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The results given in the below were obtained during kinetic studies of the following reaction : $2A + B\rightarrow C +D$

Experiment $\frac{(A)}{mol L^{-1}}$ $\frac{(B)}{mol L^{-1}}$ $\frac{(initial rate)}{mol L^{-1} min^{-1}}$

$I$ $0.1$ $0.1$ $6.00\times 10^{-3}$

$II$ $0.1$ $0.2$ $2.40\times 10^{-2}$

$III$ $0.2$ $0.1$ $1.20\times 10^{-2}$

$IV$ X $0.2$ $7.20\times 10^{-2}$

$V$ $0.3$ Y $2.880\times 10^{-1}$

X and Y in the given table are respectively :
Option: 1 $0.4 , 0.4$

Option: 2 $0.4 , 0.3$

Option: 3 $0.3 , 0.3$

Option: 4 $0.3 , 0.4$

Answers (1)

best_answer

The rate law is given as:

Rate law = k[A][B]²

Case I:

$\mathrm{Thus,\: \frac{7.2\: x\: 10^{-2}}{6\: x\: 10^{-3}}\: =\: \frac{k[X][2B]^{2}}{k[A][B]^{2}}}$

$\mathrm{Thus,\:12=\: \frac{[X]\: x\: [4B]}{[A]\: x\; [B]^{2}}}$

$\mathrm{\Rightarrow \: \frac{X}{A}\: =\: 3}$

$\mathrm{\Rightarrow \: \frac{X}{A}\: =\: 3}$

Case II:

$\mathrm{Thus,\: \frac{2.88\: x\: 10^{-1}}{6\: x\: 10^{-3}}\: =\: \frac{k[3A][Y]^{2}}{k[A][B]^{2}}}$

$\mathrm{\Rightarrow\:48\: =\: \frac{3Y^{2}}{B^{2}}}$

$\\\mathrm{Thus,\: Y\: =\: 4B}\\\mathrm{Y\: =\: 0.4}$

Thus, X and Y in the given table are: 0.3 and 0.4 respectively.

Therefore, Option(4) is correct.

Posted by

Kuldeep Maurya

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