The bond dissociation energy of X_2, Y_2 and X Y are in the ratio of 1: 0.5: 1. Δ H for the formation of X Y is -200 kJ / mol . The bond dissociation energy of X_2 will be

Correct answer: D. 800 kJ / mol Let the bond dissociation energy of X_2=a kJ / mol i.e. BE(X_2)=a kJ / mol Then, BE(Y_2)=0.5 a and BE(X Y)=a kJ / mol Given, (1)/(2) X_2+(1)/(2) Y_2 X Y, Δ H=-200 kJ / mol aligned Δ_r H & =BE( Rreactants )-BE( Product ) Δ_r H & =[(1)/(2) BE(X_2)+(1)/(2) BE(Y_2)]-BE(X Y) -200 & =(a)/(2)+(0.5 a)/(2)-a a & =(200)/(0.25)=800 kJ / mol aligned

BITSAT 2024

Exam: BITSAT
Year: 2024
Subject: Chemistry
Chapter: Thermodynamics (C)

BITSAT 2024 · Chemistry · Thermodynamics (C)

The bond dissociation energy of X_2, Y_2 and X Y are in the ratio...

The bond dissociation energy of

X_2, Y_2

and

X Y

are in the ratio of

1: 0.5: 1. \Delta H

for the formation of

X Y

-200 \mathrm{ kJ} / \mathrm{mol}

. The bond dissociation energy of

X_2

will be

Options

A $200 \mathrm{ kJ} / \mathrm{mol}$
B $100 \mathrm{ kJ} / \mathrm{mol}$
C $400 \mathrm{ kJ} / \mathrm{mol}$
D $800 \mathrm{ kJ} / \mathrm{mol}$

Correct answer

D. $800 \mathrm{ kJ} / \mathrm{mol}$

Step-by-step solution

Let the bond dissociation energy of

X_2=a \mathrm{ kJ} / \mathrm{mol}\text{ i.e. } \mathrm{BE}\left(X_2\right)=a \mathrm{ kJ} / \mathrm{mol}

Then,

\mathrm{BE}\left(\mathrm{Y}_2\right)=0.5 a

and

\mathrm{BE}(X Y)=a \mathrm{ kJ} / \mathrm{mol}

Given,

\frac{1}{2} X_2+\frac{1}{2} Y_2 \longrightarrow X Y, \Delta H=-200 \mathrm{ kJ} / \mathrm{mol}

\begin{aligned} \Delta_r H & =\mathrm{BE}(\text{ Rreactants })-\mathrm{BE}(\text{ Product }) \\ \Delta_r H & =\left[\frac{1}{2} \mathrm{BE}\left(X_2\right)+\frac{1}{2} \mathrm{BE}\left(Y_2\right)\right]-\mathrm{BE}(X Y) \\ \therefore-200 & =\frac{a}{2}+\frac{0.5 a}{2}-a \\ a & =\frac{200}{0.25}=800 \mathrm{ kJ} / \mathrm{mol} \end{aligned}

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