Isomerization of a Phosphorylated Glucose Molecule
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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.
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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).
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 ...
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