Question

The enthalpy change of the reaction, $H _{(a q)}^{+}+ OH _{(a q)} \rightarrow H _{2} O _{(l)}$ is $-57.3 \,kJ\, mol^{-1}$. If the enthalpies of formation of $H _{(a q)}^{+}$ and $H _{2} O _{(l)}$ are zero and $-285.84 \,kJ\, mol^{-1}$ respectively, then the enthalpy of formation of $OH _{(a q)}^{-}$ is

• A. $-228.54\,kJ\, mol^{-1}$
• B. $-333.14\,kJ\, mol^{-1}$
• C. $228.54\,kJ\, mol^{-1}$
• D. $333.14 \,kJ\, mol^{-1}$

Answer 1

Answer: (A)

$H _{(a q)}^{+}+ OH _{(a q)}^{-} \rightarrow H _{2} O _{(l)} ;$

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

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

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

$-57.3=-285.84-\left(0+\Delta H_{f}\left( OH ^{-}\right)\right)$

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