For this problem set you may refer to the free energies of hydrolysis given in T

Question

For this problem set you may refer to the free energies of hydrolysis given in Tables 13-4 and 13-6. In addition

fructose-1,6-bisphosphate + H2O ? fructose-6-P + Pi ?Go' = -16.3 kJ/mol -3.9 kcal/mol

1. For the following values of the equilibrium constant of a reaction, calculate ?Go at the given temperature.

a) K = 6.8 at 25 C. (answer -4.8 kJ/mol) c) K = 4.0 x 10-11 at 37 C. b) K = 8 x 106 at 25 C. d) K = 1 at 0 C.

Typical concs. of glycolytic intermediates in erythrocytes:

glucose 5.0 mM

fructose-1,6-bis-P 31 ?M

1,3-bis-P-glycerate 1 ?M

2-P-glycerate 30 ?M

lactate 2.9 mM

glucose-6-P 83 ?M

glyceraldehyde-3-P 19 ?M

2,3-bis-P-glycerate 4 ?M

P-enolpyruvate 23 ?M

ATP 1.85 mM

fructose-6-P 14 ?M

dihydroxyacetone-P 138 ?M

3-P-glycerate 118 ?M

pyruvate 51 ?M

ADP 138 ?M

Pi 1.0 mM

Explanation / Answer

a) K = 6.8 T = 25oC = 298 K R = 8.314 J/mol.K

Delta G = -RTlnK

Delta G = -(8.314) (298) ln(6.8)

Delta G = -4749.313 J/mol

= -4.75 kJ

-------------------

b)

K = 4.0 x 10^-11 T = 37 oC = 310 K R = 8.314 J/mol.K

Delta G = -RTlnK

Delta G = -(8.314) (310) ln(4.0 x 10^-11)

Delta G = -61707.04 J/mol

= -61.71 kJ

--------------------------------------------

c)

K = 8.0 x 10^6 T = 25 oC = 298 K R = 8.314 J/mol.K

Delta G = -RTlnK

Delta G = -(8.314) (298) ln(8.0 x 10^6)

Delta G = -39380.888 J/mol

= -39.4 kJ

-----------------------------------------

d)

K = 1 T = 0 oC = 273 K R = 8.314 J/mol.K

Delta G = -RTlnK

Delta G = -(8.314) (273) ln(1)

Delta G = 0 J/mol

= 0 kJ

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