Given the following reaction at 298 K: ATP( a q ) + H2O( l ) ADP( a q ) + Pi( a

Question

Given the following reaction at 298 K:
ATP(aq) + H2O(l) ADP(aq) + Pi(aq) Grxn = -30.5 kJ

Part A

In a particular cell, the concentrations of ATP, ADP, and Pi are 2.8×103 M , 1.6×103 M , and 5.1×103 M , respectively.
Calculate the free energy change for the hydrolysis of ATP under these conditions. (Assume a temperature of 298 K.)

Given the following reaction at 298 K:
ATP(aq) + H2O(l) ADP(aq) + Pi(aq) Grxn = -30.5 kJ

Part A

In a particular cell, the concentrations of ATP, ADP, and Pi are 2.8×103 M , 1.6×103 M , and 5.1×103 M , respectively.
Calculate the free energy change for the hydrolysis of ATP under these conditions. (Assume a temperature of 298 K.)

Explanation / Answer

Given reaction is ATP(aq) + H2O(l)    ADP(aq) + Pi(aq)      Grxn = -30.5 kJ

  Relation between G and Go is given by

   G = Go + RT In Keq

Keq:

Given that concentrations of ATP, ADP, and Pi are 2.8×103 M , 1.6×103 M , and 5.1×103 M , respectively.

Keq = [ADP(aq)] [Pi(aq)] / [ATP(aq)]

      = 1.6×103 M x 5.1×103 M / 2.8×103 M

      = 0.002914

G

Given that Go = - 30.5 kJ/mol = - 30.5 x 103 J/mol

Temperature T = 298 K

R = universal gas constant = 8.314 J/K/mol

G =  Go + RT In Keq

     = - 30.5 x 103 J/mol + (8.314 J/K/mol) (298 K) In (0.002914)

     = - 44964 J/mol

    = - 44.964 kJ/mol

   G = - 44.964 kJ/mol

Therefore,

free energy change for the hydrolysis of ATP = - 44.964 kJ/mol

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