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?
am I right?
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ethanol
I need help with B)balancing the equation. Thanks.
Answer this Question
To answer part A, we need to calculate the change in enthalpy (ΔH) for fermentation and respiration.
Fermentation converts glucose (C6H12O6) into ethanol (CH3CH2OH) and carbon dioxide (CO2). The balanced equation for fermentation is:
C6H12O6 → 2CH3CH2OH + 2CO2
To calculate the ΔH of fermentation, we need to find the enthalpy change of each component and apply the stoichiometric coefficients.
Assuming standard conditions, the enthalpy change of formation for glucose (C6H12O6) is 0 kJ/mol, since it is in its standard state. Ethanol (CH3CH2OH) has a standard enthalpy of formation of -277.6 kJ/mol, and carbon dioxide (CO2) has a standard enthalpy of formation of -393.5 kJ/mol.
Using these values, we can calculate the ΔH of fermentation:
ΔH fermentation = (2 × ΔHCH3CH2OH) + (2 × ΔHCO2) - (1 × ΔHC6H12O6)
= (2 × -277.6 kJ/mol) + (2 × -393.5 kJ/mol) - (0 kJ/mol)
= -555.2 kJ/mol - 787.0 kJ/mol
= -1342.2 kJ/mol
Therefore, the ΔH of fermentation is -1342.2 kJ/mol.
Now let's move on to respiration. Respiration is the combustion of glucose and ethanol to form water vapor (H2O) and carbon dioxide (CO2). The balanced equation for respiration of glucose is:
C6H12O6 + 6O2 → 6CO2 + 6H2O
The enthalpy change of formation for water vapor (H2O) and carbon dioxide (CO2) remains the same as mentioned earlier. Since glucose (C6H12O6) is in its standard state, its enthalpy change of formation is also 0 kJ/mol. Oxygen (O2) is also in its standard state, so its enthalpy change of formation is 0 kJ/mol.
Using these values, we can find the ΔH of respiration:
ΔH respiration = (6 × ΔHCO2) + (6 × ΔHH2O) - (1 × ΔHC6H12O6) - (6 × ΔHO2)
= (6 × -393.5 kJ/mol) + (6 × -241.8 kJ/mol) - (0 kJ/mol) - (6 × 0 kJ/mol)
= -2361 kJ/mol
Thus, the ΔH of respiration is -2361 kJ/mol.
For part B, the combustion reaction of ethanol is as follows:
C2H5OH + 3O2 → 2CO2 + 3H2O
To balance the equation, we can start by counting the carbon atoms first. On the reactant side, there are two carbon atoms in ethanol (C2H5OH), and on the product side, there are two carbon atoms in carbon dioxide (2CO2). So, carbon is already balanced.
Next, we can count the hydrogen atoms. On the reactant side, there are six hydrogen atoms in ethanol (C2H5OH), and on the product side, there are six hydrogen atoms in water (3H2O). So, hydrogen is also balanced.
Lastly, we can count the oxygen atoms. On the reactant side, there are three oxygen molecules (3O2), and on the product side, there are four oxygen atoms in carbon dioxide (2CO2) and six oxygen atoms in water (3H2O). So, oxygen is not balanced.
To balance the oxygen, we need to add a coefficient of 3/2 in front of O2 on the reactant side:
C2H5OH + (3/2)O2 → 2CO2 + 3H2O
Therefore, the balanced combustion reaction of ethanol is:
C2H5OH + (3/2)O2 → 2CO2 + 3H2O
Regarding the ΔH for combustion per mole of C, you are correct that ethanol has a higher ΔH for combustion per mole of C compared to sugar. Ethanol contains two carbon atoms per molecule, while glucose (sugar) contains six carbon atoms per molecule. Therefore, the combustion of ethanol releases more energy per mole of carbon compared to sugar.
I hope this explanation helps! Let me know if you have any further questions.
i think it's
C2H6OH(l) + 3O2(g) ---> 2CO2(g) + 3H2O(g)