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Chemical Kinetics and Nuclear Chemistry question

2016 · 9 Apr · Shift 1 · Q15
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Chemical Kinetics and Nuclear Chemistry question

2016 · 9 Apr · Shift 1 · Q15

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
The reaction of ozone with oxygen atoms in the presence of chlorine atoms can occur by a two step process shown below : O3O_3O3​(g) + Cl ∙{^ \bullet }∙(g) →\to→ O2O_2O2​(g) + ClO ∙{^ \bullet }∙ (g) . . . . . .(i) ki = 5.2 × 109 L mol−1 s−1 ClO ∙{^ \bullet }∙(g) + O ∙{^ \bullet }∙(g) →\to→ O2O_2O2​(g) + Cl ∙{^ \bullet }∙ (g) . . . . . . (ii) kii = 2.6 × 1010 L mol−1 s−1 The closest rate constant for the overall reaction O3O_3O3​(g) + O ∙{^ \bullet }∙(g) →\to→ 2 O2O_2O2​(g) is :
  1. A
    5.2 × 109 L mol−1 s−1
  2. B
    2.6 × 1010 L mol−1 s−1
  3. C
    3.1 × 1010 L mol−1 s−1
  4. D
    1.4 × 1020 L mol−1 s−1
View written solutionFree

Correct answer: D

  1. Write the two elementary steps

    (i)O3+Cl∙→k1O2+ClO∙,k1=5.2×109 L mol−1s−1\text{(i)}\quad O_3 + Cl^{\bullet} \xrightarrow{k_1} O_2 + ClO^{\bullet}, \qquad k_1 = 5.2\times 10^9\ \text{L mol}^{-1}\text{s}^{-1}(i)O3​+Cl∙k1​​O2​+ClO∙,k1​=5.2×109 L mol−1s−1

    (ii)ClO∙+O∙→k2O2+Cl∙,k2=2.6×1010 L mol−1s−1\text{(ii)}\quad ClO^{\bullet} + O^{\bullet} \xrightarrow{k_2} O_2 + Cl^{\bullet}, \qquad k_2 = 2.6\times 10^{10}\ \text{L mol}^{-1}\text{s}^{-1}(ii)ClO∙+O∙k2​​O2​+Cl∙,k2​=2.6×1010 L mol−1s−1

  2. Add the two steps to get the overall reaction

    On adding, the intermediate ClO∙ClO^{\bullet}ClO∙ and catalyst Cl∙Cl^{\bullet}Cl∙ cancel:

    O3+O∙→2O2O_3 + O^{\bullet} \to 2O_2O3​+O∙→2O2​

  3. Find the effective rate constant

    Since the reaction proceeds in two consecutive bimolecular elementary steps, the commonly used approximate overall rate constant for the catalytic cycle is taken as the product of the two step constants:

    koverall≈k1k2k_{\text{overall}} \approx k_1 k_2koverall​≈k1​k2​

    Substitute values:

    koverall=(5.2×109)(2.6×1010)k_{\text{overall}} = (5.2\times 10^9)(2.6\times 10^{10})koverall​=(5.2×109)(2.6×1010)

    =13.52×1019=1.352×1020= 13.52\times 10^{19} = 1.352\times 10^{20}=13.52×1019=1.352×1020

    Closest option:

    1.4×10201.4\times 10^{20}1.4×1020

  4. Check options

    • A: 5.2×1095.2\times 10^95.2×109 — only k1k_1k1​, not overall
    • B: 2.6×10102.6\times 10^{10}2.6×1010 — only k2k_2k2​, not overall
    • C: 3.1×10103.1\times 10^{10}3.1×1010 — not obtained
    • D: 1.4×10201.4\times 10^{20}1.4×1020 — matches the calculated value closely

  5. Final answer

    D  :  1.4×1020\boxed{D\;:\;1.4\times 10^{20}}D:1.4×1020​

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