Question

At a certain temperature, the first order rate constant, $k_{1}$ is found to be smaller than the second order rate constant, $k_{2}$. If the energy of activation, $E_{1}$ of the first order reaction is greater than energy of activation, $E_{2}$ of the second order reaction then with increase in temperature

• A. $k_{1}$ will increase faster than $k_{2}$, but always will remain less than $k_{2}$
• B. $k_{2}$ will increase faster than $k_{1}$
• C. $k_{1}$ will increase faster than $k_{2}$ and becomes equal to $k_{2}$
• D. $k_{1}$ will increase faster than $k_{2}$ and becomes greater than $k_{2}$

Answer 1

The Correct Option is A

$\frac{d(\ln k)}{d t}=\frac{E_{a}}{R T^{2}}$ As $E_{a}$ increases, rate of increase in $k$ also increases so $k_{1}$ will increase faster than $k_{2}$ but always will remain less than $k_{2}$.