Steady-state approximation

Real reaction mechanisms are littered with short-lived intermediates — free radicals, energised molecules, enzyme-substrate complexes — whose concentrations are essentially unmeasurable and whose exact time dependence is a nightmare of coupled differential equations. The steady-state approximation (SSA) is chemical kinetics' cleverest shortcut: instead of solving for how a reactive intermediate's concentration rises and falls, assume it stays roughly constant once formed, because it is consumed almost as fast as it forms. That one assumption converts a differential equation into simple algebra, and it is the tool that turns the Lindemann mechanism for gas-phase unimolecular reactions into a single testable rate law explaining why such reactions are first order at high pressure and second order at low pressure — a result no naive guess-the-order approach could ever produce.

The core insight

Setting $d[I]/dt\approx0$ for a fleeting, low-concentration intermediate $I$ turns an intractable differential equation into a solvable algebraic one, eliminating the unmeasurable intermediate from the final rate law.

The core idea

Where it fits. SSA is the fourth topic in the CY11001 Chemical Kinetics unit and builds directly on consecutive reactions: in $A\to B\to C$ , the intermediate $B$ rises then falls. When $B$ is much more reactive than $A$ ( $k_2\gg k_1$ ), $B$ never accumulates — it stays low and nearly constant, or "quasi-stationary." SSA formalises this quasi-stationary behaviour as something you impose on a mechanism up front, rather than something you derive by solving the full set of coupled ODEs.

The assumption itself. For any reactive intermediate $I$ in a multi-step mechanism, the steady-state approximation sets

\frac{d[I]}{dt}\approx 0

where: $[I]$ is the concentration of the intermediate species. This is justified whenever $I$ is consumed almost as fast as it is formed, so that after a brief induction period its concentration is small and roughly constant compared to the reactants and products. Setting the net rate of change of $[I]$ to zero converts a differential equation into an algebraic equation for $[I]$ in terms of the other, measurable species. That is the entire point of the trick: it eliminates the unmeasurable intermediate concentration from the final rate law, leaving an expression in terms of species you can actually monitor experimentally.

The Lindemann mechanism. The syllabus's flagship application is unimolecular gas-phase decomposition, $A\to P$ . Such reactions are not truly one elementary step; a molecule must first be energised by collision before it can react:

A+A \xrightarrow{k_1} A^*+A \qquad A^*+A \xrightarrow{k_{-1}} A+A \qquad A^* \xrightarrow{k_2} P

Here $A^*$ is the energised intermediate: formed by bimolecular collision, destroyed either by a deactivating collision or by unimolecular decomposition into products. Applying the SSA to $A^*$ and solving the resulting algebraic equation is exactly the worked derivation below.

Figure

Concentration profile showing a reactive intermediate rising briefly then staying flat and low while reactant falls and product climbs

Derivation Aof the Lindemann rate law

The goal is to eliminate the unmeasurable concentration of the energised intermediate $A^*$ from the Lindemann mechanism above and produce a single rate expression for $d[P]/dt$ purely in terms of $[A]$ and the three rate constants, one that correctly reproduces both the first-order behaviour seen at high pressure and the second-order behaviour seen at low pressure.

Step 1. Write the net rate of change of $[A^*]$ from the mechanism. $A^*$ is formed by the bimolecular activation step at rate $k_1[A]^2$ , and destroyed by two competing paths: collisional deactivation at rate $k_{-1}[A^*][A]$ and unimolecular decomposition at rate $k_2[A^*]$ . So

\frac{d[A^*]}{dt} = k_1[A]^2 - k_{-1}[A^*][A] - k_2[A^*]

Step 2. Apply the steady-state approximation. Since $A^*$ is a fleeting, low-concentration intermediate, set the left side to zero and treat what remains as algebra:

k_1[A]^2 - k_{-1}[A^*][A] - k_2[A^*] \approx 0 \;\Rightarrow\; k_1[A]^2 = [A^*]\big(k_{-1}[A]+k_2\big)

Step 3. Solve for $[A^*]$ . Dividing through isolates the intermediate's concentration entirely in terms of $[A]$ :

[A^*] = \frac{k_1[A]^2}{k_{-1}[A]+k_2}

Step 4. Substitute into the rate of product formation. Since $P$ is produced only in the final unimolecular step, $d[P]/dt = k_2[A^*]$ , so

\frac{d[P]}{dt} = \frac{k_1k_2[A]^2}{k_{-1}[A]+k_2}

where: $k_1$ is the rate constant for collisional activation, $k_{-1}$ for collisional deactivation, $k_2$ for unimolecular decomposition of $A^*$ , and $[A]$ is the concentration of the parent molecule (also the collision partner). This single expression explains why "unimolecular" reactions are first order at high pressure but second order at low pressure. At high pressure, $[A]$ is large so $k_{-1}[A]\gg k_2$ (deactivating collisions dominate); the denominator reduces to $k_{-1}[A]$ and $d[P]/dt\approx (k_1k_2/k_{-1})[A]$ , apparent first order with effective constant $k_{uni}=k_1k_2/k_{-1}$ . At low pressure, $[A]$ is small so $k_2\gg k_{-1}[A]$ (every energised molecule reacts before it can be deactivated); the denominator reduces to $k_2$ and $d[P]/dt\approx k_1[A]^2$ , apparent second order, rate-limited by the bimolecular activation step. The SSA thus produces one rate law that smoothly interpolates between two experimentally observed regimes.

Worked example

For the Lindemann mechanism, $k_1=1.5\times10^{-9}\ \text{L mol}^{-1}\text{s}^{-1}$ , $k_{-1}=2.0\times10^{9}\ \text{L mol}^{-1}\text{s}^{-1}$ , $k_2=4.0\times10^{3}\ \text{s}^{-1}$ . Find the effective rate constant $k_{uni}=\dfrac{k_1k_2[A]}{k_{-1}[A]+k_2}$ at (a) $[A]=1.0\ \text{mol L}^{-1}$ and (b) $[A]=1.0\times10^{-4}\ \text{mol L}^{-1}$ , and identify the kinetic regime in each case.

Step 1. Case (a), $[A]=1.0\ \text{mol L}^{-1}$ : $k_{-1}[A]=2.0\times10^9\times1.0=2.0\times10^9\ \text{s}^{-1}$ , which is far larger than $k_2=4.0\times10^3\ \text{s}^{-1}$ . We are deep in the high-pressure limit, so the denominator $\approx k_{-1}[A]$ :

k_{uni}\approx\frac{k_1k_2}{k_{-1}}=\frac{1.5\times10^{-9}\times4.0\times10^3}{2.0\times10^9}=3.0\times10^{-15}\ \text{s}^{-1}

This is pressure-independent, first-order behaviour; the exact formula gives essentially the same value since $k_2/(k_{-1}[A])\sim10^{-6}$ is negligible.

Step 2. Case (b), $[A]=1.0\times10^{-4}\ \text{mol L}^{-1}$ : $k_{-1}[A]=2.0\times10^9\times1.0\times10^{-4}=2.0\times10^5\ \text{s}^{-1}$ , still larger than $k_2=4.0\times10^3\ \text{s}^{-1}$ but not overwhelmingly, so use the full expression:

k_{uni}=\frac{k_1k_2[A]}{k_{-1}[A]+k_2}=\frac{6.0\times10^{-10}}{2.04\times10^5}=2.94\times10^{-15}\ \text{s}^{-1}

This is close to, but measurably below, the high-pressure value from (a) — fall-off has begun though we are not yet in the true low-pressure regime.

Answer: $k_{uni}=3.0\times10^{-15}\ \text{s}^{-1}$ at high $[A]$ (first-order, high-pressure limit) and $k_{uni}=2.94\times10^{-15}\ \text{s}^{-1}$ at low $[A]$ (mild fall-off toward the low-pressure limit).

Competitive stretch

Show, using the SSA, that the mechanism $A_2\xrightarrow{k_1}2A$ (fast, reversible with $k_{-1}$ ) followed by $A+B\xrightarrow{k_2}\text{products}$ (slow) gives a rate law that is half-order in $[A_2]$ when the reverse step dominates.

Step 1. SSA condition for the intermediate $A$ : $\dfrac{d[A]}{dt}=2k_1[A_2]-2k_{-1}[A]^2-k_2[A][B]\approx0$ .

Step 2. If reverse recombination is fast compared to the slow bimolecular step, $k_2[A][B]$ is negligible next to $k_{-1}[A]^2$ , so $2k_1[A_2]\approx2k_{-1}[A]^2$ , giving $[A]\approx(k_1/k_{-1})^{1/2}[A_2]^{1/2}$ .

Step 3. Overall rate $=k_2[A][B]=k_2(k_1/k_{-1})^{1/2}[A_2]^{1/2}[B]$ — first order in $[B]$ but half order in $[A_2]$ , a fractional order that cannot come from a single elementary step but emerges naturally here; this limit coincides with the pre-equilibrium approximation.

Common traps

Applying SSA to a species that actually accumulates. SSA only holds for a genuinely reactive, low-concentration intermediate — applying $d[I]/dt\approx0$ to the main reactant or a stable product gives nonsense.
Forgetting SSA gives an algebraic, not differential, equation. The point is to solve $d[I]/dt=0$ for $[I]$ , then substitute back — never leave $[I]$ in the final rate law.
Missing one of the two limiting regimes of Lindemann. A common exam trap is checking only the high-pressure (first-order) limit and forgetting the low-pressure (second-order) one — both come from the same denominator, approximated two different ways.
Confusing SSA with the rate-determining-step approximation. SSA zeroes an intermediate's net rate of change (species-level); RDS singles out one step as much slower than the others (step-level). Lindemann actually combines both ideas near the low-pressure limit.

Ask the tutor: if you would like a walkthrough comparing this SSA result to the exact numerical solution of the Lindemann mechanism's coupled ODEs, or want the enzyme-kinetics analogy sketched out.

How is the steady-state approximation different from just saying the reaction has reached equilibrium?

They are different ideas entirely. Equilibrium means the forward and reverse rates of one specific step are equal, so that step's net rate is zero, but the system as a whole may still be moving forward slowly through some other step. SSA instead says the net rate of change of an intermediate's concentration is zero — it is formed and destroyed by possibly several different steps whose rates need not individually balance, only their sum must. Pre-equilibrium is actually a special case of SSA, arising when one destruction pathway (the reverse step) happens to dominate all the others.

Why does the Lindemann mechanism need three steps instead of just writing A to P directly?

Because a single "unimolecular" collision cannot supply a molecule with enough energy to react on its own — energy has to arrive first, and the only realistic way a gas molecule gains energy is by colliding with another molecule. So the mechanism first energises $A$ into $A^*$ by collision, allows that collision to be reversed by a deactivating collision, and only then lets $A^*$ decompose unimolecularly. Writing $A\to P$ directly would hide exactly the physics that explains the pressure dependence of the observed order — the SSA on $A^*$ is what recovers that physics as a rate law.

If I forget which pressure limit gives which order, how do I rederive it fast under exam pressure?

Look only at the denominator $k_{-1}[A]+k_2$ from the SSA result and ask which term is bigger. High pressure means lots of $A$ molecules around, so collisional deactivation ( $k_{-1}[A]$ ) wins over unimolecular decomposition ( $k_2$ ); the denominator collapses to $k_{-1}[A]$ , cancelling one power of $[A]$ from the numerator and leaving first order. Low pressure means few collisions, so $k_2$ wins; the denominator collapses to the constant $k_2$ , leaving the full $[A]^2$ from the numerator, i.e. second order. You never need to memorise the labels — just compare the two denominator terms.

Steady-state approximation

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads