For a chemical reaction,the rate of reaction doubles when the temperature increases by $10\,^oC$. If the temperature is increased from $10\,^oC$ to $100\,^oC$,the rate of reaction will increase by a factor of .....

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 the reaction,$aA + bB \rightarrow cC + dD$,the plot of $\log k$ vs $\frac{1}{T}$ is given below. The temperature at which the rate constant of the reaction is $10^{-4} \ s^{-1}$ is ............... $K$. (Rounded-off to the nearest integer) [Given: The rate constant of the reaction is $10^{-5} \ s^{-1}$ at $500 \ K$.]

Explain: How is the value of activation energy determined based on the Arrhenius equation?

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 rate constant of a first-order reaction is $1.5 \times 10^7 \text{ s}^{-1}$ at $300 \text{ K}$ and $3.0 \times 10^7 \text{ s}^{-1}$ at $330 \text{ K}$. What is the activation energy $(E_a)$ for the reaction? $[R \times 2.303 = 19.15 \text{ J K}^{-1} \text{mol}^{-1}]$