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.

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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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
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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