The experimental data for reaction
2A + B_2 | vedclass

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Question

$\mathrm{R}=[\mathrm{A}]^{\alpha}\left[\mathrm{B}_{2}ight]^{\beta}$

$1.6 	imes 10^{-4}=\mathrm{K}[0.50]^{\alpha}[0.50]^{\beta }$  ...... $(

Vedclass · Accepted Answer

$r = K[B_2]$

Vedclass · Answer

$\mathrm{R}=[\mathrm{A}]^{\alpha}\left[\mathrm{B}_{2}ight]^{\beta}$

$1.6 	imes 10^{-4}=\mathrm{K}[0.50]^{\alpha}[0.50]^{\beta }$  ...... $(i)$

$3.2 	imes 10^{-4}=\mathrm{K}[0.50]^{\alpha}[1.00]^{\beta}$ ...... $(ii)$

$3.2 	imes 10^{-4}=\mathrm{K}[1.00]^{\alpha}[1.00]^{\beta}$ ...... $(iii)$

divide $(iii)$ and $(ii)$

$\frac{{3.2 	imes {{10}^{ - 4}}}}{{3.2 	imes {{10}^{ - 4}}}} = \frac{{{	ext{K}}{{[1.00]}^\alpha }{{[1.00]}^\beta }}}{{{	ext{K}}{{[0.50]}^\alpha }{{[1.00]}^\beta }}}$

${2^{\alpha}=1}$ 

${\alpha=0}$

divide $(ii)$ and $(i)$

$\frac{{3.2 	imes {{10}^{ - 4}}}}{{1.6 	imes {{10}^{ - 4}}}} = $ $\frac{{{	ext{K}}{{[0.50]}^\alpha }{{[1.00]}^\beta }}}{{{	ext{K}}{{[0.50]}^\alpha }{{[0.50]}^\beta }}}$

${2=2^{\beta}} $

${\beta=1}$

${	ext{R}} = {	ext{K}}\,{[{	ext{A}}]^0}{[B]^1}$

Exp.	$[A]$	$[B_2]$	Rate $(mol\,L^{-1}\,S^{-1})$
$1$	$0.50$	$0.50$	$1.6 \times {10^{ - 4}}$
$2$	$0.50$	$1.00$	$3.2 \times {10^{ - 4}}$
$3$	$1.00$	$1.00$	$3.2 \times {10^{ - 4}}$

The experimental data for reaction
$2A + B_2 \longrightarrow 2AB$

Exp. $[A]$ $[B_2]$ Rate $(mol\,L^{-1}\,S^{-1})$

$1$ $0.50$ $0.50$ $1.6 \times {10^{ - 4}}$

$2$ $0.50$ $1.00$ $3.2 \times {10^{ - 4}}$

$3$ $1.00$ $1.00$ $3.2 \times {10^{ - 4}}$

The rate law

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