Given the following data, determine the order of the reaction with respect to Cl2. 2 NO(g) + Cl:(g) → 2 NOCI(g) Еxperiment INO] (M) Rate (M/s) 3.4 x 104 8.5 x 10 |Clk] (M) 0.0300 0.0100 2 0.0150 0.0100 3. 0.0150 0.0400 3.4 x 10

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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## Determining the Order of Reaction

**Reaction:**
\[ 2 \text{NO}(g) + \text{Cl}_2(g) \rightarrow 2 \text{NOCl}(g) \]

**Objective:**  
Determine the order of the reaction with respect to \(\text{Cl}_2\).

**Data Provided:**

The table below shows the concentration of \(\text{NO}\), \(\text{Cl}_2\), and the observed reaction rate for three different experiments.

| Experiment | \([\text{NO}]\) (M) | \([\text{Cl}_2]\) (M) | Rate (M/s)   |
|------------|--------------------|----------------------|--------------|
| 1          | 0.0300             | 0.0100               | \(3.4 \times 10^{-4}\) |
| 2          | 0.0150             | 0.0100               | \(8.5 \times 10^{-5}\) |
| 3          | 0.0150             | 0.0400               | \(3.4 \times 10^{-4}\) |

**Analysis:**

- **Experiment 1 and 2:**
  - \([\text{NO}]\) is halved (from 0.0300 to 0.0150) while \([\text{Cl}_2]\) remains constant. The rate decreases from \(3.4 \times 10^{-4}\) to \(8.5 \times 10^{-5}\).
  - This indicates a proportional change, confirming reaction dependence on \([\text{NO}]\).

- **Experiment 2 and 3:**
  - \([\text{Cl}_2]\) quadruples (from 0.0100 to 0.0400) while \([\text{NO}]\) remains the same. The rate increases by a factor of 4 (from \(8.5 \times 10^{-5}\) to \(3.4 \times 10^{-4}\)).
  - This indicates a reaction order of 1 with respect to \([\text{Cl}_2]\).

By observing these changes, it can be determined that the reaction is first-order in respect to \([\text{Cl}_2]\).
Transcribed Image Text:## Determining the Order of Reaction **Reaction:** \[ 2 \text{NO}(g) + \text{Cl}_2(g) \rightarrow 2 \text{NOCl}(g) \] **Objective:** Determine the order of the reaction with respect to \(\text{Cl}_2\). **Data Provided:** The table below shows the concentration of \(\text{NO}\), \(\text{Cl}_2\), and the observed reaction rate for three different experiments. | Experiment | \([\text{NO}]\) (M) | \([\text{Cl}_2]\) (M) | Rate (M/s) | |------------|--------------------|----------------------|--------------| | 1 | 0.0300 | 0.0100 | \(3.4 \times 10^{-4}\) | | 2 | 0.0150 | 0.0100 | \(8.5 \times 10^{-5}\) | | 3 | 0.0150 | 0.0400 | \(3.4 \times 10^{-4}\) | **Analysis:** - **Experiment 1 and 2:** - \([\text{NO}]\) is halved (from 0.0300 to 0.0150) while \([\text{Cl}_2]\) remains constant. The rate decreases from \(3.4 \times 10^{-4}\) to \(8.5 \times 10^{-5}\). - This indicates a proportional change, confirming reaction dependence on \([\text{NO}]\). - **Experiment 2 and 3:** - \([\text{Cl}_2]\) quadruples (from 0.0100 to 0.0400) while \([\text{NO}]\) remains the same. The rate increases by a factor of 4 (from \(8.5 \times 10^{-5}\) to \(3.4 \times 10^{-4}\)). - This indicates a reaction order of 1 with respect to \([\text{Cl}_2]\). By observing these changes, it can be determined that the reaction is first-order in respect to \([\text{Cl}_2]\).
### Determining the Rate Law for the Reaction

**Reaction**:
\[ \text{NH}_4^+(aq) + \text{NO}_2^-(aq) \rightarrow \text{N}_2(g) + 2\text{H}_2\text{O}(l) \]

The table below shows the concentration of reactants and the corresponding reaction rate from three different experiments.

| **Experiment** | \([\text{NH}_4^+]\) (M) | \([\text{NO}_2^-]\) (M) | **Rate** \((\text{M/s})\) |
|----------------|------------------|------------------|------------------|
| 1              | 0.250            | 0.250            | \(1.25 \times 10^{-7}\) |
| 2              | 0.500            | 0.250            | \(2.50 \times 10^{-7}\) |
| 3              | 0.250            | 0.125            | \(6.25 \times 10^{-8}\)  |

**Explanation of Data**:

This data is used to determine the rate law, which relates the rate of reaction to the concentration of the reactants. The goal is to find the order of the reaction with respect to each reactant.

- **Experiment 1 vs. Experiment 2**: The concentration of \([\text{NH}_4^+]\) is doubled from 0.250 M to 0.500 M while \([\text{NO}_2^-]\) remains constant. The rate also doubles, indicating that the reaction is first order with respect to \([\text{NH}_4^+]\).

- **Experiment 1 vs. Experiment 3**: The concentration of \([\text{NO}_2^-]\) is halved from 0.250 M to 0.125 M while \([\text{NH}_4^+]\) remains constant. The rate is halved, indicating that the reaction is first order with respect to \([\text{NO}_2^-]\).

**Conclusion**:

Based on the data, the rate law for the reaction can be expressed as:

\[
\text{Rate} = k [\text{NH}_4^+]^1 [\text{NO}_2^-]^1
\]

where \(k\)
Transcribed Image Text:### Determining the Rate Law for the Reaction **Reaction**: \[ \text{NH}_4^+(aq) + \text{NO}_2^-(aq) \rightarrow \text{N}_2(g) + 2\text{H}_2\text{O}(l) \] The table below shows the concentration of reactants and the corresponding reaction rate from three different experiments. | **Experiment** | \([\text{NH}_4^+]\) (M) | \([\text{NO}_2^-]\) (M) | **Rate** \((\text{M/s})\) | |----------------|------------------|------------------|------------------| | 1 | 0.250 | 0.250 | \(1.25 \times 10^{-7}\) | | 2 | 0.500 | 0.250 | \(2.50 \times 10^{-7}\) | | 3 | 0.250 | 0.125 | \(6.25 \times 10^{-8}\) | **Explanation of Data**: This data is used to determine the rate law, which relates the rate of reaction to the concentration of the reactants. The goal is to find the order of the reaction with respect to each reactant. - **Experiment 1 vs. Experiment 2**: The concentration of \([\text{NH}_4^+]\) is doubled from 0.250 M to 0.500 M while \([\text{NO}_2^-]\) remains constant. The rate also doubles, indicating that the reaction is first order with respect to \([\text{NH}_4^+]\). - **Experiment 1 vs. Experiment 3**: The concentration of \([\text{NO}_2^-]\) is halved from 0.250 M to 0.125 M while \([\text{NH}_4^+]\) remains constant. The rate is halved, indicating that the reaction is first order with respect to \([\text{NO}_2^-]\). **Conclusion**: Based on the data, the rate law for the reaction can be expressed as: \[ \text{Rate} = k [\text{NH}_4^+]^1 [\text{NO}_2^-]^1 \] where \(k\)
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