To determine the rate law for the reaction given the experimental data, we can follow these steps:
### Step 1: Understand the General Rate Law Form
The general rate law for the reaction can be written as:
[tex]\[ \text{Rate} = k[\text{CH}_3\text{Br}]^m[\text{OH}^-]^n \][/tex]
where:
- [tex]\( k \)[/tex] is the rate constant,
- [tex]\( m \)[/tex] is the reaction order with respect to [tex]\(\text{CH}_3\text{Br}\)[/tex], and
- [tex]\( n \)[/tex] is the reaction order with respect to [tex]\(\text{OH}^-\)[/tex].
### Step 2: Analyze the Experimental Data
Using the data from the experiments:
1. Experiment 1: [tex]\([ \text{CH}_3\text{Br} ] = 0.010 \, M\)[/tex], [tex]\([ \text{OH}^- ] = 0.015 \, M\)[/tex], Rate = 0.0415 \, M/s
2. Experiment 2: [tex]\([ \text{CH}_3\text{Br} ] = 0.010 \, M\)[/tex], [tex]\([ \text{OH}^- ] = 0.030 \, M\)[/tex], Rate = 0.0830 \, M/s
3. Experiment 3: [tex]\([ \text{CH}_3\text{Br} ] = 0.030 \, M\)[/tex], [tex]\([ \text{OH}^- ] = 0.015 \, M\)[/tex], Rate = 0.125 \, M/s
### Step 3: Determine the Order with Respect to OH⁻ (n)
To find [tex]\( n \)[/tex], we compare experiments where the concentration of [tex]\(\text{CH}_3\text{Br}\)[/tex] is constant and the concentration of [tex]\(\text{OH}^-\)[/tex] changes. Compare Experiments 1 and 2:
- The concentration of [tex]\(\text{CH}_3\text{Br}\)[/tex] is constant (0.010 M).
- The concentration of [tex]\(\text{OH}^-\)[/tex] is doubled from 0.015 M to 0.030 M.
- The rate doubles from 0.0415 M/s to 0.0830 M/s.
Let’s set up the ratio:
[tex]\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[\text{OH}^-]_2}{[\text{OH}^-]_1}\right)^n \][/tex]
Substitute the given values:
[tex]\[ \frac{0.0830}{0.0415} = \left(\frac{0.030}{0.015}\right)^n \][/tex]
[tex]\[ 2 = 2^n \][/tex]
From this, it is clear that:
[tex]\[ n = 1 \][/tex]
### Step 4: Determine the Order with Respect to CH₃Br (m)
To find [tex]\( m \)[/tex], we compare experiments where the concentration of [tex]\(\text{OH}^-\)[/tex] is constant and the concentration of [tex]\(\text{CH}_3\text{Br}\)[/tex] changes. Compare Experiments 1 and 3:
- The concentration of [tex]\(\text{OH}^-\)[/tex] is constant (0.015 M).
- The concentration of [tex]\(\text{CH}_3\text{Br}\)[/tex] is tripled from 0.010 M to 0.030 M.
- The rate increases from 0.0415 M/s to 0.125 M/s.
Let’s set up the ratio:
[tex]\[ \frac{\text{Rate}_3}{\text{Rate}_1} = \left(\frac{[\text{CH}_3\text{Br}]_3}{[\text{CH}_3\text{Br}]_1}\right)^m \][/tex]
Substitute the given values:
[tex]\[ \frac{0.125}{0.0415} = \left(\frac{0.030}{0.010}\right)^m \][/tex]
[tex]\[ 3.012048192771084 = 3^m \][/tex]
From this, we solve for [tex]\( m \)[/tex]:
[tex]\[ m \approx 1 \][/tex]
### Step 5: Write the Rate Law
Given [tex]\( m \approx 1 \)[/tex] and [tex]\( n = 1 \)[/tex], the rate law for the reaction is:
[tex]\[ \text{Rate} = k[\text{CH}_3\text{Br}]^1[\text{OH}^-]^1 \][/tex]
[tex]\[ \text{Rate} = k[\text{CH}_3\text{Br}][\text{OH}^-] \][/tex]
This corresponds to the rate law from the given options:
- Option: Rate = [tex]\( k[\text{CH}_3\text{Br}][\text{OH}^-] \)[/tex]
