The bond dissociation energies of X_2 Y_2… - NEET UG 2018 Chemistry PYQ

NEET UG 2018 — Chemistry Physical Chemistry

2018mcqhard

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

Official previous-year question

Held on 30 Apr 2018 · Verified 9 Jul 2026.

Options

A
$800 \mathrm{kJ} {\mathrm{mol}}^{-1}$
B
$100 \mathrm{kJ} {\mathrm{mol}}^{-1}$
C
$200 \mathrm{kJ} {\mathrm{mol}}^{-1}$
D
$400 \mathrm{kJ} {\mathrm{mol}}^{-1}$

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Solution

Let B.E. of ${x}_{2}, {y}_{2}$ and $\mathrm{xy}$ are $x \mathrm{kJ} {\mathrm{mol}}^{-1}, 0.5x \mathrm{kJ} {\mathrm{mol}}^{-1}$ and $x \mathrm{kJ} {\mathrm{mol}}^{-1}$ respectively

$\frac{1}{2}{X}_{2}+\frac{1}{2}{Y}_{2}\rightarrow \mathrm{XY};\Delta H=-200 \mathrm{kJ} {\mathrm{mol}}^{-1}$

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

$-200=[\frac{1}{2}\times (x)+\frac{1}{2}\times (0.5x)]-[1\times (x)]$ $\Rightarrow x=800\mathrm{kJ} \mathrm{mol}$

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

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