In winemaking, the sugars in grapes undergo fermentation by yeast to yield CH3CH2OH and CO2. During cellular respiration, 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•Chemistry - Mark, Saturday, October 24, 2015 at 10:11am
i think it's

C2H6OH(l) + 3O2(g) ---> 2CO2(g) + 3H2O(g)

am I correct?..thanks!

You are correct in writing the combustion reaction for ethanol. The balanced equation is:

C2H6OH(l) + 3O2(g) ---> 2CO2(g) + 3H2O(g)

To determine which compound, sugar or ethanol, has a higher ΔH for combustion per mole of carbon, we need to calculate the enthalpies of combustion for both compounds.

The enthalpy of combustion (ΔHcomb) for a compound is the amount of heat released when one mole of the compound is burned completely in excess oxygen.

First, let's calculate the ΔHcomb for sugar (C6H12O6):

The balanced combustion equation for sugar is:

C6H12O6(s) + 6O2(g) ---> 6CO2(g) + 6H2O(g)

To calculate the ΔHcomb, we need to look up the enthalpies of formation for each product and reactant involved in the combustion reaction. The values are as follows:

Enthalpy of formation (ΔHf):
C6H12O6(s) = -1273.3 kJ/mol
CO2(g) = -393.5 kJ/mol
H2O(g) = -241.8 kJ/mol
O2(g) = 0 kJ/mol (because it is the elemental form in its standard state)

Now we can calculate ΔHcomb for sugar:

ΔHcomb = [6(ΔHf CO2) + 6(ΔHf H2O)] - [ΔHf C6H12O6 + 6(ΔHf O2)]
= [6(-393.5 kJ/mol) + 6(-241.8 kJ/mol)] - [-1273.3 kJ/mol + 6(0 kJ/mol)]
= -2392.8 kJ/mol

Therefore, the ΔHcomb for sugar is -2392.8 kJ/mol.

Now let's calculate the ΔHcomb for ethanol (C2H6OH):

Using the balanced combustion equation: C2H6OH(l) + 3O2(g) ---> 2CO2(g) + 3H2O(g)

We already have the enthalpies of formation for the products CO2(g) and H2O(g) from the previous calculation.

Enthalpy of formation (ΔHf):
C2H6OH(l) = -277.6 kJ/mol

Now we can calculate ΔHcomb for ethanol:

ΔHcomb = [2(ΔHf CO2) + 3(ΔHf H2O)] - [ΔHf C2H6OH + 3(ΔHf O2)]
= [2(-393.5 kJ/mol) + 3(-241.8 kJ/mol)] - [-277.6 kJ/mol + 3(0 kJ/mol)]
= -1367.4 kJ/mol

Therefore, the ΔHcomb for ethanol is -1367.4 kJ/mol.

Comparing the two values, we can see that the ΔHcomb for sugar is -2392.8 kJ/mol, while the ΔHcomb for ethanol is -1367.4 kJ/mol. Therefore, ethanol has a higher ΔH for combustion per mole of carbon compared to sugar.