Chlorine monoxide accumulates in the stratosphere above Antarctica each winter and plays a key role in the formation of the ozone hole above the South Pole each spring. Eventually, CIO decomposes according to the equation: 2C10 (g) > Cl₂(g) + O₂(g) The second-order rate constant for the decomposition of CIO is 6.48×10⁹ M-¹s-1 at a particular temperature. Determine the half-life of CIO when its initial concentration is 1.76×10-8 M. x 101
Chlorine monoxide accumulates in the stratosphere above Antarctica each winter and plays a key role in the formation of the ozone hole above the South Pole each spring. Eventually, CIO decomposes according to the equation: 2C10 (g) > Cl₂(g) + O₂(g) The second-order rate constant for the decomposition of CIO is 6.48×10⁹ M-¹s-1 at a particular temperature. Determine the half-life of CIO when its initial concentration is 1.76×10-8 M. x 101
General Chemistry - Standalone book (MindTap Course List)
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter13: Rates Of Reaction
Section: Chapter Questions
Problem 13.55QP: Chlorine dioxide, ClO2, is a reddish-yellow gas that is soluble in water. In basic solution it gives...
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![**Chlorine Monoxide and Its Role in Ozone Depletion**
**Introduction**
Chlorine monoxide (ClO) accumulates in the stratosphere above Antarctica each winter and plays a key role in the formation of the ozone hole above the South Pole each spring. This occurs through a decomposition reaction.
**Decomposition Reaction**
The decomposition of ClO in the stratosphere is represented by the following chemical equation:
\[ 2 \text{ClO}(g) \rightarrow \text{Cl}_2(g) + \text{O}_2(g) \]
**Rate of Decomposition**
The rate at which this decomposition occurs can be quantified through a second-order rate constant. The given second-order rate constant for the decomposition of ClO is:
\[ k = 6.48 \times 10^9 \, \text{M}^{-1}\text{s}^{-1} \]
**Problem Statement**
Determine the half-life of ClO when its initial concentration is:
\[ [\text{ClO}]_0 = 1.76 \times 10^{-8} \, \text{M} \]
**Solution**
To compute the half-life (\( t_{1/2} \)) for a second-order reaction, the formula is:
\[ t_{1/2} = \frac{1}{k [\text{ClO}]_0} \]
Given:
- \( k = 6.48 \times 10^9 \, \text{M}^{-1}\text{s}^{-1} \)
- \( [\text{ClO}]_0 = 1.76 \times 10^{-8} \, \text{M} \)
Substitute these values into the formula:
\[ t_{1/2} = \frac{1}{(6.48 \times 10^9 \, \text{M}^{-1}\text{s}^{-1})(1.76 \times 10^{-8} \, \text{M})} \]
Calculate \( t_{1/2} \):
\[ t_{1/2} = \frac{1}{1.14048 \times 10^2 \, \text{s}^{-1}} \]
\[ t_{1/2} = 8.76 \times 10^{-3} \, \text{s} \]
Thus, the half-life of ClO](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc80d6140-4b5d-446e-8970-6dff282bf772%2F77cd345a-e950-4654-be73-8edcbc84d5cf%2F47tmcu_processed.png&w=3840&q=75)
Transcribed Image Text:**Chlorine Monoxide and Its Role in Ozone Depletion**
**Introduction**
Chlorine monoxide (ClO) accumulates in the stratosphere above Antarctica each winter and plays a key role in the formation of the ozone hole above the South Pole each spring. This occurs through a decomposition reaction.
**Decomposition Reaction**
The decomposition of ClO in the stratosphere is represented by the following chemical equation:
\[ 2 \text{ClO}(g) \rightarrow \text{Cl}_2(g) + \text{O}_2(g) \]
**Rate of Decomposition**
The rate at which this decomposition occurs can be quantified through a second-order rate constant. The given second-order rate constant for the decomposition of ClO is:
\[ k = 6.48 \times 10^9 \, \text{M}^{-1}\text{s}^{-1} \]
**Problem Statement**
Determine the half-life of ClO when its initial concentration is:
\[ [\text{ClO}]_0 = 1.76 \times 10^{-8} \, \text{M} \]
**Solution**
To compute the half-life (\( t_{1/2} \)) for a second-order reaction, the formula is:
\[ t_{1/2} = \frac{1}{k [\text{ClO}]_0} \]
Given:
- \( k = 6.48 \times 10^9 \, \text{M}^{-1}\text{s}^{-1} \)
- \( [\text{ClO}]_0 = 1.76 \times 10^{-8} \, \text{M} \)
Substitute these values into the formula:
\[ t_{1/2} = \frac{1}{(6.48 \times 10^9 \, \text{M}^{-1}\text{s}^{-1})(1.76 \times 10^{-8} \, \text{M})} \]
Calculate \( t_{1/2} \):
\[ t_{1/2} = \frac{1}{1.14048 \times 10^2 \, \text{s}^{-1}} \]
\[ t_{1/2} = 8.76 \times 10^{-3} \, \text{s} \]
Thus, the half-life of ClO
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