Question

The rate of a reaction doubles when its temperature changes from 300 K to 310 K.
Activation energy of such a reaction will be: (R = $8.314\,J{K^{ - 1}}mo{l^{ - 1}}$ and log 2 = 0.301)
A.48.6 $kJ\,mo{l^{ - 1}}$
B.58.5 $kJ\,mo{l^{ - 1}}$
C.60.5 $kJ\,mo{l^{ - 1}}$
D.53.6 $kJ\,mo{l^{ - 1}}$

Answer 1

In this question, we have to find the activation energy. The molecules of reactants in a chemical system have initial energy of their own, ${E_r}$. They need to acquire threshold energy ${E_{Th}}$ before they react chemically. If the energy of the reactant is less than threshold energy then some extra energy is required to bring about the reaction (effective collision). The extra energy supplied to the reactant molecules to attain the threshold energy to undergo chemical reaction is called activation energy, ${E_a}$ .
Activation energy = Threshold energy – Average kinetic energy of the reactant molecules
${E_a} = {E_{Th}} - {E_r}$

Complete step by step answer:
Now we will solve the given numerical
Given,
When ${T_1} = 300K$
$rate = {k_1}{\left[ A \right]^n}$
$\Rightarrow r = {k_1}{\left[ A \right]^n} \to \left( 1 \right)$.
When ${T_2}$= 310 K, Rate = ${k_2}{\left[ A \right]^n}$
$\Rightarrow 2r = {k_2}{\left[ A \right]^n} \to \left( 2 \right)$
Dividing equation (2) by equation (1) we get, $\dfrac{{{k_2}}}{{{k_1}}} = \,2$

Applying the Arrhenius equation we will find the activation energy
$\log \left( {\dfrac{{{k_2}}}{{{k_1}}}} \right) = \dfrac{{{E_a}}}{{2.303\,R}}\left[ {\dfrac{{{T_2} - {T_1}}}{{{T_1}{T_2}}}} \right]$
$\Rightarrow \log \left( 2 \right) = \dfrac{{{E_a}}}{{2.303\, \times 8.314}}\left[ {\dfrac{{310 - 300}}{{310 \times 300}}} \right]$
(We have put the given values in the equation and we know log 2 = 0.301 and R= 8.314)
$\Rightarrow {E_a} = \,\dfrac{{0.301 \times 2.303 \times 8.314 \times 300 \times 310}}{{10}} \Rightarrow {E_a} = 53.6\,\,\,kJ\,\,mo{l^{ - 1}}$
Therefore, the value of activation energy is $53.6kJmo{l^{ - 1}}$
Thus, the correct option is (D).

Note:
It is to be noted that each reaction at a particular temperature has a definite value of activation energy. The value of activation energy decides the fraction of the total number of effective collisions. If the activation energy is low a large number of molecules will possess this energy. Hence, the number of effective collisions (f= fraction of the total number of effective collisions) will be more and the rate of the reaction will be high. Thus,
For fast reaction the activation energy is low. For low reaction the activation energy is high