Assume the following bi-molecular reaction is elementary as written with rate constant $k_\mathrm{f}$: $$\ce{A + A -> P}$$

This review, suggests to express the rate in terms of the production of product "P": \begin{align} \frac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} &= k_\mathrm{f}[\ce{A}]^{2},& \text{where } \frac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} &= \left(-\frac{1}{2}\right)\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t}, \end{align} which gives: $$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -2k_\mathrm{f}[\ce{A}]^{2}.\tag{1}$$

While this other review suggests that the rate be written in terms of the consumption of reactant "$\ce{A}$": $$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -k_\mathrm{f}[\ce{A}]^{2}.\tag{2}$$

Which is correct?

  • $\begingroup$ Read up on rate law, there is an issue there.. $\endgroup$ – Safdar Aug 29 '20 at 19:25
  • 1
    $\begingroup$ @Safdar I think I got it. Both expressions are correct, because 1) both rates quadruple when the reactant concentration is doubled and 2) the rate of consumption of reactants is equivalent to the rate of production of product. If we integrate both expressions and evaluate their concentration profiles w.r.t time, we will see that one has double the rate of reactant consumption compared to the other. However, vice-versa, the expression with the larger reactant consumption will have a larger rate of product formation. Therefore, as long as one is consistent, the approach is immaterial. $\endgroup$ – chr218 Aug 29 '20 at 19:50
  • $\begingroup$ Related: chemistry.stackexchange.com/questions/38167/… $\endgroup$ – Curt F. Aug 30 '20 at 21:38
  • $\begingroup$ @CurtF. Oh wow, I should have been more diligent with my research before posting this question. Should I take my question down? Also, per the the link you shared, I guess my previous comment was correct? Thanks! $\endgroup$ – chr218 Aug 31 '20 at 17:04
  • $\begingroup$ I don't think this is quite a duplicate, but the other question is related. Just wanted to make sure you saw it! $\endgroup$ – Curt F. Aug 31 '20 at 18:41

If the reaction is $$\ce{2A -> P},$$ then the rates are defined to be related as $$ -\frac{1}{2}\frac{\mathrm{d}\ce{A}}{\mathrm{d}t} = \frac{\mathrm{d}\ce{P}}{\mathrm{d}t} $$ and this is true throughout the reaction. The rate at which $\ce{P}$ is produced is half that at which $\ce{A}$ is consumed. Normally therefore a factor of 2 is expected with the rate constant. With any other usage an explanation should be added so as not to confuse things.

In the general case $$\ce{aA + bB -> pP + qQ},$$ then $$ -\frac{1}{a}\frac{\mathrm{d}\ce{A}}{\mathrm{d}t} = -\frac{1}{b}\frac{\mathrm{d}\ce{B}}{\mathrm{d}t} = \frac{1}{p}\frac{\mathrm{d}\ce{P}}{\mathrm{d}t} = \frac{1}{q}\frac{\mathrm{d}\ce{Q}}{\mathrm{d}t} = \frac{\mathrm{d}x}{\mathrm{d}t}, $$ where at any time $t$, $ax$ moles of $\ce{A}$ and $bx$ of $\ce{B}$ are consumed and $px$ of $\ce{P}$ are produced, where $x$ is the extent of reaction.

  • $\begingroup$ Using the extent of reaction, then deducing an expression for reactant consumption/product formation makes the most sense. Thanks! $\endgroup$ – chr218 Sep 2 '20 at 14:43

The first expression $$\frac{\mathrm{d}[A]}{\mathrm{d}t} = -2k_\mathrm{f}[\ce{A}]^{2}\tag{1}$$ is the correct one for the reaction given. As the OP mentions in the comments, both approaches describe the situation correctly. However, the values of $k_\mathrm{f}$ would be different by a factor of two.

You could rewrite the chemical equation by dividing by two (no longer an elementary reaction, so you have to specify that it is second order in $\ce{A}$): $$\ce{A -> 1/2 B}$$

In this case, the correct expression for the rate of $\ce{A}$ reacting would indeed be the second expression: $$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -k_\mathrm{f}[\ce{A}]^{2}\tag{2}$$

Because the two reactions are written differently, you again expect a different value for $k_\mathrm{f}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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