Question

On increasing temperature from $200K$ to $220K$, rate constant of reaction $A$ increases by $3$ times and rate constant of reaction $B$ increases by $9$ times then correct relationship between activation energy of $A$ and $B$ is:
A. ${E_A} = 3{E_B}$
B. $3{E_A} = {E_B}$
C. ${E_B} = 2{E_A}$
D. ${E_A} = 2{E_B}$

Answer 1

For solving this question we first need to understand the effects of temperature. According to the average kinetic energy of a reaction, when the temperature of a system increases, then the rate of a reaction also increases.

Complete step by step answer:

As we know that the temperature $T$ is defined as the measure of the average kinetic energy of the particles of a substance. The higher the temperature, the higher the fraction of molecules with energies then the activation energy ${E_a}$ also increases.
Based on Arrhenius equation we can easily calculate the energy of activation of reaction based on the rate constant $K$ and temperature $T$.
So, the formula of the Arrhenius equation can be written as:
$K = A{e^{ - Ea/RT}}$
Where ${E_a}$ is the Arrhenius activation energy
$A =$ pre-exponential factor
and $R =$ Ideal gas constant
So, based on the question we will first note all the given quantities:
In case of reaction $A$:-
${T_1} = 200K$​, ${T_2} = 220K$ ​
So, $3{({r_1})_A} = 2{({r_2})_A}..........\left( i \right)$
Here, ${({r_1})_A} =$ rate of reaction $A$ at ${T_1}$ ​K
${({r_2})_A} =$​ rate of reaction $A$ at ${T_2}$ ​K
As we know that the rate of a reaction is directly proportional to the rate constant.
Therefore, rate of reaction $\propto$ Rate constant
$3{({K_1})_A} = {({K_2})_A}$
Where, ${K_1}$ and ${K_2}$ are the different rate constants at temperature ${T_1}$ and ${T_2}K$
Now, we know that $\dfrac{{{{({K_1})}_A}}}{{{{({K_2})}_A}}} = \dfrac{1}{3}.........\left( {ii} \right)$
Using (i) in (ii) and putting the values
Therefore, $\dfrac{{{{({K_1})}_A}}}{{{{({K_2})}_A}}} = \dfrac{1}{3}$
$\dfrac{1}{3} = \dfrac{{A{e^{ - Ea/R{T_1}}}}}{{A{e^{ - Ea/R{T_2}}}}}$
$\dfrac{1}{3} = {e^{ - Ea/R(\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}})}}$
Here, we will substitute the values of temperature.
${e^{ - Ea/R(\dfrac{1}{{220}} - \dfrac{1}{{200}})}}$
$\dfrac{1}{3} = {e^{ + \dfrac{{{E_a}}}{R} \times \dfrac{{20}}{{220 \times 200}}}}$
${E_a} = 2200R\log \left( {\dfrac{1}{3}} \right)$
In case of reaction $B$:
${T_1} = 200K,{T_2} = 220K$
$\dfrac{{{{({K_1})}_B}}}{{{{({K_2})}_B}}} = \dfrac{1}{9}$
Now, we will follow the same steps as above:
$\log \left( {\dfrac{1}{9}} \right) = \dfrac{{{E_b}}}{R} \times \dfrac{1}{{2200}}$
${E_b} =$ activation energy of $B$
${E_b} = 2200R\log \left( {\dfrac{1}{9}} \right)$
Now, with the help of reaction A and B:
$\dfrac{{{E_a}}}{{{E_b}}} = \dfrac{{\log (1/3)}}{{\log (1/9)}} = \dfrac{1}{2}$
$2{E_a} = {E_b}$
$\therefore$ Option C is the correct answer.

Note:
We can define activation energy as the energy that is provided to compounds that result in a chemical reaction. Thus, activation energy of a reaction can be measured in joules per mole or kilojoules per mole or kilocalories per mole.