The standard enthalpy of formation of NH_3 is | vedclass

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(C) The chemical equation for the formation of $NH_3$ is: $\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} 	o NH_{3(g)}$; $\Delta H_f = -46 \, kJ \, mol^

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$352$

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(C) The chemical equation for the formation of $NH_3$ is: $\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} 	o NH_{3(g)}$; $\Delta H_f = -46 \, kJ \, mol^{-1}$.Using the bond enthalpy formula: $\Delta H_f = [\frac{1}{2} 	imes (B.E.)_{N \equiv N} + \frac{3}{2} 	imes (B.E.)_{H-H}] - [3 	imes (B.E.)_{N-H}]$.Substituting the given values: $-46 = [\frac{1}{2} 	imes 712 + \frac{3}{2} 	imes 436] - 3 	imes (B.E.)_{N-H}$.$-46 = [356 + 654] - 3 	imes (B.E.)_{N-H}$.$-46 = 1010 - 3 	imes (B.E.)_{N-H}$.$3 	imes (B.E.)_{N-H} = 1010 + 46 = 1056$.$(B.E.)_{N-H} = \frac{1056}{3} = 352 \, kJ \, mol^{-1}$.

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

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