\Delta G = \Delta H - T\Delta S = -120 - (300 \times -0.15) = -120 + 45 = -75\text{ kJ mol}^{-1}$. The magnitude is $75$.