Let $\sum_A \nu_A A \rightarrow \sum_B \nu_B B$ be a general reaction whose progress during time interval $dt$ is measured by $d\zeta$, so the amount of reactants consumed and products generated in mole would be $d N_i=\nu_i d\zeta$, wherein, $\nu_i$ would take negative values for the reactants being consumed.

From on the other hand we already know from the second law of thermodynamics that the reaction will occur in the direction in which $\sum_i \mu_i dN_i \le 0$, wherein, $\mu_i=\frac{\partial G}{\partial N_i}\bigr|_{p,T,N_j\;(j\ne i)}$ are the chemical potentials, so that we would have: $$\left(\sum_i \mu_i \nu_i\right)\; d\zeta \le 0$$ The equality implies reversibility and should denote the equilibrium (am I right?). At the equilibrium it is clear that the progression of the reaction should vanish and we should have $d\zeta=0$, so that the term $\left(\sum_i \mu_i \nu_i\right)$ should be free to gain any positive, zero or negative finite value. However, this is not what Guggenheim has written in his Thermodynamics book. He has first assumed in a given direction the reaction progresses, so that in that direction $d\zeta>0$, then has canceled out this positive quantity from the inequality and achieved: $\left(\sum_i \mu_i \nu_i\right) \le 0$, and eventually concluded that at the equilibrium the equality holds and we should have $\left(\sum_i \mu_i \nu_i\right) = 0$.

But how is it justified to be true when we already know at the equilibrium $d\zeta>0$ doesn\t hold and we should instead write: $d\zeta=0$?

Thanks for bearing with me, I'm new to chemistry.

  • $\begingroup$ please specify what is $\mu_i$ $\endgroup$
    – RE60K
    Commented Jun 29, 2014 at 8:18
  • $\begingroup$ @Aditya, their formal definition has been added. $\endgroup$
    – topology
    Commented Jun 29, 2014 at 12:08
  • $\begingroup$ Ad 1. Haven't you forgot to add a mixing entropy term? Ad 2. Maybe I misunderstand you, but I think your problem is that: It is a standard method to find a stable equilibrium point saying that the first derivative of the enthalpy etc. is zero, but even when the systems suffers a perturbation, it returns to the equilibirum. I.e. even for small, non-zero dζ the system stays in equilibrium which can only happen if (∑iμiνi)=0. $\endgroup$
    – Greg
    Commented Jun 29, 2014 at 15:27
  • $\begingroup$ @Greg, about the mixing entropy I'm not sure I have understood what you mean (sorry I'm new to chemistry), but the inequality mentioned is indeed derived from $dG\le 0$ at constant pressure and temperature. Actually I saw another book and yet the same reasoning was mentioned there, that $d\zeta$ is considered positive and so on. Hence, it seems no term has been forgotten here as the formulation is not mine. $\endgroup$
    – topology
    Commented Jun 29, 2014 at 20:05
  • $\begingroup$ @Greg, about your second point maybe you are right, I cannot judge it very easily at this point, but what you say seems more precisely addressing the question if the equilibrium achieved is a stable equilibrium. By the way, at the very equilibrium no advancement is expectable and thus we should have $d\zeta=0$. Maybe the authors are implicitly assuming that the equilibrium should also be stable, so that the term $\sum_i \mu_i \nu_i$ should be finite and preferably small, but zero is somewhat a lot idealization, or perhaps I am still wrong? $\endgroup$
    – topology
    Commented Jun 29, 2014 at 20:12

1 Answer 1


Chemical equilibrium is a stable equilibrium, i.e. the system returns to equilibrium to whatever perturbations. This also leads to the condition of equilibrium, as it should contain this information.

The short answer to your question:

Due to the stability, the system stays/returns to equilibrium at any kind of small perturbation. It also means, $\left(\sum_i \mu_i \nu_i\right)\; d\zeta = 0$ not only in the equilibrium point, but also for any small $d\zeta$. Therefore the $\left(\sum_i \mu_i \nu_i\right)$ itself must be zero.

Somewhat longer answer:

The reaction goes from one direction to the other ($\sum_A \nu_A A \rightarrow \sum_B \nu_B B$) spontaneously only till it reaches the equilibrium condition. In this direction $\sum_i \mu_i dN_i \le 0$.

We followed the reaction from a given concentration toward equilibrium. If we follow the same $\sum_A \nu_A A \rightarrow \sum_B \nu_B B$ transformation BEYOND the equilibrium point, $ \sum_i \mu_i dN_i $ must change sign at the equilibrium point, and it must be positive should be valid (since it is not a spontaneous reaction).

Therefore the condition of equilibrium is not $d\zeta = 0 $ , but sign change: $ \sum_i \mu_i dN_i =0$.


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