Lindemann mechanism

Activated reaction intermediates

The overall equation for a unimolecular reaction may be written A → P, where A is the initial reactant molecule and P is one or more products (one for isomerization, more for decomposition).

A Lindemann mechanism typically includes an activated reaction intermediate, labeled A*. The activated intermediate is produced from the reactant only after a sufficient activation energy is acquired by collision with a second molecule M, which may or may not be similar to A. It then either deactivates from A* back to A by another collision, or reacts in a unimolecular step to produce the product(s) P.

The two-step mechanism is then

A + M   ↽ − − ⇀ A ∗ + M A ∗   ⟶ P

Rate equation in steady-state approximation

The rate equation for the rate of formation of product P may be obtained by using the steady-state approximation, in which the concentration of intermediate A* is assumed constant because its rates of production and consumption are (almost) equal. This assumption simplifies the calculation of the rate equation.

For the schematic mechanism of two elementary steps above, rate constants are defined as k₁ for the forward reaction rate of the first step, k₋₁ for the reverse reaction rate of the first step, and k₂ for the forward reaction rate of the second step. For each elementary step, the order of reaction is equal to the molecularity

The rate of production of the intermediate A* in the first elementary step is simply:

d [ A ∗ ] d t = k 1 [ A ] [ M ] (forward first step)

A* is consumed both in the reverse first step and in the forward second step. The respective rates of consumption of A* are:

− d [ A ∗ ] d t = k − 1 [ A ∗ ] [ M ] (reverse first step) − d [ A ∗ ] d t = k 2 [ A ∗ ] (forward second step)

According to the steady-state approximation, the rate of production of A* equals the rate of consumption. Therefore:

k 1 [ A ] [ M ] = k − 1 [ A ∗ ] [ M ] + k 2 [ A ∗ ]

Solving for [ A ∗ ] , it is found that

[ A ∗ ] = k 1 [ A ] [ M ] k − 1 [ M ] + k 2

The overall reaction rate is

d [ P ] d t = k 2 [ A ∗ ]

Now, by substituting the calculated value for [A*], the overall reaction rate can be expressed in terms of the original reactants A and M:

d [ P ] d t = k 1 k 2 [ A ] [ M ] k − 1 [ M ] + k 2

Reaction order and rate-determining step

The steady-state rate equation is of mixed order and predicts that a unimolecular reaction can be of either first or second order, depending on which of the two terms in the denominator is larger. At sufficiently low pressures, k − 1 [ M ] ≪ k 2 so that d [ P ] / d t = k 1 [ A ] [ M ] , which is second order. That is, the rate-determining step is the first, bimolecular activation step.

At higher pressures, however, k − 1 [ M ] ≫ k 2 so that d [ P ] d t = k 1 k 2 k − 1 [ A ] which is first order, and the rate-determining step is the second step, i.e. the unimolecular reaction of the activated molecule.

The theory can be tested by defining an effective rate constant (or coefficient) k u n i which would be constant if the reaction were first order at all pressures: d [ P ] d t = k u n i [ A ] , k u n i = 1 [ A ] d [ P ] d t . The Lindemann mechanism predicts that k decreases with pressure, and that its reciprocal 1 k = k − 1 k 1 k 2 + 1 k 1 [ M ] is a linear function of 1 [ M ] or equivalently of 1 p . Experimentally for many reactions, k does decrease at low pressure, but the graph of 1 / k as a function of 1 / p is quite curved. To account accurately for the pressure-dependence of rate constants for unimolecular reactions, more elaborate theories are required such as the RRKM theory.

Example: Decomposition of dinitrogen pentoxide

The decomposition of dinitrogen pentoxide to nitrogen dioxide and nitrogen trioxide

N₂O₅ → NO₂ + NO₃

is postulated to take place via two elementary steps, which are similar in form to the schematic example given above:

1. N₂O₅ + N₂O₅ → N₂O₅* + N₂O₅
2. N₂O₅* → NO₂ + NO₃

Using the quasi steady-state approximation, the rate equation is calculated to be

Rate = k 2 [ N 2 O 5 ] ∗ = k 1 k 2 [ N 2 O 5 ] 2 k − 1 [ N 2 O 5 ] + k 2

Experiment has shown that the rate is observed as first-order in the original concentration of N₂O₅ sometimes, and second order at other times.

If k 2 ≫ k − 1 ( ≫ means "much larger than"), then the rate equation may be simplified by assuming that k − 1 [ N 2 O 5 ] + k 2 ≃ k 2 . Then the rate equation is

Rate = k 1 [ N 2 O 5 ] 2

which is second order. In contrast, if k 2 ≪ k − 1 ( ≪ means "much less than"), then the rate equation may be simplified by assuming k − 1 [ N 2 O 5 ] + k 2 ≃ k − 1 [ N 2 O 5 ] . Then the rate equation is

Rate = k 1 k 2 [ N 2 O 5 ] k − 1

which is first order.