Question

What is the activation energy of a reaction if its rate doubles when the temperature is raised from the temperature is raised from ${20^ \circ }c$ to ${35^ \circ }c$? (R = 8.314 J$mo{l^{ - 1}}{K^{ - 1}}$)
(A) 342 kJ$mo{l^{ - 1}}$
(B) 269 KJ$mo{l^{ - 1}}$
(C) 34.7 kJ$mo{l^{ - 1}}$
(D) 15.1 KJ $mo{l^{ - 1}}$

Answer 1

The main theory behind this question is Arrhenius activation energy. It is equally important to know about the effect of temperature on reaction rate. Firstly convert the unit of temperature from Celsius to kelvin.

Complete step by step answer:
The energy difference between the threshold energy and energy actually possessed by reactant molecules is known as the activation energy.
Let me give a small gist of the effect of temperature on reaction rate: As the temperature increases from ${T_1}$ to ${T_2}$. There will be a slight shift in the distribution of energy. The number of molecules whose energy are equal to or greater than the threshold energy E at temperature ${T_1}$ is less than the number of molecules whose energy are equal to or greater than the threshold energy. If the number of molecules are less than the kinetic energy will be less.
Arrhenius proposed an empirical equation for calculating the energy of activation of a reaction known as Arrhenius equation. The Arrhenius equation is:
$k = A{e^{ - \dfrac{{Ea}}{{RT}}}}$…………………Equation 1
Where, k is rate constant
A is Arrhenius pre-exponential factor
Ea is Arrhenius activation energy
T is temperature

When temperature rises from ${T_1}$ to ${T_2}$. The integrated Arrhenius equation is used to calculate activation energy. The integrated Arrhenius equation is:
$\ln \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{Ea}}{R}[\dfrac{{{T_2} - {T_1}}}{{{T_2}{T_1}}}]$…………………Equation 2
Where, ${k_2}$ and ${k_1}$ are rate constants of temperature ${T_2}$ and ${T_1}$ respectively.
Given, ${T_1}$ =${20^ \circ }C$= 293.15k
${T_2}$=${35^ \circ }C$= 308.14k

Firstly we have to find ${k_2}$ and ${k_1}$, by using equation
$\dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{A{e^{ - \dfrac{{Ea}}{{R{T_2}}}}}}}{{A{e^{ - \dfrac{{Ea}}{{R{T_1}}}}}}}$
Substituting the values
$\dfrac{{{k_2}}}{{{k_1}}} = 2$……………...equation 3
Substituting equation 3 in equation 2
$\ln 2 = - \dfrac{{Ea}}{{8.314}}[\dfrac{{15}}{{308.14 \times 293.15}}]$
$Ea = 34.7$kJ
So, the correct answer is “Option C”.

Note: In order to not make any mistake, we should be thorough with the formulas to calculate activation energy. Should know the effect of temperature on the reaction rate, kindly go through it. The most important thing is that the unit of temperature must be changed from Celsius to Kelvin.