(8) Oxidation of glucose to CO2 and water is accompanied by a large negative fre

Question

(8) Oxidation of glucose to CO2 and water is accompanied by a large negative free energy change: C6H1206 + 6O2 6 CO2 + 6 H2O G--2880 kJ/mol Many of the reactions in living organisms require a source of free energy to drive them. So the body transfers the large amount of free energy from glucose oxidation into an "energy carrier" molecule called ATP, which can provide smaller amounts of free energy to reactions that require a boost. Thus, some of the energy from the glucose oxidation is used to synthesize ATP from ADP (a) For the conversion of ADP to ATP, H20.1 kJ/mol and S"=-34.9 JK.mol. What is the G° for this reaction at 25.0 °C? (b) At the normal human body temperature of 37.0°C, will the conversion of ADP to ATP be spontaneous? Why or why not? (c) For every molecule of glucose that is oxidized, 38 molecules of ADP are converted to ATP. What is the efficiency (percentage of free energy successfully captured) of this process?

Explanation / Answer

Ans. #a. dG0 = dH0 – TdS0

            Or, dG0 = 20.1 kJ mol-1 – 298.15 K (-34.9 J K-1 mol-1)

            Or, dG0 = 20.1 kJ mol-1 + 10405.435J mol-1

            Or, dG0 = 20.1 kJ mol-1 + 10.4 kJ mol-1

            Hence, dG0 = 30.5 kJ mol-1

#b. dG0 = dH0 – TdS0

            Or, dG0 = 20.1 kJ mol-1 – 310.15 K (-34.9 J K-1 mol-1)

            Or, dG0 = 20.1 kJ mol-1 + 10824.235 J mol-1

            Or, dG0 = 20.1 kJ mol-1 + 10.8 kJ mol-1

            Hence, dG0 = 30.9 kJ mol-1

No, the conversion of ADP to ATP is NOT spontaneous at body temperature.

Since the dG0 for the reaction is positive, the reaction in non-spontaneous at 37.00C.

Note: for human body, 37.00C is taken as standard temperature.

#c. Enthalpy of formation of ATP from ADP = +30.5 kJ/mol ; [from #a.]

Total energy required for formation of 38 ATP =

Moles of ATP formed x Molar enthalpy of formation of ATP

= 38 ATP x 30.5 kJ mol-1

= 1159.0 kJ

# Total energy released from oxidation of 1 mol glucose = 2880 kJ

Now,

% efficiency = (Energy stored in ATP / Energy released from 1mol glucose) x 100
                        = (1159.0 kJ / 2880 kJ) x 100

                        = 40.24 %

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