Question

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

Answer 1

In this question we use the equation,
$\log \dfrac{{{k}_{2}}}{{{k}_{1}}} = \dfrac{{{E}_{a}}}{2.303R}\left[ \dfrac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}{{T}_{2}}} \right]$
${{E}_{a}}$ is the activation energy, ${{\text{T}}_{\text{1}}}$ and ${{\text{T}}_{\text{2}}}$ are the initial and final temperature.
R is the gas constant, k is the rate constant.

Complete step by step answer:
- In the question, the data about the rate of the reaction and its final and initial temperature is given. We have to find the activation energy of the molecules when the temperature is raised.
- It is given in the question, that the rate is doubled when the temperature is raised from ${{T}_{1}}$ to ${{T}_{2}}$ .
- We know that rate is, $r={{k}_{1}}{{\left[ A \right]}^{n}}$
${{\left[ A \right]}^{n}}$ is the concentration of the reactant,${{k}_{1}}$ is the rate constant of the reaction at the ${{20}^{\circ }}C$
- The rate of the reaction when the temperature is raised becomes,$2r={{k}_{2}}{{\left[ A \right]}^{n}}$
Since the rate has increased by twice.
Now divide the two equations and we get, $\dfrac{2r={{k}_{2}}{{\left[ A \right]}^{n}}}{r={{k}_{1}}{{\left[ A \right]}^{n}}}$
Hence we get,$\dfrac{{{k}_{2}}}{{{k}_{1}}}=2$
- And now we take, Arrhenius equation
$\log \dfrac{{{k}_{2}}}{{{k}_{1}}}=\dfrac{{{E}_{a}}}{2.303R}\left[ \dfrac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}{{T}_{2}}} \right]$
- And we will substitute all the values in the equation.
Before substituting the values convert the temperature given to Kelvin scale.
Initial temperature, ${{T}_{1}}$ = 20 + 273 = 293K
Final temperature,${{T}_{2}}$ = 35 + 273 = 308K
$R = 8.314Jmo{{l}^{-1}}{{K}^{-1}}$
$\dfrac{{{k}_{2}}}{{{k}_{1}}} = 2$
- Substitute all the values and we get,
$\log \dfrac{{{k}_{2}}}{{{k}_{1}}}=\dfrac{{{E}_{a}}}{2.303R}\left[ \dfrac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right]$
$\log 2 = \dfrac{{{E}_{a}}}{2.303\times 8.314}\left[ \dfrac{308-293}{293\times 308} \right]$
The value of log 2 is 0.3010,
$0.3010 = \dfrac{{{E}_{a}}}{2.303\times 8.314}\left[ \dfrac{15}{293\times 308} \right]$
- Now, let’s alter the equation and write for activation energy.
Therefore, ${{E}_{a}}=\dfrac{0.3010\times 2.303\times 8.314\times 293\times 308}{15}$
The activation energy, ${{E}_{a}} = 34673.48Jmo{{l}^{-1}}$
As the activation energy is obtained in $\text{J/mol}$ so convert into $\text{kJ/mol}$.
${{E}_{a}} = 34673.48Jmo{{l}^{-1}} = 34.7kJmo{{l}^{-1}}$

So, the correct answer is (B).

Note: As the temperature increases the molecules obtaining the activation energy increases and hence effective collisions take place between the reactant molecules and the chemical reaction proceeds effectively with increased rate.