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

The equilibrium constant at $27\,^{\circ}C$ and $127\,^{\circ}C$ is $30$ and $40$ respectively. What will be the ratio of activation energy $(E_{a})_{f} / (E_{a})_{b}$?

$A$ reaction rate constant is given by $k = 1.2 \times 10^{14} e^{-(25000/RT)} \, s^{-1}$. It means

According to the collision theory of chemical reaction,which of the following is correct?

For the following reactions:
$A \xrightarrow{700 \ K}$ Product
$A \xrightarrow[\text{catalyst}]{500 \ K}$ Product
it was found that $E_{a}$ is decreased by $30 \ kJ/mol$ in the presence of a catalyst. If the rate remains unchanged,the activation energy for the catalysed reaction is (Assume pre-exponential factor is same):

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