\(\mathop {N{H_4}OH}\limits_{Weak\,base} \leftrightarrow \mathop {NH_4^ + + O{H^ - }}\limits_{\Delta H = x\,kJ\,mo{l^{ - 1}}....(ii)} \)
\({H^ + } + O{H^ - } \to {H_2}O\)
\(\Delta H = - 55.90\,kJ\,mo{l^{ - 1}}....(iii)\)
(from neutralisation of strong acid and strong base)
From equation \((i),(ii)\) and \((iii)\)
\(N{H_4}OH + HCl \to NH_4^ + + C{l^ - } + {H_2}O\)
\(\Delta H = - 51.46\,kJ\,mo{l^{ - 1}}\)
\(\therefore x + ( - 55.90) = - 51.46\)
\(x = - 51.46 + 55.90\)
\( = 4.44\,kJ\,mo{l^{ - 1}}\)
\(\therefore \) Enthalpy of ionisation of
\(N{H_4}OH = 4.44\,kJ\,mo{l^{ - 1}}\)
જો $C_v = 28 \, J\,K^{-1}\, mol^{-1}$ હોય તો $\Delta U$ અને $\Delta pV$ ગણો. $(R = 8.0\, J\, K^{-1}\, mol^{-1})$