Question

The enthalpy change of the reaction, $H_{(aq)}^{+}+O{H}_{(aq)}\to H_2{O}_{(l)}$ is $-57.3kJmo{l}^{-1}$. If the enthalpies of formation of $H_{(aq)}^{+}$ and $H_2{O}_{(l)}$ are zero and $-285.84kJmo{l}^{-1}$ respectively, then the enthalpy of formation of $O{H}_{(aq)}^{-}$ is

• A. $-228.54kJmo{l}^{-1}$
• B. $-333.14kJmo{l}^{-1}$
• C. $228.54kJmo{l}^{-1}$
• D. $333.14kJmo{l}^{-1}$

Answer 1

Answer: (A)

$H_{(aq)}^{+}+O{H}_{(aq)}^{-}\to H_2{O}_{(l)};$

$\Delta H=-57.3kJ$ mol ${}^{-1}$

$\Delta H=\Sigma \Delta H_{f(\text{ Products})}-\Sigma \Delta H_{f(\text{ Reactants})}$

$-57.3=\Delta H_f\left(H_{2}O\right)-\left[\Delta H_f\left(H^{+}\right)+\Delta H_f\left(O{H}^{-}\right)\right]$

$-57.3=-285.84-\left(0+\Delta H_f\left(O{H}^{-}\right)\right)$

$\Delta H_{f(OH)}=-285.84+57.3=-228.54kJmol{}^{-1}$