# Is the equilibrium constant calculation only applicable to homogeneous reactions?

In my text book, it states the general formula for the equilibrium constant is: Then it states that "The equilibrium constant expressions described here apply to homogeneous reactions."

Does this mean that it only applies to homogeneous reactions or that it can also be applied to homogeneous reactions among other types?

• That means that heterogeneous reactions don't work like that. – Mithoron Jul 25 '19 at 0:13

If you introduce the chemical activity $$a$$ in thermodynamics you can write the general equation.

$$\frac{a_C a_D}{ a_A a_B} = K$$

which is correct for any case. Under ideal assumptions we have a linear relationship between the activity and concentration in solution. So with an arbitrary normalizing reference concentration $$c_0$$ (e.g. 1 mol/l) the equation becomes roughly:

$$\frac{c_C / c_0 c_D / c_0}{ c_A / c_0 c_B / c_0} = K$$

As you see the $$c_0$$ cancel each other out and you end up with the textbook equation that consists only of concentrations.

So the equation applies if you can scale your activities linearly with concentrations. This is usually possible for homogeneous reactions and sometimes possible for heterogenous reactions (As can be seen from the other answer).

PS: Activities hence equilibrium constants are dimensionless. You always have a factor 1 (mol/l)$$^n$$ to cancel out the units of the concentrations, if you derive it correctly.

• In the initial equation you provide, do you mean a_C for the first a in the numerator? – Liam Jul 25 '19 at 13:11
• Oh yes, sorry ;-) – mcocdawc Jul 25 '19 at 13:18
• "This is only possible for homogeneous reactions." I disagree with this statement, see my alternative answer. – Karsten Theis Jul 25 '19 at 17:01

Does this mean that it only applies to homogeneous reactions or that it can also be applied to homogeneous reactions among other types?

The latter. If you consider the distribution of a solute S in a mixture of water and octanol, this is a heterogenous system (in this case, two immiscible liquids). The equilibrium reaction (or process) is:

$$\ce{S(oct) <=> S(aq)}$$

There is an equilibrium constant for this process,

$$K = \frac{[\ce{S(aq)}]}{[\ce{S(oct)]}}.$$

If I dissolve some S in the aqueous phase, some of it will go into the octanol phase until equilibrium is established. If I dilute the aqueous phase, some S will go back to the aqueous phase to re-establish equilibrium.

• Although the system is heterogeneous, wouldn't the reaction still be homogeneous as the reactant and product are in liquid states but with a solute mixed into them? – Liam Jul 25 '19 at 17:33
• @Liam Perhaps a more convincing example, from wikipedia's article on phase-transfer catalyst: "$\ce{C8H17Br(org) + NaCN(aq) -> C8H17CN(org) + NaBr(aq)}$ (catalyzed by a R4P+Br− PTC)". This is classified as heterogeneous catalysis. – Karsten Theis Jul 25 '19 at 17:39