The rate constant for a chemical reaction taking place at $500 \ K$ is expressed as $K = A \ e^{-1000}$. The activation energy of the reaction is:

Two reactions $R_1$ and $R_2$ have identical pre-exponential factors. Activation energy of $R_1$ exceeds that of $R_2$ by $10 \, kJ \, mol^{-1}.$ If $k_1$ and $k_2$ are rate constants for reactions $R_1$ and $R_2$ respectively at $300 \, K,$ then $\ln (k_2/k_1)$ is equal to :
$(R=8.314 \, J \, mol^{-1} \, K^{-1})$

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

What is the activation energy for a reaction if its rate doubles when the temperature is raised from $20 \,^{\circ}C$ to $35 \,^{\circ}C$ in $kJ \,mol^{-1}$? $(R = 8.314 \,J \,mol^{-1} \,K^{-1})$

Consider an endothermic reaction $X \to Y$ with the activation energies $E_b$ and $E_f$ for the backward and forward reactions,respectively. In general:

The rate constants $k_1$ and $k_2$ for two different reactions are $10^{16} \times e^{-2000/T}$ and $10^{15} \times e^{-1000/T}$ respectively. The temperature at which $k_1 = k_2$ is