Question

The standard enthalpy of formation of $NH_{3}$ is - $46.0\, kJ/mol$. If the enthalpy of formation of $H_{2}$ from its atoms is - $436\, kJ/mol$ and that of $N_{2}$ is - $712\, kJ/mol$, find the average bond enthalpy of $N - H$ bond in $NH_{3}$.

Answer 1

Given $\frac{1}{2} N _2+\frac{3}{2} H _2 \rightleftharpoons NH _3$
$\Delta H_f=-46.0 \,kJ / mol$
$H+H \rightleftharpoons H_2 ; \Delta H_f=-436 \,kJ / mol$
$N+N \rightleftharpoons N_2 ; \Delta H_f=-712 \,kJ / mol $
$\Delta H_f\left(N H_3\right)=\frac{1}{2} \Delta H_{N-N}+\frac{3}{2} \Delta H_{H-H}-\Delta H_{N-H}$
$-46=\frac{1}{2}(-712)+\frac{3}{2}(-436)-\Delta H_{N-H}$
On calculation
$\Delta H_{N-H}=-964 \,k J / m o l$