${H_2}{O_{(l)}} \to \,\,H_{(aq)}^ + + \,\,OH_{(aq)}^ - \,;\,\,\,\Delta H\,\, = \,\,57.32\,\,KJ\,;$
${H_2}_{(g)} + \,\,\frac{1}{2}\,\,{O_2}_{(g)} \to \,\,{H_2}{O_{(1)}}\,;\,\,\Delta H\,\, = \,\, - 286.20\,\,KJ$
\(\Delta {H_r} = {\mkern 1mu} {\mkern 1mu} \Delta {H_{f({H_2}O,{\kern 1pt} {\kern 1pt} {l})}} - {\mkern 1mu} {\mkern 1mu} \Delta {\mkern 1mu} {H_{f({H_2}O,{\kern 1pt} {\kern 1pt} g)}} - \frac{1}{2}{\mkern 1mu} {\mkern 1mu} \Delta {H_{f{\kern 1pt} ({O_2},{\kern 1pt} g)}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \)
\({\mkern 1mu} \therefore {\mkern 1mu} {\mkern 1mu} \, - 286.20{\mkern 1mu} {\mkern 1mu} = {\mkern 1mu} {\mkern 1mu} \Delta {H_{f({H_2}O{\kern 1pt} ({l}))}}\)
\(So,\,\,\Delta {H_{f({H_2}O\,,\,\,{l})}} = \,\, - 286.\,\,20\)
\({H_2}{O_{({l})}}\, \to \,\,H_{(aq)}^ + \, + \,\,OH_{(aq)}^ - \,\,;\,\,\,\Delta H\,\, = \,\,57.32\,\,KJ\)
\(\Delta {H_r} = \,\,\Delta H_{f({H^ + },\,\,aq)}^ \circ \,\, + \,\,\Delta H_{f(O{H^ - },\,\,aq)}^ \circ - \,\,\Delta H_{f({H_2}O\,,\,{l})}^ \circ \)
\(57.32\,\, = \,\,0\,\, + \,\,\Delta H_{f(O{H^ - },\,\,aq)}^ \circ - ( - 286.20)\)
\(\Delta H_{f(O{H^ - },\,\,aq)}^ \circ = \,\,57.32\,\, - \,286.20\,\, = \,\, - 228.88\,\,KJ\)
[આપેલ : કેલોરીમીટર પ્રણાલીની ઉષ્માક્ષમતા $20\,kJ\,K^{-1}$ છે $R = 8.3\,JK^{-1}mol^{-1}$. આદર્શ વાયુ વર્તણૂંક ધારી લો.$C$ અને $H$ ના પરમાણ્વિય દળ અનુક્રમે $12$ અને $1\,g\,mol^{-1}$ છે.]
(ઉપયોગ કરો : $\Delta_{{c}} {H}($ ગ્લુકોઝ $)=-2700\, {~kJ}\, {~mol}^{-1}$ )