For a certain reaction,the rate constant $K = 2.37 \times 10^2 \ L^2 \ mol^{-2} \ s^{-1}$. What is the order of the reaction?

The rate law for the reaction below is given by the expression $Rate = k[A][B]$.
$A + B \to \text{Product}$
If the concentration of $B$ is increased from $0.1 \ M$ to $0.3 \ M$,keeping the concentration of $A$ constant at $0.1 \ M$,the rate constant $(k)$ will be:

The mechanism of the reaction $A + 2B \to D$ is
$2B \xrightarrow{k} B_2$ $[Slow]$
$B_2 + A \to D$ $[Fast]$
The rate law expression,order with respect to $A$,order with respect to $B$ and overall order of reaction are respectively

For the reaction $2A + B \to C$,the values of initial rate at different reactant concentrations are given in the table below: The rate law for the reaction is

$[A] \ (mol \ L^{-1})$	$[B] \ (mol \ L^{-1})$	Initial Rate $(mol \ L^{-1} \ s^{-1})$
$0.05$	$0.05$	$0.045$
$0.10$	$0.05$	$0.090$
$0.20$	$0.10$	$0.72$

The rate constant is the same for $3$ reactions of order $I, II$,and $III$,respectively,with the unit of concentration being in moles per litre. If the concentration of the reactant is unity,the rates of reaction $R_1, R_2, R_3$ will be:

When the initial concentration of a reactant is doubled in a reaction,its half-life period is not affected. The order of the reaction is: