Question

8. An enzymatic hydrolysis of fructose-1-P (shown below) proceeded to equilibrium at 37 °C and the equilibrium constant for the reaction was 3943. Fructose-1-P+H2O ? fructose +Pi When the system had reached equilibrium the concentration of fructose 1-P was 9.5 x 10^-5 M. Calculate the the standard free energy of hydrolysis of fructose 1-P and the initial concentration of fructose 1-P. Assume equal amounts of fructose and Pi were generated. [fructose 1-P] = 9.5 x 10^-5 M ?G = ? T=310K K=3943 K = [Products] / [Reactants] ?g = -RT ln K = -1.987 x 10^-3 kcal/mol-K (310) ln (3943) = -.61597 ln (3943) = -5.1 kcal/mol 9. Your brewery is filtering a rough beer through a hollow fiber ceramic filter. The velocity of the beer (v) is 2 m/sec. The density (?) and viscosity (?) of the beer is 1013 kg/m³ and 1.42 mPa sec, respectively. a. The inner diameter of the tubular membrane module is 6 mm. Calculate the Reynolds number. Note that 1 Pa sec = 1 kg/m sec

          8. An enzymatic hydrolysis of fructose-1-P (shown below) proceeded to equilibrium at 37 °C and the equilibrium constant for the reaction was 3943.
Fructose-1-P+H2O ? fructose +Pi
When the system had reached equilibrium the concentration of fructose 1-P was 9.5 x 10^-5 M. Calculate the the standard free energy of hydrolysis of fructose 1-P and the initial concentration of fructose 1-P.
Assume equal amounts of fructose and Pi were generated.
[fructose 1-P] = 9.5 x 10^-5 M
?G = ?
T=310K
K=3943
K = [Products] / [Reactants]
?g = -RT ln K
= -1.987 x 10^-3 kcal/mol-K (310) ln (3943)
= -.61597 ln (3943)
= -5.1 kcal/mol
9. Your brewery is filtering a rough beer through a hollow fiber ceramic filter. The velocity of the beer (v) is 2 m/sec. The density (?) and viscosity (?) of the beer is 1013 kg/m³ and 1.42 mPa sec, respectively.
a. The inner diameter of the tubular membrane module is 6 mm. Calculate the Reynolds number.
Note that 1 Pa sec = 1 kg/m sec

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8. An enzymatic hydrolysis of fructose-1-P (shown below) proceeded to equilibrium at 37 °C and the equilibrium constant for the reaction was 3943.
Fructose-1-P+H2O ? fructose +Pi
When the system had reached equilibrium the concentration of fructose 1-P was 9.5 x 10^-5 M. Calculate the the standard free energy of hydrolysis of fructose 1-P and the initial concentration of fructose 1-P.
Assume equal amounts of fructose and Pi were generated.
[fructose 1-P] = 9.5 x 10^-5 M
?G = ?
T=310K
K=3943
K = [Products] / [Reactants]
?g = -RT ln K
= -1.987 x 10^-3 kcal/mol-K (310) ln (3943)
= -.61597 ln (3943)
= -5.1 kcal/mol
9. Your brewery is filtering a rough beer through a hollow fiber ceramic filter. The velocity of the beer (v) is 2 m/sec. The density (?) and viscosity (?) of the beer is 1013 kg/m³ and 1.42 mPa sec, respectively.
a. The inner diameter of the tubular membrane module is 6 mm. Calculate the Reynolds number.
Note that 1 Pa sec = 1 kg/m sec

Added by Kathy B.

Biology for AP Courses

Julianne Zedalis, John Eggebrecht

Instant Answer

Solved by Expert Madhur L

09/14/2023

Step 1

98 x 10^-3 kcal/mol/K Plugging in the values: ΔG = - (1.98 x 10^-3) * 310 * ln(3943) ΔG = -5.081 kcal/mol Show more…

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8. An enzymatic hydrolysis of fructose-1-P shown below proceeded to equilibrium at 37°C, and the equilibrium constant for the reaction was 3943. Fructose-1-P + H2O ↔ fructose + Pi When the system had reached equilibrium, the concentration of fructose-1-P was 9.5 x 10^-5 M. Calculate the standard free energy of hydrolysis of fructose-1-P and the initial concentration of fructose-1-P. Assume equal amounts of fructose and Pi were generated. ΔG° = -RTlnK = -1.987x10^-3 kcal/mol·K (310 K) ln(3943) = -5.1 kcal/mol

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Transcript

00:01 Hello everyone, in this question we have to answer two parts.

00:03 For question number 8 we need to calculate the standard free energy of hydrolysis that is delta g.

00:10 So temperature given is 37 degree celsius.

00:14 We will convert this into kelvin.

00:17 Then k given is 3943 that is the equilibrium constant.

00:24 Then the gas constant which we'll use here is 1 .98 into 10 raised to the power minus 3 kilocal per mole per kelvin.

00:37 Now putting these values in the formula delta g equals to minus rt ln k.

00:44 So this is equal to minus 1 .98 into 10 raised to the power minus 3 into 310 into ln 3943.

00:56 So this is equal to minus 1 .98 into 10 raised to the power minus 3 into 310 multiplied by 8 .279.

01:11 So value of delta g will be minus 5 .081 kilocal per mole.

01:21 Now next we need to calculate the concentration of product.

01:26 So we'll do this with the help of reaction question which is equal to the concentration of product divided by the concentration of reactant.

01:43 So delta g is equal to minus rt ln q.

01:49 Ln q is equal to minus putting the values of delta g that is minus 5 .081.

02:00 This will be divided by rt.

02:03 So rt is 1 .98 into 10 raised to the power minus 3 multiplied by temperature which is 310.

02:14 So we'll get the value of ln q as 8 .333...

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