Question

Bond dissociation enthalphies of $H_2\left(g\right)$ and $N_2\left(g\right)$ are 436.0 kJ $mo{l}^{-1}$ and 941.8 kJ $mo{l}^{-1}$ respectively and enthalpy of formation of $N{H}_3\left(g\right)$ is -46 kJ $mo{l}^{-1}$. What is enthalpy of atomization of $N{H}_3\left(g\right)$?

• A. 390.3 kJ $mo{l}^{-1}$
• B. 1170.9 kJ $mo{l}^{-1}$
• C. 590 kJ $mo{l}^{-1}$
• D. 720 kJ $mo{l}^{-1}$

Answer 1

The correct option is B 1170.9 kJ $mo{l}^{-1}$

$\frac{1}{2}{N}_2+\frac{3}{2}{H}_2\to N{H}_3$

${\Delta }_f{H}^o=-46kJmo{l}^{-1}$

$N{H}_3\to \frac{1}{2}{N}_2+\frac{3}{2}{H}_2$

${\Delta }_f{H}^o=46kJmo{l}^{-1}$

$\frac{1}{2}{N}_2\to N$

${\Delta }_a{H}^o=\frac{941.8}{2}=470.9kJmo{l}^{-1}$

$\frac{3}{2}{H}_2\to H$

${\Delta }_a{H}^o=\frac{3}{2}\times 436=654kJmo{l}^{-1}$

Total enthalapy of atomization

$46+470.9+654=1170.9kJmo{l}^{-1}$

option B is correct