[djvan]

I'm reviewing Organic Chemistry material for the MCAT.  It has been about five years since my last organic class; I'm rusty.  The book I'm using, ExamKrackers Organic Chemistry, states the following:

"Combustion of isomeric hydrocarbons requires equal amounts of O2 and produces equal amounts of CO2 and H2O.  Therefore heats of combustion can be used to compare relative stabilities of isomers.  The higher the heat of combustion, the higher the energy level of the molecule, the less stable the molecule."

The only way I've been able to make sense of this is by thinking that lower energy bonds require more energy (heat) to break, therefore the H[Reactants] is larger than their higher energy isomers.  E.g.

Alkane + O2  CO2 + H2O + Heat

O2, CO2 and H2O are obsolete when comparing isomers, so:

Alkane  Heat


H Combustion = Σ H [Products] - Σ H [Reactants]

So a smaller heat of combustion would result from more stable isomers due to the reactants being more stable, and requiring more heat (positive number, which we subtract in above formula) to break the bonds.

I guess similarly, higher energy isomers require less energy to break bonds, and thus have a smaller Σ H [Reactants], resulting in a higher heat of combustion?

Am I thinking about this correctly?

[rwiew]

You are correct with the lower / higher energy bonds, but you went a bit wrong about how those translate into enthalpies. So Σ H [Reactants] does not tell you how much energy is required to break bonds in a molecule (or all molecules that are reactants) per se, it only tells you about the relative stability of the molecule compared to some arbitrary zero level. (strictly speaking it is the free energy G that tells you about "stability", but we assume the entropy changes here are the same for all isomers). This is why we have to calculate ΔH (or H Combustion in this case) - you must compare the change of enthalpies of product and reactants to say how much energy was required or released for that process to occur (i.e. energy needed for bond breaking and released on bond formation). In other words, enthalpy is a property of a molecule, only the change of enthalpy tells you something about making changes to that molecule, e.g. breaking bonds. Does this make sense to you?

So now in this case, if you have a more stable isomer, its own enthalpy will be lower, hence LOWER Σ H [Reactants]. Remember now that in combustion Σ H [Products] is a smaller number than Σ H [Reactants] (products are more "stable" than reactants), hence H combustion is a negative number and reaction is exothermic (heat is released). Of course you will often find H combustion values quoted as positive, that is always to mean the modulus of the value expressed in the way as you did. 

[djvan]

I think I understand. So, when less stable isomers are "combusted", they release less heat.  Which also means they have a higher heat of combustion, because the release of less heat is represented by a more positive number, correct?

[rwiew]

No. First of all - heat of combustion is a negative number, the more negative it is the more heat is released. Less stable isomers have higher H [Reactants] hence H combustion is a more negative number = the less stable the isomer, the more heat is released. Think about it in these terms: Nature wants to minimize the amount of energy, the more energy a molecule has the less stable it is. It is this additional energy that is released on combustion (as we go to the same products in all cases) that increases the total amount of energy released (i.e. makes H combusion more negative) and hence tells us that a molecule is less stable.

[Enthalpy]

The heat of combustion, and the heat of formation, are not only a matter of intermolecular bond energy.

When the reactants or products are liquid or solid (or nearly), the bonds between the molecules also count, and much. Especially at the produced H₂0 - this makes the difference between both heats of combustions.

In addition, compounds store also internal heat. This one makes most difference between isomeric alkanes, since for instance short branches don't vibrate as much as a -CH₂- inserted in a long chain. More important than the difference between bond strengths in this case.

[rwiew]

The heat of combustion, and the heat of formation, are not only a matter of intermolecular bond energy.

When the reactants or products are liquid or solid (or nearly), the bonds between the molecules also count, and much. Especially at the produced H₂0 - this makes the difference between both heats of combustions.

In addition, compounds store also internal heat. This one makes most difference between isomeric alkanes, since for instance short branches don't vibrate as much as a -CH₂- inserted in a long chain. More important than the difference between bond strengths in this case.

Sure, thank you for pointing it out. Obviously, the problem in this thread is not about specific energy modes in molecule, but rather the differences in overall energy content between isomers. I see no problem with saying "energy required to break bonds" as this does not mean bond strengths, but of course allows for differences in vibrational and rotational energies.

[Vidya]

Look at this energy diagram:
Isomeric alkane +O2------>CO2 + H2O
The products of combustion are same in all isomeric alkane
.Stable isomeric alkane will have less energy and unstable has more energy.
SoΔH combustion of stable will be less because more energy is used in breaking the stable bonds so net output of energy will be less.Unstable has weaker bonds and it is easy to break the bonds.Less energy is required to break the bonds and hence more energy will come out as combustion energy.

[djvan]

First off, a big thanks to everyone who has taken the time to respond to my question; I never expected so many replies!

Vidya, your diagram is exactly the way I was thinking, but it seems to fall apart when you apply the diagram to the formula:

ΔH Combustion = ΣH Products - ΣH Reactants

If our reactant isomer is higher energy than its counterpart, and it requires LESS energy input, and therefore a smaller positive number, ΣH Reactants would be MORE negative (I'm speaking of molecule energy level + a positive energy required for bond breaking).  Since products are the same, let's assign ΣH Products an arbitrary value of -100, which must be a more negative number than ΣH Reactants, or else the reaction would end up being endothermic.

So...

 ΔH Combustion = (-100) - [MORE negative]    ,  where more negative represents a higher energy isomer

 ... ends up being a MORE POSITIVE ΔH Combustion.  Which exactly the opposite of what your diagram depicts, and what makes sense to me.

A more negative (higher energy molecule) plus a less positive (less energy required for breaking bonds) will always be a more negative number, which we subtract from our constant value assigned to products (a negative number), always resulting in a more positive value, and therefore less heat of combustion.

...sigh...

[mjc123]

I don't think you're clear in your own mind what you mean by ΣH Reactants

If our reactant isomer is higher energy than its counterpart,... ΣH Reactants would be MORE negative

If it's higher energy, H is more positive (or less negative). That's the definition of H reactants.

ΣH Reactants would be MORE negative (I'm speaking of molecule energy level + a positive energy required for bond breaking)

Energy required for bond breaking is irrelevant. Hreactants is an energy level, not an energy change. If the reactant is less stable, H is higher, and the energy for bond breaking is lower, but you don't add these two terms together. (If you did, you would get ΣH free atoms, relative to some arbitrary zero, which would be identical for both isomers.)
Consider a Hess's law cycle:
1. C_xH_y + (x+y/4)O₂  xCO₂ + y/2H₂O; ΔH1 = ΔHcombustion = ΣH products - ΣH reactants
2. C_xH_y + (x+y/4)O₂  xC(g) + yH(g) + (2x+y/2)O; ΔH2 = ΔHbond breaking = ΣH atoms - ΣH reactants
3. xC(g) + yH(g) + (2x+y/2)O  xCO₂ + y/2H₂O; ΔH3 = ΔH bond forming = ΣH products - ΣH atoms
I hope it is clear here that H refers to an energy level, ΔH to an energy change.
Now rxn 1 = 2 + 3, so ΔH1 = ΔH2 + ΔH3, which means
ΔHcombustion = ΣH products - ΣH reactants
or alternatively ΔHcombustion = ΔH bond forming (-ve, identical for both isomers) + ΔH bond breaking (+ve)
For the less stable isomer, ΣHreactants is higher, so ΔHcomb is more negative
or (an alternative way of stating the same thing) ΔH bond breaking is less positive, so ΔH comb is more negative.
I think you were confusing ΔH bond breaking with ΣH reactants. If so, I hope it's a bit clearer now.

[curiouscat]

I think what you are confused about is only signs.

[rwiew]

Yes, you seem to be confused about signs and what ΣH's mean.

djvan:
1) Enthalpy of a substance is always 0 or positive, NEVER NEGATIVE.
2) Due to 1, sum of enthalpies of substances, such as ΣH Reactants or Products is always 0 or positive.
3) ΔH of a reaction is a difference between TWO POSITIVE NUMBERS, if that of Reactants is higher (i.e. reactants are "less stable") than that of Products, the ΔH becomes negative, as is the case with combustion.
4) From 3 it flows directly that the less stable Reactants are, the more negative ΔH of the reaction is = more heat is released.

I really can't write this any simpler, you might need to go back to the most basic things about enthalpy and enthalpy changes if you are still not comfortable with that.

[Enthalpy]

1) Enthalpy of a substance is always 0 or positive, NEVER NEGATIVE.
[...]

Could you detail what you mean? To compute heats of reaction, we commonly use enthalpies of formation of the reactants and products versus the elements in their standard state, and these enthalpies of formation are sometimes positive, sometimes negative.

If someone were willing instead to refer the compounds' enthalpy of formation to the constituent atoms rather than molecular elements, then it would always be negative.

Heat is accounted positive when the considered substance receives it. I've never seen a different convention up to now, neither in tables nor in books.

Signs do mislead users when computing heats of reaction. If one doesn't rely on them at all but instead think every time what the effect on the reaction heat is, he makes fewer mistakes.

----------

I dislike the wording "more stable" in relation with the heat of formation of a compound, because no clear relationship exists. For instance benzene is stable but nitroglycerine is unstable.

[Corribus]

@Enthalpy (this is going to be confusing)

Enthalpy of formation is an enthalpy change. I think rview is referring to the absolute enthalpy of a system. It's a rather trivial thing to say, however, since the absolute enthalpy can't be measured directly anyway, and therefore enthalpy changes are the important thermodynamic quantities in chemistry and physics. Moreover, we define the reference points as 0 by convention. We could easily define them as -4000 (whatever), in which case enthalpy values could easily be negative. Even then, though, you're implicitly using relative enthalpy values (enthalpy compared to an arbitrary reference point), not absolute enthalpies.

[Enthalpy]

We could measure heat from zero kelvin up, and the enthalpy brought in as heat from zero kelvin to the considered temperature would be positive, relative to a reference that is less arbitrary than others, and might call such an enthalpy absolute

BUT

here we're speaking about reaction enthalpy, not just about the heat capacity of one chemical compound. This one can be and often is negative, and needs a conventional reference - often the elements in their standard state an 298K, it could have been gaseous elements at zero K, or individual atoms. Especially software that claims to compute the heat of formation generally evaluates a dilute gaseous compound at zero K.

[rwiew]

Just to clarify, enthalpy is not relative and there is not arbitrary point to which it is compared. H = U + pV, it is a state function of the system. The only "relative" thing is our mathematics and it's meaning of 0. (as in lack of internal energy, pressure and volume).

Now, of course reaction enthalpy changes can be calculated from a number of things, including both enthalpies of formation of reagents and products, and the absolute enthalpies of those, and enthalpies of combustion, and enthalpies of hydrogenation.. whatever, you're just playing around with terms.

I agree the use of enthalpies of formation is a simpler idea here. All enthalpies of formation will probably be negative here (water and carbon dioxide are, oxygen is 0, carbohydrates a few I've seen in tables are negative, I imagine this will be the case for most). So we have (enthalpies of formation of products) - (enthalpies of formation of reactants) = negative - negative = negative + positive. The less stable a carbohydrate, the LESS NEGATIVE its enthalpy of formation, i.e. the more negative the enthalpy of combustion = more heat generated. As expected. djvan are you clear on this in the end?