# The Lindemann Theory of Unimolecular Reactions

In this article we will learn about the **Lindemann Theory of Unimolecular Reactions** which is also known as **Lindemann-Hinshelwood mechanism.**

It is easy to understand a bimolecular reaction on the basis of collision theory.

When two molecules A and B collide, their relative kinetic energy exceeds the threshold energy with the result that the collision results in the breaking of comes and the formation of new bonds.

But how can one account for a unimolecular reaction – a single molecule going for a reaction?

If we assume that in such a reaction $ A \longrightarrow P $ , the molecule A acquires the necessary activation energy for colliding with another molecule, then the reaction should obey second-order kinetics and not the first-order kinetics which is observed in several unimolecular gaseous reactions. A satisfactory theory of these reactions was proposed by F. A. Lindemann in 1922.

According to Lindemann, a unimolecular reaction $ A \longrightarrow P $ proceeds via the following mechanism:

$ A + A \rightleftharpoons A^* +A $

Here the rate constants being $ k_f $ for forward reaction & $ k_b$ for backward reaction and

$ A^* \longrightarrow P $ has the rate constant $ =k_{f_2} $ . ** [note this]**

Here $ A^* $ is the energized $ A $ molecule which has acquired sufficient vibrational energy to enable it to isomerize or decompose. In other words, the vibrational energy of $ A $ exceeds the threshold energy for the overall reaction $ A \longrightarrow P $ .

It must be borne in mind that $ A^* $ is simply a molecule in a high vibrational energy level and not an activated complex. In the first step, the energized molecule $ A^* $ is produced by collision with another molecule A.

What actually happens is that the kinetic energy of the second molecule is transferred into the vibrational energy of the first.

In fact, the second molecule need not be of the same species; it could be a product molecule or a foreign molecule present in the system which, however, does not appear in the overall stoichiometric reaction $ A \longrightarrow P $ .

The rate constant for the energization step is $ k_f$ . After the production of $ A^* $ , it can either be de-energized back to $ A $ (in the reverse step) by collision in which case it vibrational energy is transferred to the kinetic energy of an $ A $ molecule or be decomposed or isomerized to products (in the second step above) in which case the excess vibrational energy is used to break the appropriate chemical bonds.

In the Lindemann mechanism, a time lag exists between the energization of $ A to A^* $ and the decomposition (or isomerization) of $ A^* $ to products.

During the time lag, $ A^* $ can be de-energized back to $ A $ .

### Mathematical Treatment

According to the steady state approximation (s.s.a.), whenever a reactive (i.e. short lived) species is produced as an intermediate in a chemical reaction, its rate of formation is equal to its rate of decomposition. Here, the energized species $ A^* $ is short lived.

Its rate of formation=$ k_f \times {[A]}^2 $ and its rate of decomposition=$ k_b \times [A] [A^*] + k_{f_2} \times [A^*] $ .

Thus

$ d[A^*]/dt = k_f \times {[A]}^2 – k_b \times [A] [A^*] – k_{f_2} \times [A^*]= 0$ …..(1)

so that

$ [A^*]= \frac {k_f \times {[A]}^2} {k_b \times [A]+ k_{f_2}}$ …..(2)

The rate of the reaction is given by

$ r = -d[A]/dt =k_{f_2 }[A^*] ….(3) $

Substituting Eq.2 in Eq.3,

$ r = \frac {k_f k_{f_2} \times {[A]}^2} {k_b \times [A]+ k_{f_2}} ….(4) $

The rate law given by Equation 4 has no definite order. We can, however, consider two limiting cases, depending upon which of the two terms in the denominator of Equation 4 is greater. If $ k_b [A] >> k_{f_2} $ , then the $ k_{f_2} $ term in the denominator can be neglected giving:

$ r = (k_fk_{f_2} /k_b) [A] …….(5)$

which is the rate reaction for a first order reaction. In a gaseous reaction, this is the high pressure limit because at very high pressures. $ [A] $ is very large so that $ k_f[A] >> k_{f_2} $ .

If $ k_{f_2} >> k_b[A] $ , then the $ k_b[A] $ term in the denominator of equation 4 can be neglected giving

$ r=k_f {[A]}^2 ……(6)$

which is the rate equation of a second order reaction. This is the low pressure limit. The experimental rate is defined as

$ r= k_{uni} [A] …..(7)$

where $ k_{uni} $ is unimolecular rate constant.

Comparing Eqs.4 & 7 we have

the rate constant of Unimolecular reaction:

$ k_{uni}= \frac {k_fk_{f_2}[A]}{k_b[A]+k_{f_2}} $

or $ k_{uni}= \frac {k_fk_{f_2}}{k_b+k_{f_2}/[A]} $

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Just wanted to tell you there’s a typo in Eq (1) — there should be parentheses around $ k_{-1} \times [A] [A^*] + k_2 \times [A^*]$, so the full expression becomes “$ k_1\times [A]^2 -(k_{-1} \times [A] [A^*] + k_2 \times [A^*])=0$

(Or else the sign in front of $ k_2$ should be negative.)

Just wanted to tell you there’s a typo in Eq (1) — there should be parentheses around $ k_{-1} \times [A] [A^*] + k_2 \times [A^*]$, so the full expression becomes “$ k_1\times [A]^2 -(k_{-1} \times [A] [A^*] + k_2 \times [A^*])=0$

(Or else the sign in front of $ k_2$ should be negative.)

tooo easy i like it

tooo easy i like it

exaplain the limits and drawbacks of lindemann theory

calculated value of collision frequency is many times greater then the actual value of k

exaplain the limits and drawbacks of lindemann theory

calculated value of collision frequency is many times greater then the actual value of k

explain the flaws of theory and modification.

explain the flaws of theory and modification.

raely a good explation i also like this explation

raely a good explation i also like this explation

i con”t memory that but i did andrestanding very easily

i con”t memory that but i did andrestanding very easily

good

good

Sir, I think the eqⁿ 1 needs to be corrected! It should be

d[A*]/dt = kf [A]² – kb [A*][A] – kf2 [A*] = 0

(Not “+ kf2 [A*]”)

Yes, you are right. I missed the symbol.

## Fixed.

Sir, I think the eqⁿ 1 needs to be corrected! It should be

d[A*]/dt = kf [A]² – kb [A*][A] – kf2 [A*] = 0

(Not “+ kf2 [A*]”)

Yes, you are right. I missed the symbol.

## Fixed.