Why is $\sum\limits_j {{v_j}{\mu _j}} = 0$ sometimes can't be achieved?

Question

In callen's thermodynamics and an introduction he says

Consider the chemical reaction $$0 = \sum\limits_1^r {{v_j}{A_j}} $$ where the $v_j$ are the stoichiometric coefficients defined in Section 2.9. However the changes in the mole numbers must be in proportion to the stoichiometric coefficients, so that $$\frac{{{\rm{d}}N}}{{{v_1}}} = \frac{{dN}}{{{v_2}}} = ... = dN$$ If the chemical reaction is carried out at constant temperature and pressure (as in an open vessel) the condition of equilibrium then implies $$dG = dN\sum\nolimits_j {{v_j}{\mu _j}} = 0$$ or $$\sum\limits_j {{v_j}{\mu _j}} = 0$$ If the initial quantities of each of the chemical components is ${N_j}^0$ the chemical reaction proceeds to some extent and the mole numbers asume the new values $${N_{\rm{j}}} = {N_{\rm{j}}}^0 + \int {d{N_j} = {N_{\rm{j}}}^0 + {v_j}\Delta N} $$ where $\Delta N$ is the factor of proportionality. The chemical potentials in equation 6.51 are functions of T, P, and the mole numbers, and hence of the single unknown parameter $\Delta N$. Solution of equation 6.51 for $\Delta N$ determines the equilibrium composition of the system. The solution described is appropriate only providing that there is a sufficient quantity of each component present so that none is depleted before equilibrium is reached. That is, none of the quantities $N_j$, in equation 6.52 can become negative.

However I was wondering why is $\sum\limits_j {{v_j}{\mu _j}} = 0$ sometimes can't be achieved.The thermodynamic system always ends in equilibrium. When it is in equilibrium( the chemical reaction is carried out at constant temperature and pressure ), $dG=0$ that seems to imply that when in equilibrium,$\sum\limits_j {{v_j}{\mu _j}} = 0$ always holds. So Where does my reasoning go wrong?

Answer 1

Brian Bi gave a good physical example of a situation where the equality $\sum_j v_j\mu_j = 0$ does not hold. I want to discuss the issue from a more formal point of view. The equality $\sum_j v_j\mu_j = 0$ is a condition for a smooth minimum of the Gibbs potential provided that all $N_i$ are positive. In fact, this is the condition that the derivative of $G$ with respect to some variable is equal to zero at the minimum point. But functions do not always have a minimum where their derivative is zero. If the problem has boundaries, then even a smooth function can have a minimum at the boundary, and its derivative will not be equal to zero at the minimum point. A simple example: $y(x) = x^3$ on the interval $[-1,1]$. The minimum of this function is found at $x = -1$ and $y'(-1)\neq 0$. In the case of chemical reactions, there are natural boundaries $N_i = 0$. Therefore, depending on the initial conditions, it is possible that the Gibbs potential has a minimum at the boundary where some of the $N_i$ are equal to zero and the equality $\sum_j v_j\mu_j = 0$ does not hold.

Answer 2

The best known example is when a solid dissolves in water (though it can be any liquid, really): we model this as the reaction X(s) -> X(aq). If there is a sufficiently large amount of water compared with the amount of X, then equilibrium is reached where no solid remains. But it is not the case that the chemical potentials of X(s) and X(aq) are equal; if an infinitesimal amount of X were added to the system, it too would dissolve, because X(aq) still has a lower chemical potential than X(s). Only if you start out with enough X(s) to saturate the solution, there will be some X(s) left at equilibrium, and the chemical potentials of the solid and aqueous phases will be equal.