If $1.0$ mole of $\mathrm{I}_{2}$ is introduced into $1.0$ litre flask at $1000 \mathrm{~K}$, at quilibrium…

If $1.0$ mole of $\mathrm{I}_{2}$ is introduced into $1.0$ litre flask at $1000 \mathrm{~K}$, at quilibrium $\left(\mathrm{K}_{\mathrm{c}}=10^{-6}ight)$, which one is correct
  1. $\left[\mathrm{I}_{2}(\mathrm{~g})ight]>\left[\mathrm{I}^{-}(\mathrm{g})ight]$
  2. $\left[\mathrm{I}_{2}(\mathrm{~g})ight] < \left[\mathrm{I}^{-}(\mathrm{g})ight]$
  3. $\left[\mathrm{I}_{2}(\mathrm{~g})ight]=\left[\mathrm{I}^{-}(\mathrm{g})ight]$
  4. $\left[\mathrm{I}_{2}(\mathrm{~g})ight]=\frac{1}{2}\left[\mathrm{I}^{-}(\mathrm{g})ight]$

Solution

$\mathrm{I}_{2} ightleftharpoons 2 \mathrm{I}^{-}$
$1-x \quad 2 x$
$\mathrm{K}_{\mathrm{c}}=\frac{(2 \mathrm{x})^{2}}{(1-\mathrm{x})}=10^{-6}$
Soln. shows that $(1-x)>2 x$
$\therefore\left[\mathrm{I}_{2}(\mathrm{~g})ight]>\left[\mathrm{I}^{-}(\mathrm{g})ight]$ ^

Asked in: JEE-TOPICTESTS-CHEMISTRY

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