For an $n$-th order reaction, how to determine the time until all the reactants run out?

My approach is the following. The reaction kinetics is modeled by the differential equation

$$\frac{\mathrm d[A]}{\mathrm dt}= - k{[A]^n}$$

where $n$ is the reaction order, which can take values in the nonnegative integers and even in the rational and negative numbers. Suppose we have a chemical reaction similar to the following one

$$\ce{A -> B}$$

In this problem, I'll take an example $5$ equation with similar form with the equation above, and different order $n$. Assuming the $n$ value of the $5$ example is $1,2,0,-1$ and $\frac{1}{2}$, we will get these equations for the five example after we input the $n$ value and integrate the kinetic differential equation.

(reaction with 1st order) $\ln[A] = -kt + \ln[A]_0 $

(reaction with 2nd order) $\frac{1}{[A]} = \frac{1}{[A]_0} +kt$

(reaction with 0th order) $[A] = [A]_0 -kt$

(reaction with $\frac{1}{2}$ order) $2\left(\sqrt {[A]} - \sqrt {[A]_0}\right) = -kt$

(reaction with $-1$ order) $\frac{1}{2}[A]^2 = -kt + \frac{1}{2}\left([A]_0\right)^2$

what I was thinking is, if I want to determine the time until the reaction complete, then I will have to define $t$ when $[A]=0$ (note: I will call the time that needed to complete the reaction $t_\text{complete}$) which is not defined for reaction with 1st and 2nd order because it will involve $\infty$ value, in fact as much as I know, $t_\text{complete}$ will not be defined for $n \ge 1$, but we can calculate $t_\text{complete}$ for $n \lt 1$.

so for 0th order, the $t_\text{complete} = \frac{[A]_0}{k}$

for $\frac{1}{2}$ order, $t_\text{complete} = \frac{2\sqrt{[A]_0}}{k}$

for $-1$ order, $t_\text{complete} = \frac{([A]_0)^2}{2k}$.

my question. is all the reaction with $n \ge 1$ never complete? and is all the reaction with $n \lt 1$ is the only reaction that can complete? and I know what I've done is just manipulating the rate law equation, but if this is correct, can the $t_\text{complete}$ that I got above determine the time that needed for order $n \lt 1$ to complete?

  • 3
    $\begingroup$ These laws are made by observing large amount of molecules and using statistics. Once you go down to only a few you cannot (or should not) use statistics any more. You might want to look into explanations about this exact problem for radioactive decay which follows a first order rate law. $\endgroup$ – DSVA Dec 24 '17 at 12:37
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    $\begingroup$ The math is right, but it hardly matters. Time until completion is not a thing at all. $\endgroup$ – Ivan Neretin Dec 24 '17 at 13:37
  • $\begingroup$ As @Ivan Neretin correctly points out the completion time is not very useful. Instead we define a time for the population to be halved or more usefully when it has reached 1/e of the initial value which is called the lifetime and its reciprocal is the rate constant $k$. $\endgroup$ – porphyrin Jan 1 '18 at 16:38

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