For a first order reaction $A \rightarrow P$,the temperature $(T)$ dependent rate constant $(k)$ was found to follow the equation $\log k = -(2000) \frac{1}{T} + 6.0$. The pre-exponential factor $A$ and the activation energy $E_{a}$,respectively,are

$A$ plot of $ \frac{1}{T} $ vs. $ \ln k $ for a reaction gives the slope $ -1 \times 10^{4} \ K $. The energy of activation for the reaction is (Given $ R = 8.314 \ J \ K^{-1} \ mol^{-1} $)

What happens to the most probable kinetic energy and the energy of activation with an increase in temperature?

Consider the following transformation involving first order elementary reactions in each step at constant temperature as shown below.
$A + B \underset{\text{Step } 3}{\overset{\text{Step } 1}{\rightleftharpoons}} C \xrightarrow{\text{Step } 2} P$
Some details of the above reaction are listed below.

Step	Rate constant $(s^{-1})$	Activation energy $(kJ \ mol^{-1})$
$1$	$k_1$	$300$
$2$	$k_2$	$200$
$3$	$k_3$	$Ea_3$

If the overall rate constant of the above transformation $(k)$ is given as $k = \frac{k_1 k_2}{k_3}$ and the overall activation energy $(E_a)$ is $400 \ kJ \ mol^{-1}$,then the value of $Ea_3$ is $\qquad$ $kJ \ mol^{-1}$ (nearest integer).

The rate of a reaction doubles,when the temperature is changed from $300 \ K$ to $310 \ K$. Activation energy of the reaction is....... $(R=8.314 \ J \ K^{-1} \ mol^{-1}, \log 2=0.301)$

For a reaction,the activation energy $E_{a} = 0$ and the rate constant at $200 \ K$ is $1.6 \times 10^{6} \ s^{-1}$. The rate constant at $400 \ K$ will be (given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$):