Pre-Equilibrium Approximation

Nitric oxide reacts with oxygen to form NO₂ faster when you cool the mixture down. That is backwards from almost every reaction you meet in introductory kinetics, and it is a clue about mechanism. The trick is a fast, reversible step that builds up an intermediate, followed by a slow step that consumes it. This pattern—rapid pre-equilibrium followed by a rate-limiting step—shows up constantly in organic, inorganic, and biochemical mechanisms. It lets chemists convert a messy multi-step mechanism into one clean, testable rate law, and it explains why some reactions have negative or unusual apparent activation energies. This article builds the pre-equilibrium approximation from the reversible-reaction framework you already know. You will see the key assumption stated precisely. Then the composite rate constant is derived step by step. Finally, two worked numerical examples show how to extract elementary rate constants from bulk kinetic data.

Key takeaways

Pre-equilibrium applies when a reversible first step is much faster than a subsequent slow step.
The fast step is treated as if at equilibrium, so $K_1 = k_1/k_{-1}$ fixes the intermediate's concentration.
Substituting into the slow step's rate law gives an observed rate constant that is a composite, $k_{\text{obs}} = k_2 K_1$ .
The apparent activation energy becomes $E_{a,\text{obs}} = E_{a,2} + \Delta H_1^{\ominus}$ , which can be negative.
The approximation is a special, algebraically simpler case of the steady-state approximation.
Key idea: when a reversible step re-equilibrates much faster than it is drained by a slow step, its equilibrium constant—not its forward rate constant—controls the overall rate.

The core insight

When an intermediate is formed and destroyed by a fast, reversible step, and consumed only slowly by a subsequent step, the fast step stays essentially at equilibrium throughout the reaction, so the intermediate's concentration is set by $K_1$ rather than by the kinetics of its formation, and the observed rate law is obtained by substituting this equilibrium concentration into the rate expression for the slow, rate-determining step.

The physical picture

Consider a mechanism with an intermediate $I$ formed reversibly from reactants and then consumed to give products:

\text{A} + \text{B} \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} \text{I} \xrightarrow{k_2} \text{P}

If $k_{-1} \gg k_2$ , then $I$ reacts back to A + B many times for every one time it proceeds forward to products. The first step effectively equilibrates on a timescale much shorter than the timescale over which P accumulates. Picture two people passing a ball back and forth rapidly (the fast equilibrium) while, only occasionally, someone in the crowd steals it away permanently (the slow step). The stealing rate depends on how many people are holding the ball at any instant—but that population is set by the fast game, not by the thief. This is different from the more general steady-state approximation, which only requires that $I$ 's concentration change slowly (small $d[I]/dt$ ), a weaker and broader condition. Pre-equilibrium is the special case where the forward and reverse steps of the first reaction are individually fast compared with the drain step, so equilibrium—not merely a constant low concentration—is achieved. The practical payoff: instead of solving coupled differential equations, you can write $K_1 = [\text{I}]/([\text{A}][\text{B}])$ directly, because equilibrium algebra replaces kinetic algebra for that step.

Figure

Energy profile: fast reversible step feeding a slow rate-determining step

Figure

Concentration vs time: I equilibrates quickly, P grows slowly

The core relations

K_1 = \frac{k_1}{k_{-1}} = \frac{[\text{I}]}{[\text{A}][\text{B}]}

where $K_1$ is the equilibrium constant of the fast reversible step.

\text{rate} = k_2[\text{I}] = k_2 K_1 [\text{A}][\text{B}]

where $k_2$ is the rate constant of the slow, rate-determining step.

k_{\text{obs}} = k_2 K_1 = \frac{k_2 k_1}{k_{-1}}

where $k_{\text{obs}}$ is the single composite rate constant seen in an experimental rate law.

E_{a,\text{obs}} = E_{a,2} + \Delta H_1^{\ominus}

where $\Delta H_1^{\ominus} = E_{a,1} - E_{a,-1}$ is the standard reaction enthalpy of the fast pre-equilibrium step.

The core relations — validity condition

k_{-1} \gg k_2[\text{other reactant}]

where this inequality must hold for the equilibrium assumption to be self-consistent.

Derivation AFrom coupled rate laws to a single composite rate constant

Goal. Starting from the full mass-action rate equations for A + B ⇌ I → P, derive the observed rate law and show it reduces to $k_{\text{obs}} = k_2 K_1$ under the pre-equilibrium condition.

Step 1. Write the exact rate equations. The intermediate I is formed and consumed by three elementary steps, so

\frac{d[\text{I}]}{dt} = k_1[\text{A}][\text{B}] - k_{-1}[\text{I}] - k_2[\text{I}]

This is exact; no approximation has been made yet.

Step 2. Compare the two loss channels for I. The intermediate disappears either by reverting to A + B (rate constant $k_{-1}$ ) or by moving on to product (rate constant $k_2$ ). The pre-equilibrium condition is precisely the statement

k_{-1} \gg k_2

so that reversion back to reactants happens vastly more often than conversion to product. Physically, I samples the fast equilibrium many times before it "leaks" toward P.

Step 3. Impose quasi-equilibrium on the fast step. Because loss through $k_{-1}$ dominates the right-hand side, the fast forward and reverse fluxes very nearly cancel at every instant:

k_1[\text{A}][\text{B}] \approx k_{-1}[\text{I}]

This is stronger than steady state ( $d[\text{I}]/dt \approx 0$ overall); here the two large terms cancel each other while the small $k_2[\text{I}]$ term is neglected in comparison, consistent with Step 2.

Step 4. Solve for the intermediate's concentration. Rearranging Step 3,

[\text{I}] \approx \frac{k_1}{k_{-1}}[\text{A}][\text{B}] = K_1[\text{A}][\text{B}]

This says [I] tracks the instantaneous concentrations of A and B exactly as an equilibrium constant would predict, even while A and B are slowly being consumed overall.

Step 5. Insert into the rate law for product formation. The rate of appearance of P is controlled only by the slow step,

\frac{d[\text{P}]}{dt} = k_2[\text{I}] = k_2 K_1 [\text{A}][\text{B}]

Step 6. Identify the composite rate constant. Comparing with an experimental second-order rate law, rate $= k_{\text{obs}}[\text{A}][\text{B}]$ , gives directly

\boxed{k_{\text{obs}} = k_2 K_1 = \dfrac{k_2 k_1}{k_{-1}}}

Step 7. Translate into activation energies. Writing each rate constant in Arrhenius form, $k_i = A_i e^{-E_{a,i}/RT}$ , and $K_1 = e^{-\Delta G_1^{\ominus}/RT}$ with $\Delta G_1^{\ominus} \approx \Delta H_1^{\ominus} - T\Delta S_1^{\ominus}$ , taking $d(\ln k_{\text{obs}})/dT$ term by term yields

E_{a,\text{obs}} = E_{a,2} + \Delta H_1^{\ominus}

If the pre-equilibrium is exothermic enough ( $\Delta H_1^{\ominus}$ sufficiently negative), $E_{a,\text{obs}}$ can even become negative, so the overall reaction slows down as temperature rises—exactly the anomaly seen in $2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2$ .

Worked examples

Example 1 — Extracting $k_{\text{obs}}$ from elementary constants (foundational). The gas-phase reaction $2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2$ is believed to proceed by

\text{NO} + \text{NO} \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} \text{N}_2\text{O}_2, \qquad \text{N}_2\text{O}_2 + \text{O}_2 \xrightarrow{k_2} 2\text{NO}_2

Suppose at 300 K the elementary rate constants are $k_1 = 1.2\times10^{9}\ \text{dm}^3\text{mol}^{-1}\text{s}^{-1}$ , $k_{-1} = 4.0\times10^{9}\ \text{s}^{-1}$ , and $k_2 = 3.0\times10^{6}\ \text{dm}^3\text{mol}^{-1}\text{s}^{-1}$ . Step 1: check the pre-equilibrium condition. We need $k_{-1} \gg k_2 \times$ (typical [O₂]). With [O₂] ~ 1 mol dm⁻³ or less under normal lab conditions, $k_2[\text{O}_2] \le 3.0\times10^{6}\ \text{s}^{-1}$ , which is indeed far smaller than $k_{-1} = 4.0\times10^{9}\ \text{s}^{-1}$ (ratio ~1300). The approximation is well justified. Step 2: compute $K_1$ .

K_1 = \frac{k_1}{k_{-1}} = \frac{1.2\times10^{9}}{4.0\times10^{9}} = 0.30\ \text{dm}^3\text{mol}^{-1}

Step 3: compute $k_{\text{obs}}$ .

k_{\text{obs}} = k_2 K_1 = (3.0\times10^{6}\ \text{dm}^3\text{mol}^{-1}\text{s}^{-1})(0.30\ \text{dm}^3\text{mol}^{-1}) = 9.0\times10^{5}\ \text{dm}^6\text{mol}^{-2}\text{s}^{-1}

Step 4: sanity check units. The overall rate law is rate $= k_{\text{obs}}[\text{NO}]^2[\text{O}_2]$ , third order overall, so $k_{\text{obs}}$ must carry units of $\text{dm}^6\text{mol}^{-2}\text{s}^{-1}$ —consistent with what we obtained. This matches the experimentally observed third-order rate law for this reaction (second order in NO, first order in O₂), and the magnitude $\sim10^{6}\ \text{dm}^6\text{mol}^{-2}\text{s}^{-1}$ is the right order for a termolecular-looking gas reaction proceeding through a bound intermediate.

Example 2 — Working backward: apparent activation energy (applied, harder). For the same mechanism, experiments give the elementary Arrhenius parameters: $E_{a,1} = 0\ \text{kJ mol}^{-1}$ (barrierless radical-radical association), $E_{a,-1} = 21\ \text{kJ mol}^{-1}$ (N₂O₂ bond dissociation), and $E_{a,2} = 25\ \text{kJ mol}^{-1}$ . An experimentalist measures $k_{\text{obs}}$ at two temperatures and wants to predict the apparent activation energy without measuring a full Arrhenius plot. Step 1: find $\Delta H_1^{\ominus}$ .

\Delta H_1^{\ominus} = E_{a,1} - E_{a,-1} = 0 - 21\ \text{kJ mol}^{-1} = -21\ \text{kJ mol}^{-1}

This is negative, confirming that dimerization of NO to N₂O₂ is exothermic, as expected for bond formation. Step 2: apply the composite activation energy relation.

E_{a,\text{obs}} = E_{a,2} + \Delta H_1^{\ominus} = 25\ \text{kJ mol}^{-1} + (-21\ \text{kJ mol}^{-1}) = 4\ \text{kJ mol}^{-1}

Step 3: interpret and cross-check against a known fact. $E_{a,\text{obs}} = 4\ \text{kJ mol}^{-1}$ is small and positive here—but note that with only slightly more exothermic pre-equilibrium (say $\Delta H_1^{\ominus} = -30\ \text{kJ mol}^{-1}$ ), $E_{a,\text{obs}}$ would flip negative. This is exactly the experimentally documented behavior of $2\text{NO}+\text{O}_2 \rightarrow 2\text{NO}_2$ : its rate constant decreases modestly as temperature increases, corresponding to a small negative observed activation energy in the range $-4$ to $-6\ \text{kJ mol}^{-1}$ depending on the data set used, a hallmark signature that a pre-equilibrium step lies upstream of the true rate-determining step. Step 4: quantify the temperature sensitivity. Using $k_{\text{obs}}(T_2)/k_{\text{obs}}(T_1) = \exp\left[-\dfrac{E_{a,\text{obs}}}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)\right]$ with $E_{a,\text{obs}} = 4000\ \text{J mol}^{-1}$ , going from 280 K to 320 K gives

\ln\frac{k_{\text{obs}}(320)}{k_{\text{obs}}(280)} = -\frac{4000}{8.314}\left(\frac{1}{320}-\frac{1}{280}\right) = -481.3 \times (-4.46\times10^{-4}) = 0.215

so $k_{\text{obs}}$ increases by only about 24% over that 40 K range—a far weaker temperature dependence than a normal single-step reaction with $E_a \sim 50$ – $100\ \text{kJ mol}^{-1}$ would show, confirming the diagnostic value of a small or negative $E_{a,\text{obs}}$ .

Limiting cases & common traps

Situation	What happens
$k_{-1} \gg k_2$	Pre-equilibrium valid; use $k_{\text{obs}} = k_2K_1$
$k_{-1} \sim k_2$ or $k_{-1} \ll k_2$	Pre-equilibrium fails; use the full steady-state treatment instead
$\Delta H_1^{\ominus}$ strongly negative	$E_{a,\text{obs}}$ can become negative; rate falls as $T$ rises
Reactant concentration very high, raising the effective drain rate $k_2[\text{X}]$	Can violate $k_{-1}\gg k_2[\text{X}]$ even if $k_{-1}\gg k_2$ alone looked fine
Only one elementary rate constant known	Cannot get $k_{\text{obs}}$ ; need $k_1$ , $k_{-1}$ , and $k_2$ , or $K_1$ and $k_2$

The chemist behind it

The pre-equilibrium idea grew out of early twentieth-century efforts to rationalize rate laws that did not match simple bimolecular collision pictures, particularly reactions with fractional or unusual orders and anomalous temperature dependence. Physical chemists studying nitrogen oxide chemistry and ester hydrolysis mechanisms in the 1920s–1930s needed a way to reconcile fast acid-base or association equilibria with a separate slow chemical step, and the algebraic shortcut of substituting an equilibrium constant into a rate law became a standard tool taught alongside the steady-state method developed at the same time.

Origins · 1 minVictor Henri and how this was discovered

Victor Henri

1872–1940 · pioneered treating enzyme-substrate binding as a rapid pre-equilibrium, laying groundwork for Michaelis-Menten kinetics

Ask the tutor: How does the pre-equilibrium approximation simplify into the Michaelis-Menten rate law when an enzyme's substrate-binding step is fast?

How is pre-equilibrium different from the steady-state approximation?

Steady state assumes only that $d[\text{I}]/dt \approx 0$ overall—the intermediate's concentration is small and roughly constant, but the forward and reverse fluxes of the first step need not individually dominate. Pre-equilibrium is stricter: it requires $k_{-1} \gg k_2$ , so the first step's forward and reverse rates nearly cancel independently of the slow step. Every pre-equilibrium result can be derived as a limiting case of the full steady-state expression, but not every steady-state system satisfies the pre-equilibrium condition.

Why can the apparent activation energy be negative?

The composite rate constant $k_{\text{obs}} = k_2K_1$ combines a genuine Arrhenius rate constant $k_2$ with an equilibrium constant $K_1$ , which itself falls with temperature if the pre-equilibrium step is exothermic (van 't Hoff behavior). If that fall is steeper than the rise in $k_2$ , the product $k_{\text{obs}}$ decreases with temperature overall, mathematically equivalent to a negative $E_{a,\text{obs}} = E_{a,2}+\Delta H_1^{\ominus}$ . This is a real, measurable phenomenon, not an artifact.

Can pre-equilibrium apply to more than two steps?

Yes. If several fast, reversible steps precede one slow, rate-determining step, each fast equilibrium contributes its own equilibrium constant to the composite expression, and $k_{\text{obs}}$ becomes a product of the slow step's rate constant with all the preceding $K_i$ values. This generalization is exactly how enzyme kinetics treats multi-step substrate binding before catalysis, and how acid-catalyzed mechanisms treat rapid protonation before a slow bond-breaking step.

How do I know experimentally that a mechanism follows pre-equilibrium rather than a single elementary step?

Key signatures include a rate law whose order does not match simple collision theory (for example, third order overall from formally bimolecular steps), an unusually small or negative apparent activation energy, and successful independent measurement of $K_1$ (from spectroscopy or equilibrium studies) that, combined with an independently measured $k_2$ , reproduces the observed $k_{\text{obs}}$ . Isotope-exchange experiments on the proposed intermediate provide further supporting evidence.

Where does pre-equilibrium lead next in kinetics?

The same logic—fast reversible binding followed by a slow catalytic step—underlies the Michaelis-Menten treatment of enzyme kinetics, where substrate binding to the enzyme is often (though not always) fast compared with the catalytic turnover step. It also connects to transition-state theory, which treats the formation of the activated complex itself as a quasi-equilibrium with the reactants, extending the same algebraic trick to the rate-determining step itself.