Question

Calculate the equilibrium constant for the enzymatic hydrolysis of 0.02 M glucose-6-phosphate to glucose and inorganic phosphate given that 0.04 % of the original glucose-6-phosphate remained after reaching equilibrium and the activity of water is unity

Answer 1

The reaction under consideration is the conversion of Glucose-6-phosphate to Glucose and Inorganic phosphate. Hence, the chemical reaction can be represented as:

Thus, one mole of glucose-6-phosphate is converted to one mole of Glucose and one mole of inorganic phosphate.

The equilibrium constant is given as the ratio of the product of concentrations of product and the product of concentrations of the substrate.

The starting concentration of glucose-6-phosphate is 0.02 M. At equilibrium, 0.04% of this remains.

Hence, the concentration of Glucose-6-phosphate that has been used up is:

Now, one mole of glucose-6-phosphate gives one mole of glucose and one mole of Pi. Hence at equilibrium, the concentration of Glucose and Pi will be:

Thus, the equilibrium constant becomes: