An isomer, A, dimerize as follows: 2 A - A2 This process is known to be second order in the reactant, and the rate constant, k, is know to be 0.276 L mol-is-1 at 25 °C. In an experiment, A was placed in a container at 25 °C where [A], = 8.602 x 10-2 M. Calculate the concentration of A2 (in units of M) after 3.98 min. Answer: 0.01289

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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### Example Problem: Second Order Reaction of an Isomer

**Problem Statement:**
An isomer, A, dimerizes as follows:
\[ 2A \rightarrow A_2 \]

This process is known to be second order in the reactant, and the rate constant, \(k\), is known to be 0.276 L mol\(^{-1}\)s\(^{-1}\) at 25 °C. In an experiment, A was placed in a container at 25 °C where \([A]_0 = 8.602 \times 10^{-2} \, \text{M}\). Calculate the concentration of \(A_2\) (in units of M) after 3.98 minutes.

**Incorrect Student Answer:**
\[ \textrm{Answer: } 0.01289 \]

This answer was marked incorrect.

**Correct Answer:**
The correct concentration of \(A_2\) (in units of M) after 3.98 minutes is:
\[ 0.0365612 \]

### Detailed Explanation
For a second-order reaction, the rate law is given by:
\[ \frac{1}{[A]} = kt + \frac{1}{[A]_0} \]

**Where:**
- \([A]\) is the concentration of \(A\) at time \(t\)
- \(k\) is the rate constant
- \([A]_0\) is the initial concentration of \(A\)
- \(t\) is the time the reaction has been running

**Calculations:**
Given data:
- Rate constant, \(k = 0.276 \, \text{L mol}^{-1} \text{s}^{-1} \)
- Initial concentration, \([A]_0 = 8.602 \times 10^{-2} \, \text{M} \)
- Time, \(t = 3.98 \, \text{min} = 3.98 \times 60 \, \text{s} = 238.8 \, \text{s} \)

Plugging these values into the second-order rate equation:
\[ \frac{1}{[A]} = (0.276 \, \text{L mol}^{-1} \text{s}^{-1})(238.8 \, \text{s}) + \frac{1}{8.602
Transcribed Image Text:### Example Problem: Second Order Reaction of an Isomer **Problem Statement:** An isomer, A, dimerizes as follows: \[ 2A \rightarrow A_2 \] This process is known to be second order in the reactant, and the rate constant, \(k\), is known to be 0.276 L mol\(^{-1}\)s\(^{-1}\) at 25 °C. In an experiment, A was placed in a container at 25 °C where \([A]_0 = 8.602 \times 10^{-2} \, \text{M}\). Calculate the concentration of \(A_2\) (in units of M) after 3.98 minutes. **Incorrect Student Answer:** \[ \textrm{Answer: } 0.01289 \] This answer was marked incorrect. **Correct Answer:** The correct concentration of \(A_2\) (in units of M) after 3.98 minutes is: \[ 0.0365612 \] ### Detailed Explanation For a second-order reaction, the rate law is given by: \[ \frac{1}{[A]} = kt + \frac{1}{[A]_0} \] **Where:** - \([A]\) is the concentration of \(A\) at time \(t\) - \(k\) is the rate constant - \([A]_0\) is the initial concentration of \(A\) - \(t\) is the time the reaction has been running **Calculations:** Given data: - Rate constant, \(k = 0.276 \, \text{L mol}^{-1} \text{s}^{-1} \) - Initial concentration, \([A]_0 = 8.602 \times 10^{-2} \, \text{M} \) - Time, \(t = 3.98 \, \text{min} = 3.98 \times 60 \, \text{s} = 238.8 \, \text{s} \) Plugging these values into the second-order rate equation: \[ \frac{1}{[A]} = (0.276 \, \text{L mol}^{-1} \text{s}^{-1})(238.8 \, \text{s}) + \frac{1}{8.602
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