Question

For the reaction : $X_2+X^{-}\rightleftharpoons X_3^{-}(X=iodine)$. This reaction is set up in aqueous medium. We start with 1 mol of $X_2$ and 0.5 mol of $X^{-}$ in 1L flask. After equilibrium is reached, excess of $AgN{O}_3$ gave 0.25 mol of yellow ppt. Equilibrium constant is :

• A. 1.33
• B. 2.66
• C. 2
• D. 3

Answer 1

The correct option is A 1.33

The equilibrium reaction is $X_2+X^{-}\rightleftharpoons X_3^{-}.$

1 0.5 0 ------ initial

1-x 0.5-x x ------- final

Thus the unreacted moles of $X^{-}$ are 0.5-x. They corresponds to 0.25 moles of yellow precipitate formed with silver nitrite solution.

Hence, $x=\mathrm{0.25.}$

Th expression for the equilibrium constant is $K_C=\frac{\left[X_3^{-}\right]}{\left[X_2\right]\left[X^{-}\right]}=\frac{0.25}{0.75\times 0.25}=\mathrm{1.33.}$