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:

+ 1056 k J {m o l}^{- 1}

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- 1102 k J {m o l}^{- 1}

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- 964 k J {m o l}^{- 1}

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+ 352 k J {m o l}^{- 1}

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Solution

The correct option is D $+ 352 k J {m o l}^{- 1}$
$H_{r e a c t i o n} = \sum {B E}_{r e a c t a n t s} - \sum {B E}_{p r o d u c t s}$
$- 46 = (\frac{(}{7} 12) + \frac{3}{2} (436)) - 3 x$
$x = + 352 k J {m o l}^{- 1}$

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The standard enthalpy of formation of $N H_{3}$ is $46.0 k J m o l^{- 1}$ . If the enthalpy of formation of $H_{2}$ from its atom 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