Question

The energy of activation for a reaction is 50 $ {\text{kJ/mol}} $ . The presence of a catalyst lowers the energy of activation by 25%. What will be the effect on the rate of the reaction at $ {\text{30}}^\circ {\text{C}} $ if the other things remain the same?
(A) $ 142.75 $
(B) $ 242.75 $
(C) $ 342.75 $
(D) $ 442.75 $

Answer 1

Hint : The term catalyst refers to a substance which can increase the rate of a reaction without itself being consumed that is without itself undergoing any chemical change. The effect on the rate of the reaction in the presence of catalyst can be evaluated by comparing the activation energies in the presence and absence of catalyst.

Complete Step By Step Answer:
Given that the energy of activation for a reaction is 50 $ {\text{kJ/mol}} $ .
In the presence of a catalyst, the energy of activation is reduced by 25% of 50 $ {\text{kJ/mol}} $ which means it is reduced by 12.5 $ {\text{kJ/mol}} $ . Thus, the energy of activation is reduced from 50 $ {\text{kJ/mol}} $ to $ \left( {{\text{50 - 12}}{\text{.5}}} \right){\text{ = 37}}{\text{.5kJ/mol}} $ .
Also given, the temperature $ {\text{T = 30}}^\circ {\text{C = }}\left( {{\text{30 + 273}}} \right){\text{K = 303K}} $ and $ {\text{R = 8}}{\text{.314J}}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}} $ .
Now, the relationship between the temperature with the rate of a reaction and also with the rate constant ‘k’ is given by the Arrhenius expression. It is usually expressed in the form:
 $ k= Ae^{\dfrac{-E_{a}}{RT}} $
where k is the rate constant, A is the pre-exponential factor and represents a constant called the frequency factor, $ {{\text{E}}_{\text{a}}} $ represents the energy of activation, R is the gas constant and T is the absolute temperature. The two quantities A and $ {{\text{E}}_{\text{a}}} $ are also collectively named as the Arrhenius parameters.
Let $ {{\text{k}}_{\text{1}}} $ and $ {{\text{E}}_{{\text{a1}}}} $ represent the rate constant and the energy of activation of the reaction in the absence of the catalyst and let $ {{\text{k}}_2} $ and $ {{\text{E}}_{{\text{a2}}}} $ represent the rate constant and the energy of activation of the reaction in the presence of the catalyst. So we will have: $ E_{a1}= 50kJ/mol $ and $ E_{a2}= 37.5kJ/mol $ . Converting kilojoules into joules we get, $ E_{a1}= 50\times 10^{3}J/mol $ and $ E_{a2}= 37.5\times 10^{3}J/mol $ .
If we put the value of $ {{\text{E}}_{{\text{a1}}}} $ in the Arrhenius expression, we will get the expression for the absence of the catalyst:
 $ k_{1}= Ae^{\dfrac{-50\times 10^{3}}{RT}} $ ……… (1)
and if we put the value of $ {{\text{E}}_{{\text{a2}}}} $ in the Arrhenius expression, we will get the expression for the presence of the catalyst:
 $ k_{2}= Ae^{\dfrac{-37.5\times 10^{3}}{RT}} $ ……… (2)
Divide equation (1) by equation (2),
 $
\dfrac{k_{1}}{k_{2}}= \dfrac{Ae^{\dfrac{-50\times 10^{3}}{RT}}}{Ae^{\dfrac{-37.5\times 10^{3}}{RT}}}
\Rightarrow \dfrac{k_{1}}{k_{2}}=e^{\dfrac{\left ( -50+37.5 \right )\times 10^{3}}{RT}}
\Rightarrow \dfrac{k_{1}}{k_{2}}=e^{\dfrac{-12.5\times 10^{3}}{RT}}
\Rightarrow \dfrac{k_{2}}{k_{1}}=e^{\dfrac{12.5\times 10^{3}}{RT}}
  $
Take log on both sides and substitute the values of R and T,
 $
  2.303log\dfrac{k_{2}}{k_{1}}= \dfrac{12.5\times 10^{3}}{8.314\times 303}
\Rightarrow log\dfrac{k_{2}}{k_{1}}= \dfrac{12.5\times 10^{3}}{8.314\times 303\times 2.303}
\Rightarrow \dfrac{k_{2}}{k_{1}}=Antilog\left ( \dfrac{12.5\times 10^{3}}{8.314\times 303\times 2.303} \right )
\Rightarrow \dfrac{k_{2}}{k_{1}}=142.75
  $
So, the rate of the reaction will increase $ 142.75 $ times.
So, the correct option is (A).

Note :
The term $ e^{\dfrac{-E_{a}}{RT}} $ denotes the fraction of molecules having energy equal to or greater than the energy of activation $ {{\text{E}}_{\text{a}}} $ . It is also called the Boltzmann factor. As the exponential factor is dimensionless, the pre-exponential factor A has the same units as that of rate constant. This is why it is also called the frequency factor.