7. Sulfuryl chloride, SO₂Cl2, is a colorless, corrosive liquid whose vapor decomposes in a first-order reaction to sulfur dioxide and chlorine. At 320 °C, the rate constant is 2.20 x 10-5 s¹. SO₂Cl2 → SO₂ + Cl₂ a. What is the half-life of SO₂Cl2 at this temperature? b. How long would it take for 30.0% of the SO₂Cl2 to decompose?

Chemistry: The Molecular Science
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Chapter11: Chemical Kinetics: Rates Of Reactions
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**Sulfuryl Chloride Decomposition: Kinetics and Calculations**

Sulfuryl chloride, \( \mathrm{SO_2Cl_2} \), is a colorless, corrosive liquid known to decompose through a first-order reaction to form sulfur dioxide and chlorine. At a temperature of \( 320 \, ^\circ \mathrm{C} \), the reaction follows this rate equation with a rate constant \( k = 2.20 \times 10^{-5} \, \mathrm{s^{-1}} \).

The decomposition reaction is represented as:

\[ \mathrm{SO_2Cl_2 \rightarrow SO_2 + Cl_2} \]

### Questions:

**a. What is the half-life of \( \mathrm{SO_2Cl_2} \) at this temperature?**

**b. How long would it take for 30.0% of the \( \mathrm{SO_2Cl_2} \) to decompose?**

### Explanation:

#### Part (a): Calculation of Half-life

For a first-order reaction, the half-life (\( t_{1/2} \)) can be calculated using the formula:

\[ t_{1/2} = \frac{0.693}{k} \]

Given:
\[ k = 2.20 \times 10^{-5} \, \mathrm{s^{-1}} \]

Substitute the value of \( k \):

\[ t_{1/2} = \frac{0.693}{2.20 \times 10^{-5} \, \mathrm{s^{-1}}} \]

Perform the calculation to find the half-life.

#### Part (b): Time for 30.0% Decomposition

To determine the time required for 30% decomposition, we utilize the first-order decay formula given by:

\[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]

Where:
- \( [A]_0 \) is the initial concentration.
- \( [A] \) is the concentration after time \( t \).
- \( k \) is the rate constant.
- \( t \) is the time.

Given that 30% of \( \mathrm{SO_2Cl_2} \) decomposes, 70% (\( [A]/[A]_0 = 0.70 \)) of it remains. Thus,

\[
Transcribed Image Text:**Sulfuryl Chloride Decomposition: Kinetics and Calculations** Sulfuryl chloride, \( \mathrm{SO_2Cl_2} \), is a colorless, corrosive liquid known to decompose through a first-order reaction to form sulfur dioxide and chlorine. At a temperature of \( 320 \, ^\circ \mathrm{C} \), the reaction follows this rate equation with a rate constant \( k = 2.20 \times 10^{-5} \, \mathrm{s^{-1}} \). The decomposition reaction is represented as: \[ \mathrm{SO_2Cl_2 \rightarrow SO_2 + Cl_2} \] ### Questions: **a. What is the half-life of \( \mathrm{SO_2Cl_2} \) at this temperature?** **b. How long would it take for 30.0% of the \( \mathrm{SO_2Cl_2} \) to decompose?** ### Explanation: #### Part (a): Calculation of Half-life For a first-order reaction, the half-life (\( t_{1/2} \)) can be calculated using the formula: \[ t_{1/2} = \frac{0.693}{k} \] Given: \[ k = 2.20 \times 10^{-5} \, \mathrm{s^{-1}} \] Substitute the value of \( k \): \[ t_{1/2} = \frac{0.693}{2.20 \times 10^{-5} \, \mathrm{s^{-1}}} \] Perform the calculation to find the half-life. #### Part (b): Time for 30.0% Decomposition To determine the time required for 30% decomposition, we utilize the first-order decay formula given by: \[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \] Where: - \( [A]_0 \) is the initial concentration. - \( [A] \) is the concentration after time \( t \). - \( k \) is the rate constant. - \( t \) is the time. Given that 30% of \( \mathrm{SO_2Cl_2} \) decomposes, 70% (\( [A]/[A]_0 = 0.70 \)) of it remains. Thus, \[
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