IIT JEE , AIEEE Forum - Articles Category , Community shelf - for beginners.........effects of dilution on an equilibrium......

nivedh_89

In the section that introduced the LeChâtelier principle(in most text bks....

), it is mentioned that diluting a weak acid such as acetic acid CH₃COOH ("HAc") will shift the dissociation equilibrium to the right:

HAc + H₂O

H₃O⁺ + Ac^-

Thus a 0.10M solution of acetic acid is 1.3% ionized, while in a 0.01M solution, 4.3% of the HAc molecules will be dissociated. This is because as the solution becomes more dilute, the product [H₃O⁺][Ac^-] decreases more rapidly than does the [HAc] term. At the same time the concentration of H₂O becomes greater, but because it is so large to start with (about 55.5M), any effect this might have is negligible, which is why no [H₂O] term appears in the equilibrium expression.

For a reaction such as CH₃COOH(l) + C₂H₅OH(l)

CH₃COOC₂H₅(l) + H₂O(l) (in which the water concentration does change), dilution will have no effect on the equilibrium; the situation is analogous to the way the pressure dependence of a gas-phase reaction depends on the number of moles of gaseous components on either side of the equation.

Problem Example 5

The biochemical formation of a disaccharide (double) sugar from two monosaccharides is exemplified by the reaction

fructose + glucose-6-phosphate

sucrose-6-phosphate

(Sucrose is ordinary table sugar.) To what volume should a solution containing 0.050 mol of each monosaccharide be diluted in order to bring about 5% conversion to sucrose phosphate?

Solution:

The initial and final numbers of moles are as follows:

	fructose	glucose-6-P	sucrose-6-P
initial moles:	0.05	0.05	0
equilibrium moles:	0.0485	0.0485	0.0015

Substituting into the expression for K_c in which the solution volume is the unknown, we have

Solving for V gives a final solution volume of 78 mL.

Community Contributions - Articles by goIITians