Ammonia contains 3 N – H bonds while hydrazine contains a N – N single bond and 4 N – H bonds. Use the given data to determine the enthalpy change as indicated. Given : $Δ_{f} H o f N_{2} H_{4} (g) i s 76 k J m o l^{- 1}$ .
Bond Bond energy $(k J m o l^{- 1})$
H – H 436
N – N 193
$N \equiv N$ 941

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The bond enthalpy of N – H bond is 386 kJ $m o l^{- 1} .$

The bond enthalpy of N – H bond is 424 kJ $m o l^{- 1} .$

The enthalpy of formation of $N H_{3} (g)$ is – 33.5 kJ $m o l^{- 1} .$

The enthalpy of formation of $N H_{3} (g)$ is – 147.5 kJ $m o l^{- 1} .$

answer is B, C.

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Detailed Solution

$N \equiv N (g) + 2 H_{2} (g) \to H_{2} N - N H_{2} (g) Δ_{f} H (N_{2} H_{4}, g) = 76 = \in_{N \equiv N} + 2 \in_{H - H} - 4 \in_{N - H} - \in_{N - N} 76 = 941 + (2 \times 436) - 4 \times \in_{N - H} - 193 \in_{N - H} = 386 k J m o l^{- 1} \frac{1}{2} N \equiv N (g) + \frac{3}{2} H_{2} (g) \to N H_{3} (g) Δ_{f} H (N H_{3}, g) = (\frac{1}{2} \times 941) + (\frac{3}{2} \times 436) - 3 \in_{N - H} Δ_{f} H (N H_{3}, g) = 470.5 + 654 - (3 \times 386) = - 33.5$