What is standard $\mathrm{N} \equiv \mathrm{N}$ bond enthalpy from following reaction?…

What is standard $\mathrm{N} \equiv \mathrm{N}$ bond enthalpy from following reaction? $\mathrm{N}_{2_{(\mathrm{g})}}+3 \mathrm{H}_{2_{(\mathrm{g})}} \longrightarrow 2 \mathrm{NH}_{3_{(\mathrm{g})}} \Delta \mathrm{H}^{\circ}=-83 \mathrm{KJ}$ $\left(\Delta \mathrm{H}^{\circ}{ }_{(\mathrm{H}-\mathrm{H})}=435 \mathrm{~kJ}, \Delta \mathrm{H}^{\circ}{ }_{(\mathrm{N}-\mathrm{H})}=389 \mathrm{~kJ}\right)$
  1. 435 kJ
  2. $2334 \mathrm{~kJ}$
  3. $946 \mathrm{~kJ}$
  4. $1305 \mathrm{~kJ}$

Solution

$\mathrm{N}_{2(8)}+3 \mathrm{H}_{2(8)} \longrightarrow 2 \mathrm{NH}_{3(8)} \quad \Delta \mathrm{H}^{\circ}=-83 \mathrm{~kJ}$ $\Delta \mathrm{H}^{0}=\sum \Delta \mathrm{H}^{0}$ (reactant bonds) $-\sum \Delta \mathrm{H}^{0}$ (product bonds) $\left.\therefore \Delta H^{0}=\left[\Delta H^{0}{ }_{(N=N)}^{0}+3 \Delta H^{0}{ }^{0} H-H\right)\right]-\left[6 \Delta H^{0}{ }^{0}(N-H)\right]$ $\therefore-83=\Delta H_{(N=N)}^{0}+3(435)-6(389)$ $\therefore-83=\Delta H_{(N=N)}^{0}+1305-2334$ $\therefore \Delta H^{0}(N \sim N)=-83-1305+2334=946 \mathrm{~kJ}$

Asked in: MHT CET 2020 (19 Oct Shift 2)

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