Serial reactions, or consecutive reactions, are two or more reactions in which the product of the first reaction becomes the reactant in the next. The simplest case of a serial reaction involves a reagent A turns into B which in turn, again in the reaction environment, turns into P. In the simplest case, all reactions are irreversible reactions of the first order, so we can be write

$$ \mathrm{A} \xrightarrow{k_1} \mathrm{B} \xrightarrow{k_2} \mathrm{P} $$

the rates of these reactions will be

$$ \begin{equation*} \begin{cases} r_\mathrm{A} = -k_1\ c_\mathrm{A} \\ r_\mathrm{B} = k_1\ c_\mathrm{A} -\ k_2\ c_\mathrm{B} \\ r_\mathrm{P} = k_2\ c_\mathrm{B} \\ \end{cases} \end{equation*} $$

Where $k$ are the kinetics constants, and $c$ is the concentration of the various substances. Plotting $r = f (t)$, I get this

enter image description here

The maximum rate of P formation is reached when $r_\mathrm{B}$ is zero. My hypothesis is that the maximum rate of formation of P must be reached when the concentration of B is maximum, therefore at the minimum of the $r_\mathrm{B}(t)$ curve.. Is my guess right, or is the graph right?

  • $\begingroup$ I hope this would be helpful. $\endgroup$ Jul 22 at 18:30
  • $\begingroup$ Remember that a condition for a maximum of a differentiable function is the first derivative is zero. So rB must be zero for B having maximal concentration. $\endgroup$
    – Poutnik
    Jul 23 at 7:31

1 Answer 1


Simple anwser.

Your answer is correct. The concentration of species B reach a maximum when the rate of formation of the intermediate must equals the rate of dissapeance.

Since $r_B$ can be written as

$$ r_B = \frac{d|B|}{dt} = k_1|A| - k_2|B|$$

this function has an stationary point at $t_{max}$ when $r_B = 0$. $r_B$ is positive for $t < t_{max}$ and negative after reaching the maximum $t > t_{max}$.

This result is intuitive if you consider that below $t_{max}$ the concentration of B increases ($r_B > 0$) and drecreases above it $t_{max}$.

The rate equations have analytic solutions and are easily integrated numerically. But you must be awere that the analitical solutions of the rate equations differ for $k_1 = k_2$ and $k_1 \neq k_2$.

You can find a detailed discussion of this problem in the textbooks:

  • J. H. Espenson. Chemical Kinetics and Reaction Mechanisms, second edition. Mc Graw-Hill (1995).
  • J. I. Steinfeld, J. S. Francisco, W. L. Hase. Chemical Kinetics and Dynamics, second ed., Prentice Hall (1999).

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