# If a reaction is reversible, when can it be said to have reached equilibrium?

Apr 30, 2017

When the rate of the forward reaction is EQUIVALENT to the rate of the reverse reaction.............

#### Explanation:

For the reaction,

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

There is certainly a $\text{forward rate} = {k}_{f} \left[A\right] \left[B\right]$

And a backwards rate, $\text{reverse rate} = {k}_{r} \left[C\right] \left[D\right]$.

And by definition, $\text{chemical equilibrium}$ specifies NOT the cessation of chemical change, BUT $\text{EQUALITY of FORWARD}$ and $\text{REVERSE RATES.}$

And thus at equilibrium, ${k}_{f} \left[A\right] \left[B\right] = {k}_{r} \left[C\right] \left[D\right]$, AND

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

And, clearly, if the forward rate is FASTER than the reverse rate, products are favoured at equilibrium..........and vice versa.

The quotient, ${k}_{f} / {k}_{r}$ is better known as ${K}_{\text{eq}}$, the thermodynamic equilibrium constant, which is a constant according to temperature.

And so.........${k}_{f} / {k}_{r} = {K}_{\text{eq}} = \frac{\left[C\right] \left[D\right]}{\left[A\right] \left[B\right]}$, and this is an equation with which you will get VERY familiar.........