Glucose-1-phosphate can be converted to fructose-6-phosphate by coupling the two

Question

Glucose-1-phosphate can be converted to fructose-6-phosphate by coupling the two reactions below Glucose-1-phosphate rightarrow Glucose-6-phosphate delta G degree = -7.3 kJ/mol Fructose-6-phosphate rightarrow Glucose-6-phosphate delta G degree = -1.6 kJ/mol Determine the delta G degree for the conversion of glucose-1-phosphate to fructose-6-phosphate Determine the equilibrium constant K_eq, for the conversion of glucose-1-phosphate to fructose-6-phosphate at 25 degree C. If the concentration of glucose-1-phosphate is 0.5 M and fructose-6-phospahte is 1.5 M, then what is the value of delta G degree at 25 degree C? For a reaction A reversible B + C the equilibrium constant is determine to be 1 times 10^4. Determine the for the reaction in KJ/mol.

Explanation / Answer

1)

consider the reactions

a) glucose-1-phosphate --> glucose-6-phosphate dGo a = -7.3

b) fructose-6-phosphate --> glucose-6-phosphate dGo b = -1.6

c) glucose-1-phosphate --> fructose-6-phosphate dGo c = ??

we can see that

c = a - b

so

dGo c = dGo a - dGo b

dGo c = -7.3 + 1.6

dGo c = -5.7

so

dGo for the conversion of glucose-1-phosphate to fructose-6-phosphate is -5.7 kJ /mol

2)

we know that

dGo = -RTlnKeq

so

-5.7 x 1000 = -8.314 x 298 x lnKeq

Keq = 9.98

so

the value of equilibrium constant is 9.98

3)

now

we know that

dG = dGo + RT lnQ

the reaction quotient is given by

Q = [fructose-6-phosphate] / [glucose-1-phosphate]

Q = 1.5 / 0.5

Q = 3

so

dG = ( -5.7 x 1000) + ( 8.314 x 298 x ln 3)

dG = -2.978 x 1000

dG = -2.978 kJ/mol

4)

A ---> B

given

Keq = 1 x 10^4

now

dGo = -RTlnKeq

dGo = -8.314 x 298 x ln 1 x 10^4

dGo = -22.82 x 1000

dGo = -22.82 kJ/mol

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