What is the activation energy $(kJ \, mol^{-1})$ for a reaction if its rate constant doubles when the temperature is raised from $300 \, K$ to $400 \, K$ ? $(R = 8.314 \, J \, mol^{-1} \, K^{-1})$

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 $\qquad$ $[K = \text{equilibrium constant of the reaction}, k_f = \text{rate constant of forward reaction}, k_b = \text{rate constant of backward reaction}]$

The decomposition of $N_2O$ into $N_2$ and $O_2$ in the presence of argon follows first-order kinetics. If the rate constant is given by $K = 5.0 \times 10^{11} \, e^{-2000/T}$,what is the activation energy of the reaction?

Define the following terms:
$1.$ Collision
$2.$ Frequency of collision

Consider the following energy profile for the reaction $X + Y \rightarrow R + S$. Which of the following deductions about the reaction is not correct?

$A$ catalyst lowers the activation energy for a certain reaction from $83.314 \, kJ \, mol^{-1}$ to $75 \, kJ \, mol^{-1}$ at $500 \, K$. What will be the rate of reaction as compared to the uncatalysed reaction? Assume other things being equal.