The reaction mechanism for the decomposition of $\ce{A2}$ is thought to be:

\begin{align} \ce{A2 &<=>[$k_1$][$k_1'$] A + A} & &\text{ (fast)} \\[0.2cm] \ce{A + B &<=>[$k_2$] P} & &\text{ (slow)} \end{align}

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

\begin{align} \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = \color{red}{2} k_1[\ce{A2}] - \color{red}{2}k_1'[\ce{A}]^2 - k_2[\ce{A}][\ce{B}] \approx 0 \end{align}

In the equation above, why is the coefficient of $2$ present for both the forward and reverse reactions (highlighted in red)?

  • 2
    $\begingroup$ Welcome to Chemistry! A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors. $\endgroup$ Commented May 24, 2021 at 13:55
  • $\begingroup$ (𝒅[𝑨])/𝒅𝒕=𝟐𝒌_𝟏 [𝑨𝟐 ]−𝟐𝒌_(−𝟏) [𝑨]^𝟐−𝒌_𝟐 [𝑨][𝑩] for reaction mechanism 1. 𝑨_𝟐→𝑨+𝑨 𝒌_𝟏 and 2. 𝑨+𝑨→𝑨_𝟐 𝒌_(−𝟏) and 3. 𝑨+𝑩→𝑷 𝒌_𝟐 $\endgroup$
    – Tara
    Commented May 24, 2021 at 14:02
  • 3
    $\begingroup$ It would be really great if you could edit your question. On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax). If you want to know more, please have a look here and here. $\endgroup$ Commented May 24, 2021 at 14:05
  • $\begingroup$ I did it for you this time, but please do take a look at the guide which Martin linked. $\endgroup$ Commented May 24, 2021 at 14:09
  • $\begingroup$ Thank you, this is my first question, so I leran. $\endgroup$
    – Tara
    Commented May 24, 2021 at 14:18

1 Answer 1


$$\ce{A_n ->[$k$] nA}$$

For the reaction given above, rate of formation of $\ce{A}$ is defined as:

$$\frac{1}{n}\frac{\mathrm d\ce{[A]}}{\mathrm dt} = k\ce{[A_n]}$$

In this case, we have three different reactions that involve A. They are:

\begin{align} \ce{A_n &->[$k_1$] nA} \tag{1}\label{1}\\ \ce{nA &->[$k_1'$] A_n} \tag{2}\label{2}\\ \ce{A + B &->[$k_2$] P} \tag{3}\label{3}\\ \end{align}

Writing the rate equations for $\eqref{1}$, $\eqref{2}$, $\eqref{3}$ and summing them up to get total rate, we get:

\begin{align} \frac{1}{n}\frac{\mathrm d\ce{[A]1}}{\mathrm dt} &= k_1\ce{[A_n]} \tag{4}\label{4}\\ -\frac{1}{n}\frac{\mathrm d\ce{[A]2}}{\mathrm dt} &= k'_1 [\ce{A}]^n\tag{5}\label{5}\\ -\frac{\mathrm d\ce{[A]3}}{\mathrm dt} &= k_2\ce{[A][B]} \tag{6}\label{6}\\ \frac{\mathrm d\ce{[A]_\mathrm{tot}}}{\mathrm dt} &= \frac{\mathrm d\ce{[A]1}}{\mathrm dt} + \frac{\mathrm d\ce{[A]2}}{\mathrm dt}+\frac{\mathrm d\ce{[A]3}}{\mathrm dt} \tag{7}\label{7} \\ \end{align}

Substituting values from $\eqref{4}$, $\eqref{5}$, $\eqref{6}$ into $\eqref{7}$, we get:

$$\frac{\mathrm d\ce{[A]_\mathrm{tot}}}{\mathrm dt} = \color{red}{n}k_1\ce{[A_n]} - \color{red}{n}k'_1\ce{[A]^n} - k_2\ce{[A][B]} $$

Here, $n = 2$, therefore

$$\frac{\mathrm d\ce{[A]_\mathrm{tot}}}{\mathrm dt} = \color{red}{2}k_1\ce{[A_n]} - \color{red}{2}k'_1\ce{[A]^2} - k_2\ce{[A][B]} $$

  • 1
    $\begingroup$ Thank you very much! But I still a little be confused, shouldn't it be [A] to the power of n in the equation (5)? That gives quadratic equation for the intermediate for the given mechanism.. $\endgroup$
    – Tara
    Commented May 24, 2021 at 16:08
  • $\begingroup$ I tried this way to solve next problem. The suggested reaction mechanism is the same, but first reaction is switched. In the textbook,(which is Atkins PhysChem), coefficient 2 appears only for forward reaction. Is it mistake or the I need to use different approach? $\endgroup$
    – Tara
    Commented May 24, 2021 at 19:34
  • $\begingroup$ @Tara Please use the answer field only for actual answers. If you are new to the site, I suggest you take the tour and learn more about it. I have converted your answer to a comment here; if you have more follow up questions, it might be better to ask again and link here for context, see also How to Ask. $\endgroup$ Commented May 24, 2021 at 19:40

Your Answer

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

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