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. ${\text{E}}_{\text{a}}^{{}^{\prime }}>{\text{E}}_{\text{a}}^{\prime \prime }$
• B. ${\text{E}}_{\text{a}}^{{}^{\prime }}={\text{E}}_{\text{a}}^{\prime \prime }$
• C. ${\text{E}}_{\text{a}}^{{}^{\prime }}<{\text{E}}_{\text{a}}^{\prime \prime }$
• D. ${\text{E}}_{\text{a}}^{{}^{\prime }}<\text{ 4}{\text{E}}_{\text{a}}^{\prime \prime }$

Answer 1

Answer: (C)

Rate constant related with activation energy by Arrhenius equation ( $k=A{e}^{-\frac{E_a}{RT}}$ ). When activation energy decreases, rate constant increases. As $k^{{}^{\prime }}>k^{{}^{\prime \prime }},$ $E_a^{{}^{\prime }}<E_a^{\prime \prime }$