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NCERT Exemplar · Q23

Q.The value of ΔfH° for NH3 is -91.8 kJ mol^-1. Calculate enthalpy change for the following reaction :
2NH3(g) → N2(g) + 3H2(g)

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The standard enthalpy of formation tells us the energy change when one mole of a compound forms from its elements. Reversing this process for two moles of ammonia gives an enthalpy change of +183.6 kJ\boxed{+183.6 \text{ kJ}}.

The standard enthalpy of formation, ΔfH∘\Delta_f H^\circ, measures the enthalpy change when one mole of a substance is formed from its constituent elements in their standard states. For ammonia, the formation reaction is:

12N2(g)+32H2(g)⟶NH3(g)ΔfH∘=−91.8 kJ mol−1\frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g) \longrightarrow \text{NH}_3(g) \qquad \Delta_f H^\circ = -91.8 \text{ kJ mol}^{-1}

The negative sign tells us that forming ammonia from nitrogen and hydrogen releases energy—it's exothermic.

Now we need the enthalpy change for the decomposition of ammonia back into its elements. This is precisely the reverse of the formation reaction, scaled up for two moles.

Step-by-step calculation

  1. Write the formation reaction for one mole of NH₃

12N2(g)+32H2(g)⟶NH3(g)ΔH1=−91.8 kJ\frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g) \longrightarrow \text{NH}_3(g) \qquad \Delta H_1 = -91.8 \text{ kJ}

  1. Recognize that the target reaction is the reverse

    The question asks for:

2NH3(g)⟶N2(g)+3H2(g)2\text{NH}_3(g) \longrightarrow \text{N}_2(g) + 3\text{H}_2(g)

This decomposes ammonia into nitrogen and hydrogen—exactly opposite to formation.

  1. Apply Hess's law for the reverse reaction

    When you reverse a chemical equation, the sign of ΔH\Delta H flips. The decomposition of one mole of ammonia is:

NH3(g)⟶12N2(g)+32H2(g)ΔH=+91.8 kJ\text{NH}_3(g) \longrightarrow \frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g) \qquad \Delta H = +91.8 \text{ kJ}

  1. Scale to two moles of NH₃

    The target reaction involves two moles of ammonia, so multiply the enthalpy change by 2:

    ΔHreaction=2×(+91.8 kJ)=+183.6 kJ\Delta H_{\text{reaction}} = 2 \times (+91.8 \text{ kJ}) = +183.6 \text{ kJ} …

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