Ask Question
1 November, 12:56

24. The standard free energy of the aldolase reaction (in the reverse aldol direction) is 23.8 kJ/mol. The concentrations of the three metabolites in the hepatocyte are: fructose 1,6-bisphosphate (1.4 X 10-5 M), glyceraldehyde 3-phosphate (3 X 10-6 M) and dihydroxyacetone phosphate (1.6 x 10-5 M). Calculate the actual free energy of the reaction at 37 degrees C.

+2
Answers (1)
  1. 1 November, 13:40
    0
    The actual free energy = 78.54 X 10⁻² J

    Explanation:

    Given standard free energy = 23.8 kJ/mol

    Free energy due to the presence of fructose 1,6-bisphosphate (1.4 X 10⁻⁵ M)

    = 1.4 X 10⁻⁵ M * (23800 J/mol)

    = 33.32 X 10⁻² J

    Free energy due to the presence of glyceraldehyde 3-phosphate (3 X 10⁻⁶ M)

    = 3 X 10⁻⁶ M * (23800 J/mol)

    =7.14 X 10⁻² J

    Free energy due to the presence of dihydroxyacetone phosphate (1.6 x 10⁻⁵M)

    = 1.6 x 10⁻⁵M * (23800 J/mol)

    = 38.08 X 10⁻² J

    The actual free = 33.32 X 10⁻² J + 7.14 X 10⁻² J + 38.08 X 10⁻² J

    The actual free energy = 78.54 X 10⁻² J
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “24. The standard free energy of the aldolase reaction (in the reverse aldol direction) is 23.8 kJ/mol. The concentrations of the three ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers