Chapter 5: Problem 60
Calculate the standard free energy change for the ionization: \(\mathrm{HF}(\mathrm{aq}) \rightarrow \mathrm{H}^{+}(\mathrm{aq})\) \(+\mathrm{F}^{-}(\mathrm{aq})\) from the following data: \(\mathrm{HF}(\mathrm{aq}) \rightarrow \mathrm{HF}(\mathrm{g}) ; \Delta G^{\circ}=23.9 \mathrm{~kJ}\) \(\mathrm{HF}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{F}(\mathrm{g}) ; \Delta G^{\circ}=555.1 \mathrm{~kJ}\) \(\mathrm{H}(\mathrm{g}) \rightarrow \mathrm{H}^{+}(\mathrm{g})+\mathrm{e} ; \Delta G^{\circ}=1320.2 \mathrm{~kJ}\) \(\mathrm{F}(\mathrm{g})+\mathrm{e} \rightarrow \mathrm{F}^{-}(\mathrm{g}) ; \Delta G^{\circ}=-347.5 \mathrm{~kJ}\) \(\mathrm{H}^{+}(\mathrm{g})+\mathrm{F}^{-}(\mathrm{g}) \stackrel{\mathrm{aq} .}{\longrightarrow} \mathrm{H}^{+}(\mathrm{aq})+\mathrm{F}^{-}(\mathrm{aq})\) \(\Delta G^{\circ}=-1513.6 \mathrm{~kJ}\) (a) \(-38.1 \mathrm{~kJ}\) (b) \(+38.1 \mathrm{~kJ}\) (c) \(-1489.7 \mathrm{~kJ}\) (d) \(-1513.6 \mathrm{~kJ}\)
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