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Question: In which of the following reactions, the ratio of final rate to initial rate will be maximum? (A) ...

In which of the following reactions, the ratio of final rate to initial rate will be maximum?
(A) Ea=40kJ/mol{E_a} = 40kJ/mol , Temperature rise 200210K200 - 210K
(B) Ea=90kJ/mol{E_a} = 90kJ/mol , Temperature rise 300320K300 - 320K
(C) Ea=80kJ/mol{E_a} = 80kJ/mol , Temperature rise 300310K300 - 310K
(D) All will have same rate

Explanation

Solution

Hint- In this we will put light on a very basic yet very important topic of chemistry. In this question we will learn about chemical kinetics and the effect of temperature and catalyst on the rate of a reaction. We will also learn about the catalyst and activation energy. And this information will further lead us in approaching our solution. Below here these concepts are explained properly.

Complete step by step solution:

Chemical kinetics is the branch of physical chemistry that deals with concern, the rate of chemical reaction is to resemble thermodynamics, it deals with the direction in which an activity occurs but in itself tells nothing about its rate.
Effect of temperature and catalyst on rate of a reaction- The reaction rate decreases with a decrease in temperature. Catalysts can lower the activation energy and increase the reaction rate without being consumed in the reaction. Differences in the inherent structures of reactant can lead to differences in reaction rates.
Catalyst- A catalyst is a substance that speeds up a chemical reaction, without being utilized in the reaction. It increases the rate of reaction by decreasing the activation energy for a reaction.
Activation Energy- Activation energy is the minimum energy required for a chemical reaction to proceed.
k=AeEaRTk = A_e^{ - \dfrac{{{E_a}}}{{RT}}}
Let 22 denotes the final state and 11 denotes initial state.
k2=AeEaRT2,k1=AeEaRT1{k_2} = A_e^{ - \dfrac{{{E_a}}}{{R{T_2}}}},\,{k_1} = A_e^{ - \dfrac{{{E_a}}}{{R{T_1}}}}
The ratio of final rate and the initial rate=k2k1=eEaR[1T21T1] = \dfrac{{{k_2}}}{{{k_1}}} = {e^{ - \dfrac{{{E_a}}}{R}\left[ {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right]}}
=eEaR[1T11T2]= {e^{ \dfrac{{{E_a}}}{R}\left[ {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right]}}
=eEaRT2T1T1T2= {e^{\dfrac{{{E_a}}}{R}\,\,\dfrac{{{T_2} - {T_1}}}{{{T_1}{T_2}}}}}
Higher the value ofEa(T2T1)T1T2\dfrac{{{E_a}\left( {{T_2} - {T_1}} \right)}}{{{T_1}\,{T_2}}} , higher will be the ratio, the highest value is obtained for option BB.
So, the correct answer is option B.

Note- Chemical kinetics, a branch of physical chemistry, is also called reaction kinetics and helps us to understand the rates of reactions and how it is influenced by the certain condition. When we talk about the applications, Chemical kinetics guide us in the controlling of reaction conditions and also in improving the reaction rate, to increase the production of the products.
It also guides us in suppressing or slowing down the rate of side reactions.