The Lindemann-Hinshelwood Mechanism

Introduction

Some gas reactions produce products when a single energised molecule decomposes or rearranges. Why, then, can their measured rate constants depend on pressure? The molecule often gains energy through collisions with another particle before its unimolecular change. The Lindemann-Hinshelwood mechanism is a compact model for this activation and deactivation. Its steady-state rate law shows a low-pressure collision-controlled limit and a high-pressure limit in which increasing the bath-gas density no longer increases the effective first-order rate.

Core explanation

Write the simple scheme A + M → A + M with rate constant k₁, A + M → A + M with k₋₁, and A → P with rate constant k₂. Here A denotes an energised form of A, and M is a bath-gas collision partner. M can be another A molecule or a different gas. It appears in both activation and deactivation steps but is not consumed by the net A → P chemistry. The asterisk does not mean a different chemical formula; it indicates sufficient internal energy for the product-forming process in this simplified model.

Assume A remains at a small, approximately steady concentration compared with changes in the A reservoir. Its formation rate is k₁[A][M]. Its loss rate is k₋₁[A ][M] + k₂[A ]. Setting formation approximately equal to loss gives [A ] ≈ k₁[A][M]/(k₋₁[M] + k₂). Product appears at rate r = k₂[A ] = k₁k₂[A][M]/(k₋₁[M] + k₂). Define an effective first-order constant k eff = r/[A] = k₁k₂[M]/(k₋₁[M] + k₂). Although the product step A → P is unimolecular, the observed k eff changes with bath-gas concentration.

At low pressure, [M] is small and k₋₁[M] ≪ k₂. The denominator is approximately k₂, so r ≈ k₁[A][M] and k eff ≈ k₁[M]. Activation collisions are scarce, and an energised A usually reacts before another collision deactivates it. If M = A, the limiting rate can look second order in A, r ≈ k₁[A]². If M is a separately controlled inert bath gas, rate is first order in A and also first order in [M] in this limit.

At high pressure, k₋₁[M] ≫ k₂. Then r ≈ (k₁k₂/k₋₁)[A] and k eff approaches k∞ = k₁k₂/k₋₁, independent of further increases in [M] in this simple model. Many collisions form A , but many also deactivate it before product formation. Their balance fixes an approximately pressure-independent energised fraction. Between the limits, the effective rate constant bends through a fall-off region. The concentration [M] at which the two denominator terms are equal is k₂/k₋₁ in this algebraic model; at that point k eff is half of the high-pressure limit.

Pressure is used as shorthand because, for an ideal gas at fixed T, [M] = p M/(RT) in molar concentration units consistent with R. Total pressure may not be a sufficient variable if the mixture composition changes: different collision partners can transfer energy with different efficiencies. Thus adding a bath gas can change the rate without participating in the net chemical equation. Temperature also changes collision frequencies, energy-transfer efficiency and the fraction of A molecules with enough internal energy.

The scheme is deliberately coarse. Real energised molecules have an energy distribution, not a single A state. The unimolecular rate can depend on internal energy, and collisions may transfer only part of that energy. Advanced RRK/RRKM and master-equation approaches account for these features. A primary ACS study of the Hinshelwood-Lindemann picture discusses how thermal unimolecular kinetics depends on collision frequency and energy-specific decomposition. A primary pressure-dependent kinetics study explores collision-efficiency assumptions in more detailed calculations. The simple model remains valuable because its limiting behaviour can be derived transparently.

Step-by-step reasoning

1. Identify the collision partner M, the energised intermediate A and the product-forming step. 2. Write formation and both loss rates for A with their correct concentration factors. 3. Apply the steady-state approximation to solve for [A ]. 4. Substitute into r = k₂[A ] and simplify k eff = r/[A]. 5. Compare k₋₁[M] with k₂ to find low- and high-pressure limits. 6. State whether M is A or a separate bath gas before assigning observed reaction order.

Visual explanation

Draw a box labeled A . Put one incoming arrow from A + M marked “activation collision, k₁,” and two outgoing arrows: one back to A + M marked “deactivation, k₋₁,” and one toward P marked “reaction, k₂.” Below, plot k eff against [M]: a straight rising segment near the origin bends toward a horizontal line labeled k∞. Mark the concentration k₂/k₋₁ at the halfway height. This graph links the denominator in the rate law with a measurable pressure trend.

Real-world analogy

Imagine a player who must receive a pass before attempting a shot. At low pass frequency, more passes directly raise the number of shots. At very high pass frequency, extra passes also interrupt or reset the player, so shot production stops rising proportionally. The analogy captures competing activation and deactivation encounters, but molecules exchange energy statistically rather than following game strategy.

Real-world example

In a gas-phase decomposition experiment, researchers hold the reacting A concentration and temperature as steady as practical while varying a bath-gas pressure. A rising k eff at low pressure followed by a plateau supports collisional activation and fall-off behaviour. Changing the bath gas can shift the curve because some collision partners transfer internal energy more effectively. A mechanistic assignment still requires checking products and competing channels; similar pressure trends can emerge from more complicated networks.

Why?

Why does a product-forming unimolecular step show second-order behaviour at low pressure when M = A? The bottleneck is not decomposition of an already energised A . It is creating A by a collision between two A molecules. When almost every created A reacts before deactivation, product rate follows the collision frequency, proportional to [A]². At high pressure, activation and deactivation collisions balance, so the net energised fraction becomes nearly pressure independent and rate becomes first order in A.

Common misconception

“Unimolecular reaction” means no collision can affect its rate. It describes the product-forming elementary event, while preparation of an energised reactant can involve collisions. Another mistake is assuming M must be chemically consumed; it transfers energy and reappears. Finally, the high-pressure plateau does not mean collisions have stopped. It results from activation and deactivation both becoming rapid relative to decomposition in the simple scheme.

Worked example

Suppose k₁ = 2.0 L mol⁻¹ s⁻¹, k₋₁ = 4.0 L mol⁻¹ s⁻¹, k₂ = 8.0 s⁻¹, and [M] = 1.0 mol L⁻¹. Then k eff = k₁k₂[M]/(k₋₁[M]+k₂) = (2.0 × 8.0 × 1.0)/(4.0 × 1.0 + 8.0) = 16/12 = 1.33 s⁻¹. The high-pressure limit is k∞ = k₁k₂/k₋₁ = 4.0 s⁻¹, so this condition has reached one-third of that limit. The halfway concentration would be k₂/k₋₁ = 2.0 mol L⁻¹. If [A] = 0.10 mol L⁻¹ at the stated condition, product rate is 0.133 mol L⁻¹ s⁻¹.

Quick check

1. In the low-pressure limit, what is k eff approximately? Answer: k eff ≈ k₁[M], because k₂ dominates the denominator.

Exam focus

Write the three elementary steps and perform the steady-state algebra rather than memorising a pressure formula. Explain which denominator term dominates in each limit and preserve the distinction between [M] and [A]. If M = A, identify the low-pressure second-order form; if M is an independently varied bath gas, identify first order in each concentration at low pressure. State that a real fall-off curve may need energy-resolved theory.

Advanced insight

The high-pressure limit corresponds to rapid energy randomisation relative to reaction, but detailed models resolve many vibrational energy grains and collision-by-collision transfer probabilities. The master equation follows populations across those grains, while RRKM theory supplies an energy-specific unimolecular decay rate for a given molecular structure. Broadening of an observed fall-off curve relative to the simplest Lindemann form can reveal limitations of the single-A model. Temperature and bath-gas identity should therefore be fitted together when quantitative combustion or atmospheric chemistry predictions are needed.

Summary

The Lindemann-Hinshelwood mechanism explains pressure-dependent rates for a product-forming unimolecular step by adding collisional activation and deactivation. Steady-state treatment gives k eff = k₁k₂[M]/(k₋₁[M]+k₂). At low pressure, the effective rate grows with [M]; at high pressure, it approaches k₁k₂/k₋₁. The simple model captures fall-off behaviour and observed order changes, while real energy distributions motivate more detailed theories.

Practice questions

1. For the simple mechanism, derive [A ] under steady state. Answer: Set k₁[A][M] = k₋₁[A ][M] + k₂[A ], giving [A ] = k₁[A][M]/(k₋₁[M]+k₂).

2. If [M] is much smaller than k₂/k₋₁, what is the pressure trend of k eff? Answer: It is approximately proportional to [M], hence to bath-gas partial pressure at fixed temperature for an ideal gas.

3. What is k eff at [M] = k₂/k₋₁ relative to its high-pressure limit? Answer: It is one-half of k∞ in the simple algebraic model.

4. Why is the model's single A state an approximation? Answer: Real molecules occupy a distribution of internal energies and collision partners transfer variable amounts of energy, so one energised population cannot describe all microscopic states.

5. Why does changing M sometimes change the fall-off curve even at equal total pressure? Answer: Different bath-gas molecules can have different collision and energy-transfer efficiencies, so equal pressure does not guarantee equal activation or deactivation rates.