Determine the rate law for the reaction, 2IC1 + H2 = I, + 2HCi, from the following initial rate data: Initial Rate(Ms') 2.04x102 4.08x102 2.55x103 5.09x103 [H,], [ICI], 0.250 0.500 0.500 0.500 0.125 0.125 0.125 0.250 A) R = k[IC1]? В) R = k[H2]? C) R = k[ICI][H_]? D) R = k[ICl][H,] E) R = k[ICI]°[H¿] 000
Determine the rate law for the reaction, 2IC1 + H2 = I, + 2HCi, from the following initial rate data: Initial Rate(Ms') 2.04x102 4.08x102 2.55x103 5.09x103 [H,], [ICI], 0.250 0.500 0.500 0.500 0.125 0.125 0.125 0.250 A) R = k[IC1]? В) R = k[H2]? C) R = k[ICI][H_]? D) R = k[ICl][H,] E) R = k[ICI]°[H¿] 000
Chemistry
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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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![### Determining the Rate Law
#### Reaction:
\[ 2ICl + H_2 \rightarrow I_2 + 2HCl \]
#### Initial Rate Data:
| \([ICl]_0\) (M) | \([H_2]_0\) (M) | Initial Rate (M/s) |
|:---------------:|:---------------:|:--------------------:|
| 0.250 | 0.500 | \(2.04 \times 10^{-2}\) |
| 0.500 | 0.500 | \(4.08 \times 10^{-2}\) |
| 0.125 | 0.125 | \(2.55 \times 10^{-3}\) |
| 0.125 | 0.250 | \(5.09 \times 10^{-3}\) |
Based on the given initial rate data, determine the rate law for the reaction.
#### Rate Law Options:
A) \[ R = k[ICl]^2 \]
B) \[ R = k[H_2]^2 \]
C) \[ R = k[ICl][H_2]^2 \]
D) \[ R = k[ICl][H_2] \]
E) \[ R = k[ICl]^2[H_2] \]
### Explanation of Data:
The given data provides the initial concentrations of \(ICl\) and \(H_2\) along with the corresponding initial reaction rates. To determine the rate law, analyze the changes in reactant concentrations and their effect on the reaction rate.
1. **Comparing Rows 1 and 2:**
- \( [ICl]_0 \) doubles (0.250 M to 0.500 M)
- \( [H_2]_0 \) is constant (0.500 M)
- Initial rate also doubles (\(2.04 \times 10^{-2}\) to \(4.08 \times 10^{-2}\))
2. **Comparing Rows 3 and 4:**
- \( [ICl]_0 \) is constant (0.125 M)
- \( [H_2]_0 \) doubles (0.125 M to 0.250 M)
- Initial rate approximately](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4246633b-29b8-4fd8-93c1-8799be822ca9%2Fa91a0541-bbca-43c0-9a55-38eef1daf8d1%2Fkbhhl5xj_processed.png&w=3840&q=75)
Transcribed Image Text:### Determining the Rate Law
#### Reaction:
\[ 2ICl + H_2 \rightarrow I_2 + 2HCl \]
#### Initial Rate Data:
| \([ICl]_0\) (M) | \([H_2]_0\) (M) | Initial Rate (M/s) |
|:---------------:|:---------------:|:--------------------:|
| 0.250 | 0.500 | \(2.04 \times 10^{-2}\) |
| 0.500 | 0.500 | \(4.08 \times 10^{-2}\) |
| 0.125 | 0.125 | \(2.55 \times 10^{-3}\) |
| 0.125 | 0.250 | \(5.09 \times 10^{-3}\) |
Based on the given initial rate data, determine the rate law for the reaction.
#### Rate Law Options:
A) \[ R = k[ICl]^2 \]
B) \[ R = k[H_2]^2 \]
C) \[ R = k[ICl][H_2]^2 \]
D) \[ R = k[ICl][H_2] \]
E) \[ R = k[ICl]^2[H_2] \]
### Explanation of Data:
The given data provides the initial concentrations of \(ICl\) and \(H_2\) along with the corresponding initial reaction rates. To determine the rate law, analyze the changes in reactant concentrations and their effect on the reaction rate.
1. **Comparing Rows 1 and 2:**
- \( [ICl]_0 \) doubles (0.250 M to 0.500 M)
- \( [H_2]_0 \) is constant (0.500 M)
- Initial rate also doubles (\(2.04 \times 10^{-2}\) to \(4.08 \times 10^{-2}\))
2. **Comparing Rows 3 and 4:**
- \( [ICl]_0 \) is constant (0.125 M)
- \( [H_2]_0 \) doubles (0.125 M to 0.250 M)
- Initial rate approximately
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