$\therefore H^{+}$ના મોલ $=\frac{0.01}{2} \times 0=0.01$
$NaOH$ ના મોલ $=0.1 \times 0.15=0.015$
$\therefore OH ^{-}$ ના મોલ $=0.015\,mol$
$=57.3 \times 0.01$
$=0.573\, kJ$
$(i)$ $N_2H_4$$_{(l)}$ $+$ $2H_2O_2$$_{(l)}$ $\rightarrow$ $N_2$$_{(g)}$ $+$ $4H_2O$$_{(l)}$; $\Delta r{H_1}^ \circ = - 818 \,kJ/mol$
$(ii)$ $N_2H_4$$_{(l)}$ $+$ $O_2$$_{(g)}$ $\rightarrow$ $N_2$$_{(g)}$ $+$ $2H_2O$$_{(l)}$; $\Delta r{H_2}^ \circ = - 622 \,kJ/mol$
$(iii)$ ${H_2}_{(g)}\,\, $+$ \,\,\frac{1}{2}\,{O_2}_{(g)}\,\, \to \,\,{H_2}O_{(l)}\,\,\,;\,\,{\Delta }r{H_3}^ \circ \, = \,\, - 285\,\,kJ/mol$
${C_p}$ of ${H_2}{O_{\left( g \right)}}$ $ = 33.58\,J\,{K^{ - 1}}\,mo{l^{ - 1}}$
${C_p}$ of ${H_{2\left( g \right)}}$ $ = 28.84\,J\,{K^{ - 1}}\,mo{l^{ - 1}}$
${C_p}$ of ${O_{2\left( g \right)}}$ $ = 29.37\,J\,{K^{ - 1}}\,mo{l^{ - 1}}$
$HCl + NaOH \rightarrow NaCl + H _{2} O \Delta H =-57.3\, kJ\,mol ^{-1}$
$CH _{3} COOH + NaOH \rightarrow CH _{3} COONa + H _{2} O$
$\Delta H =-55.3\,kJ\,mol ^{-1}$
વિદ્યાર્થી દ્વારા ગણતરી કરેલ $CH_3COOH$ની આયનિકરણ એન્થાલ્પી $......\,kJ\, mol ^{-1}$ છે.