Question
$$\Delta H_f^ \circ $$ of water is $$ - 285.8\,kJ\,mo{l^{ - 1}}.$$ If enthalpy of neutralisation of monoacidic strong base is $$ - 57.3\,kJ\,mo{l^{ - 1}}.\,\,\Delta H_f^ \circ $$ of $$O{H^ - }\,ion$$ will be
A.
$$ - 114.25\,kJ\,mo{l^{ - 1}}$$
B.
$$114.25\,kJ\,mo{l^{ - 1}}$$
C.
$$228.5\,kJ\,mo{l^{ - 1}}$$
D.
$$ - 228.5\,kJ\,mo{l^{ - 1}}$$
Answer :
$$ - 228.5\,kJ\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& {H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( l \right); \cr
& \Delta H = - 285.8\,kJ\,\,\,....\left( {\text{i}} \right) \cr
& {H^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right) \to {H_2}O\left( l \right); \cr
& \Delta H = - 57.3\,kJ\,\,\,....\left( {{\text{ii}}} \right) \cr
& \frac{1}{2}{H_2}\left( g \right) + aq \to {H^ + }\left( {aq} \right) + {e^ - };\Delta H = 0 \cr
& {\text{(by convention)}}\,\,\,...\left( {{\text{iii}}} \right) \cr
& \left( {\text{i}} \right){\text{ - }}\left( {{\text{ii}}} \right){\text{ - }}\left( {{\text{iii}}} \right)\,\,{\text{gives,}} \cr} $$
$$\frac{1}{2}{H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) + {e^ - } + aq \to $$ $$O{H^ - }\left( {aq} \right)$$
$$\Delta H = - 285.8 + 57.3 = - 228.5\,kJ$$