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Question

$R\,\, = \,k{[A]^x}\,{[B]^y}$

$6.93\, 	imes \,{10^{ - 3}}\, = \,k\,{(0.1)^x}\,{(0.2)^y}$        ...... $(i)$

$6.93\, 	imes

Vedclass · Accepted Answer

$10$

Vedclass · Answer

$R\,\, = \,k{[A]^x}\,{[B]^y}$

$6.93\, 	imes \,{10^{ - 3}}\, = \,k\,{(0.1)^x}\,{(0.2)^y}$        ...... $(i)$

$6.93\, 	imes \,{10^{ - 3}}\, = \,k\,{(0.1)^x}\,{(0.25)^y}$        ...... $(ii)$

$1.386 	imes \,{10^{ - 2}}\, = \,k\,{(0.2)^x}\,{(0.3)^y}$         . .... $(iii)$

$ \Rightarrow \,y = 0$  (from $(i)$ and $(ii)$), zero order w.r.t. $B$

$x=1$  (from $(i)$ and $(ii)$)

$ \Rightarrow \,$ First order wrt $A$

$ \Rightarrow 6.93	imes 10^{-3}\,=\,k(0.1)$

$ \Rightarrow k=6.93	imes 10^{-3}\,min^{-1}$

Experiment	$[A]$ (in $mol\, L^{-1})$	$[B]$ (in $mol\, L^{-1})$	Initial rate of reaction (in $mol\, L^{-1}\,min^{-1})$
$I$	$0.10$	$0.20$	$6.93 \times {10^{ - 3}}$
$II$	$0.10$	$0.25$	$6.93 \times {10^{ - 3}}$
$III$	$0.20$	$0.30$	$1.386 \times {10^{ - 2}}$

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