213. What is the effect of temperature on the rate constant of reaction? How can this effect of temperature on rate constant be respresented quantitatively?

The rate constant of reaction increases with increase of temperature. This increase is generally two fold to five fold for 10 K rise in temperature. This is explained on the basis of collision theory. The main parts of collision theory are as follows:

(i) For a reaction to occur, there must be collision between the reacting species.

(ii) Only a certain fraction of total collisions are effective in forming the products.

(iii) For effective collisions, the molecule must possess the sufficient energy (equal or greater
than threshold energy) as well as proper orientation.
On the basis above conclusions, the rate of reaction is given by
Rate =f x 2 (where f is the effective collisions and is total number of collisions per unit volume per second).

Quantitatively, the effect of temperature on the rate of a reaction and hence on the rate constant k was proposed by Arrhenius. The equation, called Arrhenius equation is usually written in the form

$\mathrm{K}={\mathrm{Ae}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT}}$            ...(i)
.
where A is a constant called frequency factor (because it gives the frequency of binary collisions of the reacting molecules per second per litre, E₀ is the energy of activation, R is a gas constant and T is the absolute temperature. The factor e^–E_a/RT gives the fraction of molecules having energy equal to or greater than the activation energy, E_a.

The energy of activation (E_a) is an important quantity and it is characteristic of the reaction. Using the above equation, its value can be calculated.
Taking logarithm or both sides of equation (i), we get,
$\mathrm{In} \mathrm{k} = \mathrm{In} \mathrm{A} - \frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_1}$

If the value of the rate constant at temper-atures T₁ and T₂ are k₁ and k₂ respectively, then we have

 $$\begin{gathered} \mathrm{In} {\mathrm{k}}_1 = \mathrm{In} \mathrm{A} - \frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_1} ...\left(\mathrm{ii}\right) \\ \mathrm{In} {\mathrm{k}}_2 = \mathrm{In} \mathrm{A} - \frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_2} ...\left(\mathrm{iii}\right) \end{gathered}$$

Subtracting eqn. (i) from eqn. (ii), we get

   $$\begin{gathered} \mathrm{In} {\mathrm{k}}_2 - \mathrm{In} {\mathrm{k}}_1 = \frac{-{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_2}+\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_1} \\ = \frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_1}+\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_2} \end{gathered}$$
     

$$\begin{gathered} \mathrm{or} \mathrm{In}\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1} = \frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}+\left(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right) \\ = \frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{{\mathrm{T}}_2-{\mathrm{T}}_1}{{\mathrm{T}}_1{\mathrm{T}}_2}\right) \end{gathered}$$

or       $\log \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1} = \frac{{\mathrm{E}}_{\mathrm{a}}}{{\displaystyle \frac{2.303}{\mathrm{R}}}}\left(\frac{{\mathrm{T}}_2-{\mathrm{T}}_1}{{\mathrm{T}}_1 {\mathrm{T}}_2}\right)$

Thus knowing the values of the constant k₁ and k₂ at two different temperature T₁ and T₂, the value of E_a can be calculated.