# How do the rates of the forward and reverse reactions compare at a state of dynamic chemical equilibrium?

May 4, 2018

They are equal, and NONzero.

At dynamic chemical equilibrium, the rates of the forward and reverse reactions are equal to each other, i.e.

$a A + b B \stackrel{{k}_{1} \text{ }}{r i g h t \le f t h a r p \infty n s} c C + \mathrm{dD}$
${\text{ "" "" "" }}^{{k}_{- 1}}$

For this, assuming the equilibrium consists of elementary reactions, the forward rate law and reverse rate law are:

${r}_{1} \left(t\right) = {k}_{1} {\left[A\right]}^{a} {\left[B\right]}^{b}$
${r}_{- 1} \left(t\right) = {k}_{- 1} {\left[C\right]}^{c} {\left[D\right]}^{d}$

At equilibrium, ${r}_{1} \left(t\right) = {r}_{- 1} \left(t\right)$, so:

${k}_{1} {\left[A\right]}^{a} {\left[B\right]}^{b} = {k}_{- 1} {\left[C\right]}^{c} {\left[D\right]}^{d}$

From this, we obtain:

$K \equiv {k}_{1} / \left({k}_{- 1}\right) = \frac{{\left[C\right]}^{c} {\left[D\right]}^{d}}{{\left[A\right]}^{a} {\left[B\right]}^{b}}$

We know that rate constants are temperature-dependent, and thus, so is $K$.

It is also important to note that the rates of the forward and reverse reactions MUST be nonzero to have a dynamic chemical equilibrium.