$(i)$ $NH_3$ $_{(g)} + aq$ $\rightarrow$ $NH_3$ $_{(aq)}$, $\Delta H$ $= -8.4 \,Kcal.$
$(ii)$ $HCl_{(g)} + aq$ $\rightarrow$ $HCl{(aq)}$, $\Delta H =$ ${-1}7.3\, Kcal.$
$(iii)$ $NH_3$ $_{(aq)} + HCl_{(aq)}$ $\rightarrow$ $NH_4Cl $ $_{(aq)}$, $\Delta H = -12.5\, Kcal$.
$(iv)$ $NH_4Cl$ $_{(s)} + aq$ $\rightarrow$ $NH_4Cl$ $_{(aq)}$, $\Delta H = +3.9 \,Kcal.$
$(A)$ $2 CO ( g )+ O _2( g ) \rightarrow 2 CO _2( g ) \quad \Delta H _1^\theta=- x\,kJ\,mol { }^{-1}$
$(B)$ $C$ (graphite) $+ O _2$ (g) $\rightarrow CO _2$ (g) $\Delta H _2^\theta=- y\,kJ\,mol -1$
$C$(ગ્રેફાઈટ) $+$ $\frac{1}{2} O _2( g ) \rightarrow CO ( g )$ પ્રક્રિયા માટે $\Delta H ^\theta$ શોધો.
${C_{\left( {graphite} \right)}} + {O_{2\left( g \right)}} \to C{O_{2\left( g \right)}}\,;\,\Delta H = -393.5\,kJ$
${C_2}{H_{4\left( g \right)}} + 3{O_{2\left( g \right)}} \to 2C{O_{2\left( g \right)}} + 2{H_2}{O_{\left( l \right)}}\,;\,\Delta H = - 1410.9\,kJ$
${H_{2\left( g \right)}} + 1/2{O_{2\left( g \right)}} \to {H_2}{O_{\left( l \right)}}\,;\,\Delta H = - 285.8\,kJ$