$2 A+B \rightarrow C+D$ પ્રક્રિયાના ગતિકી અભ્યાસ દરમિયાન નીચેના પરિણામો મળ્યાં છે :

પ્રયોગ	$[ A ] / mol L ^{-1}$	$[ B ] / mol L ^{-1}$	$D$ ની બનાવટનો પ્રારંભિક વેગ $/ mol \,L ^{-1} \,min ^{-1}$
$I$	$0.1$	$0.1$	$6.0 \times 10^{-3}$
$II$	$0.3$	$0.2$	$7.2 \times 10^{-2}$
$III$	$0.3$	$0.4$	$2.88 \times 10^{-1}$
$IV$	$0.4$	$0.1$	$2.40 \times 10^{-2}$

પ્રક્રિયાનો વેગ નિયમ અને વેગ અચળાંક નક્કી કરો.

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

Therefore, rate of the reaction is qiven by

Rate $=k[ A ]^{x}[ B ]^{y}$

According to the question,

$6.0 \times 10^{-3}=k[0.1]^{x}[0.1]^{y}$          ........$(i)$

$7.2 \times 10^{-2}=k[0.3]^{x}[0.2]^{y}$           ..........$(ii)$

$2.88 \times 10^{-1}=k[0.3]^{x}[0.4]^{y}$          .........$(iii)$

$2.40 \times 10^{-2}=k[0.4]^{x}[0.1]^{y}$          ...........$(iv)$

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

$\frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}}=\frac{k[0.4]^{x}[0.1]^{y}}{k[0.1]^{x}[0.1]^{y}}$

$\Rightarrow 4=\frac{[0.4]^{x}}{[0.1]^{x}}$

$ \Rightarrow 4 = {\left( {\frac{{0.4}}{{0.1}}} \right)^x}$

$\Rightarrow(4)^{1}=4^{x}$

$\Rightarrow x=1$

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

$\frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}}=\frac{k[0.3]^{x}[0.4]^{y}}{k[0.3]^{x}[0.2]^{y}}$

$ \Rightarrow 4 = {\left( {\frac{{0.4}}{{0.2}}} \right)^y}$

$\Rightarrow 4=2^{y}$

$\Rightarrow 2^{2}=2^{y}$

$\Rightarrow y=2$

Therefore, the rate law is

Rate $=k[ A ][ B ]^{2}$

$\Rightarrow \quad k=\frac{\text { Rate }}{[ A ][ B ]^{2}}$

From experiment $I$, we obtain

$k=\frac{6.0 \times 10^{-3} \,mol\, L ^{-1} \,min ^{-1}}{\left(0.1\, mol\, L ^{-1}\right)\left(0.1 \,mol\, L ^{-1}\right)^{2}}$

$=6.0 L ^{2} \,mol ^{-2}\, min ^{-1}$

From experiment $II$, we obtain

$k=\frac{7.2 \times 10^{-2} \,mol \,L ^{-1} \,min ^{-1}}{\left(0.3\, mol\, L ^{-1}\right)\left(0.2 \,mol\, L ^{-1}\right)^{2}}$

$=6.0\, L ^{2}\, mol ^{-2}\, min ^{-1}$

From experiment $III$, we obtain

$k=\frac{2.88 \times 10^{-1}\, mol\, L ^{-1}\, min ^{-1}}{\left(0.3 \,mol\, L ^{-1}\right)\left(0.4\, mol\, L ^{-1}\right)^{2}}$

$=6.0\, L ^{2}\, mol ^{-2}\, min ^{-1}$

From experiment $IV ,$ we obtain

$k=\frac{2.40 \times 10^{-2}\, mol\, L ^{-1} \,min ^{-1}}{\left(0.4\, mol\, L ^{-1}\right)\left(0.1 \,mol\, L ^{-1}\right)^{2}}$

$=6.0\, L ^{2}\, mol ^{-2}\, min ^{-1}$

Therefore, rate constant, $k=6.0 \,L ^{2}\, mol ^{-2}\, min ^{-1}$