The standard enthalpy of formation of NH_3 is | vedclass

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(C) The formation reaction is: $\frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) \rightarrow NH_3(g)$; $\Delta H_f = -46.0 \, kJ/mol$. Bond dissociation energi

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$+352$

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(C) The formation reaction is: $\frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) ightarrow NH_3(g)$; $\Delta H_f = -46.0 \, kJ/mol$. Bond dissociation energies are the negative of the enthalpy of formation from atoms: $BE(N \equiv N) = +712 \, kJ/mol$ $BE(H-H) = +436 \, kJ/mol$. Using the formula: $\Delta H_f = [\frac{1}{2} BE(N \equiv N) + \frac{3}{2} BE(H-H)] - [3 	imes BE(N-H)]$. Substituting the values: $-46.0 = [\frac{1}{2}(712) + \frac{3}{2}(436)] - 3 	imes BE(N-H)$. $-46.0 = [356 + 654] - 3 	imes BE(N-H)$. $-46.0 = 1010 - 3 	imes BE(N-H)$. $3 	imes BE(N-H) = 1010 + 46 = 1056$. $BE(N-H) = 1056 / 3 = 352 \, kJ/mol$.

The standard enthalpy of formation of $NH_3$ 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 $NH_3$ is......$kJ/mol$

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