For the reaction $A \to \text{Products}$,the rate of reaction becomes one-fourth when the concentration of $A$ is halved. What is the order of the reaction?

$A_2 + B_2 \to 2AB$; $R.O.R = k[A_2]^a[B_2]^b$

Initial $[A_2]$	Initial $[B_2]$	$R.O.R. (r) \ M s^{-1}$
$0.2$	$0.2$	$0.04$
$0.1$	$0.4$	$0.04$
$0.2$	$0.4$	$0.08$

Order of reaction with respect to $A_2$ and $B_2$ are respectively:

The following mechanism has been proposed for the reaction of $NO$ with $Br_2$ to form $NOBr$:
$NO_{(g)} + Br_{2(g)} \rightleftharpoons NOBr_{2(g)}$
$NOBr_{2(g)} + NO_{(g)} \longrightarrow 2NOBr_{(g)}$
If the second step is the rate-determining step,the order of the reaction with respect to $NO_{(g)}$ is:

For the reaction $A + B \xrightarrow{K} C$,identify the incorrect order of reaction indicated against the rate expression.

For the reaction,$2A + B \to \text{products}$,when the concentrations of $A$ and $B$ both were doubled,the rate of the reaction increased from $0.3 \ mol \ L^{-1} \ s^{-1}$ to $2.4 \ mol \ L^{-1} \ s^{-1}$. When the concentration of $A$ alone is doubled,the rate increased from $0.3 \ mol \ L^{-1} \ s^{-1}$ to $0.6 \ mol \ L^{-1} \ s^{-1}$. Which one of the following statements is correct?

For the reaction $A + B \to \text{products}$,it is found that the rate of the reaction is proportional to the concentration of $A$,but it is independent of the concentration of $B$. Then: