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Heat of Formation

N2g + 3H2g⟶ 2...

Question

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

Δ_{f} H N H_{3} = - 46 k J m o l^{- 1}

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Δ_{f} H N H_{3} = + 46 k J m o l^{- 1}

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Δ_{f} H N H_{3} = - 96 k J m o l^{- 1}

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None of these

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Solution

The correct option is B $Δ_{f} H N H_{3} = - 46 k J m o l^{- 1}$
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:

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

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