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The question is long because I wanted to include the whole thought process.

Given the hypothetical reaction:

$$\ce{ A(s) + B(aq) <=> C(aq) + D(aq)}$$

One would obtain the equilibrium constant:

$$K_c =\dfrac{[\ce{C}][\ce{D}]}{[\ce{B}]}$$

The "concentration" of $\ce{A}$ is excluded from the equilibrium constant because its activity is taken to be unity. As such, changing the amount of solid does not affect the position of equilibrium and does not lead to changes in the concentrations of $\ce{B}$, $\ce{C}$ and $\ce{D}$. So far this makes sense to me.

However, I run into problems when I consider this from a kinetics perspective. Let us consider a water-filled container containing $\ce{B(aq)}$, $\ce{C(aq)}$, $\ce{D(aq)}$ and some excess $\ce{A(s)}$, and the system is at equilibrium.

The system would be in dynamic equilibrium, such that

$$\text{rate}_\text{forward}=k_\text{forward}\ce{[B]}$$ $$\text{rate}_\text{reverse}=k_\text{forward}\ce{[C][D]}$$ $$K_c=\frac{k_{f}}{k_{r}}=\frac{[\ce{C}][\ce{D}]}{[\ce{B}]}$$

Scenario (1): If I were to now remove the excess $\ce{A(s)}$ from the container, the concentrations of $\ce{[B]}$, $\ce{[C]}$ and $\ce{[D]}$ wouldn't change, because the expression for $\text{rate}_\text{f}$ does not contain $\ce{A(s)}$ at all. Therefore, the rate of the forward reaction is not affected by the removal of $\ce{A(s)}$ and the system continues to remain in dynamic equilibrium.

Scenario (2): However, if I set up a new water-filled container containing only $\ce{B(aq)}$, there would be no $\ce{A(s)}$ to react with, and the system would not approach the same equilibrium as above. This is despite: $\text{rate}_{f}=k_{f}\ce{[B]}$, and since $\ce{[B]}>0$, the forward reaction seems like it should proceed. Of course, this is impossible, so logically I would think that the participation of the solid should be accounted for by the pre-exponential factor.

If so, then going back to Scenario (1), would removal of $\ce{A(s)}$ not also affect the pre exponential factor for $\text{k}_f$, thereby shifting the position of equilibrium?

Essentially, how do I resolve this apparent paradox: when the system is at equilibrium, removing $\ce{A(s)}$ doesn't affect $\text{rate}_f$, but when an initial state devoid of $\ce{A(s)}$ is set up, the equation for $\text{rate}_f$ should equal 0.

In true stackexchange fashion, I have tried to resolve this issue myself. Perhaps the answer lies in some hidden way that $\ce{A(s)}$ is accounted for in the pre-exponential factor. Another way I tried to rationalise is by considering that removing $\ce{A(s)}$ from the system at equilibrium does not actually remove all the $\ce{A(s)}$ in the system, if one assumes that a small amount of dissolved $\ce{A(aq)}$ might have entered the solution. But then how would this explanation work for completely insoluble substances? And how would this explanation work if the system was gaseous instead, such that $\ce{ A(s) + B(g) <=> C(g) + D(g)}$, and $\ce{A(s)}$ is a completely non subliming substance like a chunk of metal?

*Important note: please don't answer using considerations of activities and chemical potentials. That would be missing the point. The issue I am having is when considering this problem from a kinetics perspective. Also, I would appreciate an answer considering both the mathematics and the possible conceptual errors.

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    $\begingroup$ Systems with more than one phase are called "heterogeneous". The kinetics that describe them can be much more complicated than the basic equations you have written, which assume homogeneous systems (though they can be applied to heterogeneous systems that meet certain criteria). Simply search for "kinetics of heterogeneous systems" and you will find more information than can be written in an answer here. $\endgroup$
    – Andrew
    Oct 11 '21 at 12:54
  • $\begingroup$ @Andrew thanks for the insightful comment. It seems I do not yet have the knowledge to sufficiently grasp the problem. However, (this may sound ignorant), it seems to me that heterogenous systems need not necessarily invoke such complications; a reaction A(s) + B(aq) -> C(aq) is adequately modeled by rate=k[B]. Using this same model equation, can it explain why rate=0 when the initial state of the system has no A(s) at all? $\endgroup$
    – Heat
    Oct 11 '21 at 19:01
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    $\begingroup$ Why do you believe that "a reaction A(s) + B(aq) -> C(aq) is adequately modeled by rate=k[B]."? This is only true in a small subset of cases where other criteria are met, for example the surface area of the phase interface remains constant, and even in those cases, properties of A (such as surface area) are included in k such that k goes to zero when A is absent. $\endgroup$
    – Andrew
    Oct 11 '21 at 19:34
  • $\begingroup$ Thanks. I guess that is the essence of my question - yes, in that case k would go to zero, such that rate=0 when no initial A(s) is present. But this would contradict scenario (1) in my question, where removal of A(s) does not cause rate of forward rxn to drop to 0. I guess we then return to your first comment asking me to learn kinetics of heterogenous systems; I could be patient and slowly build up my learning to reach that stage. On the other hand I also see this as an opportunity for some expert here to provide an incisive answer to resolve this question's mathematics and concepts $\endgroup$
    – Heat
    Oct 11 '21 at 20:23
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    $\begingroup$ It would not contradict scenario (1), since in scenario (1), the forward rate does indeed drop to 0. What you're missing is that such a scenario can only occur if the surface area of A also appears in kr, so both forward and reverse reactions go to 0 and the equilibrium concentrations do not change, which was explained in the answer you got to this question yesterday (see first sentence). $\endgroup$
    – Andrew
    Oct 11 '21 at 23:21
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Both the forward and the reverse rate are proportional to the surface area of A. Steelwool rust faster than a block of steel. Crystals grow faster with microcrystals instead of a single crystal in the mother liquor. Once you update your rate expressions to include the surface area of A in contact with the solution, it should all make more sense.

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  • $\begingroup$ Thanks. But unless I am missing something, that would not fully resolve the issue, since in Scenario (1), after removal of A(s), the surface area should be 0 and it would seem rate(forward)=0<rate(backward), which contradicts the fact that dynamic equilibrium should not be upset. Perhaps it would help if it you could show me how surface area is accounted for in the rate equation mathematically. Thanks! $\endgroup$
    – Heat
    Oct 11 '21 at 6:23
  • $\begingroup$ If you remove all of A, there is no longer an equilibrium, and both rates are zero. $\endgroup$ Oct 11 '21 at 11:48

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