An endothermic reaction with high activation energy for the forward reaction is given by the diagram:

$A$ reaction takes place in three steps with individual rate constant and activation energy,

Step	Rate constant and Activation energy
$Step \ 1$	$k_1, E_{a_1} = 180 \ kJ \ mol^{-1}$
$Step \ 2$	$k_2, E_{a_2} = 80 \ kJ \ mol^{-1}$
$Step \ 3$	$k_3, E_{a_3} = 50 \ kJ \ mol^{-1}$

Overall rate constant,$k = (k_1 k_2 / k_3)^{2/3}$. The overall activation energy of the reaction will be ........ $kJ \ mol^{-1}$.

The activation energy for the forward reaction is $150 \ kJ \ mol^{-1}$ and that of the reverse reaction is $260 \ kJ \ mol^{-1}$. The enthalpy change for the reaction is:

For the reaction $A_2 + B_2 \rightleftharpoons 2 AB$,the activation energies $(E_a)$ for the forward and backward reactions are $180 \ kJ \ mol^{-1}$ and $200 \ kJ \ mol^{-1}$ respectively. If a catalyst lowers the $E_a$ for both reactions by $100 \ kJ \ mol^{-1}$,which of the following statements is correct?

Given below is an expression for the rate constant of a first order reaction occurring at a certain temperature, $T (\text{K})$.
$\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$
The energy of activation in $\text{kcal mol}^{-1}$ for the reaction is :
(Given : $k$ is $\text{s}^{-1}$, $R = 1.987 \text{ cal mol}^{-1} \text{ K}^{-1}$)

$A \rightarrow B$
The rate constants of the above reaction at $200 \, K$ and $300 \, K$ are $0.03 \, min^{-1}$ and $0.05 \, min^{-1}$ respectively. The activation energy for the reaction is $.... \, J$ (Nearest integer).
(Given: $\ln 10 = 2.3$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$,$\log 5 = 0.70$,$\log 3 = 0.48$,$\log 2 = 0.30$)