Explore BrainMass
Share

Explore BrainMass

    Isomerization of a Phosphorylated Glucose Molecule

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    The isomerization of a phosphorylated glucose molecule from 1-p to 6-p form with standard free energy change=-1.74kcal/mole. Find the equilibrium concentration ratio of glucose-p in the two isometric states.

    © BrainMass Inc. brainmass.com October 9, 2019, 4:15 pm ad1c9bdddf
    https://brainmass.com/biology/cell-biochem/isomerization-phosphorylated-glucose-molecule-27177

    Solution Preview

    The relationship between the free energy change (DG) and the equilibrium concentration ratio (K) is:

    DG = -RT ln K

    T = absolute temperature. 298 K is standard but 310 K is ...

    Solution Summary

    Answer is found using the formula DG = -RT ln K, representing the relationship between the free energy change (DG) and the equilibrium concentration ratio (K).

    $2.19