Class 12 Chemistry

The number of reacting species taking part in an elementary reaction, which must collide simultaneously to bring about a chemical reaction is called molecularity of a reaction.

Example of unimolecular reaction: `NH_4NO_2→N_2+2H_2O`

Example of bimolecular reaction: `2HI→H_2+I_2`

Example of trimolecular reaction: `2NO+O_2→2NO_2`

The probablity of more thatn three molecules to collide and react simultaneously is very small. So, reactions with the molecularity three are very rare. Such reactions are slow to proceed. So, complex reactions involving more than three molecules in the stoichiometric equation must take place in more than one step.

Let us take example of following reaction:

`KC\lO_3+6Fe\SO_4+3H_2SO_4``→KCl+3Fe_2(SO_4)_3+3H_2O`

This reaction appears to be of tenth order. But in reality, it is a second order reaction because the reaction takes place in several steps. The overall rate of such a reaction is controlled by the slowest step in the reaction. Such a step is called the **rate determining step**.

Order of a reaction is applicable to elementary as well as complex reactions. But molecularity is applicable only to elementary reactions. Molecularity has no meaning for complex reaction.

Let us consider following reaction:

`R→P`

Rate `=-(d[R])/(dt)=k[R]^0`

Or, `=-(d[R])/(dt)=k\xx1`

Or, `d[R}=-k dt`

Integrating both sides

`[R]=-kt+I` ……….. (1)

Here, I the constant of integration

At `t=0`, the concentration of reactant `R=[R]_0`

Substituting in equation (1)

`[R]_0=-k\xx0+I`

Or, `[R]_0=I`

Substituting in equation (1)

`[R]=-kt+[R]_0`……..(2)

On further simplifying the equation, we get the rate constant k as follows:

`k=([R]_0-[R])/t` …………. (3)

Let us consider following reaction:

`R→P`

Rate `=-(d[R])/(dt)=k[R]`

Or, `=-(d[R])/([R]=kdt` ……….(1)

Integrating this equation, we get

`In[R]=-kt+I` …………(2)

When t = 0, R = [R]_{0}

So, equation (2) can be written as follows:

`In[R]_0=-k\xx0+I`

Or, `In[R]_0=I`

Substituting the value of I in equation (2), we get:

`In[R]=-kt+In[R]_0` ………….(3)

Rearranging this equation, we get the following:

`In([R])/([R]_0)=-kt`

Or, `k=1/tIn([R]_0)/([R])`

At time t_{1}, equation (2) can be written as follows:

`In[R]_1=kt_1-In[R]_0` ……..(4)

At time t_{1}, we get:

`In[R]_2=kt_2-In[R]_0` ………….(5)

Subtracting equation (5) from (4) we get

`In[R]_1-In[R]_2=-kt_1-(-kt_2)`

Or, `In([R]_1)/([R]_2)=k(t_2-t_1)`

Equation (3) can also we written as follows:

Taking antilog of both sides

`[R]=[R]_0e^(-kt)`

The time in which concentration of a reactant is reduced to one halg of its initial concentration is called half life of the reaction.

We know, for zero order reaction:

`k=([R]_0-[R])/(t)`

When `t=1/2` the rate constant is givne as follows:

`k=([R]_o-1/2[R]_0)/(t_(1/2))`

Or, `t_(1/2)=([R]_0)/(2k)`

For First order reaction,

`k=(2.303)/t\lo\g([R]_0)/([R])`

At `t_(1/2)`, the above equation becomes

`[R]=([R]_0)/([R])`

Or, `k=(2.303)/(t_(1/2))\lo\g\([R]_0)/([R]/2)`

Or, `t_(1/2)=(2.303)/(k)\lo\g2`

Or, `t_(1/2)=(2.303)/k\xx0.301`

Or, `t_(1/2)=(0.693)/k`

For zero order reaction `t_(1/2)∝[R]_0`

For first order reaction t_{1/2} is independent of [R]_{0}

**Pseudo First Order Reaction:** When one reactant is present in large excess, its concentration remains more or less constant throughout the reaction. In this case, the reaction behaves as first order reaction. Such reactions are called pseudo first order reaction. Following reaction shows an example of pseudo first order reaction.

`CH_3CO\OC_2H_5+H_2O→CH_3CO\OH+C_2h_5OH`

In this reaction, water is present in excess amount. So, the concentration of water does not change much throughout the reaction. So, rate of the equation can be given as follows:

Rate `=k’[CH_3CO\OC_2H_5][H_2O]`

Since we are taking the concentration of H_{2}O as constant

So, `k-k’[H_2]`

So, the rate equation can be written as follows:

Rate `=k[CH_3CO\OC_2H_5]`

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