Can we measure $\Delta G$ when a reaction hasn't reach equilibrium?

Question

Suppose we have the reaction $$\mathrm{A + B \leftrightharpoons C}$$ The system starts with reactants (or products) and reaches equilibrium. Can we measure the change in Gibbs energy $ΔG$ at a moment where the system isn't in equilibrium? I mean the initial state will have some $G$ equals to $G_1$ (Gibbs free energy of the reactants). As the reaction keeps going then at some point we will have both products and reactants (still not in equilibrium). Is now Gibbs free energy still defined? Can we know measure $ΔG$ as:

$$ΔG=G_2 -G_1$$

I am asking because we say that thermodynamic variables can only calculated at equilibrium. So the above states ($1$ and $2$) must be equilibrium states. But they aren't as the reaction still keeps going.

What I want to clarify is why we are free to draw the Gibbs free energy of the system as a function of the reaction coordinate and calculate difference of Gibbs free energy at different point of the extent of reaction. How we can measure a thermodynamic potential at a point (in the extent of reaction coordinate) where the system is not in equilibrium. I looked also in this post $\Delta G$ and reaction coordinate where in an answer there is a diagram of Gibbs free energy of the system as function of the reaction coordinate.

This post imported from StackExchange Physics at 2025-01-23 14:41 (UTC), posted by SE-user Antonios Sarikas

Comments

Yes, if you accept the idea that, even if the system is not quite at equilibrium, you can still calculate its Gibbs free energy. Some people would balk at this, but, personally, not me. We do this all the time when we use the open system, time dependent version of the first law of thermodynamics, with respect to U.

This post imported from StackExchange Physics at 2025-01-23 14:41 (UTC), posted by SE-user Chet Miller

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@ChetMiller Can we say therefore that the thermodynamic potentials such as $G$, $U$, $H$ etc. can be well defined (we can measure them) even if the system is not in equilibrium? Because for example we can measure $U$ in principle (the sum of kinetic and potential energy) even if the system hasn't fixed temperature pressure etc .

This post imported from StackExchange Physics at 2025-01-23 14:41 (UTC), posted by SE-user Antonios Sarikas

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@Chet Miller so you are talking about the realm of non equilibrium thermodynamics?

This post imported from StackExchange Physics at 2025-01-23 14:41 (UTC), posted by SE-user Bob D

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I suppose that this approximation would be considered that.

This post imported from StackExchange Physics at 2025-01-23 14:41 (UTC), posted by SE-user Chet Miller

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@BobD Of course, as the minimum is approached, the rate of entropy generation decreases (I'm guessing, in proportion to the square of the overall reaction rate), so that, in the vicinity of the minimum in G, entropy generation due to deviation from equilibrium strongly approaches zero.

This post imported from StackExchange Physics at 2025-01-23 14:41 (UTC), posted by SE-user Chet Miller

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@BobD Using Nonequilibrium thermodynamics, Bird, Stewart, and Lightfoot, Transport Phenomena give the rate of entropy generation from homogeneous chemical reaction as $$\sum_{I=1}^N{\frac{\mu_ir_i}{T}}$$where $\mu_i$ is the chemical potential of species i, and $r_i$ is the molar rate of production of species i per unit volume. See Chapter 24, Section 24.1, The equation of change for entropy

This post imported from StackExchange Physics at 2025-01-23 14:41 (UTC), posted by SE-user Chet Miller

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Just to offer as many points of view as possible to the OP, let me refer to Astarita: Thermodynamics: An Advanced Textbook for Chemical Engineers, especially chapters 2–3. The answer there is: it depends. There are systems for which the free energy doesn't depend on rates of change, even out of equilibrium; and systems for which it does (for example it could depend on the rate of change of volume). In the former case you can measure it out of equilibrium to get its equilibrium value; in the latter case you can't.

This post imported from StackExchange Physics at 2025-01-23 14:41 (UTC), posted by SE-user pglpm

Answer 1

My answer is based on Astarita: Thermodynamics: An Advanced Textbook for Chemical Engineers (Springer 1990), especially chapters 2 and 3 (Homogeneous reactions). There, the answer to your question is: it depends.

The free energy $G$ (and also entropy and other kinds of free energy) is defined also out of equilibrium.

There are systems for which the free energy depends, out of equilibrium, only on the same variables that define the equilibrium of the system (Astarita calls these the "site"). For example let's say temperature $T$ and volume $V$. Then $G(t) = G[T(t), V(t)]$ at every time $t$. Let's say that at some time $t_0$ the system passes through values $T(t_0)=T_0$ and $V(t_0) = V_0$, without being in equilibrium. Then you can measure the free energy at time $t_0$, even if the system is not at equilibrium then, and the value you find is the same the system would have in equilibrium at $T_0$ and $V_0$.

There are systems for which the free energy depends also on non-equilibrium variables. For example we could have $G(t) = G[T(t), V(t), \dot{V}(t)]$, where $\dot{V}(t)$ is the instantaneous rate of change of volume at time $t$. In this case, even if $T(t_0)=T_0$ and $V(t_0) = V_0$ at some time $t_0$, the free energy doesn't have, at that time, the value it would have in equilibrium at $T_0$ and $V_0$ (which is $G[T=T_0, V=V_0, \dot{V}=0]$), because it depends on $\dot{V}(t_0)$, which is different from zero if the system is out of equilibrium.

What I wrote is just an example; the dependence could be on other non-equilibrium quantities and rates of change.

It's worth taking a look at the book, because it discusses such non-equilibrium matters, especially in regard to chemical reactions, at great lengths.

This post imported from StackExchange Physics at 2025-01-23 14:41 (UTC), posted by SE-user pglpm

Answer 2

I am asking because we say that thermodynamic variables can only calculated at equilibrium. So the above states ($1$ and $2$) must be equilibrium states. But they aren't as the reaction still keeps going.

They are in chemical equilibrium as long as the forward and reverse reaction rates are equal.

It's the same for thermal equilibrium. Two objects are in thermal equilibrium with one another if they are at the same temperature. But even though they are in thermal equilibrium, at the microscopic level energy continues to transfer between particles of the two objects. It is just that the rate of energy transfer from particles in object 1 to particles in object 2 equals the rate of energy transfer from particles of object 2 to particles of object 1 for a net transfer of energy of zero resulting in thermal equilibrium.

Hope this helps.

This post imported from StackExchange Physics at 2025-01-23 14:41 (UTC), posted by SE-user Bob D

Comments

I am not asking when the system reaches equilibrium. I am asking if we can measure the thermodynamic variables when the system hasn't reached equilibrium (when the rates are not equal). I will edit the post because I may confused you. I apologise.

This post imported from StackExchange Physics at 2025-01-23 14:41 (UTC), posted by SE-user Antonios Sarikas

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OK, let me know when it is edited.

This post imported from StackExchange Physics at 2025-01-23 14:41 (UTC), posted by SE-user Bob D

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Ok I edited. To avoid writing on comments we can continue on chat if you want.

This post imported from StackExchange Physics at 2025-01-23 14:41 (UTC), posted by SE-user Antonios Sarikas

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@AntoniosSarikas Thanks for the invite, but since your question seems to deal with the subject of non equilibrium thermodynamics, which I am not familiar with, I'll pass. Good luck.

This post imported from StackExchange Physics at 2025-01-23 14:41 (UTC), posted by SE-user Bob D