# Why do we use dG < 0 to describe a spontaneous process?

With something like a reaction or phase change, all sources use the criterion that $dG < 0$ for the reaction to be spontaneous and then substitute an appropriate expression for $dG$ to the specific application. For phase change my book stats $(\mu'-\mu'')dn' < 0$ where $'$ is phase A and $''$ is phase B, $\mu$ is the chemical potential and $dn' > 0$ means gaining of molecules in phase A.

However, reactions require activation energy and phase change require nucleation (I think this is kind of like an activation energy); both of these processes require an increase in $G$ before a larger decrease. Why do we not use $\Delta G$ (between initial and final equilibrium states) instead?

Edit:

I am confused by the use of dG (a rate differential) as opposed to $\Delta G$ (a net change between end states) because I'm not sure how one would integrate this for say, a phase change, since it must pass through a potential barrier where G must increase. Is the integral just $G_2-G_1$ between the end states and thermodynamically the process ignores the barrier (since thermodynamics deals with equilibrium only)? A typical thermodynamic process I'm imagining is quasistatic compression, where we can get work from the integral of PdV and P is defined as it goes through an infinite number of equilibrium states; I'm not sure if it's possible to draw a parallel here for free energy and phase change/reacting systems.

Edit 2:

From this graph it looks like G increases then decreases to stable equilibrium after phase change. If we find dG/dr from this graph and integrate dG(r), should we get the same $\Delta G$ as from thermodynamics (before and after phase equilibrium)? If so, this seems to me like finding work of a rapid piston-cylinder compression of an insulated system; PdV is undefined since we are not passing through a set of equilibrium states but we can still find work from $\Delta U$. Difference is that in this case, our initial state is technically not in equilibrium/stable since the system favours a phase change, so we are beginning from a state that technically does not exist on the phase diagram (like if we have a supercooled/superheated liquid).

• Activation barriers are kinetic phenomena. In pure thermodynamic considerations they are not included. There only the free energy of the initial and the final state are considered. The way which your system takes to get from initial to final is of no concern. – Philipp Oct 16 '14 at 20:35
• definitely +1 to Philipp here ... I would say though you can compare thermodynamically the transition complex .. but as I said in the other question it is a mix up between Kinetics and thermodynamics :) – AngusTheMan Oct 16 '14 at 20:45
• @Philipp I suspected that I was mixing up concepts. My confusion is that thermodynamics deals with equilibrium initial/final states (compressing piston from two equilibrium states etc.). However, phase transitions for instance only happen when the system is dis-equilibrium (i.e. supercooling liquid), how do we define an initial equilibrium state for such a case? – Yandle Oct 16 '14 at 21:46
• @Yandle When judging the spontaneity of a process from the thermodynamics perspective you compare $G$ for an initial equilibrium state to $G$ for a final equilibrium state. "Usual" thermodynamics is not able to describe non-equilibrium states . – Philipp Oct 16 '14 at 22:00
• @Philipp I am confused by the fact that dG = 0 when a system is in equilibrium, but dG is not 0 when the system say, has too much of phase A than phase B (chemical potentials do not balance). In that sense, I am confused about how the initial state is in equilibrium. – Yandle Oct 16 '14 at 22:03

Maybe its best to see how we generally derive the Gibbs energy... If you think about this then the answer should be clear ;)

The Gibbs potential is a function of $P$, $T$ and $N$. That is to say: \begin{equation} G=G(T,P,N) \end{equation} Where $T$ is the temperature, $P$ is the pressure and $N$ is the number of particles. \begin{equation} dG=\frac{\partial G}{\partial P}\bigg|_{T,N}dP+\frac{\partial P}{\partial T}\bigg|_{P,N}dT+\sum _{i=1}^{I}\frac{\partial G}{\partial N_i}\bigg|_{P,T,N_j\neq i}dN_i \end{equation} Use some Maxwell relations (Look these up): \begin{equation} dG=VdP-SdT+\sum ^{I}_{i=1}\mu _i dN_i \end{equation} By definition we state the following formula at constant temperature and pressure: \begin{equation} dG=\sum _{i=1}^{I}\mu _idN_i \end{equation} Therefore you are actually just expressing the Gibbs energy in a different way when your book quotes the chemical potential. In fact the very definition of the chemical potential is:

\begin{equation} \mu _i=\bigg(\frac{\partial G}{\partial N_i}\bigg)_{P,T,N_{j\neq i}} \end{equation}

I don't know your mathematical ability but if it helps...its pretty clear that this is just a simple Legendre transform of the internal energy to switch between $U$ and $G$. (As an aside), with a little product ruling we get: \begin{equation} dG=\sum _{i=1}^{I}\big(\mu _idN_i+N_id\mu _i\big) \end{equation} This leaves us with the Gibbs-Duhem relation: \begin{equation} \sum ^{I}_{i=1}N_id\mu_i=-SdT+VdP \end{equation} Which is a handy result.

Also just to add on the end here about nucleation points... lets imagine a precipitation in solution....As the size of the nucleus grows (as more bits "clump" together) the Gibbs energy of the particle will rise ... this is due to two competing effects:

1) the interface interaction (increasing energy)

2) the volume free energy (decreasing energy)

The total energy is just the resultant of these processes ... at small size the surface interactions outcompete the volume energy liberation .... as the radius of the particle increases we see a switch.

• I follow the math, but it's the logic I'm having trouble with. I edited my original question to make it more clear. – Yandle Oct 17 '14 at 5:50
• Is this first or second year university chemistry? (I read about chemistry in my spare time and don't think I've ever come across this so I'm a little curious). – Sherlock Holmes Oct 17 '14 at 6:35
• @SherlockHolmes This is usually introduced in 1st year and expanded upon in second year...at least in the UK. – AngusTheMan Oct 17 '14 at 10:02
• A really good book on classical thermodynamics is by Finn ... I read that before I tacked statistical mechanics ... It helped a lot .. – AngusTheMan Oct 17 '14 at 10:07