Question

Rate constant $ K=1.2\times {{10}^{3}}mo{{l}^{-1}}L{{s}^{-1}} $ and $ {{E}_{a}}=2.0\times {{10}^{2}}kJmo{{l}^{-1}} $ , when $ T\to \infty $ , then?
(A) $ A=2.0\times {{10}^{2}}\text{ mo}{{\text{l}}^{-1}}\text{L}{{\text{s}}^{-1}} $
(B) $ A=1.2\times {{10}^{3}}\text{ mo}{{\text{l}}^{-1}}\text{L}{{\text{s}}^{-1}} $
(C) $ A=2.4\times {{10}^{2}}\text{ mo}{{\text{l}}^{-1}}\text{L}{{\text{s}}^{-1}} $
(D) $ A=1.2\times {{10}^{3}}\text{ mo}{{\text{l}}^{-1}}\text{L}{{\text{s}}^{-1}} $

Answer 1

Hint :The pre-exponential factor, also known as frequency factor, is an important component of the Arrhenius equation which represents the frequency at which reactant molecules collide at a standard concentration. The units of pre-exponential factors are completely dependent on the order of the reaction.

Complete Step By Step Answer:
Arrhenius equation: It is the equation which states that the rate constant of the reaction is a function of its activation energy and temperature which is associated with the reaction. The function is expressed as follows:
 $ K=A{{e}^{\left( -\dfrac{{{E}_{a}}}{RT} \right)}} $
Where, $ K $ is the rate constant, $ A $ is the pre-exponential factor, $ {{E}_{a}} $ is the activation energy and $ T $ is the temperature.
Now, it is given to us that the temperature of the system is approaching infinity i.e., $ T\to \infty $ . Therefore, if we substitute the given data and apply limits, then the expression will be as follows:
 $ 1.2\times {{10}^{3}}=\underset{T\to \infty }{\mathop{\lim }}\,A{{e}^{\left( -\dfrac{2.0\times {{10}^{2}}\times {{10}^{3}}}{8.314T} \right)}} $
Applying limits in the expression:
 $ \Rightarrow 1.2\times {{10}^{3}}=A{{e}^{\left( 0 \right)}} $
Because the value for $ \dfrac{1}{\infty } $ is equal to zero.
 $ \therefore A=1.2\times {{10}^{3}}\text{ mo}{{\text{l}}^{-1}}\text{L}{{\text{s}}^{-1}}\ \ \ \ [\because {{a}^{o}}=1] $
Hence, the value for pre-exponential factor under the given conditions is $ 1.2\times {{10}^{3}}\text{ mo}{{\text{l}}^{-1}}\text{L}{{\text{s}}^{-1}} $ .
Thus, option (B) is the correct answer.

Additional Information:
As the unit of the pre-exponential factor $ A $ is $ \text{mo}{{\text{l}}^{-1}}\text{L}{{\text{s}}^{-1}} $ which indicates that the reaction is of second order which means rate of the reaction is proportional to the concentration of two reactant molecules.

Note :
Pre-exponential factor is a function of temperature because the factors of the transition state and the collision of the reactant molecules are responsive to the change in temperature. As the temperature of the system increases, the molecules will collide faster which leads to an increase in the pre-exponential factor $ A $ .