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 ΔH of fermentation, we need to determine the enthalpy change for the reaction:
C6H12O6 → 2CH3CH2OH + 2CO2
Using the balanced equation, we can calculate the enthalpy change for fermentation using the bond energies of the reactants and products.
The bond energy for C-C bond is 347 kJ/mol, the bond energy for C-H bond is 414 kJ/mol, and for C-O bond is 360 kJ/mol.
ΔH = (2 * [bond energy of C-C]) + (2 * [bond energy of C-H]) + (4 * [bond energy of C-O]) - [bond energy of C6H12O6]
ΔH = (2 * 347) + (2 * 414) + (4 * 360) - [bond energy of C6H12O6]
To calculate the ΔH of respiration (combustion), we need to determine the enthalpy change for the reaction:
C6H12O6 + 6O2 → 6CO2 + 6H2O
Using the balanced equation, we can calculate the enthalpy change for respiration using the bond energies of the reactants and products.
The bond energy for O=O bond is 498 kJ/mol, and for O-H bond is 463 kJ/mol.
ΔH = (6 * [bond energy of C-O]) + (12 * [bond energy of O-H]) - (6 * [bond energy of C-C]) - (6 * [bond energy of O=O])
ΔH = (6 * 360) + (12 * 463) - (6 * 347) - (6 * 498)
B) The combustion reaction for ethanol is:
C2H5OH + 3O2 → 2CO2 + 3H2O
To determine which has a higher ΔH for combustion per mole of C, sugar or ethanol, we need to compare the enthalpy changes calculated in part A.
If the enthalpy change for combustion per mole of C is greater for sugar (ΔH sugar > ΔH ethanol), then sugar has a higher ΔH for combustion per mole of C.
A) To calculate ΔH (change in enthalpy) for fermentation and respiration, we need to compare the enthalpies of the reactants and products.
For fermentation, the reaction can be represented as follows:
C6H12O6 → 2 CH3CH2OH + 2 CO2
The ΔH of fermentation can be calculated by subtracting the enthalpies of the reactants from the enthalpies of the products. We can use standard enthalpy values for the enthalpies of formation of the compounds involved.
The enthalpy change for the combustion of 1 mole of C6H12O6 to yield 2 moles of CH3CH2OH is given by:
ΔH of fermentation = (2 * ΔHf(CH3CH2OH)) + (2 * ΔHf(CO2)) - (ΔHf(C6H12O6))
For respiration (combustion), the reaction can be represented as follows:
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O
The ΔH of respiration (combustion) can be calculated in a similar manner:
ΔH of respiration = (6 * ΔHf(CO2)) + (6 * ΔHf(H2O)) - (ΔHf(C6H12O6))
To get the values of ΔHf (standard enthalpy of formation), we can refer to tables where the values for each compound are given. By subtracting the sum of the enthalpies of the reactants from the sum of the enthalpies of the products, we can obtain the ΔH for both fermentation and respiration.
B) The combustion reaction for ethanol can be represented as follows:
C2H5OH + 3 O2 → 2 CO2 + 3 H2O
To determine which has a higher ΔH for combustion per mole of C, we need to compare the enthalpies of combustion for sugar and ethanol, considering the molar ratios of C in each compound.
The enthalpy change for the combustion of 1 mole of sugar (C6H12O6) can be calculated using the balanced equation for respiration (combustion) mentioned earlier.
The enthalpy change for the combustion of 1 mole of ethanol (C2H5OH) can be calculated using the balanced equation for ethanol combustion.
By comparing the enthalpy changes per mole of C for both sugar and ethanol combustion, we can determine which one has a higher ΔH for combustion per mole of C.