Why does increasing the available surface area of a pure solid not effect equilibrium?

Question

Take a reverseable reaction of a gas and a solid, $$\ce{A(g) + B(s) <=> C(g)}$$

The rate of the forward reaction will depend on the surface area of the solid available. If I take that system at equilibrium and add more of the solid, especially if it's in a form with a large surface area this should change the rate of the forward reaction, but this doesn't change the equilibrium. The closest to an answer I've found is that the rate of the reverse reaction increases by an equal amount so that $K$ remains constant, but no justification for that is given.

Increasing the rate of the forward reaction increases the concentration of $C$, which should increase the rate of the reverse reaction, that makes intuitive sense. But I can't see any obvious reason why that should be equal. Presumably my textbook is being a bit reductionist here.

Is it making a simplifying assumption and not bothering to mention that assumption? (the reactions are being considered as independent of surface area? There are unconsidered phenomenon involved where the activity of a pure solid can't be treated as unity and we're only dealing with an approximation?).

Or are there more complex factors at play that result in the rate of the reverse reaction increasing by the same factor as the rate of forward reaction, and the book just doesn't bother to discuss them?

Answer 1

You can change the amount of $\ce{B}$ available and the forward rate, but that doesn't change the rate constant. And the equilibrium constant is directly related to the rate constants of the forward and reverse reactions:

$$K = \frac{k}{k'}, \text{where }k\text{ is forward rate cosntant, }k'\text{ is reverse rate constant}$$

As @MaxW points out, you'll get to equilibrium faster, but it's the same equilibrium.

Edit:

Here's another way to think about it. Consider the reaction: $$\ce{H2O_{(l)}<=>H2O_{(g)}}$$

Does the equilibrium between liquid water and water vapor change if you increase the surface area of a pool of water? No, because even though evaporation is faster, there's also more area for condensation to take place.

In your specific case, the reverse reaction from $\ce{C}$ needs to interact with some amount of solid to redeposit $\ce{B}$. Otherwise, you're making $\ce{B_{(g)}}$ and that's not the same thing...

Edit 2 (for OP's comments): Notice that your reaction is actually extremely unrealistic. Gases don't just form solid or liquids on their own; they need some nucleation seed. Otherwise, there would be no point in seeding rain clouds.

As @IvanNeretin points out, invoking microscopic reversibility, you'd need to invoke these nucleation points in your reaction, forward and backward. These points are of course whatever surface is available on $\ce{B}$.

This makes the reaction: $$\ce{A_{(g)} + B_{(g)} + n B_{*} \overset{k}{\underset{k'}{<=>}} n B_{*} + C_{(g)}}$$

$\ce{B_{*}}$ refers to the surface. Notice that it cancels out so it doesn't affect the equilibrium constant, but having more surface area will increase the forward rate if you increase the surface. It will also increase the reverse rate, which is why the equilibrium constant can stay the same.

Answer 2

If you pulverize B (s) to have more surface area, the system will still reach the same equilibrium. All it'll do is make the system reach that equilibrium faster. But it's still the same equilibrium.