Question

In the Arrhenius equation,$\text{ k = A e}{{\text{ }}^{\dfrac{-{{\text{E}}_{\text{a}}}}{\text{RT}}}}$ the factor $\text{ e}{{\text{ }}^{\dfrac{-{{\text{E}}_{\text{a}}}}{\text{RT}}}}$ is known as:
A) Frequency factor
B) Activation factor
C) Pre-exponential factor
D) Boltzmann factor

Answer 1

the Arrhenius equation established the relation between the rate constant and the activation energy.
$\text{ k = A e}{{\text{ }}^{\dfrac{-{{\text{E}}_{\text{a}}}}{\text{RT}}}}$
The kinetic theory explains that the molecules which have the most probable velocity undergo the reaction. The velocity of the molecule depends on the Boltzmann constant$\text{ }{{\text{k}}_{\text{B}}}\text{ }$.

Complete answer:
Arrhenius proposed an empirical relationship between the energy of activation ‘A’ and rate constant ‘k’ of a reaction at the temperature ‘T.The relation is as follows:
$\text{ k = A e}{{\text{ }}^{\dfrac{-{{\text{E}}_{\text{a}}}}{\text{RT}}}}$
Where $\text{ }{{\text{E}}_{\text{a}}}\text{ }$ is the Arrhenius activation energy and $\text{ A }$ is called the Arrhenius pre-exponential factor. Since the exponential factor in the equation is dimensionless, the pre-exponential factor A has the same unit as the rate constant k.
The Arrhenius equation can be expressed in terms of the Boltzmann constant as follows,
$\text{ k = A e}{{\text{ }}^{\dfrac{-{{\text{E}}_{\text{a}}}}{{{\text{k}}_{\text{B}}}\text{T}}}}\text{ }$
Where, ${{\text{k}}_{\text{B}}}$ is the Boltzmann constant.
Now, according to the kinetic theory of gases, the most probable velocity of the reactant molecules increases with the temperature rise. The temperature rise, therefore, increases the fraction of the molecules having high velocities. This means that more reactant molecules now can react with others to form a product. The relation of rate with the temperature is given by the factor,
$\text{ exp }\left( -\dfrac{m{{c}^{2}}}{2kT} \right)\text{ = exp }\left( -\dfrac{{{\text{E}}_{\text{a}}}}{{{\text{k}}_{\text{B}}}\text{T}} \right)\text{ }$
The exponent has a negative sign and temperature T is in the denominator. The factor, therefore, increases with the increase in temperature .This factor is known as the Boltzmann Factor.
The greater the value of temperature the greater is the value of$\text{ }{{\text{E}}_{\text{a}}}\text{ }$. Hence, the rapid increase in the Boltzmann constant will increase in temperature.
Thus, $\text{ e}{{\text{ }}^{\dfrac{-{{\text{E}}_{\text{a}}}}{\text{RT}}}}\text{ }$ is called the Boltzmann factor.

Hence, (D) is the correct option.

Note:
For first-order reaction, the Arrhenius equation, the rate constant has the units of frequency. Hence, $\text{ A }$is also called the frequency factor. $\text{ }{{\text{E}}_{\text{a}}}\text{ }$And $\text{ A }$are called the Arrhenius parameter. The kinetic theory is used to relate the kinetic of reaction and energy barrier.