Enthalpy of formation of $\mathrm{CO}(g), \mathrm{CO}_2(g), \mathrm{N}_2 \mathrm{O}(g)$ and $\mathrm{N}_2…
Enthalpy of formation of $\mathrm{CO}(g), \mathrm{CO}_2(g), \mathrm{N}_2 \mathrm{O}(g)$ and $\mathrm{N}_2 \mathrm{O}_4(g)$ are $-110,-393,81,9.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. Calculate $\Delta_r H$ for the following reaction.
$$
\mathrm{N}_2 \mathrm{O}_4(g)+3 \mathrm{CQ}(g) \longrightarrow \mathrm{N}_2 \mathrm{O}(g)+3 \mathrm{CO}_2(g)
$$
$-569 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$+569 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$+778 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$-778 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
Given,
Enthalpy of formation of $\mathrm{CO}(g)=-110 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Enthalpy of formation of $\mathrm{CO}_2(g)=-393 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Enthalpy of formation of $\mathrm{N}_2 \mathrm{O}(g)=81 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Enthalpy of formation of $\mathrm{N}_2 \mathrm{O}_4(g)=9.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$
\mathrm{N}_2 \mathrm{O}_4(g)+3 \mathrm{CO}(g) \longrightarrow \mathrm{N}_2 \mathrm{O}(g)+3 \mathrm{CO}_2(g)
$
$
\begin{aligned}
\Delta_r H & =\Delta H_{\text {Product }}-\Delta H_{\text {Reactant }} \\
& =[(81)+3(-393)]-[(9.7)+3(-110)] \\
& =[81-1179]-[9.7-330] \\
& =(-1098)-(-320.3)=-777.7 \\
& \approx-778 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$