Question

The enthalpy change for the reaction is:
$Xe{F}_4⟶X{e}^{+}+F^{-}+F_2+F;\Delta H=?$

The average $Xe--F$ bond energy is $34kcalmo{l}^{-1}$ . ${IE}_1$ of Xe is $279kcalmo{l}^{-1}$ and electron affinity of F is $85kcalmo{l}^{-1}$ and $e_{F--F}=38kcalmo{l}^{-1}$.

• A. $345kcalmo{l}^{-1}$
• B. $292kcalmo{l}^{-1}$
• C. $174kcalmo{l}^{-1}$
• D. None of these

Answer 1

The correct option is B $292kcalmo{l}^{-1}$

Bond energy $={BE}_{X_e-F}=34kCal/mol$
So, $X_e{F}_4=4(X_e-F)$ bonds
$\Rightarrow$ total bond energy$=4\times 34kCal/mol\Rightarrow {BE}_{X_e-F_4}=136kCal/mol$
Now this amount of bond energy is liberated during formation of $X_e{F}_4$ ${\Delta }_f{H}_{X_e{F}_4}=-136kCal/mol$
$X_e+4F\to X_e{F}_4$ ${\Delta }_f{H}_{X_e{F}_4}=-136kCal/mol$ ...(1)
Now $I{E}_1$ of $X_e=279kCal/mol$
So $X_e\to {X_e}^{+}+e^{-}$ $\Delta H=+I{E}_1$ ...(2)
Now, $EA$ of $F=85kCal/mol$
$F\to F^{-}-e^{-}$ $\Delta H=-EA$ ...(3)
(energy is released in this process)
and $2F\to F_2$ $\Delta K=-e_{F-F}$ ...(4)
Now $\left(1\right)$ becomes$\Rightarrow X_e{F}_4\to X_e+4F$ ${\Delta }_f{H}_{X_e{F}_4}=136kCal/mol$ ...(5)
Now, $\left(2\right)+\left(3\right)+\left(4\right)+\left(5\right)$ gives $X_e{F}_4+X_e+3F\to X_e+4F+X_e^{+}+e^{-}+F^{-}-e^{-}+F_2$
$\Delta H={\Delta }_f{H}_{X_e{F}_4}+\left(H{E}_1\right)+(-EA)+(-e_{F-F})=136+279+(-85)+(-38)=292$
$\Rightarrow X_e{F}_4=X_e^{+}+F^{-}+F_2+F$
$\Delta H=292kCal/mol$