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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CH3CH2OH (l)+ 3O2 (g)..>2CO2(g)+3H2O (g)

ethanol

A) To calculate the enthalpy change (ΔH) of fermentation, we need to know the energy released (or absorbed) during the process. Given that fermentation produces ethanol (CH3CH2OH) and carbon dioxide (CO2) from glucose (C6H12O6), we can consider the reaction:

C6H12O6 (s) → 2 CH3CH2OH (l) + 2 CO2 (g)

The enthalpy change of this reaction can be obtained by subtracting the enthalpies of the reactants from the enthalpies of the products. However, since we are not given specific values for enthalpies, we need to find the standard enthalpy of formation (ΔHf) for each compound involved in the reaction.

For glucose:
C6H12O6 (s) → 6 C (graphite) + 6 H2 (g) + 3 O2 (g)
ΔHf = ΔHf(C6H12O6) - [6ΔHf(C) + 6ΔHf(H2) + 3ΔHf(O2)]

For ethanol:
CH3CH2OH (l) → C (graphite) + 2 H2 (g) + O2 (g)
ΔHf = ΔHf(CH3CH2OH) - [ΔHf(C) + 2ΔHf(H2) + ΔHf(O2)]

For carbon dioxide:
CO2 (g) → C (graphite) + O2 (g)
ΔHf = ΔHf(CO2) - [ΔHf(C) + ΔHf(O2)]

Once we have the ΔHf values for each compound, we can substitute them into the equation and calculate ΔH of fermentation.

For the enthalpy change of respiration (combustion), we need to consider the combustion of sugar (glucose) and ethanol. The combustion reaction for ethanol is:

CH3CH2OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (g)

To calculate the ΔH of combustion for sugar and ethanol, we need to know the ΔHf values for each compound involved in the reaction, similar to what we did for fermentation. Once we have the ΔHf values, we can substitute them into the equation and calculate ΔH of combustion.

B) The combustion reaction for ethanol is indeed as follows:

CH3CH2OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (g)

To determine whether sugar or ethanol has a higher ΔH of combustion per mole of carbon, we need to compare the coefficients of carbon (C) in each reaction.

For sugar (glucose):
C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (g)

For ethanol:
CH3CH2OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (g)

Comparing the coefficients, we see that one mole of sugar (glucose) produces 6 moles of CO2 for every 6 moles of carbon, while one mole of ethanol produces 2 moles of CO2 for every 1 mole of carbon. Therefore, sugar has a higher ΔH of combustion per mole of carbon compared to ethanol.