In winemaking, the sugars in grapes undergo fermentationby yeast to yield CH3CH2OH and CO2. During cellularrespiration, sugar and enthanol are "burned" to water vapor andCO2.
A) Using C6H12O6 for sugar, calculate ÄH of fermentation and of respiration(combustion).
B) Write a combustion reaction for ethanol. Which has ahigher ÄH for combustion per mole of C, sugaror ethanol?
A) To calculate the enthalpy change (ΔH) of fermentation and respiration (combustion), we need to consider the balanced chemical equations for these reactions:
Fermentation:
C6H12O6 (sugar) → 2CH3CH2OH (ethanol) + 2CO2
Respiration (combustion):
C6H12O6 (sugar) + 6O2 → 6CO2 + 6H2O
To calculate the ΔH of fermentation, we need to know the enthalpies of formation (ΔHf) of the reactants and products. The enthalpy change of a reaction is given by the difference in the sum of the ΔHf of the products and the sum of the ΔHf of the reactants. We can look up the ΔHf values for C6H12O6, CH3CH2OH, and CO2 in a reference table or database.
Let's assume:
ΔHf(C6H12O6) = -1274 kJ/mol (negative sign indicates it's an exothermic reaction)
ΔHf(CH3CH2OH) = -277 kJ/mol
ΔHf(CO2) = -393 kJ/mol
Now we can calculate ΔH of fermentation:
ΔH(fermentation) = [2 × ΔHf(CH3CH2OH) + 2 × ΔHf(CO2)] - ΔHf(C6H12O6)
ΔH(fermentation) = [2 × (-277) + 2 × (-393)] - (-1274)
ΔH(fermentation) = -220 kJ/mol
To calculate the ΔH of respiration (combustion), we use the same approach:
ΔH(respiration) = [6 × ΔHf(CO2) + 6 × ΔHf(H2O)] - ΔHf(C6H12O6)
Assuming:
ΔHf(H2O) = -286 kJ/mol
ΔH(respiration) = [6 × (-393) + 6 × (-286)] - (-1274)
ΔH(respiration) = -2808 kJ/mol
So, the ΔH of fermentation is -220 kJ/mol, and the ΔH of respiration (combustion) is -2808 kJ/mol.
B) The combustion reaction for ethanol is:
C2H5OH (ethanol) + 3O2 → 2CO2 + 3H2O
To determine which has a higher ΔH for combustion per mole of carbon (C), we need to compare the enthalpy changes (ΔH) of combustion per mole of carbon in sugar and ethanol.
In sugar (C6H12O6), there are 6 carbon atoms per molecule, whereas in ethanol (C2H5OH), there are 2 carbon atoms per molecule.
To calculate ΔH of combustion per mole of carbon, we divide the ΔH of combustion by the number of moles of carbon in the reactant.
For sugar:
ΔH(sugar) = -2808 kJ/mol (from previous calculation)
ΔH(sugar per C) = ΔH(sugar) / 6
For ethanol:
ΔH(ethanol) = ΔHf(C2H5OH) - [2 × ΔHf(CO2) + 3 × ΔHf(H2O)]
Assuming:
ΔHf(C2H5OH) = -277 kJ/mol (from previous calculation)
ΔH(ethanol per C) = ΔH(ethanol) / 2
Now, we can compare ΔH(sugar per C) and ΔH(ethanol per C) to determine which has a higher enthalpy change for combustion per mole of carbon.