| Expt | [A] | [B] | \[~rate(mol\text{ }d{{m}^{-3}}\text{ }{{s}^{-1}})\] |
| 1 | 0.2 | 0.2 | 2 |
| 2 | 0.2 | 0.4 | 4 |
| 3 | 0.6 | 04 | 36 |
A) \[r=k{{[A]}^{2}}[B]\]
B) \[r=k[A]{{[B]}^{2}}\]
C) \[r=k{{[A]}^{3}}[B]\]
D) \[r=k{{[A]}^{2}}{{[B]}^{2}}\]
Correct Answer: A
Solution :
: \[A+B\xrightarrow{{}}Products\] Rate \[=k{{[A]}^{\alpha }}{{[B]}^{\beta }}\] From expt. No. 1 \[2=k{{[0.2]}^{\alpha }}{{[0.4]}^{\beta }}\] ...(1) From expt. No. 2 \[4=k{{[0.2]}^{\alpha }}{{[0.4]}^{\beta }}\] ...(2) dividing eq. (1) and (2) we get \[\frac{2}{4}=\frac{k}{k}\frac{{{[0.2]}^{\alpha }}{{[0.2]}^{\beta }}}{{{[0.2]}^{\alpha }}{{[0.4]}^{\beta }}}\] or \[\frac{1}{2}={{\left[ \frac{1}{2} \right]}^{\beta }}\]\[\Rightarrow \]\[\beta =1\] From expt. No. 3 \[36=k{{[0.6]}^{\alpha }}{{[0.4]}^{\beta }}\] ...(3) dividing eqn. (2) and (3) we get \[\frac{4}{36}=\frac{k{{[0.2]}^{\alpha }}{{[0.4]}^{\beta }}}{k{{[0.6]}^{\alpha }}{{[0.4]}^{\beta }}}\] or \[\frac{1}{9}={{\left[ \frac{1}{3} \right]}^{\alpha }}\]\[\Rightarrow \]\[\alpha =2\] \[\Rightarrow \] Rate law for the given reaction is \[r=k{{[A]}^{2}}{{[B]}^{1}}\]
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