| Expt. Initial concentration \[(mol\,{{L}^{-1}})\] | Initial rate of formation of D \[(mol\,{{L}^{-1}}{{\min }^{-1}})\] | ||
| [A] | [B] | ||
| 1. | 0.1 | 0.1 | \[6.0\times {{10}^{-3}}\] |
| 2. | 0.3 | 0.2 | \[7.2\times {{10}^{-2}}\] |
| 3. | 0.3 | 0.1 | \[2.88\times {{10}^{-1}}\] |
| 4. | 0.4 | 0.1 | \[2.4\times {{10}^{-2}}\] |
A) \[Rate=k[A][B]\]
B) \[Rate=k[A]{{[B]}^{2}}\]
C) \[Rate=k{{[A]}^{2}}{{[B]}^{2}}\]
D) \[Rate=k{{[A]}^{2}}[B]\]
Correct Answer: B
Solution :
Let the order of reaction with respect to A is\[x\] and B is y. \[Rate=k{{[A]}^{x}}{{[B]}^{y}}\] 1.\[Rate=6.0\times {{10}^{-3}}={{(0.1)}^{x}}{{(0.1)}^{y}}\] ...(i) 2. \[7.2\times {{10}^{-2}}={{(0.3)}^{x}}{{(0.2)}^{y}}\] ...(ii) 3. \[2.88\times {{10}^{1}}={{(0.3)}^{x}}{{(0.4)}^{y}}\] ...(iii) 4. \[2.4\times {{10}^{2}}={{(0.4)}^{x}}{{(0.1)}^{y}}\] ...(iv) On dividing Eq. (i) by (iv), we get \[\frac{6.0\times {{10}^{-3}}}{2.4\times {{10}^{-2}}}={{\left( \frac{0.1}{0.4} \right)}^{x}}{{\left( \frac{0.1}{0.1} \right)}^{y}}\] \[\therefore \] \[x=1\] On dividing Eq. (ii) by (iii), we get \[\frac{7.2\times {{10}^{-2}}}{28.8\times {{10}^{-2}}}={{\left( \frac{0.3}{0.3} \right)}^{x}}{{\left( \frac{0.2}{0.4} \right)}^{y}}\] \[\frac{1}{4}={{\left( \frac{1}{2} \right)}^{y}}\] \[y=2\] Thus, the rate law is \[rate=k\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{ }\!\![\!\!\text{ B}{{\text{ }\!\!]\!\!\text{ }}^{2}}\]
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