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Exercises · 5.13

Q.Given N2(g)+3H2(g)→2NH3(g)N_2(g) + 3H_2(g) \rightarrow 2NH_3(g); ΔrH=−92.4\Delta_r H = -92.4 kJ mol−1^{-1}. What is the standard enthalpy of formation of NH3NH_3 gas?

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The standard enthalpy of formation of a compound is the enthalpy change when one mole of it is formed from its elements in their standard states. For NH₃, the given reaction produces two moles, so the formation enthalpy is half of ΔrH\Delta_r H: -46.2 kJ mol⁻¹.

The key here is to not confuse the enthalpy change of a reaction with the standard enthalpy of formation. They are related, but not the same thing.

The standard enthalpy of formation, ΔfH∘\Delta_f H^\circ, is defined for the formation of exactly one mole of a compound from its constituent elements in their standard states. For ammonia (NH₃), the formation reaction would be:

12N2(g)+32H2(g)→NH3(g)\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightarrow NH_3(g)

Notice the coefficients: they are fractions, because we only want one mole of product.

The reaction you are given produces two moles of NH₃:

N2(g)+3H2(g)→2NH3(g)ΔrH=−92.4 kJ mol−1N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \quad \Delta_r H = -92.4 \text{ kJ mol}^{-1}

The ΔrH\Delta_r H here is the enthalpy change for the reaction as written — for the formation of two moles of NH₃. To get the enthalpy change per mole of NH₃, you simply divide by 2.

  1. Identify the target. We need ΔfH∘\Delta_f H^\circ for NH₃(g). That is the enthalpy change for forming 1 mole of NH₃ from N₂ and H₂ in their standard states.

  2. Relate the given reaction to the target. The given reaction forms 2 moles of NH₃. So the enthalpy change for forming 1 mole is half of the given ΔrH\Delta_r H.

  3. Perform the calculation. …

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