Answer to Question #105977 in Chemistry for Meen

Question

There are two forms of glucose, designated alpha (a) and beta (b), which are in equilibrium in aqueous solution. The equilibrium constant for the reaction is 1.5 at 30°C.

a-D-glucose(aq) <=> b-D-glucose(aq) K=1.5

(a) If you begin with a fresh 1.0 M solution of a-D-glucose in water, what will be its concentration when equilibrium is reached?

(b) Calculate the percentage of a-glucose and of b-glucose present at equilibrium in aqueous solution at 30°C.

Answer 1

If the initial concentration of a-gluocse is 1.0 M, in equilibrium solution, the concneration of b-glucose is x while the concentration of a-gluocse is (1-x). As a result, the reaction equilibrium can be described as following:

K = x / (1-x)

where K - the equilibrium constant.

From here:

x = K/ (1+K) = 1.5 / (1 + 1.5) = 0.6 M

1 - x = 0.4 M

(a) Concentration of a-glucose in equilibrium solution is 0.4 M

(b) As 1 M is 100%, the solution consists of 0.4 M × 100% = 40% of a-glucose and 0.6 M × 100% = 60% of b-glucose.

Answer: (a) 0.4 M; (b) 40% and 60%.