1) What is the order of the reaction with respect to NO and respect to O2? 2) Write the rate law for this reaction including the value of the rate constant (k)
1) What is the order of the reaction with respect to NO and respect to O2? 2) Write the rate law for this reaction including the value of the rate constant (k)
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
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![### Reaction Kinetics: Determining Reaction Order and Rate Laws
The following reaction illustrates the formation of a highly reactive oxygen atom when NO (a common air pollutant from car emissions) is oxidized:
\[ \text{NO } (g) + \text{O}_2 \, (g) \rightarrow \text{NO}_2 \, (g) + \text{O} \, (g) \]
Kinetic data from experimental trials of this reaction are summarized in the table below:
| NO (M) | O₂ (M) | Rate (M/s) |
|--------|--------|------------|
| 1.00 | 1.00 | 3 x 10⁻⁴ |
| 1.01 | 2.00 | 6 x 10⁻⁴ |
| 1.00 | 2.97 | 9 x 10⁻⁴ |
| 4.98 | 3.00 | 45 x 10⁻⁴ |
| 10.0 | 6.00 | ? |
#### Questions:
1. **What is the order of the reaction with respect to NO and with respect to O₂?**
2. **Write the rate law for this reaction including the value of the rate constant (k).**
3. **What is the missing rate value in the final row of the table?**
#### Determining Reaction Order:
1. **NO Order:**
- Compare the rates when \([O_2]\) is constant:
\[ \frac{Rate_2}{Rate_1} = \frac{6 \times 10^{-4}}{3 \times 10^{-4}} = 2 \]
\[ \frac{[NO_2]}{[NO_1]} = \frac{1.01}{1.00} \approx 1 \]
- Since \(\frac{Rate_2}{Rate_1} = 2\) when \([NO]\) is approximately constant, the reaction is first-order with respect to \([O_2]\).
2. **O₂ Order:**
- Compare the rates when \([NO]\) is constant:
\[ \frac{Rate_3}{Rate_2} = \frac{9 \times 10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fabf7f939-657a-4cb9-aa9d-1310fbdc7d61%2F567ae279-8a00-4c6e-bcfb-17880cbcb02e%2F8lq613_processed.png&w=3840&q=75)
Transcribed Image Text:### Reaction Kinetics: Determining Reaction Order and Rate Laws
The following reaction illustrates the formation of a highly reactive oxygen atom when NO (a common air pollutant from car emissions) is oxidized:
\[ \text{NO } (g) + \text{O}_2 \, (g) \rightarrow \text{NO}_2 \, (g) + \text{O} \, (g) \]
Kinetic data from experimental trials of this reaction are summarized in the table below:
| NO (M) | O₂ (M) | Rate (M/s) |
|--------|--------|------------|
| 1.00 | 1.00 | 3 x 10⁻⁴ |
| 1.01 | 2.00 | 6 x 10⁻⁴ |
| 1.00 | 2.97 | 9 x 10⁻⁴ |
| 4.98 | 3.00 | 45 x 10⁻⁴ |
| 10.0 | 6.00 | ? |
#### Questions:
1. **What is the order of the reaction with respect to NO and with respect to O₂?**
2. **Write the rate law for this reaction including the value of the rate constant (k).**
3. **What is the missing rate value in the final row of the table?**
#### Determining Reaction Order:
1. **NO Order:**
- Compare the rates when \([O_2]\) is constant:
\[ \frac{Rate_2}{Rate_1} = \frac{6 \times 10^{-4}}{3 \times 10^{-4}} = 2 \]
\[ \frac{[NO_2]}{[NO_1]} = \frac{1.01}{1.00} \approx 1 \]
- Since \(\frac{Rate_2}{Rate_1} = 2\) when \([NO]\) is approximately constant, the reaction is first-order with respect to \([O_2]\).
2. **O₂ Order:**
- Compare the rates when \([NO]\) is constant:
\[ \frac{Rate_3}{Rate_2} = \frac{9 \times 10
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