Question

Calculate the bond enthalpy of N-H in NH3​

Answer 1

1145is the enthalpy of NH3

Answer 2

Answer : The bond enthalpy of N-H in $NH_3$ is, 557.6 KJ/mole​

Explanation :

The balanced chemical reaction are,

(1) $\frac{1}{2}N_2(g)+\frac{2}{3}O_2(g)\rightarrow NH_3(g)$    $\Delta H_1=-46KJ/mole$

(2) $\frac{1}{2}H_2(g)\rightarrow H(g)$    $\Delta H_2=218KJ/mole$

(3) $\frac{1}{2}N_2(g)\rightarrow N(g)$    $\Delta H_3=973KJ/mole$

The bond dissociation reaction of ammonia will be,

$NH_3(g)\rightarrow 3H+N$    $\Delta H=?$

Now adding reverse reaction of reaction 1, thrice of reaction 2 and reaction 3, we get the bond enthalpy of N-H.

The expression of bond dissociation enthalpy is,

$\Delta H=\frac{\Delta H_1+3\times \Delta H_2+\Delta H_3}{3}$

$\Delta H=\frac{46+3\times 218+973}{3}=557.6KJ/mole$

Therefore, the bond enthalpy of N-H in $NH_3$ is, 557.6 KJ/mole​