In solution, the sugar -glucose undergoes a process called intramolecular rearra

Question

In solution, the sugar -glucose undergoes a process called intramolecular rearrangement to produce -glucose. The molecular formula for both is C6H12O6 (molecular weight 180.2 g/mol). This equilibrium reaction can be simply written as -glucose -glucose In the diagram below, 2 molecules of -glucose and 4 molecules of -glucose are contained in an aqueous solution. Each glucose molecule is represented by a single a symbol ( or ). 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.

Explanation / Answer

Let's write again the overall reaction:

2-C6H12O6 <--------> 4-C6H12O6

This can be re.written as:

-C6H12O6 <--------> 2-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 = [-C6H12O6]4 / [-C6H12O6]2

Kc = 44 / 22

Kc = 64

Hope this helps

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