Question

The rate of reaction quadruples when the temperature changes from $300K$ to $310K$.The activation energy of this reaction is:
(Assume activation energy and pre-exponential factor are independent of temperature; $\ln 2=0.693$ $R=8.314Jmo{{l}^{-1}}{{K}^{-1}}$)

Answer 1

Hint:The activation energy of a reaction is the minimum energy which is required by the reactant in order to get converted into the desired product.
With increase in activation energy the rate of reaction also increases, as more molecules or the reacting species get energized enough to get converted to the product.

Formula used:
\[\ln (\dfrac{{{K}_{2}}}{{{K}_{1}}})=\dfrac{{{E}_{a}}}{R}[\dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}}]\]
${{K}_{1}}$ and ${{K}_{2}}$ are the ratio of initial and final rate of the reaction, ${{E}_{a}}$ is the activation energy $R$ is the gas constant and ${{T}_{1}}$ is the initial temperature and ${{T}_{2}}$ is the final temperature at which the system is present.

Complete step-by-step solution :We know that the chemical reactions take place more quickly comparatively, at higher temperatures. Some common examples would be Milk turns sour much more quickly if it is not stored at refrigerator and is left at room temperature; similarly butter or oil goes rancid more rapidly in the time of summer as compared to the winters; and the eggs hard-boil more rapidly at the sea level as compared to the mountains. The reason for this could be justified by Thermal energy which brings out a relation between direction and motion at the molecular level. As we increase the temperature, the molecules begin to move at a faster rate and collide with each other more vigorously, which increases the likelihood of the bond cleavages as well as the rearrangements.Arrhenius proposed an equation by the combination of the concepts of Boltzmann distribution law along with the activation energy which is one of the most important relationships in the field of physical chemistry:
$k=A{{e}^{-\dfrac{{{E}_{a}}}{RT}}}$
Where $A$ is the pre exponential factor, $e$ is the exponential,${{E}_{a}}$ is the activation energy $R$ is the gas constant and $T$ is the temperature at which the system is present. This equation is also called the Arrhenius equation. In order to solve the given question, we will use a modified version of this equation, which is independent of the value of the pre-exponential factor, or the exponent. It can be expressed as,
\[\ln (\dfrac{{{K}_{2}}}{{{K}_{1}}})=\dfrac{{{E}_{a}}}{R}[\dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}}]\]
Now, it is mentioned that the rate of the reaction $(\dfrac{{{K}_{2}}}{{{K}_{1}}})$ quadruples, meaning it becomes four times, so the ratio of the rate of the reaction becomes $\dfrac{4}{1}$, and the value of gas constant $R$ is given as $8.314$ and the initial temperature ${{T}_{1}}$ is given as $300K$, and final temperature ${{T}_{2}}$ is given as $310K$.
Now after substituting these values in the above equation we get,
\[\ln (\dfrac{4}{1})=\dfrac{{{E}_{a}}}{8.314}[\dfrac{1}{300}-\dfrac{1}{310}]\]
After solving the above equation, we get,
${{E}_{a}}=107.2KJmo{{l}^{-1}}$ which was the required answer.

Note:For the calculation of activation energy by using the value of, ratio of rate constant, we use the modified version of the Arrhenius equation which is independent of the exponent.
The value of gas constant we use depends on the units of temperature and pressure, since the temperature was in kelvin, the value of gas constant became $8.314Jmo{{l}^{-1}}{{K}^{-1}}$, which is also provided in the question.