Question

Equation ${\text{log K = log A - }}\dfrac{{{\text{Ea }}}}{{{\text{2}}{\text{.303RT}}}}$, Where ${\text{Ea }}$ is the activation energy. When a graph is plotted for ${\text{logK}}$ vs $\dfrac{{\text{1}}}{{\text{T}}}$ , a straight line with a slope of ${\text{ - 4250K}}$ is obtained. Calculate ‘${\text{Ea }}$’ for the reaction. (${\text{R = 8}}{\text{.314J}}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$).

Answer 1

We have to know that the chemical kinetics is one of the topics used to study the kinetics nature of the chemical reaction. It is used to optimize the chemical reaction for industrial purposes. The reactant to the product so many parameters are required. All are optimised by using this chemical kinetics. Chemical kinetics is used as a mechanism of reactant to product in the chemical reaction. Activation energy of reaction is nothing but the amount of energy required for collision of reactant molecules in chemical reaction.
Activation energy = Threshold energy - energy of molecules
Colloidal molecules have a minimum amount of energy to complete the reaction that energy is called as threshold energy.
Formula used:
The slope of the straight line is equal to the ratio between activation energy and gas constant.
${\text{m = - }}\dfrac{{{\text{Ea}}}}{{{\text{2}}{\text{.303R}}}}$
Here
The activation energy is ${\text{Ea }}$.
The gas constant is ${\text{R}}$.
The slope of the straight line is ${\text{m}}$.

Complete answer:
The Arrhenius equation is
${\text{log K = log A - }}\dfrac{{{\text{Ea }}}}{{{\text{2}}{\text{.303RT}}}}$
The linear graph equation is
${\text{y = c + xm}}$
Compare Arrhenius equation with linear graph equation to give,
${\text{y = log K }}$
${\text{c = log A }}$
${\text{m = - }}\dfrac{{{\text{Ea }}}}{{{\text{2}}{\text{.303R}}}}$
${\text{x = }}\dfrac{{\text{1}}}{{\text{T}}}$
When a graph is plotted for ${\text{logK}}$ vs $\dfrac{{\text{1}}}{{\text{T}}}$, a straight line with a slope of ${\text{ - 4250K}}$ is obtained. Calculate ‘${\text{Ea }}$’ for the reaction. (${\text{R = 8}}{\text{.314J}}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$).
So, we want
${\text{m = - }}\dfrac{{{\text{Ea }}}}{{{\text{2}}{\text{.303R}}}}$
The gas constant ${\text{R}}$ is ${\text{8}}{\text{.314J}}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$.
The slope of the straight line ${\text{m}}$ is ${\text{ - 4250K}}$.
The activation energy ${\text{Ea }}$is calculated as,
${\text{Ea}} = - {\text{m}} \times {\text{2}}{\text{.303}} \times {\text{R}}$
Now we can substitute the known values we get,
$\Rightarrow {\text{Ea}} = - \left( { - {\text{4250K}}} \right) \times {\text{2}}{\text{.303}} \times {\text{8}}{\text{.314J}}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$
${\text{Ea}} = {\text{81375}}{\text{.35Jmo}}{{\text{l}}^{{\text{ - 1}}}}$
On simplification we get,
$\Rightarrow {\text{Ea}} = {\text{81}}{\text{.375kJmo}}{{\text{l}}^{{\text{ - 1}}}}$
The activation energy ‘${\text{Ea }}$’ for the reaction is ${\text{81}}{\text{.375kJmo}}{{\text{l}}^{{\text{ - 1}}}}$ .
Equation ${\text{log K = log A - }}\dfrac{{{\text{Ea }}}}{{{\text{2}}{\text{.303RT}}}}$
Where ${\text{Ea }}$is the activation energy. When a graph is plotted for ${\text{logK}}$ vs $\dfrac{{\text{1}}}{{\text{T}}}$, a straight line with a slope of ${\text{ - 4250K}}$ is obtained. The activation energy ‘${\text{Ea }}$’ for the reaction is ${\text{81}}{\text{.375kJmo}}{{\text{l}}^{{\text{ - 1}}}}$, if the value of the gas constant is ${\text{R = 8}}{\text{.314J}}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$.

Note:
We have to know that the rate of the reaction is an important factor for the study of reaction. The rate of reaction is an important concept for chemical kinetics. Rate of the reaction depends on the concentration of the reactant. The rate of reaction is also calculated by using the concentration of the product in the chemical reaction. Depending on the concentration, the sign of the rate will change.