Smallest Heat of Combustion: Cyclopentane

  • Thread starter Thread starter ilovephysicssomuch
  • Start date Start date
  • Tags Tags
    Heat
AI Thread Summary
The discussion centers on identifying which isomer of C5H10 has the smallest heat of combustion, with initial speculation favoring cyclopentane due to its lack of substituents. Participants clarify that the stability of the compound is crucial, noting that less stable compounds typically have smaller heats of combustion. The conversation shifts to comparing various isomers, ultimately suggesting that 1,1-Dimethylcyclopropane may have the smallest heat of combustion due to its instability from close methyl group interactions. The importance of bond strain and enthalpy of formation in determining combustion heat is emphasized, with cyclopentane being less strained than other isomers. Overall, the discussion highlights the relationship between molecular stability and heat of combustion in hydrocarbons.
ilovephysicssomuch
Messages
23
Reaction score
0
Which isomer of C5H10 would you expect to have the smallest heat of combustion?
A. cyclopentane
D. cis-1,2-Dimethylcyclopentane
E. trans-1,2-Dimethylcyclopentane

.. I'm thinking A because there are no groups.. can I please be confirmed?o:)
 
Physics news on Phys.org
I will confirm to you that you are guessing, then I'll throw you a bone. Somewhere in your text or class notes, you've been given information on heats of formation of bonds (C-C σ, C-C &pi, C-H), or heats of formation of hydrocarbons (alkenes, alkanes). Give you something to think on?
 
Well...

cis-1,2-dimethylcyclopropane has a larger negative heat of combustion than trans-1,2-Dimethylcyclopropane.

The c-c bonds in cyclopropane are bent... which makes for weaker bonds.. so now its E
 
haha I am just messing with you.. It's Definitely A.. According to research =P
 
Ok.. Final Answer...

I changed my mind.. to have the smallest heat of combustion, the compound must be the least stable..
and if we had 5 choices:
A) Cyclopentane
B) Methylcyclopropane
C)1,1-Dimethylcyclopropane
D) cis-1,2-dimethylcyclopropane
E) trans-1,2-Dimethylcyclopropane

C) would have the smallest heat of combustion because it is least stable...
due to the interaction of the two methyl groups... that are located CLOSEST together...
hmmm.. is that a good reasoning? So.. C)1,1-Dimethylcyclopropane
 
EASY TO BREAK = SMALLEST HEAT OF COMBUSTION? is that true?:confused:
 
You're thinking about the right things, but you came up just slightly backward for a result. Everything winds up at the same enthalpy when it's combusted, carbon dioxide plus water in the same stoichiometry. Now how do the starting enthalpy levels compare to one another? Methylcyclobutane has a lot of ring strain, enough so that it has a positive enthalpy of formation relative to hydrogen and carbon. Cyclopentane is slightly strained, but not much, and has a negative enthalpy of formation. Formation of water and carbon dioxide from oxygen and these two isomers is accompanied by an enthalpy change that is the difference between h-carbon plus oxygen and water plus carbon dioxide. Given that the enthalpies of formation of water and carbon dioxide are large and negative, which is the larger difference?
 
Back
Top