Get the answers you need from a community of experts on IDNLearn.com. Join our community to receive timely and reliable responses to your questions from knowledgeable professionals.
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]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for visiting IDNLearn.com. We’re here to provide dependable answers, so visit us again soon.
