Question

In solution, the sugar α-glucose undergoes a process called intramolecular rearrangement to produce β-glucose. The molecular formula for both is C6H12O6 (molecular weight 180.2 g/mol). This equilibrium reaction can be simply written as α-glucose ⇌ β-glucose In the diagram below, 2 molecules of α-glucose and 4 molecules of β-glucose are contained in an aqueous solution. Each glucose molecule is represented by a single a symbol (α or β). The net volume of the solution is 10.0 mL. Assume that the solution is at equilibrium. Write an equilibrium constant expression, and calculate a numerical value for the equilibrium constant.

Answer 1

Let's write again the overall reaction:

2α-C6H12O6 <--------> 4β-C6H12O6

This can be re.written as:

α-C6H12O6 <--------> 2β-C6H12O6

Now, in order to do this, I need either the innitial moles of glucose or the innitial mass to do this. I don't have these values to know which quantity should I use, but let's assume that the solution is in equilibrium, and in equilibrium we have those molecules.

Kc = [β-C6H12O6]⁴ / [α-C6H12O6]²

Kc = 4⁴ / 2²

Kc = 64

Hope this helps