Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1638 Upvotes
Verified Answer
The correct answer is:
None of these
None of these
$$
\begin{aligned}
& X Y \longrightarrow X_{(g)}+Y_{(g)} ; \quad \Delta \mathrm{H}=+\mathrm{a} \mathrm{kJ} / \mathrm{mole} \ldots \ldots \ldots \ldots . .(\mathrm{i}) \\
& \mathrm{X}_2 \longrightarrow 2 \mathrm{X} ; \quad \Delta \mathrm{H}=+\mathrm{a} \mathrm{kJ} / \mathrm{mole} \ldots \ldots \ldots \ldots \ldots \text { (ii) } \\
& \mathrm{Y}_2 \longrightarrow 2 \mathrm{Y} ; \quad \Delta \mathrm{H}=+0.5 \mathrm{a} \mathrm{kJ} / \mathrm{mole} \ldots \ldots \ldots \ldots . \text { (iii) } \\
& \frac{1}{2} \times\left. \text { (ii) }+\frac{1}{2} \times \text { (iii) }-\right.\text { (i), Gives } \\
& \frac{1}{2} \mathrm{X}_2+\frac{1}{2} \mathrm{Y}_2 \longrightarrow \mathrm{XY} ; \quad \Delta \mathrm{H}=\left(+\frac{\mathrm{a}}{2}+\frac{0.5}{2} \mathrm{a}-\mathrm{a}\right) \mathrm{kJ} / \mathrm{mole} \\
& +\frac{\mathrm{a}}{2}+\frac{0.5 \mathrm{a}}{2}-\mathrm{a}=-200 \\
& \mathrm{a}=800 .
\end{aligned}
$$
\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}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.