The rate constant for this second-order reaction is 0.780 M-·s- at 300 °C. A → products How long, in seconds, would it take for the concentration of A to decrease from 0.740 M to 0.260 M?
The rate constant for this second-order reaction is 0.780 M-·s- at 300 °C. A → products How long, in seconds, would it take for the concentration of A to decrease from 0.740 M to 0.260 M?
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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![### Second Order Reaction Kinetics
#### Problem Statement
The rate constant for this second-order reaction is \(0.780 \, \text{M}^{-1} \cdot \text{s}^{-1}\) at 300 °C.
\[
\text{A} \rightarrow \text{products}
\]
How long, in seconds, would it take for the concentration of A to decrease from 0.740 M to 0.260 M?
---
#### Incorrect Attempt
\[
t = \textbf{1.28} \quad \text{(Incorrect)}
\]
---
### Explanation
To solve this problem, we use the integrated rate law for a second-order reaction, which is given by:
\[
\frac{1}{[A]} = \frac{1}{[A]_0} + kt
\]
where:
- \([A]_0\) is the initial concentration of A.
- \([A]\) is the concentration of A at time \( t \).
- \( k \) is the rate constant.
Given data:
- \([A]_0\) = 0.740 M
- \([A]\) = 0.260 M
- \( k \) = 0.780 M\(^{-1}\) s\(^{-1}\)
Substitute these values into the integrated rate law formula to solve for \( t \):
\[
\frac{1}{0.260} - \frac{1}{0.740} = 0.780 \cdot t
\]
---
By solving the equation, you can find the correct time \( t \) that it takes for the concentration of A to decrease to the given value.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff00f8b75-2bac-4335-a200-62a4fa07a89d%2F275d6e01-902f-4582-b192-821242e1e55c%2Ffztrlir_processed.png&w=3840&q=75)
Transcribed Image Text:### Second Order Reaction Kinetics
#### Problem Statement
The rate constant for this second-order reaction is \(0.780 \, \text{M}^{-1} \cdot \text{s}^{-1}\) at 300 °C.
\[
\text{A} \rightarrow \text{products}
\]
How long, in seconds, would it take for the concentration of A to decrease from 0.740 M to 0.260 M?
---
#### Incorrect Attempt
\[
t = \textbf{1.28} \quad \text{(Incorrect)}
\]
---
### Explanation
To solve this problem, we use the integrated rate law for a second-order reaction, which is given by:
\[
\frac{1}{[A]} = \frac{1}{[A]_0} + kt
\]
where:
- \([A]_0\) is the initial concentration of A.
- \([A]\) is the concentration of A at time \( t \).
- \( k \) is the rate constant.
Given data:
- \([A]_0\) = 0.740 M
- \([A]\) = 0.260 M
- \( k \) = 0.780 M\(^{-1}\) s\(^{-1}\)
Substitute these values into the integrated rate law formula to solve for \( t \):
\[
\frac{1}{0.260} - \frac{1}{0.740} = 0.780 \cdot t
\]
---
By solving the equation, you can find the correct time \( t \) that it takes for the concentration of A to decrease to the given value.
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