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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?

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palipali964

Answer:

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

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