For an endothermic reaction,where $\Delta H$ represents the enthalpy of the reaction in $kJ/mol$,the minimum value for the energy of activation will be

The following graph is obtained for a reaction $(A \rightarrow P)$. The activation energy ($E_{a}$ in $kJ \ mol^{-1}$) and heat of reaction ($|\Delta H|$ in $kJ \ mol^{-1}$) for this reaction are respectively ($x=$ reaction coordinate; $y=E$ in $kJ \ mol^{-1}$)

Consider the following reaction that proceeds from $A$ to $B$ in three steps as shown in the energy profile diagram. Choose the correct values for the following parameters:
$1$. Number of intermediates
$2$. Number of activated complexes
$3$. Rate determining step

The energies of activation for forward and reverse reactions for $A_2 + B_2 \rightleftharpoons 2AB$ are $180 \, kJ \, mol^{-1}$ and $200 \, kJ \, mol^{-1}$ respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by $100 \, kJ \, mol^{-1}.$ The enthalpy change of the reaction $(A_2 + B_2 \rightarrow 2AB)$ in the presence of a catalyst will be (in $kJ \, mol^{-1}$)

The energy profile diagram for a multi-step reaction,$A$ $\xrightarrow{1} B$ $\xrightarrow{2} C$ $\xrightarrow{3} D,$ is given below. The rate-determining step of the reaction is:

What is the slope of the straight line for the graph drawn between $\ln k$ and $\frac{1}{T}$,where $k$ is the rate constant of a reaction at temperature $T$?