The ΔG°’ value for the malate DH reaction is +7.10 kcal/mol, and the ΔG°’ value for the citrate synthase reaction is -7.53 kcal/mol. Determine Keq for the combined, coupled reaction (e.g., L-malate <--> citrate, plus ancillary cofactors, reactants, and side-products), and compare this value to Keq for the malate DH reaction alone. By what factor do you have to multiply the Keq of the malate DH reaction to equal the Keq for the coupled reaction. Assume a temperature of 37 °C and report your answer to the nearest ones.
The reaction is: L-malate <------> citrate
With the given ΔG, let's calculate ΔG of the rxn:
ΔGrxn = -7.53 - 7.10 = -14.63 kJ/mol or -14630 J/mol
Now to get K:
Keq = exp(-ΔG/RT)
Keq= exp(14630 / 8.3144 * 37+273)
Keq = 281.82
The Keq for the malate only:
Keq = exp(-7100 / 8.3144 * 310)
Keq = 0.0636
Factor = 281.82 / 0.0636 = 4429
Hope this helps
The ΔG°’ value for the malate DH reaction is +7.10 kcal/mol, and the ΔG°’ value for...
Imagine that for the unimolecular reaction A ⟺
B, the value of ΔG° is -1.17
kcal/mol. If equimolar initial amounts of A and B react at 25 °C,
what fraction of the molecules will be B molecules once the system
has come to equilibrium?
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