Given

N₂(g) + 3H₂(g) → 2NH₃(g); = –92.4 kJ mol^–1

What is the standard enthalpy of formation of NH₃ gas?

Given: For the given reaction,

N₂(g) + 3H₂(g) → 2NH₃(g)

Enthalpy of reaction (Δ_rH^°) = –92.4 kJ mol^–1

To calculate standard enthalpy of formation of NH₃ gas, first we will divide the whole reaction by 2, so that we get 1 mol of NH₃

∴ N₂ (g) + H₂(g) → NH₃(g)

Hence, standard enthalpy of NH₃ is equal to the 1/2 Δ_rH^°

As, Δ_rH^° = –92.4 kJ mol^–1

∴ Standard enthalpy of NH₃ =

⇒Standard enthalpy of NH₃ = -46.2 kJ mol^-1

Thus, standard enthalpy of formation of NH₃ gas = -46.2 kJ mol^-1

Note: Standard enthalpy of formation is the amount of heat absorbed or evolved when 1 mole of the substance is directly obtained from its constituent elements.