Question

A reaction can take place by two paths. ${{K}_{1}}$ and ${{K}_{2}}$ are the rate constants for the two paths and ${{E}_{1}}$ and ${{E}_{2}}$ are their respective activation energies. At temperature; ${{T}_{a}}$: ${{K}_{1}}>{{K}_{2}}$, ${{E}_{1}}<{{E}_{2}}$. Its temperature is raised to ${{T}_{b}}$, the rate constants change to ${{K}_{1}}'$ and ${{K}_{2}}'$. Which relation is correct between ${{K}_{1}}$, ${{K}_{2}}$, ${{K}_{1}}'$ and ${{K}_{2}}'$ (considering activation energy does not change with the temperature)?
A. $\dfrac{{{K}_{1}}'}{{{K}_{1}}}>\dfrac{{{K}_{2}}'}{{{K}_{2}}}$
B. $\dfrac{{{K}_{1}}'}{{{K}_{1}}}=\dfrac{{{K}_{2}}'}{{{K}_{2}}}$
C. $\dfrac{{{K}_{1}}'}{{{K}_{1}}}<\dfrac{{{K}_{2}}'}{{{K}_{2}}}$

Answer 1

Hint: You can solve this question by simply applying the Arrhenius equation, which is a formula for the temperature dependence of the reaction rates. It is used to determine the rate of the chemical reactions and for calculating the energy of activation.

Complete step by step solution:
First of all; we should know that the Arrhenius equation which was given by Svante Arrhenius in $1889$ is helpful in calculating the rate of reaction and plays an important role in chemical kinetics. The equation is given as:
$K=A{{e}^{-\dfrac{{{E}_{a}}}{RT}}}$ , where
K is the rate constant,
A is the pre-exponential energy,
${{E}_{a}}$ is the activation energy,
R is the gas constant and T is the temperature.
Given that,
A reaction is taking place by two paths where ${{K}_{1}}\text{ and }{{K}_{2}}$ are the rate constants for the two paths and ${{E}_{1}}\text{ and }{{E}_{2}}$ are the activation energies of the reaction in two paths.
At temperature ${{T}_{a}}$, ${{K}_{1}}>{{K}_{2}}$ and ${{E}_{1}}<{{E}_{2}}$
The temperature is raised where it becomes ${{T}_{b}}$ and the rate constants become $K_{1}^{'}\text{ and }K_{2}^{'}$. We have to find out the relation between these constants keeping the activation energies constant at both temperatures.
So, at temperature ${{T}_{a}}$ applying the Arrhenius equation we get:
${{K}_{1}}=A{{e}^{-\dfrac{{{E}_{1}}}{R{{T}_{a}}}}}$
${{K}_{2}}=A{{e}^{-\dfrac{{{E}_{2}}}{R{{T}_{a}}}}}$
Similarly, at temperature ${{T}_{b}}$we will get:
$K_{1}^{'}=A{{e}^{-\dfrac{{{E}_{1}}}{R{{T}_{b}}}}}$
$K_{2}^{'}=A{{e}^{-\dfrac{{{E}_{2}}}{R{{T}_{b}}}}}$
By taking the ratio of ${{K}_{1}}\text{ and }K_{1}^{'}$, we will get:
$\dfrac{{{K}_{1}}}{K_{1}^{'}}={{e}^{-\dfrac{{{E}_{1}}}{R}\left( \dfrac{1}{{{T}_{a}}}-\dfrac{1}{{{T}_{b}}} \right)}}$ which can also be written as
$\dfrac{K_{1}^{'}}{{{K}_{1}}}={{e}^{\dfrac{{{E}_{1}}}{R}\left( \dfrac{1}{{{T}_{a}}}-\dfrac{1}{{{T}_{b}}} \right)}}$
Similarly, the ratio of ${{K}_{2}}\text{ and }K_{2}^{'}$ will be:
$\dfrac{{{K}_{2}}}{K_{2}^{'}}={{e}^{-\dfrac{{{E}_{2}}}{R}\left( \dfrac{1}{{{T}_{a}}}-\dfrac{1}{{{T}_{b}}} \right)}}$ or $\dfrac{K_{2}^{'}}{{{K}_{2}}}={{e}^{\dfrac{{{E}_{2}}}{R}\left( \dfrac{1}{{{T}_{a}}}-\dfrac{1}{{{T}_{b}}} \right)}}$
Considering ${{E}_{1}}<{{E}_{2}}$:
${{e}^{\dfrac{{{E}_{1}}}{R}\left( \dfrac{1}{{{T}_{a}}}-\dfrac{1}{{{T}_{b}}} \right)}}<{{e}^{\dfrac{{{E}_{2}}}{R}\left( \dfrac{1}{{{T}_{a}}}-\dfrac{1}{{{T}_{b}}} \right)}}$
Therefore,
$\dfrac{K_{1}^{'}}{{{K}_{1}}}<\dfrac{K_{2}^{'}}{{{K}_{2}}}$

Hence, the correct option is C.

Note: With the help of the Arrhenius equation, we can find out the values of the temperature, frequency, presence of the catalyst, effect of energy barrier and orientation of a collision. Be careful while changing the signs.