Question

Compare the following two ways to break down glucose, C6H12O6.

a) Glucose is metabolized via complete oxidation to carbon dioxide and water in the glycolytic pathway: C6H12O6(s) + 6O2(g) ---> 6CO2(g) + 6H2O(l) Calculate ΔHø(298K) for this reaction.

b) Glucose is broken down into ethanol and carbon dioxide in alcoholic fermentation by microorganisms: C6H12O6(s) ---> 2 C2H5OH(l) + 2CO2(g) Calculate the amount of heat liberated in a yeast brew upon fermentation of 1 mol of glucose at 25˚C and 1 atm.

c) Compare the amount of heat liberated by the complete combustion of 1.0 g glucose to that liberated by fermentation of 1.0 g glucose.

Answer 1

a) Glucose is metabolized via complete oxidation to carbon dioxide and water in the glycolytic pathway:

C6H12O6(s) + 6O2(g) ---> 6CO2(g) + 6H2O(l) Calculate ΔHø(298K) for this reaction.

Hrxn = Hproducts - Hreactants

HRxn = (6*−393.51 + 6*−285.830) - (-1271) = -2805.04 kJ/mol rxn

HRxn = -2805.04 kJ/mol rxn

b) Glucose is broken down into ethanol and carbon dioxide in alcoholic fermentation by microorganisms: C6H12O6(s) ---> 2 C2H5OH(l) + 2CO2(g) Calculate the amount of heat liberated in a yeast brew upon fermentation of 1 mol of glucose at 25˚C and 1 atm.

HRxn = Hproducts - Hreactants

HRxn = 2*(-277.0) + 2 *(-393.51 )) - (1271) = 2612.02 kJ/mol rxn

c)

1mol of glucose = 180 g/mol

then

mol = 1/180 = 0.0055555 mol of glucose

combustion --> 2612.02 kJ/mol *0.0055555 mol = 14.51107 kJ

yeast --> 2805.04kJ/mol *0.0055555 mol = 15.581 kJ

clearly... yeast is more exothermic