Chemistry for Engineering Students
Chemistry for Engineering Students
4th Edition
ISBN: 9781337398909
Author: Lawrence S. Brown, Tom Holme
Publisher: Cengage Learning
bartleby

Videos

Textbook Question
Chapter 11, Problem 11.36PAE

The reaction

NO(g) + O,(g) — NO,(g) + 0(g)

plays a role in the formation of nitrogen dioxide in automobile engines. Suppose that a series of experiments measured the rate of this reaction at 500 K and produced the following data;

    [NO]

(mol L ’)[OJ

(mol L 1)Rate = -A[NO]/Af (mol L_1 s-1) 0.002 0.005 8.0 X 10"'7 0.002 0.010 1.6 X 10-'6 0.006 0.005 2.4 X IO-'6

Derive a rate law for the reaction and determine the value of the rate constant.

Expert Solution & Answer
Check Mark
Interpretation Introduction

Interpretation: Given the experimental data obtained for a reaction at 500K, derive the rate law for the reaction and find the value of the rate constant.

Concept Introduction: Orders of reaction are constantly determined by doing experiments. Consequently without experimental information, we can't conclude anything about the order of a reaction just by having a look at the equation for the reaction. By doing experiments involving a reaction between A and B, the rate of the reaction is identified to be related to the concentrations of A and B as follows:

rate=k[A]a[B]b ->

This is the Rate Equation.

Where,

Rate is in the units of mol dm-3s-1

k is the rate constant

A, B- concentrations in mol dm-3

a - Order of reaction with respect to A

b- Order of reaction with respect to B

If temperature is given, the rate is usually considered to be a function of the initial concentrations of the reactants A and B.

Answer to Problem 11.36PAE

Solution: The rate law of the reaction is Rate=k[NO][O2]  and the value of rate constant is 0.8

Explanation of Solution

Given information: Reaction: NO(g)+O2(g)NO2(g)+O(g), Temperature =500K

Experimental Data

Chemistry for Engineering Students, Chapter 11, Problem 11.36PAE

Step 1: For the reaction:

NO(g)+O2(g)NO2(g)+O(g)

The rate law can be determined using the rate equation as follows:

Rate= k[NO]a[O2]b 

Where,

a= Order of the reaction with respect to NO

b= Order of the reaction with respect to O2

Step 2: From the first, second and third rows of the given experimental data,

k ( 0.002 )a (0.005)b = 8.0× 10 17(1) k ( 0.002 ) a (0.010) b  = 1.6× 10 16 (2) k ( 0.006 ) a (0.005) b  = 2.4× 10 16 (3)

Step 3: Divide (2) by (1), we get

1.6× 10 16 8× 10 17 = k ( 0.002 ) a (0.010) b k ( 0.002 ) a (0.005) b 16× 10 17 8× 10 17 = k ( 0.002 ) a (0.010) b k ( 0.002 ) a (0.005) b 2= (2) b ( 2 ) 1 = ( 2 ) b b=1

Step 4: Divide (3) by (1), we get

2.4× 10 16 8× 10 17 = k ( 0.006 ) a (0.005) b k ( 0.002 ) a (0.005) b 24× 10 17 8× 10 17 = k ( 0.006 ) a (0.005) b k ( 0.002 ) a (0.005) b 3= (3) a ( 3 ) 1 = ( 3 ) a a=1

Step 5: Rate Equation = >Rate= k[A]a[B]b Rate= k[NO]a[O2]bk[NO][O2]

Step 6: Substitute a=1, b=1 values in (1)

k(0.002)a(0.005)b = 8×107k (0.002)(0.005) = 8×107k(2× 10 3×5× 10 3)=8×107k=8× 10 7 10 5=0.08

Conclusion

It does not make a difference what the number of reactants there are. The concentration of every reactant will be present in the rate equation, raised to some power. These powers resemble the individual orders with respect to each reactant. The sum of these powers results in the overall order of the reaction. The rate constant will be a constant value for a given reaction only if the concentration of the reactants is changed without changing any other factors.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 11 Solutions

Chemistry for Engineering Students

Ch. 11 - List two types of chemical compounds that must be...Ch. 11 - Prob. 11.3PAECh. 11 - Prob. 11.4PAECh. 11 - Prob. 11.5PAECh. 11 - Prob. 11.6PAECh. 11 - Asphalt is composed of a mixture of organic...Ch. 11 - Prob. 11.8PAECh. 11 - Prob. 11.9PAECh. 11 - For each of the following, suggest appropriate...Ch. 11 - Prob. 11.11PAECh. 11 - Rank the following in order of increasing reaction...Ch. 11 - Prob. 11.13PAECh. 11 - Candle wax is a mixture of hydrocarbons. In the...Ch. 11 - Prob. 11.15PAECh. 11 - The reaction for the Haber process, the industrial...Ch. 11 - 11.17 Ammonia can react with oxygen to produce...Ch. 11 - The following data were obtained in the...Ch. 11 - Prob. 11.19PAECh. 11 - Experimental data are listed here for the reaction...Ch. 11 - Azomethane, CH3NNCH3, is not a stable compound,...Ch. 11 - Prob. 11.22PAECh. 11 - A reaction has the experimental rate equation Rate...Ch. 11 - Second-order rate constants used in modeling...Ch. 11 - For each of the rate laws below, what is the order...Ch. 11 - 11.26 The reaction of C(Xg) with NO2(g) is second...Ch. 11 - Prob. 11.27PAECh. 11 - Prob. 11.28PAECh. 11 - The hypothetical reaction, A + B —*C, has the rate...Ch. 11 - The rate of the decomposition of hydrogen...Ch. 11 - Prob. 11.31PAECh. 11 - 11.32 The following experimental data were...Ch. 11 - The following experimental data were obtained for...Ch. 11 - 11.34 Rate data were obtained at 25°C for the...Ch. 11 - 11.35 For the reaction 2 NO(g) + 2 H?(g) — N,(g) +...Ch. 11 - The reaction NO(g) + O,(g) — NO,(g) + 0(g) plays a...Ch. 11 - Prob. 11.37PAECh. 11 - Prob. 11.38PAECh. 11 - The decomposition of N2O5 in solution in carbon...Ch. 11 - In Exercise 11.39, if the initial concentration of...Ch. 11 - 11.41 For a drug to be effective in treating an...Ch. 11 - Amoxicillin is an antibiotic packaged as a powder....Ch. 11 - As with any drug, aspirin (acetylsalicylic acid)...Ch. 11 - 11.44 A possible reaction for the degradation of...Ch. 11 - The initial concentration of the reactant in a...Ch. 11 - A substance undergoes first-order decomposition....Ch. 11 - Prob. 11.47PAECh. 11 - 11.48 The following data were collected for the...Ch. 11 - The rate of photodecomposition of the herbicide...Ch. 11 - Prob. 11.50PAECh. 11 - 11.51 Peroxyacetyl nitrate (PAN) has the chemical...Ch. 11 - Hydrogen peroxide (H20i) decomposes into water and...Ch. 11 - 11.53 The reaction in which CO, decomposes to CO...Ch. 11 - use the kineticmolecular theory to explain why an...Ch. 11 - The following rate constants were obtained in an...Ch. 11 - The table below presents measured rate constants...Ch. 11 - Prob. 11.57PAECh. 11 - Prob. 11.58PAECh. 11 - Can a reaction mechanism ever be proven correct?...Ch. 11 - Prob. 11.60PAECh. 11 - Describe how the Chapman cycle is a reaction...Ch. 11 - Prob. 11.62PAECh. 11 - The following mechanism is proposed for a...Ch. 11 - 11.64 HBr is oxidized in the following reaction: 4...Ch. 11 - Prob. 11.65PAECh. 11 - Prob. 11.66PAECh. 11 - What distinguishes homogeneous and heterogeneous...Ch. 11 - Prob. 11.68PAECh. 11 - In Chapter 3, we discussed the conversion of...Ch. 11 - The label on a bottle of 3% (by volume) hydrogen...Ch. 11 - Prob. 11.71PAECh. 11 - Prob. 11.72PAECh. 11 - Prob. 11.73PAECh. 11 - 11.74 The AQI includes six levels, including...Ch. 11 - Prob. 11.75PAECh. 11 - Prob. 11.76PAECh. 11 - Prob. 11.77PAECh. 11 - Prob. 11.78PAECh. 11 - Prob. 11.79PAECh. 11 - Prob. 11.80PAECh. 11 - Prob. 11.81PAECh. 11 - Prob. 11.82PAECh. 11 - Bacteria cause milk to go sour by generating...Ch. 11 - Prob. 11.84PAECh. 11 - Prob. 11.85PAECh. 11 - Prob. 11.86PAECh. 11 - Prob. 11.87PAECh. 11 - Prob. 11.88PAECh. 11 - Prob. 11.89PAECh. 11 - 11.90 Draw a hypothetical activation energy...Ch. 11 - Prob. 11.91PAECh. 11 - Prob. 11.92PAECh. 11 - 11.93 On a particular day, the ozone level in...Ch. 11 - Prob. 11.94PAECh. 11 - The following is a thought experiment. Imagine...Ch. 11 - The following statements relate to the reaction...Ch. 11 - Prob. 11.97PAECh. 11 - Experiments show that the reaction of nitrogen...Ch. 11 - Substances that poison a catalyst pose a major...Ch. 11 - Prob. 11.100PAECh. 11 - Prob. 11.101PAECh. 11 - 11.102 Suppose that you are studying a reaction...Ch. 11 - Prob. 11.103PAECh. 11 - Prob. 11.104PAECh. 11 - Prob. 11.105PAECh. 11 - Prob. 11.106PAECh. 11 - 11.1047 Fluorine often reacts explosively. What...Ch. 11 - Prob. 11.108PAECh. 11 - Prob. 11.109PAECh. 11 - When formic acid is heated, it decomposes to...
Knowledge Booster
Chemistry
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
  • 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?
    Diethylhydrazine reacts with iodine according to the following equation: Â (C2H5)2(NH)2(l)+I2(aq)(C2H5)2N2+2HI(aq)The rate of the reaction is followed by monitoring the disappearance of the purple color due to iodine. The following data are obtained at a certain temperature. (a) What is the order of the reaction with respect to diethylhydrazine, iodine, and overall? (b) Write the rate expression of the reaction. (c) Calculate k for the reaction. (d) What must [(C2H5)2] be so that the rate of the reaction is 5.00104mol/Lh when [ I2 ]=0.500M?
    Hydrogen iodide, HI, decomposes in the gas phase to produce hydrogen, H2, and iodine, I2. The value of the rate constant, k, fur the reaction was measured at several different temperatures and the data are shown here: Temperature (K) k (M -1 5-1) 555 6.23107 575 2.42106 645 1.44104 700 2.01103 What is the value of the activation energy (in kJ/mol) for this reaction?
    Recommended textbooks for you
  • Chemistry for Engineering Students
    Chemistry
    ISBN:9781337398909
    Author:Lawrence S. Brown, Tom Holme
    Publisher:Cengage Learning
    Chemistry: Principles and Reactions
    Chemistry
    ISBN:9781305079373
    Author:William L. Masterton, Cecile N. Hurley
    Publisher:Cengage Learning
    Chemistry by OpenStax (2015-05-04)
    Chemistry
    ISBN:9781938168390
    Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
    Publisher:OpenStax
  • 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
    Chemistry
    ISBN:9781305957404
    Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
    Publisher:Cengage Learning
  • Chemistry for Engineering Students
    Chemistry
    ISBN:9781337398909
    Author:Lawrence S. Brown, Tom Holme
    Publisher:Cengage Learning
    Chemistry: Principles and Reactions
    Chemistry
    ISBN:9781305079373
    Author:William L. Masterton, Cecile N. Hurley
    Publisher:Cengage Learning
    Chemistry by OpenStax (2015-05-04)
    Chemistry
    ISBN:9781938168390
    Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
    Publisher:OpenStax
    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
    Chemistry
    ISBN:9781305957404
    Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
    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