A very useful concept in the study of the kinetics of a reaction is the half-life. This concept is used in many other fields. Perhaps the best known is the study of radioactive decay processes. These processes are, from our point of view, first order reactions, since the rate of disintegration of a radionuclide only depends on the amount of it present in the sample.
We will define the half-life of a reactant as the time required for half of its initial concentration to have reacted. It is usually represented as $t_{1/2}$.
In the case of a first order reaction, whose rate equation we know to be:
$$v = k \cdot [A]$$
Then, according to the definition of $t_{1/2}$, we have that:
$$t = t_{1/2} \Longrightarrow[A] = \frac{[A]_0}{2}$$
Using the logarithmic formula for first order reactions:
$$\ln[A] = \ln[A]_0 - k \cdot t$$
we will have to:
$$\ln \frac{[A]_0}{2} = \ln [A]_0 - k \cdot t_{1/2}$$
hence:
$$\ln [A]_0 - \ln 2 - \ln [A]_0 = -\ k \cdot t_{1/2}$$
and we are left with:
$$t_{1/2} = \frac{-\ \ln 2}{-\ k} = \frac{\ln 2}{k}$$
From this expression, we see that the half-life is, in this case, independent of the initial concentration. On the other hand, it is evident that the specific rate constant has dimensions of $(time)^{-1}$ and is also independent of the initial concentration.
Similarly, we can apply the concept of half-life to the case of a second-order reaction with respect to a reactant. For this type of reaction, we have:
$$\frac{1}{[A]} - \frac{1}{[A]_0} = k \cdot t$$
then, substituting the values of the definition of half-life, we will have:
$$\frac{1}{\frac{[A]_0}{2}} - \frac{1}{[A]_0} = k \cdot t_{1/2}$$
that is to say:
$$\frac{2}{[A]_0} - \frac{1}{[A]_0} = k \cdot t_{1/2}$$
now we simplify
$$\frac{1}{[A]_0} = k \cdot t_{1/2}$$
and clear:
$$t_{1/2} = \frac{1}{k \cdot [A]_0}$$
As we can see, in the case of second order reactions, the half-life does depend on the initial concentration of the reference reagent.