Question

The enthalpy of formation of ammonia when calculated from the following bond energy data is ($B.E.$ of $N - H$, $H - H$, $N \equiv N$ is $389 \,kj \,mol^{-1}$, $435\, kj\, mol^{-1}$, $945.36\, kj\, mol^{-1}$ respectively)

• A. $- 41.82\, kj\, mol^{-1}$
• B. $+ 83.64\, kj\, mol^{-1}$
• C. $- 945.36\, kj\, mol^{-1}$
• D. $- 833\, kj\, mol^{-1}$

Answer 1

Answer: (A)

$N_{2} +3H_{2} \rightarrow 2NH_{3}$
$\Delta H = \Delta H\left(N \equiv N\right)+3\times\Delta H \left(H-H\right)-2 \times 3\Delta H\left(N-H\right)$
$= 945.36 + 3 \times 435.0 - 6 \times 389.0 = - 83.64 \,kj$
Heat of formation of $NH_{3}= \frac{-83.64}{2} = -41.82 \,kj/mol$