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 for the reaction:

N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)

Given: For the given reaction:

N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)

Enthalpy of formation (ΔH_{f (CO)}) = -110 kJ mol^-1

ΔH_{f (CO2)} = -393 kJ mol^-1

ΔH_{f (N2O)} = 81kJ mol^-1

ΔH_{f (N2O4)} = 9.7 kJ mol^–1

Now, to calculate Δ_rH for the reaction, we apply the formula given below:

Δ_rH = (Sum of enthalpy of formation of products) – (sum of enthalpy of reactants)

or, Δ_rH = ∑ Δ_fH (Products) - ∑ Δ_fH (Reactants)

In the given reaction,

N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)

Products are N₂O and CO₂

Reactants are N₂O₄ and CO

∴ Δ_rH = [Δ_fH (N₂O) + 3Δ_fH (CO₂)] - [Δ_fH(N₂O₄) + 3Δ_fH (CO)]

By substituting the values given, we get

⇒Δ_rH = [81 kJ/mol + 3 × (-393 kJ/mol)] – [9.7 kJ/mol + 3 × (-110 kJ/mol)]

⇒Δ_rH = -777.7 kJ/mol

Thus, the value of Δ_rH for the reaction is -777.7 kJ/mol.

Note: Enthalpy of reaction (Δ_rH) is actually the difference between the enthalpies of the products and the reactants when the quantities of the reactants indicated by the chemical equation have completely reacted.