The oxidation of nitrogen monoxide by oxygen at 25 °C 2 NO + 022 NO2 is second order in NO and first order in O2. Complete the rate law for this reaction in the box below. Use the form k[A]"B]".. , where '1' is understood for m, n . (don't enter 1) and concentrations taken to the zero power do not appear. Rate = In an experiment to determine the rate law, the rate of the reaction was determined to be 6.91×10o-4 Ms1 when [NO] = 5.26x103 M and [02] = 2.53x103 M. From this experiment, the rate constant is
The oxidation of nitrogen monoxide by oxygen at 25 °C 2 NO + 022 NO2 is second order in NO and first order in O2. Complete the rate law for this reaction in the box below. Use the form k[A]"B]".. , where '1' is understood for m, n . (don't enter 1) and concentrations taken to the zero power do not appear. Rate = In an experiment to determine the rate law, the rate of the reaction was determined to be 6.91×10o-4 Ms1 when [NO] = 5.26x103 M and [02] = 2.53x103 M. From this experiment, the rate constant is
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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![**The Oxidation of Nitrogen Monoxide by Oxygen at 25°C**
**Reaction:**
\[ 2 \text{NO} + \text{O}_2 \rightarrow 2 \text{NO}_2 \]
This reaction is second order in \(\text{NO}\) and first order in \(\text{O}_2\).
**Rate Law Completion:**
To write the rate law for this reaction, use the form \( k[\text{A}]^m[\text{B}]^n \), where '1' is understood for \( m, n \) (don't enter 1), and concentrations taken to the zero power do not appear.
**Rate =**
\[ k[\text{NO}]^2[\text{O}_2] \]
**Experimental Data:**
In an experiment to determine the rate law, the rate of the reaction was determined to be \( 6.91 \times 10^{-4} \, \text{M s}^{-1} \) when \([\text{NO}] = 5.26 \times 10^{-3} \, \text{M} \) and \([\text{O}_2] = 2.53 \times 10^{-3} \, \text{M} \). From this experiment, the rate constant is \( \boxed{k} \, \text{M}^{-2}\text{s}^{-1} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc4eca0ea-f525-4d8f-9259-04e892974916%2F6d4445c7-fbaa-4bb7-8418-4962efcb5ac1%2Fsijhn2k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**The Oxidation of Nitrogen Monoxide by Oxygen at 25°C**
**Reaction:**
\[ 2 \text{NO} + \text{O}_2 \rightarrow 2 \text{NO}_2 \]
This reaction is second order in \(\text{NO}\) and first order in \(\text{O}_2\).
**Rate Law Completion:**
To write the rate law for this reaction, use the form \( k[\text{A}]^m[\text{B}]^n \), where '1' is understood for \( m, n \) (don't enter 1), and concentrations taken to the zero power do not appear.
**Rate =**
\[ k[\text{NO}]^2[\text{O}_2] \]
**Experimental Data:**
In an experiment to determine the rate law, the rate of the reaction was determined to be \( 6.91 \times 10^{-4} \, \text{M s}^{-1} \) when \([\text{NO}] = 5.26 \times 10^{-3} \, \text{M} \) and \([\text{O}_2] = 2.53 \times 10^{-3} \, \text{M} \). From this experiment, the rate constant is \( \boxed{k} \, \text{M}^{-2}\text{s}^{-1} \).
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