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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?

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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:

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

Separating:

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

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.

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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

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  • $\begingroup$ 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 Dec 13 '15 at 17:32
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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.

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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 $[A]_0=0$).

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  • $\begingroup$ Good clearification! And thanks for catching the typo :) $\endgroup$ – Kjetil Sonerud May 28 '14 at 8:39

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