The decomposition of a hydrocarbon follows the equation $K = (4.5 \times 10^{11} \, s^{-1}) e^{-28000 \, K/T}$. What will be the value of activation energy in $KJ \, mol^{-1}$?

For the reaction $A \rightarrow B$,the rate constant $k$ (in $s^{-1}$) is given by $\log_{10} k = 20.35 - \frac{2.47 \times 10^{3}}{T}$. The energy of activation in $kJ \, mol^{-1}$ is ..... . (Nearest integer) [Given: $R = 8.314 \, J \, K^{-1} \, mol^{-1}$]

Explain: How is the value of activation energy determined based on the Arrhenius equation?

Reactant $(A)$ produces two products. If $Ea_2 = 2 Ea_1$,then $K_1$ and $K_2$ are related as:
$A \xrightarrow{K_1} B$,activation energy: $Ea_1$
$A \xrightarrow{K_2} C$,activation energy: $Ea_2$

The temperature dependence of the rate constant $k$ is expressed as $k = A e^{-E_a / RT}$. When a plot between $\log k$ and $1/T$ is plotted,we get the graph as shown. What is the value of the slope in the graph?

$A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$
The curves $M$ and $N$ represent the variation of energy with reaction coordinate for the reaction in absence and presence of catalyst.
Which value represents the activation energy $(E_a)$ for the backward reaction in the presence of catalyst?