Reactant $A$ shows two reactions:
$A \xrightarrow{K_1} B$,activation energy $= Ea_1$
$A \xrightarrow{K_2} C$,activation energy $= Ea_2$
If $Ea_1 = \frac{Ea_2}{3}$,then the relation between $K_1$ and $K_2$ is:

The activation energy of a first-order reaction at $25\,^{\circ}C$ is $30\,kJ/mol$. In the presence of a catalyst,the activation energy of the same reaction at $25\,^{\circ}C$ becomes $24\,kJ/mol$. The rate of the reaction in the presence of the catalyst will be how many times the original rate?

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

The activation energy of a forward reaction is $50 \, kcal$. What will be the activation energy of its reverse reaction?

At $27^{\circ}C$ in the presence of a catalyst, the activation energy of a reaction is lowered by $10 \ kJ \ mol^{-1}$. The logarithm ratio of $\frac{k(\text{catalysed})}{k(\text{uncatalysed})}$ is .... (Consider that the frequency factor for both the reactions is the same)

For the reaction $A_2 + B_2 \rightleftharpoons 2AB$,the activation energies for the forward and backward reactions are $180 \, kJ \, mol^{-1}$ and $200 \, kJ \, mol^{-1}$ respectively. In the presence of a catalyst,the activation energy for both (forward and backward) reactions decreases by $100 \, kJ \, mol^{-1}$. What will be the enthalpy change $(\Delta H)$ for the reaction $(A_2 + B_2 \rightarrow 2AB)$ in the presence of a catalyst in $kJ \, mol^{-1}$?