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 Δ_rH for the reaction:

N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)

Answer:

Δ_rH for a reaction is defined as the difference between Δ_fH value of products and Δ_fH value of reactants.

Δ_rH = $\sum$ Δ_fH (products) - $\sum$ Δ_fH (reactants)

For the given reaction,

N₂O_4(g)+ 3CO_(g) → N₂O_(g) + 3CO_2(g)

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

Substituting the values of Δ_fH for N₂O, CO₂, N₂O₄,and CO from the question, we get:

Δ_rH = [{81 KJ mol^-1 + 3(-393) KJ mol^-1} - {9.7KJ mol^-1 + 3(-110)KJ mol^-1}]

Δ_rH = -777.7 KJ mol^-1

Hence, the value of Δ_rH for the reaction is -777.7 KJ mol^-1.

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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 Δ_rH for the reaction:

N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)

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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 ΔrH for the reaction: N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)

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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 Δ_rH for the reaction:

N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)