Gibb\'s free energy for the equilibria of 3 phosphohexoses in water at 25-degree
ID: 174618 • Letter: G
Question
Gibb's free energy for the equilibria of 3 phosphohexoses in water at 25-degree celsius are:
calculate the equilibrium concentration of the 3 phosphates(Glucose-1-phosphate, Glucose-6-phosphate and Fructose-6-phosphate) that were produced after an initial 0.1M solution containing only glucose-6-phosphate was brought into contact with a preparation containing 2 enzymes that catalyze these phosphate interconversions. You may assume that there was no change in volume so that the total phosphate concentration remains constant.
Explanation / Answer
Calculation of equlibrium constant from gibbs relation is
1] for GLucose 1 - phosphate to Glucose 6 phosphate
Given G=-7.197 KJ / mol =7197 J /mol
R =- 8.314 J /mol K
G =-RT ln K
-7197 = (-8.314J/mol K ) (298) ln K
=- 2477.7 J /mol K ln K
ln K =-7197/ -2477.7
ln k =2.90
K =e^ (2.90)
=18.25
Calculations of equilibrium concentration o.1 M solution of glu 1 phosphate & 0.1 M glucose 6 phosphate
K = [Glu 6 P] /[Glu 1 P] = (1+ x) /(1- x)
18.25 (1-x) =( 1+x )
18.25 -18.25x =1+x
18.25-1 =18.25 x + x
17.25 = 19.25 x
x = 17.25/19.25
x = 0.89 Molar
conc of glucose 1 p =1- x = 1- 0.89 = 0.11 molar
[ Glu 6 phosphate ] =1+ x = 1+ 0.89 = 1.89 molar
2] glucose 6 - Fru 6 phosphate
delta r G = +2.1 KJ/mol =2100 j/ mole
G = - RT ln K
+2100 J / mole =( -8.314) 298 ln K
= -2477.7 ln K
ln K = 2100 / - 2477.7
=- 0.847
K = e ^ (-o.847)
=1/ e ^ 0.847
=1/ 2.33
K = 0.429
Calculation of equlibrium concentration
K = [Fru 6 p ] / [ Glu 6 p] = (1+x) / ( 1- x)
0.429 (1- x ) = (1+x )
0.4290- 0.429x = 1+ x
0.4290 - 1 = x +0.429
- 0.571 = 1.429 x
-0.571 /1.429 = x
x = - 0.39
[ fructose 6 - phosphate ] =1+ x
= 1+ (-0.39) = 0.61 molar
[ fru.6 p ] =0.61 molar
[ glu 6 p ] = 1- x
= 1- ( -0.39 )
=1.39 molar
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