Chemical Principles: The Quest for Insight
Chemical Principles: The Quest for Insight
7th Edition
ISBN: 9781464183959
Author: Peter Atkins, Loretta Jones, Leroy Laverman
Publisher: W. H. Freeman
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Chapter 7, Problem 7B.2AST
Interpretation Introduction

Interpretation:

The data given for the decomposition of N2O5 at 25°C follows first order kinetics has to be confirmed and the rate constant for the decomposition of N2O5 has to be determined.

Concept Introduction:

According to the integrated rate law for the first order reaction, the concentration of reactant is the exponential function of time.  The two equations that represent the integrated rate law for the first order kinetics is shown below.

    ln[A]t=ln[A]0krt

The plot of natural logarithm of concentration of reactant against time is the straight line and the slope of the graph is kr.  Therefore, the rate constant for the first order reaction is equal to the negative of slope of concentration against time.

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Predict the products of this organic reaction: O N IN A N + H2O + HCI ? Specifically, in the drawing area below draw the skeletal ("line") structure of the product, or products, of this reaction. If there's more than one product, draw them in any arrangement you like, so long as they aren't touching. If there aren't any products because this reaction won't happen, check the No reaction box under the drawing area. 田 C + Explanation Check Click and drag to start drawing a structure. C © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center

Chapter 7 Solutions

Chemical Principles: The Quest for Insight

Ch. 7 - Prob. 7A.3ECh. 7 - Prob. 7A.4ECh. 7 - Prob. 7A.7ECh. 7 - Prob. 7A.8ECh. 7 - Prob. 7A.9ECh. 7 - Prob. 7A.10ECh. 7 - Prob. 7A.11ECh. 7 - Prob. 7A.12ECh. 7 - Prob. 7A.13ECh. 7 - Prob. 7A.14ECh. 7 - Prob. 7A.15ECh. 7 - Prob. 7A.16ECh. 7 - Prob. 7A.17ECh. 7 - Prob. 7A.18ECh. 7 - Prob. 7B.1ASTCh. 7 - Prob. 7B.1BSTCh. 7 - Prob. 7B.2ASTCh. 7 - Prob. 7B.2BSTCh. 7 - Prob. 7B.3ASTCh. 7 - Prob. 7B.3BSTCh. 7 - Prob. 7B.4ASTCh. 7 - Prob. 7B.4BSTCh. 7 - Prob. 7B.5ASTCh. 7 - Prob. 7B.5BSTCh. 7 - Prob. 7B.1ECh. 7 - Prob. 7B.2ECh. 7 - Prob. 7B.3ECh. 7 - Prob. 7B.4ECh. 7 - Prob. 7B.5ECh. 7 - Prob. 7B.6ECh. 7 - Prob. 7B.7ECh. 7 - Prob. 7B.8ECh. 7 - Prob. 7B.9ECh. 7 - Prob. 7B.10ECh. 7 - Prob. 7B.13ECh. 7 - Prob. 7B.14ECh. 7 - Prob. 7B.15ECh. 7 - Prob. 7B.16ECh. 7 - Prob. 7B.17ECh. 7 - Prob. 7B.18ECh. 7 - Prob. 7B.19ECh. 7 - Prob. 7B.20ECh. 7 - Prob. 7B.21ECh. 7 - Prob. 7B.22ECh. 7 - Prob. 7C.1ASTCh. 7 - Prob. 7C.1BSTCh. 7 - Prob. 7C.2ASTCh. 7 - Prob. 7C.2BSTCh. 7 - Prob. 7C.1ECh. 7 - Prob. 7C.2ECh. 7 - Prob. 7C.3ECh. 7 - Prob. 7C.4ECh. 7 - Prob. 7C.5ECh. 7 - Prob. 7C.6ECh. 7 - Prob. 7C.7ECh. 7 - Prob. 7C.8ECh. 7 - Prob. 7C.9ECh. 7 - Prob. 7C.11ECh. 7 - Prob. 7C.12ECh. 7 - Prob. 7D.1ASTCh. 7 - Prob. 7D.1BSTCh. 7 - Prob. 7D.2ASTCh. 7 - Prob. 7D.2BSTCh. 7 - Prob. 7D.1ECh. 7 - Prob. 7D.2ECh. 7 - Prob. 7D.3ECh. 7 - Prob. 7D.5ECh. 7 - Prob. 7D.6ECh. 7 - Prob. 7D.7ECh. 7 - Prob. 7D.8ECh. 7 - Prob. 7E.1ASTCh. 7 - Prob. 7E.1BSTCh. 7 - Prob. 7E.1ECh. 7 - Prob. 7E.2ECh. 7 - Prob. 7E.3ECh. 7 - Prob. 7E.4ECh. 7 - Prob. 7E.5ECh. 7 - Prob. 7E.6ECh. 7 - Prob. 7E.7ECh. 7 - Prob. 7E.8ECh. 7 - Prob. 7E.9ECh. 7 - Prob. 1OCECh. 7 - Prob. 7.1ECh. 7 - Prob. 7.2ECh. 7 - Prob. 7.3ECh. 7 - Prob. 7.4ECh. 7 - Prob. 7.5ECh. 7 - Prob. 7.6ECh. 7 - Prob. 7.7ECh. 7 - Prob. 7.9ECh. 7 - Prob. 7.11ECh. 7 - Prob. 7.14ECh. 7 - Prob. 7.15ECh. 7 - Prob. 7.17ECh. 7 - Prob. 7.19ECh. 7 - Prob. 7.20ECh. 7 - Prob. 7.23ECh. 7 - Prob. 7.25ECh. 7 - Prob. 7.26ECh. 7 - Prob. 7.29ECh. 7 - Prob. 7.30ECh. 7 - Prob. 7.31E
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Kinetics: Initial Rates and Integrated Rate Laws; Author: Professor Dave Explains;https://www.youtube.com/watch?v=wYqQCojggyM;License: Standard YouTube License, CC-BY