Enthalpies of formation of CO(g), CO 2 (g), N 2 O(g) and N 2 O 4 (g) are –110, – 393, 81 and 9.7 kJ mol –1 respectively. Find the value of for the reaction: N 2 O 4 (g) + 3CO(g) → N 2 O(g) + 3CO 2 (g)

Question

Philoid · Accepted Answer

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.

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)

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)