Hello I need help with these question: based on the example I included, what other trials and different concentrations could I have used to end up with the same overall reaction order of 2, and what would be the rate law formula.
Hello I need help with these question: based on the example I included, what other trials and different concentrations could I have used to end up with the same overall reaction order of 2, and what would be the rate law formula.
Chemistry
10th Edition
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
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Hello I need help with these question: based on the example I included, what other trials and different concentrations could I have used to end up with the same overall reaction order of 2, and what would be the rate law formula.
![## Determination of Reaction Order
### Part (a)
The equation given is:
\[
\frac{262}{131} = \frac{k[0.10M]^a [0.10M]^x}{k[0.20M]^a [0.10M]^x}
\]
Simplifying the equation:
\[
\frac{2}{1} = \frac{0.5^a}{0.5^x} = 4
\]
From this, we can determine that:
\[
\left(\frac{1}{262}\right) \div \left(\frac{1}{131}\right) = 0.5^a
\]
Simplifying further:
\[
\frac{0.5}{0.5} = 0.5^a
\]
Hence:
\[
1 = a
\]
### Part (b)
The equation here is:
\[
\frac{262}{131} = \frac{k[0.10M]^a [0.10M]^b}{k[0.10M]^a [0.20M]^b}
\]
Simplifying the equation:
\[
\frac{2}{1} = \frac{0.5^b}{0.5^b} = 4
\]
Thus:
\[
\left(\frac{1}{262}\right) \div \left(\frac{1}{131}\right) = 0.5^b
\]
Simplifying further:
\[
\frac{0.5}{0.5} = 0.5^b
\]
Hence:
\[
1 = b
\]
### Conclusion
The overall order of the reaction is:
\[
\text{Overall order} = 2
\]
This analysis determines the individual reaction orders with respect to each reactant, concluding with an overall reaction order of 2.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F08bb2a94-da17-43f2-90ab-4ad444df37b3%2Fc9b01834-c1c9-46b3-a0c9-603dac568d6d%2F53ubpgm_processed.png&w=3840&q=75)
Transcribed Image Text:## Determination of Reaction Order
### Part (a)
The equation given is:
\[
\frac{262}{131} = \frac{k[0.10M]^a [0.10M]^x}{k[0.20M]^a [0.10M]^x}
\]
Simplifying the equation:
\[
\frac{2}{1} = \frac{0.5^a}{0.5^x} = 4
\]
From this, we can determine that:
\[
\left(\frac{1}{262}\right) \div \left(\frac{1}{131}\right) = 0.5^a
\]
Simplifying further:
\[
\frac{0.5}{0.5} = 0.5^a
\]
Hence:
\[
1 = a
\]
### Part (b)
The equation here is:
\[
\frac{262}{131} = \frac{k[0.10M]^a [0.10M]^b}{k[0.10M]^a [0.20M]^b}
\]
Simplifying the equation:
\[
\frac{2}{1} = \frac{0.5^b}{0.5^b} = 4
\]
Thus:
\[
\left(\frac{1}{262}\right) \div \left(\frac{1}{131}\right) = 0.5^b
\]
Simplifying further:
\[
\frac{0.5}{0.5} = 0.5^b
\]
Hence:
\[
1 = b
\]
### Conclusion
The overall order of the reaction is:
\[
\text{Overall order} = 2
\]
This analysis determines the individual reaction orders with respect to each reactant, concluding with an overall reaction order of 2.
![**Prelaboratory Exercise**
1. **Given the following data, determine the orders with respect to the concentrations of substances A and B in the reaction:**
\[
A + B \rightarrow \text{products}
\]
| [A]initial | [B]initial | Time for reaction |
|------------|------------|-------------------|
| 0.10 M | 0.10 M | 262 s |
| 0.20 M | 0.10 M | 131 s |
| 0.30 M | 0.10 M | 87 s |
| 0.20 M | 0.20 M | 66 s |
| 0.10 M | 0.20 M | 131 s |
*Instructions*: Use the above data to calculate the reaction orders with respect to substances A and B. Analyze how the change in concentration affects the reaction time.
2. **What must be the units of the specific rate constant in Prelaboratory Question 1, given that the rate of reaction would be measured in M/s?**
*Hint*: Consider the general rate law and the units involved in the calculations to determine the units of the rate constant.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F08bb2a94-da17-43f2-90ab-4ad444df37b3%2Fc9b01834-c1c9-46b3-a0c9-603dac568d6d%2F7nf1kxf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Prelaboratory Exercise**
1. **Given the following data, determine the orders with respect to the concentrations of substances A and B in the reaction:**
\[
A + B \rightarrow \text{products}
\]
| [A]initial | [B]initial | Time for reaction |
|------------|------------|-------------------|
| 0.10 M | 0.10 M | 262 s |
| 0.20 M | 0.10 M | 131 s |
| 0.30 M | 0.10 M | 87 s |
| 0.20 M | 0.20 M | 66 s |
| 0.10 M | 0.20 M | 131 s |
*Instructions*: Use the above data to calculate the reaction orders with respect to substances A and B. Analyze how the change in concentration affects the reaction time.
2. **What must be the units of the specific rate constant in Prelaboratory Question 1, given that the rate of reaction would be measured in M/s?**
*Hint*: Consider the general rate law and the units involved in the calculations to determine the units of the rate constant.
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