Question

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

Part A In a particular cell, the concentrations of ATP, ADP, and Pi are 2.8×10−3 M , 1.6×10−3 M , and 5.1×10−3 M , respectively. Calculate the free energy change for the hydrolysis of ATP under these conditions. (Assume a temperature of 298 K.)

Answer 1

Given reaction is ATP(aq) + H2O(l)    → ADP(aq) + Pi(aq)      ΔG∘rxn = -30.5 kJ

  Relation between ∆G and ∆G^o is given by

   ∆G = ∆G^o + RT In Keq

Keq:

Given that concentrations of ATP, ADP, and Pi are 2.8×10⁻³ M , 1.6×10⁻³ M , and 5.1×10⁻³ M , respectively.

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

      = 1.6×10⁻³ M x 5.1×10⁻³ M / 2.8×10⁻³ M

      = 0.002914

∆G

Given that ∆G^o = - 30.5 kJ/mol = - 30.5 x 10³ J/mol

Temperature T = 298 K

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

∆G =  ∆G^o + 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