ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM
7th Edition
ISBN: 9781319399849
Author: ATKINS
Publisher: MAC HIGHER
expand_more
expand_more
format_list_bulleted
Question
Chapter 7, Problem 7B.17E
(a)
Interpretation Introduction
Interpretation:
The time taken by given second order reaction to take place has to be determined.
Concept Introduction:
The equation that represents the integrated rate law for the second order kinetics is shown below. The unit for the rate constant for the second order reaction is
(b)
Interpretation Introduction
Interpretation:
The time taken by given second order reaction to takes place has to be determined.
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 decomposition of XY is second order in XY and has a rate constant of 7.41 × 10−3 L·mol−1·s−1 at a certain temperature, the half-life for this reaction at an initial concentration of 0.101 mol·L−1 1336. A) If the initial concentration of XY is 0.225 mol·L−1, how long will it take for the concentration to decrease to 6.95 × 10−2 mol·L−1 ?, B) If the initial concentration of XY is 0.080 mol·L−1, what is the concentration of XY after 75 s ?
Lanthanum(III) phosphate crystallizes as a hemihydrate,
LAPO4 · H20. When it is heated, it loses water to give
anhydrous lanthanum(III) phosphate:
2(LAPO4 · H2O(s)) → 2 LaPO4 (s) + H20(g)
This reaction is first order in the chemical amount of
LAPO, · H2O. The rate constant varies with tempera-
ture as follows:
Temperature (°C)
k (s-1)
205
2.3 x 10-4
219
3.69 x 10-4
246
7.75 x 10-4
260
12.3 x 10-4
Compute the activation energy of this reaction.
6. The rate constant for the reaction, 2 N₂O5 (g) → 4 NO2 (g) + O2 (g), doubles when the
temperature is raised from 295.65 K to 300.62 K.
(a) Determine the activation energy (in kJ/mol) for the reaction, assuming that the pre-
exponential factor, A, in the Arrhenius equation is independent of temperature.
(b) At what temperature would you predict this rate constant to increase by another factor
of 10 relative to its value at 300.62 K?
Chapter 7 Solutions
ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM
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
Similar questions
- The label on a bottle of 3% (by volume) hydrogen peroxide, H2O2, purchased at a grocery store, states that the solution should be stored in a cool, dark place. H2O2decomposes slowly over time, and the rate of decomposition increases with an increase in temperature and in the presence of light. However, the rate of decomposition increases dramatically if a small amount of powdered MnO- is added to the solution. The decomposition products are H2O and O2. MnO2 is not consumed in the reaction. Write the equation for the decomposition of H2O2. What role does MnO2 play? In the chemistry lab, a student substituted a chunk of MnO2 for the powdered compound. The reaction rate was not appreciably increased. WTiat is one possible explanation for this observation? Is MnO2 part of the stoichiometry of the decomposition of H2O2?arrow_forwardAt 573 K, gaseous NO2(g) decomposes, forming NO(g) and O2(g). If a vessel containing NO2(g) has an initial concentration of 1.9 102 mol/L, how long will it take for 75% of the NO2(g) to decompose? The decomposition of NO2(g) is second-order in the reactant and the rate constant for this reaction, at 573 K, is 1.1 L/mol s.arrow_forwardThe decomposition of iodoethane in the gas phase proceeds according to the following equation: C2H5I(g)C2H4(g)+HI(g) At 660. K, k = 7.2 104 sl; at 720. K, k = 1.7 102 sl. What is the value of the rate constant for this first-order decomposition at 325C? If the initial pressure of iodoethane is 894 torr at 245C, what is the pressure of iodoethane after three half-lives?arrow_forward
- The Raschig reaction produces the industrially important reducing agent hydrazine, N2H4, from ammonia, NH3, and hypochlorite ion, OCl−, in basic aqueous solution. A proposed mechanism is Step 1: Step 2: Step 3: What is the overall stoichiometric equation? Which step is rate-limiting? What reaction intermediates are involved? What rate law is predicted by this mechanism?arrow_forwardAssume that the formation of nitrogen dioxide: 2NO(g) + O2(g) 2NO2(g) is an elementary reaction. (a) Write the rate law for this reaction. (b) A sample of air at a certain temperature is contaminated with 2.0 ppm of NO by volume. Under these conditions, can the rate law be simplified? If so, write the simplified rate law. (c) Under the conditions described in part (b), the half-life of the reaction has been estimated to be 6.4 × 103 min. What would the half-life be if the initial concentration of NO were 10 ppm?arrow_forwardAssume that the formation of nitrogen dioxide, 2 NO(g) + 02(g) – 2 NO2(g) is an elementary reaction. (a) Write the rate law for this reaction. (Rate expressions take the general form: rate = k. [A]ª . [B]b.) chemPad О Help Greek - rate=k•[NO]2.[02] rate=k*[NO]^2*[O_2] Correct. (b) A sample of air at a certain temperature is contaminated with 1.9 ppm of NO by volume. Under these conditions, can the rate law be simplified? If so, write the simplified rate law. If not, repeat your answer from above. (Rate expressions take the general form: rate = k . [A]ª . [B]b. Use k' for the new rate constant as needed.) chemPad O Help Greek - rate=k':[NO]2 rate=k*[NO]^2 Correct. (c) Under the conditions described in (b), the half-life of the reaction has been estimated to be 6.7x103 min. What would the half-life be if the initial concentration of NO were 12.4 ppm? 4.0 |1030192 X min Supporting Materials Periodic Table Constants and E Supplemental Dataarrow_forward
- The following kinetic data are collected for the initial rates of a reaction 2 X + Z→ products: Experiment [X ]o(M) [Z]o(M) Rate (M/s) 0.25 0.25 4.0 x 10! 0.50 0.50 3.2 x 102 0.50 0.75 7.2 x 102 (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of X is 0.75 M and that of Z is 1.25 M? 2. 3.arrow_forward1 Rate constants for the first-order decomposition of acetonedicarboxylic acid CO(CH2COOH)2(aq) → CO(CH3)2(aq) + 2 CO2(g) acetonedicarboxylic acidacetone are k = 4.75 ×10–4 s–1 at 293 K and k = 1.63 ×10–3 at 303 K. What is the activation energy, Ea, for this reaction? Select one: a. 71KJ/mol b. 81KJ/mol c. 51KJ/mol d. 91kJ/molarrow_forwardUnder certain conditions the rate of this reaction is zero order in hydrogen iodide with a rate constant of 0.0069 M · s¯¹: 2 HI(g) → H₂(g) +1₂(g) Suppose a 5.0 L flask is charged under these conditions with 400. mmol of hydrogen iodide. How much is left 3.0 s later? You may assume no other reaction is important. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits. 0 00 X x10 ? olo Ararrow_forward
- Under certain conditions the rate of this reaction is zero order in hydrogen iodide with a rate constant of 0.0085 M ·s 2 HI (g) – H, (g) + I,(3) Suppose a 2.0 L flask is charged under these conditions with 350. mmol of hydrogen iodide. How much is left 5.0 s later? You may assume no other reaction is important. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits. Oarrow_forwardConsider the following reaction: 1. 2 N,O5 (g) → 4 NO, (g) + O, (g) The initial concentration of N2O5 was 0.48 mol/L, and 25 minutes after initiating the reaction, all of the N,Os has been consumed. (a) Calculate the average rate of the reaction over this 25-minute time interval. (b) Is it correct to assume that the rate law is Rate = k[N,O5]² based on the balanced chemical equation? Briefly explain your answer.arrow_forwardUnder certain conditions the rate of this reaction is zero order in hydrogen iodide with a rate constant of 0.010 M·s ': 2 HI (g) → H, (g)+I, (g) Suppose a 4.0 L flask is charged under these conditions with 100. mmol of hydrogen iodide. How much is left 0.6 s later? You may assume no other reaction is important. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits. Ox10arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Chemistry: Matter and ChangeChemistryISBN:9780078746376Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl WistromPublisher:Glencoe/McGraw-Hill School Pub Co
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher: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: Matter and Change
Chemistry
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Glencoe/McGraw-Hill School Pub Co
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
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