Question

The rate constant $\left( k ^{\prime}\right)$ of one reaction is double the rate constant $\left( k ^{\prime \prime}\right)$ of another reaction. Then the relationship between the corresponding activation energies of the two reactions ( $E _{ a }^{\prime}$ and $E _{ a }$ ") will be (Assume the pre-exponential factor \& temperature to be same)

• A. $\text{E}_{\text{a}}^{'}> \, \text{E}_{\text{a}}^{′′}$
• B. $\text{E}_{\text{a}}^{'}= \, \text{E}_{\text{a}}^{′′}$
• C. $\text{E}_{\text{a}}^{'} < \, \text{E}_{\text{a}}^{′′}$
• D. $\text{E}_{\text{a}}^{'} < \text{ 4}\text{E}_{\text{a}}^{′′}$

Answer 1

Answer: (C)

Rate constant related with activation energy by Arrhenius equation ( $k=Ae^{- \frac{E_{a}}{RT}}$ ). When activation energy decreases, rate constant increases. As $k^{'}>k^{''}, \, $ $E_{a}^{'} < \, E_{a}^{′′}$