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Exercises · 5.12

Q.Enthalpies of formation of CO(g)CO(g), CO2(g)CO_2(g), N2O(g)N_2O(g) and N2O4(g)N_2O_4(g) are –110, –393, 81 and 9.7 kJ mol−1^{-1} respectively. Find the value of ΔrH\Delta_r H for the reaction: N2O4(g)+3CO(g)→N2O(g)+3CO2(g)N_2O_4(g) + 3CO(g) \rightarrow N_2O(g) + 3CO_2(g).

Mizoram MbseTextbookSubjective· 3mImportance★★★★★est
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The reaction enthalpy is found by applying Hess’s law: ΔrH=∑ΔfH∘(products)−∑ΔfH∘(reactants)\Delta_r H = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants}). Substituting the given formation enthalpies gives ΔrH=−777.7 kJ mol−1\Delta_r H = -777.7\ \text{kJ mol}^{-1}.

The key idea here is that enthalpy is a state function. That means the change in enthalpy for a reaction depends only on the initial and final states, not on the path taken. So if we know the standard enthalpies of formation of every reactant and product, we can calculate the reaction enthalpy directly — no need to imagine how the reaction actually happens at the molecular level.

The standard enthalpy of formation ΔfH∘\Delta_f H^\circ of a compound is the enthalpy change when one mole of that compound is formed from its elements in their standard states. By definition, elements in their standard states have ΔfH∘=0\Delta_f H^\circ = 0. The reaction enthalpy is then:

ΔrH∘=∑νΔfH∘(products)−∑νΔfH∘(reactants)\Delta_r H^\circ = \sum \nu \Delta_f H^\circ(\text{products}) - \sum \nu \Delta_f H^\circ(\text{reactants})

where ν\nu are the stoichiometric coefficients. This is a direct consequence of Hess’s law — you can think of it as “un-forming” the reactants (reverse of formation, so sign flips) and then forming the products.

Let’s apply it step by step.

  1. Write down the given data clearly.

    We have:

    • ΔfH∘(CO(g))=−110 kJ mol−1\Delta_f H^\circ(CO(g)) = -110\ \text{kJ mol}^{-1}
    • ΔfH∘(CO2(g))=−393 kJ mol−1\Delta_f H^\circ(CO_2(g)) = -393\ \text{kJ mol}^{-1}
    • ΔfH∘(N2O(g))=81 kJ mol−1\Delta_f H^\circ(N_2O(g)) = 81\ \text{kJ mol}^{-1}
    • ΔfH∘(N2O4(g))=9.7 kJ mol−1\Delta_f H^\circ(N_2O_4(g)) = 9.7\ \text{kJ mol}^{-1}

    All values are per mole of the compound.

  2. Identify the stoichiometric coefficients.

    The reaction is:

N2O4(g)+3CO(g)→N2O(g)+3CO2(g)N_2O_4(g) + 3CO(g) \rightarrow N_2O(g) + 3CO_2(g)

So:

  • Reactants: 11 mole of N2O4N_2O_4, 33 moles of COCO
  • Products: 11 mole of N2ON_2O, 33 moles of CO2CO_2
  1. Apply the formula.

ΔrH∘=[1⋅ΔfH∘(N2O)+3⋅ΔfH∘(CO2)]−[1⋅ΔfH∘(N2O4)+3⋅ΔfH∘(CO)]\Delta_r H^\circ = \left[1 \cdot \Delta_f H^\circ(N_2O) + 3 \cdot \Delta_f H^\circ(CO_2)\right] - \left[1 \cdot \Delta_f H^\circ(N_2O_4) + 3 \cdot \Delta_f H^\circ(CO)\right]

  1. Substitute the numbers. …

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