Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: happychappie1700 on February 19, 2019, 10:15:52 AM
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Dear Colleagues,
The following question asks me: Calculate the Enthalpy of Reaction for ( using Hess's Law ):
C6H12O6 ----->2C2H5OH+2CO2
Using the following data ( Enthalpy of Combustion ): Glucose=-2820 Ethanol -1367 The answer is:
(https://img.photobucket.com/albums/v725/kavanagh/P1060174_zpsqt0mr2m7.jpg)
The answer then states that there is a linear method available for obtaining the same answer:
ΔrH=ΣΔcHreactant-ΣΔcHreactant
Which is : -2820 - ( 2 x -1367 ) = - 86 KJ mol
My confusion is: if the question is asking for the Enthalpy of Reaction ( Delta H ) - assuming its for the whole reaction, then why is the other product CO2 not included in the calculation ? ( why is only 2C2H5OH included in the calculation ? ).
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I would try writing out each chemical reaction and adding them as appropriate.
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What do you think the combustion enthalpy of CO2 is? Can you combust CO2?
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What do you think the combustion enthalpy of CO2 is? Can you combust CO2?
I presume you can't combust Carbon Dioxide, but why is Carbon Dioxide taken into account in the following question / equation:
Calculate the standard enthalpy of combustion of Methanol:
2CH3OH + 3O2 -----> 4CO2+2H2O
Using Enthalpy of Formation Data:
CH3OH = -238.6 KJ/mol
O2 = 0
CO2 = -393.5 KJ/mol
H20 = -241.8 KJ/mol
Equation = ΔrH=ΣΔfHproducts -ΣΔfHreactants
= [ 4 (-393.5) + 2 (-241)] - [2 (-238.6) + (0)] = -1277 KJ / Mol
Why is the CO2 figure of -393.5 included in this calculation ( why is it not ignored as in the first question about the fermentation of glucose )?
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Because CO2 has an enthalpy of formation, but not an enthalpy of combustion.
Consider the equations
2C + 4O2 + 4H2 :rarrow: 2CH3OH + 3O2 ΔH1 = 2ΔHfCH3OH
2C + 4O2 + 4H2 :rarrow: 2CO2 + 4H2O ΔH2 = 2ΔHfCO2 + 4ΔHfH2O
2CH3OH + 3O2 :rarrow: 2CO2 + 4H2O ΔH3 = ΔH2 - ΔH1 = 2ΔHcombCH3OH
(note you balanced the equation wrongly, and the final answer has to be divided by 2)
You can see, I hope, how the enthalpy of formation of CO2 must be included here.
But consider the right hand side of your combustion cycle:
2C2H5OH + 2CO2 + 6O2 :rarrow: {4CO2 + 6H2O} + 2CO2
The 2CO2 on the LHS is unchanged, so there is no enthalpy change associated with it.