1. An enzymatic hydrolysis of fructose-1-P,
Fructose-1-P H2O 34 fructose Pi was allowed to proceed to equilibrium at 25°C. The original concentration of fructose-1-P was 0.2 M, but when the system had reached equilibrium the concentration of fructose-1-P was only 6.52 10 5 M. Calculate the equilibrium constant for this reaction and the free energy of hydrolysis of fructose-1-P.
2. The equilibrium constant for some process Azy B is 0.5 at 20°C and 10 at 30°C. Assuming that H° is independent of temperature, calculate H° for this reaction. Determine G° and S°
at 20° and at 30°C. Why is it important in this problem to assume that H° is independent of temperature?
3. The standard-state free energy of hydrolysis for acetyl phosphate is G° 42.3 kJ/mol.
Acetyl-P H2O 88n acetate Pi Calculate the free energy change for acetyl phosphate hydrolysis
in a solution of 2 mM acetate, 2 mM phosphate, and 3 nM acetyl
phosphate.
Fructose-1-P H2O 34 fructose Pi was allowed to proceed to equilibrium at 25°C. The original concentration of fructose-1-P was 0.2 M, but when the system had reached equilibrium the concentration of fructose-1-P was only 6.52 10 5 M. Calculate the equilibrium constant for this reaction and the free energy of hydrolysis of fructose-1-P.
2. The equilibrium constant for some process Azy B is 0.5 at 20°C and 10 at 30°C. Assuming that H° is independent of temperature, calculate H° for this reaction. Determine G° and S°
at 20° and at 30°C. Why is it important in this problem to assume that H° is independent of temperature?
3. The standard-state free energy of hydrolysis for acetyl phosphate is G° 42.3 kJ/mol.
Acetyl-P H2O 88n acetate Pi Calculate the free energy change for acetyl phosphate hydrolysis
in a solution of 2 mM acetate, 2 mM phosphate, and 3 nM acetyl
phosphate.
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Here is what I calculated from your first question:
What you are concerned about only are the molar concentrations of reactants and products after the reaction is completed or after time approaches infinity so to speak. Of course in this case where you have an enzymatic reaction going on, this type of reaction is usually rapid and catalytic meaning usually zero order kinetics. Anyway, after the reaction has been completed, you are left with 6.52x10^-5 M fructose-1-P (reactant) and the product, fructose is at (0.2 - 6.52x10-5) M. The amount of water is in excess and be held as a constant as the product and reactant value meaning this value for water does not enter into the equilibrium ratio equation. So you can write the equilibrium equation as follows which are the values after the reaction has completed:
[fructose]/[fructose-1-P] = K = (0.2 - 6.52x10^-5)/6.52x10^-5) = 0.1999348 / 6.52x10^-5) = 3066.48
The free energy can be calculated from thermodynamic arguments as follows:
What you are concerned about only are the molar concentrations of reactants and products after the reaction is completed or after time approaches infinity so to speak. Of course in this case where you have an enzymatic reaction going on, this type of reaction is usually rapid and catalytic meaning usually zero order kinetics. Anyway, after the reaction has been completed, you are left with 6.52x10^-5 M fructose-1-P (reactant) and the product, fructose is at (0.2 - 6.52x10-5) M. The amount of water is in excess and be held as a constant as the product and reactant value meaning this value for water does not enter into the equilibrium ratio equation. So you can write the equilibrium equation as follows which are the values after the reaction has completed:
[fructose]/[fructose-1-P] = K = (0.2 - 6.52x10^-5)/6.52x10^-5) = 0.1999348 / 6.52x10^-5) = 3066.48
The free energy can be calculated from thermodynamic arguments as follows:
keywords: heat,transfer,on,thermodynamic,question,A question on thermodynamic heat transfer