What is the value of the slope of the plot $\ln K$ versus $\frac{1}{T}$ for a reaction having $E_{a} = 33.256 \ J \ mol^{-1}$?

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 $T$?

$A$ reaction is catalysed by $X$. Here $X$:

The following figure shows a graph of $\log_{10}K$ vs $\frac{1}{T}$,where $K$ is the rate constant and $T$ is the temperature. The straight line $BC$ has a slope,$\tan \theta = -\frac{1}{2.303}$,and an intercept of $5$ on the $Y$-axis. Thus,$E_a$,the energy of activation,is ....... $cal$.

For the decomposition reaction of $N_2O_5$,the slope of the graph of $\log K$ versus $1/T$ is $-1.2 \times 10^4 \ K$. Calculate the activation energy $(E_a)$ of the reaction.

The correct reaction profile diagram for a positive catalyst reaction.