Steady State Approximation

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The steady state approximation posits that after an initial induction periods, an interval during which the concentration of reaction intermediates rises from zero, and during major parts of the reaction, the rates of change of concentration of all reaction intermediates ($I$) are nearly equal to zero.

$$\frac{\mathrm{d}[\ce{I}]}{\mathrm{d}t} \approx 0 $$


Illustrative example

Consider the following sequence of consecutive elementary reactions:

$$\ce{A} \ce{->[k_a]} \ce{I} \ce{->[k_b]} \ce{P} $$

One can write the rate of change of $\ce{I}$ as follows:

$$\frac{\mathrm{d}[\ce{I}]}{\mathrm{d}t} = k_a[\ce{A}] -k_b[\ce{I}] \approx 0$$

Thus, one obtains $\ce{[I]} = \frac{k_a}{k_b}[\ce{A}]$

The rate of formation of product, $$\frac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} = k_b[\ce{I}]$$

However, using the previously obtained result one obtains $$\frac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} \approx k_a[\ce{A}]$$

This brief example, I believe captures the essence of the steady state approximation. Intuitively, this would be a reasonable approximation to make if $k_b > k_a$. Which means that the first step is the bottleneck in the process, because once the intermediate appears, it immediately decomposes to yield products.


A more concrete example

Consider the following reaction:

$$ 2 \ce{N2O5}(g) \ce{->} 4\ce{NO2}(g) + \ce{O2}(g) $$

With the proposed mechanism,

$$ \ce{N2O5}(g) \ce{<=>[k_a][k'_a]} \ce{NO2}(g) + \ce{NO3}(g) \tag{i, reversible}$$

$$\ce{NO2}(g) + \ce{NO3}(g) \ce{->[k_b]} \ce{NO2}(g) + \ce{O2}(g) + \ce{NO}(g) \tag{ii}$$

$$\ce{NO}(g)+ \ce{N2O5}(g) \ce{->[k_c]} \ce{NO2}(g)+ \ce{NO2}(g)+\ce{NO2}(g) \tag{iii}$$

First step would be to identify the intermediates, i.e anything that is not a reactant or a product. In this case the intermediates would be $\ce{NO}$ and $\ce{NO3}$ because they are produced and consumed during the course of the reaction.

Second, write down the rates of change of concentration of intermediates and set them equal to zero.

$$\frac{\mathrm{d}[\ce{NO}]}{\mathrm{d}t} = \overbrace{k_b[\ce{NO2}][NO3]}^{\text{produces intermediate, step ii }} - \overbrace{k_c[NO][N2O5]}^{\text{consumes intermediate, step iii}} \approx 0 \tag{1}$$

$$\frac{\mathrm{d}[\ce{NO3}]}{\mathrm{d}t} = \overbrace{k_a[\ce{N2O5}]}^{\text{produces intermediate, step i }} - \overbrace{k_a'[\ce{NO2}][\ce{NO3}]}^{\text{consumes intermediate, step i, rev}} - \overbrace{k_b[\ce{NO2}][\ce{NO3}]}^{\text{consumes intermediate, step ii}} \approx 0 \tag{2}$$

Solving (1) and (2) one can obtains

$$[\ce{NO3}] = \frac{k_a[\ce{N2O5}]}{(k_a'+k_b)[\ce{NO2}]} \tag{3}$$

and $$ [\ce{NO}] = \frac{k_b[\ce{NO2}][\ce{NO3}]}{k_c[\ce{N2O5}]} = \frac{k_ak_b}{(k_a'+k_b)k_c} \tag{4}$$

Thus, the rate of change of $\ce{N2O5}$ is

$$\frac{\mathrm{d}[\ce{N2O5}]}{\mathrm{d}t} = -k_a[\ce{N2O5}] + k_a[\ce{NO2}][\ce{NO3}] -k_c[\ce{N2O5}][\ce{NO}] \tag{5}$$

Substituting the results from (3) and (4) into (5)

$$\frac{\mathrm{d}[\ce{N2O5}]}{\mathrm{d}t} = -k_a[\ce{N2O5}] + \frac{k_a'k_a[\ce{N2O5}]}{(k_a'+k_b)} - \frac{k_ak_b[\ce{N2O5}]}{(k_a'+k_b)} $$

Which simplifies, into a first order expression with respect to $\ce{N2O5}$

$$\frac{\mathrm{d}[\ce{N2O5}]}{\mathrm{d}t} = - \underbrace{\left( \frac{2k_ak_b}{k_a'+k_b} \right)}_{k_r}[\ce{N2O5}]$$

$k_r$ can be thought of as a "combined" rate constant for the first order decay of $\ce{N2O5}$ which depends on $k_a$, $k_a'$, $k_b$ but not on $k_c$

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Updated on November 27, 2022

Comments

  • Caroline
    Caroline 12 months

    Is there anyone who can explain the steady state approximation to me? We’re given a series of reaction steps, and we need to deduce the rate law for the formation of the product.

    It would be super helpful if someone could help me out with an example!

    Have a nice day :)

    • getafix
      getafix about 7 years
      Could you be a little more specific please? Also, welcome to chemistry.SE Feel free to take the tour to learn more about the site. cheers!
    • orthocresol
      orthocresol about 7 years
    • getafix
      getafix about 7 years
      I'm sure a simple google search would yield some results. I have, nevertheless, posted an answer, which might prove useful to you, and perhaps to future visitors.