# How do we define the thermodynamic equilibrium for a chemical reaction?

Oct 26, 2016

Good question, I will make an attempt to explain.

#### Explanation:

We write the general equilibrium as:

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

As with any reaction there is a rate forward, $\text{rate forward} = {k}_{f} \left[A\right] \left[B\right]$, and of course a rate backward, $\text{rate backward} = {k}_{r} \left[C\right] \left[D\right]$.

Now, by definition, the chemical condition of equilibrium is defined when the forward and reverse rates are equal:

$\text{i.e." " rate forward "=" rate backward}$,

$\text{i.e.}$ ${k}_{f} \left[A\right] \left[B\right] = {k}_{r} \left[C\right] \left[D\right]$

And upon rearrangement,

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

And the quotient ${k}_{f} / {k}_{r} = {K}_{c}$, otherwise known as the equilibrium constant for the reaction. Generally ${K}_{c}$ is a constant for a given temperature. ${K}_{c}$ can be large (i.e. the products are favoured at equilibrium) or small (the reactants are favoured). ${K}_{c}$ may be formally related to the thermodynamic properties of the reaction, though I am not going to do it here.

Now ${K}_{c}$ must be measured, and as a constant it cannot be altered. However, a chemist or engineer can certainly manipulate the equilibrium. For instance, if we remove (somehow) the products of the reaction, $C$ and $D$, the equilibrium will have to re-establish itself, and it does this by moving to the right as written to satisfy the equilibrium equation, and to re-establish equilibrium concentrations of $C$ and $D$. On the other hand, if we pump more reactant into the equilibrium, the equilibrium will move in a forward direction to cope with increased $\left[A\right]$ and $\left[B\right]$.

There should be many answers here that deal with equilibria. If there is a specific problem or query, ask, and someone will help you.