For the reaction $2 A + B \longrightarrow D + E$,the following mechanism has been proposed: $A + B \longrightarrow C + D$ (slow) and $A + C \longrightarrow E$ (fast). Determine the rate law.

Why is the probability of a reaction with molecularity higher than $3$ very rare?

The rate of reaction,$A + B \rightarrow \text{product}$,is $7.2 \times 10^{-2} \ mol \ dm^{-3} \ s^{-1}$ at $[A] = 0.4 \ mol \ dm^{-3}$ and $[B] = 0.1 \ mol \ dm^{-3}$. The reaction is first order in $A$ and second order in $B$. Calculate the rate constant.

For the reaction $2A + B \rightarrow \text{product}$,the rate of the reaction is $15 \times 10^{-2} \ mol \ dm^{-3} \ sec^{-1}$ when $[A] = 0.3 \ mol \ dm^{-3}$ and $[B] = 0.05 \ mol \ dm^{-3}$. What is the value of the rate constant if the reaction is first order in both the reactants?

$A$ chemical reaction takes place in two steps as follows:
$(i) \ NO_2Cl_{(g)} \xrightarrow{K_1} NO_{2(g)} + Cl_{(g)}$
$(ii) \ NO_2Cl_{(g)} + Cl_{(g)} \xrightarrow{K_2} NO_{2(g)} + Cl_{2(g)}$
Identify the reaction intermediate.

The following results have been obtained during the kinetic studies of the reaction:
$2 A + B \rightarrow C + D$

Experiment	$[A] / mol \, L^{-1}$	$[B] / mol \, L^{-1}$	Initial rate of formation of $D / mol \, L^{-1} \, min^{-1}$
$I$	$0.1$	$0.1$	$6.0 \times 10^{-3}$
$II$	$0.3$	$0.2$	$7.2 \times 10^{-2}$
$III$	$0.3$	$0.4$	$2.88 \times 10^{-1}$
$IV$	$0.4$	$0.1$	$2.40 \times 10^{-2}$

Determine the rate law and the rate constant for the reaction.