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Chemical kinetics, involving the study of rate of a reaction and rate law, is a very important part of physical chemistry. A knowledge of free energy changes involved in a given process tells us whether the process will occur or not. But to decide the speed of a reaction, we need the help of chemical kinetics and rate law.

Steady state approximation technique is used to decide the exact rate of a reaction.

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## Reaction Mechanism

The mechanism of a reaction is the explanation of how a reaction takes place.For example, we might find out which bonds are broken, what happens in the transition state, and whether the reaction takes place in one or more than one stage. Often it is possible to think of several mechanisms for a single reaction.

Working out a mechanism for a reaction can be very challenging, and fascinating. A reaction that appears straight froward from its chemical equation may have a very complex mechanism.

For example,
CH3COCH3 + I2 $\rightarrow$ CH3COCH3 + HI

The equation actually shows that one mole of propanone will react with one mole of Iodine to give one mole of iodopropane. But, this actually involves a very complex three step mechanism.

By looking at the reaction, we will not be able to decide the rate of the reaction.

The chemical equation tells us nothing definite about the rate or mechanism of a reaction. It is also seen that, when the rate is decided, it is not always that the three steps leading to the reaction takes place at the same rate. One step may be very fast, while the other may take much longer time to complete. So, "The slowest step in a reaction determines the rate of the reaction."

Each step in a chemical reaction is called an elementary step. The rate law of a reaction is determined by the rate determining step, so there may be substances in the elementary step, that do not appear in the rate law. Also, many substances may be made and used during the course of the reaction. These compounds, which appear in between the reaction, are intermediates.

Intermediates are a structure formed in one elementary step but consumed during a later step of the mechanism.
1. These intermediates, or change in the intermediates, though the reaction is actually not taken into account, when writing the rate law.
2. Steady state approximation deals with the fact that there is no change in state variables, like entropy, temperature, pressure etc, in the intermediate step.
3. So, the steady state approximation actually deals with the fact that the the change in the intermediates does not affect the rate of a reaction. It is very important to keep in mind that a balanced chemical equation and a reaction mechanism are two different expressions.
4. A rate law can be written from a mechanism, never from a balanced chemical equation. Because, many reaction pathways occur very rapidly, it can be difficult to determine mechanisms.
5. So, when an intermediate is formed in a reaction, it is consumed immediately.
6. Thus, it does not affect the rate of the reaction. Therefore, it is not necessary to take this into account in the rate law.

## Approximations for Complex Reactions

After a reaction has been found complex, we always look into the kinetic data, and a mechanism is suggested for a complex reaction.

In dealing with the progress and mechanism of a chemical reaction, we come across many concepts like reaction coordinate and activated complex. Reaction coordinate is defined as a function of all the coordinates in space of all species taking part in a reaction. It is a measure of the progress of a given reaction along the path of minimum potential energy at any instant, beginning with the reactant molecules and terminating in the product molecules. At a point where the reaction coordinate has the maximum potential energy, is located the activated complex.

At this point, the atoms of the reactants have a configuration of maximum energy of mutual repulsion. Two approximations are generally used for elucidating the mechanism of a complex reaction. These are
1. Equilibrium approximation

## Equilibrium Approximation

Consider a reaction in which reactant R changes to product, p through the formation of a series of consecutive intermediates, I1, I2, I3 ..........In, as follows.

k0 k1 k2                                       kn
Reactants $\to$ I1 $\to$ I2 $\to$ I3 $\to$ ...........................In $\to$ Products

The whole reaction sequence may be described in terms of a single coordinate composing the coordinates of the individual steps.

In order to derive the rate equation for the reaction, we assume that a rate determining step exists. This rate determining step is the slowest in a sequence of steps. It is further assumed that all the steps preceding the rate determining step are in equilibrium.

In cases where the reactants are investigated under such conditions that the slowest rate-determining step does not exist, one assumes the steady state approximation, for the transient, or short lived, intermediate species. In such mechanisms, as shown below.

k0 k1 k2 kn
Reactants $\to$ I1 $\to$ I2 $\to$ I3 $\to$ ........................... In $\to$ Products

The rate of formation of an intermediate is equal to the rate of its decomposition so that

$\frac{d[I1]}{dt}$ = $\frac{d[I2]}{dt}$ = 0

Steady state approximation is used to check the consistency of the rate law.

In the following reaction,

A + B $\overset{k_{1}}{\rightarrow}$ X slow step 1
A + B $\overset{k_{2}}{\rightarrow}$ C fast step 2

The intermediate X is very reactive. In such cases, the concentration of X does not build up significantly during the course of the reaction. This implies that,

k1[A] [B] = k2[X] [D] …… (37)

or, the formation of C, given by $\frac{\mathrm{d}[C]}{\mathrm{d} x}$
$\frac{\mathrm{d}[C]}{\mathrm{d} x}$ = k1 [A][B] .....(38)

The basic assumption in steady-state approximation is that, the rate of change of the intermediate is almost zero,

$\left ( \frac{\mathrm{d}[X]}{\mathrm{d}{t}} \cong 0\right )$

Below are some problem.

### Solved Example

Question: For the thermal decomposition of Ozone to oxygen, the following mechanism has been suggested.

k1
O3 ⇔  O2  + O

k-1

k2
O3 + O →  2O2

Use the steady state approximation and other suitable approximations to account for observed rate law:

r = -k [O3]2 / [O2].

Solution:

Ozone decomposes in steps 1 and 2 and is formed in step -1, the reverse step.

So,

r = $\frac{d[O_3]}{dt}$  = -k1[O3]  + k-1[O2] [O] - k2 [O3][O]  --------------(1)

Using steady state approximation for O atom, which is an intermediate species, we have:

$\frac{d[O]}{dt}$  = k1[O3] - k-1[O2] [O] - k2 [O3][O]

Therefore, k1[O3]  = {k-1[O2] + k2 [O3]}[O]
[O]  = $\frac{k_1[O_3]}{k_-1[O_2] + k_2 [O_3]}$

Substituting for [O] in the rate expression(1), we have:

r =  $\frac{ -k_1k_-1[O_2][O_3] - k_1k_2[O_3]^2 + k_1k_-1[O_2][O_3] - k_1k_2[O_3]^2}{k_-1 [O_2] + k_2[O_3]}$

The first and the third terms in the numerator of this equation cancel and the second and fourth terms are added algebraically giving

r = - $\frac{2k_1k_2[O_3]^2}{k_-1[O_2] + k_2[O_3]}$

If we make further approximation that in the above equation, k-1[O2] ≥ k2[O3], then, the second term in the denominator is dropped to give

r =  - $\frac{2k_1k_2[O_3]^2}{k_-1[O_2]}$ = -k $\frac{[O_3]^2}{[O_2]}$

which agrees with the observed rate law if  k = 2k1k2 / k-1.

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