Living organisms use energy from the metabolism of food to create an energy-rich

Question

Living organisms use energy from the metabolism of food to create an energy-rich molecule called adenosine triphosphate (ATP). The ATP then acts as an energy source for a variety of reactions that the living organism must carry out to survive. ATP provides energy through its hydrolysis, which can be symbolized as follows:
ATP(aq)+H2O(l)ADP(aq)+Pi(aq)
Grxn= -30.5 kJ
where ADP represents adenosine diphosphate and Pi represents an inorganic phosphate group (such as HPO24).

a) Calculate the equilibrium constant, K, for the above reaction at 298 K

B)The free energy obtained from the oxidation (reaction with oxygen) of glucose (C6H12O6) to form carbon dioxide and water can be used to re-form ATP by driving the above reaction in reverse. Calculate the standard free energy change for the oxidation of glucose. Grxn=

c) Estimate the maximum number of moles of ATP that can be formed by the oxidation of one mole of glucose. nATP=

Explanation / Answer

Ans. #A.         dG0 = - RT lnK         - equation 1

            Where,

                        dG0 = Standard Gibb's free energy

                        R = Universal gas constant = 0.008315 kJ mol-1 K-1

T = Temperature in kelvin

K = Equilibrium constant

Putting the values in equation 1-

            dG0 = - (0.008315 kJ mol-1 K-1) x 310 K x ln K                  ; [370C = 310 K]

            or, - 30.5 kJ mol-1 = - 2.57765 kJ mol-1 x ln K

            or, ln K = 30.5 kJ mol-1 / 2.57765 kJ mol-1 = 11.8325

            or, 2.303 log K = 11.8325

            or, log K = 11.8325 / 2.303 = 5.137863656

            or, K = antilog 5.137863656 = 1.37 x 105

Therefore, equilibrium constant K = 1.37 x 105

#B. Balanced reaction:           C6H12O6 + 6O2 --> 6CO2 + 6H2O

Using Hess’s law, the standard free energy change for oxidation of glucose is –

            dG0 = (6 x dGf0 of CO2) + (6 x dGf0 of H2O)

                        = 6 x (-394.39 kJ/mol) + 6 x (- 237.14 kJ/mol) = -3789.18 kJ/mol

Note: The negative sign indicates the release of energy during oxidation of glucose.

#C. Maximum number of ATP formed from complete oxidation of 1 mol glucose =

                                    dG0 of glucose oxidation / dG0 of ATP formation

                                    = (3789.18 kJ/mol) / (30.5 kJ/mol)

                                    = 124.24

                                    = 124

Therefore, moles of ATP formed =124

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