$A$ reaction is catalysed by $X$. Here $X$:

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)

According to the collision theory of chemical reactions:

Given the activation energy for the forward direction $(E_{af})$ is $50 \ kJ$ and for the backward direction $(E_{ab})$ is $30 \ kJ$. The reaction would be:

In a reaction, for every $10^{\circ} C$ rise of temperature, the rate is doubled. If the temperature is increased from $10^{\circ} C$ to $100^{\circ} C$, the rate of the reaction will become $:-$ (in $\times$)

Activation energy of any reaction depends on