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Question: The reaction of ozone with oxygen atoms in the presence of chlorine atoms can occur by a two step pr...

The reaction of ozone with oxygen atoms in the presence of chlorine atoms can occur by a two step process shown below: [JEE(Main)-2016]

O3(g)+Cl(g)→O2(g)+ClO(g)O_3(g) + Cl (g) \rightarrow O_2(g) + ClO (g) ......(i)

ki=5.2×109Lmol−1s−1k_i = 5.2 \times 10^9 L mol^{-1} s^{-1}

ClO(g)+O(g)→O2(g)+Cl(g)ClO (g) + O (g) \rightarrow O_2(g) + Cl(g) ......(ii)

kii=2.6×1010Lmol−1s−1k_{ii} = 2.6 \times 10^{10} L mol^{-1} s^{-1}

The closest rate constant for the overall reaction O3(g)+O(g)→2O2(g)O_3(g) + O(g) \rightarrow 2O_2(g) is:

A

3.1×1010Lmol−1s−13.1 \times 10^{10} L mol^{-1} s^{-1}

B

2.6×1010Lmol−1s−12.6 \times 10^{10} L mol^{-1} s^{-1}

C

5.2×109Lmol−1s−15.2 \times 10^9 L mol^{-1} s^{-1}

D

1.4×1020Lmol−1s−11.4 \times 10^{20} L mol^{-1} s^{-1}

Answer

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

Explanation

Solution

The overall reaction proceeds through two elementary steps. The rate of a multi-step reaction is determined by its slowest step (rate-determining step). Comparing the given rate constants, ki=5.2×109Lmol−1s−1k_i = 5.2 \times 10^9 L mol^{-1} s^{-1} and kii=2.6×1010Lmol−1s−1k_{ii} = 2.6 \times 10^{10} L mol^{-1} s^{-1}. Since ki<kiik_i < k_{ii}, the first step (O3(g)+Cl(g)→O2(g)+ClO(g)O_3(g) + Cl (g) \rightarrow O_2(g) + ClO (g)) is the rate-determining step. Therefore, the rate constant for the overall reaction is approximately equal to the rate constant of this slowest step, which is ki=5.2×109Lmol−1s−1k_i = 5.2 \times 10^9 L mol^{-1} s^{-1}.