MCQ
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 $-712\, kJ/mol$, the average bond enthalpy of $N - H$ bond in $NH_3$ is......$kJ/mol$
  • A
    $- 1102$
  • $- 964$
  • C
    $+352$
  • D
    $+1056$

Answer

Correct option: B.
$- 964$
b
Given $\frac{1}{2}{N_2} + \frac{3}{2}{H_2} \leftrightarrow N{H_3};$

$\Delta {H_f} =  - 46.0\,kJ/mol$

$H + H \leftrightarrow {H_2};\,\Delta {H_f} =  - 436\,kJ/mol$

$N + N \leftrightarrow {N_2};\,\Delta {H_f} =  - 712\,kJ/mol$

$\Delta {H_f}(N{H_3}) = \frac{1}{2}\Delta {H_{N - N}} + \frac{3}{2}\Delta {H_{H - H}} - \Delta {H_{N - H}}$

$ - 46 = \frac{1}{2}( - 712) + \frac{3}{2}( - 436) - \Delta {H_{N - H}}$

On calculation

$\Delta {H_{N - H}} =  - 964\,kJ/mol$

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