For a reaction $A + B \rightarrow \text{Product}$,the rate of reaction doubles when the concentration of $A$ is doubled. The rate becomes four times when the concentrations of both $A$ and $B$ are doubled. What is the order of the reaction with respect to $A$ and $B$ respectively?

The hypothetical reaction : $2A + B \to C + D$ is catalyzed by $E$ as indicated in the possible mechanism below -
Step-$1$ : $A + E \rightleftharpoons AE$ (fast)
Step-$2$ : $AE + A \to A_2 + E$ (slow)
Step-$3$ : $A_2 + B \to C + D$ (fast)
What rate law best agrees with this mechanism?

If the decomposition of hydrogen peroxide is a first-order reaction,its rate law equation can be represented as:

The unit of the rate constant for a second-order reaction is ....

For the reaction $A + B \rightarrow$ products,it is observed that
$(i)$ on doubling the initial concentration of $A$ only,the rate of reaction is also doubled and
$(ii)$ on doubling the initial concentration of both $A$ and $B$,there is a change by a factor of $8$ in the rate of the reaction.
The rate of this reaction is given by

For a particular reaction,the rate expression is given as $r = k[A][B]^{0.5}$. If the volume of the vessel is reduced to one-fourth of the initial volume,the rate of reaction would