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)

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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
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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