[enz1]

Hi all,

Under constant temperature and pressure G = H + TS, where TS reflects the energy stored in disordered particles (thus having no ability to be transformed into work) and H reflects the energy stored with the potential to do work (by uniform motion).

Why is then that dG = dw' (w' = non-expansion work) instead of dH = dw' ?

[Juan R.]

1)
Gibbs is defined as a double Legendre transformation from U:

U(S,V,N) --> G(T,p,N)

G == U + pV - TS     (notice the minus sign)

differentiating

dG = dU + pdV + Vdp - TdS - SdT

dG = (TdS - pdV + mudN) + pdV + Vdp - TdS - SdT

dG = - SdT + Vdp + mudN

the expansion work term (-pdV) has gone.

2)
Enthalpy is defined as a single Legendre transformation from U:

U(S,V,N) --> H(S,p,N)

H == U + pV

differentiating

dH = dU + pdV + Vdp

dH = (TdS - pdV + mudN) + pdV + Vdp

dH = TdS + mudN + Vdp

the expansion work term (-pdV) has gone too.

3)
Notice that you can also first obtain H from U and then G from H

U(S,V,N) --> H(S,p,N) --> G(T,p,N)

In that case H == (U + pV)

G == H - TS

and the conclusions are the same than in 1) because

G == H - TS = U + pV - TS

[tamim83]

Why is then that dG = dw' (w' = non-expansion work) instead of dH = dw' ?

Remember that this is an inequality, that is dG < dw'.  The Gibbs free energy describes the maximum non-expansion work that can be produced by a chemical transformation.  

In order to calculate this, you need to consider the contributions from making and breaking bonds (which is included in the enthalpy term) and the change in entropy of the system.  Both contributions will contribute to the calculation of maximum non-expansion work.  

As an aside, you also need to consider both contributions when determining whether or not a chemical reaction is spontaneous.  Enthalpy alone will not cut it.  There are some "endothermic reactions" that are indeed spontaneous.  

[enz1]

Thanks a lot for the answers.

Regards