The conversion of molecules $X$ to $Y$ follows second order kinetics. If the concentration of $X$ is increased to three times,how will it affect the rate of formation of $Y$?

For the elementary reaction $M \rightarrow N$,the rate of disappearance of $M$ increases by a factor of $8$ upon doubling the concentration of $M$. The order of the reaction with respect to $M$ is :

Consider the kinetic data given in the following table for the reaction $A + B + C \rightarrow$ Product.

Experiment No.	$[A] \ (mol \ dm^{-3})$	$[B] \ (mol \ dm^{-3})$	$[C] \ (mol \ dm^{-3})$	Rate of reaction $(mol \ dm^{-3} \ s^{-1})$
$1$	$0.2$	$0.1$	$0.1$	$6.0 \times 10^{-5}$
$2$	$0.2$	$0.2$	$0.1$	$6.0 \times 10^{-5}$
$3$	$0.2$	$0.1$	$0.2$	$1.2 \times 10^{-4}$
$4$	$0.3$	$0.1$	$0.1$	$9.0 \times 10^{-5}$

The rate of the reaction for $[A]=0.15 \ mol \ dm^{-3}, [B]=0.25 \ mol \ dm^{-3}$ and $[C]=0.15 \ mol \ dm^{-3}$ is found to be $Y \times 10^{-5} \ mol \ dm^{-3} \ s^{-1}$. The value of $Y$ is . . . . . .

What is the order and molecularity of the following elementary reaction? $2NO_2(g) \rightarrow 2NO(g) + O_2(g)$, rate $= k[NO_2]^2$

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:

For the reaction ${H_{2(g)} + I_{2(g)} \to 2HI_{(g)}}$,the possible mechanism is given below:
$I_2 \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} 2 I$ (fast step)
${2I + H_2 \xrightarrow{K_2} 2HI}$ (slow step)
Determine the rate law for the reaction.