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(B) For a reaction,the rate law can be expressed as $Rate = k[A]^x[B]^y$.$1$. When the concentration of $B$ is doubled,the half-life does not change.

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$L \ mol^{-1} \ s^{-1}$

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(B) For a reaction,the rate law can be expressed as $Rate = k[A]^x[B]^y$.$1$. When the concentration of $B$ is doubled,the half-life does not change. This indicates that the reaction is of first order with respect to $B$ $(y=1)$,as the half-life of a first-order reaction is independent of the initial concentration of the reactant.$2$. When the concentration of $A$ is doubled,the rate increases by two times. This indicates that the reaction is of first order with respect to $A$ $(x=1)$.$3$. The overall order of the reaction is $n = x + y = 1 + 1 = 2$.$4$. The unit of the rate constant for a reaction of order $n$ is $(mol \ L^{-1})^{1-n} \ s^{-1}$. For $n=2$,the unit is $(mol \ L^{-1})^{1-2} \ s^{-1} = (mol \ L^{-1})^{-1} \ s^{-1} = L \ mol^{-1} \ s^{-1}$.

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

Consider the reaction,$2A + B \rightarrow$ products. When concentration of $B$ alone was doubled,the half-life did not change. When the concentration of $A$ alone was doubled,the rate increased by two times. The unit of rate constant for this reaction is

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

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

$0.2$ $0.2$ $0.04$

$0.1$ $0.4$ $0.04$

$0.2$ $0.4$ $0.08$

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For the gaseous reaction,$N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2$,the rate can be expressed as:
$-\frac{d[N_2O_5]}{dt} = K_1[N_2O_5]$
$+\frac{d[NO_2]}{dt} = K_2[N_2O_5]$
$+\frac{d[O_2]}{dt} = K_3[N_2O_5]$
The correct relation between $K_1, K_2$ and $K_3$ is:

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