# What are some common mistakes students make with dynamic equilibrium?

May 26, 2016

There is a persistent belief that equilibrium represents the cessation of chemical change, which of course is erroneous.

#### Explanation:

For the given reaction, there is a both a forward rate, ${k}_{f}$, and a reverse rate, ${k}_{r}$:

$A + B r i g h t \le f t h a r p \infty n s C + D$

Depending on the equilibrium, the products or reactants may be favoured. As chemists, as physical scientists, we recognize that at equilibrium, the forward and reverse rates are EQUAL, i.e.:

${k}_{f} \left[A\right] \left[B\right] = {k}_{r} \left[C\right] \left[D\right]$.

Given this equality, we may write: $\frac{\left[C\right] \left[D\right]}{\left[A\right] \left[B\right]}$ $=$ ${k}_{f} / {k}_{r}$.

Now, clearly by definition, ${k}_{f} / {k}_{r}$ is a constant. And, by definition, ${k}_{f} / {k}_{r} = {K}_{c}$, where ${K}_{c}$ is the equilibrium constant for the reaction. The value of ${K}_{c}$ must be established must be established by measurement, i.e. experiment.

So what's the lesson to learn? That equilibrium does NOT represent the cessation of chemical change, but rather equality of forward and reverse rates. Should the equilibrium be perturbed somehow (say be removing the products), the equilibrium will shift towards the PRODUCT side so as to re-establish the equilibrium. This is the origin of Le Chatelier's principle: $\text{when a system at equilibrium}$ $\text{is subject to an external perturbation, the equilirium will }$ $\text{move so as to offset the external perturbation}$. Note that here, $\text{offset is not equivalent to counteract!}$

So the message? Equilibrium means equality of the forward and reverse rates.