Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

In presence of a catalyst, the activation energy is lowered by 3kcal at. Hence, the rate of reaction will increase by:
(A) 32 times
(B) 243 times
(C) 3 times
(D) 148 times

seo-qna
Last updated date: 15th Jun 2024
Total views: 411.9k
Views today: 7.11k
Answer
VerifiedVerified
411.9k+ views
Hint: Presence of catalyst increases the rate of reaction and lowers the activation energy without itself being consumed in the reaction. Rate of reaction is directly proportional to the negative exponential of Activation energy.

Complete answer:
Catalyst provide surface area for the reaction, and when surface area for reaction is increased , more molecules come together and rate of reaction increases. Catalyst itself do not get consumed in the reaction and is let at the end of the reaction as it is.
Any reaction proceeds through consuming some amount of activation energy. Catalyst provide a new reaction pathway, in which activation energy is lowered compared to the reaction where catalyst is not used.A catalyst increases the rate of a reaction by lowering the activation energy so that more reactant molecules collide with enough energy to surmount the smaller energy barrier.
Relation of activation energy and rate constant is given by Arrhenius equation,
\[k=A{{e}^{\frac{-{{E}_{a}}}{RT}}}\]
Where, k is the rate constant,
${{E}_{a}}$ is the activation energy,
R is the universal gas constant=2cal
T is the temperature in Kelvin= ${{27}^{\circ }}C$+273 = 300K
When compared with rate of reaction, with and without the use of catalyst, the expression will be written as,
     \[\dfrac{{{k}_{1}}}{{{k}_{2}}}=\frac{{{e}^{\dfrac{-{{E}_{a1}}}{RT}}}}{{{e}^{\dfrac{-{{E}_{a2}}}{RT}}}}\]
Where ${{E}_{a1}}$ = activation energy without the use of catalyst
${{E}_{a2}}$= activation energy with the use of catalyst
${{k}_{1}}$= rate of reaction without the use of catalyst
${{k}_{2}}$= rate of reaction with the use of catalyst
                    \[\dfrac{{{k}_{1}}}{{{k}_{2}}}={{e}^{\dfrac{^{{{E}_{a1}}-{{E}_{a2}}}}{RT}}}\]
Given in the question value of difference of the two activation energies is 3kcal, putting the values in the equation, we get,
     \[\begin{align}
  & \dfrac{{{k}_{2}}}{{{k}_{1}}}={{e}^{\dfrac{3000}{2\times 300}}} \\
 & \Rightarrow \dfrac{{{k}_{2}}}{{{k}_{1}}}={{e}^{5}} \\
 & \Rightarrow {{k}_{2}}=148{{k}_{1}} \\
\end{align}\]
The rate of reaction is increased by 148 times.
So, the correct answer is “Option D”.

Note: Other than presence of catalyst, there are many other factors which effect the rate of reaction, they are, concentration of reaction, physical state of reactants an surface area, optimum temperature and sometimes optimum pH.