The value of equilibrium constant of the reaction $\mathrm{HI}(\mathrm{g}) \rightleftharpoons \frac{1}{2}…

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
  1. $\frac{1}{8}$
  2. $\frac{1}{16}$
  3. $\frac{1}{64}$
  4. 16

Solution

$\mathrm{HI} \rightleftharpoons \frac{1}{2} \mathrm{H}_2+\frac{1}{2} \mathrm{I}_2, \mathrm{~K}_1=8.0$ or $2 \mathrm{HI} \rightleftharpoons \mathrm{H}_2+\mathrm{I}_2, \mathrm{~K}_2=64$ $\text {or } \mathrm{H}_2+\mathrm{I}_2 \rightleftharpoons 2 \mathrm{HI}, \mathrm{K}_3=\frac{1}{64}$

Asked in: NEET 2008 (Mains)

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