Answer
a) $\Delta H^{\circ}_{r} = - 65.2 kJ/mol $
b) $\Delta H^{\circ}_{r} = - 9.0 kJ/mol$
Work Step by Step
a)
Standard enthalpy of reaction = (sum of standard enthalpies of formation of products) – (sum of standard enthalpies of formation of reactants).
$\Delta H^{\circ}_{r} = ∑H^{\circ}_f{products} - ∑ H^{\circ}_f{reactants}$.
$\Delta H^{\circ}_{r} = [\Delta H^{\circ}_f{F^{-}_{(aq)}} + \Delta H^{\circ}_f{H_{2}O_{(l)}}] – [\Delta H^{\circ}_f{HF_{aq}} + \Delta H^{\circ}_f{ OH^{-}_{aq}}]$
$\Delta H^{\circ}_{r} = [ -329.1 + - 285.8] – [-320.1 + -229.6] kJ/mol$
$\Delta H^{\circ}_{r} = - 614.9 – (-549.7) = - 65.2 kJ/mol $
b)
Given standard enthalpy change for the reaction,
$ H^{+}_{(aq)} + OH^{-}_{(aq)} → H_{2}O_{l}$ as - 56.2 kJ/mol.
$\Delta H^{\circ}_{r} = [\Delta H^{\circ}_f{H_{2}O_{(l)}}] – [\Delta H^{\circ}_f{H^{+}_{(aq)}} +\Delta H^{\circ}_f{ OH^{-}_{aq}}]$
From this, we can calculate the enthalpy of formation of $ H^{+}_{(aq)}$.
$\Delta H^{\circ}_f{H^{+}_{(aq)}} = [\Delta H^{\circ}_f{H_{2}O_{(l)}}] – [\Delta H^{\circ}_{r} +\Delta H^{\circ}_f{ OH^{-}_{aq}}]$
$\Delta H^{\circ}_f{H^{+}_{(aq)}} = [- 285.8] – [- 56.2+- 229.6] = 0$
Therefore standard enthalpy change for the reaction,
$ HF_{aq} → H^{+}_{(aq)} + F^{-}_{(aq)} $ is,
$\Delta H^{\circ}_{r} = [\Delta H^{\circ}_f{H^{+}_{(aq)}} +\Delta H^{\circ}_f{ F^{-}_{aq}}] – [\Delta H^{\circ}_f{HF_{(aq)}}] $
$\Delta H^{\circ}_{r} = [0 + - 329.1] – [ - 320.1] = - 9.0 kJ/mol$