Question:

Phosphoglucoisomerase catalyzes the following reaction: \[ \text{Glu-6-P} \rightleftharpoons \text{Fru-6-P} \] If 0.05% of the original concentration of Glu-6-P remains at equilibrium, then the equilibrium constant of this reaction is ............

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The equilibrium constant is calculated by dividing the concentration of products by the concentration of reactants at equilibrium.
Updated On: Dec 11, 2025
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Correct Answer: 1999

Solution and Explanation

Step 1: Understanding equilibrium constant.
At equilibrium, the concentration of products and reactants is related by the equilibrium constant \( K_{\text{eq}} \). The reaction is: \[ \text{Glu-6-P} \rightleftharpoons \text{Fru-6-P} \] Let the initial concentration of Glu-6-P be \( [\text{Glu-6-P}]_0 \). At equilibrium, the concentration of Glu-6-P is \( 0.05 [\text{Glu-6-P}]_0 \), and the concentration of Fru-6-P is \( 0.95 [\text{Glu-6-P}]_0 \). The equilibrium constant \( K_{\text{eq}} \) is given by: \[ K_{\text{eq}} = \frac{[\text{Fru-6-P}]}{[\text{Glu-6-P}]} = \frac{0.95 [\text{Glu-6-P}]_0}{0.05 [\text{Glu-6-P}]_0} = 19 \] Step 2: Conclusion.
The equilibrium constant \( K_{\text{eq}} \) is 20.
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