Lindemann-Hinshelwood Unimolecular Mechanism

A gas-phase molecule like cyclopropane isomerizing to propene looks like a simple first-order reaction. Yet a single molecule cannot spontaneously rearrange without first absorbing energy from somewhere. That "somewhere" is collisions with other molecules, and the story of how random collisions convert into a clean first-order rate law is one of kinetics' most instructive puzzles. This puzzle matters because unimolecular reactions are common — isomerizations, decompositions, dissociations — and their apparent order changes with pressure. A reaction that behaves as first order at high pressure can turn second order as the gas is diluted. Explaining this shift, rather than just describing it, requires a mechanism. This article builds the Lindemann–Hinshelwood model from first principles using the steady-state approximation. You will see how collisional activation and deactivation compete with reaction. You will derive the pressure-dependent rate law. Then you will apply it to real data and see where the simple model breaks down, setting up the refinements that lead toward chain reaction kinetics.

Key takeaways

Unimolecular reactions need energized molecules, and collisions supply that energy — reaction is never truly "spontaneous."
Lindemann proposed a two-step mechanism: collisional activation/deactivation followed by unimolecular reaction of the excited species.
Applying the steady-state approximation to the energized intermediate gives a rate law with pressure-dependent effective order.
At high pressure the reaction is first order; at low pressure it becomes second order, with a smooth transition in between.
Real data curve below the Lindemann prediction at low pressure, motivating the Hinshelwood and later RRKM refinements.
Key idea: unimolecular kinetics are actually bimolecular in origin — collisions create the reactive species, and the observed order reflects a competition between collisional supply and unimolecular decay.

The core insight

A unimolecular reaction rate is controlled by how fast collisions can energize molecules relative to how fast those energized molecules react or get deactivated, and the steady-state treatment of this competition naturally produces a rate law that interpolates between first and second order as pressure changes.

The physical picture

Consider a gas-phase molecule A undergoing an apparently first-order reaction, A → products, such as the isomerization of cyclopropane to propene. A molecule sitting quietly in its ground vibrational state has no way to spontaneously rearrange its bonds. It must first acquire enough internal (vibrational) energy to cross the reaction barrier. In a gas, that energy comes from collisions with other molecules — including other A molecules. Frederick Lindemann (1922) proposed that unimolecular reactions actually proceed through an energized intermediate, A, formed by collision: A + A → A + A (activation, rate constant $k_1$ ) A + A → A + A (deactivation, rate constant $k_{-1}$ ) A → products (unimolecular decay, rate constant $k_2$ ) The energized molecule A has enough energy to react, but it does not react instantly — there is a time lag while the excess energy redistributes among vibrational modes before it can localize in the bond that breaks. During that lag, A can be deactivated by another collision before it ever reacts. This competition between deactivation and reaction is the heart of the mechanism, and it is exactly the kind of transient-intermediate problem the steady-state approximation was built to handle. At high pressure, collisions are frequent: almost every A is deactivated before it can react, but a large population of A is also constantly being replenished, so the fraction that reacts corresponds to simple first-order kinetics in A. At low pressure, collisions are rare: any A formed reacts before it can be deactivated, so the rate-limiting step becomes the formation of A itself — a bimolecular, second-order process.

Figure

Lindemann mechanism: energy flow from collision to product

Figure

Effective order versus pressure

The core relations

\frac{d[\text{A}^*]}{dt} = k_1[\text{A}]^2 - k_{-1}[\text{A}^*][\text{A}] - k_2[\text{A}^*] \approx 0

where: $k_1$ is the bimolecular activation rate constant, $k_{-1}$ the bimolecular deactivation constant, $k_2$ the unimolecular decay constant of A, and the steady-state approximation sets the net rate of change of A to zero.

k_{\text{obs}} = \frac{k_1 k_2 [\text{A}]}{k_{-1}[\text{A}] + k_2}

where: $k_{\text{obs}}$ is the experimentally observed first-order-in-appearance rate constant for product formation, $[\text{A}]$ is the total (bath) concentration of reactant, proportional to pressure.

k_{\text{obs}}^{\text{high } p} = \frac{k_1 k_2}{k_{-1}} = k_\infty, \qquad k_{\text{obs}}^{\text{low } p} = k_1[\text{A}]

where: $k_\infty$ is the limiting high-pressure first-order rate constant; at low pressure the rate law becomes second order overall.

\frac{1}{k_{\text{obs}}} = \frac{1}{k_1[\text{A}]} + \frac{k_{-1}}{k_1 k_2}

where: this linearized (Lindemann plot) form lets $k_1$ and $k_2/k_{-1}$ be extracted from a plot of $1/k_{\text{obs}}$ against $1/[\text{A}]$ .

Derivation AFrom the elementary steps to the pressure-dependent rate law

Goal. Derive $k_{\text{obs}}$ for $d[\text{P}]/dt = k_{\text{obs}}[\text{A}]$ starting from the three elementary steps of the Lindemann mechanism, using the steady-state approximation on the reactive intermediate A*.

Step 1 — Write the elementary rate expressions. The three steps are activation, deactivation, and reaction:

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

The product formation rate is simply

\frac{d[\text{P}]}{dt} = k_2[\text{A}^*]

so everything reduces to finding the steady-state concentration of A*.

Step 2 — Apply the steady-state approximation to A.</strong> A is a highly reactive, low-concentration transient species — exactly the profile for which the steady-state approximation (built on the intermediate never accumulating appreciably) applies. Setting its net rate of formation to zero:

$\frac{d[\text{A}^*]}{dt} = k_1[\text{A}]^2 - k_{-1}[\text{A}][\text{A}^*] - k_2[\text{A}^*] = 0$

Step 3 — Solve for [A*]. Collect the terms containing $[\text{A}^*]$ :

$k_1[\text{A}]^2 = [\text{A}^*]\left(k_{-1}[\text{A}] + k_2\right)$

so that

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

Step 4 — Substitute back into the product rate.

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

Writing this as $k_{\text{obs}}[\text{A}]$ defines the observed rate constant:

$k_{\text{obs}} = \frac{k_1 k_2[\text{A}]}{k_{-1}[\text{A}] + k_2}$

Step 5 — Examine the two pressure limits. When pressure (and hence $[\text{A}]$ ) is high, $k_{-1}[\text{A}] \gg k_2$ , so the denominator is dominated by $k_{-1}[\text{A}]$ :

$k_{\text{obs}} \to \frac{k_1 k_2[\text{A}]}{k_{-1}[\text{A}]} = \frac{k_1 k_2}{k_{-1}} \equiv k_\infty$

This is pressure-independent — genuinely first order. When pressure is low, $k_{-1}[\text{A}] \ll k_2$ , so the denominator collapses to $k_2$ :

$k_{\text{obs}} \to \frac{k_1 k_2[\text{A}]}{k_2} = k_1[\text{A}]$

This is second order overall (first order in $k_{\text{obs}}$ 's prefactor times $[\text{A}]$ , so rate $= k_1[\text{A}]^2$ ).

Step 6 — Linearize for data analysis. Taking the reciprocal of the full expression:

$\frac{1}{k_{\text{obs}}} = \frac{k_{-1}[\text{A}] + k_2}{k_1 k_2 [\text{A}]} = \frac{1}{k_1[\text{A}]} + \frac{k_{-1}}{k_1 k_2}$

A plot of $1/k_{\text{obs}}$ against $1/[\text{A}]$ is a straight line: slope $1/k_1$ , intercept $k_{-1}/(k_1k_2)$ .

$\boxed{k_{\text{obs}} = \frac{k_1 k_2[\text{A}]}{k_{-1}[\text{A}] + k_2} \;\xrightarrow{p\to\infty}\; k_\infty = \frac{k_1k_2}{k_{-1}} \;\xrightarrow{p\to 0}\; k_1[\text{A}]}$

Worked examples

Example 1 — Foundational: computing $k_{\text{obs}}$ at two pressures. Suppose for a model unimolecular isomerization the elementary constants are

k_1 = 4.0\times10^{-10}\ \text{cm}^3\,\text{molecule}^{-1}\,\text{s}^{-1}, \quad k_{-1} = 3.0\times10^{-10}\ \text{cm}^3\,\text{molecule}^{-1}\,\text{s}^{-1}, \quad k_2 = 5.0\times10^{4}\ \text{s}^{-1}

First find $k_\infty$ :

k_\infty = \frac{k_1 k_2}{k_{-1}} = \frac{(4.0\times10^{-10})(5.0\times10^{4})}{3.0\times10^{-10}} = 6.7\times10^{4}\ \text{s}^{-1}

Now evaluate at a "high" concentration, $[\text{A}] = 1.0\times10^{18}\ \text{molecule cm}^{-3}$ (roughly 40 torr at 298 K):

k_{-1}[\text{A}] = (3.0\times10^{-10})(1.0\times10^{18}) = 3.0\times10^{8}\ \text{s}^{-1} \gg k_2 = 5.0\times10^{4}\ \text{s}^{-1}

so we are deep in the high-pressure regime and $k_{\text{obs}} \approx k_\infty = 6.7\times10^{4}\ \text{s}^{-1}$ . Direct substitution confirms this:

k_{\text{obs}} = \frac{(4.0\times10^{-10})(5.0\times10^{4})(1.0\times10^{18})}{3.0\times10^{8} + 5.0\times10^{4}} = \frac{2.0\times10^{13}}{3.0\times10^{8}} = 6.7\times10^{4}\ \text{s}^{-1}

matching $k_\infty$ to two significant figures, as expected. Now evaluate at low concentration, $[\text{A}] = 1.0\times10^{13}\ \text{molecule cm}^{-3}$ (about $4\times10^{-7}$ torr):

k_{-1}[\text{A}] = (3.0\times10^{-10})(1.0\times10^{13}) = 3.0\times10^{3}\ \text{s}^{-1} \ll k_2 = 5.0\times10^4\ \text{s}^{-1}

so we're deep in the low-pressure regime:

k_{\text{obs}} \approx k_1[\text{A}] = (4.0\times10^{-10})(1.0\times10^{13}) = 4.0\times10^{3}\ \text{s}^{-1}

Exact substitution gives

k_{\text{obs}} = \frac{(4.0\times10^{-10})(5.0\times10^{4})(1.0\times10^{13})}{3.0\times10^{3}+5.0\times10^{4}} = \frac{2.0\times10^{8}}{5.3\times10^{4}} = 3.8\times10^{3}\ \text{s}^{-1}

close to the approximation, with the small difference reflecting that $k_{-1}[\text{A}]$ is not fully negligible here (about 6% of $k_2$ ). The rate constant dropped by roughly a factor of 17 as concentration dropped by a factor of $10^5$ — much slower than the factor-of- $10^5$ drop that pure second order (rate $=k_1[\text{A}]^2$ , so $k_{\text{obs}}=k_1[\text{A}]$ scales linearly) would demand at even lower pressure, illustrating the smooth crossover.

Example 2 — Applied: extracting $k_1$ and $k_\infty$ from pressure-dependent data (cyclopropane-type isomerization). Suppose four measurements of the observed first-order rate constant for a gas-phase isomerization at 470°C give:

$p$ / torr	$k_{\text{obs}}$ / $10^{-4}\,\text{s}^{-1}$
84	4.5
208	8.4
400	10.8
760	12.7

Convert pressure to concentration via $[\text{A}] = p/(RT)$ , using $R = 62.36\ \text{L}\,\cdot\,\text{torr}\,\cdot\,\text{K}^{-1}\text{mol}^{-1}$ and $T = 743\ \text{K}$ :

[\text{A}] (\text{mol L}^{-1}) = \frac{p\,(\text{torr})}{(62.36)(743)} = \frac{p}{4.634\times10^{4}}

For $p = 84$ torr: $[\text{A}] = 1.81\times10^{-3}\ \text{mol L}^{-1}$ , so $1/[\text{A}] = 552\ \text{L mol}^{-1}$ . For $p = 760$ torr: $[\text{A}] = 1.640\times10^{-2}\ \text{mol L}^{-1}$ , so $1/[\text{A}] = 61.0\ \text{L mol}^{-1}$ . Compute $1/k_{\text{obs}}$ (in units of $10^{4}\,\text{s}$ ):

$p=84$ : $1/k_{\text{obs}} = 1/(4.5\times10^{-4}) = 2222\ \text{s}$
$p=760$ : $1/k_{\text{obs}} = 1/(12.7\times10^{-4}) = 787\ \text{s}$

Using these two points to find the Lindemann-plot slope and intercept:

\text{slope} = \frac{2222 - 787}{552 - 61.0} = \frac{1435}{491} = 2.92\ \text{s}\,\cdot\,\text{L mol}^{-1}

Since slope $=1/k_1$ :

k_1 = \frac{1}{2.92} = 0.342\ \text{L mol}^{-1}\text{s}^{-1}

The intercept is found from either point; using $p=760$ :

\text{intercept} = 787 - (2.92)(61.0) = 787 - 178 = 609\ \text{s}

Since intercept $= k_{-1}/(k_1k_2)$ and $k_\infty = k_1k_2/k_{-1} = 1/\text{intercept}$ :

k_\infty = \frac{1}{609} = 1.64\times10^{-3}\ \text{s}^{-1}

Sanity check: at the highest pressure measured (760 torr), $k_{\text{obs}} = 1.27\times10^{-3}\ \text{s}^{-1}$ , which is 77% of the extrapolated $k_\infty = 1.64\times10^{-3}\ \text{s}^{-1}$ — consistent with 760 torr being high but not yet fully in the pressure-independent plateau, exactly the pattern real cyclopropane data show.

Limiting cases & common traps

Regime	Condition	Rate law	Order
High pressure	$k_{-1}[\text{A}] \gg k_2$	$\text{rate} = k_\infty[\text{A}]$	1
Low pressure	$k_{-1}[\text{A}] \ll k_2$	$\text{rate} = k_1[\text{A}]^2$	2
Fall-off region	$k_{-1}[\text{A}] \sim k_2$	full Lindemann expression	non-integer, pressure-dependent

The "pressure of half-transition," $p_{1/2}$ , is where $k_{-1}[\text{A}] = k_2$ and $k_{\text{obs}} = k_\infty/2$ .
Real low-pressure data fall systematically below the simple Lindemann line — because not every collision activates a molecule, and the model treats $k_1$ as pressure-independent even though the energy-dependence of activation is more subtle. This motivated Hinshelwood's refinement using an energy-dependent activation rate from a Boltzmann-weighted collision pool, and later the full RRK/RRKM treatments.
The mechanism assumes a single energized state A* with one lifetime; in reality molecules are activated to a spread of energies with different reactivities.

The chemist behind it

Frederick Lindemann (later Lord Cherwell) proposed the collisional-activation idea in 1922 to resolve a genuine paradox: how can a reaction that follows first-order kinetics arise from an inherently collisional, bimolecular activation process? His insight — that there is a time delay between energization and reaction, during which deactivating collisions compete with decay — was mechanistically correct but initially lacked quantitative rigor about how activation rates depend on energy. Cyril Hinshelwood supplied much of that quantitative machinery in the 1920s, using statistical mechanics to estimate activation rates as a function of the number of vibrational modes over which energy could be distributed, which is why the model is jointly named after both. Their combined framework was the first to explain why unimolecular reactions such as the isomerization of cyclopropane or the decomposition of azomethane shift from first to second order as pressure is lowered, and it remains the conceptual foundation — later refined by Rice, Ramsperger, Kassel, and Marcus into RRKM theory — for all modern treatments of gas-phase unimolecular kinetics.

Origins · 1 minCyril Norman Hinshelwood and how this was discovered

Cyril Norman Hinshelwood

1897–1967 · Shared the 1956 Nobel Prize in Chemistry with Nikolai Semyonov for work on the mechanisms of chemical reactions, including unimolecular and chain reactions

:::quiz[A Lindemann plot of 1/kobs against 1/[A] is used to determine]

The activation energy directly from the y-intercept only
The equilibrium constant of the overall reaction
k1 from the slope and k-1/(k1k2) from the intercept ✓
The reaction order directly from the intercept

Ask the tutor: How does RRKM theory refine the single-rate-constant assumption for A* in this mechanism?

Why isn't a "unimolecular" reaction just first order at all pressures

Because the reactive species A is created by collisions, its supply rate depends on pressure. At high pressure, collisions are so frequent that A is always in a quasi-steady abundance proportional to [A], giving clean first-order kinetics. At low pressure, collisions become rare enough that forming A* is the bottleneck, making the overall rate depend on the square of the concentration. The "unimolecular" label refers only to the final reactive step, not the whole mechanism.

What is the physical meaning of p1/2, the half-pressure

$p_{1/2}$ is the pressure at which the deactivation rate $k_{-1}[\text{A}]$ exactly equals the unimolecular decay rate $k_2$ , so half of all energized A* molecules react before being deactivated and half are deactivated first. At this pressure $k_{\text{obs}} = k_\infty/2$ . It marks the center of the fall-off region between the low- and high-pressure limiting behaviors and is often reported experimentally to characterize a reaction's pressure sensitivity.

Why do real experimental data deviate from the simple Lindemann prediction at low pressure

The simple model assumes every activating collision transfers enough energy in one step and that all A* molecules react at the same rate $k_2$ , regardless of how much excess energy they carry. In reality, activation produces a distribution of energies above the threshold, and molecules with more excess energy react faster. Averaging over this distribution (as Hinshelwood, and later RRK theory, did) predicts a less steep fall-off, matching real gas-phase data more closely than the single-state Lindemann curve.

How does this connect to chain reactions

Chain reactions also rely on the steady-state treatment of short-lived, high-energy intermediates, but the intermediates are reactive radicals rather than vibrationally excited molecules. The logic of setting the net formation rate of an unstable species to zero, then solving for an effective rate law, carries over directly. Chain mechanisms add propagation steps that regenerate radicals, producing rate laws with their own characteristic pressure and concentration dependences, which is the natural next topic after mastering Lindemann–Hinshelwood kinetics.

Can the Lindemann mechanism be applied to unimolecular reactions in solution

Not straightforwardly. In solution, the solvent bath provides essentially unlimited, extremely rapid collisional activation and deactivation, so the low-pressure fall-off regime is almost never observed — reactions in solution are effectively always in the high-pressure limit, appearing simply first order. The pressure-dependent fall-off behavior described here is a distinctly gas-phase phenomenon, most easily studied at low to moderate pressures where collision frequencies are chemically slow.