Question

If the bond dissociation energies of $X{Y}_1{X}_2$ and $Y_2$ (all diatomic molecules) are in the ratio of $1:1:0.5$ and ${\delta }_{1}H$ for the formation of $XY$ is $-200kJmo{l}^{-1}$. The bond dissociation energy of $X_2$ will be:

• A. $200kJ;/mol$
• B. $100kJ/mol$
• C. $800kJ/mol$
• D. $300kJ/mol$

Answer 1

The correct option is C $800kJ/mol$

Solution:- (C) $800kJ/mol$

Given that the bond dissociation energy of $XY,X_2$ and $Y_2$ are in the ratio $1:1:0.5$.

Let the bond dissociation energy of $XY,X_2$ and $Y_2$ be $2x,2x$ and $xkJ/mol$ respectively.

Now,

$\frac{1}{2}{X}_2+\frac{1}{2}{Y}_2⟶XY;\Delta H=-200kJ/mol$

For the above reaction-

$\Delta H=\frac{1}{2}{B.E.}_{(X-X)}+\frac{1}{2}{B.E.}_{(Y-Y)}-{B.E.}_{(X-Y)}$

$\Rightarrow -200=\frac{1}{2}\times 2x+\frac{1}{2}\times x-2x$

$\Rightarrow -200=\frac{3}{2}x-2x$

$\Rightarrow x=2\times 200=400$

Therefore,

Bond dissociatio energy of $X_2=2\times 400=800kJ/mol$

Hence the bond dissociation energy of $X_2$ is $800kJ/mol$.