When temperature is increased from $27\,{}^{\circ}C$ to $127\,{}^{\circ}C$,the rate of reaction becomes doubled. The activation energy $E_a$ will be ........ $kCal$.

What is the value of the slope if $\log_{10} K$ ($y$-axis) is plotted versus $1/T$ ($x$-axis) for the Arrhenius equation?

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 $T$?

In the Arrhenius equation,the pre-exponential factor represents ...

The decomposition of $N_2O$ into $N_2$ and $O_2$ in the presence of argon follows first-order kinetics. If the rate constant is given by $K = 5.0 \times 10^{11} \, e^{-2000/T}$,what is the activation energy of the reaction?

If a reaction follows the Arrhenius equation,the plot $\ln k$ vs $1/(RT)$ gives a straight line with a gradient of $(-y) \ unit$. The energy required to activate the reactant is