The bond dissociation energies of X_ 2 , Y_ 2 and XY are in the r… - Vidyadip

MCQ

The bond dissociation energies of $\mathrm{X}_{2}, \mathrm{Y}_{2}$ and $\mathrm{XY}$ are in the ratio of $1: 0.5: 1 . \Delta \mathrm{H}$ for the formation of $\mathrm{XY}$ is $-200\; \mathrm{kJ} \mathrm{mol}^{-1} .$ The bond dissociation energy of $X_2$ will be......$\mathrm{kJ} \mathrm{mol}^{-1}$

A
$200$
B
$100$
✓
$800$
D
$400$

✓

Answer

Correct option: C.

$800$

c
let B.E. of $\mathrm{x}_{2}, \mathrm{y}_{2} \;and\; \mathrm{xy}$ are $\mathrm{x} \;\mathrm{kJ}\; mol^{-1}$

$0.5 \;\mathrm{x} \mathrm{kJ} \mathrm{mol}^{-1}$ and $\mathrm{x}\; \mathrm{kJ} \mathrm{mol}^{-1}$ respectively

$\frac{1}{2} \mathrm{x}_{2}+\frac{1}{2} \mathrm{y}_{2} \rightarrow \mathrm{xy} ; \Delta \mathrm{H}=-200 \mathrm{kJmol}^{-1}$

$\Delta \mathrm{H}=-200=\Sigma(\mathrm{B} . \mathrm{E})$ Reactant $-\Sigma(\mathrm{B} . \mathrm{E})$ product

$=\left[\frac{1}{2} \times(\mathrm{x})+\frac{1}{2} \times(0.5 \mathrm{x})\right]-[1 \times(\mathrm{x})]$

B.E of $\mathrm{X}_{2}=\mathrm{x}=800 \mathrm{kJ} \mathrm{mol}^{-1}$

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