1. Tardigrade
  2. Question
  3. Chemistry
  4. Given the bond energies of N≡ N,H-H and N-H bonds are 945,436 and 391kJmol- 1 respectively, the enthalpy of the reaction N2 (. g .)+3H2 (. g .) arrow 2(NH)3 (. g .) is :

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Q. Given the bond energies of $N\equiv N,H-H$ and $N-H$ bonds are $945,436$ and $391kJmol^{- 1}$ respectively, the enthalpy of the reaction $N_{2 \left(\right. g \left.\right)}+3H_{2 \left(\right. g \left.\right)} \rightarrow 2\left(NH\right)_{3 \left(\right. g \left.\right)}$ is :

NTA AbhyasNTA Abhyas 2020

Solution:

$N\equiv N_{\left(\right. g \left.\right)}+3\left(\right.H-H\left.\right) \rightarrow 2\left(NH\right)_{3}\left(\right.6N-H\left(\left.\right)_{g}$
$\Delta H_{reaction}=\displaystyle \sum _{}^{}BE_{reactants}-\displaystyle \sum _{}^{}BE_{products}$ $=2253-2346$
$=-93kJ$