The following data were collected for the reaction [tex]CH_3Br (aq) + OH^{-} (aq) \rightarrow CH_3OH (aq) + Br^{-} (aq)[/tex].

\begin{tabular}{|c|c|c|c|}
\hline
Experiment & {[tex]$[CH_3Br]$[/tex] (M)} & {[tex]$[OH^{-}]$[/tex] (M)} & Initial Rate (M/s) \\
\hline
1 & 0.010 & 0.015 & 0.0415 \\
\hline
2 & 0.010 & 0.030 & 0.0830 \\
\hline
3 & 0.030 & 0.015 & 0.125 \\
\hline
\end{tabular}

What is the rate law for the reaction?

A. Rate [tex]$= k[CH_3Br][OH^{-}]^2$[/tex]

B. Rate [tex]$= k[CH_3Br]^2[OH^{-}]$[/tex]

C. Rate [tex]$= k[CH_3Br]^2[OH^{-}]^2$[/tex]

D. Rate [tex]$= k[CH_3Br][OH^{-}]$[/tex]



Answer :

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]