For the reaction A + B to products, what wil | vedclass

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Question

Rate $=\mathrm{K}[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$

From exp. $( 1)$

$5 	imes {10^{ - 4}} = {	ext{K}}{\left[ {2.5 	imes {{10

Vedclass · Accepted Answer

$2$ with respect to $A$ and $1$ with respect to $B$

Vedclass · Answer

Rate $=\mathrm{K}[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$

From exp. $( 1)$

$5 	imes {10^{ - 4}} = {	ext{K}}{\left[ {2.5 	imes {{10}^{ - 4}}} ight]^x}{\left[ {3 	imes {{10}^{ - 5}}} ight]^y}$  ...... $(i)$

From exp. $( 2)$

$4 	imes 10^{-3}=\mathrm{K}\left[5 	imes 10^{-4}ight]^{\mathrm{x}}\left[6 	imes 10^{-5}ight]^{\mathrm{y}}$    ...... $(ii)$

Dividing $(ii)$ by $(i)$, 

$\frac{4 	imes 10^{-3}}{5 	imes 10^{-4}}=2^{x} \cdot 2^{y}=8$

From exp. $( 3)$

$1.6 	imes {10^{ - 2}} = {	ext{K}}{\left[ {1 	imes {{10}^{ - 3}}} ight]^{	ext{x}}}{\left[ {6 	imes {{10}^{ - 5}}} ight]^y}$   ...... $(iii)$

Dividing $(iii)$ by $(ii)$.

${\frac{1.6 	imes 10^{-2}}{4 	imes 10^{-3}}=2^{x}=4}$

Or    $x=2$,   $y=1$

Hence, order with respect to $\mathrm{A}$ is $2$ and with respect to $\mathrm{B}$ is $1$

Exp.	$[A]\,(mol\,L^{-1})$	$[B]\,(mol\,L^{-1})$	Initial rate $(mol\,L^{-1}\,s^{-1})$
$1.$	$2.5\times 10^{-4}$	$3\times 10^{-5}$	$5\times 10^{-4}$
$2.$	$5\times 10^{-4}$	$6\times 10^{-5}$	$4\times 10^{-3}$
$3.$	$1\times 10^{-3}$	$6\times 10^{-5}$	$1.6\times 10^{-2}$

$p ( mm Hg )$	$50$	$100$	$200$	$400$
Relative $t _{1 / 2}( s )$	$4$	$2$	$1$	$0.5$

${N_2}$ in $cc$	$6.25$	$9.50$	$11.42$	$13.65$	$35.05$
Time (minutes)	$10$	$15$	$20$	$25$	Finally

Exp	$[A]_0$	$[B]_0$	initial rate
$1$	$0.012$	$0.035$	$0.10$
$2$	$0.024$	$0.035$	$0.80$
$3$	$0.012$	$0.070$	$0.10$
$4$	$0.024$	$0.070$	$0.80$

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