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I know that this is a very basic question, but I am new to chemistry and I cannot figure it out on my own:

A combustion is always exothermic, so ΔH is negative. It says in my textbook that more stable molecules have a lower heat of combustion (I am assuming here that this is basically identical to stating that it has a lower enthalpy of combustion).

What I do not understand is the following: more stable molecules have a lower enthalpy of formation than less stable molecules. When undergoing combustion, stable molecules should be expelling less heat than less stable ones. So shouldn't the enthalpy of combustion of more stable molecules be higher (i.e. less negative)?

N.B. The question is related to the following: I was asked to determine which of the following two compounds has a lower enthalpy of combustion. I understand that B is more stable than A, but I am not able to translate that to "a lower enthalpy of combustion" because of my lack of understanding of the above…

enter image description here

enter image description here

Many thanks!

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I have redrawn your two structures A and B in their most stable conformations. Note that A has all equatorial methyl groups while B has two equatorial groups and one that is axial. The estimated difference in the heats of formation and combustion is 1.8 kcal/mol, the same as the difference in energy between the equatorial and axial conformations of methyl cyclohexane. Because A and B produce the same amount of CO2 and H2O upon combustion, the more stable A's heat of combustion is 1.8 kcal/mol less than that for B. Compare your A and B with their enantiomers S and T at this ChemSE site at which time I addressed a similar question.

enter image description here

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  • $\begingroup$ Thank you! So A is more stable, and therefore its enthalpy of combustion is lower, correct? Just summarising to see if I fully understood it... $\endgroup$ – dalta Jul 17 at 9:24
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    $\begingroup$ Correct. The sum of the heats of formation and combustion in this example are the same. If you have a more negative heat of formation, i.e., more stable, you have a smaller heat of combustion. $\endgroup$ – user55119 Jul 17 at 12:21
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The following schematic (not meant to represent the actual changes in $H$) illustrates the problem: enter image description here

The first step (from leftmost reagents to middle products) represents the formation reaction (not balanced), the second step (from middle to oxidation products on the right) the combustion reaction.

It pays to be careful using words such as "greater" or "less" when comparing signed quantities. For instance, stating "$H$ for B is less than for A" might not make it clear whether one means "more negative" or "smaller in magnitude".

In this schematic it is clear that $\Delta \Delta _\text{f} H(A-B)>0$, that is, A has greater (more positive) enthalpy of formation than B. On the other hand $\Delta \Delta _{\text{comb}} H(A-B)<0$, that is, the heat of combustion is more negative for A than for B. Note that where we place A and B is not important for the purposes of this illustration, where we compare the relative enthalpies qualitatively (not quantitatively). What's important is that $H_A>H_B$.

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