$ \Delta ng = 1$
$\Delta H = \Delta U + \Delta ngRT $
$ 174 = \Delta U + 1 \times 8.314 \times {10^{ - 4}} \times 1250$
$\Delta U = 163.6\,kJ.$
$CaO_{(s)}\,\, + \,\,{H_2}O_{(l)}\,\, \to \,\,Ca{(OH)_2}_{(s)}\,;\,\,\,........(i)$ $\,\Delta {H_{1.8\,^oC}} = \,\, - \,\,15.26\,\,K\,cal$
$H_2O_{(l)}\,$ $ \to $ ${H_{2{(g)}}}$ $+$ $\frac{1}{2}O_{2(g)}$ $\,\Delta {H_{1.8\,^oC}} = \,\, - \,\,68.37\,\,K\,cal$
$Ca_{(s)} + \frac{1}{2}O_{2(g)} = CaO_{(s)}$ $\,\Delta {H_{1.8\,^oC}} = \,\, \,\,-151.80\,\,K\,cal$
$(i)\, {\Delta _f}{H^o}$ of $N_2O$ is $82\, kJ\, mol^{-1}$ છે,
$(ii)$ $N \equiv N,N = N,O = O$ અને $N = O$ બંધઊર્જા અનુક્રમે $946, 418, 498$ અને $607\, kJ\, mol^{- 1}$ છે. તો $N_2O$ ની સંસ્પંદન ઊર્જા ......$kJ$