Equilibriums and Overshooting
Not really. Overshooting doesn't happen, rather the concentrations gradually approaches the equilibrium concentration as an asymptote.

You'll soon see what I mean.

As an example, for simplicity, lets consider:

$$\ce{P &lt;=&gt;[k_1][k_2] Q}$$

I will assume here that the forward and backward reactions are first order, so we get the rate to be:

$$\text{Rate}_{\text{forward}}=k_1\ce{[P]}$$

and

$$\text{Rate}_{\text{backward}}=k_2\ce{[Q]}$$

As a result the net rate would be:

$$\frac{d\ce{[P]}}{dt}=-k_1\ce{[P]}+k_2\ce{[Q]}$$

Let's take the initial concentration of $\ce{P}$ to be $\ce{[P]_0}$, and the initial concentration of $\ce{Q}$ to be zero.

Suppose at any time $t$, the concentration of the reactant is $\ce{[P]}$. Thus, $\ce{[Q]} = \ce{[P]_0} - \ce{[P]}$. Now, we get the rate law to be:

$$\frac{d\ce{[P]}}{dt} = k_2\ce{[P]_0} - (k_1 + k_2)\ce{[P]}$$

Integrating this from $\ce{[P]_0}$ to $\ce{[P]}$:

$$\int_{\ce{[P]_0}}^{\ce{[P]}}\frac{d\ce{[P]}}{k_2{[P]_0}-(k_1 + k_2)\ce{[P]}}=\int_0^tdt$$

Solving the integral, we get:

$$\frac{-1}{(k_1 + k_2)}\log_e{\left(\frac{(k_1 + k_2)\ce{[P]} - k_2\ce{[P]_0}}{k_1\ce{[P]_0}}\right)}=t$$

$$\frac{(k_1 + k_2)\ce{[P]} - k_2\ce{[P]_0}}{k_1\ce{[P]_0}}=e^{-(k_1 + k_2)t}$$

Finally, on simplifying it, we get our integrated rate equation as:

$$\ce{[P]}=\ce{[P]_0}\frac{k_2 + k_1e^{-(k_1+k_2)t}}{k_1+k_2}$$

Let's plot this on a graph to understand what this is actually saying:


(purple-$\ce{[P]}$ and orange-$\ce{[Q]}$)

You can clearly observe that the concentrations never overshoot the asymptotes (black dotted line). The asymptotes denote the equilibrium concentrations, so it's evident that the reaction tends toward the equilibrium concentration.