What is enthalpy of formation of $\mathrm{NH}_3$ if bond enthalpies are as $(\mathrm{N} \equiv…

What is enthalpy of formation of $\mathrm{NH}_3$ if bond enthalpies are as $(\mathrm{N} \equiv \mathrm{N})=941 \mathrm{~kJ},(\mathrm{H}-\mathrm{H})=436 \mathrm{~kJ},(\mathrm{~N}-\mathrm{H})=389 \mathrm{~kJ}$ ?
  1. $-84.5 \mathrm{~kJ}$
  2. $-21.25 \mathrm{~kJ}$
  3. $-42.5 \mathrm{~kJ}$
  4. $-63.45 \mathrm{~kJ}$

Solution

$\begin{aligned} & \frac{1}{2} \mathrm{~N}_{2(\mathrm{~g})}+\frac{3}{2} \mathrm{H}_{2(\mathrm{~g})} \longrightarrow \mathrm{NH}_{3(\mathrm{~g})} \\ & \Delta \mathrm{H}_{\text {reacion }}=\Delta \mathrm{H}_{\mathrm{f}\left(\mathrm{NH}_3\right)}=\frac{1}{2} \mathrm{BE}_{(\mathrm{NaN})}+\frac{3}{2} \mathrm{BE}(\mathrm{H}-\mathrm{H})-3 \mathrm{BE}_{(\mathrm{N}-\mathrm{H})} \\ & \Delta \mathrm{H}_{\mathrm{f}\left(\mathrm{NH}_3\right)}=\frac{1}{2} \times 941+\frac{3}{2} \times 436-3 \times 389 \\ & =470.5+654-1167 \\ & =-42.5 \mathrm{~kJ}\end{aligned}$

Asked in: MHT CET 2021 (22 Sep Shift 2)

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