Q65 / 199 · NEET 2008 (Mains) · Chemistry
The value of equilibrium constant of the reaction
$$
\mathrm{HI}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{I}_2 \text { is } 8.0
$$
The equilibrium constant of the reaction $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$ will be
- A$\frac{1}{8}$
- B$\frac{1}{16}$
- Correct answerC$\frac{1}{64}$
- D16
Solution
$$
\mathrm{HI} \rightleftharpoons \frac{1}{2} \mathrm{H}_2+\frac{1}{2} \mathrm{I}_2, \mathrm{~K}_1=8.0
$$
or $2 \mathrm{HI} \rightleftharpoons \mathrm{H}_2+\mathrm{I}_2, \mathrm{~K}_2=64$
$$
\text { or } \mathrm{H}_2+\mathrm{I}_2 \rightleftharpoons 2 \mathrm{HI}, \mathrm{K}_3=\frac{1}{64}
$$
Hence, the correct answer is option: 'C'.