Question

The value of the standard free energy (∆G°’) for the hydrolysis of ATP (ATP + H2O -> ADP + Pi)...

The value of the standard free energy (∆G°’) for the hydrolysis of ATP (ATP + H2O -> ADP + Pi) is relatively difficult to determine because of the small concentration of ATP remaining at equilibrium. The value can be determined from the equilibrium constant of two related chemical reactions.

Glucose-6-phosphate + H2O -> glucose + Pi ; k’eq = 270

ATP + glucose -> ADP + glucose-6-phosphate ; K’eq = 890

From this information, calculate the standard free energy of ATP hydrolysis ((∆G°’) at 298K.

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Answer #1

the relation between Gibbs free energy and equilibrium constant is

media%2Fb03%2Fb03ff2e7-5d63-4e01-b6d7-b2

So, for first equation \bigtriangleup G^{\circ}=-RTlnKeq , delta Go= -8.314 J/mol*298K*ln270 = 13.870KJ

Applying the same formula for second equation, we get, delta Go = 16.8KJ

Adding equation 1 and 2 we get,

glucose-6-phosphate+H2O+ATP+Glucose ------> glucose +Pi+ADP+glucose-6-phosphate. delta Go=30.67KJ

Overall reaction is ATP + H2O -----> ADP + Pi. delta Go = 30.67KJ is the final answer

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