The mechanism for the reaction is given below:
$2P + Q \to S + T$
$P + Q \to R + S$ (slow)
$P + R \to T$ (fast)
The rate law expression for the reaction is:

The rate constant for the reaction $A \longrightarrow B$ is $2 \times 10^{-4} \ L \ mol^{-1} \ min^{-1}$. The concentration of $A$ at which the rate of the reaction is $(1 / 12) \times 10^{-5} \ M \ sec^{-1}$ is :-

Select the incorrect option :

The reaction between $X$ and $Y$ is first order with respect to $X$ and zero order with respect to $Y$.

$Experiment$	$[X] / (mol \ L^{-1})$	$[Y] / (mol \ L^{-1})$	$\text{Initial rate} / (mol \ L^{-1} \ min^{-1})$
$I$	$0.1$	$0.1$	$2 \times 10^{-3}$
$II$	$0.2$	$0.2$	$4 \times 10^{-3}$
$III$	$0.4$	$0.4$	$M \times 10^{-3}$
$IV$	$0.1$	$0.2$	$2 \times 10^{-3}$

Examine the data of the table and calculate the ratio of the numerical value of $M$ to $0.2$.

For a reaction between $A$ and $B$,the order with respect to $A$ is $2$ and the order with respect to $B$ is $3$. If the concentrations of both $A$ and $B$ are doubled,the rate will increase by a factor of:

For the reaction $A + 2B \to C$,the rate is given by $R = k[A][B]^2$. The order of the reaction is: