For reaction $\ce{2A→B}$ (elementary step), according to the rate law, rate $= k [A]^2$.

In some calculations, we use $k[A]^2$ as the production rate of B. Why isn't it ${1 \over 2} k[A]^2$?

In this case, is the consumption rate of A also $k[A]^2$? If so, why are they the same when in fact two molecules of A produce one molecule of B?

I'm really confused about this part. Can anyone explain why coefficient doesn't play a role in this calculation? Thanks!


2 Answers 2


This confusion pops up often around here.

The problem usually comes down to insufficient consideration of units.

When you say "rate", what do you mean? It has units of mol per volume per second, e.g. $\mathrm{\frac{mol}{L\cdot s}}$. But that isn't enough information. You need to know moles of what.

First example: if the rate is defined as loss of A

For example, if you say that the "rate" means the disappearance of A, then the units of the rate are really $\mathrm{\frac{mol~of~A}{L\cdot s}}$. If so, then as long as:

  • The reaction is $\ce{2 A->B}$; and
  • The rate as you have defined it can be expressed by $r_A = k_1~[\ce{A}]^2$ ...

Then the rate of appearance of B will be $r_B = \frac{1}{2} k_1~[\ce{A}]^2$

Second example: if the "rate" is defined as appearance of B

If conversely you suppose that the "rate" refers to the appearance of B, then the units of rate are really $\mathrm{\frac{mol~of~B}{L\cdot s}}$, and if so, then as long as

  • The reaction is $\ce{2 A->B}$; and
  • The rate as you have defined it can be expressed by $r_B = k_2~[\ce{A}]^2$ ...

Then the rate of disappearance of A will be $r_A = 2 k_2~[\ce{A}]^2$

Third example: a totally separate definition of rate

You say its an elementary reaction. Let's say that we didn't know that and wrote the reaction as $\ce{200 A -> 100 B}$. The rate, according to this new example definition, is the rate of the reaction. The units of the rate are $\mathrm{\frac{mol~of~"reaction"}{L\cdot s}}$. If the reaction is still 2nd-order then the rate of the reaction will be $$r_r = k_3~[\ce{A}]^2$$

According to our reaction, 200 moles of A disappear for each mole of reaction, so $$r_A = 200 k_3~[\ce{A}]^2$$ and 100 moles of B appear, so $$r_B = 100 k_3~[\ce{A}]^2$$

Final thoughts

You will notice that I used different subscripts to distinguish between the $k$ values for the different assumptions. That's because they aren't equal to each other. The value of the "rate constant" $k$ depends on how you define the rate.

No matter how we define "the rate" and what we define as the stoichiometry (e.g. $\ce{2A -> B}$ or $\ce{A -> 1/2B}$ or $\ce{200 A -> 100 B}$, the rates must be equal. Therefore:

$$r_A = 200 k_3~[\ce{A}]^2 = 2 k_2~[\ce{A}]^2 = k_1~[\ce{A}]^2$$

That is,

$$ 200 k_3 = 2 k_2 = k_1$$

But no matter which way we decide to write it, it is still a second-order reaction because "the rate", no matter how we define it, is proportional to $[\ce{A}]^2$.

  • 2
    $\begingroup$ So it's just depends on how I define rate and the k I mentioned actually has different values, right? Clear now. Thank you very much! $\endgroup$
    – Eiswein.Y
    Sep 30, 2015 at 21:17
  • $\begingroup$ With respect to what mention in the above "Final thoughts" : IF we define the stoichiometry ( e.g.4A⟶2B), what is the rate of disappearance of A in our reaction? is it equal 4K3[A]^2 which is wrong or how we can define it according this new stoichiometry / $\endgroup$ Jan 24, 2017 at 14:55
  • 1
    $\begingroup$ You can't express a reaction as $\ce{4A -> 2B}$ because you can divide by 2 to get $\ce{2A -> B}$. If you could multiple by any integer then you'd have an infinite number of rate equations. $\endgroup$
    – MaxW
    Jan 24, 2017 at 18:42
  • 2
    $\begingroup$ @CurtF. What does it mean "each mole of reaction"? $\endgroup$
    – Apurvium
    Oct 5, 2021 at 8:15
  • 1
    $\begingroup$ @Apurvium: If "the reaction" is 200 A -> 100 B, then each mole "of reaction" refers to 200 mol of A and 100 mol of B. If the reaction is 0.5 A -> B, then each mole of "reaction" refers to 0.5 mol of A and 1 mol of B. $\endgroup$
    – Curt F.
    Oct 5, 2021 at 18:30

The rate of production of $\ce{B}$ is given by:

$$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = k[\ce{A}]^2$$

The rate of consumption of $\ce{A}$ is given by:

$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -2k[\ce{A}]^2$$

The reason for having the coefficients this way is because of the way the rate of the reaction, $r$, is defined (source: IUPAC Gold Book):

$$r = -\frac{1}{a}\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = \frac{1}{b}\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}$$

where $a$ and $b$ indicate the stoichiometric coefficients in the reaction $a \ce{A ->} b \ce{B}$ (2 for A, 1 for B). If you assert that the rate of the reaction is equal to$k[\ce{A}]^2$, then you will obtain the first two equations above. You may choose to define it otherwise - read Curt's answer. But ultimately the most important thing is that you respect the fact that

$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -2 \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}$$

This equation says that for every molecule of $\ce{B}$ that is produced, two molecules of $\ce{A}$ are consumed. If you went against this, your reaction wouldn't make any sense at all. (Note that this, in general, only applies to a reaction that does not involve significant buildup of any intermediates. For a single-step reaction such as the one you proposed, this holds true as there are no intermediates.)

Do see this previous question for more discussion about the reaction you have mentioned.

  • 2
    $\begingroup$ Nice IUPAC reference! I should have known IUPAC had an official declaration on the appropriate way to represent stoichiometry. $\endgroup$
    – Curt F.
    Sep 30, 2015 at 20:29
  • 1
    $\begingroup$ @Apurvium You need to be careful here. The integrated rate law you quoted assumes that the stoichiometric coefficient of A is 1. If you take the correct definition of k as shown in the second equation in my answer, and integrate that, you should find the integrated rate law gets an extra factor of 2. $\endgroup$ Oct 5, 2021 at 7:41
  • 1
    $\begingroup$ @Apurvium exactly the same as your previous question. The equation t_1/2 = ln 2 / k is only valid for A -> B. If you have 2A -> 2B then there is an extra factor of 2 in that equation. Again this is derived from the integrated rate law. I suggest going through its derivation again. $\endgroup$ Oct 7, 2021 at 7:59
  • 1
    $\begingroup$ @Apurvium I don't get your question, then. The half-life is constant, regardless of what stoichiometric coefficient you use. That was precisely my point: $k$ can change (by a factor of $n$) but $t_{1/2}$ doesn't change (because to get it you would divide by that same factor of $n$). $\endgroup$ Oct 7, 2021 at 9:13
  • 1
    $\begingroup$ @Apurvium Yes, and this is entirely natural: because the physical description of the system can't change, regardless of how you (an external observer) choose to describe the system. $\endgroup$ Oct 7, 2021 at 11:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.