In the Arrhenius equation,$k = A e^{-E_a/RT}$,the rate constant $k$ becomes equal to the frequency factor $A$ when:

The rate constant of a reaction is increased $4$ times after the addition of a catalyst to the reaction mixture at the same temperature of $27^{\circ} C$. The change in the activation energy of this reaction is (Take $\ln(1/4) = -1.386, R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

The temperature at which the rate constants of the two gaseous reactions given below become equal is . . . . . . $K$. (Nearest integer).
$X \longrightarrow Y \quad k_1 = 10^6 e^{\frac{-30000}{T}}$
$P \longrightarrow Q \quad k_2 = 10^4 e^{\frac{-24000}{T}}$
Given: $\ln 10 = 2.303$

According to the collision theory of reaction rates,the rate of reaction increases with temperature due to:

The activation energy for a reaction at the temperature $T \ K$ was found to be $2.303 \ RT \ J \ mol^{-1}$. The ratio of the rate constant to the Arrhenius factor is

What is the value of the slope of the plot $\ln K$ versus $\frac{1}{T}$ for a reaction having $E_{a} = 33.256 \ J \ mol^{-1}$?