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

##### 1 Answer
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.........