Question

Please show calculations, thank you!

QUESTION 5 Glucose (C&H1206) 2 Lactate (C H,Os) o Compare anaerobic and aerobic catabolism of glucose under standard conditions. The net reactions are shown on the right: He' ~-194 kJ/mol (99%) TAS® ~ +2 kJ/mol ( 1%) ΔGo, ~-196 kJ/mol (a) List possible reason(s) for the aerobic catabolism of glucose yielding so much more ATP than anaerobic catabolism of glucose. Glucose (CH 602 - 6 CO2+ 6 H2O o ΔΗΡ. ~-2800 kJ/mol (98%) T So, ~ +50 kJ/mol (296) ΔGo, ~-2850 kJ/mol (b) The concentration of a molecule is not expected to Mean bond enthalpy kJ/mol, 25 Chatm change how much energy is required to break individual bonds in the molecule. So it makes sense that the AH is largely independent of the concentration, that is, ΔΗ ~ ΔΗ®. Therefore,aGo'- ΔΗ-TAS®. Bond C-C C-H C-O C-O O-H O-O 412 74 60 463 Estimate AH for the oxidation for 1mol of glucose to 6 CO2 and 6 H2O from the mean bond enthalpies given in the table. Use the open chain form of glucose to work the problem. (C) Assuming that 45kJ of energy is required for generating 1mol of ATP under cellular conditions, estimate the maximum moles of ATP that can be synthesized from the oxidation of glucose to CO, and H2O if Δ.G ~ ΔH ~ 2800kJ/mol. Based on this, what is the efficiency of aerobic catabolism?

Answer 1

a) The oxidation of glucose molecules in cells of organisms for energy for various cellular functions.It occurs both aerobically (in the presence of oxygen) and anaerobically (in the absence of oxygen).Aerobic oxidation occurs in two steps.During the first step glucose is converted to pyruvate along with NADH and two molecules of ATP.In presence of oxygen ,then the free energy contained in NADH is further released via re-oxidization of the mitochondrial electron chain and releases 30 ATP per mol of glucose.This step does not occur in the absence of oxygen instead NADH is reoxidized by reducing pyruvate to lactate.This is the reason aerobic catalysis of glucose produces more ATP than anaerobic catalysis.

b)

On the reactant side the glucose molecule contains 5 C-C bonds 7 C-H bonds 1 C=O bond 5 C-O bonds and 5 O-H bonds while oxygen molecule has one O=O bond.On products side one CO₂ molecule contains two C=O bonds and one H₂O molecule contains two O-H bonds.During the course of reaction the bonds on reactant side are broken absorbing energy while products are formed by releasing energy.

the amount of energy required to break the reactants=(amount of energy required to break 5 C-C bonds)+(amount of energy required to break 7 C-H bonds)+(amount of energy required to break 1 C=O bond)+(amount of energy required to break 5 C-O bonds)+(amount of energy required to break 5 O-H bonds)+(amount of energy required to break one O=O bond)

=5×(amount of energy required to break one C-C bonds)+7×(amount of energy required to break C-H bonds)+(amount of energy required to break 1 C=O bond)+5×(amount of energy required to break C-O bonds)+5(amount of energy required to break O-H bonds)+(amount of energy required to break one O=O bond)  

=5×(Bond enthalpy of C-C bonds)+7×(Bond enthalpy of C-H bonds)+(Bond enthalpy of 1 C=O bond)+5×(Bond enthalpy ofC-O bonds)+5(Bond enthalpy of O-H bonds)+(Bond enthalpy of O=O bond)

=5×(348)+7×(412)+(743)+5×(360)+5(463)+(497)

= 9979 kJ/mol

energy required to break a bond=energy released during bond formation

amount of energy released by formation of bonds = 6(energy of formation of C=O)+62(energy of formation of O-H bond)

= 6(743)+62(463)

= 10014 kJ/mol

H of the reaction = amount of energy released by formation of bonds - the amount of energy required to break the bonds in reactants

= (9979 - 10014) kJ/mol

= -35 kJ/mol

c)

energy required for generating one mole of ATP= 45kJ

H of oxidation of glucose = 2800 kJ/mol

number of moles ATP molecules produced =  

ΔΗ oforidationofglucose energyrequiredforgeneratingonemoleofATP

   =

2800 45

= 62.22 moles