Question
On the basis of the following thermochemical data :
$$\left( {{\Delta _f}{G^ \circ }H_{\left( {aq} \right)}^ + = 0} \right)$$
$${H_2}O\left( \ell \right) \to {H^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right)\,;$$ $$\Delta H = 57.32\,kJ$$
$${H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( \ell \right)\,;$$ $$\Delta H = - 286.20\,kJ$$
The value of enthalpy of formation of $$O{H^ - }$$ ion at $${25^ \circ }C$$ is:
A.
$$- 228.88\,kJ$$
B.
$$+228.88\,kJ$$
C.
$$-343.52\,kJ$$
D.
$$-22.88\,kJ$$
Answer :
$$- 228.88\,kJ$$
Solution :
$$\eqalign{
& {\text{Given, for reaction}} \cr
& \left( {\text{i}} \right)\,\,{H_2}O\left( \ell \right) \to {H^ + }\left( {aq.} \right) + O{H^ - }\left( {aq.} \right)\,;\Delta {H_r} = 57.32\,kJ \cr
& \left( {{\text{ii}}} \right)\,{H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( \ell \right)\,;\,\Delta {H_r} = - 286.20\,kJ \cr
& {\text{For reaction (i)}} \cr
& \Delta {H_r} = \Delta {H^ \circ }_f\left( {{H^ + }.aq} \right) + \Delta {H^ \circ }_f\left( {O{H^ - }.aq} \right) - \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right) \cr
& 57.32 = 0 + \Delta {H^ \circ }_f\left( {O{H^ - },aq} \right) - \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right)\,\,\,\,\,...\left( {{\text{iii}}} \right) \cr
& {\text{For reaction (ii)}} \cr
& \Delta {H_r} = \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right) - \Delta {H^ \circ }_f\left( {{H_2},g} \right) - \frac{1}{2}\Delta {H^ \circ }_f\left( {{O_2},g} \right) \cr
& - 286.20 = \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right) \cr
& {\text{On replacing this value in equ}}{\text{. (iii) we have}} \cr
& 57.32 = \Delta {H^ \circ }_f\left( {O{H^ - },aq} \right) - \left( { - 286.20} \right) \cr
& \Delta {H^ \circ }_f = - 286.20 + 57.32 \cr
& = - 228.88\,kJ \cr} $$