Question

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

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

Answer 1

The correct option is A 390.3 $kJmo{l}^{-1}$

Given $B.E$ of $H_2=436$kj/mol

Bond enthalpy of $N_2$ $=941.3$kJ/mol

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

${\Delta H}^0\left({NH}_3\right)=B.E$ of reactant $-B.E$ of producrs

$=\frac{1}{2}\left(N_{2}Bond\right)+\frac{3}{2}\left(H_{2}Bond\right)-3(N-HBond)$

$-46=\frac{1}{2}\times 941.3+\frac{3}{2}\times 436-3(N-HBond)$

$3(B.E$ of $N-H$ Bond$)$ $=470.65+654+46$

$B.E$ of $NH$ bond $=\frac{1170.65}{3}=390.2$KJ/mol