Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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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
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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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