\(\Delta H=2 \times-46.0 \mathrm{\,kJ} \,\mathrm{mol}^{-1}\)
Let \(x\) be the bond enthalpy of \(N-H\) bond then
\(\Delta H=\sum\) Bond energies of products
\(-\sum\) Bond energies of reactants
\(2 \times-46=712+3 \times(436)-6 x\)
\(-92=2020-6 x\)
\(6 x=2020+92\)
\(\Rightarrow 6 x=2112\)
\(\Rightarrow x=+352\; k J / m o l\)
$(A)$ $2 CO ( g )+ O _2( g ) \rightarrow 2 CO _2( g ) \quad \Delta H _1^\theta=- x\,kJ\,mol { }^{-1}$
$(B)$ $C$ (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.
(આપેલ : $\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )