For the reaction glutamate- (aq) + oxaloacetate2- (aq) double arrow -ketoglutara

Question

For the reaction glutamate- (aq) + oxaloacetate2- (aq) double arrow -ketoglutarate2- (aq) + aspartate- (aq) the equilibrium constant at standard temperature is Keq = 2.82. Suppose the concentrations of glutamate and oxaloacetate were 10 mM and the concentration of -ketoglutarate was 25 mM when the system was at equilibrium. An additional 25 mM of -ketoglutarate is added to the solution. After the -ketoglutarate is added, when the system has settled down and no net reaction is occurring, what is the value of G' for this reaction?

Explanation / Answer

Glutamate + oxaloacetate àà aspartate + -ketoglutarate          K’eq = 2.82

K’eq =[ aspartate][-ketoglutarate]/[Glutamate][ oxaloacetate]

2.82 = (10 mM/L * 10 mM/L)/(25 mM/L * [oxaloacetate])______ at standard condition

[oxaloacetate] = 2.82*25/100

[oxaloacetate] = 0.705 mM/L

Now 25 mM of -ketoglutarate was added so new equilibrium constant will be

K’eq = (10 mM/L * 10 mM/L)/(50 mM/L * 0.705 mM/L)

K’eq = 2.836

DeltaGo’= -RT ln K’eq

R=8.315 J/moloK

T=298oK

DeltaGo =8.315 * 298 * ln 2.836

DeltaGo = - 2583.68 J/Mol

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