Question

$I_2\left(aq\right)+I^{-}\left(aq\right)\rightleftharpoons I_3^{-}\left(aq\right).$ We started with $1$ mole of $I_2$ and $0.5$ mole of $I^{-}$ in one litre flask. After equilibrium is reached excess of $AgN{O}_3$ gave $0.25$ mole of yellow precipitate. Equilibrium constant is:

• A. 1.33
• B. 2.66
• C. 2
• D. 3

Answer 1

The correct option is A 1.33

The given reaction is :-

$I_2\left(aq\right)+I^{-}\left(aq\right)\rightleftharpoons {I_3}^{-1}\left(aq\right)$

Initial moles : $1$ $0.5$ $0$

At eqm : $(1-x)$ $(0.5-x)$ $x$

Now, at equation with excess of ${AgNO}_3$ $0.25$ moles of yellow ppt. is obtained.

$\Rightarrow 0.5-x=0.25\Rightarrow x=0.25$ ($\because$ $I^{-}$ reacts with ${AgNO}_3$ to give yellow ppt.)

Now, At eqm, $\left[I_2\right]=1-0.25=0.75$

$\left[I^{-}\right]=0.25$

$\left[{I_3}^{-}\right]=0.25$

$\therefore K_C=\frac{\left[{I_3}^{-}\right]}{\left[I_2\right]\left[I^{-}\right]}=\frac{0.25}{0.75\times 0.25}=\frac{4}{3}=1.33$