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
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–Ea/RT gives the fraction of molecules having energy equal to or greater than the activation energy, Ea.
The energy of activation (Ea) 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,
If the value of the rate constant at temper-atures T1 and T2 are k1 and k2 respectively, then we have
Subtracting eqn. (i) from eqn. (ii), we get
Thus knowing the values of the constant k1 and k2 at two different temperature T1 and T2, the value of Ea can be calculated.
This reaction follows second order kinetics.
So that, the rate equation for this reaction will
Rate, R = k[X]2 .............(1)
Let initial concentration is x mol L−1,
Plug the value in equation (1)
Rate, R1 = k .(a)2
Given that concentration is increasing by 3 times so new concentration will 3a mol L−1
Plug the value in equation (1) we get
Rate, R2 = k (3a)2
We have already get that R1 = ka2 plus this value we get
R2 = 9 R1
So that, the rate of formation will increase by 9 times.
Rate = k[A]2
If concentration of X is increased to three times,
Rate = k[3A]2
or Rate = 9 k A2
Thus, rate will increase 9 times.