What is the equilibrium constant for the enzymatic hydrolysis of 0.1 M glucose-6-phosphate to glucose and inorganic phosphate given that 0.05% of the original glucose-6-phosphate remained after reaching equilibrium and the activity of water is unity?

1.999 or 2

Explanation:

The equilibrium constant is concentration of the products over reactants at equilibrium (see below the formula).

Formula: Keq = [Products]/[Reactants]

Keq = [ (100% - 0.05%) / 100*0.1 M) ] / [ (0.05%/100*0.1 M) ]

= 0.09995/0.00005

= 1.999 which is ~2