Use the References to access inmportant values if needed for this question. The gas phase decomposition of nitrogen dioxide at 383 °C NO2(g)NO(g) + ½ O2(g) is second order in NO2 with a rate constant of 0.540 Ms. If the initial concentration of NO, is 0.446 M, the concentration of NO2 will be M after 20.7 seconds have passed.
Use the References to access inmportant values if needed for this question. The gas phase decomposition of nitrogen dioxide at 383 °C NO2(g)NO(g) + ½ O2(g) is second order in NO2 with a rate constant of 0.540 Ms. If the initial concentration of NO, is 0.446 M, the concentration of NO2 will be M after 20.7 seconds have passed.
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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![**Gas Phase Decomposition of Nitrogen Dioxide**
In this exercise, we will explore the gas phase decomposition of nitrogen dioxide at a temperature of 383 °C. The decomposition reaction is represented by the following chemical equation:
\[ \text{NO}_2(\text{g}) \rightarrow \text{NO}(\text{g}) + \frac{1}{2} \text{O}_2(\text{g}) \]
This reaction is second-order with respect to \(\text{NO}_2\), and the rate constant (\(k\)) for this reaction is \(0.540 \, \text{M}^{-1} \text{s}^{-1}\).
**Problem Statement:**
Determine the concentration of \(\text{NO}_2\) after 20.7 seconds, given that the initial concentration of \(\text{NO}_2\) is \(0.446 \, \text{M}\).
**Calculation:**
For a second-order reaction, the relationship between the concentration of reactant and time is given by:
\[ \frac{1}{[\text{NO}_2]} = \frac{1}{[\text{NO}_2]_0} + kt \]
Here:
- \([\text{NO}_2]\) is the concentration of \(\text{NO}_2\) at time \(t\)
- \([\text{NO}_2]_0\) is the initial concentration of \(\text{NO}_2\)
- \(k\) is the rate constant
- \(t\) is the time
Given data:
- \([\text{NO}_2]_0 = 0.446 \, \text{M}\)
- \(k = 0.540 \, \text{M}^{-1} \text{s}^{-1}\)
- \(t = 20.7 \, \text{s}\)
To find the concentration of \(\text{NO}_2\) after 20.7 seconds, substitute the given values into the second-order rate equation:
\[ \frac{1}{[\text{NO}_2]} = \frac{1}{0.446 \, \text{M}} + (0.540 \, \text{M}^{-1} \text{s}^{-1})(20.7 \, \text{s}) \]
Calculate the term on the right-hand side:
\[ \frac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F848b3b37-0e47-4626-8c62-5eaa520ec29b%2F36460baf-e3e2-4b25-adb4-fd9b1a2a48e3%2Flqy06eg.jpeg&w=3840&q=75)
Transcribed Image Text:**Gas Phase Decomposition of Nitrogen Dioxide**
In this exercise, we will explore the gas phase decomposition of nitrogen dioxide at a temperature of 383 °C. The decomposition reaction is represented by the following chemical equation:
\[ \text{NO}_2(\text{g}) \rightarrow \text{NO}(\text{g}) + \frac{1}{2} \text{O}_2(\text{g}) \]
This reaction is second-order with respect to \(\text{NO}_2\), and the rate constant (\(k\)) for this reaction is \(0.540 \, \text{M}^{-1} \text{s}^{-1}\).
**Problem Statement:**
Determine the concentration of \(\text{NO}_2\) after 20.7 seconds, given that the initial concentration of \(\text{NO}_2\) is \(0.446 \, \text{M}\).
**Calculation:**
For a second-order reaction, the relationship between the concentration of reactant and time is given by:
\[ \frac{1}{[\text{NO}_2]} = \frac{1}{[\text{NO}_2]_0} + kt \]
Here:
- \([\text{NO}_2]\) is the concentration of \(\text{NO}_2\) at time \(t\)
- \([\text{NO}_2]_0\) is the initial concentration of \(\text{NO}_2\)
- \(k\) is the rate constant
- \(t\) is the time
Given data:
- \([\text{NO}_2]_0 = 0.446 \, \text{M}\)
- \(k = 0.540 \, \text{M}^{-1} \text{s}^{-1}\)
- \(t = 20.7 \, \text{s}\)
To find the concentration of \(\text{NO}_2\) after 20.7 seconds, substitute the given values into the second-order rate equation:
\[ \frac{1}{[\text{NO}_2]} = \frac{1}{0.446 \, \text{M}} + (0.540 \, \text{M}^{-1} \text{s}^{-1})(20.7 \, \text{s}) \]
Calculate the term on the right-hand side:
\[ \frac
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