For the non-stoichiometric reaction 2A + B to | vedclass

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Question

Let order with respect to $A$ and $B$ are $\alpha$ and $\beta$ respectively.

$	herefore \frac{ dc }{ dt }= k [ A ]^\alpha[ B ]^\beta$

$1.2 	im

Vedclass · Accepted Answer

$\frac{dc}{dt} = k[A]$

Vedclass · Answer

Let order with respect to $A$ and $B$ are $\alpha$ and $\beta$ respectively.

$	herefore \frac{ dc }{ dt }= k [ A ]^\alpha[ B ]^\beta$

$1.2 	imes 10^{-3}= k [0.1]^\alpha[0.1]^\beta \quad...(i)$

$1.2 	imes 10^{-3}= k [0.1]^\alpha[0.2]^\beta \quad...(ii)$

$2.4 	imes 10^{-3}= k [0.2]^\alpha[0.1]^\beta \quad...(iii)$

Dividing $(ii)$ by $(i): 2^\beta=1 \quad 	herefore \beta=0$

Dividing $(iii)$ by $(i): 2^\alpha=2 \quad 	herefore \alpha=1$

Rate law will be: $\frac{ dc }{ dt }= k [ A ]$

Initial Conc. $(A)$	Initial Conc. $(B)$	Initial rate of formation of $C\,(mol\, L^-S^-)$
$0.1\,M$	$0.1\,M$	$1.2 \times 10^{-3}$
$0.1\,M$	$0.2\,M$	$1.2 \times 10^{-3}$
$0.2\,M$	$0.1\,M$	$2.4 \times 10^{-3}$

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

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