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Q.

The standard enthalpy of formation of $N H_{3}$ is $- 46 K J m o l^{- 1} .$ If the enthalpy of formation of $H_{2}$ from its atoms is $- 436 K J m o l^{- 1}$ and that of $N_{2}$ is $- 712 K J m o l^{- 1},$ the average bond enthalpy of N-H bond in $N H_{3}$ is (in $K J m o l^{- 1}$ )

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answer is 352.

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Detailed Solution

$\frac{1}{2} N_{2} + \frac{3}{2} H_{2} \to N H_{3}$

$Δ H_{f} (N H_{3}) = (\frac{1}{2} B . E N_{2} + \frac{3}{2} B . E H_{2}) - 3 B . E (N - H)$

$- 46 = \frac{1}{2} (712) + \frac{3}{2} (436) - 3 B . E_{N - H}$

$B . E_{N - H} = 352 K J / m o l e$

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The standard enthalpy of formation of NH3 is −46 KJmol−1. If the enthalpy of formation of H2 from its atoms is −436KJmol−1 and that of N2 is −712KJ mol−1, the average bond enthalpy of N-H bond in NH3 is (in KJ mol−1 )