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

Book Online Demo

Check Your IQ

Try Test

Courses

Offline Centres

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

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

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

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

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

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

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE