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The following data were obtained for the reaction :

$A+2B\rightarrow C$

Experiment [A]/M [B]/M Initial rate of formation of C/M min^-1

$1$ $0.2$ $0.3$ $4.2\times 10^{-2}$

$2$ $0.1$ $0.1$ $6.0\times 10^{-3}$

$3$ $0.4$ $0.3$ $1.68\times 10^{-1}$

$4$ $0.1$ $0.4$ $2.40\times 10^{-2}$

(a) Find the order of reaction with respect to A and B.

(b) Write the rate law and overall order of reaction.

(c) Calculate the rate constant (k).

Answers (1)

$A+2B\rightarrow C$

Supose order w.r.t A is $\alpha$ and w.r.t B is $\beta$ .

$Rate\: law[r=k[A]^\alpha [B]^\beta ]$

$\left ( Rate \right )exp_1,\: r_1=4.2\times 10^{-2}=k[0.2]^\alpha [0.3]^\beta\: \: \: \: \: \: \: \: \: \: ---(1)$

$r_2=6.0\times 10^{-3}=k[0.1]^\alpha [0.1]^\beta\: \: \: \: \: \: \: \: \: \: ---(2)$

$r_3=1.68\times 10^{-1}=k[0.4]^\alpha [0.3]^\beta\: \: \: \: \: \: \: \: \: \: ---(3)$

$r_4=2.40\times 10^{-2}=k[0.1]^\alpha [0.4]^\beta\: \: \: \: \: \: \: \: \: \: ---(4)$

$\frac{r_1}{r_3}, we\: \: \: get.$

$\frac{r_1}{r_3}=\frac{4.2\times10^{-2}}{1.68\times10^{-1}} =\frac{k[0.2]^\alpha[0.3]^\beta}{k[0.4]^\alpha[0.3]^\beta}$

$\Rightarrow \frac{25}{100}=\left [ \frac{1}{2} \right ]^\alpha$

$\Rightarrow \left ( \frac{1}{2} \right )^2=\left [ \frac{1}{2} \right ]^\alpha$

$\Rightarrow \alpha =2$

similarly for $\beta$ , $\frac{r_2}{r_4}=\frac{6.0\times10^{-3}}{2.40\times10^{-2}}=\frac{k[0.1]^\alpha[0.1]^\beta}{k[0.1]^\alpha[0.4]^\beta}$

$\Rightarrow\frac{[0.1]^\beta}{[0.4]^\beta}=\frac{6.0\times10^{-3}}{2.40\times10^{-2}}=\frac{1}{4}\Rightarrow \beta =1$

(a) Hence, order w.r.t A is 2, w.r.t B is 1

(b) Rate Law:

$r=k[A]^2[B]$

overall order $=2+1=3$

(c) $k=\frac{r_1}{[0.2]^\alpha[0.3]^\beta}$ using eqⁿ (1)

$=\frac{4.2\times10^{-2}}{[0.2]^2[0.3]}=3.5mol^{-2}L^{-2}min^{-1}$

Posted by

Sumit Saini

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