The reaction of ozone with oxygen atoms in the presence of chlorine atoms can occur by a two step process shown below: (i) O3(g)+Cl∙(g)→O2(g)+ClO∙(g) kI=5.2×109Lmol−1s−1(ii) ClO∙(g)+O∙(g)→O2(g)+Cl∙(g) kII=2.6×1010Lmol−1s−1The closest rate constant for the overall reaction O3(g)+O∙(g)→2O2(g) is

5.2 × 10 9 L mol − 1 s − 1

5.2 × 10 9 L mol − 1 s − 1

3.1 × 10 10 L mol − 1 s − 1

2.6 × 10 10 L mol − 1 s − 1

5.2 × 10 9 L mol − 1 s − 1

1.4 × 10 20 L mol − 1 s − 1

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The reaction of ozone with oxygen atoms in the presence of chlorine atoms can occur by a two step process shown below:
(i) $O_{3} (g) + {Cl}^{∙} (g) \to O_{2} (g) + {ClO}^{∙} (g)$ $k_{I} = 5.2 \times 10^{9} L {mol}^{- 1} s^{- 1}$
(ii) ${ClO}^{∙} (g) + O^{∙} (g) \to O_{2} (g) + {Cl}^{∙} (g) k_{II} = 2.6 \times 10^{10} L {mol}^{- 1} s^{- 1}$
The closest rate constant for the overall reaction $O_{3} (g) + O^{∙} (g) \to 2 O_{2} (g)$ is

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$3.1 \times 10^{10} L {mol}^{- 1} s^{- 1}$

$2.6 \times 10^{10} L {mol}^{- 1} s^{- 1}$

$5.2 \times 10^{9} L {mol}^{- 1} s^{- 1}$

$1.4 \times 10^{20} L {mol}^{- 1} s^{- 1}$

answer is C.

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Slow step is the rate-determining step. Hence, the rate constant will be $5.2 \times 10^{9} L {mol}^{- 1} s^{- 1}$ .

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