Given the following information,

$$\begin{gathered} {\mathbf{H}}^{+}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_2\mathbf{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }{\mathbf{H}}_3{\mathbf{O}}^{+}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\Delta H}\mathbf{=}\mathbf{-}\mathbf{720}\mathbf{\text{kJ}} \\ {\mathbf{H}}^{+}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_2\mathbf{O}\mathbf{\left(}\mathit{l}\mathbf{\right)}\mathbf{\to }{\mathbf{H}}_3{\mathbf{O}}^{+}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\Delta {\rm H}}\mathbf{=}\mathbf{-}\mathbf{1090}\mathbf{\text{kJ}} \\ {\mathbf{H}}_2\mathbf{O}\mathbf{\left(}\mathit{l}\mathbf{\right)}\mathbf{\to }{\mathbf{H}}_2\mathbf{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\Delta H}\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{7}\mathbf{\text{kJ}} \end{gathered}$$

Calculate the heat of solution of the hydronium ion:

${\mathbf{H}}_3{\mathbf{O}}^{+}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\stackrel{{\mathbf{H}}_2\mathbf{O}}{\mathbf{\to }}{\mathbf{H}}_3{\mathbf{O}}^{+}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}$

A hydrogen ion bound to a water molecule, ${\mathbf{\text{H}}}_{\mathbf{\text{3}}}\mathbf{\text{O+}}$, the form in which hydrogen ions are found in aqueous solution.

Write the given equations.

${\text{H}}^{\text{+}}\left(g\right){\text{+H}}_{\text{2}}\text{O}\left(g\right)\to {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(g\right)\Delta H_1=-720\text{kJ}$ ….. (1)

${\text{H}}^{\text{+}}\left(g\right){\text{+H}}_{\text{2}}\text{O}\left(l\right)\to {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)\Delta H_2=-1090\text{kJ}$ ….. (2)

${\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_2\text{O}\left(g\right)\Delta H_3=40.7\text{kJ}$ ….. (3)

The desired equation is as follows:

$H_3{O}^{+}\left(g\right)\stackrel{H_{2}O}{\to }{H}_3{O}^{+}\left(aq\right)$ ….. (4)

Reverse equation (1).

${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(g\right)\to {\text{H}}^{\text{+}}\left(g\right){\text{+H}}_2\text{O}\left(g\right)\Delta H_5=720\text{kJ}$ ….. (5)

Reverse equation (3).

${\text{H}}_2\text{O}\left(g\right)\to {\text{H}}_2\text{O}\left(l\right)\Delta H_6=-40.7\text{kJ}$ ….. (6)

Add equations (2), (5), and (6).

$$\begin{gathered} \frac{{\displaystyle {\text{H}}^{+}\left(g\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_{\text{3}}{\text{O}}^{+}\left(aq\right) \\ {\text{H}}_{\text{3}}{\text{O}}^{+}\left(g\right)\to {\text{H}}^{+}\left(g\right)+{\text{H}}_2\text{O}\left(g\right) \\ {\text{H}}_2\text{O}\left(g\right)\to {\text{H}}_{\text{2}}\text{O}\left(l\right)}}{{\displaystyle {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(g\right)\to {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)}} \end{gathered}$$

Hence, the desired equation (4) adds the three equations.

Calculate the value of $\Delta H$for the desired equation as follows:

$\begin{array}{rcl}\Delta H& =& \Delta H_2+\Delta H_5+\Delta H_6\\ & =& (-1090\text{kJ})+\left(720\text{kJ}\right)+(-40\text{kJ})\\ & =& -410.7\text{kJ}\\ & \approx & -411\text{kJ}\end{array}$

Hence, the heat of solution of the hydronium ion is $-411\text{kJ}$.