Q73 / 225 · JEE Main 2005 · Chemistry

If the bond dissociation energies of $X Y, X_2$ and $Y_2$ (all diatomic molecules) are in the ratio of 1:1:0.5 and $\Delta_t \mathrm{H}$ for the formation of $X Y$ is $-200 \mathrm{~kJ} \mathrm{~mole}^{-1}$. The bond dissociation energy of $\mathrm{X}_2$ will be

  • A
    $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • B
    $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • C
    $300 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • Correct answer
    D
    None of these

Solution

$$ \begin{aligned} & X Y \longrightarrow X_{(g)}+Y_{(g)} ; \quad \Delta \mathrm{H}=+\mathrm{a} \mathrm{kJ} / \mathrm{mole} \ldots \ldots \ldots \ldots . .(\mathrm{i}) \\ & \mathrm{X}_2 \longrightarrow 2 \mathrm{X} ; \quad \Delta \mathrm{H}=+\mathrm{a} \mathrm{kJ} / \mathrm{mole} \ldots \ldots \ldots \ldots \ldots \text { (ii) } \\ & \mathrm{Y}_2 \longrightarrow 2 \mathrm{Y} ; \quad \Delta \mathrm{H}=+0.5 \mathrm{a} \mathrm{kJ} / \mathrm{mole} \ldots \ldots \ldots \ldots . \text { (iii) } \\ & \frac{1}{2} \times\left. \text { (ii) }+\frac{1}{2} \times \text { (iii) }-\right.\text { (i), Gives } \\ & \frac{1}{2} \mathrm{X}_2+\frac{1}{2} \mathrm{Y}_2 \longrightarrow \mathrm{XY} ; \quad \Delta \mathrm{H}=\left(+\frac{\mathrm{a}}{2}+\frac{0.5}{2} \mathrm{a}-\mathrm{a}\right) \mathrm{kJ} / \mathrm{mole} \\ & +\frac{\mathrm{a}}{2}+\frac{0.5 \mathrm{a}}{2}-\mathrm{a}=-200 \\ & \mathrm{a}=800 . \end{aligned} $$

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

Relevant Questions