The data in the table were collected at 580 K for the reaction 2 HI(g) → H₂(g) + L₂(g) Time (s) [HI] (mol.L-¹) 0 1.00 1000. 0.11 2000. 0.061 3000. 0.041 4000. 0.031 From the graph, determine the rate constant, k, for the loss of HI. k= TOOLS x10 Using a graphing calculator or software, plot the data in an appropriate fashion to determine the order of the reaction: O 0 O 1 2 Determine the units of the rate constant. mol-L-¹.1 L.mol-¹.s-¹

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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**Kinetics - Determining the Order of Reaction and Rate Constant Calculation**

This educational segment aims to elucidate the method of determining the order of a chemical reaction and its corresponding rate constant, \( k \), based on experimental data.

### Reaction Data

The data was collected at 580 K for the reaction:

\[ 2 \, \text{HI(g)} \rightarrow \text{H}_2(g) + \text{I}_2(g) \]

**Table of Data:**

| Time (s) | [HI] (mol·L\(^{-1}\)) |
|----------|-----------------------|
| 0        | 1.00                  |
| 1000     | 0.11                  |
| 2000     | 0.061                 |
| 3000     | 0.041                 |
| 4000     | 0.031                 |

### Data Analysis

1. **Plotting the Data:**
   Using a graphing calculator or software, plot the concentration of HI against time to determine the reaction order.

2. **Determination of Reaction Order:**
   Plot the data in an appropriate graph:
   - Zero-order: [HI] vs. Time
   - First-order: ln[HI] vs. Time
   - Second-order: \( \frac{1}{[\text{HI}]} \) vs. Time

   Based on the provided question, it is suggested that the reaction order is 2. Therefore, a second-order reaction graph \( \frac{1}{[\text{HI}]} \) vs. Time should be linear.

3. **Graph Interpretation:**
   Upon plotting, if \( \frac{1}{[\text{HI}]} \) vs. Time yields a straight line, it confirms the reaction is second-order.

### Rate Constant Calculation

From the graph, determine the rate constant, \( k \), for the loss of HI.

\[ k = \text{(Slope of the line in the second-order plot)} \]

**Example Calculation:**
Assuming the line fits the second-order reaction model perfectly, the slope \( k \) can be determined directly from the graph.

### Units of Rate Constant

For a second-order reaction, the units of the rate constant \( k \) are \( \text{L·mol}^{-1}\text{s}^{-1} \).

### Conclusion

By plotting the concentration
Transcribed Image Text:--- **Kinetics - Determining the Order of Reaction and Rate Constant Calculation** This educational segment aims to elucidate the method of determining the order of a chemical reaction and its corresponding rate constant, \( k \), based on experimental data. ### Reaction Data The data was collected at 580 K for the reaction: \[ 2 \, \text{HI(g)} \rightarrow \text{H}_2(g) + \text{I}_2(g) \] **Table of Data:** | Time (s) | [HI] (mol·L\(^{-1}\)) | |----------|-----------------------| | 0 | 1.00 | | 1000 | 0.11 | | 2000 | 0.061 | | 3000 | 0.041 | | 4000 | 0.031 | ### Data Analysis 1. **Plotting the Data:** Using a graphing calculator or software, plot the concentration of HI against time to determine the reaction order. 2. **Determination of Reaction Order:** Plot the data in an appropriate graph: - Zero-order: [HI] vs. Time - First-order: ln[HI] vs. Time - Second-order: \( \frac{1}{[\text{HI}]} \) vs. Time Based on the provided question, it is suggested that the reaction order is 2. Therefore, a second-order reaction graph \( \frac{1}{[\text{HI}]} \) vs. Time should be linear. 3. **Graph Interpretation:** Upon plotting, if \( \frac{1}{[\text{HI}]} \) vs. Time yields a straight line, it confirms the reaction is second-order. ### Rate Constant Calculation From the graph, determine the rate constant, \( k \), for the loss of HI. \[ k = \text{(Slope of the line in the second-order plot)} \] **Example Calculation:** Assuming the line fits the second-order reaction model perfectly, the slope \( k \) can be determined directly from the graph. ### Units of Rate Constant For a second-order reaction, the units of the rate constant \( k \) are \( \text{L·mol}^{-1}\text{s}^{-1} \). ### Conclusion By plotting the concentration
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