$A$ large increase in the rate of a reaction for a rise in temperature is due to:

Activation energy for the acid catalysed hydration of sucrose is $6.22 \ kJ \ mol^{-1}$,while the activation energy is only $2.15 \ kJ \ mol^{-1}$ when hydrolysis is catalysed by the enzyme sucrase. Explain.

Consider the following transformation involving first order elementary reactions in each step at constant temperature as shown below.
$A + B \underset{\text{Step } 3}{\overset{\text{Step } 1}{\rightleftharpoons}} C \xrightarrow{\text{Step } 2} P$
Some details of the above reaction are listed below.

Step	Rate constant $(s^{-1})$	Activation energy $(kJ \ mol^{-1})$
$1$	$k_1$	$300$
$2$	$k_2$	$200$
$3$	$k_3$	$Ea_3$

If the overall rate constant of the above transformation $(k)$ is given as $k = \frac{k_1 k_2}{k_3}$ and the overall activation energy $(E_a)$ is $400 \ kJ \ mol^{-1}$,then the value of $Ea_3$ is $\qquad$ $kJ \ mol^{-1}$ (nearest integer).

For an exothermic reaction $A \rightarrow B$,the activation energy is $15 \, K \, cal/mol$ and the heat of reaction is $5 \, K \, cal/mol$. The activation energy for the reverse reaction $B \rightarrow A$ will be ......... $K \, cal/mol$.

According to the collision theory of reaction rates,the rate of reaction increases with temperature due to:

Assertion $(A)$ : $A$ catalyst increases the rate of a reaction.
Reason $(R)$ : In presence of a catalyst,the activation energy of the reaction increases.
The correct answer is