Chemical Principles: The Quest for Insight
7th Edition
ISBN: 9781464183959
Author: Peter Atkins, Loretta Jones, Leroy Laverman
Publisher: W. H. Freeman
expand_more
expand_more
format_list_bulleted
Videos
Question
Chapter 7, Problem 7B.8E
(a)
Interpretation Introduction
Interpretation:
The time taken by
Concept Introduction:
According to the integrated rate law for the first order reaction, the concentration of reactant is the exponential function of time. The equation that represents the integrated rate law for the first order kinetics is shown below.
The half-life of the particular
(b)
Interpretation Introduction
Interpretation:
The time taken by
Concept Introduction:
Same as part (a).
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
The equilibrium constant for the binding of a drug molecule to a protein was measured as 200. In a separate experiment, the rate constant for the binding process, which is second order overall, was found to be 1.5 × 108 dm3 mol–1 s–1. What is the rate constant for the first-order dissociation of the drug molecule from the protein–drug complex?
A second-order-reaction of the type A +2 B→ P was carried out in a
solution that was initially 0.075 mol dm3 in A and 0.030 mol dm3 in B. After
1.0 h the concentration of A had fallen to 0.045 mol dm3. (a) Calculate the
rate constant. (b) What is the half-life of the reactants?
The rate constant for the decomposition of a certain substance is
1.70 x 10-2 dm³ mol-¹ s¹ at 24°C and 2.01 x 10-2 dm³ mol-¹ s¹ at 37°C.
Evaluate the Arrhenius parameters of the reaction.
A reaction 2 AP has a third-order rate law with k = 3.50 x 104 dmº
mol2 s¹. Calculate the time required for the concentration of A to change
from 0.077 mol dm3 to 0.021 mol dm³.
The degradation of the antibiotic clindamycin stored at 343 K in aqueous solution at pH 4 is found to be first order with a rate constant of 2.49 x 10−7 s −1. Over the temperature range 320 K to 360 K theactivation energy was found to be 123.3 kJ mol−1.
(a) Calculate the rate constant at 325 K.(b) The threshold for product safety is 1% degradation. At 295 K the time taken for 1% of the antibiotic to degrade is found to be close to 0.01/ k. Comment on the shelf-life of the drug.
Chapter 7 Solutions
Chemical Principles: The Quest for Insight
Ch. 7 - Prob. 7A.1ASTCh. 7 - Prob. 7A.1BSTCh. 7 - Prob. 7A.2ASTCh. 7 - Prob. 7A.2BSTCh. 7 - Prob. 7A.3ASTCh. 7 - Prob. 7A.3BSTCh. 7 - Prob. 7A.4ASTCh. 7 - Prob. 7A.4BSTCh. 7 - Prob. 7A.1ECh. 7 - Prob. 7A.2E
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
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Similar questions
- Explain why half-lives are not normally used to describe reactions other than first order.arrow_forwardThe rate law for the reaction A + 3B → 2C + D was found to be v = k[A]2[B]2. (a) What are the units of k? (b) Express the rate law in terms of the rates of formation and consumption of A and C, respectively.arrow_forwardThe activation energy for the isomerization reaction CH3NC(g) → CH3CN(g) is 161 kJ mol−¹, and the reaction rate constant at 600 K is 0.41 s¯¹. (a) What is the value of the pre-exponential factor A for this reaction? (b) What is the value of the rate constant of the reaction at 1000 K?arrow_forward
- (d) The first-order rate constants for the fluorescence and phosphorescence of molecule X are 4.5 × 10" s' and 0.50 s', respectively. Briefly explain the large order of magnitude difference in the rate constant values. (e) Explain how long (in s) it takes for 1.0% of fluorescence and phosphorescence to occur following termination of excitation.arrow_forwardA second-order reaction of the type A + 2 B → P was carried out in a solution that was initially 0.050 mol dm−3 in A and 0.030 mol dm−3 in B. After 1.0 h the concentration of A had fallen to 0.040 mol dm−3. (a) Calculate the rate constant. (b) What is the half-life of each reactant?arrow_forward(iv) The activation energy, Ea, and pre-exponential factor, A, for the decomposition of N2O5: N2O5→ 2 NO2 + ½ O2 are: E = 102.2 kJ mol-1 and A = 2.81 x 1013 s-1. (a) Using these data calculate the rate constant of the reaction at 300 K. (b) Assuming the reaction is first order calculate the rate of the reaction, at 300 K, when the concentration of N2O5 is 0.015 mol L-1.arrow_forward
- (a) Define the following terms:(i) Collision frequency(ii) Rate constant (k)(b) The rate constant of a first order reaction increases from 4 X 10-2 to 24 X 10-2 when the temperature changes from 300 K to 350 K. Calculate the energy of activation (JEa). (log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021, log 6 = 0.7782)arrow_forwardAcetone is one of the most important solvents in organicchemistry, used to dissolve everything from fats and waxes toairplane glue and nail polish. At high temperatures, it decom-poses in a first-order process to methane and ketene(CH2=C=O). At 600C, the rate constant is 8.7x10^-3s⁻¹.(a) What is the half-life of the reaction?(b) How long does it take for 40.% of a sample of acetone todecompose?(c) How long does it take for 90.% of a sample of acetone todecompose?arrow_forward(a) Explain the meaning of the sentence:“The velocity laws of reactions are empirical” (b) It can be determined that the velocity law of a generic reaction A +B→Pév = k[A]x[B]y. Plot on a graph the variation of [A] and [P] with time, and explain why v= d[P]/dt = -d[A]/dt. (c) Explain the meaning of the terms k, x and y, in the velocity law, presented in item b: (d) Explain what it means in practice for a velocity law to be of zero order. Plot ,A as a function of t, for the zero-order reaction A→Products. (e) Explain what elementary reactions are and how they can be classified.arrow_forward
arrow_back_ios
arrow_forward_ios
Recommended textbooks for you
Principles of Modern ChemistryChemistryISBN:9781305079113Author:David W. Oxtoby, H. Pat Gillis, Laurie J. ButlerPublisher:Cengage Learning
Chemistry for Engineering StudentsChemistryISBN:9781337398909Author:Lawrence S. Brown, Tom HolmePublisher:Cengage Learning
ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning
Chemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning
Principles of Modern Chemistry
Chemistry
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:Cengage Learning
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Kinetics: Initial Rates and Integrated Rate Laws; Author: Professor Dave Explains;https://www.youtube.com/watch?v=wYqQCojggyM;License: Standard YouTube License, CC-BY