# Explain what it means for a system to be in equilibrium. How do we express this in a chemical reaction (give an example)?

May 15, 2017

A system in equilibrium indicates that the rate of some forward process within the system is equal to the rate of the corresponding backward process.

This is true whether it is a phase equilibrium, thermal equilibrium, dynamic reaction equilibrium, etc.

An example of a chemical reaction equilibrium is:

${\text{N"_2"O"_4(g) stackrel(k_1" ")(rightleftharpoons) 2"NO}}_{2} \left(g\right)$
${\textcolor{w h i t e}{a a a a a a a}}^{{k}_{- 1}}$

where the $r i g h t \le f t h a r p \infty n s$ indicates that the forward process

${\text{N"_2"O"_4(g) stackrel(k_1" ")(->) 2"NO}}_{2} \left(g\right)$,

and the backward process

$2 {\text{NO"_2(g) stackrel(k_(-1)" ")(->) "N"_2"O}}_{4} \left(g\right)$,

are equal in rate.

These processes tend to be considered elementary reactions, in which the stoichiometric coefficients correspond to the reactant order.

Thus, the forward rate law is

${r}_{1} \left(t\right) = {k}_{1} \left[{\text{N"_2"O}}_{4}\right]$,

and the backward rate law is

${r}_{- 1} \left(t\right) = {k}_{- 1} {\left[{\text{NO}}_{2}\right]}^{2}$,

where ${k}_{1}$ and ${k}_{- 1}$ are the rate constants for the forward and backward reactions, respectively.

For this dynamic reaction equilibrium, since the forward and backward processes have the same rate by definition, we therefore have that ${r}_{1} \left(t\right) = {r}_{- 1} \left(t\right)$, and so:

${k}_{1} {\left[{\text{N"_2"O"_4] = k_(-1)["NO}}_{2}\right]}^{2}$

By rearranging, we get:

$\frac{{k}_{1}}{{k}_{- 1}} = \left(\left[{\text{NO"_2]^2)/(["N"_2"O}}_{4}\right]\right)$

for which we define the equilibrium constant:

$K = \frac{{k}_{1}}{{k}_{- 1}} = \left(\left[{\text{NO"_2]^2)/(["N"_2"O}}_{4}\right]\right)$

Even if you didn't follow all that, make sure you remember that in general, for the reaction

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

we have the general equilibrium constant mass action expression:

$\boldsymbol{K = \frac{{\left[C\right]}^{c} {\left[D\right]}^{d}}{{\left[A\right]}^{a} {\left[B\right]}^{b}}}$

As you proceed in further studies on chemical equilibrium, you'll be writing this many times...