Enthalpies of formation $C O (g), C O_{2} (g), N_{2} O (g)$ and $N_{2} O_{4} (g)$ are $- 110, - 393, 81$ and $9.7$ KJ/mol respectively. Find the value of $Δ$ H for the reaction.

$N_{2} O_{4} (g) + 3 C O \to N_{2} O (g) + 3 C O_{2} (g)$ .

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Solution

$\begin{matrix} Δ r_{H} = 3 Δ_{C O_{2}} + 3 Δ_{N_{2} O} - 3 Δ_{C O} - Δ_{N_{2} O_{4}} = 3 \times (- 393) + (81) - 3 \times (- 110) - (9.7) = - (777.7) K J \end{matrix}$

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