I want to understand the assumptions that have to be fulfilled in order for the steady state approximation to work. I did some research myself and was able to deduce assumptions of the form $k_1\gg k_2$ for some systems, by solving the system exactly and taking the limiting case in which the exact system reduces to the approximative system. (Later I found this answer doing the same thing Meaning of steady state (kinetics)) This approach will not work for most systems though, as no closed exact solution exists for most non-linear differential equation systems.

I want to develop a clear chemical intuition in which case it is justified to use the approximation. Most literature seems to just tell the reader that it's fine to use the approximation in the reaction thats discussed. This is unsatisfying. From a mathematical standpoint this approximation is all but trivial.

  • $\begingroup$ This steady-state approximation is always wrong. It is only justified by common sense. In chemistry, you will never be and stay exactly at a point where the rate of the reaction producing a compound is equal to the rate of destruction of this compound. But you may imagine being "not too far" from this ideal position. That is the trick used for understanding the kinetics of some chain reactions. $\endgroup$
    – Maurice
    Feb 2, 2020 at 14:32
  • $\begingroup$ Obviously it's never exact but I want to develop an understanding to what kind of processes it's an acceptable approximation. Surely there must be papers comparing numerical solutions to steady state approximations and surely there must be 1 or 2 handy rules for estimating weather it's OK to use steady state approximation based on the underlying chemistry. $\endgroup$
    – user86728
    Feb 5, 2020 at 21:33
  • 3
    $\begingroup$ Whether it’s okay is going to depend also on what you are using the result for, so one can’t make general rules. A good way to get the feel for the inaccuracy is to plot product vs time for SS approx and actual. To simulate actual, you don’t need an explicit solution, just iterate the differential equations at short time intervals $\endgroup$
    – Andrew
    Feb 5, 2020 at 22:07
  • $\begingroup$ Related: chemistry.stackexchange.com/q/110794 $\endgroup$
    – Karsten
    Feb 6, 2020 at 13:55
  • $\begingroup$ In addition to k1>>k2, there is also the condition k2*t>>1, because it only holds after a brief induction period (initially, the rate of change of the intermediate is greater than 0, but by the time k2*t>>1, it has become 0). You can see that from this image:chem.libretexts.org/@api/deki/files/50967/… Note that initially, d[B]/dt>0, but it quickly becomes 0. This condition often isn't mentioned because the induction period is typically very short. $\endgroup$
    – atbm
    Feb 7, 2020 at 16:59

2 Answers 2


No very satisfying justifications are available unless you do a case-by-case analysis. But PSSA does not come from common sense. In general, pseudostationary state approximation (PSSA) is justified by the theory of differential equations.

In fact, the pseudostationary state concentrations represent the zeroth approximation to a set of particular solutions of the correct chemical kinetic equations, which are called principal solutions. After the initial period, the concentrations of the intermediates approach their principal solution rather than their pseudostationary state values.

Therefore, the accuracy or usefulness of the pseudostationary state assumption depends upon the magnitude of the differences between the principal solution and the pseudostationary state concentrations.

OK, that sounds useless. But after doing some maths (like finding the principle solutions), some (obviously very ambiguous) rules are indeed justified mathematically.

  • Induction period should be short. (As you can see PSSA solutions cannot satisfy the initial conditions)

  • Destruction of the intermediates should be fast.

For example, in $\ce{A ->[$k_1$] B ->[$k_2$] C}$ if $k_2\gg k_1$ then the conditions above are met while if $k_1\gg k_2$, they aren't.

By the way, PSSA originates from Michaelis-Menton theory for enzyme kinetics. PSSA in it is well justified while the conditions above are not necessarily met. Consider

$$ \ce{E + S <=>[$k_1,k_{-1}$] ES ->[$k_2$] P}. $$

Let $[\ce{S}]=s,[\ce{E}]=e,[\ce{ES}]=c,[\ce{P}]=p,e+c:=e_0,s+p+c:=s_0$. Furthermore, let $\tau=k_1e_0t,u(\tau)=s/s_0,v(\tau)=c/e_0$. The kinetic equations can be written as

$$ \begin{cases} \frac{\mathrm du}{\mathrm d\tau}=-u+(u+K-\lambda)v\\ \color{red}{\varepsilon}\frac{\mathrm dv}{\mathrm d\tau}=u-(u+K)v \end{cases} $$

where $\lambda=k_2/\left(k_1s_0\right),K=\left(k_{-1}+k_2\right)/\left(k_1s_0\right),\color{red}{\varepsilon=e_0/s_0\ll 1}$. Therefore, PSSA can be applied (i.e. setting $u-(u+K)v=0$, justified by perturbation methods) without loss of much accuracy. This example indicates that more often than not, a case-by-case analysis is required.

Finally, I think to use PSSA safely, one should do some experiments to check (numerical simulations, real experiments, etc.).

  • $\begingroup$ Thank you for you input. This is close to what I personally would have answered. There is no general justification and care should be taken when applying PSSA. Eventhough many introductionary courses just take it as given, that the approximation is fine, there are cases where it is probably not. $\endgroup$
    – user86728
    Feb 11, 2020 at 14:21

In the scheme $ A \overset {k_1} \longrightarrow B \overset {k_2}\longrightarrow C$ the exact solution for B is $B=\displaystyle \frac{k_1A_0}{k_2-k_1}\left( e^{-k_1t}-e^{-k_2t} \right) $. To find out how quickly the steady state is reached and by how much it deviates from the true value, the steady state is compared to the exact value by taking a ratio of concentrations.

Species B follows $\displaystyle \frac{dB}{dt}+ k_2B=k_1A_0e^{-k_1t}$ and letting $dB/dt=0$ forms our steady state value $B_{ss}$. The ratio to measure the extent of deviation will be defined as $\displaystyle R=\frac{B}{B_{ss}} = \frac{k_2}{k_2-k_1}\left( 1-e^{-(k_2-k_1)t} \right)$ and as a measure at steady state we let $R=1$.

Initially when $t=0,\; R_0=1$ and at long times $\displaystyle R_{\infty} = \frac{k_2}{k_2-k_1}$, so the difference between $R_0$ and $R{\infty}$ indicates the extend or deviation of steady state from the true value.

If, for example, $k_2 = 5k_1$ then $R_\infty-R_0=1/4$ or a 25% deviation. When $dB/dt=0$, species B has reached its maximum value which takes time $\displaystyle t_m =\frac{1}{k_2-k_1}\ln(k_2/k_1)$ and in this example this takes $\ln(5)/(4k_1)$ seconds, i.e. this time is needed before steady state can be invoked.

This analysis indicates that for steady state to apply $k_2 \gg k_1$, i.e. the steady state applies under conditions that B disappears rapidly once it is formed. For other schemes the same general principles apply but the analysis will be far more complicated.


Your Answer

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