# In dynamic equilibrium, how does the equilibrium constant relate to rate constants?

Mar 29, 2017

For dynamic equilibrium, we have that the rate of the forward reaction is equal to the backwards reaction. Thus, for the generic reaction

$a A + b B r i g h t \le f t h a r p \infty n s c C + \mathrm{dD}$,

the forward reaction is

$a A + b B \to c C + \mathrm{dD}$

and the backward reaction is

$c C + \mathrm{dD} \to a A + b B$.

Their rate laws, assuming elementary reactions, are:

${r}_{f w d} \left(t\right) = {k}_{f w d} {\left[A\right]}^{a} {\left[B\right]}^{b}$

${r}_{r e v} \left(t\right) = {k}_{r e v} {\left[C\right]}^{c} {\left[D\right]}^{d}$

but since ${r}_{f w d} \left(t\right) = {r}_{r e v} \left(t\right)$, we must have that:

${k}_{f w d} {\left[A\right]}^{a} {\left[B\right]}^{b} = {k}_{r e v} {\left[C\right]}^{c} {\left[D\right]}^{d}$

or

$\boldsymbol{{k}_{f w d} / {k}_{r e v} = \frac{{\left[C\right]}^{c} {\left[D\right]}^{d}}{{\left[A\right]}^{a} {\left[B\right]}^{b}}}$

By definition, we thus have that

$\boldsymbol{K = \frac{{k}_{f w d}}{{k}_{r e v}}}$

is the equilibrium constant for the forward and reverse reactions, and is the ratio of the rate constants for the reverse reaction and forward reaction.

Since the reverse reaction has the products as the reactants and the forward reaction has the reactants as the reactants, the equilibrium constant gives products over reactants raised to their stoichiometric coefficients.