# Solids and Liquids in Kinetics and Equilibrium

1. Mar 13, 2016

### UMath1

I just now read in my textbook that solids and liquids are not included in equilibrium expressions, are they included in rate laws? If not, why? Would a greater molar amount of solid or liquid increase the rate of reaction? And why isn't surface area included in Rate Laws?

Also, for the equilibrium expression I understand that the concentration of a pure liquid or solid remains constant but why does it that means it has no bearing on k? Why doesn't increasing the concentration of a reactant solid increase the forward rate of reaction?

2. Mar 13, 2016

### Bystander

What's the natural logarithm of one?

3. Mar 13, 2016

### lavoisier

* * *Hi Bystander, I was writing my reply as yours came up. I hope you don't mind my going for a more didactic / lengthy explanation. * * *
@UMath1 : Here's a simple way of looking at this; see also the caveat below.
Suppose that A and B can react in a solvent X to form C.
A is a solid and is only partially soluble in X (solubility = S), whereas B is completely soluble.
If the reaction is bimolecular, the reaction rate (mol/L of C formed per unit time) will be proportional to the product of the concentrations of the two reactants:

r = k * [A] * [B]

As long as there is solid A, its concentration in X, i.e. in the phase/solvent where the reaction takes place will be constant and = S. Therefore:

r = k * S * [B] = k' * [B]

If you add more A, S won't change, so the reaction shouldn't change rate. The reaction rate is independent of the number of moles of A.

I think this addresses your main point of why the concentration of an undissolved solid or liquid 'in itself' doesn't affect the rate.
You need to consider the concentrations in the phase where the reaction takes place.

Suppose you want to react NaHCO3 with an aqueous solution of HCl.
If you throw finely powdered NaHCO3 into the acid, you'll get a very fast reaction.
If you throw in the same mass of NaHCO3, but in the form of a compact, solid block, it will react overall much more slowly.
Still, the solubility of NaHCO3 in water is the same.
The reason why the first reaction is faster is that a finely subdivided solid dissolves faster than a compact block of the same solid.
So here you're looking at two consecutive kinetic processes: 1) dissolution of NaHCO3 into H2O and 2) reaction between HCO3- and HCl.
As you know, a sequence of reactions can only be as fast as the slowest one.
In the second case, the solid block takes more time to dissolve, thus slowing down the overall sequence.

I'm sure physical chemistry experts can give you even better examples, and I seem to remember there are cases, in catalysis for instance, where reactions occur in a special phase between liquid and solid, and there indeed I believe the surface area should be taken into account in the rate law.

4. Mar 13, 2016

### Staff: Mentor

Surface area is definitely important. Dissolving Zn in an acid will be much faster in the case of a large surface sheets than in the case a sphere.

Why we don't care? Because typically we deal with the same form of the reagent, so the difference is small (but in some cases it matters and should be taken into account).

Why it doesn't matter for the equilibrium then? Because surface area changes BOTH forward and backward reaction in exactly the same way, so their ratio stays constant.