Calculate the standard free energy change for the ionization: from the following data:<br><br><br><b — Thermodynamics and Thermochemistry Chemistry Question
Question
Calculate the standard free energy change for the ionization: $\text{HF(aq)} \to \text{H}^+\text{(aq)} + \text{F}^-\text{(aq)}$ from the following data:<br>$\text{HF(aq)} \to \text{HF(g)}; \Delta G^\circ = 23.9\text{ kJ}$<br>$\text{HF(g)} \to \text{H(g)} + \text{F(g)}; \Delta G^\circ = 555.1\text{ kJ}$<br>$\text{H(g)} \to \text{H}^+\text{(g)} + \text{e}^-; \Delta G^\circ = 1320.2\text{ kJ}$<br>$\text{F(g)} + \text{e}^- \to \text{F}^-\text{(g)}; \Delta G^\circ = -347.5\text{ kJ}$<br>$\text{H}^+\text{(g)} + \text{F}^-\text{(g)} \xrightarrow{\text{aq}} \text{H}^+\text{(aq)} + \text{F}^-\text{(aq)}; \Delta G^\circ = -1513.6\text{ kJ}$
💡 Solution & Explanation
Simply sum the free energy changes of all the consecutive reaction steps to obtain the overall ionization free energy change: $\Delta G^\circ = 23.9 + 555.1 + 1320.2 - 347.5 - 1513.6 = +38.1\text{ kJ}$.