# Can anyone explain, why there are coefficients 2 in the equation for rate of change [A]?

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)?

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• (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 – Tara May 24 at 14:02
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• Thank you, this is my first question, so I leran. – Tara May 24 at 14:18

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

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

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

• 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.. – Tara May 24 at 16:08
• 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? – Tara May 24 at 19:34