Question

The standard enthalpy of formation of $NH_3$ is $-46.0\, kJ \,mol^{-1}$. If the enthalpy of formation of $H_2$ from its atoms is $-436\, kJ \,mol^{-1}$ and that of $N_2$ is $-712\, kJ\, mol^{-1}$, the average bond enthalpy of $N-H$ bond in $NH_3$ is

• A. $-964 \,kJ\, mol^{-1}$
• B. $+352 \,kJ\, mol^{-1}$
• C. $+\,1056 \,kJ\, mol^{-1}$
• D. $-\, 1102 \,kJ\, mol^{-1}$

Answer 1

Answer: (B)

Enthalpy of formation of $NH_3 = -46 \,kJ/mole$
$\therefore N_{2} + 3H_{2} \to 2NH_{3} \,\Delta H_{f} = - 2 \,x \,46\, kJ \,mol$
Bond breaking is endothermic and Bond formation is exothermic
Assuming ‘x’ is the bond energy of $N-H$ bond $(kJ\, mol^{-1})$
$\therefore 712 + (3 \,x \,436)- 6x = -46 \,x \,2$
$\therefore x = 352\, kJ/mol$