The enthalpy change $(Δ H)$ for the process $N_{2} H_{4} (g) \to 2 N (g) + 4 H (g) is 1724 kJ {mol}^{- 1}$ if the bond energy of $N - H$ bond in ammonia is $391 kJ {mol}^{- 1}$ what is the bond energy of $N - H$ bond is $N_{2} H_{4}$

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$160 kJ {mol}^{- 1}$

$391 kJ {mol}^{- 1}$

$1173 kJ {mol}^{- 1}$

$320 kJ {mol}^{- 1}$

answer is A.

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Detailed Solution

$\begin{matrix} H - \overset{H}{N} - \overset{H}{N} - H (So, 4 N - H \end{matrix}$ bond present

means their energy of $N - N in N_{2} H_{4}$ so the bond energy of $N - N in N_{2} H_{4}$

$= 1724 - 1564 = 160 KJ / mol .$

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