(3) (16 pts total) Sugars can be digested aerobically or anaerobically (e.g., by

Question

(3) (16 pts total) Sugars can be digested aerobically or anaerobically (e.g., by yeast). During aerobic digestion, sugars react with oxygen to form carbon dioxide and water. During anaerobic digestion, sugars are decomposed by enzymes, acting as catalysts, in the presence of water to create carbon dioxide and alcohols. (a) (4 pts) Write the balanced chemical equation for aerobic digestion of one mole of sucrose, C12H201s). (b) (4 pts) Evaluate Go (k) and AE° (V) for the aerobic reaction in (a) for one mole of sucrose at 25°C. (INFO not in Appendix C: AH (sucrose, s)-2221.2 kJ/mol; S° (sucrose, s)- 392.4 J/mol.K) (c) (4 pts) Write the balanced chemical equation for anaerobic digestion of one mole of sucrose. Assume that ethanol, C2H5OH(/), is the only alcohol product for anaerobic digestion. (d) (4 pts) Evaluate G(k) and AE° (V) for the anaerobic reaction in (c) for one mole of sucrose at 25°C

Explanation / Answer

a)

balanced reaction for digestion:

C12H22O11 + O2 = CO2 + H2O

balance C, then H, finally O

C12H22O11 + 12O2 = 12CO2 + 11H2O

add phases

C12H22O11(s) + 12O2(g) = 12CO2(g) + 11H2O(l)

b)

d G= Gproducts- Greactants

dG = (12CO2+ 11H2O) - (C12H22O11)

dGsucrose = (-2221.2) - 298*(392.4/1000) = -2338.13

dG = (12-393.5+ 11*284.8) - (-2338.13)

dG = 5089.43 KJ/mol

ii)

dG = -n*F*E°cell

E°cell = -dG/(nF) = -5089430/(24*96500) = -2.197 V

c)

anaerobic:

C12H22O11 = C2H5OH + CO2

we must add H2O

C12H22O11 + H2O = 4C2H5OH + 4CO2

add phases

C12H22O11(s) + H2O(l) = 4C2H5OH(g) + 4CO2(g)

dG = Gprod - Greact = (4*-167.9 + 4*-394.4) - (-2338.13 + -237.1)

dG = 326.03 kJ/mol

E = -dG/(nF) = -326030/(24*96500) =

E = -0.1407

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