Question

Bond dissociation enthalpies of $H_2\left(g\right)$ and $N_2\left(g\right)$ are 436.0 KJ ${mol}^{-1}$ and 941.8 ${mol}^{-1}$ and enthalpy of formation of ${NH}_3\left(g\right)$ is -46 KJ ${mol}^{-1}$. What is enthalpy of atomization of ${NH}_3\left(g\right)?$ What is average bond enthalpy of $N-H$ bond?

Answer 1

$N_2+3H_2⟶2N{H}_3$ $\Delta H=-46kJ/mol$

${\Delta }_f{H}_{\left(N{H}_3\right)}=\left(\text{BE of reactants- BE of products}\right)$

$\Rightarrow -46kJ/mol=(\frac{1}{2}\text{BE of}N\equiv N+\frac{3}{2}\text{BE of H-H bond})-\left(\text{3 BE of N-H bond}\right)$

$=\frac{1}{2}\left(941.8\right)+\frac{3}{2}\left(436.0\right)-3\left(\text{BE of N-H}\right)$

$\Rightarrow \left(\text{BE of N-H}\right)=\frac{1}{3}(470.9+654+46)=390.3kJ/mol$

Enthalpy of atomisation of $N{H}_3=3\times \text{BE of N-H bond}$

$=3\times 390.3$

$=1170.9kJ/mol$