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

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?

• Forget the concentration of C. It doesn't change at all. Just why would the rate of the forward reaction increase? Because of the increased surface area? That's right, but... wait, the reverse reaction occurs on the same surface, so its rate would also increase by the same factor. – Ivan Neretin Nov 7 '16 at 17:21
• Again that argument is reductionist and boils down to "because it does". C could decompose regardless of A being present and can do so anywhere in the volume. Which means the increased surface area from adding more of A shouldn't nessicarily increase the rate of the reverse reaction by the same factor. Which means something else is going on to make that true. – Random Hero Nov 7 '16 at 17:42
• There is a subtle factor that you're overlooking, and that is how fast does the system come to equilibrium. So with lots of surface area for B the system will come to equilibrium faster, but the value of the equilibrium constant will be the same. – MaxW Nov 7 '16 at 19:09

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.

• None of that describes the physical process of why however, which is what I'm interested in. If the case of the evaporation of water it easy to see what going on physically. The equilibrium is a result of the vapor pressure which is constant wrt surface area. In a closed system that evaporation will increase the ambient pressure until it matches the vapor pressure, at which point equilibrium is reached. – Random Hero Nov 7 '16 at 23:17
• That's not any different from your example! The two gases maintain an equilibrium. – Zhe Nov 7 '16 at 23:38
• Just to note, the catalysis reaction: $$\ce{A_{(g)} + B_{(g)} + n B_{*} \overset{k}{\underset{k'}{<=>}} n B_{*} + C_{(g)}}$$ has nothing to do with $$\ce{A(g) + B(s) <=> C(g)}$$ which implies consuming the macroscopic amount of solid. Such reactions are not "extremely unrealistic". They include routine commercial metal purification, rectification of uranium isotopes and restoration of filament in halogen lamp. – sa7 Nov 8 '16 at 13:41
• That's not true. If your want to create a solid from a gas, it does not just happen. A solid implies interactions that a simple reaction equation does not capture. This is not a catalytic reaction. The surface on the left is not the surface on the right. "Cancelling" them is even dubious, but I'm trying to express the properties of a solid that the OP's equation cannot convey. – Zhe Nov 8 '16 at 14:15
• One last comment. If you think the original reaction is fine as written, consider running the reverse reaction on a single molecule of $\ce{C}$. Do you get back solid $\ce{B}$? If you think the answer is yes, please define what a single molecule solid looks like. – Zhe Nov 8 '16 at 17:42

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.