Conceptual tussle with kinetics of equilibrium and non-equilibrium processes

Prologue:

Recently started with Chemical Kinetics at class. Came across a couple of points in Levine's Physical Chemistry (the book was recommended on Chem.SE; so I thought I'd give it a read :D) that I'm having serious issues wrapping my head around. This query is about the kinetics of equilibrium (and non-equilibrium) processes, and the fact that this has puzzled me for quite a while... I attribute to my (seriously questionable) understanding of equilibrium processes (courtesy: High-school education). It would be prudent to keep my shortcomings in Kinetics (newbie) and Equilibrium (possibly flawed understanding) in mind while addressing my query in your answer or the comments section. Thanks!

According to Physical Chemistry (Levine,I.N), Chapter 15, "Kinetics":

Processes in systems in equilibrium are reversible and are comparatively easy to treat. This chapter and the next deal with non-equilibrium processes, which are irreversible and hard to treat. The rate of a reversible process is infinitesimal. Irreversible processes occur at nonzero rates.

Which, if I understood correctly, can be paraphrased as:

Equilibrium processes are reversible and have infinitesimal (practically "non-existent") rates.

Non-equilibrium process are irreversible and have non-zero ("finite" or "significant"?) rates.

From my "understanding" of Equilibrium; chemical equilibrium is an example of dynamic equilibrium. When a chemical reaction, $$\ce{A + B -> C + D}$$

has proceeded to equilibrium. The "forward" rate of reaction (rate at which $$C$$ + $$D$$ is spit out) must equal the "backward" rate of reaction (rate at which $$A$$ + $$B$$ is spit out). There is no net formation of products/reactants, but both the forward and backward rates are positive and finite.

My question?

Why does Levine say that equilibrium processes have "infinitesimal" (almost non-existent) reaction rates? I thought the forward and backward reaction rates in a reaction at equilibrium are finite...

• I'm currently trying to wrap my head around @Curt's answer here [ chemistry.stackexchange.com/questions/47262/… ], since it might help here. It sounds amazing (and correct)...but it contradicts everything I've learnt at school so far. :'( – paracetamol Aug 8 '17 at 16:57
• I cannot find the corresponding passage in the 6th edition of Levine, so I can only guess, but my assumption is that the author is referring to the net rate of reaction being infinitesimal. After all, the rate of reaction is defined to be proportional to $\mathrm d[\ce{A}]/\mathrm dt$ and if the system is at equilibrium, the concentrations of all components are constant (or at most changing infinitesimally due to some reversible change in external conditions) and hence the rate is zero (or at most infinitesimal). – orthocresol Aug 8 '17 at 17:11
• @ortho It's on page 474 (section 15.1), and yeah...she could be referring to the net reaction rate. Bear with me here, but...wouldn't the reaction then no longer be in equilibrium? I mean, the net rate ought to be zero, right? Thanks! (re-reading your comment...didn't see the edit) – paracetamol Aug 8 '17 at 17:14

OK, I think I am correct about this. If not then I suppose somebody will come along and correct me anyway.

The author here is referring to the net rate being infinitesimal. To understand where the infinitesimal comes from, we need to figure out what is being meant by an "equilibrium process", or a "reversible process". On page 43, Levine describes a reversible process as

one where the system is always infinitesimally close to equilibrium, and an infinitesimal change in conditions can reverse the process to restore both system and surroundings to their initial states.

Often these are exemplified using thermodynamics processes, e.g. expansion/compression of a gas, but since we're talking about chemical kinetics, we could use an example where the temperature of the system is being increased in a reversible fashion. This means that the temperature is being increased in infinitesimal steps of $\mathrm{d}T$. The total change in temperature of the process is calculated via integration:

$$\Delta T = \int_{T_1}^{T_2}\mathrm{d}T = T_2 - T_1$$

and likewise for other properties. If something (e.g. the pressure $p$) depends on temperature, i.e. $p = p(T)$, then we can find the overall change in pressure

$$\Delta p = \int_{p(T_1)}^{p(T_2)} p(T)\,\mathrm{d}T$$

If the system is at equilibrium ($Q = K$), then an infinitesimal increase in temperature leads to an infinitesimal change in the equilibrium constant $K$, and hence an infinitesimal change in $Q$ (essentially this is due to Le Chatelier's principle, but more properly it is explained by the system moving in the direction of decreasing Gibbs free energy). This leads to an infinitesimal change in the concentrations of the reacting species, and hence the overall rate of reaction is infinitesimal.

If nothing is changing, the system would still be considered as being at equilibrium. The net rate in this case is, of course, zero. However, such a system is not particularly interesting, as there is no process actually occurring.

So, the bottom line is that for a system in equilibrium, its properties must be changing at most infinitesimally.

• Yup, this works, thanks! (I'll be keeping an eye out for other answers too...but I guess I can safely "accept" this for now :D) – paracetamol Aug 8 '17 at 17:54
• Quick question: On page 516, Levine says the rate conversion rate $J$ (and hence the reaction rate $r$) becomes zero at equilibrium...which would indicate she's talking about the net rate. However, given the way the conversion rate $J$ is defined (equation), I find this contradictory. Could you take a look at it? – paracetamol Aug 8 '17 at 20:59

I don't have Levine, but I'd assume using the term "infinitesimal" justifies an assumption of continuity and hence allows taking the derivative, you know: dX/dy. I could be wrong (and anyway did you really mean to suggest that we would be able to tell you what Levine was thinking when that was written? (I forgot my crystal ball...)). By definition the net reaction "at equilibrium" is zero. For equilibrium processes, the net reaction (rate) is infinitesimal (because what we mean by "at equilibrium" for changes in the (macroscopic) system is "near equilibrium", yeah we're totally sloppy but we don't have much choice because the concept is really hard to pin down into a simple definition.) As far as what Levine means by Non-equilibrium processes being irreversible is probably (just my guess here) about the path that the process takes. Most meaningful Thermodynamic values are path independent (for beginning students). Path independence means that no matter how you get from state A to state B, the Thermodynamic result is the same. In real world (irreversible) processes, the path you take matters. Consider a simple physical system, a ball and a hill. The potential energy difference between the ball at the bottom of the hill and the ball at the top is easy to calculate (P.E.= mg*(h2-h1)). The actual energy you use to move the ball up depends on the actual path you take. The actual path is irreversible, the ideal difference is reversible (although I didn't explain why - it has to do with it being an extremal value - a maximum or minimum).

• Alphonse I can concur with your answer (though the last bit wasn't really necessary, I guess?), yes, improved formatting would help. You didn't have to "delete" it :/ – paracetamol Aug 8 '17 at 18:04
• There is "delete" button there for you. Answer is not a comment. – Mithoron Aug 8 '17 at 18:15