Inconsistency in Elementary Bi-Molecular Reaction Rate Expressions

Question

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?

Answer 1

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.

Answer 2

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 $k_\mathrm{f}$.