Question

If the bond dissociation energy of $XY,X_2,Y_2$ (all are gaseous diatomic molecules) are in the ratio of 1:1:0.5 and enthalpy for the formation of $XY$ is -200kJ/mol. The bond dissociation energy of $X_2$ will be:

• A. 200 KJ/ mol
• B. 100 KJ/ mol
• C. 800 KJ/ mol
• D. 300 KJ/ mol

Answer 1

Answer: (C)

800 KJ/ 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 energy of $XY,X_2$ and $Y_2$ be $2x,2x$ and $x$ respectively.

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

For the above reaction,

$\Delta H={\sum B.E.}_{\left(\text{reactant}\right)}-{\sum B.E.}_{\left(\text{product}\right)}$

$\Rightarrow -200=(\frac{1}{2}\times 2x+\frac{1}{2}\times x)-\left(2x\right)$

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

$\Rightarrow x=(-200)\times (-2)=400$

Therefore,

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

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