In kinetics, we use the steady state approximation to relate the concentration of intermediates to the concentration of reactants. For example, given the mechanism $$\ce{A->[k_1]B}$$ $$\ce{B->[k_2]C}$$ using the steady state approximation for B gives $k_1[A]=k_2[B]$, so $[B]=\frac{k_1}{k_2}[A]$. Since B is at steady state, $[B]$ should be a constant through time. But the equation says that $[B]$ is a function of $[A]$ (and $[A]$ changes with time), so how can $[B]$ be constant? Am I misunderstanding the meaning of steady state here?


1 Answer 1


Good question. Let's examine your system in more detail.

$$ \frac{dA}{dt} = - k_1 A $$ $$ \frac{dB}{dt} = k_1 A - k_2 B $$ $$ \frac{dC}{dt} = k_2 B $$

  1. The third equation is redundant because $C$ can always be found from the conservation equation $C = A_0 - A - B$.

  2. This system can be solved exactly without any further assumptions. So thank you for picking a system that has this nice feature. An exact solution will let us compare to any approximate solutions obtained via simplifying assumptions. Assuming that the initial conditions are $A(0) = A_0$ and $B(0) = 0$, then the exact solution is

$$ A(t) = A_0 e^{-k_1 t} $$ $$ B(t) = \frac{k_1}{k_2-k_1} A_0 \left (e^{-k_1 t} - e^{-k_2 t} \right )$$

  1. What about the quasi-steady state approximation (QSSA)? The QSSA is an assumption that $B$ does not really vary with $t$ as far as $C$ is concerned. That is, predicting the rate of formation of $C$ can be done by assuming therate of formation of $B$ is zero. So if $k_1 \gg k_2$, then $A$ is more or less "instantly" and quantitively converted to $B$, after which $B$ converts to $C$ via the usual first order kinetics with rate constant $k_2$. So this case, $k_1 \gg k_2$, is already a clue that the QSSA is invalid for this mechanism.

  2. Why is the QSSA invalid for this case of $k_1 \gg k_2$? Usually, the QSSA is invoked on catalytic reaction schemes. In these schemes, it is assumed that the concentration of catalyst is very small relative to the concentration of substrate. Thus there is an upper bound on the possible concentration of the intermediate, a bimolecular complex of reactant and catalyst: the total amount of catalyst added limits how much bimolecular complex there can be at any time. This assumption that substrate is in excess relative to catalyst must be satisfied for the QSSA to make sense.

  3. What about the other case?If $k_2 \gg k_1$, then at time scales where $k_2 t \gg 1$, then $B \approx \frac{k_1}{k_2}A_0 e^{-k_1 t}$. The maximum concentration of $B$ in this case is the ratio of rate constants, a ratio that we've already assumed to be much less than one. And since the maximum concentration of $C$ is not similarly bounded, this is a case where the QSSA makes sense. This implies that the conservation equation can be written as $A_0 = A + B + C \approx A + C$. Proceeding with the QSSA assumption to find the rate of formation of $C$:

$$\frac{dB}{dt} \approx 0$$ $$B = \frac{k_1}{k_2}A$$ $$A_0 = A + B + C \approx A + C$$ $$\frac{dC}{dt} = +k_2 B \approx k_1 A = k_1 (A_0 - C) $$

That last equation is now a single-variable first-order linear ODE and could be solved for $C(t)$ if desired. But the main thing is that with the QSSA, $B$ was eliminated from the expression for $C$

  1. I can't think of a physical reason why a system where $k_1 = k_2$ exactly couldn't exist. But someone else will have to come along and explain how to get the math to work out.

    If $k_1 = k_2 = k$, then the system of equations is : $$ \frac{dA}{dt} = - kA$$ $$ \frac{dB}{dt} = k(A - B)$$

    The solution is: $$A = A_0 e^{-k t}$$ $$B = A_0\; k t\; e^{-k t}$$ In this case, there is still no generally valid way to apply the QSSA. The derivative of $B$ is $B'(t) = k A_0 e^{-k t} \left(1 - kt \right)$ and is equal to zero only when at the precise time where $kt=1$. It isn't approximately equal to zero in any other time range, except as $t \rightarrow \infty$. So, the QSSA isn't valid to the case when $k_1 $ is exactly equal to $k_2$, and it isn't valid when $k_1 \neq k_2$ either.

  2. Bottom line: using the QSSA in this problem is equivalent to assuming $k_2 \gg k_1$. You might not have thought so without going through this whole analysis. The key fact is that the QSSA requires that variation in the concentration of the intermediate be small relative to the variation in concentration of the products or substrates that you care about. In mechanisms involving catalysts, this happens any time that catalyst levels are small relative to substrate(s). However, in other mechanisms, this happens only with arbitrary restrictions on the kinetic parameters that may not be obvious a priori.

  • $\begingroup$ It seems to me that the conditions in (5) or the condition in (4) give a "quasi-steady sate." The OP's problem was expecting an "absolute" steady state. $\endgroup$
    – MaxW
    Commented Nov 15, 2015 at 23:03
  • $\begingroup$ I'm not 100% sure but I don't think so. In 5, B is varying exponentially. That isn't steady state! $\endgroup$
    – Curt F.
    Commented Nov 16, 2015 at 0:54
  • $\begingroup$ Ah, yes... (5) further depends on $k_1 \ll 1$ and $k_2 \ll 1$ so that the exponential can be approximated with an expansion of $e$. $\endgroup$
    – MaxW
    Commented Nov 16, 2015 at 1:01
  • 1
    $\begingroup$ @CurtF. So you're saying that we can't use steady state approximation in this case? However, many problems I've seen that involve steady state approx do not involve a catalyst. Ex. $$\ce{2NO<=>N2O2}$$ $$\ce{N2O2+H2->N2O + H2O}$$ Also, this wikipedia article link says the condition for steady state approx is $k_1<<k_2$, so it seems like steady state approx is applicable here... $\endgroup$
    – carbenoid
    Commented Nov 17, 2015 at 3:06
  • 1
    $\begingroup$ In the previous comment, I meant $$\ce{N2O2 + H2 -> N2O + H2O}$$. Sorry, I didn't know SE wouldn't let me edit my comment after 5 min. :( $\endgroup$
    – carbenoid
    Commented Nov 17, 2015 at 3:13

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