$$ln[A]_t = -kt + ln[A]_0$$
where:
* $[A]_t$ is the concentration of the reactant A at time t
* $k$ is the rate constant
* $[A]_0$ is the initial concentration of the reactant A
We are given that the successive half-lives of the reaction are 10 min and 40 min. The half-life of a first-order reaction is given by:
$$t_{1/2} = \frac{ln2}{k}$$
where:
* $t_{1/2}$ is the half-life of the reaction
* $k$ is the rate constant
We can use the given half-lives to calculate the rate constant:
$$k = \frac{ln2}{t_{1/2}}$$
$$k = \frac{ln2}{40 \ min} = 1.15 \times 10^{-2} min^{-1}$$
We are also given that the initial concentration of the reactant A was 0.10 M. We can use this information to calculate the concentration of A at any time t:
$$ln[A]_t = -kt + ln[A]_0$$
$$ln[A]_t = -1.15 \times 10^{-2} min^{-1} \times t + ln(0.10 M)$$
$$[A]_t = e^{-1.15 \times 10^{-2} min^{-1} \times t + ln(0.10 M)}$$
This is the integrated rate law for the reaction of A to Products.
