The rate constant of a reaction varies with temperature according to the equation: $\log K = \text{constant} - \frac{E_a}{2.303 RT}$. If a plot of $\log K$ versus $1/T$ yields a straight line with a slope of $-5632$,then the activation energy of the reaction is .......... $kJ \ mol^{-1}$.

Catalyst $A$ reduces the activation energy for a reaction by $10 \ kJ \ mol^{-1}$ at $300 \ K$. The ratio of rate $\frac{k_{T, \text{Catalysed}}}{k_{T, \text{Uncatalysed}}}$ is $e^{x}$. Find the value of $x$ [nearest integer].
[Assume that the pre-exponential factor is same in both the cases.
Given $R = 8.31 \ J \ K^{-1} \ mol^{-1}$]

For a reaction,$A \rightarrow B$,the average energies of $A$ and $B$ are $30 \ kcal/mol$ and $60 \ kcal/mol$ respectively. The energy of activation for the backward reaction is $93 \ kcal/mol$. The energy of activation for the forward reaction is:

In an exothermic reaction $A \to B$,the evolved heat is $280 \ kJ \ mol^{-1}$ and the activation energy is $200 \ kJ \ mol^{-1}$. The activation energy of the reverse reaction $B \to A$ is $.......... \ kJ \ mol^{-1}$.

The rate of gas phase chemical reactions generally increases rapidly with a rise in temperature. This is mainly because

Which plot of $\ln k$ vs $\frac{1}{T}$ is consistent with the Arrhenius equation?