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

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

Create a free account to view solution

View Solution Free
Topic: Chemical Equilibrium·Practice all Chemical Equilibrium questions

More Chemical Equilibrium Questions

For the reaction : PCl5 (g) ⇋ PCl3 (g) + Cl2 (g)The forward reaction at constant temperature is favoured by...The equilibrium constants KP1 and KP2 for the reactions X ⇋ 2Y and Z ⇋ P + Q, respectively are in the ratio ...The equilibrium N2 (g) + O2 (g) ⇋ 2NO (g) is estabilished in a reaction vessel of 2.5 L capacity. The amounts of N...For the equilibrium A(g) $\rightleftharpoons$nB(g), the equilibrium constant KP is related with the degree of dissociati...Observe the following equilibrium in a 1 L flask.$$A(g) \rightleftharpoons B(g)$$At $T(K)$, the equilibrium concentratio...