$N_{2} (g) + 3 H_{2} (g) ⟶ 2 N H_{3} : Δ H = - 92 k J$ is Haber`s process for manufacture of $N H_{3}$ . What is the heat of formation of $N H_{3}$ ? (in kJ/mol)

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Solution

As given,
$N_{2} (g) + 3 H_{2} (g) ⟶ 2 N H_{3} : Δ H = - 92 k J$
so per mole formation enthalpy of ammonia is

$1 / 2 N_{2} (g) + 3 / 2 H_{2} (g) ⟶ N H_{3} : Δ H = - 92 / 2 = - 46 k J$

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The enthalpy of formation of $N H_{3}$ (g) is $- 46.2$ kJ mol $^{- 1}$ . The heat of the following reaction is:

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