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In chemical kinetics, since we often have to deal with complex reactions, approximations are introduced in order to make the calculations easier. A widely used simplification is the steady state approximation, which can be applied when an intermediate is present in the reaction. This can be applied when the concentration of the intermediate is considered essentially constant over time. This is true when dealing with very reactive intermediates that transform as soon as they form, so their concentration is practically constant.

Consider a reaction where

$$\ce{A ->[$k_1$] B ->[$k_2$] P}$$

A point where the concentration of the intermediate B is definitely constant is at the maximum point of the curve $ [\ce{B}] = f (t) $, obtained by setting

$$ \begin{equation} \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = k_1 [\ce{A}] - k_2 [\ce{B}] = 0 \end{equation} $$

By applying the approximation of the steady state, the concentration of $\ce{B}$ is given by the expression (after separation of variables and integration)

$$[\ce{B}] = [\ce{A}]_0 \frac{k_1}{k_2} \mathrm{e}^{-k_1t} \tag{1}$$

The exact solution of $(1)$ is

$$[\ce{B}] = \frac{k_1[\ce{A}]_0}{k_2-k_1} \left( \mathrm{e}^{-k_1t} - \mathrm{e}^{-k_2t} \right) \tag{2}$$

$(1)$ and $(2)$ are equivalent when $k_2 \gg k_1$ and $\mathrm{e}^{-k_2t} = 0$. In other words, $\ce{B}$ does not accumulate. If we attempt a physical interpretation of these two conditions:

  • $k_2 \gg k_1$ guarantees that the intermediate reacts as soon as it is formed, which is intuitive

  • $\mathrm{e}^{-k_2t} = 0$: this condition is valid when $t \gg k^{-1}_2$, that is, after an induction period equal to $t \approx k^{-1}_2$.

Could anyone help me understand why the condition $t \gg k^{-1}_2$ is required?

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  • $\begingroup$ Are you confused with the term ``induction period'', or with the mathematical condition? $\endgroup$ Jul 23 at 14:48
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    $\begingroup$ The statement is contradictory, but $t \approx k_2^{-1}$ is correct up to log factors. From Atkins: The induction period is an interval during which the concentrations of intermediates rise from zero, and during the major part of the reaction, the rates of change of concentrations of all reaction intermediates are negligibly small. By this definition and taking the derivative, we obtain that both the concentration of $[\ce{B}]$ and its derivative attains maxima at $\sim \ln k_2/k_2$ provided that $k_2 \gg k_1$. After this $[\ce{B}]$ and $\mathrm d[\ce{B}]/\mathrm dt$ decrease. $\endgroup$ Jul 23 at 15:00
  • $\begingroup$ @JeffreyWang I'm not seeing how the units work for the $\ln k_2 / k_2$ term. What is the implicit ratio or dividend of $k_2$ inside the $\ln$ that makes the units go away? $\endgroup$
    – Curt F.
    Jul 28 at 17:41
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    $\begingroup$ @CurtF. The exact point is $\ln \frac{k_2}{k_1}/(k_2-k_1)=(\ln k_2-\ln k_1)/(k_2-k_1) \approx \ln k_2/k_2$. $\endgroup$ Jul 28 at 18:50

2 Answers 2

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Let's look at your exact solution again.

$$ [\mathrm{B}] = \dfrac{k_1[\mathrm{A}]_0}{k_2-k_1} \left( e^{-k_1\ t} - e^{-k_2\ t} \right) \qquad $$

Let's think carefully about the criteria for the steady-state approximation, to well, approximate, this exact solution.

  1. $k_2 \ggg k_1$. This is a criterion you mentioned, and for the right reasons. If $k_2 \ggg k_1$, then $k_2 - k_1 \approx k_2$. Another way of saying $k_2 \ggg k_1$ is $k_2 \approx M k_1$, where $M$ is a large number like 100 or whatever.

  2. I think you had the right thought about the other term, but a bit too harsh. We don't need $e^{-k_2t} = 0$, we need $e^{-k_1t} \ggg e^{-k_2t}$. If you write it that way, then maybe you can see that this will always be true according to the first criterion, unless $t$ is extremely small.

  3. Thus, now we are asking, how large does $t$ have to be for the approximation $e^{-k_1t} - e^{-k_2t} \approx e^{-k_1t}$ to hold? If $t$ is 0, then we have the "approximation" $e^0 - e^0 \approx e^0$, which reduces to $1 - 1 = 1$, which isn't very good. But if $k_2 t \approx 1$ or equivalently $t \approx k_2^{-1}$, then the approximation is $e^{-\frac{k_2 t}{M}} - e^{-1} \approx e^{-\frac{k_2 t}{M}}$ or $e^{-\frac{1}{M}} - e^{-1} \approx e^{-\frac{1}{M}}$, which is better. If $M$ is 100, then it boils down to $0.62 \approx 1$, which is a much better approximation than $0 = 1$.

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    $\begingroup$ Hi Curt, can I suggest $e^{-k_2t/M}$ $e^{-k_2t/M}$ over $e^{-\frac{k_2t}{M}}$ $e^{-\frac{k_2t}{M}}$ for inline math: the latter is a bit hard to read, imo. (though to be totally honest I personally like $\exp(-k_2t/M)$ $\exp(-k_2t/M)$ myself since that doesn't lead to any cramped characters) $\endgroup$ Aug 12 at 2:03
  • $\begingroup$ A minor nitpick in addition to @orthocresol 's comment: ISO 80000-2 suggests to use upright style for operators and constants with a fixed meaning to differentiate them from variables, so please feel free to consider using \mathrm e $\mathrm e$ for Euler number if you find this suggestion reasonable. $\endgroup$
    – andselisk
    Aug 12 at 5:49
  • $\begingroup$ Both of those changes sound good to me. Feel free to edit away! $\endgroup$
    – Curt F.
    Aug 13 at 0:00
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Approximate solutions are equivalent to solutions are when

$k_2 \ggg k_1$: this condition ensures that the intermediate reacts as soon as it is formed

$e^{-k_2\ t} = 0$: for this to be true, this expression must be checked

$$e^{-k_1\ t} - e^{-k_2\ t} \approxeq e^{-k_1\ t}$$

this expression is verified only after $t$ takes values greater than or equal to the period of introduction. The introduction period is the period necessary for intermediate to reach its maximum concentration. According to the steady state approximation, once the introduction period has elapsed, the concentration of intermediate remains constant and equal to its maximum concentration: therefore, the steady state approximation cannot be applied for values of $t$ less than $t_\mathrm{max}$. The value of $t_\mathrm{max}$ is given by

$$t_{\mathrm{max}} = \dfrac{\ln k_2 - \ln k_1}{k_2 - k_1} \qquad (\bigstar)$$

since we are in the hypothesis $k_2 \ggg k_1$, $(\bigstar)$ boils down to the form $$t_\mathrm{max} = \dfrac{\ln k_2}{k_2}$$ for the induction period to have elapsed, an equal time interval must have elapsed $t\approx k^{- 1}_2 $

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