Empirical equivalence of shifted chemical potential $\mu_i$

It is often said that, in classical thermodynamics, entropy $$S$$ and energy $$U$$ are defined only up to an additive constant proportional to the total amount of substance $$N=\sum_i N_i$$ (where the sum is over all of the system's components). That is, substituting $$S\rightarrow S+c\cdot N$$ and $$U\rightarrow U+d\cdot N$$ for $$c,d\in \mathbb{R}$$ everywhere in the theory leaves all empirical predictions unchanged.

As Gibbs already noted (cf. Gibbs "The Scientific Papers, Vol I", pp. 95-6), this implies that, for any system of uniform temperature, the chemical potentials for the system's components are only defined up to an additive factor of $$d-c\cdot T$$, where $$T$$ is the temperature. That is, shifting $$\mu_i \rightarrow \mu_i + d - c\cdot T$$ for all $$i$$ leaves all empirical predictions unchanged. (To derive this, note that, where $$F$$ is the Helmholtz free energy, $$F=U-TS$$ and $$\left(\frac{\partial F}{\partial N_i}\right)_{T,V,N_{i\neq j}}=\mu_i$$.)

Now, another well-known result is that, for any chemical reaction at equilibrium, we have $$\sum_i \mu_ib_i=0,$$ where $$b_i$$ are the stochiometric coefficients. What's confusing me is that this equation is NOT generally invariant under a shift $$\mu_i \rightarrow \mu_i + d - c\cdot T$$. For, generally, $$\sum_i b_i \cdot (d-c\cdot T)\neq 0.$$ For instance, in $$2H_2 + O_2 = 2H_2O$$, simply choose $$d=1$$ and $$c=0$$; then $$\sum_i b_i \cdot (d-c\cdot T)=1$$. Indeed, more generally, it would seem that the shifts $$S\rightarrow S+c\cdot N$$ and $$U\rightarrow U+d\cdot N$$ induce shifts in the locations of the minima of $$F$$, since they induce the shift $$F\rightarrow F+N(d-cT)$$.

How is this compatible with the claim that the shifts $$S\rightarrow S+c\cdot N$$ and $$U\rightarrow U+d\cdot N$$ leave the empirical predictions unchanged?

• It looks to me like you've shown that the more general formula must be $\sum_i\mu^\prime_ib_i=\sum_ib_i(d-cT)$, where the transformed $\mu^\prime\equiv\mu+d-cT$. Sep 28, 2023 at 18:06
• @Chemomechanics Thanks! Yes, it would seem so. I'm a bit worried though: when people say that entropy is only defined up to a term $+c\dot N$, do they assume that total amount of substance $N$ is fixed? Because otherwise the locations of entropy's maxima would generally shift by introduction of the term --- and if they shift, thermodynamics' empirical predictions would change, no? But in chemical reactions, total $N$ doesn't generally remain fixed. So I'm worried that entropy has its "gauge freedom" only if we exclude chemical reactions.
– Jens
Sep 28, 2023 at 19:21
• When there are multiple different chemical species, modifying the definition of $U$ or $S$ in a way which treats them the same seems wrong; $U'=U-d\sum_i N_i$ says we adopt a definition of energy with a term in which all particles contribute the same way to energy, irrespective of their type. This is unphysical, H$_3$O$^+$ should contribute differently than H$_2$O, already due to different mass, and here also different charge. It gets obviously wrong when $d$ gets very very large, then it overpowers the original $U$ and all particle types become almost the same in $U'$. Dec 18, 2023 at 11:44
• The possible modification seems rather $U' = U - \sum_i d_i N_i$, with each chemical species having its own $d_i$. Then all $d_i$'s may be related in a way which preserves standard results. Dec 18, 2023 at 11:50
• Similarly for entropy, unless all species have the same concentration, it seems wrong to change entropy via a term which is as if they all contributed the same amount. Dec 18, 2023 at 11:52