Section $3$: This same idea applies to more complex reactions. For example, if you are given that the reaction:

$A+BC+D$ is at equilibrium, adding more $C$ to the right side will make the reaction proceed to the left, to restore its balance. The concentrations of all four ingredients will now be different than they were at the beginning.

For any chemical reaction, this is defined as the product of the concentrations of the products divided by the product of the concentrations of the reactants. For the reaction above, the mass action ratio $($M$.$A$.$R$.)$ would be:

Mass Action Ratio $=\frac{[C][D]}{[A][B]}$

Answer the following question:

If you began this reaction with A and $B$ at $1M($and no $C$ or $D),$ and the M$.$A$.$R$.$ at equilibrium was $2\mathrm{.}4,$ what do you think it would be in a reaction where the starting concentrations of A and $B$ were $0\mathrm{.}1M?$