[Jonnopiii]

Hi,

I'm currently doing an experimental report which involves the burning of alcohols and measuring mass lost to find the heat of combustion for each alcohol. After looking over some data I have come across that Alkanes have higher heats of combustion than alcohols but I have no idea why. I thought I understood by using the equation ΔH = H products – H reactants ; that if the reactants took more energy to break the bonds, the result would be a greater magnitude in energy released (noting that combustion is an exothermic reaction so ΔH will be negative). So if an alcohol was in the reactants, which have much higher boiling points than alkanes and thus is harder to break the bonds, shouldn't the magnitude of the ΔH value produced be greater for alcohols than that of alkanes? Please feel free to let me know the reason as to why for a respective number of carbons, the alkane will have a higher heat of combustion than the alcohol.

Thanks

[thetada]

I thought I understood by using the equation ΔH = H products – H reactants ; that if the reactants took more energy to break the bonds, the result would be a greater magnitude in energy released (noting that combustion is an exothermic reaction so ΔH will be negative).

It's the other way round. Exothermic reactions occur when the amount of energy released by the formation of bonds in the products is greater than the energy required to break the bonds in the reactants. So the more energy required to break the bonds in the reactants, the more likely the reaction is to be endothermic. You may be using the equation incorrectly. One way to determine enthalpy changes is to use average bond enthalpy. In that case, the enthalpy change of combustion is equal to the sum of the individual bond enthalpies of the reactants minus the sum of the individual bond enthalpies of the products (ie reactants - products). A good explanation is available here:

 http://www.chemguide.co.uk/physical/energetics/bondenthalpies.html.

So if an alcohol was in the reactants, which have much higher boiling points than alkanes and thus is harder to break the bonds, shouldn't the magnitude of the ΔH value produced be greater for alcohols than that of alkanes? Please feel free to let me know the reason as to why for a respective number of carbons, the alkane will have a higher heat of combustion than the alcohol.

What kind of bonds are you breaking in each case? What bonds break when the alkane / alcohol is combusted? What bonds break when the alkane / alcohol boils?

[Corribus]

First, combustion happens in the gas phase, so intermolecular forces don't matter a whole lot. It's also important to be explicit about what you're comparing. Do you mean combustion per molecule, per carbon atom, or what? (e.g., heat per mole, heat per gram)

Finally, you may find this thread relevant:

http://www.chemicalforums.com/index.php?topic=77107.msg281123#msg281123

[Enthalpy]

An alcohol R-OH releases less heat when burning than the alkane R-H with the same R because it has already formed a C-O and an H-O strong bonds, prior to combustion. So to say, an alcohol is an alkane already partly burnt.

As the combustion products are the same for R-OH and R-H, the difference comes from the fuels only. You can compare their enthalpy of formation with a table:
https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation
for instance ethane and ethanol.

It would be useful to observe that the heat of vaporisation, resulting from the intermolecular force, is small in comparison.