| (i) \[OC{{l}^{\Theta }}+{{H}_{2}}OHOCl+\overset{\Theta }{\mathop{O}}\,H\,\,(fast)\] |
| (ii) \[{{I}^{\Theta }}+HOCl\xrightarrow{{{k}_{3}}}HOI+C{{l}^{\Theta }}\,\,(slow)\] |
| (iii) \[\overset{\Theta }{\mathop{O}}\,H+HOI{{H}_{2}}O+O{{I}^{\Theta }}\,(fast)\] |
| The rate of consumption of \[{{I}^{\Theta }}\] in the following equation is: |
A) \[\frac{{{k}_{3}}{{k}_{1}}}{{{k}_{2}}}\frac{[OC{{l}^{\Theta }}][{{I}^{\Theta }}]}{[\overset{\Theta }{\mathop{O}}\,H]}\]
B) \[\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{3}}}\frac{[OC{{l}^{\Theta }}][{{I}^{\Theta }}]}{[\overset{\Theta }{\mathop{O}}\,H]}\]
C) \[\frac{{{k}_{1}}{{k}_{3}}}{{{k}_{2}}}\frac{[OC{{l}^{\Theta }}][{{I}^{\Theta }}]}{[\overset{\Theta }{\mathop{O}}\,H]}\]
D) \[\frac{{{k}_{2}}{{k}_{3}}}{{{k}_{1}}}\frac{[OC{{l}^{\Theta }}][{{I}^{\Theta }}]}{[\overset{\Theta }{\mathop{O}}\,H]}\]
Correct Answer: C
Solution :
[c] Step (ii) is the rate-determining step. \[\therefore \] Rate \[{{k}_{3}}[{{I}^{\Theta }}]\,\,[HOCl]\] ....(l) \[(HOCl)\] produced in step (i) is consumed in step (ii), so it is a reactive intermediate. Similary, \[(HOI)\] produced in step (ii) is consumed in step (iii). so it is also a reactive intermediate. The concentration of both reactive species is determined from their equilibrium constant value. \[\therefore \] From step (i) \[\frac{{{k}_{1}}}{{{k}_{2}}}=\frac{[HOCl]\,[\overset{\Theta }{\mathop{O}}\,H]}{[OC{{l}^{\Theta }}]}\Rightarrow [HOCl]=\frac{{{k}_{1}}}{{{k}_{2}}}\frac{[OC{{l}^{\Theta }}]}{[\overset{\Theta }{\mathop{O}}\,H]}\] ?.(2) Substitute the value of \[[HOCl]\]from Eq. (2) in Eq. Rate \[=\frac{{{k}_{1}}{{k}_{3}}}{{{k}_{2}}}\frac{[OC{{l}^{\Theta }}][{{I}^{\Theta }}]}{[\overset{\Theta }{\mathop{O}}\,H]}\] Hence answer is (3).
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