For calculating the ΔG’, we can simply use the equation [ΔG°' = ΔG' + RT ln K] where K is the equilibrium constant.
so,
ΔG°' = ΔG' + RT ln K
=> ΔG' = ΔG°' - RT ln K
=> ΔG' = 410 - (1.987*298)[ln(83/14)] = 410-1053.86 =
-643.86 cal/mol (Answer)
In a cell, the concentration of glucose-6-phosphate is 83 μM and of fructose-6-phosphate is 14 μM....
The second step of glycolysis is the conversion of glucose-6-phosphate to fructose-6-phosphate. What is the ΔG for this reaction at 25oC, when the glucose-6-phosphate] is 60 mM and the [fructose-6-phosphate] is 40 mM. Note: the ΔGo’ for this reaction is +1.7 kJ/mol (at 25oC). 0.780 kJ/mol 1.26 kJ/mol 1.47 kJ/mol 2.70 kJ/mol 0.695 kJ/mol 2.62 kJ/mol Answer not listed
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Consider the given interconversion, which occurs in glycolysis. fructose 6-phosphate glucose 6-phosphate K1.97 What is AG for the reaction (K measured at 25 °C)? AG -1680.9 kJ/mol If the concentration of fructose 6-phosphate is adjusted to 1.4 M and that of glucose 6-phosphate is adjusted to 0.75 M, what is AG? AG =-4402.59 kJ/mol Which statements are consistent with the conditions at which AG is measured? The temperature is 273 K. The initial concentrations of reactant and product are 1...
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