Slope of the straight line obtained by plotting $\log_{10} k$ against $\frac{1}{T}$ represents which term?

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

Consider the following plots of log of rate constant $k$ $(\log k)$ vs $\frac{1}{T}$ for three different reactions. The correct order of activation energies of these reactions is

The activation energy of one of the reactions in a biochemical process is $532611 \, J \, mol^{-1}$. When the temperature falls from $310 \, K$ to $300 \, K$,the change in rate constant observed is $k_{300} = x \times 10^{-3} \, k_{310}$. The value of $x$ is $.....$ [Given: $\ln 10 = 2.3$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$]

For a hypothetical reaction $A \to B$,the activation energies for forward and backward reactions are $19 \ kJ/mol$ and $9 \ kJ/mol$ respectively. The heat of reaction is $.... \ kJ$.

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: