Question

The ΔG°' for the reaction ATP + H₂O <----- ADP + P_i + H⁺ is -30.5 kJ mol^-1. Other organophosphate species also undergo hydrolysis of the phosphate moiety via a similar reaction. ------>

Consider the following reaction that is not spontaneous when the concentrations of the reactants and products are all 1 M.

   ATP + Acetic acid <----- ADP + Acetyl phosphate
------>
What concentration of acetic acid would be necessary to make the reaction just begin to be spontaneous? (All other species in the reaction remain at 1 M concentration. The temperature is T = 25 °C. The hydrolysis reaction for acetyl phosphate is Acetyl phosphate + H₂O <------ Acetic acid + P_i + H⁺; ΔG°' = -42.2 kJ mol^-1.) ------>

[acetic acid] = _____M

Answer 1

Ans. Given, the standard free energy change of phosphate hydrolysis for-

            ATP + H₂O ------> ADP + Pi                       ; dG^0’ = -30.5 kJ/ mol           - Reaction 1

            Ace-P + H₂O ----> Acetic acid + Pi + H⁺ ; dG^0’ = -61.9 kJ/ mol           - Reaction 2

# When a reaction is reversed, the sign of standard free energy change is also reversed.

So,

Acetic acid + Pi + H⁺ ------> Ace-P + H₂O       ; dG^0’ = +42.2 kJ/mol          - Reaction 3

#. The net coupled reaction for the formation of acetyl phosphate (Ace-P) can be written as follow-

            Acetic acid + Pi + H⁺ ---> Ace-P + H₂O            ; dG^0’ = +42.2 kJ/mol ; Reaction 3

ATP + H₂O ------> ADP + Pi                                   ; dG^0’ = -30.5 kJ/ mol ; Reaction 1

Net coupled Rxn: ATP + Acetic acid + H⁺ ---------> Ace-P + ADP            ; dG^0’_net = ?

The dG^0’ for the net coupled reaction is given by-

            dG^0’_net = dG^0’ of Rxn 3 + dG^0’ of Rxn 1

            Or, dG^0’_net = 42.2 kJ/mol + (-30.5 kJ/mol) = +11.7 kJ/mol

# Hence, dG^0’ for the (net) reaction = +11.7 kJ / mol

# Calculating desired [acetic acid]:

# Using the equation dG^0’ = - RT lnKeq              - equation 1

Where, dG^0’ = standard/ theoretical free energy change

T = temperature in kelvin = (⁰C + 273.15) K

Keq = equilibrium constant under given condition

R = 0.0083146 kJ mol^-1 K^-1

#Let the [acetic acid] = X M , while all other concentrations remain constant at 1.0 M.

ATP + Acetic acid + H⁺ ---------> Ace-P + ADP

Equilibrium constant for the reaction, Keq = [Ace-P] [ADP] / ([ATP] [acetic acid] [H⁺])

            Or, Keq = 1/ X

Putting the value of Keq in equation 1-

            11.7 kJ mol^-1 K^-1 = - (0.0083146 kJ mol^-1 K^-1) x 298.15K x ln (1/X)

            Or, (11.7 kJ mol^-1 K^-1) / (2.479 kJ mol^-1 K^-1) = 2.303 log (1/X)

            Or, 4.72 / 2.303 = log (1/X)

            Or, (1/X) = antilog (2.05)

            Or, (1/X) = 112.202

            Or, X = 1 / 112.202

Hence, X = 0.0089

Therefore, required [Acetic acid] = X M = 0.0098 M