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The reaction of ozone with oxygen atoms ...

The reaction of ozone with oxygen atoms in the presence of chlorine atoms can occur by a two step process shown below:
`O_(3)(g)+Cl^(.)rarrO_(2)(g)+ClO^(.)(g)…(i)`
`k_(i)=5.2xx10^(9)" L mol"^(-1)s^(-1)`
`ClO^(.)(g)+O^(.)(g)rarrO_(2)(g)+Cl^(.)(g).......(ii)`
`k_(ii)=2.6xx10^(10)"L mol"^(-1)s^(-1)`
The closest rate constant for the overall reaction
`O_(3)(g)+O^(.)(g)rarr2O_(2)(g)` is :

A

`1.4xx10^(20)" L mol"^(-1)s^(-1)`

B

`3.1xx10^(20)" L mol"^(-1)s^(-1)`

C

`5.2xx10^(20)" L mol"^(-1)s^(-1)`

D

`2.6xx10^(20)" L mol"^(-1)s^(-1)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the closest rate constant for the overall reaction \( O_3(g) + O(g) \rightarrow 2O_2(g) \), we will follow these steps: ### Step 1: Write down the two given reactions and their rate constants. 1. The first reaction is: \[ O_3(g) + Cl^{\cdot} \rightarrow O_2(g) + ClO^{\cdot} \quad (i) \] with a rate constant \( k_i = 5.2 \times 10^9 \, \text{L mol}^{-1} \text{s}^{-1} \). 2. The second reaction is: \[ ClO^{\cdot} + O^{\cdot} \rightarrow O_2(g) + Cl^{\cdot} \quad (ii) \] with a rate constant \( k_{ii} = 2.6 \times 10^{10} \, \text{L mol}^{-1} \text{s}^{-1} \). ### Step 2: Add the two reactions to find the overall reaction. When we add the two reactions, we can cancel out the intermediates: - \( Cl^{\cdot} \) appears on both sides of the first reaction. - \( ClO^{\cdot} \) appears on both sides of the second reaction. Thus, the overall reaction becomes: \[ O_3(g) + O^{\cdot}(g) \rightarrow 2O_2(g) \] ### Step 3: Determine the rate constant for the overall reaction. The rate constant for the overall reaction, \( k_{overall} \), can be calculated by multiplying the rate constants of the individual steps: \[ k_{overall} = k_i \times k_{ii} \] Substituting the values: \[ k_{overall} = (5.2 \times 10^9) \times (2.6 \times 10^{10}) \] ### Step 4: Perform the multiplication. Calculating: \[ k_{overall} = 5.2 \times 2.6 \times 10^{9 + 10} = 13.52 \times 10^{19} \] ### Step 5: Express the result in scientific notation. To express \( 13.52 \times 10^{19} \) in proper scientific notation: \[ k_{overall} = 1.352 \times 10^{20} \, \text{L mol}^{-1} \text{s}^{-1} \] ### Step 6: Round to the appropriate significant figures. Rounding \( 1.352 \) to two significant figures gives us: \[ k_{overall} \approx 1.4 \times 10^{20} \, \text{L mol}^{-1} \text{s}^{-1} \] ### Final Answer: The closest rate constant for the overall reaction \( O_3(g) + O(g) \rightarrow 2O_2(g) \) is: \[ \boxed{1.4 \times 10^{20} \, \text{L mol}^{-1} \text{s}^{-1}} \]
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The reaction of O_(3) with chlorine atom is given as : O_(3)(g)+Cl(g)rarrO_(2)(g)+ClO(g),k_(1)=5.2xx10^(9)L"mol"^(-1)sec^(-1) ClO(g)+O(g)rarrCl(g)+O_(2)(g),k_(2)=2.6xx10^(10)L"mol"^(-1)sec^(-1) Which of theses values is closest to the rate constant of the overall reaction ? O_(3)(g)+O(g)rarr2O_(2)(g)

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Knowledge Check

  • For a reaction, 2N_(2)O_(5)(g) to 4NO_(2)(g) + O_(2)(g) rate of reaction is:

    A
    Rate `= - (d[N_(2)O_(5)])/(dt) = - (1)/(4) (d[NO_(2)])/(dt) = (1)/(2)(d[O_(2)])/(dt)`
    B
    Rate `= - (1)/(2) (d[n_(2)O_(5)])/(dt) = (1)/(4)(d[NO_(2)])/(dt) = (d[O_(2)])/(2)`
    C
    Rate `= - (1)/(4) (d[N_(2)O_(5)])/(dt) = (1)/(2) (d[NO_(2)])/(dt) = (d[O_(2)])/(dt)`
    D
    Rate `= - (1)/(2) (d[N_(2)O_(5)])/(dt) = (1)/(2) (d[NO_(2)])/(dt) = (1)/(2) (d[O_(2)])/(dt)`
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