Question

The oxidation of glucose (C6H12O6) in body tissue produces CO2 and H2O. In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol (C2H5OH) and CO2. Using data given in Appendix C, compare the equilibrium constants for the following reactions: C6H12O6(s)+6O2(g)⥫⥬6CO2(g)+6H2O(l)C6H12O6(s)⥫⥬2C2H5OH(l)+2CO2(g) Compare the maximum work that can be obtained from these processes under standard conditions.

Answer 1

a) The Keq for aerobic oxidation (Keq = very large) is greater than Keq (Keq = 8.96x10^39) for anaerobic decomposition.

b) we can see that the work for aerobic oxidation is greater (ΔG° = -2878.9 kJ) than the work for anaerobic decomposition (ΔG° = 227.9 kJ)

Step-by-step explanation:

a)

having the reaction:

C6H12O6 + 6O2 = 6CO2 + 6H2O

we will calculate the Gibbs energy:

ΔG° = ΔG°products - ΔG°reactants

The products are the CO2 and H2O, the reactants are C6H12O6 and O2

ΔG° = ((6*(-394.4 kJ)) + (6*(-237.13 kJ))) - (-910.4 kJ + 6*(0)) = -2878.8 kJ

Keq = e^(-ΔG°)/R*T = e^-(-2878.8x10^3 J)/(8.314 * 298)) = this value is very large

For the next reaction:

C6H12O6 = 2C2H5OH + 2CO2

ΔG° = ((2*(-174.76 kJ)) + (2*(-394.4 kJ))) - (-910.4 kJ) = -227.9 kJ

Keq = e^-(-227.9x10^3/(8.314*298)) = 8.96x10^39

The Keq for aerobic oxidation is greater than Keq for anaerobic decomposition.

b)

The work obtained from aerobic oxidation is equal to ΔG° = -2878.9 kJ and the work obtained from anaerobic is equal to ΔG° = 227.9 kJ

we can see that the work for aerobic oxidation is greater than the work for anaerobic decomposition

Answer 2

Step-by-step explanation:

Solved solution is in the attach document.