Question

The rate constant ${{k}_{1}}$ of a reaction is found to be double that of rate constant ${{k}_{2}}$ of another reaction. The relationship between corresponding activation energies of the two reactions at same temperature $({{E}_{1}}\text{ }and\text{ }{{E}_{2}})$ can be represented as:
A. ${{E}_{1}}>{{E}_{2}}$
B. ${{E}_{1}}<{{E}_{2}}$
C. ${{E}_{1}}={{E}_{2}}$
D. ${{E}_{1}}\approx {{E}_{2}}$

Answer 1

Hint As we know that rate of a reaction is the speed with which a chemical reaction takes place. It is found that the rate constant doubles for the reaction, this means that the rate constant for the first reaction will be equal to twice the rate constant for the second reaction.

Complete Step by step solution:
- As we know that the expression for the Arrhenius equation is given by the equation:
$k=A{{e}^{\dfrac{-Ea}{RT}}}$
- It is found that when the activation energy increases, the rate constant of the reaction decreases. The rate constant of first reaction is found to double that of the rate constant of the second reaction.
- Therefore, the activation energy of the first reaction is less than that of the second reaction that is ${{E}_{1}}<{{E}_{2}}$.

- Hence, we can conclude that the correct option is (b), that is the relationship between corresponding activation energies of the two reactions at same temperature $({{E}_{1}}\text{ }and\text{ }{{E}_{2}})$can be represented as: ${{E}_{1}} <{{E}_{2}}$.

Note: As we know that activation energy can be calculated using different methods. The Arrhenius equation can be used to measure it, and also when the rate constant and the two temperatures are known. We must remember to convert the temperature into kelvin while calculating the activation energy.