For a $1^{\text{st}}$ order reaction $ A \rightarrow B$, the corresponding rate law is: $$\text{rate} = k[A]^{1}$$ and the integrated rate law for this is $$ \ln \left(\frac{A}{A-x}\right) = kt$$

Going by this equation, every first-order reaction must get completed only at infinity. Is this true? If not, then what is the correct explanation?


5 Answers 5


To add to the other answers that only address the mathematical behavior of the first order rate equation close to $t=\infty$, let me address what actually happens physically.

Before you reach $[A]_t=0$, the first order rate equation actually breaks down. It isn't valid anymore, because one of the assumptions on which it is based, namely that the reaction rate is deterministic, breaks down.

When you get to really low quantities of (one of) your reactants, reaction events, i.e. two molecules meeting, are blind luck. Actually, for high quantities this is also the case, but then the behavior is a physical version of the law of large numbers: a probabilistic system behaves deterministic on average.

You can simulate this basically by flipping coins whether a reaction occurs or not. If you do this for a simple $A$ to $B$ reaction you get graphs like the following:

enter image description here

Here the chance of reaction per timestep for each molecule $A$ is 10%. You can see that the curves at low number of molecules $N$, are jumpy (and actually would look different everytime I rerun my calculation), while for $N=1000$ you already see the fairly smooth and familiar first order behavior. Just note that close to t=10 also the $N=1000$ case is behaving jumpy again as you can see from the zoom below.

enter image description here

  • $\begingroup$ Some first order reactions don't involve two molecules. The archetype being radioactive decay (OK strictly a nuclear not a chemical reaction, but the same principles apply). $\endgroup$
    – matt_black
    Commented Dec 13, 2015 at 17:32

No, because matter is not continuously divisible

The flaw is the mathematical reasoning is that it assumes that matter is continuously divisible. In reality it consists of finite numbers of atoms or molecules.

In a truly first order, irreversible, reaction there will eventually be just one last molecule or atom and it will react leaving nothing. The mathematical view is a statistical approximation that works when there are enough units not to worry about the lumpiness.

The classic example (though strictly a nuclear not a chemical process) is radioactive decay. The probability of any atom decaying is constant so the rate of decay depends only on the number of atoms present. But we eventually get down to just one atom. And that atom will decay. But when only small numbers are left the smooth pattern of decay will be lumpy and will no longer obey the smooth mathematics of the first order equation.

But you won't have to wait an infinite amount of time to get to zero.

If you start with a mole (about 6 x 1023 atoms) of substance you will have about one atom or molecule left after about 79 half-lives. So you can be fairly sure (never exactly sure due to the probabilistic nature of quantum things) that there will be no reaction still occurring after about 80 half-lives.


As far as I can see, the expression you have come up with is not entirely correct. If we assume that the reaction is $\ce{A -> B}$ (i.e. saying that it is approximately irreversible, or that $k_\mathrm{reverse} \ll k_\mathrm{forward}$), we have:

$$ \mathrm{rate} = \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -k[\ce{A}] $$


$$ \frac{\mathrm{d}[\ce{A}]}{[\ce{A}]} = -k\,\mathrm{d}t $$

Integration gives:

$$ \ln\frac{[\ce{A}]}{[\ce{A}]_0} = -kt $$ $$ [\ce{A}] = [\ce{A}]_0 \mathrm e^{-kt} $$

Here, $[\ce{A}]_0$ is the initial concentration of $\ce{A}$. It is correct that mathematically, $[\ce{A}] > 0$ for all $t<\infty$, from the last equation. In practise, however, the amount of $[\ce{A}]$ after a certain time interval $t$ will be negligibly small. How long it takes depends on $k$; if $k$ is large, the time for the reaction to go essentially to completion will be short.


To add to Kjetil's answer, the expression given by the OP and the expression derived by Kjetil are (almost) equivalent. If we define the OP's $A$ as $[A]_0$, the original concentration of $A$ and the OP's $A-x$ as $[A]_t$, the concentration of $A$ at time $t$, then we rearrange and take the natural log of both sides...

$$[A]_t=[A]_0e^{-kt}\ \ \ \ \ (1)$$ $$\ln{\left(\dfrac{[A]_t}{[A]_0}\right)}=\ln{\left(e^{-kt}\right)}\ \ \ \ \ (2)$$ $$ln{\left(\dfrac{[A]_t}{[A]_0}\right)}=-kt\ \ \ \ \ (3)$$ $$ln{\left(\dfrac{[A]_0}{[A]_t}\right)}=kt\ \ \ \ \ (4)$$

The fundamental differences between the original equation and the one with the natural logarithms is that we have to make some assumptions. Our derivation produces an undefined term if $[A]_0=0$ (equation 2), so we must assume that $[A]_0>0$, which is logical. We also must assume $[A]_t\ne 0$ (equation 4). Only equation 1 works with $[A]_t=0$, which as was pointed out is only technically true at $t=\infty$ (or $[A]_0=0$).

  • $\begingroup$ Good clearification! And thanks for catching the typo :) $\endgroup$ Commented May 28, 2014 at 8:39

It is interesting to derive the first order rate expression to see why the decay continues to infinity. (It was probably first derived by von Schweidler in 1905!) The basic assumption is that the probability of reaction $p$ in a time interval $\Delta t$ is independent of the past history and depends only on the length of the time interval $\Delta t$, and when this is sufficiently short the probability is proportional to this interval as $p=k\Delta t$ where $k$ is the proportionality constant which depends on the particular atom of molecule. (This will become the decay rate constant)

The probability of no reaction in the short time interval $\Delta t$ is $1-p=1-k\Delta t$. If it survives this interval the probability of surviving the next interval is also $1-k\Delta t$ and so by the law of probabilities, the chance of surviving both intervals is $(1-k\Delta t)^2$ and so for $n$ such intervals the probability is $(1-k\Delta t)^n$. The total time is $t=n\Delta t$ and so by substituting we obtain $(1-kt/n)^n$. The chance that the molecule or atom will remain unchanged at time $t$ is just that when $\Delta t$ tends to zero, i.e. is made infinitesimally small and then we can recognize that $e^x=\lim_{n\to \infty}(1+x/n)^n$ and so we obtain the limiting value as $p=e^{-kt}$. Instead of considering just one atom let there be $N_0$ initially then the fraction remaining at unreacted at time $t$ is $N/N_0=e^{-kt}$.

So even for a single molecule the probability of remaining unreacted can extend to infinity. In practice this is only a matter of sensitivity of your experiment. If you do time-resolved single-photon counting to measure the decay of an excited state the decays are routinely measured down to 5 orders of magnitude, transient absorption decays, remarkably, have been shown to follow a decay to at least 9 orders of magnitude in intensity. But eventually either no matter how many times the experiment is repeated or how many molecule we start with the noise inherent in the experiment will dominate and we can go no further.

Turning the equation round we have $t=\frac{1}{k}\ln(p)$ so that the chance $p$ of waiting for a long time for the molecule to react becomes increasingly small as $t$ increases. Thus if the decay has a rate constant of 1/nanosecond the chance of waiting even as short a time as a microsecond without it decaying first is tiny, $p=e^{-10^{-6}\cdot 10^9}\approx 10^{-435}$! This is of course why we say that first order reactions comes to a halt after a given time, the chance of observing become so small and is lost in noise even if we start with a mole's worth of molecules: $10^{-435+23}$ is still utterly minute.

  • $\begingroup$ In your example, the reaction has been running for approx. 1443 half life periods. So the probability you calculate is $0.5^{1443} \approx 10^{-435}$. It is like flipping a coin 1443 times and see only heads. Not very likely, as can be intuitively grasped. $\endgroup$ Commented Apr 21, 2022 at 12:47
  • $\begingroup$ @Snijderfrey, nice example but 'not very likely' is a bit of an understatement :) $\endgroup$
    – porphyrin
    Commented Apr 21, 2022 at 17:32
  • $\begingroup$ Yes, certainly an understatement. If we had 1443 people flipping a coin every second and have them record their result, they would still have a ridiculously low chance to observe the described outcome within the lifespan of the universe. And the reasoning would not be much different with only 100 half life periods. The numbers involved are next to impossible to conceive. $\endgroup$ Commented Apr 21, 2022 at 18:25

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