The value of $Δ_{f} H^{⊖}$ for $N H_{3}$ is -91.8kJ $m o l^{- 1}$ . Calculate enthalpy change for the following reaction.
$2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$

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Solution

Ans. Given, $\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \to N H_{3} (g); Δ_{r} H^{⊖} = - 91.8 k J m o l^{- 1}$
( $Δ_{f} H^{⊖}$ means enthalpy of formation of 1 mole of $N H_{3}$ )
$∴$ Enthalpy change for the formation of 2 moles of $N H_{3}$
$N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g); Δ_{f} H^{⊖} = 2 \times - 91.8 = - 183.6 k J m o l^{- 1}$
And for the reverse reaction,
$2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g); Δ_{f} H^{⊖} = + 183.6 k J m o l^{- 1}$
Hence, the value of $Δ_{f} H^{⊖} for N H_{3} is 183.6 k J m o l^{- 1}$

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