For cyclic process : \(\Delta U = 0 \Rightarrow q = - w\)
For isothermal process : \(\Delta U = 0 \Rightarrow q = - w\)
For adiabatic process : \(q = 0 \Rightarrow \Delta U = W\)
For isochoric process : \(w = 0 \Rightarrow \Delta U = q\)
$N \equiv N\,\left( {946\,kJ\,mo{l^{ - 1}}} \right)\,;\,N = N\,\left( {418\,kJ\,mo{l^{ - 1}}} \right)$
$O = O\,\left( {498\,kJ\,mo{l^{ - 1}}} \right)\,;\,N = O\,\left( {607\,kJ\,mo{l^{ - 1}}} \right)$;
$CaO_{(s)}\,\, + \,\,{H_2}O_{(l)}\,\, \to \,\,Ca{(OH)_2}_{(s)}\,;\,\,\,........(i)$ $\,\Delta {H_{1.8\,^oC}} = \,\, - \,\,15.26\,\,K\,cal$
$H_2O_{(l)}\,$ $ \to $ ${H_{2{(g)}}}$ $+$ $\frac{1}{2}O_{2(g)}$ $\,\Delta {H_{1.8\,^oC}} = \,\, - \,\,68.37\,\,K\,cal$
$Ca_{(s)} + \frac{1}{2}O_{2(g)} = CaO_{(s)}$ $\,\Delta {H_{1.8\,^oC}} = \,\, \,\,-151.80\,\,K\,cal$
${H_2}O(l)$ $\rightleftharpoons$ ${H_2}O(g)$
$(i)$ $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$