Minimization of the Gibbs energy when number of molecules varies

[murillo137]

TL;DR Summary

Why exactly is the Gibbs energy minimized at equilibrium, even when the number of particles/molecules varies?

I have a rather basic question regarding the chemical potential ($\mu$) in thermodynamics and its relation to the Gibbs free energy ($G$). All thermodynamics textbooks I've looked at (Landau & Lifshitz, Kittel...) derive the fact that, at constant temperature $T$ and pressure $p$, the Gibbs free energy is minimized. The proof is also in Wikipedia (https://en.wikipedia.org/wiki/Gibbs_free_energy). It uses the definition $G = U +pV - TS$, such that at constant $T$ and $p$:
$$dG = dU + pdV -TdS$$.
The First Law of thermodynamics reads $dU = \delta Q - pdV + \sum_{i} \mu_{i} dN_{i}$. The Second Law reads $TdS \geq \delta Q$, giving:

$TdS \geq dU + pdV - \sum_{i} \mu_{i} dN_{i}$

$dU + pdV - TdS \leq \sum_{i} \mu_{i} dN_{i}$

$dG \leq \sum_{i} \mu_{i} dN_{i}$

If the number of particles stays constant, then clearly $dG\leq 0$, such that $G$ is minimized. My issue is, usually textbooks seem to then just assume that $G$ is minimized, even when the number of particles can change (e.g. in a chemical reaction), such that at equilibrium $dG=0=\sum_{i} \mu_{i} dN_{i}$. For a reaction like $A + B \to C + D$, one can thus derive a relation between chemical potentials:

$\mu_{A} + \mu_{B} - \mu_{C} - \mu_{D} = 0$

My issue is that in the proof the $G$ is minimized, it seems necessary to assume a constant number of particles, yet textbooks derive the relationship between the chemical potentials assuming that this is true regardless of the fact that naturally the number of particles of different types changes in a chemical reaction. How does one reconcile this? Can anyone give a rigorous proof of the minimization of $G$, even in the case where chemical reactions can occur?

 

[Chestermiller]

TL;DR Summary: Why exactly is the Gibbs energy minimized at equilibrium, even when the number of particles/molecules varies?

I have a rather basic question regarding the chemical potential ($\mu$) in thermodynamics and its relation to the Gibbs free energy ($G$). All thermodynamics textbooks I've looked at (Landau & Lifshitz, Kittel...) derive the fact that, at constant temperature $T$ and pressure $p$, the Gibbs free energy is minimized. The proof is also in Wikipedia (https://en.wikipedia.org/wiki/Gibbs_free_energy). It uses the definition $G = U +pV - TS$, such that at constant $T$ and $p$:
$$dG = dU + pdV -TdS$$.
The First Law of thermodynamics reads $dU = \delta Q - pdV + \sum_{i} \mu_{i} dN_{i}$. The Second Law reads $TdS \geq \delta Q$, giving:

$TdS \geq dU + pdV - \sum_{i} \mu_{i} dN_{i}$

$dU + pdV - TdS \leq \sum_{i} \mu_{i} dN_{i}$

$dG \leq \sum_{i} \mu_{i} dN_{i}$

If the number of particles stays constant, then clearly $dG\leq 0$, such that $G$ is minimized. My issue is, usually textbooks seem to then just assume that $G$ is minimized, even when the number of particles can change (e.g. in a chemical reaction), such that at equilibrium $dG=0=\sum_{i} \mu_{i} dN_{i}$. For a reaction like $A + B \to C + D$, one can thus derive a relation between chemical potentials:

$\mu_{A} + \mu_{B} - \mu_{C} - \mu_{D} = 0$

My issue is that in the proof the $G$ is minimized, it seems necessary to assume a constant number of particles, yet textbooks derive the relationship between the chemical potentials assuming that this is true regardless of the fact that naturally the number of particles of different types changes in a chemical reaction. How does one reconcile this? Can anyone give a rigorous proof of the minimization of $G$, even in the case where chemical reactions can occur?

See Smith and Van Ness, Chapters 10-12, Introduction to Chemical Engineering Thermodynamics.

 

[murillo137]

See Smith and Van Ness, Chapters 10-12, Introduction to Chemical Engineering Thermodynamics.

Thanks for your reply! I've checked out the reference, unfortunately it's still not clear to me.

Specifically, in Section 12.4, they use the following form of the first law:

$dU = dQ + dW = dQ − PdV$

They have also used this form in Chapter 10, to derive the relation between the chemical potentials of each component in a chemical reaction.

Is this form also valid when the number of particles of each type can vary, though? That would mean to me, naively, that for a system at constant volume and without heat exchange to its environment, it internal energy $U$ must be constant. But if chemical reactions occur in the system, doesn't the internal energy change? If an exothermal reaction occurs, for example, won't the internal energy of the system increase, although no heat has been transferred and no mechanical work has been done? Or am I misunderstanding something?

 

[Lord Jestocost]

TL;DR Summary: Why exactly is the Gibbs energy minimized at equilibrium, even when the number of particles/molecules varies?

In such a case, the system minimizes – so to speak - its “grand free energy” or “grand potential” at fixed temperature $T$ and fixed chemical potential $\mu$.

 

[murillo137]

In such a case, the system minimizes – so to speak - its “grand free energy” or “grand potential” at fixed temperature $T$ and fixed chemical potential $\mu$.

I don't think the grand potential comes into play here. I'm talking about systems of constant pressure and temperature, which are also closed in the sense that no matter can come in or out, but where chemical reactions can occur, such that the relative amounts of each chemical component can vary.

 

[Chestermiller]

I must have a different edition than you. Section 12.4 in my book is about low pressure VLE using activity coefficient data.

 

[murillo137]

I must have a different edition than you. Section 12.4 in my book is about low pressure VLE using activity coefficient data.

Oh, sorry! I meant the section titled "EQUILIBRIUM AND PHASE STABILITY" in chapter 12. In chapter 10, I also referred to the section 10.2 entitled "THE CHEMICAL POTENTIAL AND EQUILIBRIUM", where the authors also state that $dG = − S dT + V dP$ for the system, because it's closed, even though the ratio of each chemical component can still vary.

 

[Chestermiller]

Oh, sorry! I meant the section titled "EQUILIBRIUM AND PHASE STABILITY" in chapter 12. In chapter 10, I also referred to the section 10.2 entitled "THE CHEMICAL POTENTIAL AND EQUILIBRIUM", where the authors also state that $dG = − S dT + V dP$ for the system, because it's closed, even though the ratio of each chemical component can still vary.

See Eqn 10.2, where n is the total number of moles and G is the Gibbs energy per mole. The most important eqns are on the bottom of page 322 and pages 323-324.

 

[murillo137]

See Eqn 10.2, where n is the total number of moles and G is the Gibbs energy per mole. The most important eqns are on the bottom of page 322 and pages 323-324.

Thanks, indeed I had seen this equation 10.2 (unfortunately, the other pages you mentioned are very different in my edition). My issue is, the authors assume both this equation 10.2 $(d ( nG ) = ( nV ) dP − ( nS ) dT + \sum_{i} μ_{i} dn_{i})$ and equation 6.7 $(d ( nG ) = − nS dT + nV dP)$ are valid at the same time. Here is the first paragraph of Section 10.2 (THE CHEMICAL POTENTIAL AND EQUILIBRIUM):

"
Practical applications of the chemical potential will become clearer in later chapters that treat
chemical and phase equilibria. However, at this point one can already appreciate its role in
these analyses. For a closed, single-phase PVT system containing chemically reactive species,
Eqs. (6.7) and (10.2) must both be valid, the former simply because the system is closed and
the second because of its generality. In addition, for a closed system, all differentials dni in
Eq. (10.2) must result from chemical reaction. Comparison of these two equations shows that
they can both be valid only if:

$\sum_{i} μ_{i} dn_{i} = 0$
"

Of course Eqn. 10.2 is the most general, I just don't see why it's valid to assume Eqn. 6.7 is true as well. I get that the system is closed, but chemical reactions can still change the internal energy and thus $d ( nU ) \neq T d ( nS ) − P d ( nV )$

 

[Chestermiller]

It seems to me that Ean 6.7 can’t apply to a closed system with chemical reaction unless the reaction is at equilibrium.

 

[murillo137]

It seems to me that Ean 6.7 can’t apply to a closed system with chemical reaction unless the reaction is at equilibrium.

Yeah, I was also thinking the same. But then I'm not sure how to derive the relation between the chemical potentials $(\mu_A + \mu_B = \mu_C + \mu_D)$ in a reaction $A+B \to C+D$

 

[Chestermiller]

Yeah, I was also thinking the same. But then I'm not sure how to derive the relation between the chemical potentials $(\mu_A + \mu_B = \mu_C + \mu_D)$ in a reaction $A+B \to C+D$

$$d(nG)=\mu_Cdn_c+\mu_Ddn_D-\mu_Adn_A+\mu_Bdn_B$$ and $$dn_C=d\xi$$$$dn_D=d\xi$$
$$dn_A=-d\xi$$
$$dn_B=-d\xi$$where $\xi$ is the extent of the reaction.

 

[murillo137]

Thanks, this part was clear to me. But in the end, it all boils down to the statement that at equilibrium, $\sum_{i} \mu_{i} dn_{i} = 0$, but I don't really see the justification for this. Textbooks seem to go in circles, "proving" this statement with the minimization of $G$, even though they only prove that $G$ is minimized for a fixed number of particles. I guess my question is simply, why is $\sum_{i} \mu_{i} dn_{i} = 0$ at equilibrium? In a cosmology textbook I have been studying, this condition is defined as "chemical equilibrium" (i.e. reactions occur much faster than other parameters evolve), as opposed to the usual thermal equilibrium. Maybe these conditions are different?

Thanks in any case for all the help so far!

 

[Chestermiller]

Do you know what the expression for mu is for the case of a gas species in ideal gas mixture?

 

[murillo137]

Do you know what the expression for mu is for the case of a gas species in ideal gas mixture?

I don't... but I was anyway hoping for a more general derivation of the relation, since I need it in the context of cosmology, were the gases can be highly relativistic. Or maybe this formula would also apply in that case?

 

[Chestermiller]

I don't... but I was anyway hoping for a more general derivation of the relation, since I need it in the context of cosmology, were the gases can be highly relativistic. Or maybe this formula would also apply in that case?

I can’t help you there.

 

[Tazerfish]

Hm. Can't the minimisation of Gibbs free energy be derived from entropy minimisation directly, without considering the internal "details" at all?

Sure, the extra heat/enthalpy must be liberated somehow, and the specific entropy of your substances can't change without some sort of reactions; nevertheless, I don't recall the derivation of Gibbs free energy minimisation dealing with these "details".

Also, with rarified gases moving at relativistic speeds—are you sure you don't need entirely different tools?