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Phosphoglucose isomerase was incubated with 0.2 M of glucose 6-phosphate. On reaching
equilibrium, 55% of glucose 6-phosphate was converted to fructose 6-phosphate. The equilibrium constant for this reaction is _______.
    Correct answer is between '1.20,1.24'. Can you explain this answer?
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    Phosphoglucose isomerase was incubated with 0.2 M of glucose 6-phospha...
    Equilibrium Constant Calculation for Phosphoglucose Isomerase Reaction

    Given:
    - Phosphoglucose isomerase was incubated with 0.2 M of glucose 6-phosphate.
    - At equilibrium, 55% of glucose 6-phosphate was converted to fructose 6-phosphate.

    To find:
    - The equilibrium constant for this reaction.

    Solution:

    1. Write the reaction equation:
    Glucose 6-phosphate ⇌ Fructose 6-phosphate

    2. Write the equilibrium constant expression:
    K_eq = [Fructose 6-phosphate] / [Glucose 6-phosphate]

    3. Substitute the given information:
    55% of glucose 6-phosphate was converted to fructose 6-phosphate.
    So, [Fructose 6-phosphate] = 0.55 x 0.2 M = 0.11 M
    And, [Glucose 6-phosphate] = (0.2 - 0.11) M = 0.09 M

    4. Calculate the equilibrium constant:
    K_eq = [Fructose 6-phosphate] / [Glucose 6-phosphate] = 0.11 M / 0.09 M = 1.222

    5. Round off the answer to two significant figures:
    K_eq = 1.22

    Therefore, the equilibrium constant for the reaction catalyzed by phosphoglucose isomerase is between 1.20 and 1.24.
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    Phosphoglucose isomerase was incubated with 0.2 M of glucose 6-phospha...
    Phosphoglucose isomerase was incubated with 0.2 M of glucose 6-phosphate. On reachingequilibrium, 55% of glucose 6-phosphate was converted to fructose 6-phosphate. The equilibriumconstant for this reaction is
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    Phosphoglucose isomerase was incubated with 0.2 M of glucose 6-phosphate. On reachingequilibrium, 55% of glucose 6-phosphate was converted to fructose 6-phosphate. The equilibrium constant for this reaction is _______.Correct answer is between '1.20,1.24'. Can you explain this answer?
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    Phosphoglucose isomerase was incubated with 0.2 M of glucose 6-phosphate. On reachingequilibrium, 55% of glucose 6-phosphate was converted to fructose 6-phosphate. The equilibrium constant for this reaction is _______.Correct answer is between '1.20,1.24'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Phosphoglucose isomerase was incubated with 0.2 M of glucose 6-phosphate. On reachingequilibrium, 55% of glucose 6-phosphate was converted to fructose 6-phosphate. The equilibrium constant for this reaction is _______.Correct answer is between '1.20,1.24'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Phosphoglucose isomerase was incubated with 0.2 M of glucose 6-phosphate. On reachingequilibrium, 55% of glucose 6-phosphate was converted to fructose 6-phosphate. The equilibrium constant for this reaction is _______.Correct answer is between '1.20,1.24'. Can you explain this answer?.
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