At a certain temperature the rate of this reaction is second order in SO3 with a rate constant of 0.013M¹s¹: 2SO3(g) -2SO₂ (g) + O₂(g) Suppose a vessel contains SO3 at a concentration of 1.36M. Calculate how long it takes for the concentration of SO3 to decrease by 91.%. You may assume no other reaction is important. Round your answer to 2 significant digits. →

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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**Chemical Reaction Kinetics Problem**

At a certain temperature, the rate of this reaction is second order in \( \text{SO}_3 \), with a rate constant of \( 0.013 \, M^{-1} \, s^{-1} \):

\[ 2 \text{SO}_3 (g) \rightarrow 2 \text{SO}_2 (g) + \text{O}_2 (g) \]

Suppose a vessel contains \( \text{SO}_3 \) at a concentration of \( 1.36 \, M \). Calculate how long it takes for the concentration of \( \text{SO}_3 \) to decrease by 91.1%. You may assume no other reaction is important.

**Instructions:**
1. **Understand the Reaction Order:** This reaction is second order, meaning the rate law can be expressed as: \[ \text{Rate} = k[\text{SO}_3]^2 \]
2. **Initial and Final Concentrations:** 
   - Initial concentration (\( [A]_0 \)) = \( 1.36 \, M \)
   - Final concentration (\( [A] \)) after 91.1% decrease = \( 1.36 \, M \times (1 - 0.911) = 1.36 \, M \times 0.089 = 0.12104 \, M \)
3. **Second-Order Reaction Formula:** Use the integrated rate law for a second-order reaction: \[ \frac{1}{[A]} - \frac{1}{[A]_0} = kt \]

**Given Data:**
   - Rate constant (\( k \)) = \( 0.013 \, M^{-1} \, s^{-1} \)
   - Initial concentration (\( [A]_0 \)) = \( 1.36 \, M \)
   - Final concentration (\( [A] \)) = \( 0.12104 \, M \)

**Calculation:**
1. Plug in the values into the second-order integrated rate equation:
   \[ \frac{1}{0.12104} - \frac{1}{1.36} = 0.013 \times t \]
2. Solve for \( t \):
   \[ \frac{1}{0.12104} = 8.26 \,
Transcribed Image Text:**Chemical Reaction Kinetics Problem** At a certain temperature, the rate of this reaction is second order in \( \text{SO}_3 \), with a rate constant of \( 0.013 \, M^{-1} \, s^{-1} \): \[ 2 \text{SO}_3 (g) \rightarrow 2 \text{SO}_2 (g) + \text{O}_2 (g) \] Suppose a vessel contains \( \text{SO}_3 \) at a concentration of \( 1.36 \, M \). Calculate how long it takes for the concentration of \( \text{SO}_3 \) to decrease by 91.1%. You may assume no other reaction is important. **Instructions:** 1. **Understand the Reaction Order:** This reaction is second order, meaning the rate law can be expressed as: \[ \text{Rate} = k[\text{SO}_3]^2 \] 2. **Initial and Final Concentrations:** - Initial concentration (\( [A]_0 \)) = \( 1.36 \, M \) - Final concentration (\( [A] \)) after 91.1% decrease = \( 1.36 \, M \times (1 - 0.911) = 1.36 \, M \times 0.089 = 0.12104 \, M \) 3. **Second-Order Reaction Formula:** Use the integrated rate law for a second-order reaction: \[ \frac{1}{[A]} - \frac{1}{[A]_0} = kt \] **Given Data:** - Rate constant (\( k \)) = \( 0.013 \, M^{-1} \, s^{-1} \) - Initial concentration (\( [A]_0 \)) = \( 1.36 \, M \) - Final concentration (\( [A] \)) = \( 0.12104 \, M \) **Calculation:** 1. Plug in the values into the second-order integrated rate equation: \[ \frac{1}{0.12104} - \frac{1}{1.36} = 0.013 \times t \] 2. Solve for \( t \): \[ \frac{1}{0.12104} = 8.26 \,
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