# Lindemann mechanism

In chemical kinetics, the Lindemann mechanism, sometimes called the Lindemann-Hinshelwood mechanism, is a schematic reaction mechanism. Frederick Lindemann discovered the concept in 1921 and Cyril Hinshelwood developed it.[1]

It breaks down a stepwise reaction into two or more elementary steps, then it gives a rate constant for each elementary step. The rate law and rate equation for the entire reaction can be derived from this information.

Lindemann mechanisms have been used to model gas phase decomposition reactions. Although the net formula for a decomposition may appear to be first-order (unimolecular) in the reactant, a Lindemann mechanism may show that the reaction is actually second-order (bimolecular).[2]

## Activated reaction intermediates

A Lindemann mechanism typically includes an activated reaction intermediate, labeled A* (where A can be any element or compound). The activated intermediate is produced from the reactants only after a sufficient activation energy is applied. It then either deactivates from A* back to A, or reacts with another (dis)similar reagent to produce yet another reaction intermediate or the final product.

## The steady-state approximation

In some cases, one of the elementary steps is much slower than the other steps. This slow step is called the rate-determining step because it is the only step that affects the rate. In layman's terms, a rate-determining step could be compared to traveling through a traffic jam: the time it takes to complete a journey is most severely affected by the time spent waiting in the traffic jam, which is the slow step of the journey.

In the steady-state approximation, it is assumed that each of the elementary steps influences the rate, so there is no "fast" or "slow" step. Therefore, all of the steps must be accounted for in calculating the rate equation. It is also assumed that the concentration of intermediate A* remains constant over time because the concentration of A* builds up very slowly but decays very quickly over the course of a reaction, and the concentration of A* never becomes large. This assumption simplifies the calculation of the rate equation.

## Examples

### General schematic example

The schematic reaction A + M → P is assumed to consist of two elementary steps:

1. A + M → A* + M (forward reaction rate = k1; reverse reaction rate = k-1)
2. A* → P (forward reaction rate = k2)

Assuming that the concentration of intermediate A* is held constant according to the quasi steady-state approximation, what is the rate of formation of product P?

First, find the rates of production and consumption of intermediate A*. The rate of production of A* in the first elementary step is simply:

$\frac{d[A^*]}{dt} = k_1 [A] [M]$ (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:

$\frac{-d[A^*]}{dt} = k_{-1} [A^*] [M]$ (reverse first step)
$\frac{-d[A^*]}{dt} = k_2 [A^*]$ (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

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

The overall reaction rate is

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

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

$\frac{d[P]}{dt} = \frac{k_1 k_2 [A] [M]}{k_{-1} [M] + k_2}$[3]

### Specific practical example

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

$Rate = k_2 [N_2 O_5]^* = \frac{k_1 k_2 [N_2 O_5]^2}{k_{-1} [N_2 O_5 ] + k_2}$

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 $k_2 \gg k_{-1}$ ($\gg$ means "much larger than"), then the rate equation may be simplified by assuming that $k_{-1} [N_2 O_5 ] + k_2 \simeq k_2$. Then the rate equation is

$Rate = k_1 [N_2 O_5 ]^2$

which is second order. In contrast, if $k_2 \ll k_{-1}$ ($\ll$ means "much less than"), then the rate equation may be simplified by assuming $k_{-1} [N_2 O_5 ] + k_2 \simeq k_{-1} [N_2 O_5 ]$. Then the rate equation is

$Rate = \frac{k_1 k_2 [N_2 O_5]}{k_{-1}}$

which is first order.[4]

## References

1. ^ [1] Access date 8 December 2007.
2. ^ [2] "Gas phase decomposition by the Lindemann mechanism" by S. L. Cole and J. W. Wilder. SIAM Journal on Applied Mathematics, Vol. 51, No. 6 (Dec., 1991), pp. 1489-1497.
3. ^ The question is taken from GRE Chemistry Test Practice Book, based on the 2000 exam, question 98.
4. ^ "Lindemann Mechanism" by W. R. Salzman at the University of Arizona, 2004. Access date 8 December 2007.