At a constant temperature,the activation energy of a reaction is found to be $2.303 \, RT \, J \, mol^{-1}$. The ratio of the rate constant to the Arrhenius constant will be $......$.

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 $\ln(1/4) = -1.386, R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

Assertion $(A)$ : $A$ catalyst increases the rate of a reaction.
Reason $(R)$ : In presence of a catalyst, the activation energy of the reaction increases.
The correct answer is

The influence of temperature on the rate of reaction can be found out by

The plot of $\log k_f$ versus $1 / T$ for a reversible reaction $A_{(g)} \rightleftharpoons P_{(g)}$ is shown. Pre-exponential factors for the forward and backward reactions are $10^{15} \ s^{-1}$ and $10^{11} \ s^{-1}$,respectively. If the value of $\log K$ for the reaction at $500 \ K$ is $6$,the value of $|\log k_b|$ at $250 \ K$ is $\qquad$ $[K = \text{equilibrium constant of the reaction}, k_f = \text{rate constant of forward reaction}, k_b = \text{rate constant of backward reaction}]$

For a reaction having three steps, the overall rate constant is $K = \frac{k_1 k_2}{k_3}$. The values of $E_{a1}$, $E_{a2}$ and $E_{a3}$ (activation energies for each step) are $40$, $50$ and $60 \text{ kJ mol}^{-1}$ respectively. The overall activation energy $E_a$ of the reaction is: