For the reaction between A and B,the initial | vedclass

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(C) The rate law is given by $r = k[A]^x[B]^y$.From the data:$1) \ 5.0 	imes 10^{-5} = k(0.2)^x(0.3)^y$$2) \ 5.0 	imes 10^{-5} = k(0.2)^x(0.1)^y$$3)

Vedclass · Accepted Answer

$\frac{3}{2}, 0$

Vedclass · Answer

(C) The rate law is given by $r = k[A]^x[B]^y$.From the data:$1) \ 5.0 	imes 10^{-5} = k(0.2)^x(0.3)^y$$2) \ 5.0 	imes 10^{-5} = k(0.2)^x(0.1)^y$$3) \ 1.4 	imes 10^{-4} = k(0.4)^x(0.05)^y$Dividing $(1)$ by $(2)$:$1 = (0.3/0.1)^y = 3^y \implies y = 0$.Dividing $(3)$ by $(1)$:$\frac{1.4 	imes 10^{-4}}{5.0 	imes 10^{-5}} = \frac{k(0.4)^x(0.05)^0}{k(0.2)^x(0.3)^0}$$2.8 = (0.4/0.2)^x = 2^x$Taking $\log$ on both sides: $x \log 2 = \log 2.8 \implies x = \frac{\log 2.8}{\log 2} \approx 1.48 \approx \frac{3}{2}$.Thus,the order with respect to $A$ is $\frac{3}{2}$ and with respect to $B$ is $0$.

$[A] / mol \ L^{-1}$	$0.2, 0.2, 0.4$
$[B] / mol \ L^{-1}$	$0.3, 0.1, 0.05$
$r_0 / mol \ L^{-1} s^{-1}$	$5.0 \times 10^{-5}, 5.0 \times 10^{-5}, 1.4 \times 10^{-4}$

$A / mol \ L^{-1}$	$0.20$	$0.20$	$0.40$
$B / mol \ L^{-1}$	$0.30$	$0.10$	$0.05$
$r_0 / mol \ L^{-1} \ s^{-1}$	$5.07 \times 10^{-5}$	$5.07 \times 10^{-5}$	$1.43 \times 10^{-4}$

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