The data for the reaction A + B to C is

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Question

Let $r=(1)^{x}(2)^{y}$

$x = \frac{{\log \left( {\frac{{{r_1}}}{{{r_2}}}} \right)}}{{\log \left( {\frac{{{a_1}}}{{{a_2}}}} \right)}}$ $ = \frac

Vedclass · Accepted Answer

$r = k\, [A]^3$

Vedclass · Answer

Let $r=(1)^{x}(2)^{y}$

$x = \frac{{\log \left( {\frac{{{r_1}}}{{{r_2}}}} \right)}}{{\log \left( {\frac{{{a_1}}}{{{a_2}}}} \right)}}$ $ = \frac{{\log \frac{{0.1}}{{0.1}}}}{{\log \left( {\frac{{0.012}}{{0.024}}} \right)}}$ $ = \frac{{\log \left( {\frac{1}{8}} \right)}}{{\log \left( {\frac{1}{2}} \right)}}$

$x=3$

$y = \frac{{\log \frac{{{r_1}}}{{{r_3}}}}}{{\log \left( {\frac{{{b_1}}}{{{b_2}}}} \right)}}$ $ = \frac{{\log \left( {\frac{{0.1}}{{0.1}}} \right)}}{{\log \left( {\frac{{0.035}}{{0.070}}} \right)}}$ $ = \frac{{\log (1)}}{{\log \left( {\frac{1}{2}} \right)}}$

$y=0$

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

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

$A$ $(mol/l)$	$B$ $(mol/l)$	Rate
$0.05$	$0.05$	$1.2\times 10^{-3}$
$0.10$	$0.05$	$2.4\times 10^{-3}$
$0.05$	$0.10$	$1.2\times 10^{-3}$

The data for the reaction $A + B \to C$ is

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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Exp $[A]_0$ $[B]_0$ initial rate

$1$ $0.012$ $0.035$ $0.10$

$2$ $0.024$ $0.035$ $0.80$

$3$ $0.012$ $0.070$ $0.10$

$4$ $0.024$ $0.070$ $0.80$

Calculate the order of the reaction in $A$ and $B$

$A$

$(mol/l)$

$B$

$(mol/l)$
Rate

$0.05$ $0.05$ $1.2\times 10^{-3}$

$0.10$ $0.05$ $2.4\times 10^{-3}$

$0.05$ $0.10$ $1.2\times 10^{-3}$