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

$I_2\left(aq\right)+I^{-}\left(aq\right)\rightleftharpoons I_3^{-}\left(aq\right)$
Initial $I_2=1$ mol and $I^{-}=0.5$ mol
$K_c=\frac{\left[I_3^{-}\right]}{\left[I_2\right]\left[I^{-}\right]}$
$I^{-}+AgN{O}_3\to \underset{yellowprecipitate}{AgI}\downarrow +N{O}_3^{-}$

0.25 moles of yellow precipitate corresponds to 0.25 moles of iodide ions at equilibrium.
0.5 - 0.25 = 0.25 moles of iodide ions have reacted with 0.25 moles of iodine to form 0.25 of iodate ions.
1 - 0.25 = 0.75 moles of iodine remains at equilibrium.
Since total volume is $1L$, the number of moles is equal to molar concentration.
$K_c=\frac{\left[I_3^{-}\right]}{\left[I_2\right]\left[I^{-}\right]}=\frac{0.25}{0.75\times 0.25}=1.33$