# Why are solids and liquids not included in the equilibrium constant? What about in a reaction rate calculation?

Take for instance the reaction

$$\ce{H2(g) + I2(s) <=> 2HI(g)}$$

The equilibrium constant would not include the solid $$\ce{I2}$$, but why is this? I have read that its concentration is a constant, but what does that mean? If you increase the amount of $$\ce{I2}$$ the reaction should shift towards the right so logically it does appear to play a part in the equilibrium.

For liquids, I can understand why you don't include it when you have something like water which (usually) does not play a part in the reaction, but why do you still not include when the liquid takes part in the reaction?

Also, do you include solids/liquids for reaction rates? Increasing the amount of a solid should increase the surface area and thus make for a faster reaction, so it seems like you should, but I am not too sure.

It very much depends on what definition of the equilibrium constant you are looking. The most common usage of the same has quite a variety of possible setups, see goldbook:

Equilibrium Constant
Quantity characterizing the equilibrium of a chemical reaction and defined by an expression of the type $$K_x = \Pi_B x_B^{\nu_B},$$ where $\nu_B$ is the stoichiometric number of a reactant (negative) or product (positive) for the reaction and $x$ stands for a quantity which can be the equilibrium value either of pressure, fugacity, amount concentration, amount fraction, molality, relative activity or reciprocal absolute activity defining the pressure based, fugacity based, concentration based, amount fraction based, molality based, relative activity based or standard equilibrium constant (then denoted $K^\circ$ ), respectively.

The standard equilibrium constant is always unitless, as it is defined differently (goldbook)

Standard Equilibrium Constant $K$, $K^\circ$ (Synonym: thermodynamic equilibrium constant) Quantity defined by $$K^\circ = \exp\left\{-\frac{\Delta_rG^\circ}{RT}\right\}$$ where $\Delta_rG^\circ$ is the standard reaction Gibbs energy, $R$ the gas constant and $T$ the thermodynamic temperature. Some chemists prefer the name thermodynamic equilibrium constant and the symbol $K$.

It is worthwhile to note, that the first definition is always an approximation to the standard definition and in the standard definition all compounds regardless of their state are included in the equilibrium.

If you are looking at a reaction that takes place in gas phase, solid materials will play a constant role, since their partial pressure will solely depend on their vapour pressure and can therefore be regarded as constant. Therefore it can be seen as a part of the equilibrium constant. (The same applies to liquids while deriving for gas phase.) If you would increase the amount of solid in the system you would still not change it's concentration in gas phase.

For the reaction $$\ce{H2 (g) + I2 (s) <=> 2HI (g)}$$ you can form the standard equilibrium constant with activities/ fugacities $$K^\circ = \frac{a^2(\ce{HI})}{a(\ce{H2})\cdot{}a(\ce{I2})},$$ with $$a=\frac{f}{p^\circ}.$$ The activity for a pure solid is normally defined as one ($a(\ce{I2})=1$) and therefore $$K^\circ\approx K = \frac{a^2(\ce{HI})}{a(\ce{H2})}$$

For concentration dependent equilibrium constants, the following assumption is to be considered: $c(\ce{H2O})\approx55.6~\mathrm{mol/L}$ and is usually always much larger than that of any other component of the system in the range where the equilibrium approximation can be used. Also in most solution based reactions, the solute itself does not directly influence the reaction. Its concentration will therefore not change (significantly) and can be included in the equilibrium constant.

You are correct in assuming that the kinetics of a surface reaction will only depend of the actual area of the surface (and of course the reactants forming the products). But here again this area is most likely to be considered as constant and will result in a scalar for the reaction rate (in the area where an accurate description of the reaction is possible).

As of Greg's comment, He is of course right. In an equilibrium process the surface area will affect both directions of the reaction, hence it will not show up in the reaction rate at all. (The only effect it might have is, that the equilibrium state will be reached faster.)

• So from what I understand, solids and liquids are considered to be constants because their "concentration" depends on their partial pressure which is independent of the amount present, right? And for the rate expression it is also ignored for the same reason and any effects due to surface area are included in the rate constant? Jul 2, 2014 at 4:33
• Yes you are correct. As long as the surface does not change, everything related to it is constant and can be included in the rate constant. Jul 2, 2014 at 5:36
• There is! Tabulated are mostly standard values at standard temperature and pressure. Have a look at the second definition and you will see the temperature dependency of $K$. Jul 2, 2014 at 6:12
• It is not true. The reaction is both direction will scale with the surface, therefore the it will not appear in the equilibrium constant.
– Greg
Jul 2, 2014 at 7:37
• @Greg I hope my edit fixes this. Jul 2, 2014 at 7:43

[OP] The equilibrium constant would not include the solid $$\ce{I2}$$, but why is this?

Let me explain this with a different example. If you have a saturated solution (e.g. lemonade with too much sugar in it) it is at equilibrium. If you add more sugar, the lemonade does not get sweeter. That tells you that the amount of solid does not matter (as long as there is some), so it also tells you it should not be a variable in the equilibrium constant.

[OP] I have read that its concentration is a constant, but what does that mean? If you increase the amount of $$\ce{I2}$$ the reaction should shift towards the right so logically it does appear to play a part in the equilibrium.

No, increasing the amount of a solid does not disturb an equilibrium.

[OP] Also, do you include solids/liquids for reaction rates? Increasing the amount of a solid should increase the surface area and thus make for a faster reaction, so it seems like you should, but I am not too sure.

Yes, more solid means a faster forward reaction. At the same time, it means a faster reverse reaction (more surface area for iodine to deposit on). For that reason, the equilibrium constant does not change. In the example above with the sugar, finely granulated sugar dissolves faster than coarsly granulated sugar, but the solubility (and the the equilibrium constant) is the same.

[OP] Why are solids and liquids not included in the equilibrium constant?

They are, in a way. Concentrations enter the equilibrium constant as relative to a standard concentration. For solutes, the standard concentration is $$\pu{1 mol L-1}$$. Speaking loosely, you take the concentration in $$\pu{1 mol L-1}$$ and drop the units. For pure substances in the liquid and solid state and for solvents, the standard concentration is $$c^\bullet$$, the concentration of the pure substance. For pure substances, this always gives a value of 1, so you might as well leave it out. Even for solvents, it will be very close to 1 for dilute solutions. For gases, on the other hand, the concentration depends on the pressure, and you have to agree on whether to use partial pressure (with a standard of 1 atm) or concentration (with a standard of $$\pu{1 mol L-1}$$) when defining the equilibrium constant.

Using concentrations is an approximation in any case. For more sophisticated treatments, you would use the activity, which is also defined as 1 for pure solids and liquids.

Solids and liquids, as long as there is some available, are fully available to the reaction in a macroscopic amount that does not limit the reaction.

This is different from a solution of a certain molarity or a gas of a certain pressure. The sparseness of solutes or gases means that the availability is limited by the concentration or the pressure. For solids and liquids, the availability is considered constant as long as some macroscopic amount is present.

It's plain and simple. We use equilibrium expression to see whose concentration has what effect on the rates of reactions. If you have a pure solid, and you increase its amount you do change the rates of reactions. But a pure solid has no concentration because it is not dissolved in any solvent (that's why we call it pure solid). So, basically you just increase the amount of your pure solid, and that will have an effect on the concentration of some other reactant or product. So, the change in concentration of that reactant or product will be seen in equilibrium constant(because that reactant or product does have a concentration). And in the equilibrium constant we see the effect of change of concentration of any product or reactant, not amount. So, the change in amount of our pure solid will have an effect of change in concentration of some other reactant or product. And we will take into account the effect of increasing the amount of solid, by seeing the change in concentration of any other resulting reactant or product in our equilibrium or rate equation. Hopefully, this intuition helps!

• "We use the equilibrium constant to see whose concentration has what effect on the rates of reactions." I'd be careful about tying the equilibrium constant to kinetics, it may confuse less educated readers. Mar 24 at 20:10