Q8 / 100 · BITSAT 2024 (Memory Based Paper 3) · Chemistry

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 $X Y$ is $-200 \mathrm{~kJ} / \mathrm{mol}$. The bond dissociation energy of $X_2$ will be

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

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

Hence, the correct answer is option: 'D'.

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