Answer :

Given: For the given reaction,

N2(g) + 3H2(g) 2NH3(g)


Enthalpy of reaction (ΔrH°) = –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 ΔrH°


As, ΔrH° = –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.


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