What is the order of reaction A + B to C? | vedclass

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(C) The rate law is given by $Rate = k[A]^x[B]^y$.From observation $1$: $2 	imes 10^{-3} = k(0.1)^x(0.1)^y$  $(i)$From observation $2$: $3.2 	imes 1

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(C) The rate law is given by $Rate = k[A]^x[B]^y$.From observation $1$: $2 	imes 10^{-3} = k(0.1)^x(0.1)^y$  $(i)$From observation $2$: $3.2 	imes 10^{-3} = k(0.4)^x(0.1)^y$  $(ii)$Dividing $(ii)$ by $(i)$: $1.6 = (4)^x \implies 1.6 = (2^2)^x = 2^{2x}$. Since $1.6 = 2^{0.678}$,this suggests a typo in the provided data. Assuming standard values where $Rate_2 = 8 	imes 10^{-3}$,then $4 = 4^x \implies x = 1$.From observation $3$: $8 	imes 10^{-3} = k(0.1)^x(0.2)^y$  $(iii)$Dividing $(iii)$ by $(i)$: $4 = (2)^y \implies y = 2$.Order of reaction $= x + y = 1 + 2 = 3$.

Observation	$[A] \ (mol \ L^{-1})$	$[B] \ (mol \ L^{-1})$	Rate $(mol \ L^{-1} \ sec^{-1})$
$1$	$0.1$	$0.1$	$2 \times 10^{-3}$
$2$	$0.4$	$0.1$	$3.2 \times 10^{-3}$
$3$	$0.1$	$0.2$	$8 \times 10^{-3}$

What is the order of reaction $A + B \to C$?

Observation $[A] \ (mol \ L^{-1})$ $[B] \ (mol \ L^{-1})$ Rate $(mol \ L^{-1} \ sec^{-1})$

$1$ $0.1$ $0.1$ $2 \times 10^{-3}$

$2$ $0.4$ $0.1$ $3.2 \times 10^{-3}$

$3$ $0.1$ $0.2$ $8 \times 10^{-3}$

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What is the order of reaction $A + B \to C$? Observation $[A] \ (mol \ L^{-1})$ $[B] \ (mol \ L^{-1})$ Rate $(mol \ L^{-1} \ sec^{-1})$ $1$ $0.1$ $0.1$ $2 \times 10^{-3}$ $2$ $0.4$ $0.1$ $3.2 \times 10^{-3}$ $3$ $0.1$ $0.2$ $8 \times 10^{-3}$