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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!

Curt F.
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Eiswein.Y
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  • This is similar but probably not a duplicate of http://chemistry.stackexchange.com/questions/32304/law-of-mass-action – Curt F. Sep 30 '15 at 19:56

2 Answers2

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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 $[\ce{A}]^2$.

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Curt F.
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    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! – Eiswein.Y Sep 30 '15 at 21:17
  • 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 / – Adnan AL-Amleh Jan 24 '17 at 14:55
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    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. – MaxW Jan 24 '17 at 18:42
  • NO + 1/2Cl2 -----> NOCl how to relate the rate of the reaction with rate of disappearance NO/ – Adnan AL-Amleh Jan 24 '17 at 19:24
  • I'm confused by all the new comments on this old answer of mine. First, I think my answer does answer the question. Second, you absolutely can define the reaction as $\ce{4A -> 2}$ B, in which case the rate of disappearance of A will be $-4 k_4\ce{[A]}^2$. There are in fact an infinite number of possible rate equations and an infinite number of corresponding rate constants. You just have to make sure your reaction definition is consistent with your rate equations. – Curt F. Jan 24 '17 at 20:21
  • Just a doubt ..In many textbooks they dont explicitly mention whether the rate in the rate law expression is the rate of disappearance of reactant A or appearance product B..Do we by default assume its the rate of disappearance of A..Like in the OP's question what could his rate expression probably refer to? – rohit_r Apr 25 '18 at 11:33
  • @CurtF. here you said" No matter how we define "the rate" and what we define as the stoichiometry (e.g. 2A⟶B or A⟶12B or 200A⟶100B, the rates must be equal. Therefore:

    rA=200k3 [A]2=2k2 [A]2=k1 [A]2 " this seems true that rate of consumption of reactant must not change as reaction doesn't change . But what about the net rate of reaction ? (which is defined as rate of consumption of reactants divided by their stoichiometric coefficients and rate of production of products divided by their stoichiometric coefficients ) .That will change here then .But it shouldnt as reaction remains same

     – Raghav Madan Oct 03 '21 at 00:25
  • @CurtF. (continuation to prev comment ) reaction remains same since only stoichiometric coefficients are changed . So shouldnt both the overall rate and the rate of production and consumption be same ? – Raghav Madan Oct 03 '21 at 00:31
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    @CurtF. What does it mean "each mole of reaction"? – Apurvium Oct 05 '21 at 08:15
  • @RaghavMadan: it will change because when you change "the reaction", you are changing what the "net rate of reaction" means. For 200 A -> 100 B, if the rate of "the reaction" is 1 mol / second then the rate of disappearance of A is 200 mol / s. If you define the reaction as 2A -> B, then if the rate of this very different "the reaction" is 1 mol / second, then the rate of disappearance of A is 2 mol / s. – Curt F. Oct 05 '21 at 18:29
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    @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. – Curt F. Oct 05 '21 at 18:30
  • @CurtF. ok , so net rate of reaction changes on changing to stoichiometric coefficient .But rate of consumption and production must remain same as the reaction is same .Rate of consumption /production is a more important and usefull quantity than overall rate of reaction . To keep one of them constant the other has to change .So we prefer to keep rate of consumption and production constant instead of overall rate .Am i right here ? – Raghav Madan Oct 06 '21 at 13:51
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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.

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orthocresol
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    Nice IUPAC reference! I should have known IUPAC had an official declaration on the appropriate way to represent stoichiometry. – Curt F. Sep 30 '15 at 20:29
  • @orthocresol Which $k$ to be used in the equation, $kt=\ln \frac{[A]_0}{[A]_t}$ for first order reaction, $\ce{2A -> B}$? All are constant but $k$. – Apurvium Oct 05 '21 at 04:00
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    @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. – orthocresol Oct 05 '21 at 07:41
  • @orthocresol As $k$ changes, rate of reaction and half-life of reaction also changes. For radioactive disintegration, the stoichiometry is always one and half-life is constant for a particular nuclear reaction. It does not make sense to have different half-lives here. But what does it mean if we multiply any nuclear reaction by some factor? – Apurvium Oct 07 '21 at 05:40
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    @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. – orthocresol Oct 07 '21 at 07:59
  • @orthocresol I got it but my question was about radioactive disintegration for which half-life is always constant for a radioactive species. – Apurvium Oct 07 '21 at 08:55
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    @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 $n$). – orthocresol Oct 07 '21 at 09:13
  • @orthocresol Finally I derived it! But I was trying to reason out like this: For the reaction $\ce{2A->2B}$, the rate of reaction becomes half but rate of disappearance of A remains the same as for $\ce{A->B}$. If we start with equal number of moles of A, say $a$ moles, then the time taken to reduce the number of moles of A to $a/2$ will remain same in both the cases. Am I correct? – Apurvium Oct 07 '21 at 10:55
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    @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. – orthocresol Oct 07 '21 at 11:20
  • @orthocresol Does any book explain these confusions in detail? – Apurvium Oct 07 '21 at 11:48
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    @Apurvium Not that I know of, but it's not that hard to figure it out for yourself, to be honest. You must think carefully about which quantities are human-defined and are thus arbitrary (e.g. the rate constant $k$: you can't measure $k$, you can only measure something else from which you calculate $k$ — but how you do this depends on your definition, i.e. the stoichiometric coefficients) and what quantities are directly observable and thus cannot be changed (e.g. half-life, or in general, any time $t$ taken for $\ce{[A]}$ to decrease by a given factor). – orthocresol Oct 07 '21 at 12:55