The velocity constant of a reaction at $290 \ K$ was found to be $3.2 \times 10^{-3}$. At $300 \ K$ it will be

For reaction $A \to B$,rate constant $K_1 = A_1 e^{-E_{a_1}/RT}$ and for the reaction $X \to Y$,rate constant $K_2 = A_2 e^{-E_{a_2}/RT}$. If $A_1 = 10^8, A_2 = 10^{10}$ and $E_{a_1} = 600 \ cal \ mol^{-1}$,$E_{a_2} = 1800 \ cal \ mol^{-1}$,then the temperature at which $K_1 = K_2$ is (given: $R = 2 \ cal \ K^{-1} \ mol^{-1}$):

Explain: How is the value of activation energy determined based on the Arrhenius equation?

For an endothermic reaction $X \rightarrow Y$,the activation energies for the forward and backward reactions are $E_f$ and $E_b$ respectively. Then,in general:

Reactant $A$ converts to product $D$ through the given mechanism (with the net evolution of heat) :

$A \rightarrow B$	$slow ; \Delta H=+ve$
$B \rightarrow C$	$fast ; \Delta H=-ve$
$C \rightarrow D$	$fast ; \Delta H=-ve$

Which of the following represents the above reaction mechanism ?

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