Draw an energy profile diagram for a three-step reaction in which the first step is the slowest and the last step is the fastest. (Assume that the reaction is exothermic)

Consider the following reversible reaction,
$A_{(g)} + B_{(g)} \rightleftharpoons AB_{(g)}.$
The activation energy of the backward reaction exceeds that of the forward reaction by $2RT$ (in $J \ mol^{-1}$). If the pre-exponential factor of the forward reaction is $4$ times that of the reverse reaction,the absolute value of $\Delta G^{\ominus}$ (in $J \ mol^{-1}$) for the reaction at $300 \ K$ is. . . . . (Given; $\ln(2)=0.7, RT=2500 \ J \ mol^{-1}$ at $300 \ K$ and $G$ is the Gibbs energy)

The temperature-dependent equation for the rate constant is written as:

The activation energy of a reaction is $0$. The rate constant of the reaction:

For a reaction,the rate constants $K_1$ and $K_2$ are given by $10^{16} \cdot e^{-2000/T}$ and $10^{15} \cdot e^{-1000/T}$ respectively. At what temperature will $K_1 = K_2$?

The following energy profile diagram is given for the reaction $A + B \to C + D$. The enthalpy change and activation energy for the reverse reaction $C + D \to A + B$ are respectively: