# The standard enthalpy of formation of $NH_3$ is $-46\: kJ/mol.$ 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

$\begin {array} {1 1} (A)\;-964\: kJ/mol & \quad (B)\;352\: kJ/mol \\ (C)\; 1056\: kJ/mol & \quad (D)\;-1102 \: kJ/mol \end {array}$

$\large\frac{1}{2}$$N_2(g) + H_2(g) \rightarrow NH_3(g); \Delta H_f^{\circ} = -46 \: kJ/mol 2H(g) \rightarrow H_2(g);\: \: \: \: \: \Delta H_f^{\circ} = -436\: kJ/mol 2N(g) \rightarrow N_2(g);\: \: \: \: \: \Delta H_f^{\circ} = -712\: kJ/mol Let x is bond energy of N-H bond \large\frac{1}{2}$$ \times(-712) + 3(-436)/2 – 3x = -46$
$\Rightarrow 3x = 1056$
$\Rightarrow x = 352\: kJ/mol$
Ans : (B)