Vapour pressure and equilibrium constant

Question

Consider a case I where I have a container halfway filled with a pure volatile solvent. Now we know that liquid molecules will start getting converted into vapour phase, and then dynamic equilibrium will be reached. The equilibrium constant here is given by $K= [P_0]$, where $P_0$ is the partial pressure of the vapours. ($[P]$ due to pure liquids is considered as 1, so denominator is 1 and hence $K = [P_0]$).

Now we know $K$ only depends on temperature and is independent of any impurities being added, right?

Now, consider the case if I add some impurities to the pure solvent. We know there will be a lowering of vapour pressure and now if I write the equilibrium constant equation, $K = [P_1]$ , where $P_1$ denotes the new vapour pressure when impurities are present, and also $P_1 < P_0$.

So the problem is: $K$ has changed ! I mean previously $K$ was $P_0$ and now $K$ is $P_1$.

So $K$ value has changed by adding impurities. But my book says impurities has no effect on equilibrium constant. Where am I wrong?

Answer 1

Equilibrium constants contain activities of all components involved. For gases, these are proportional to partial pressures, so we use those instead. For liquids and solids, they are constant, so we tend to disregard them altogether. But the activity of a liquid ceases to be constant as you start adding impurities. All of a sudden, you have to use $K=\dfrac{P_{gas}}{a_{liquid}}$.