Find out the enthalpy change of the following reaction using the given data:

N2+O2⟶2NO

  • Bond dissociation energy of N₂ = 958.38 kJ/mol

  • Bond dissociation energy of O₂ = 498 kJ/mol

  • Bond formation energy of NO = –626 kJ/mol

Solution:

To find the enthalpy change (ΔH) of the reaction, we can use the following formula:

ΔH=∑(Bond dissociation energies of reactants)−∑(Bond formation energies of products)

  1. Bond Dissociation Energies (Reactants):

    • For N₂: To break the N≡N triple bond in one molecule of nitrogen, 958.38 kJ/mol is required.

    • For O₂: To break the O=O double bond in one molecule of oxygen, 498 kJ/mol is required.

    The total bond dissociation energy for the reactants is:

    958.38 kJ/mol (N₂)+498 kJ/mol (O₂)=1456.38 kJ/mol

  2. Bond Formation Energies (Products):

    • For each molecule of NO, 626 kJ/mol is released. Since the reaction produces 2 moles of NO, the total bond formation energy is:

 2×(−626)=−1252 kJ/mol

Now, applying the formula for ΔH:

ΔH=1456.38 kJ/mol−(−1252 kJ/mol)=1456.38+1252=2708.38 kJ/mol

Conclusion: The enthalpy change for the reaction is +2708.38 kJ/mol, indicating that the reaction is endothermic, meaning it absorbs energy from the surroundings.

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