The following results have been obtained during the kinetic studies of the reaction:
$2 A+B \rightarrow C+D$
Experiment
$[ A ] / mol L ^{-1}$
$[ B ] / mol L ^{-1}$
Initial rate of formation of $D / mol \,L ^{-1} \,min ^{-1}$
$I$
$0.1$
$0.1$
$6.0 \times 10^{-3}$
$II$
$0.3$
$0.2$
$7.2 \times 10^{-2}$
$III$
$0.3$
$0.4$
$2.88 \times 10^{-1}$
$IV$
$0.4$
$0.1$
$2.40 \times 10^{-2}$
Determine the rate law and the rate constant for the reaction.
The following results have been obtained during the kinetic studies of the reaction:
$2 A+B \rightarrow C+D$
| Experiment | $[ A ] / mol L ^{-1}$ | $[ B ] / mol L ^{-1}$ | Initial rate of formation of $D / mol \,L ^{-1} \,min ^{-1}$ |
| $I$ | $0.1$ | $0.1$ | $6.0 \times 10^{-3}$ |
| $II$ | $0.3$ | $0.2$ | $7.2 \times 10^{-2}$ |
| $III$ | $0.3$ | $0.4$ | $2.88 \times 10^{-1}$ |
| $IV$ | $0.4$ | $0.1$ | $2.40 \times 10^{-2}$ |
Determine the rate law and the rate constant for the reaction.
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, $k=6.0 \,L ^{2}\, mol ^{-2}\, min ^{-1}$
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The initial rate of the reaction is recorded as $r_1$ when the reaction starts with $1.5 \mathrm{~atm}$ pressure of $\mathrm{A}$ and $0.7 \mathrm{~atm}$ pressure of B. After some time, the rate $r_2$ is recorded when the pressure of $C$ becomes $0.5 \mathrm{~atm}$. The ratio $r_1: r_2$ is $\qquad$ $\times 10^{-1}$.
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