-1 -1 At a certain temperature this reaction follows second-order kinetics with a rate constant of 4.91 M 'S : 2HI (g) → H, (g) +I, (g) Suppose a vessel contains HI at a concentration of 1.33M. Calculate the concentration of HI in the vessel 2.40 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits. OM ?
-1 -1 At a certain temperature this reaction follows second-order kinetics with a rate constant of 4.91 M 'S : 2HI (g) → H, (g) +I, (g) Suppose a vessel contains HI at a concentration of 1.33M. Calculate the concentration of HI in the vessel 2.40 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits. OM ?
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 Educational Module
#### Reaction Overview
At a certain temperature, this reaction follows second-order kinetics with a rate constant of \(4.91 \, M^{-1} \cdot s^{-1}\):
\[ 2\text{HI (g)} \rightarrow \text{H}_2 \text{(g)} + \text{I}_2 \text{(g)} \]
#### Problem Statement
Suppose a vessel contains HI at a concentration of \(1.33 \, M\). Calculate the concentration of HI in the vessel \(2.40\) seconds later. You may assume no other reaction is important.
Round your answer to 2 significant digits.
#### Inputs and Interface
- **Initial Concentration (\( [HI]_0 \)**): \(1.33 \, M\)
- **Time (t): 2.40 \, \text{s}**
- **Rate Constant (k):** \(4.91 \, M^{-1} \cdot s^{-1}\)
- **Required Output:** Concentration of HI \( [HI] \, \text{(M)} \) after \(2.40 \, \text{s}\)
#### Solving the Problem
To calculate the concentration of HI after a given time period for a second-order reaction, the integrated rate law for second-order kinetics is used:
\[
\frac{1}{[HI]_t} = \frac{1}{[HI]_0} + kt
\]
Where:
- \([HI]_t\) is the concentration of HI at time t
- \([HI]_0\) is the initial concentration of HI
- \(k\) is the rate constant
- \(t\) is the elapsed time
Rearranging for \([HI]_t\):
\[
[HI]_t = \frac{1}{ \frac{1}{[HI]_0} + kt }
\]
Substituting the given values:
\[
[HI]_t = \frac{1}{ \frac{1}{1.33} + (4.91 \times 2.40) }
\]
Perform the calculations to find the concentration of HI:
\[
[HI]_t = \frac{1}{ 0.7519 + 11.784 } = \frac{1}{ 12.5359 } \approx 0.079](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe3bd40d5-d460-480c-9c59-6ef7376a4584%2F8156b39a-5cec-4adb-8247-9addcad30e6a%2F7tw7wf7_processed.png&w=3840&q=75)
Transcribed Image Text:### Second-Order Reaction Kinetics Educational Module
#### Reaction Overview
At a certain temperature, this reaction follows second-order kinetics with a rate constant of \(4.91 \, M^{-1} \cdot s^{-1}\):
\[ 2\text{HI (g)} \rightarrow \text{H}_2 \text{(g)} + \text{I}_2 \text{(g)} \]
#### Problem Statement
Suppose a vessel contains HI at a concentration of \(1.33 \, M\). Calculate the concentration of HI in the vessel \(2.40\) seconds later. You may assume no other reaction is important.
Round your answer to 2 significant digits.
#### Inputs and Interface
- **Initial Concentration (\( [HI]_0 \)**): \(1.33 \, M\)
- **Time (t): 2.40 \, \text{s}**
- **Rate Constant (k):** \(4.91 \, M^{-1} \cdot s^{-1}\)
- **Required Output:** Concentration of HI \( [HI] \, \text{(M)} \) after \(2.40 \, \text{s}\)
#### Solving the Problem
To calculate the concentration of HI after a given time period for a second-order reaction, the integrated rate law for second-order kinetics is used:
\[
\frac{1}{[HI]_t} = \frac{1}{[HI]_0} + kt
\]
Where:
- \([HI]_t\) is the concentration of HI at time t
- \([HI]_0\) is the initial concentration of HI
- \(k\) is the rate constant
- \(t\) is the elapsed time
Rearranging for \([HI]_t\):
\[
[HI]_t = \frac{1}{ \frac{1}{[HI]_0} + kt }
\]
Substituting the given values:
\[
[HI]_t = \frac{1}{ \frac{1}{1.33} + (4.91 \times 2.40) }
\]
Perform the calculations to find the concentration of HI:
\[
[HI]_t = \frac{1}{ 0.7519 + 11.784 } = \frac{1}{ 12.5359 } \approx 0.079
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