[MackTuesday]

I want to teach myself chemical thermodynamics and Wikipedia isn't too strong on pedagogy.

From reviews on Amazon it appears that the Atkins textbook on physical chemistry is losing favor to the one by McQuarrie.  I might buy it except for one thing.  I don't need a quantum mechanical treatment of chemical thermodynamics as long as classical theory is enough to give me an effective qualitative understanding of the concepts.

Here's an example of the things that are holding me up:  When the enthalpy of a reaction is negative, it's exothermic, which means the system releases heat. If the system is closed, that must mean the internal energy increases because no heat escapes. But if the system does no work, enthalpy change *is* internal energy change and internal energy change *is* heat change. dH = dU + d(pV) = dU = δQ - δW = δQ.  So enthalpy decreases, but heat increases?  How can that possibly be?

The math isn't giving me problems.  The concepts are.

So what do you think?  Atkins or McQuarrie?  Or something else?

[Yggdrasil]

I don't have a textbook recommendation (I haven't used either of the two, but I've used McQuarrie's statistical mechanics text which is good), but I can answer your question.

In general, there are three ways of changing the enthalpy of a system.  First, you can change the internal energy of the system (by doing work or transferring heat).  Second, you can change (PV).  For most ideal gas problems, these are the only two you need to consider.   However, a third way that is not often mentioned in introductory thermodynamics is by changing the chemical composition of the system (adding or subtracting molecules, or changing one type of molecule to another type of molecule via a reaction).  Therefore, the equation dH = dU + d(pV) is valid only for systems that are fixed in chemical composition.  A more complete equation is:



Where µ_j represent the chemical potentials of each chemical species in the system and N_j represent the amounts of each chemical species in the system.  In the case of an exothermic reaction, the enthalpy of the system decreases because you move from reactants with a greater chemical potential energy to lesser chemical potential energy.  Because energy is conserved, the chemical potential energy of the reactants is converted to heat which gets transferred to the surroundings.

[MackTuesday]

So chemical potential energy is included in enthalpy.  That helps!  Thank you.

If you're still feeling charitable I have a couple more questions.  If not I certainly can't blame you.

What if the system is rigid and adiabatic?  (By that I mean d(PV) = 0 and there's no heat transfer across the system boundary.)  Is it possible for enthalpy to change in that case?  It seems like we must have dU = -Σ(µ∙dN) and dH = 0, right?  Again, no work is done, so dU = δQ - δW = δQ, and the energy never changes to something other than Q or µN, right?

Or maybe I'll just buy that book so I can stop worrying and start understanding.

[Yggdrasil]

Well, if a system is rigid, it means that the volume cannot change (therefore work cannot be done), but it does not imply that Δ(PV) = 0 as the pressure can still change.  The condition for Δ(PV) = 0 is for the reaction to occur isothermally (assuming an ideal gas where the number of molecules in the system stays constant).

But, what will happen if an exothermic reaction between ideal gases occurs in a rigid, adiabatic container?  In this case, you are correct that dU = -Σ(µ∙dN) since w and q are zero.  Because the internal energy of an ideal gas depends only on its temperature, the fact that the internal energy has changed implies that the temperature of the system has changed.  So, as perhaps is expected, the system will increase its temperature.

How do we calculate the change in enthalpy?  Well, that's a bit more complicated because you would have to account both for the change in chemical potential of the system as well as the Δ(PV), but it should work out to the same value as expected under the standard constant pressure, constant temperature conditions.

[MackTuesday]

Oh, rigid doesn't mean d(PV) = 0.  Of course.  Somehow I got it mixed up with PdV.

I'm gonna go ahead and buy that McQuarrie book.  You've been very helpful, Yggdrasil.  Thanks a lot.

[McCoy]

So what do you think?  Atkins or McQuarrie?  Or something else?

Atkin, at least in my opinion. Or get that Mcquarrie thing and tell us what you think.
Besides, I think PChem by Engel and Reid, Laidler PChen are quite good too.

[MackTuesday]

I got McQuarrie and I'm going through it.  I haven't actually gotten to the thermodynamics stuff yet, but so far I like the book's presentation of quantum mechanics and its use in developing atomic and molecular orbitals.  I just thought I'd come back and let people know what I thought of the book.  I don't regret the purchase at all.

[phillyj]

Atkin, at least in my opinion. Or get that Mcquarrie thing and tell us what you think.
Besides, I think PChem by Engel and Reid, Laidler PChen are quite good too.

Engel is no good. I used for thermodynamics and it is being used for quantum. I didn't like how things were presented. I have used Laidler's text and it's not bad.

Out of all the books I have went through/read a bit (Levine, Laider, Engel, Atkins & De Julio [The Elements of Physical Chemistry],  and McQuarrie), the best is McQuarrie. It gives clear information. The Atkins book is pretty easy to understand also.

[FreeTheBee]

Both Atkins and McQuarrie treat classical thermodynamics. At my university we used Atkins for physical chemistry courses and McQuarrie for thermodynamics. Both books are good, but personally I prefer McQuarrie. The book is very complete and clearly written in my opinion.