If the half-life of a reaction is halved when the initial concentration of the reactant is doubled,what is the order of the reaction?

For the reaction $2NO_{(g)} + O_{2_{(g)}} \rightarrow 2NO_{2_{(g)}}$,the volume of the reaction vessel is suddenly reduced to half of its original volume. If the reaction is first order with respect to $O_2$ and second order with respect to $NO$,the rate of the reaction will:

In a reaction,the concentration of reactant is increased two times and three times,then the increases in the rate of reaction were four times and nine times respectively. The order of reaction is:

For a reaction $A + B \rightarrow \text{Product}$,the order with respect to $A$ is $2$ and with respect to $B$ is $3$. If the concentration of both is doubled,by how much will the rate increase?

For the reaction $A_2 + B_2 \to 2AB$,the experimental data is given below. Determine the order of the reaction.

Experiment No.	$[A_2] \text{ (M)}$	$[B_2] \text{ (M)}$	Rate $(M \cdot s^{-1})$
$1$	$0.1$	$0.1$	$1.6 \times 10^{-4}$
$2$	$0.1$	$0.2$	$3.2 \times 10^{-4}$
$3$	$0.2$	$0.1$	$3.2 \times 10^{-4}$

Calculate the order of the reaction with respect to $A$ and $B$ based on the following data:

$[A] \ (mol/L)$	$[B] \ (mol/L)$	Rate $(mol/L \cdot s)$
$0.05$	$0.05$	$1.2 \times 10^{-3}$
$0.10$	$0.05$	$2.4 \times 10^{-3}$
$0.05$	$0.10$	$1.2 \times 10^{-3}$