Question

The standard enthalpy of formation of $N{H}_3$ 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 $N{H}_3$ is:

• A. $-1102$ kJ/mol
• B. $-964$ kJ/mol
• C. $+352$ kJ/mol
• D. $+1056$ kJ/mol

Answer 1

The correct option is C $+352$ kJ/mol

$\frac{1}{2}{N}_2+\frac{3}{2}{H}_2\to N{H}_3$
$\Delta H_r^0=\Delta H_{f\left(N{H}_3\right)}^0-\frac{3}{2}\Delta H_{f\left(H_2\right)}^0-\frac{1}{2}\Delta H_{f\left(N_2\right)}^0$

where $\Delta H_r^0=$ the average bond enthalpy of N−H bond in NH${}_3$

$3\times \Delta H_r^0=46-(\frac{3}{2}\times (-436\left)\right)-(\frac{1}{2}\times (-712\left)\right)$
$3\times \Delta H_r^0=46+654+356$

$\Delta H_r^0=\frac{1056}{3}$
$\Delta H_r^0=+352$ kJ/mol.

Hence, the correct option is $C$