The bond dissociation energies of $X_2 , Y_ 2$ and XY are in the ratio of 1 : 0.5 : 1. ΔH for the formation of XY is –200 kJ $mol^{–1}$. The bond dissociation energy of $X_2$ will be
$800 kJ mol^{–1}$
$100 kJ mol^{–1}$
$200 kJ mol^{–1}$
$400 kJ mol^{–1}$
Explanation:
The reaction for $Δ_fH^°(XY)$
$\dfrac{1}{2}X_2(g)+\dfrac{1}{2}Y_2(g)\rightarrow XY(g)$
Bond energies of $X_2, Y_2$ and XY are X, $\dfrac{X}{2}$, X respectively.
$\therefore \triangle H=\left(\dfrac{X}{2}+\dfrac{X}{4}\right)-X=-200$
On solving, we get
$\Rightarrow-\dfrac{X}{2}+\dfrac{X}{4}=-200$
$\Rightarrow X=800kJ/mole$
The reaction for $Δ_fH^°(XY)$
$\dfrac{1}{2}X_2(g)+\dfrac{1}{2}Y_2(g)\rightarrow XY(g)$
Bond energies of $X_2, Y_2$ and XY are X, $\dfrac{X}{2}$, X respectively.
$\therefore \triangle H=\left(\dfrac{X}{2}+\dfrac{X}{4}\right)-X=-200$
On solving, we get
$\Rightarrow-\dfrac{X}{2}+\dfrac{X}{4}=-200$
$\Rightarrow X=800kJ/mole$