Question
One of the key steps in glycolytic pathway in mitochondria in cells is the conversion (isomerization) of glucose-6-phosphate to fructose-6-phosphate Glucose-6-phosphate Fructose-6-phosphate At 298 K (25 degree C) the change in Gibbs Free Energy for the reaction is delta G degree = 1.7 kJ/mol Calculate the equilibrium constant, K_eq for this reaction Which of the compounds in the reaction will be present in a higher concentration? The next step in glycolysis is the conversion of fructose-6-phosphate to 1.6-fructose-bisphosphate. By itself, this is not a spontaneous reaction, but in the cell it occurs with a change in free energy of delta G degree = -14.2 kJ/mol. What is the value of equilibrium constant for this 2^nd reaction? What chemical reaction is responsible for driving this non-spontaneous reaction?
Explanation / Answer
G6P= Glucose 6 phosphate F6P= Fructose 6 phosphate and 1,6F = 1,6 fructose
For the reaction G6P-<-->F6P
deltaGO= -1.7*1000 j/mol
deltaGo=-RT lnK= -1.7*1000
where R= 8.314 J/mole.K and K= equilibrium constant
T=298.15K
lnK= 1.7*1000/(8.314*298.15)
K= 1.98 = [F6P]/ [G6P]
Since K>1, G6P concentration is less than f6P at equilibrium
2
For the reaction F6P<-----> 1,6F
deltaGo = -14.2*1000 j/mole
lnK= 14.2*1000/(8.314*298.15)
K=307. 52
K= [1,6F]/ [F6P] >1
Hence 1,6F is more at equilibrium than F6P