The Kinetic Basis of the Equilibrium Constant

Reaction equilibrium was one of the most mystifying concepts when I was first learning chemistry. The rules for deriving the equilibrium formula for a reaction seemed counter intuitive although straightforward and easy to learn. Here is a simple hypothetical reaction between gases A and B that we will examine to determine the equilibrium constant:

$$2A + B \rightleftharpoons A_2B$$

Any astute chemistry student will know that the equilibrium constant is written as follows:

$$K = \frac{[A_2B]}{[A]^2[B]}$$

An interesting observation is that what used to be addition in the reaction equation is now multiplication, and the right side of the reaction is divided by the left.

This transformation might seem arbitrary but it is a fairly obvious consequence of reaction kinetics. Recall that we can write a rate equations for the forward reaction like so:

$$r_f = k_f[A]^2[B]$$ $$r_r = k_r[A_2B]$$

where $r_f$ and $r_r$ are the forward and reverse reaction rates and $k_f$ and $k_r$ are the rate constants for those reactions. We can make the same observation for these equations as we did when calculating the equilibrium constant: adding reagents together manifests itself in the rate equation as multiplication.

Now why is this the case? Well, what the rate equation is trying to describe is the amount that the particles collide with one another to combine and form the product. Since all of the particles must be near one another this problem can be seen as the probability that 2 As and 1 B are near one another. Now since concentration is similar to the probability that the particles, $[A]^2[B]$ is similar to the probability that the reagents will be near one another. An additional constant is needed to account for the difficulty for particles to combine once they are actually near one another.

The equilibrium constant can be seen as the tendency of the reagents and products of the reaction to be on the right side of the equation. Since the reaction rate defines how fast the reagents turn into their products one can observe that a good notion of the equilibrium constant might be to simply divide the reverse reaction rate by the forward reaction rate like so:

$$\frac{r_r}{r_f} = \frac{k_r[A_2B]}{k_f[A]^2[B]}$$

You might notice that this is very similar to the equilibrium constant that we wrote down earlier. This is no coincidence. In fact, the two rate constants are the same, which means that $K = \frac{r_r}{r_f}$. To figure out why $k_r = k_f$ we need to first understand the principle of microscopic reversibility. This states that all processes are symmetrical with respect to time; the consequence of this is that the forward and reverse reactions proceed by the same mechanism and therefore take the same time and proceed at the same rate with the same reaction coefficients at equilibrium.

I brushed over quite a few details that might be a problem in more difficult chemical reactions but the concept is quite simple. At least for me figuring out this connection made equilibrium constants much more intuitive and helped me understand how to apply them better.