For a first-order gaseous reaction,a plot of $\log \, k$ versus $1/T$ gives a straight line with a slope of $-8000$. Calculate the activation energy $(E_a)$ of the reaction in $cal$.

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 ?

For a first-order reaction at $300 \, ^\circ C$,the activation energy is $35 \, kcal \, mol^{-1}$ and the frequency factor is $1.45 \times 10^{11} \, s^{-1}$. Calculate the rate constant.

From the given data for the reaction $H_2 + I_2 \rightarrow 2HI$,calculate the activation energy $(E_a)$:
$T_1 = 769 \ K, \ 1/T_1 = 1.3 \times 10^{-3} \ K^{-1}, \ \log_{10} K_1 = 2.9$
$T_2 = 667 \ K, \ 1/T_2 = 1.5 \times 10^{-3} \ K^{-1}, \ \log_{10} K_2 = 1.1$

In a bimolecular reaction,the steric factor $P$ was experimentally determined to be $4.5$. The correct option$(s)$ among the following is(are)
$[A]$ The activation energy of the reaction is unaffected by the value of the steric factor
$[B]$ Experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation
$[C]$ Since $P=4.5$,the reaction will not proceed unless an effective catalyst is used
$[D]$ The value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally

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