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Physical · Kinetics

Steady state vs pre-equilibrium

3 min read· unreviewed

Many mechanisms pass through a reactive intermediate that never builds up. Take the simplest case:

A⇌  k−1    k1  I→  k2  P\mathrm{A} \xrightleftharpoons[\;k_{-1}\;]{\;k_1\;} \mathrm{I} \xrightarrow{\;k_2\;} \mathrm{P}

The exact rate equations are coupled, so we approximate. There are two standard approximations. The pre-equilibrium approximation is a special case of the steady-state one, so knowing which you are using, and why, is most of the work.

The steady-state approximation

If I is consumed about as fast as it is made, its concentration stays small and nearly constant after a short induction period. Set its net rate of change to zero:

d[I]dt=k1[A]−k−1[I]−k2[I]≈0\frac{\mathrm{d}[\mathrm{I}]}{\mathrm{d}t} = k_1[\mathrm{A}] - k_{-1}[\mathrm{I}] - k_2[\mathrm{I}] \approx 0

so

[I]=k1[A]k−1+k2[\mathrm{I}] = \frac{k_1[\mathrm{A}]}{k_{-1} + k_2}

The rate of forming P is then

rate=k2[I]=k1k2k−1+k2 [A]\text{rate} = k_2[\mathrm{I}] = \frac{k_1k_2}{k_{-1}+k_2}\,[\mathrm{A}]

This is the steady-state result. Notice what it does not assume: nothing about which step is faster.

The pre-equilibrium approximation

Now suppose I falls back to A much faster than it goes on to P, so k−1≫k2k_{-1} \gg k_2. Then A and I stay close to equilibrium, with [I]=K1[A][\mathrm{I}] = K_1[\mathrm{A}] where K1=k1/k−1K_1 = k_1/k_{-1}, and

rate=k2K1[A]=k1k2k−1[A]\text{rate} = k_2K_1[\mathrm{A}] = \frac{k_1k_2}{k_{-1}}[\mathrm{A}]

This is exactly the steady-state result with k2k_2 dropped from the denominator. The pre-equilibrium form is a limit, not a rival theory.

The opposite limit, k2≫k−1k_2 \gg k_{-1}, gives rate =k1[A]= k_1[\mathrm{A}]. Almost every I that forms goes on to P, so the first step controls the rate.

Using the approximation in exam problems

  • Apply the steady-state condition to intermediates only, never to reactants or products.
  • With two intermediates, write one steady-state equation for each and solve them together.
  • Check the answer against the limits. Does it collapse to something sensible when one rate constant dominates?

The Lindemann mechanism for unimolecular gas reactions is the same algebra with a collision partner M; the next page derives it, and Michaelis–Menten, step by step.

Try it

Switch the approximations below and watch the derivation and the limiting cases change.

Reversible formation of an intermediate
  1. Step 1A⇌k−1k1I\mathrm{A} \xrightleftharpoons[k_{-1}]{k_{1}} \mathrm{I}
  2. Step 2I→k2P\mathrm{I} \xrightarrow{k_{2}} \mathrm{P}
Steady state:
Pre-equilibrium:

  1. Rate from the elementary steps

    d[P]dt=k2[I]\frac{\mathrm{d}[\mathrm{P}]}{\mathrm{d}t} = k_{2}[\mathrm{I}]

  2. Steady state on I\mathrm{I}

    d[I]dt=k1[A]−k−1[I]−k2[I]=0\frac{\mathrm{d}[\mathrm{I}]}{\mathrm{d}t} = k_{1}[\mathrm{A}] - k_{-1}[\mathrm{I}] - k_{2}[\mathrm{I}] = 0

  3. Solve for [I][\mathrm{I}]

    [I]=k1[A]k−1+k2[\mathrm{I}] = \dfrac{k_{1}[\mathrm{A}]}{k_{-1} + k_{2}}

Rate law

d[P]dt=k1k2[A]k−1+k2\frac{\mathrm{d}[\mathrm{P}]}{\mathrm{d}t} = \dfrac{k_{1}k_{2}[\mathrm{A}]}{k_{-1} + k_{2}}

Limiting cases

  • If k−1≫k2k_{-1} \gg k_{2}:
    d[P]dt≈k1k2[A]k−1\frac{\mathrm{d}[\mathrm{P}]}{\mathrm{d}t} \approx \dfrac{k_{1}k_{2}[\mathrm{A}]}{k_{-1}}
  • If k2≫k−1k_{2} \gg k_{-1}:
    d[P]dt≈k1[A]\frac{\mathrm{d}[\mathrm{P}]}{\mathrm{d}t} \approx k_{1}[\mathrm{A}]

Quick check

  1. 1 of 3

    For A ⇌ I → P, suppose k2≫k−1k_2 \gg k_{-1}. Which rate law does the steady-state result reduce to?

  2. 2 of 3

    In the pre-equilibrium limit, kobs=k1k2/k−1k_\mathrm{obs} = k_1k_2/k_{-1}. What is the apparent activation energy?

  3. 3 of 3

    The rate of 2NO + O₂ → 2NO₂ falls as temperature rises. How can a mechanism explain a negative apparent activation energy?

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