[happychappie1700]

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:

The answer then states that there is a linear method available for obtaining the same answer:
Δ_rH=ΣΔ_cH_reactant-ΣΔ_cH_reactant

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 ? ).

[Babcock_Hall]

I would try writing out each chemical reaction and adding them as appropriate.

[mjc123]

What do you think the combustion enthalpy of CO₂ is? Can you combust CO₂?

[happychappie1700]

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 )?

[mjc123]

Because CO₂ has an enthalpy of formation, but not an enthalpy of combustion.
Consider the equations
2C + 4O₂ + 4H₂  2CH₃OH + 3O₂ ΔH1 = 2ΔHfCH₃OH
2C + 4O₂ + 4H₂  2CO₂ + 4H₂O   ΔH2 = 2ΔHfCO₂ + 4ΔHfH₂O
2CH₃OH + 3O₂  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  {4CO₂ + 6H₂O} + 2CO₂
The 2CO₂ on the LHS is unchanged, so there is no enthalpy change associated with it.