Enthalpies of formation of $CO_{(g)}$,$CO_{2(g)}$,$N_2O_{(g)}$ and $N_2O_{4(g)}$ are $-110$,$-393$,$81$ and $9.7 \, kJ \, mol^{-1}$ respectively. Find the value of $\Delta_r H$ for the reaction:
$N_2O_{4(g)} + 3 CO_{(g)} \rightarrow N_2O_{(g)} + 3 CO_{2(g)}$

(D) $\Delta_r H$ for a reaction is defined as the difference between the sum of $\Delta_f H$ values of products and the sum of $\Delta_f H$ values of reactants.
$\Delta_r H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants})$
For the given reaction:
$N_2O_{4(g)} + 3 CO_{(g)} \rightarrow N_2O_{(g)} + 3 CO_{2(g)}$
$\Delta_r H = [\Delta_f H(N_2O) + 3 \Delta_f H(CO_2)] - [\Delta_f H(N_2O_4) + 3 \Delta_f H(CO)]$
Substituting the given values:
$\Delta_r H = [81 + 3(-393)] - [9.7 + 3(-110)] \, kJ \, mol^{-1}$
$\Delta_r H = [81 - 1179] - [9.7 - 330] \, kJ \, mol^{-1}$
$\Delta_r H = -1098 - (-320.3) \, kJ \, mol^{-1}$
$\Delta_r H = -777.7 \, kJ \, mol^{-1}$