Is there a link between where the functional group is located in an organic compound and the compound's heat of combustion? It has been mentioned before that it is difficult to accurately get the enthalpy of a reaction experimentally; thus, enthalpies of formation are used instead when calculating it.This is why I looked up the enthalpy of formation of liquid at STP data for two compounds, 1-Heptanol and 2-Heptanol, which are $\pu{403.4 kJ mol^{-1}}$ and $\pu{416.9 kJ mol^{-1}}$ respectively.
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1You can find some calculation here. – Mathew Mahindaratne Apr 11 '21 at 07:22
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1Sure it has an effect. Why it should not be? – Alchimista Apr 11 '21 at 09:20
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@MathewMahindaratne Thank you for the notice, I updated the post. – abtoiew Apr 11 '21 at 09:34
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@Alchimista Why would there be an effect? I don't think I understand it. – abtoiew Apr 11 '21 at 09:35
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Because two different molecules are different in basically every property, except composition in isomers like in this case. Even the heats of combustion of conformers - imagine to burn them separately - are different! – Alchimista Apr 11 '21 at 09:39
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@Alchimista Then the explanation is "because everything else is different". I don't get it. Is it because the bonds are stronger that the enthalpy is higher in 2-Heptanol? For example, I also checked out their boiling poins and 1-Heptanol has a higher boiling point: 175°C versus 160°C; which makes me think that because a 1-Heptanol can approach another one's functional group to form a bond from a range of angles broader than that of 2-Heptanol, its boiling point is higher. I can't find an explanation like this one for the enthalpies though. – abtoiew Apr 11 '21 at 10:00
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2Deciding the order of heat of combustion of isomeric alkanes; Heats of combustion and stability of rings; What causes the difference in heat energy released by combustion for the butanol isomers? – andselisk Apr 11 '21 at 10:44
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@abtoiew see the link posted below. For the aspect of boiling points, in the specific case the reason is as you sketched. In general it will depend on shape and size. – Alchimista Apr 11 '21 at 11:04
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@MathewMahindaratne The source you linked actually did help a lot! Do you have any other sources where I can find heat of combustion values for other organic compounds like Aliphatic Aldehydes with more than 7 Carbons. – abtoiew Apr 11 '21 at 12:36
1 Answers
The exhaustive combustion of an organic compound consisting of carbon and hydrogen only will yield carbon dioxide and water. Thus, if the heat of formation for isomers differ from each other, then the heat of combustion will differ, too.
This entry in the English Wikipedia states, for example $\pu{−146.9 kJ mol^{-1}}$ for n-pentane, $\pu{−154.4 kJ mol^{-1}}$ about isopentane (methylbutane), and $\pu{−167.8 kJ mol^{-1}}$ for neopentane / dimethylpropane; all of same number of atoms $\ce{C5H12}$ (pentane), but with different connection of methyl groups to a backbone in common. The very table continues for other branched isomers of alkanes, too.
Possibly there is a misunderstanding between the computation of heat of formations from paper (or, in a computer program), on one hand, and the experimental determination of these properties, too.
For the former, the approach typically taught in schools to determine the enthalpy of formation is
- to imagine the reaction of $\ce{C_xH_y}$ with $\ce{O2}$ to yield $\ce{CO2}$ and $\ce{H2O}$
- to look up the standard heat of formation of $\ce{CO2}$ and $\ce{H2O}$, hopefully with a discern between water as gas and water as liquid
- to apply Hess's law to compute $\Delta_fH^\circ$ for $\ce{C_xH_y}$ with the convention that $\Delta_fH^\circ$ for elements in their most stable form (here, oxygen) is set zero
This approach however is an approximation which does not account how the atoms are connected with each other and their neighbours, or bond order; as if e.g., a carbon in acetylene does not differ from the carbon in methane.
If one leaves the imaginary reaction to $\ce{CO2}$ and $\ce{H2O}$, there are empirical increment systems based on experimentation, set to improve this approximation when computing this on paper.$^*$ With the advent of quantum chemistry, it is up to you to select a theory suitable enough to compute this in silico, too (example by MOPAC).
Among the classic approaches to determine experimentally the heat of formation via the heat of reaction (again, Hess' law is applied) is bomb calorimetry. Here, exhaustive combustion a known quantity of your compound takes place in an atmosphere of pure high-pressure oxygen. It is one of the experiments students may encounter early when studying physical chemistry.
$^*$ Publications like 1993ChemRev2419, 2011JPhysChemA10576, or 2012JPhysChemA7196 review / present extensions, or are basis for more recent work, too (cited by below the landing page of the publication) in addition to the increment system above mentioned Wikipedia entry currently presents. Of course, one equally may adjust existing / establish a new increment systems, e.g., as combination of own work (bomb calorimetry) and using data already recorded by others and presented in primary references (articles) and tables (e.g., CRC Handbook of Chemistry and Physics, NIST Webbook Chemistry)
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The increment systems you linked work for compounds that contain C and H only. Though, what would I do if I needed to improve the approximation when calculating the enthalpy of formation of a compound that has C, H, and O like Aldehydes? – abtoiew Apr 11 '21 at 20:39
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1Either a) identify other increment systems including elements other than C and H in the literature (examples 1993ChemRev2419, 2011JPhysChemA10576, 2012JPhysChemA7196 review / present extensions; and are basis for more recent work, too [cited by]). Or b) establish a new increment system, maybe as combination of own work (bomb calorimetry) and using data already recorded by others and presented in primary references (articles) and tables (e.g., CRC Handbook of Chemistry and Physics) – Buttonwood Apr 12 '21 at 17:16