91Ó°ÊÓ

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.

${\text{H}}^{\text{+}}\left(g\right){\text{+H}}_{\text{2}}\text{O}\left(g\right)→{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(g\right)\mathrm{Δ}{H}_1=−720\text{kJ}$ ….. (1)

${\text{H}}^{\text{+}}\left(g\right){\text{+H}}_{\text{2}}\text{O}\left(l\right)→{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)\mathrm{Δ}{H}_2=−1090\text{kJ}$ ….. (2)

${\text{H}}_2\text{O}\left(l\right)→{\text{H}}_2\text{O}\left(g\right)\mathrm{Δ}{H}_3=40.7\text{kJ}$ ….. (3)

The desired equation is as follows:

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

Reverse equation (1).

${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(g\right)→{\text{H}}^{\text{+}}\left(g\right){\text{+H}}_2\text{O}\left(g\right)\mathrm{Δ}{H}_5=720\text{kJ}$ ….. (5)

Reverse equation (3).

${\text{H}}_2\text{O}\left(g\right)→{\text{H}}_2\text{O}\left(l\right)\mathrm{Δ}{H}_6=−40.7\text{kJ}$ ….. (6)

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

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

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

Calculate the value of $\mathrm{Δ}H$for the desired equation as follows:

$\begin{array}{rcl}\mathrm{Δ}H& =& \mathrm{Δ}{H}_2+\mathrm{Δ}{H}_5+\mathrm{Δ}{H}_6\\ & =& (−1090\text{kJ})+\left(720\text{kJ}\right)+(−40\text{kJ})\\ & =& −410.7\text{kJ}\\ & ≈& −411\text{kJ}\end{array}$

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