Question

When dissolved in water, glucose (corn syrup) and fructose (fruit sugar) exist in equilibrium as follow: fructose ⇌ glucose A chemist prepared a 0.244 M solution of fructose at 25°C. At equilibrium it was found that its concentration had decreased to 0.163 M. Calculate the equilibrium constant for the reaction.

Answer 1

Answer:

Kc = 0.497M

Explanation:

Given,

Initial concentration of fructose = 0.244M at 25°C and at equilibrium it's concentration = 0.163M

Fructose ---> glucose

<---

Equilibrium constant, Kc = ?

Let's tabulate the changes in equilibrium concentration.

Fructose. Glucose.

Initial. 0.244M. 0

Change. -X. +X

Equilibrium. 0.244-X. X

We can find X from;

[Fructose] = 0.163M at equilibrium

0.244 - X. = 0.163

X = 0.081

Therefore,

Equilibrium constant, Kc

Kc = [glucose]/[fructose]

= X/(0.244-X)

Where, our X = 0.081

= 0.081/ (0.244-0.081)

Kc = 0.497M