[disillusioned19]

I'm looking for a derivation which tells me that the Gibbs free energy, as well as being an indicator of spontaneous change, also tells me the maximum non-expansion work that can be done by a system.

In Atkins' Physical Chemistry, it states that the enthalpy can be written as dH = dq + dw + d(pV) where dw is non-expansion work. I thought, however, that dq was the definition of dH at constant pressure, so how come we see both terms in this description?

Perhaps my concept of enthalpy is wrong: I always thought that enthalpy was a means of keeping track where heat transferred from a system went, i.e. either into internal energy or expansion work ( or even non-expansion work). Therefore, I'd expect to see dU instead of dq, however no such luck!

Thanks in advance.

[Zeppos10]

this reply is a bit late, but anyway:

we know ∆U = Q + W = Q + W' + W" where W" = useful work = shaft work = technical work. (indicated as w_e by Atkins in eqn 2.3.6 in my edition) W' is work done against the atmosphere = -p'∆V, if p' is the external=atmospheric pressure. Now define H=U+p'V (!!!!) and the first law becomes: ∆H= Q+W" . (this differs from Atkins)
If no useful work is done (W"=0) it follows: ∆H=Q.
Note how important the 'constant pressure' condition is: it makes p=p', if you didn't do that at the beginning.

this does not answer your question about G. it helps you to understand H as a physical quantity: that the first step.

[Bryan Sanctuary]

First to correct.  The Gibbs energy is NOT an indicator of spontaneous change.  It is only that at constant T and P.  If one or the other is not constant, then the Gibbs energy does not tell us about spontaneity.  (At constant T and V, for example, the Helmholtz energy must be used, like in a bomb calorimeter). 

The second thing you have a bit wrong is that the enthalpy is equal to the heat ONLY at constant pressure:

dH = dU + d(PV) = dq + dw + d(PV) 

or at constant P

dH = dq_P +(-PdV) + PdV, or dH = dq_P.

I put a subscript P on dq because it is the heat at constant pressure. 

So you have to be careful and always be aware which variables are fixed when using thermo properties.

Let us suppose you are doing an experiment on a bench top, so P and T are constant.  The Gibb's energy is given by

G = H-TS

and the enthalpy is given by

H = U +PV

So put them together:

G = U + PV -TS

Now suppose we are at constant P and T and look at a change:

dG = dU+PdV-TdS

But you want no PV work, so

dG = dU-TdS=dq_P+dw_{non-PV work)}-TdS

Now the most efficient path between any two states is the reversible path.  For a reversible path you must have

dq_P=TdS

and so you find:

dG = dw_{non-PV work)}  (only at constant T and P)

which I hope answers your question.

This is discussed in section 3.7 of my Physical Chemistry text book,

http://www.mchmultimedia.com/store/Thermodynamics.html