Question

Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, – 393, 81 and 9.7 kJ mol-1 respectively. Find the value of ∆rH for the reaction.
$N_2O_4(g) + 3CO(g) → N_2O(g) + 3CO_2(g)$

Answer 1

$∆_rH$ for a reaction is defined as the difference between $∆_fH$ value of products and $∆_fH$ value of reactants.
$△_rH$ =$Σ△_fH$ (product) - $Σ△_fH$ (reactants)
For the given reaction,
$N_2O_4(g) + 3CO(g) → N_2O(g) + 3CO_2(g)$
$△_rH$ = $△_fH (NO_2)$+ $3△_fH (CO_2)$ - $△_fH (N_2O_4)$ + $3△_fH (CO)$
Substituting the values of $∆_fH$ for $N_2O, CO_2, N_2O_4,$ and $CO$ from the question, we get:
$△_rH = [{81\ kJ mol^{-1} + 3(-393)\ kJ mol^{-1}} - {9.7\ kJ mol^{-1} + 3(-110)\ kJ mol^{-1}}]$
$△_rH = -777.7\ kJ mol^{-1}$

Hence, the value of $∆_r H$ for the reaction is $-777.7\ kJ mol^{-1}$.