For the reaction $A + 2B \rightleftharpoons 2C + D$,if the concentration of $A$ is increased four times and the concentration of $B$ is reduced to half,the rate of the reaction will be:

$A_2 + 2 \, B \to 2 \, AB$

$[A_2]$	$[B]$	$-d[A_2]/dt$
$0.1$	$0.2$	$1 \times 10^{-2} \, M \, s^{-1}$
$0.2$	$0.2$	$2 \times 10^{-2} \, M \, s^{-1}$
$0.2$	$0.4$	$8 \times 10^{-2} \, M \, s^{-1}$

The order of reaction with respect to $A_2$ and $B$ are respectively:

The rate constant for the reaction,$2N_2O_5 \to 4NO_2 + O_2$ is $3.0 \times 10^{-4} \ s^{-1}$. If the reaction starts with $1.0 \ mol \ L^{-1}$ of $N_2O_5$,calculate the rate of formation of $NO_2$ at the moment when the concentration of $O_2$ is $0.1 \ mol \ L^{-1}$.

For a chemical reaction $A \rightarrow B$,the rate of the reaction is $2 \times 10^{-3} \ mol \ dm^{-3} \ s^{-1}$,when the initial concentration is $0.05 \ mol \ dm^{-3}$. The rate of the same reaction is $1.6 \times 10^{-2} \ mol \ dm^{-3} \ s^{-1}$ when the initial concentration is $0.1 \ mol \ dm^{-3}$. The order of the reaction is

The following data was obtained for the chemical reaction given below at $975 \ K$: $2 NO_{(g)} + 2 H_{2(g)} \rightarrow N_{2(g)} + 2 H_{2}O_{(g)}$

Experiment	$[NO] \ (mol \ L^{-1})$	$[H_{2}] \ (mol \ L^{-1})$	Rate $(mol \ L^{-1} \ s^{-1})$
$1$	$8 \times 10^{-5}$	$8 \times 10^{-5}$	$7 \times 10^{-9}$
$2$	$24 \times 10^{-5}$	$8 \times 10^{-5}$	$2.1 \times 10^{-8}$
$3$	$24 \times 10^{-5}$	$32 \times 10^{-5}$	$8.4 \times 10^{-8}$

The order of the reaction with respect to $NO$ is ..... .

The three experimental data for determining the differential rate of reaction $Cl_{2(g)} + 2NO_{(g)} \rightarrow 2NOCl_{(g)}$ at $310 \ K$ temperature are provided. $(a)$ Derive the differential rate of reaction. $(b)$ Calculate the order of reaction. $(c)$ Calculate the value of the rate constant.