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?
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?
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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,
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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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