Does the "extent of reaction" have a physical meaning?

Question

I am going to argue that it really doesn't have a useful or "comprehensible" physical meaning.

Lets say we have the following reaction: $$3A\rightarrow2B$$

Chemists define the extent of reaction as:

$$\xi=\frac{n_{A_f}-n_{A_i}}{v_A}$$ where $n_{A_f}$ is the moles of $A$ once the reaction reaches equilibrium, $n_{A_i}$ is the moles of $A$ before the reaction began, and $v_A$ is stoichiometric coefficient of A for this reaction.

The numerator makes physical sense, it is simply how many moles of A are consumed over the course of reaction. The denominator makes sense because it states the amount of molecules needed to be consumed to generate the product.

But once you put the numerator over the denominator, it starts to lose any physical meaning to me and simply becomes a useful mathematical conversion factor. If there is an exact physical meaning here, what is it?

P.S. It has the units of moles which is weird because it doesn't equal the change in moles for this reaction or the amount of moles consumed and produced on either side.

Answer 1

The stoichiometric coefficients $\nu_i$ are arbitrary anyway; the only thing that matters is the ratio of them. $\ce{3A -> 2B}$ and $\ce{6A -> 4B}$ both describe the same physical process occurring (unless it is implying something about mechanisms, but in this context of thermodynamics, that's not the case). Clearly, even for a single unchanging system, its value of $\xi$ will vary depending on how you define the stoichiometric coefficients. So, I wouldn't expect $\xi$ to have any physical meaning. Personally (in my limited knowledge of thermodynamics) I just view it as a useful, but ultimately arbitrary, definition.

[In the same vein, I would say that the magnitude of the equilibrium constant $K$ doesn't have any physical meaning either, since it also depends on how you define the stoichiometric coefficients. But I digress.]

Answer 2

Suppose that there is a reaction at constant temperature and pressure $\ce{A <=> B }$ then the free energy change is given by considering the changes in the chemical potential $\mu_A,~\mu_B$ and number of moles $dn_A,~ dn_B$ as reaction proceeds, i.e. $dG=\mu_Adn_A+\mu_Bdn_B$. The quantity $dn_A$ is negative and $dn_B$ positive.

The extent of reaction $\xi$ (unit mole) is defined so that it is zero at the start of a reaction (all reactants) and 1 when one mole of reactants has converted into products. When $\ce{A <=> B }$ the change in extent of reaction is $d\xi = dn_B=-dn_A$ and so $dG=(\mu_B-\mu_A)d\xi$. The reaction will proceed until the change in free energy is zero, $\left (\frac{\partial G}{\partial \xi} \right)_{T,p} = \mu_A-\mu_B=0$ or $\mu_A=\mu_B$.

As an example of using $\xi$, the chemical potential for a perfect gas is, $\mu =\mu^{\mathrm {o}}+RT\ln(p)$ then $$ \left(\frac{\partial G}{\partial \xi} \right)_{T,p} =\mu_B^{\mathrm {o}}-\mu_A^{\mathrm {o}}+RT\ln \left(\frac{p_B}{p_A} \right)$$

Often the (confusing) notation $$\Delta_rG =\left(\frac{\partial G}{\partial \xi} \right)_{T,p}$$ is used, and is confusing because in thermodynamics using $\Delta$ usually means a simple difference not a derivative. At equilibrium $$\Delta_rG = \left(\frac{\partial G}{\partial \xi} \right)_{T,p}=0 $$ and so the familiar expression $\Delta G= -RT\ln(K_p) $ is obtained where the equilibrium constant is $K_p= (p_B/p_A)_{eq}$

Notes: The extent of reaction does simplify some calculations but the same results are obtained if $dn_A, ~dn_B$ etc. are used.Then it is necessary to express all the changes in $dn_i$ terms of say $dn_A$ and then calculate $\left(\frac{\partial G}{\partial n_A} \right)_{T,p}$ etc.

In the definition of chemical potential the pressure is understood to be divided by 1 bar so that $\ln(p)$ is dimensionless.

Some authors definitions of $\xi$ refer to the change from the initial to equilibrium amount.