$A$ અને $B$ વચ્ચેની પ્રક્રિયામાં પ્રારંભિક પ્રક્રિયા વેગ $\left(r_{0}\right)$ $A$ અને $B$ ની જુદી જુદી સાંદ્રતાએ માપવામાં આવ્યા હતા. જે નીચે આપેલા છે :

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

$A$ અને $B$ ના સંદર્ભમાં પ્રક્રિયા ક્રમ શું હશે ?

Let the order of the reaction with respect to $A$ be $x$ and with respect to $B$ be $y$.

Therefore,

$r _{0}=k[ A ]^{x}[ B ]^{y}$

$5.07 \times {10^{ - 5}} = k{[0.20]^x}{[0.30]^y}$      .......$(i)$

$5.07 \times 10^{-5}=k[0.20]^{x}[0.10]^{y}$        .........$(ii)$

$1.43 \times 10^{-4}=k[0.40]^{x}[0.05]^{y}$        ..........$(iii)$

Dividing equation $(i)$ by $(ii),$ we obtain

$\frac{5.07 \times 10^{-5}}{5.07 \times 10^{-5}}=\frac{k[0.20]^{x}[0.30]^{y}}{k[0.20]^{x}[0.10]^{y}}$

$\Rightarrow 1=\frac{[0.30]^{y}}{[0.10]^{y}}$

$ \Rightarrow {\left( {\frac{{0.30}}{{0.10}}} \right)^0} = {\left( {\frac{{0.30}}{{0.10}}} \right)^y}$

$\Rightarrow y=0$

Dividing equation $(iii)$ by $(ii),$ we obtain

$\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}=\frac{k[0.40]^{x}[0.05]^{y}}{k[0.20]^{x}[0.30]^{y}}$

$\Rightarrow \frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}=\frac{[0.40]^{x}}{[0.20]^{x}}$      $\left[\begin{array}{l}\text { since } y=0 \\ {[0.05]^{y}=[0.30]^{y}=1}\end{array}\right]$

$\Rightarrow 2.821=2^{x}$

$\Rightarrow \log 2.821=x \log 2$       (Taking log on both sides)

$\Rightarrow x=\frac{\log 2.821}{\log 2}$

$=1.5$ (approximately)

Hence, the order of the reaction with respect to $A$ is $1.5$ and with respect to $B$ is zero