Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 11, Problem 111QRT
Interpretation Introduction

Interpretation:

The mechanisms which are compatible with the rate law have to be chosen.

  Rate=k[Cl2]3/2[CO]

Concept Introduction:

Steady-state approximation:

Once the steady state condition is reached, the rate of formation of the intermediate is equal to the rate of destruction of the intermediate.

Expert Solution & Answer
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Explanation of Solution

(a) The accepted mechanism:

  12Cl2ClfastCl+Cl2Cl3fastCl3+COCOCl2+ClslowCl12Cl2fast

The rate determining step is the slowest step in the mechanism.  Thus, the rate of the reaction is given below.

  Rate=k3[Cl3][CO]

Applying the steady-state approximation on the intermediates (Cl3,Cl), the concentration of the intermediates can be obtained in terms of reactants.

At steady state condition, the rate of formation of the intermediate is equal to the rate of destruction of the intermediate.

Cl3 is created in the forward reaction of step 2 and destroyed in the reverse reaction of step 2 and in the reaction of step 3.  Thus,

  Rateforwardreaction2=k2[Cl][Cl2]Ratereversereaction2=k2[Cl3]Ratereaction3=k3[Cl3][CO]

Now,

  Rateforwardreaction2=Ratereversereaction2+Ratereaction3k2[Cl][Cl2]=k2[Cl3]+k3[Cl3][CO]

Because the rate of step 3 is presumed to be much smaller than the rate of step 2, the second term is presumed to be negligibly small compared to the first term.

  Rateforwardreaction2Ratereversereaction2k2[Cl][Cl2]=k2[Cl3][Cl3]=k2k2[Cl][Cl2]

Cl is created in the forward reaction of step 1 and destroyed in the reverse reaction of step 1, in the forward reaction of step 2.  Thus,

  Rateforwardreaction1=k1[Cl2]1/2Ratereversereaction2=k1[Cl]Rateforwardreaction2=k2[Cl][Cl2]

Now,

  Rateforwardreaction1=Ratereversereaction1+Rateforwardreaction2k1[Cl2]1/2=k1[Cl]+k2[Cl][Cl2]=(k1+k2[Cl2])[Cl][Cl]=k1[Cl2]1/2(k1+k2[Cl2])

Now, the rate law can be modified as given below.

  Rate=k3(k2k2[Cl][Cl2])[CO]Rate=k3(k2k2[Cl2])(k1[Cl2]1/2(k1+k2[Cl2]))[CO]

Assumption:

It is conceivable that the recombination of Cl atoms might be faster than the Cl2 reaction with Cl.  If the second step is much slower than the fast reverse first step, the rate law can be more simplified as:

  Rate=k3(k2k2[Cl][Cl2])[CO]Rate=k3(k2k2[Cl2])(k1[Cl2]1/2(k1+k2[Cl2]))[CO]Rate=k3(k2k2)(k1k1)[Cl2]3/2[CO]Rate=k'[Cl2]3/2[CO]

Therefore, the rate law of the reaction is Rate=k'[Cl2]3/2[CO].  This is consistent with the observed rate law.  If the last assumption is not correct, then this rate law is not consistent with the observed rate law.

(b) The accepted mechanism:

  Cl2+COCCl2+OslowO+Cl2Cl2OfastCl2O+CCl2COCl2+Cl2fast

The rate determining step is the slowest step in the mechanism.  Thus, the rate of the reaction is given below.

  Rate=k[Cl2][CO]

This rate law is not consistent with the observed rate law.

(c) The accepted mechanism:

  12Cl2ClfastCl+COCOClfastCOCl+Cl2COCl2+ClslowCl12Cl2fast

The rate determining step is the slowest step in the mechanism.  Thus, the rate of the reaction is given below.

  Rate=k3[COCl][Cl2]

Applying the steady-state approximation on the intermediates (COCl,Cl), the concentration of the intermediates can be obtained in terms of reactants.

At steady state condition, the rate of formation of the intermediate is equal to the rate of destruction of the intermediate.

Cl is created in the forward reaction of step 1 and destroyed in the reverse reaction of step 1 and in the forward reaction of step 2.  Thus,

  Rateforwardreaction1=k1[Cl2]1/2Ratereversereaction2=k1[Cl]Rateforwardreaction2=k2[Cl][Cl2]

Now,

  Rateforwardreaction1=Ratereversereaction1+Rateforwardreaction2k1[Cl2]1/2=k1[Cl]+k2[Cl][Cl2]=(k1+k2[Cl2])[Cl][Cl]=k1[Cl2]1/2(k1+k2[Cl2])

COCl is created in the forward reaction of step 2 and destroyed in the reverse reaction of step 2, in the reaction of step 3.  Thus,

  Rateforwardreaction2=k2[Cl][CO]Ratereversereaction2=k2[COCl]Ratereaction3=k3[COCl][Cl2]

Now,

  Rateforwardreaction2=Ratereversereaction2+Ratereaction3k2[Cl][CO]=k2[COCl]+k3[COCl][Cl2]

Because the rate of step 3 is presumed to be much smaller than the rate of step 2, the second term is presumed to be negligibly small compared to the first term.

  Rateforwardreaction2Ratereversereaction2k2[Cl][CO]=k2[COCl][COCl]=k2k2[Cl][CO]

Now, the rate law can be modified as given below.

  Rate=k3[COCl][Cl2]Rate=k3(k2k2[Cl][CO])[Cl2]Rate=k3(k2k2)(k1[Cl2]1/2(k1+k2[Cl2]))[CO][Cl2]

Assumption:

It is conceivable that the recombination of Cl atoms might be faster than the CO reaction with Cl.  If the second step is much slower than the fast reverse first step, the rate law can be more simplified as:

  Rate=k3(k2k2)(k1[Cl2]1/2(k1+k2[Cl2]))[CO][Cl2]Rate=k3(k2k2)(k1k1)[Cl2]3/2[CO]Rate=k'[Cl2]3/2[CO]

Therefore, the rate law of the reaction is Rate=k'[Cl2]3/2[CO].  This is consistent with the observed rate law.  If the last assumption is not correct, then this rate law is not consistent with the observed rate law.

(d) The accepted mechanism:

  Cl2+COCOCl+ClfastCOCl+Cl2COCl2+ClslowCl+ClCl2fast

The rate determining step is the slowest step in the mechanism.  Thus, the rate of the reaction is given below.

  Rate=k2[COCl][Cl2]

Applying the steady-state approximation on the intermediates (COCl,Cl), the concentration of the intermediates can be obtained in terms of reactants.

At steady state condition, the rate of formation of the intermediate is equal to the rate of destruction of the intermediate.

COCl is created in the forward reaction of step 1 and destroyed in the reverse reaction of step 1 and in the reaction of step 2.  Thus,

  Rateforwardreaction1=k1[Cl2][CO]Ratereversereaction1=k1[COCl][Cl]Ratereaction2=k2[COCl][Cl2]

Now,

  Rateforwardreaction1=Ratereversereaction1+Ratereaction2k1[Cl2][CO]=k1[COCl][Cl]+k2[COCl][Cl2]

Because the rate of step 2 is presumed to be much smaller than the rate of step 1, the second term is presumed to be negligibly small compared to the first term.

  Rateforwardreaction2Ratereversereaction2k1[Cl2][CO]=k1[COCl][Cl][COCl]=k1k1[Cl2][CO][Cl]

Cl is created in the forward reaction of step 1, also in the reaction of step 2 and destroyed in the reverse reaction of step 1 and in the reaction of step 3.  Thus,

  Rateforwardreaction1=k1[Cl2][CO]Ratereversereaction1=k1[COCl][Cl]Ratereaction2=k2[COCl][Cl2]Ratereaction3=k3[Cl]2

Now,

  Rateforwardreaction1+Ratereaction2=Ratereversereaction1+Ratereaction3k1[Cl2][CO]+k2[COCl][Cl2]=k1[COCl][Cl]+k3[Cl]2

This cannot be solved without any assumption.

So, option (a) and (c) is correct option.

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Chapter 11 Solutions

Chemistry: The Molecular Science

Ch. 11.3 - Prob. 11.6PSPCh. 11.3 - Prob. 11.7PSPCh. 11.4 - Prob. 11.6ECh. 11.4 - Prob. 11.7CECh. 11.4 - Prob. 11.8PSPCh. 11.4 - Prob. 11.8CECh. 11.5 - Prob. 11.9PSPCh. 11.5 - The frequency factor A is 6.31 108 L mol1 s1 and...Ch. 11.6 - Prob. 11.10CECh. 11.7 - Prob. 11.11ECh. 11.7 - The Raschig reaction produces the industrially...Ch. 11.7 - Prob. 11.12ECh. 11.8 - The oxidation of thallium(I) ion by cerium(IV) ion...Ch. 11.9 - Prob. 11.11PSPCh. 11.9 - Prob. 11.14CECh. 11 - An excellent way to make highly pure nickel metal...Ch. 11 - Prob. 1QRTCh. 11 - Prob. 2QRTCh. 11 - Prob. 3QRTCh. 11 - Prob. 4QRTCh. 11 - Prob. 5QRTCh. 11 - Prob. 6QRTCh. 11 - Prob. 7QRTCh. 11 - Prob. 8QRTCh. 11 - Prob. 9QRTCh. 11 - Prob. 10QRTCh. 11 - Prob. 11QRTCh. 11 - Cyclobutane can decompose to form ethylene: The...Ch. 11 - Prob. 13QRTCh. 11 - Prob. 14QRTCh. 11 - For the reaction 2NO2(g)2NO(g)+O2(g) make...Ch. 11 - Prob. 16QRTCh. 11 - Prob. 17QRTCh. 11 - Ammonia is produced by the reaction between...Ch. 11 - Prob. 19QRTCh. 11 - Prob. 20QRTCh. 11 - The reaction of CO(g) + NO2(g) is second-order in...Ch. 11 - Nitrosyl bromide, NOBr, is formed from NO and Br2....Ch. 11 - Prob. 23QRTCh. 11 - Prob. 24QRTCh. 11 - Prob. 25QRTCh. 11 - For the reaction these data were obtained at 1100...Ch. 11 - Prob. 27QRTCh. 11 - Prob. 28QRTCh. 11 - Prob. 29QRTCh. 11 - Prob. 30QRTCh. 11 - Prob. 31QRTCh. 11 - Prob. 32QRTCh. 11 - For the reaction of phenyl acetate with water the...Ch. 11 - When phenacyl bromide and pyridine are both...Ch. 11 - The compound p-methoxybenzonitrile N-oxide, which...Ch. 11 - Prob. 36QRTCh. 11 - Radioactive gold-198 is used in the diagnosis of...Ch. 11 - Prob. 38QRTCh. 11 - Prob. 39QRTCh. 11 - Prob. 40QRTCh. 11 - Prob. 41QRTCh. 11 - Prob. 42QRTCh. 11 - Prob. 43QRTCh. 11 - Prob. 44QRTCh. 11 - Prob. 45QRTCh. 11 - Prob. 46QRTCh. 11 - Prob. 47QRTCh. 11 - Prob. 48QRTCh. 11 - Prob. 49QRTCh. 11 - Prob. 50QRTCh. 11 - Prob. 51QRTCh. 11 - Prob. 52QRTCh. 11 - For the reaction of iodine atoms with hydrogen...Ch. 11 - Prob. 54QRTCh. 11 - The activation energy Ea is 139.7 kJ mol1 for the...Ch. 11 - Prob. 56QRTCh. 11 - Prob. 57QRTCh. 11 - Prob. 58QRTCh. 11 - Prob. 59QRTCh. 11 - Prob. 60QRTCh. 11 - Prob. 61QRTCh. 11 - Prob. 62QRTCh. 11 - Prob. 63QRTCh. 11 - Which of the reactions in Question 62 would (a)...Ch. 11 - Prob. 65QRTCh. 11 - Prob. 66QRTCh. 11 - Prob. 67QRTCh. 11 - Prob. 68QRTCh. 11 - Prob. 69QRTCh. 11 - Prob. 70QRTCh. 11 - Prob. 71QRTCh. 11 - For the reaction the rate law is Rate=k[(CH3)3CBr]...Ch. 11 - Prob. 73QRTCh. 11 - Prob. 74QRTCh. 11 - Prob. 75QRTCh. 11 - For this reaction mechanism, write the chemical...Ch. 11 - Prob. 77QRTCh. 11 - Prob. 78QRTCh. 11 - Prob. 79QRTCh. 11 - When enzymes are present at very low...Ch. 11 - Prob. 81QRTCh. 11 - The reaction is catalyzed by the enzyme succinate...Ch. 11 - Prob. 83QRTCh. 11 - Many biochemical reactions are catalyzed by acids....Ch. 11 - Prob. 85QRTCh. 11 - Prob. 86QRTCh. 11 - Prob. 87QRTCh. 11 - Prob. 88QRTCh. 11 - Prob. 89QRTCh. 11 - Prob. 90QRTCh. 11 - Prob. 91QRTCh. 11 - Prob. 92QRTCh. 11 - Prob. 93QRTCh. 11 - Prob. 94QRTCh. 11 - Nitryl fluoride is an explosive compound that can...Ch. 11 - Prob. 96QRTCh. 11 - Prob. 97QRTCh. 11 - For a reaction involving the decomposition of a...Ch. 11 - Prob. 99QRTCh. 11 - Prob. 100QRTCh. 11 - Prob. 101QRTCh. 11 - This graph shows the change in concentration as a...Ch. 11 - Prob. 103QRTCh. 11 - Prob. 104QRTCh. 11 - Prob. 105QRTCh. 11 - Prob. 106QRTCh. 11 - Prob. 107QRTCh. 11 - Prob. 108QRTCh. 11 - Prob. 109QRTCh. 11 - Prob. 110QRTCh. 11 - Prob. 111QRTCh. 11 - Prob. 112QRTCh. 11 - Prob. 113QRTCh. 11 - Prob. 114QRTCh. 11 - Prob. 115QRTCh. 11 - Prob. 116QRTCh. 11 - Prob. 118QRTCh. 11 - Prob. 119QRTCh. 11 - In a time-resolved picosecond spectroscopy...Ch. 11 - If you know some calculus, derive the integrated...Ch. 11 - If you know some calculus, derive the integrated...Ch. 11 - (Section 11-5) A rule of thumb is that for a...Ch. 11 - Prob. 11.BCPCh. 11 - Prob. 11.CCPCh. 11 - Prob. 11.DCP
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