2. The intracellular concentration of glucose-6-phosphate and fructose-6-phosphate are measured inside the cell of a patient and the following values were obtained: [glucose-6-phosphate] = 8 x 10-5 M [fructose-6-phosphate] = 1.5 x 10-5 M %3D %3D Is this reaction at equilibrium?. What is the value of Q?. What do you predict the sign for AG will be? Calculate AG for this reaction..

Chemistry by OpenStax (2015-05-04)
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Chapter16: Thermodynamics
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Problem 48E: At mom temperature, the equilibrium constant (Kw) for the self-ionization of water is 1.001014....
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2. The intracellular concentration of glucose-6-phosphate and fructose-6-phosphate
are measured inside the cell of a patient and the following values were obtained:
(glucose-6-phosphate] = 8 x 10-5 M
[fructose-6-phosphate] = 1.5 x 10-5 M
%3D
%3D
Is this reaction at equilibrium?
What is the value of Q?.
What do you predict the sign for AG will be?
Calculate AG for this reaction.
Transcribed Image Text:2. The intracellular concentration of glucose-6-phosphate and fructose-6-phosphate are measured inside the cell of a patient and the following values were obtained: (glucose-6-phosphate] = 8 x 10-5 M [fructose-6-phosphate] = 1.5 x 10-5 M %3D %3D Is this reaction at equilibrium? What is the value of Q?. What do you predict the sign for AG will be? Calculate AG for this reaction.
Expert Solution
Step 1

If the value of G is negative will represent the reaction conditions are favorable. If the value of G is positive will represent the reaction conditions are unfavorable. When the reaction is an equilibrium value of G is zero. Unit of G is kJ per mol.

 
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