Question

$\Delta H_{\text{f}}^o$ for $N{F}_3-113kJ/mol$. Bond energy for $N-F$ bond is $27.5kJ/mo$l. Calculate the bond energies of $N_2$ and $F_2$ if their magnitudes are in the ratio $6:1$ are, respectively

• A. $822.5,137.1kJ/mol$
• B. $979.8,163.3kJ/mol$
• C. $949.32,157.22kJ/mol$
• D. $762.6,127.1kJ/mol$

Answer 1

The correct option is C $949.32,157.22kJ/mol$

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

$\Delta H_f^o=-113kJ/mol$

Bond Bond dissociation enthalapy

$N-F$ $\Delta H_1=27.5$

$N-N$ $\Delta H_2=6x$

$F-F$ $\Delta H_3=x$

$275\times 3+(-113)=6x\times \frac{1}{2}+\frac{3}{2}x$

$712=\frac{9}{2}x$

$1424=9x$

$x=\frac{1424}{9}=158.22kJmo{l}^{-1}$

Bond dissociation enthalapy of $N_2=6\times 158.22$

$=949.33kJmo{l}^{-1}$

Bond dissociation enthalapy of $F_2=158.22kJmo{l}^{-1}$

option C