Heat of combustion..

[2312_Lib]

Just want to make sure this is right...

A more stable compound will have a lower heat of combustion --> which is a more negative deltaH
and will release LESS heat

because it is more thermodynamically favored, SO that means that MORE heat will be released during formation

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[Osminog]

Heat of combustion is the total energy released as heat when a compound combusts in O2.

If you're standing on a hardwood floor for 5 hours, you will feel some relief when you get a chance to sit down. If you're standing on hot coals for 5 hours, you will feel a lot of relief when you get a chance to sit down.

In both cases, you prefer to sit down (after all, combustion is an exothermic reaction), but you release more energy when you start from a relatively unstable/energetically costly state (standing on hot coals) than when start from a relatively stable/energetically favorable state (standing on a hardwood floor).

Furthermore, the original bond that is relatively unstable (standing on hot coals) would be more energetically costly to re-form than the relatively stable bond (standing on a hardwood floor). After all, it requires much more energy to get back up and stand on hot coals than to get back up and stand on a hardwood floor.

 

[PositivePhotoresist]

Another way to think about this is utilizing delta G = delta H - T delta S. Assuming similar entropy values and temperatures between two compounds, suppose we had a compound with a heat of combustion of -100 kJ/mol and another with -200 kJ/mol. If you plug these values in, you’d see that the Gibbs free energy of combustion of compound 2 is more negative. That means that it is more likely to combust because it is a more spontaneous event. This implies that the initial state of this compound is more unstable than the other one.
Generally though, your rationale is correct. Thermodynamics does connect with stability. However, there are some instances where we have to account for entropic costs.

 

[2312_Lib]

Another way to think about this is utilizing delta G = delta H - T delta S. Assuming similar entropy values and temperatures between two compounds, suppose we had a compound with a heat of combustion of -100 kJ/mol and another with -200 kJ/mol. If you plug these values in, you’d see that the Gibbs free energy of combustion of compound 2 is more negative. That means that it is more likely to combust because it is a more spontaneous event. This implies that the initial state of this compound is more unstable than the other one.
Generally though, your rationale is correct. Thermodynamics does connect with stability. However, there are some instances where we have to account for entropic costs.

Thank you! I just wanted to confirm my thinking in a different perspective!