Expression which acquire constant value at equilibrium
- The measurable properties of the system become constant at equilibrium. This is so because the concentration of the substances become constant. For example, in case of evaporation of water, vapour pressure of water becomes constant.
- Equilibrium can be established only in case of closed system. The system should neither gain matter from the surroundings nor lose matter to the surroundings. For example, evaporation of water occurs only in closed containers and not in open vessels.
- The equilibrium is always dynamic in nature. This means that the process does not stop but the changes take place in the forward and the backward directions with the same rate.
Rate of dissolution = Rate of precipitation
Fig: 7.2 - Dynamic nature of solid in liquid equilibrium
- The measurable/visible properties of the system must remain unchanged with time. Once, an equilibrium is reached the properties of the system do not change with time.
- When equilibrium is attained, there exists an expression of concentrations of the substances involved in equilibrium, which becomes constant at a given temperature. For example, for equilibrium:

- The magnitude of the constant value of the concentration related expression (called equilibrium constant) gives an indication of the extent to which the reaction proceeds before acquiring equilibrium. For example, for the dissolution of CO2 in water:

Problem
1. The solubility of iodine in water is 1.1 x 10-3 mol L-1 at 288 K. 0.2 g of iodine is stirred in 100 cm3 of water till equilibrium is reached, what will be the mass of iodine found in solution and the mass that is left undissolved. After equilibrium is reached with 0.2g of iodine and 100 cm3 of water, we add 150 cm3 of water to the system. How much iodine will be dissolved and how much will be left undissolved and what will be the concentration of iodine in solution?
Solution
Solubility of iodine = 1.1 x 10-3 mol L-1
This means that 1.1 x 10-3 mol of iodine is present in 1 L of water.Molecular mass of iodine = 254
Amount of iodine dissolved per litre of water=1.1 x 10-3 x 254= 0.279 g
= 0.0279 g
Amount of iodine added = 0.2 gAmount of undissolved iodine in 100 cm3of water = 0.2 - 0.0279
= 0.172 gWhen 150 cm3 of water is added after the equilibrium, the total volume becomes 100 + 150 = 250 cm3
Amount of undissolved iodine = 0.2 - 0.07 = 0.130 g
Molar concentration = 0.07 g x1000 mL =0.0011 mol L-1254 g mol-1x 250 mL


