# Low or high enthalpy of combustion in relation to stability

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…

Many thanks!

The following schematic (not meant to represent the actual changes in $$H$$) illustrates the problem:
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 $$H_A>H_B$$.