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BITSAT 2009
Exam
BITSAT
Year
2009
Subject
Chemistry
Chapter
Chemical Kinetics

BITSAT 2009 · Chemistry · Chemical Kinetics

The chemical reaction 2 O_ 3 3 O_ 2 proceeds as follows: rate law...

The chemical reaction 2O3⟶3O22 \mathrm{O}_{3} \longrightarrow 3 \mathrm{O}_{2} proceeds as follows: The chemical reaction 2 O_ 3 3 O_ 2 proceeds as follows: rate law expression should be:
rate law expression should be:

Options

  1. A\mathrm{r}=\mathrm{k}[O3]2\left[\mathrm{O}_{3}\right]^{2}
  2. Br=k\mathrm{r}=\mathrm{k}\left[\mathrm{O}_{3}\right]^{2}[O2]−1\left[\mathrm{O}_{2}\right]^{-1}
  3. Cr=k\mathrm{r}=\mathrm{k}\left[\mathrm{O}_{3}\right]^{2}[O2]2\left[\mathrm{O}_{2}\right]^{2}
  4. Dr=k\mathrm{r}=\mathrm{k}\left[\mathrm{O}_{3}\right][O2]2\left[\mathrm{O}_{2}\right]^{2}

Correct answer

B. r=k\mathrm{r}=\mathrm{k}\left[\mathrm{O}_{3}\right]^{2}[O2]−1\left[\mathrm{O}_{2}\right]^{-1}

Step-by-step solution

O3⟶ Fast O2+O\mathrm{O}_{3} \stackrel{\text{ Fast }}{\longrightarrow} \mathrm{O}_{2}+\mathrm{O}O+O3⟶ Slow 2O2k=[O2][O][O3] Rate =\mathrmk′[O3][O]\begin{array}{l}\mathrm{O}+\mathrm{O}_{3} \stackrel{\text{ Slow }}{\longrightarrow}2 \mathrm{O}_{2}\\ \mathrm{k}=\frac{\left[\mathrm{O}_{2}\right][\mathrm{O}]}{\left[\mathrm{O}_{3}\right]}\\ \text{ Rate }=\mathrmk^{\prime}\left[\mathrm{O}_{3}\right][\mathrm{O}] \end{array}$put[O]from(ii)put [O] from (ii)\mathrm{r}=\frac{\mathrmk^{\prime}\left[\mathrm{O}_{3}\right] \mathrm{k}\left[\mathrm{O}_{3}\right]}{\left[\mathrm{O}_{2}\right]}=\mathrm{k}\left[\mathrm{O}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{-1}$ Note: Intermediates are never represented in rate law equation.
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