# Why is chemical equilibrium dynamic?

Mar 8, 2018

Because there are many factors that can change the Products/Reactants ratio!

#### Explanation:

Chemical equilibrium refers to the balance between products and reactants after a given reaction has reached a state of order, in which both reactants and products are forming at a constant rate. It is dynamic because there are many factors that affect what that ratio will be, as defined by LeChatelier.

Heat

Affects the solubility of the products / reactants, yet also will change the equilibrium constant if the reaction is exothermic or endothermic. This is the case because at different temperatures, more products or reactants may exist in solution.

Concentrations

If there is a very large concentration of either a product or reactant, it will shift the equilibrium one way or the other. This is because the addition of reactants or products will always yield a reaction one way or the other, unless the reaction can no longer dissolve any more of it. In that case, it becomes a precipitate which does not effect equilibrium

Pressure

This affects gaseous products and reactants. The side where there is more gas is the side in which the equilibrium will shift to if the pressure is decreased, because more can fit now. Vice versa as well.

Conclusion

There are many factors that shift equilibrium (Kc) so we must always be wary of these factors before we jump to a conclusion about what the equilibrium is, because it is very dynamic!

Mar 8, 2018

Good question....we can approach the answer on the basis of $\text{dynamic}$ rates of reaction....

#### 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, $\text{rate backward} = {k}_{r} \left[C\right] \left[D\right]$, i.e. dynamic processes...

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.