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Sagot :
To determine the rate law for the reaction [tex]\( A + B \rightarrow C \)[/tex], we need to analyze the given experimental data by comparing how different concentrations of [tex]\(A\)[/tex] and [tex]\(B\)[/tex] affect the initial reaction rate. Here's a step-by-step explanation:
1. Examine the experimental data:
- Experiment 1: [tex]\([A] = 0.451 \, M\)[/tex], [tex]\([B] = 0.885 \, M\)[/tex], rate [tex]\( = 1.13 \, M/s\)[/tex]
- Experiment 2: [tex]\([A] = 0.451 \, M\)[/tex], [tex]\([B] = 1.77 \, M\)[/tex], rate [tex]\( = 1.13 \, M/s\)[/tex]
- Experiment 3: [tex]\([A] = 1.35 \, M\)[/tex], [tex]\([B] = 0.885 \, M\)[/tex], rate [tex]\( = 10.17 \, M/s\)[/tex]
2. Determine the order with respect to [tex]\(A\)[/tex] by comparing Experiments 1 and 3 where [tex]\([B]\)[/tex] is constant:
- [tex]\( \text{Rate}_1 = k[A]_1^n[B]_1^m \)[/tex]
- [tex]\( \text{Rate}_3 = k[A]_3^n[B]_1^m \)[/tex]
Using the provided data:
- [tex]\( \text{Rate}_1 = 1.13 \, M/s \)[/tex]
- [tex]\( \text{Rate}_3 = 10.17 \, M/s \)[/tex]
- [tex]\( [A]_1 = 0.451 \, M \)[/tex]
- [tex]\( [A]_3 = 1.35 \, M \)[/tex]
Plugging these into the rate expression ratio:
[tex]\[ \frac{\text{Rate}_3}{\text{Rate}_1} = \left(\frac{[A]_3}{[A]_1}\right)^n \][/tex]
[tex]\[ \frac{10.17}{1.13} = \left(\frac{1.35}{0.451}\right)^n \][/tex]
Solving for [tex]\(n\)[/tex]:
[tex]\[ \frac{10.17}{1.13} \approx 9 \][/tex]
[tex]\[ \left(\frac{1.35}{0.451}\right)^n \approx 3^n \][/tex]
[tex]\[ n \approx 2 \][/tex]
3. Determine the order with respect to [tex]\(B\)[/tex] by comparing Experiments 1 and 2 where [tex]\([A]\)[/tex] is constant:
- [tex]\( \text{Rate}_1 = k[A]_1^n[B]_1^m \)[/tex]
- [tex]\( \text{Rate}_2 = k[A]_1^n[B]_2^m \)[/tex]
Using the provided data:
- [tex]\( \text{Rate}_1 = 1.13 \, M/s \)[/tex]
- [tex]\( \text{Rate}_2 = 1.13 \, M/s \)[/tex]
- [tex]\( [B]_1 = 0.885 \, M \)[/tex]
- [tex]\( [B]_2 = 1.77 \, M \)[/tex]
Plugging these into the rate expression ratio:
[tex]\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[B]_2}{[B]_1}\right)^m \][/tex]
[tex]\[ \frac{1.13}{1.13} = \left(\frac{1.77}{0.885}\right)^m \][/tex]
Solving for [tex]\( m \)[/tex]:
[tex]\[ 1 = 2^m \][/tex]
[tex]\[ m = 0 \][/tex]
4. Write the rate law:
The exponents [tex]\( n \)[/tex] and [tex]\( m \)[/tex] represent the orders of the reaction with respect to [tex]\( A \)[/tex] and [tex]\( B \)[/tex] respectively. Here, [tex]\( n = 2 \)[/tex] and [tex]\( m = 0 \)[/tex]. Hence, the rate law can be written as:
[tex]\[ \text{Rate} = k[A]^2 \][/tex]
5. Verify the rate law:
Given [tex]\( n = 2 \)[/tex] and [tex]\( m = 0 \)[/tex], the most appropriate rate law from the options is:
[tex]\[ \boxed{k[A]^2} \][/tex]
1. Examine the experimental data:
- Experiment 1: [tex]\([A] = 0.451 \, M\)[/tex], [tex]\([B] = 0.885 \, M\)[/tex], rate [tex]\( = 1.13 \, M/s\)[/tex]
- Experiment 2: [tex]\([A] = 0.451 \, M\)[/tex], [tex]\([B] = 1.77 \, M\)[/tex], rate [tex]\( = 1.13 \, M/s\)[/tex]
- Experiment 3: [tex]\([A] = 1.35 \, M\)[/tex], [tex]\([B] = 0.885 \, M\)[/tex], rate [tex]\( = 10.17 \, M/s\)[/tex]
2. Determine the order with respect to [tex]\(A\)[/tex] by comparing Experiments 1 and 3 where [tex]\([B]\)[/tex] is constant:
- [tex]\( \text{Rate}_1 = k[A]_1^n[B]_1^m \)[/tex]
- [tex]\( \text{Rate}_3 = k[A]_3^n[B]_1^m \)[/tex]
Using the provided data:
- [tex]\( \text{Rate}_1 = 1.13 \, M/s \)[/tex]
- [tex]\( \text{Rate}_3 = 10.17 \, M/s \)[/tex]
- [tex]\( [A]_1 = 0.451 \, M \)[/tex]
- [tex]\( [A]_3 = 1.35 \, M \)[/tex]
Plugging these into the rate expression ratio:
[tex]\[ \frac{\text{Rate}_3}{\text{Rate}_1} = \left(\frac{[A]_3}{[A]_1}\right)^n \][/tex]
[tex]\[ \frac{10.17}{1.13} = \left(\frac{1.35}{0.451}\right)^n \][/tex]
Solving for [tex]\(n\)[/tex]:
[tex]\[ \frac{10.17}{1.13} \approx 9 \][/tex]
[tex]\[ \left(\frac{1.35}{0.451}\right)^n \approx 3^n \][/tex]
[tex]\[ n \approx 2 \][/tex]
3. Determine the order with respect to [tex]\(B\)[/tex] by comparing Experiments 1 and 2 where [tex]\([A]\)[/tex] is constant:
- [tex]\( \text{Rate}_1 = k[A]_1^n[B]_1^m \)[/tex]
- [tex]\( \text{Rate}_2 = k[A]_1^n[B]_2^m \)[/tex]
Using the provided data:
- [tex]\( \text{Rate}_1 = 1.13 \, M/s \)[/tex]
- [tex]\( \text{Rate}_2 = 1.13 \, M/s \)[/tex]
- [tex]\( [B]_1 = 0.885 \, M \)[/tex]
- [tex]\( [B]_2 = 1.77 \, M \)[/tex]
Plugging these into the rate expression ratio:
[tex]\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[B]_2}{[B]_1}\right)^m \][/tex]
[tex]\[ \frac{1.13}{1.13} = \left(\frac{1.77}{0.885}\right)^m \][/tex]
Solving for [tex]\( m \)[/tex]:
[tex]\[ 1 = 2^m \][/tex]
[tex]\[ m = 0 \][/tex]
4. Write the rate law:
The exponents [tex]\( n \)[/tex] and [tex]\( m \)[/tex] represent the orders of the reaction with respect to [tex]\( A \)[/tex] and [tex]\( B \)[/tex] respectively. Here, [tex]\( n = 2 \)[/tex] and [tex]\( m = 0 \)[/tex]. Hence, the rate law can be written as:
[tex]\[ \text{Rate} = k[A]^2 \][/tex]
5. Verify the rate law:
Given [tex]\( n = 2 \)[/tex] and [tex]\( m = 0 \)[/tex], the most appropriate rate law from the options is:
[tex]\[ \boxed{k[A]^2} \][/tex]
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