All about the Light Absorption’s theory on the basis of Jablonski diagram.
According to the Grotthus – Draper Law of photo-chemical activation:
Only that light which is absorbed by a system, can bring a photo-chemical change.
However it is not true that all the kind of light(s) that are absorbed could bring a photo-chemical change. The absorption of light may result in a number of other phenomena as well.
For instance, the light absorbed may cause only a decrease in the intensity of the incident radiation. This event is governed by the Beer-Lambert Law.
Secondly, the light absorbed may be re-emitted almost instantaneously, within $10^{-8}$ seconds, in one or more steps. This phenomenon is well-known as fluorescence.
Sometimes the light absorbed is given out slowly and even long after the removal of the source of light. This phenomenon is known as phosphorescence.
In order to understand Jablonski diagram, we first need to go through some basic facts. Many molecules have an even number of electrons and thus in the ground state, all the electrons are spin paired. The quantity $ \mathbf {2S+1} $ , where $ S $ is the total electronic spin, is known as the spin multiplicity of a state. When the spins are paired $ \uparrow \downarrow $ as shown in the figure, the upward orientation of the electron spin is cancelled by the downward orientation so that total electronic spin $ \mathbf {S=0} $ . That makes spin multiplicity of the state 1.
Thus, the spin multiplicity of the molecule is 1. We express it by saying that the molecule is in the singlet ground state.
When by the absorption of a photon of a suitable energy $ h \nu $ , one of the paired electrons goes to a higher energy level (excited state), the spin orientation of the single electrons may be either parallel or anti-parallel. [see image]
• If spins are parallel, $ \mathbf {S=1} $ or $ \mathbf {2S+1=3} $ i.e., the spin multiplicity is 3. This is expressed by saying that the molecule is in the triplet excited state. • If the spins are anti-parallel, then $ \mathbf{S=0} $ so that $ \mathbf {2S+1=1} $ which is the singlet excited state, as already discussed.
See, since the electron can jump to any of the higher electronic states depending upon the energy of the photon absorbed, we get a series of singlet excited states, $ {S_n} $ and a series of triplet excited state $ {T_n}$where $ n =1, 2, 3 \ldots $ . Thus $ S_1, , S_2, , S_3, \ldots $ etc are respectively known as first singlet excited states, second singlet excited states and so on. Similarly, in $ T_1, , T_2,, ….. $, they are respectively known as first triplet excited state, second triplet excited state and so on.
Make sure, you are not confused in $ \mathbf{S}$ & $ S_n $.
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[fn id="X7dTOV1"].
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[fn id="ICGTOV1"] 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[fn id="dYbUOV1"] 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]} $
Have 50 extra minutes? Watch this lecture on Lindemann Theory of Unimolecular Reaction.
https://www.youtube.com/watch?v=k3doEU1yeG4
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