What does the equilibrium constant mean for reactions with very small concentrations?

When you calculate the equilibrium constant for a reaction with a higher concentration of reactants, the answer is fairly intuitive. For example, for the reaction:

$$\ce{A + B <=> C}$$

when we have $$\pu{0.10 M}$$ of A, $$\pu{0.30 M}$$ of B and $$\pu{0.45 M}$$ of C the equation for the equilibrium constant is:

$$K = \frac{[\pu{0.45 M}]}{[\pu{0.10 M}][\pu{0.30 M}]} = \pu{15M^{-1}}$$

This is relatively intuitive, we can see that the concentration of C is larger than the product of A and B and so saying the products are favoured makes sense.

However, an issue comes about when you consider really small amounts of all of them. For example, for the same reaction

$$\ce{A + B <=> C}$$

If we have $$\pu{2.4E-4 M}$$ of A, $$\pu{2.4E-4 M}$$ of B and $$\pu{7.5E-6 M}$$ of C we have the equation:

$$K = \frac{[\pu{7.5E-6 M} ]}{[\pu{2.4E-4 M} ][\pu{2.4E-4 M} ]} = \pu{130 M^{-1}}$$

In this case, the equilibrium constant suggests the products are favoured but if you look at how much product we have compared to reactants there's so much more reactants than products. The constant makes sense since multiplying two numbers less than 1 will get a smaller number but it's confusing when you consider the implications that has on the equilibrium. My immediate assumption is that I am misunderstanding the relationship between the equilibrium constant and the concentrations of species in the reaction and if this is the case is there a more intuitive way of looking at this?

• You have it the wrong way round. At a given temperature for any equilibrium $K$ has a fixed value, i.e. is constant. The concentrations must always match this so you cannot just guess arbitrary values. The extent of dissociation changes to ensure concentration ratios agree with $K$. Oct 11 '20 at 8:06
• @ Nathan. I think you simply forget the units. The equilibrium constant has a unit. For your system, the equilibrium constant is not $130$, as you state. It is $130~ L/mol$ Oct 17 '20 at 20:09

• if you want to find the concentrations, you should add them, why did you multiply? As @porphyrin said in the comments to the question, concentrations of chemicals will change to satisfy the value of $K_{eq}$, not the opposite Jul 15 '21 at 14:09