For a reaction,the graph of $\ln k$ (on y-axis) and $1 / T$ (on x-axis) is a straight line with a slope $-2 \times 10^4 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$

The temperature coefficient of a reaction is defined as:

The rate constant $(K')$ of one reaction is double the rate constant $(K'')$ of another reaction. What is the relationship between the corresponding activation energies of the two reactions (${E_a}'$ and ${E_a}''$)?

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 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,$A \rightarrow B$,the average energies of $A$ and $B$ are $30 \ kcal/mol$ and $60 \ kcal/mol$ respectively. The energy of activation for the backward reaction is $93 \ kcal/mol$. The energy of activation for the forward reaction is: