Question

The enthalpy change (ΔH) for the reaction N₂(g) + 3H₂(g) → 2NH₃(g) is -92 kJ/mol.

What is the enthalpy change for the decomposition of 1 mol of NH₃(g) into its elements?

• A. \( +46 \, \text{kJ/mol} \)
• B. \( -46 \, \text{kJ/mol} \)
• C. \( +92 \, \text{kJ/mol} \)
• D. \( -92 \, \text{kJ/mol}

Hint:

\textbf{Key Fact:} The enthalpy change of a reverse reaction is the negative of the forward reaction, adjusted for stoichiometry.

Answer 1

The Correct Option is A

• Given Reaction: N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = -92 kJ/mol. This ΔH corresponds to the formation of 2 moles of NH₃.
• Enthalpy for Formation of 1 Mole of NH₃: For 1 mole of NH₃, the reaction is ½N₂(g) + 1½H₂(g) → NH₃(g), so ΔH = -92 / 2 = -46 kJ/mol.
• Reverse Reaction (Decomposition): The decomposition reaction is NH₃(g) → ½N₂(g) + 1½H₂(g). Since enthalpy is a state function, the enthalpy change for the reverse reaction is the negative of the forward reaction: ΔH_decomp = -(-46) = +46 kJ/mol.
• Conclusion: The enthalpy change for the decomposition of 1 mole of NH₃ is +46 kJ/mol.