Question

For the reaction $N_2+3X_2\to 2N{X}_3$ where $X=F,Cl$ (the average bond energies are $F-F=155kJmo{l}^{-1}$, $N-F=272kJmo{l}^{-1}$, $Cl-Cl=242kJmo{l}^{-1}$ , $N-Cl=200kJmo{l}^{-1}$ and $N\equiv N=941kJmo{l}^{-1}$). The heats of formation of $N{F}_3$ and $NC{l}_3$ in $kJmo{l}^{-1}$, respectively, are closest to:

• A. $-226$ and $+467$
• B. $+226$ and $-467$
• C. $-151$ and $+311$
• D. $+151$ and $-311$

Answer 1

Answer: (A)

$-226$ and $+467$

$N_2+3X_2\to 2N{X}_3$

Energy to break 1 mole of $N_2$ bond $=941kJmo{l}^{-1}$

Energy to break 3 mole of $F_2$ bond $=3\times 155kJmo{l}^{-1}$

Energy released caused of bond formation of 2 moles of 3 bonds $N-F=2\times 3\times 272$

Heat of formation of $N{F}_3$ by Hess law $=941+3\times 155-(2\times 3\times 272)=-226kJmo{l}^{-1}$

Similarly for $NC{l}_3$,

Just replacing the necessary values in previous calculations, heat of formation of $NC{l}_3=941+3\times 242-(2\times 3\times 200)=+467kJmo{l}^{-1}$.

Hence, option $A$ is correct.