(B) The rate law is given by $r = K[A]^{x}[B]^{y}$.Using experiment $(i)$ and $(iv)$ where $[B]$ is constant:$6.0 \times 10^{-3} = K(0.1)^{x}(0.1)^{y}

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(B) The rate law is given by $r = K[A]^{x}[B]^{y}$.Using experiment $(i)$ and $(iv)$ where $[B]$ is constant:$6.0 \times 10^{-3} = K(0.1)^{x}(0.1)^{y}$$2.40 \times 10^{-2} = K(0.4)^{x}(0.1)^{y}$Dividing $(iv)$ by $(i)$ gives: $4 = (4)^{x}$,so $x = 1$.Using experiment $(ii)$ and $(iii)$ where $[A]$ is constant:$7.2 \times 10^{-2} = K(0.3)^{x}(0.2)^{y}$$2.88 \times 10^{-1} = K(0.3)^{x}(0.4)^{y}$Dividing $(iii)$ by $(ii)$ gives: $4 = (2)^{y}$,so $y = 2$.Overall order $= x + y = 1 + 2 = 3$.

Class 12 Chemistry Chemical Kinetics Rate law , Rate constant , Order of Reaction and Molecularity

Medium

During the kinetic study of the reaction $2A + B \rightarrow C + D$,the following results were obtained:

Experiment $[A] \ (M), [B] \ (M)$ and Initial rate of formation of $D$

$i. \ [A]=0.1, [B]=0.1$ $6.0 \times 10^{-3} \ M \ s^{-1}$

$ii. \ [A]=0.3, [B]=0.2$ $7.2 \times 10^{-2} \ M \ s^{-1}$

$iii. \ [A]=0.3, [B]=0.4$ $2.88 \times 10^{-1} \ M \ s^{-1}$

$iv. \ [A]=0.4, [B]=0.1$ $2.40 \times 10^{-2} \ M \ s^{-1}$

Based on the above data,the overall order of the reaction is:

Experiment	$[A] \ (M), [B] \ (M)$ and Initial rate of formation of $D$
$i. \ [A]=0.1, [B]=0.1$	$6.0 \times 10^{-3} \ M \ s^{-1}$
$ii. \ [A]=0.3, [B]=0.2$	$7.2 \times 10^{-2} \ M \ s^{-1}$
$iii. \ [A]=0.3, [B]=0.4$	$2.88 \times 10^{-1} \ M \ s^{-1}$
$iv. \ [A]=0.4, [B]=0.1$	$2.40 \times 10^{-2} \ M \ s^{-1}$