$A$ plot of $ \frac{1}{T} $ vs. $ \ln k $ for a reaction gives the slope $ -1 \times 10^{4} \ K $. The energy of activation for the reaction is (Given $ R = 8.314 \ J \ K^{-1} \ mol^{-1} $)

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

Find the correct equation among the following.

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

The first order rate constant for the decomposition of ethyl iodide by the reaction $C_{2}H_{5}I_{(g)} \rightarrow C_{2}H_{4(g)} + HI_{(g)}$ at $600 \ K$ is $1.60 \times 10^{-5} \ s^{-1}$. Its energy of activation is $209 \ kJ/mol$. Calculate the rate constant of the reaction at $700 \ K$.

The rate constants for the decomposition of acetaldehyde have been measured over the temperature range $700-1000 \ K$. The data has been analysed by plotting a $\ln \ k \ vs \ \frac{10^{3}}{T}$ graph,which gives a slope of $-18.5$. The value of activation energy for the reaction is $...... \ kJ \ mol^{-1}$. (Nearest integer) (Given: $R = 8.31 \ J \ K^{-1} \ mol^{-1}$)