Question

Given ΔG˚' for each of the following reactions: ATP->ADP + Pi ΔG˚'= -30.5 kj/mol Glucose-6-phosphate-> glucose + Pi ΔG˚'= -13.8 kj/mol Show how you would calculate the standard free energy change (ΔG˚') for the overall reaction: ATP + glucose -> glucose-6-phosphate+ ADP Then, How do you calculate the equilibrium constant K'eq for the overall reaction that is above at 25˚C?

Answer 1

Solution :-

We need to use both equations and rearrange them so that by adding them we can get the final equation.

ATP->ADP + Pi ΔG˚'= -30.5 kj/mol                                                 eq 1

Glucose-6-phosphate ----> glucose + Pi ΔG˚'= -13.8 kj/mol           eq 2

ATP + glucose -> glucose-6-phosphate+ ADP deltaG= ?

We need to reverse the eq 2 so that we can get the glucose on the left side of the equation

Glucose + Pi ----- > Glucose-6-phosphate ΔG˚'= 13.8 kJ/mol eq 3

Now we have to add eq 1 and eq 3

ATP->ADP + Pi ΔG˚'= -30.5 kj/mol

Glucose + Pi ----- > Glucose-6-phosphate ΔG˚'= 13.8 kj/mol

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ATP + glucose ---> glucose-6-phosphate+ ADP Delta G=(-30.5 kJ/mol)+(13.8kJ/mol)= -16.7 kJ/mol

Therefore the ΔG˚' for the overall reaction is -16.7 kJ/mol

Now lets calculate the Keq for the overall reaction

ΔG˚'= - RT ln K

T= 25 C + 273 = 298 K

R= 8.314 J/mol K

-16.7 kJ/mol * 1000 J / 1 kJ = -16700 J/mol

Lets put the values in the formula

-16700 J/mol = - 8.314 J per mol K * 298 K * ln K

-16700 J/mol / (- 8.314 J per mol K * 298 K) = ln K

6.74 = ln K

Anti ln [6.74 ] = K

e^(6.74) = K

846 = K

Therefore equilibrium constant for the overall reaction is Keq = 846