The enthalpy change $(\Delta H)$ for the process $\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g}) ightarrow 2…

The enthalpy change $(\Delta H)$ for the process $\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g}) ightarrow 2 \mathrm{~N}(\mathrm{~g})+4 \mathrm{H}(\mathrm{g})$ is $1724 \mathrm{KJ} \mathrm{mol}^{-1}$. If
the bond energy of $\mathbf{N}-\mathbf{H}$ bond in ammonia is $391 \mathbf{K J} \mathrm{mol}^{-1} .$ What is the bond energy of $\mathrm{N}$
$-\mathbf{N}$ bond is $\mathbf{N}_{\mathbf{2}} \mathbf{H}_{\mathbf{4}}$
  1. $160 \mathrm{KJ} \mathrm{mol}^{-1}$
  2. $391 \mathrm{KJ} \mathrm{mol}^{-1}$
  3. $1173 \mathrm{KJ} \mathrm{mol}^{-1}$
  4. $320 \mathrm{KJ} \mathrm{mol}^{-1}$

Solution



(So, $4 \mathrm{~N}-\mathrm{H}$ bond present $)$
means their energy $=391 \times 4=1564$
so the bond energy of $\mathrm{N}-\mathrm{N}$ in $\mathrm{N}_{2} \mathrm{H}_{4}$
$=1724-1564=160 \mathrm{KJ} / \mathrm{mol}$ ^

Asked in: JEE-TOPICTESTS-CHEMISTRY

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