At a certain temperature this reaction follows first-order kinetics with a rate constant of 0.0968 s 2N,O; (g) → 2N,O, (g) + O, (g) Suppose a vessel contains N,0, at a concentration of 1.07M. Calculate the concentration of N,O, in the vessel 6.40 seconds late You may assume no other reaction is important. Round your answer to 2 significant digits.
At a certain temperature this reaction follows first-order kinetics with a rate constant of 0.0968 s 2N,O; (g) → 2N,O, (g) + O, (g) Suppose a vessel contains N,0, at a concentration of 1.07M. Calculate the concentration of N,O, in the vessel 6.40 seconds late You may assume no other reaction is important. Round your answer to 2 significant digits.
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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![### Reaction Kinetics Problem
**Problem Statement:**
At a certain temperature, this reaction follows first-order kinetics with a rate constant of \(0.0968 \, \text{s}^{-1}\):
\[ 2 \text{N}_2\text{O}_5 (g) \rightarrow 2 \text{N}_2\text{O}_4 (g) + \text{O}_2 (g) \]
Suppose a vessel contains \(\text{N}_2\text{O}_5\) at a concentration of \(1.07 \, M\). Calculate the concentration of \(\text{N}_2\text{O}_5\) in the vessel 6.40 seconds later. You may assume no other reaction is important.
Round your answer to 2 significant digits.
**Calculation:**
Use the formula for first-order reactions:
\[ [\text{A}]_t = [\text{A}]_0 \cdot e^{-kt} \]
Where:
- \([\text{A}]_t\) is the concentration of the reactant at time \(t\).
- \([\text{A}]_0\) is the initial concentration of the reactant.
- \(k\) is the rate constant.
- \(t\) is the time.
Given:
- \( [\text{N}_2\text{O}_5]_0 = 1.07 \, M \)
- \( k = 0.0968 \, \text{s}^{-1} \)
- \( t = 6.40 \, \text{s} \)
Calculate \([\text{N}_2\text{O}_5]_t\):
\[
[\text{N}_2\text{O}_5]_{6.40} = 1.07 e^{-0.0968 \times 6.40}
\]
Solve for the exponent first:
\[
0.0968 \times 6.40 = 0.61952
\]
Now, find \( e^{-0.61952} \):
\[
e^{-0.61952} \approx 0.538
\]
Finally, calculate the concentration:
\[
[\text{N}_2\text{O}_5]_{6.40} = 1.07 \times 0.538 = 0.575](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F76b6cb9f-216e-4f5f-860c-25f99efb2a99%2Fefead1cb-2e73-4ac0-9f18-cd73dcfd5475%2Fhwsx4rl_processed.png&w=3840&q=75)
Transcribed Image Text:### Reaction Kinetics Problem
**Problem Statement:**
At a certain temperature, this reaction follows first-order kinetics with a rate constant of \(0.0968 \, \text{s}^{-1}\):
\[ 2 \text{N}_2\text{O}_5 (g) \rightarrow 2 \text{N}_2\text{O}_4 (g) + \text{O}_2 (g) \]
Suppose a vessel contains \(\text{N}_2\text{O}_5\) at a concentration of \(1.07 \, M\). Calculate the concentration of \(\text{N}_2\text{O}_5\) in the vessel 6.40 seconds later. You may assume no other reaction is important.
Round your answer to 2 significant digits.
**Calculation:**
Use the formula for first-order reactions:
\[ [\text{A}]_t = [\text{A}]_0 \cdot e^{-kt} \]
Where:
- \([\text{A}]_t\) is the concentration of the reactant at time \(t\).
- \([\text{A}]_0\) is the initial concentration of the reactant.
- \(k\) is the rate constant.
- \(t\) is the time.
Given:
- \( [\text{N}_2\text{O}_5]_0 = 1.07 \, M \)
- \( k = 0.0968 \, \text{s}^{-1} \)
- \( t = 6.40 \, \text{s} \)
Calculate \([\text{N}_2\text{O}_5]_t\):
\[
[\text{N}_2\text{O}_5]_{6.40} = 1.07 e^{-0.0968 \times 6.40}
\]
Solve for the exponent first:
\[
0.0968 \times 6.40 = 0.61952
\]
Now, find \( e^{-0.61952} \):
\[
e^{-0.61952} \approx 0.538
\]
Finally, calculate the concentration:
\[
[\text{N}_2\text{O}_5]_{6.40} = 1.07 \times 0.538 = 0.575
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