Question

The rate of the reaction quadruples when the temperature changes from $300$ to $310\,K$.The activation energy of this reaction is:
(Assume activation energy and pre-exponential factor are independent of temperature $In{\rm{ }}2 = 0.693;{\rm{ }}R{\rm{ }} = {\rm{ }}8.314\,J{\rm{ }}mo{l^{ - 1}}{\rm{ }}{K^{ - 1}}$)
A. $26.8{\rm{ }}\,kJ{\rm{ }}mo{l^{ - 1}}$
B. $414.4{\rm{ }}\,kJ{\rm{ }}mo{l^{ - 1}}$
C. $107.2{\rm{ }}\,kJ{\rm{ }}mo{l^{ - 1}}$
D. $53.6{\rm{ }}\,kJ{\rm{ }}mo{l^{ - 1}}$

Answer 1

Activation energy is like the steep of the mountain if it is too high then the reactants will not get converted into the product, so it should be such that the reactant will easily pass the barrier to get converted into the products.

Complete answer
Reactants will combine to form products. It will happen if the conditions are favorable, the rate of the reaction will be increased by using the catalyst which will increase the rate of the reaction by decreasing the activation energy hence reactants will easily break its bond and product will be formed by formation of the new bond. This happens on the surface of the catalyst.
Here due to the change in temperature, the value of the rate of the reaction quadruples and the effect on activation energy value need to be calculated.
$\rm{k} \, = \rm{Ae}^{\rm{-\frac{E_a}{RT}}}$
The Arrhenius equation is, $\rm{k} \, = \rm{Ae}^{\rm{-\frac{E_a}{RT}}}$
Where, ${\rm{k}}$ is the rate constant,
And ${\rm{A}}$ is the frequency factor which depends upon the two factors (a) activation energy (${{\rm{E}}_{\rm{a}}}$) and
(b) orientation factor, ${\rm{T}}$ is the temperature of the reaction and ${\rm{R}}$ is the gas constant.
Given in the question is the rate of the reaction quadruples that means ${k_2} = \,\,4\;{k_1}$when the temperature changes from initial to final.

$${T_2} = 310K\\\ {T_1} = 300K$$ $$In{\rm{ }}2 = 0.693\\\ R{\rm{ }} = {\rm{ }}8.314J{\rm{ }}mo{l^{ - 1}}{\rm{ }}{K^{ - 1}}$$

$\rm{k} \, = \rm{Ae}^{\rm{-\frac{E_a}{RT}}}$
Taking log on both sides we get,
$\Rightarrow ln{k_1} = - \dfrac{{{E_a}}}{{R{T_1}}} + lnA - - - (1)$
$\Rightarrow ln{k_2} = - \dfrac{{{E_a}}}{{R{T_2}}} + lnA - - - (2)$
Subtracting $\left( 2 \right)$ from $\left( 1 \right)$
we get,
$\Rightarrow \ln \left( {\dfrac{{{k_2}}}{{{k_1}}}} \right)\, = \,\, - \,\dfrac{{{E_a}}}{R}\left( {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right)$
$\Rightarrow \ln 4 = - \dfrac{{{E_a}}}{{8.314\;J\;{K^{ - 1}}mo{l^{ - 1}}}}\left( {\dfrac{1}{{310K}} - \dfrac{1}{{300K}}} \right)$

$$\Rightarrow {E_a} = \,\dfrac{{(\ln 4)\;(8.314\;J{K^{ - 1}}mo{l^{ - 1}})(310K)\,(300K)}}{{10K}}\\\ = \dfrac{{2 \times 0.693 \times 8.314 \times 310 \times 300}}{{10}}J\;mo{l^{ - 1}}\\\ = 107165.8\;J\;mo{l^{ - 1}}\\\ = 107.2\;kJ\;mo{l^{ - 1}}$$

**Hence the answer is (C) $107.2\;kJ\;mo{l^{ - 1}}$

Note: **
The value of R is different as per the unit so take care of it before selecting the value of R. The Arrhenius equation relates the temperature with the activation energy of the forward reaction.