Chemical EquilibriumHard

Question

I2 + I−$\rightleftharpoons$I3−. This reaction is set up in aqueous medium. We start with 1 mol of I2 and 0.5 mol of I¯ in 1 L flask. After equilibrium is reached, excess of AgNO3 gave 0.25 mol of yellow precipitate. The equilibrium constant is

Options

A.1.33
B.2.66
C.0.375
D.0.75

Solution

At equilibrium, mole of I– = mole of AgI (yellow) precipitate = 0.25

I2 + I− $\rightleftharpoons$ I3−

1 mole 0.5 mole 0

Equilibrium 1 – x 0.5 – x x

= 0.75 = 0.25 = 0.25

∴ x = 0.25

$\therefore K = \frac{\left( \frac{0.25}{1} \right)}{\left( \frac{0.75}{1} \right) \times \left( \frac{0.25}{1} \right)} = 1.33$

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