| During the kinetic study of the reaction, \[2A+B\to C+D,\] following results were obtained [AIPMT (S) 2010] |
| \[Ru\,[A]/mol\,{{L}^{-1}}[B]/mol\,{{L}^{-1}}\]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 \[2.40\times {{10}^{-2}}\] |
| Based on the above data which one of the following is correct ? |
A) \[rate=k{{[A]}^{2}}[B]\]
B) \[rate=k[A][B]\]
C) \[rate=k{{[A]}^{2}}{{[B]}^{2}}\]
D) \[rate=k[A]{{[B]}^{2}}\]
Correct Answer: D
Solution :
| [d] Let the order of reaction with respect to A is x and with respect to B is y. Thus, |
| (I) rate\[=k{{(.1)}^{x}}{{(0.1)}^{y}}=6.0\times {{10}^{-3}}\] |
| (II) rate \[=k{{(0.3)}^{x}}{{(0.2)}^{y}}=7.2\times {{10}^{-2}}\] |
| (III) rate \[=k{{(0.3)}^{x}}{{(0.30)}^{y}}=2.88\times {{10}^{-1}}\] |
| (IV) rate\[=k{{(0.4)}^{x}}{{(0.1)}^{y}}=2.40\times {{10}^{-2}}\] |
| On dividing Eq. (I) by (IV), we get |
| \[{{\left( \frac{0.1}{0.4} \right)}^{x}}{{\left( \frac{0.1}{0.1} \right)}^{y}}=\frac{6.0\times {{10}^{-3}}}{2.4\times {{10}^{-2}}}\] |
| or \[{{\left( \frac{1}{4} \right)}^{x}}={{\left( \frac{1}{4} \right)}^{1}}\] |
| \[\therefore \] \[x=1\] |
| On dividing Eq. (II) by (III), we get |
| \[{{\left( \frac{0.3}{0.3} \right)}^{x}}{{\left( \frac{0.2}{0.4} \right)}^{y}}=\frac{7.2\times {{10}^{-2}}}{2.88\times {{10}^{-1}}}\] |
| or \[{{\left( \frac{1}{2} \right)}^{x}}=\frac{1}{4}\] |
| or \[{{\left( \frac{1}{2} \right)}^{y}}={{\left( \frac{1}{2} \right)}^{2}}\] |
| \[\therefore \] \[y=2\] |
| Thus, rate law is, |
| rate \[=k{{[A]}^{1}}{{[B]}^{2}}\]or \[=k[A]{{[B]}^{2}}\] |
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