The mechanisms which are compatible with the rate law have to be chosen. R a t e = k [ C l 2 ] 3 / 2 [ C O ] 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.

Question

Want to see more full solutions like this?

Answer 1

Question

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

Explanation of Solution

(a) The accepted mechanism:

12Cl2⇌Cl fastCl+Cl2⇌Cl3 fastCl3+CO→COCl2+Cl slowCl⇌12Cl2 fast

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,

Rateforward reaction 2=k2[Cl][Cl2]Ratereverse reaction 2=k−2[Cl3]Ratereaction 3=k3[Cl3][CO]

Now,

Rateforward reaction 2=Ratereverse reaction 2+Ratereaction 3k2[Cl][Cl2]=k−2[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.

Rateforward reaction 2≅Ratereverse reaction 2k2[Cl][Cl2]=k−2[Cl3][Cl3]=k2k−2[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,

Rateforward reaction 1=k1[Cl2]1/2Ratereverse reaction 2=k−1[Cl]Rateforward reaction 2=k2[Cl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rateforward reaction 2k1[Cl2]1/2=k−1[Cl]+k2[Cl][Cl2]=(k−1+k2[Cl2])[Cl][Cl]=k1[Cl2]1/2(k−1+k2[Cl2])

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

Rate=k3(k2k−2[Cl][Cl2])[CO]Rate=k3(k2k−2[Cl2])(k1[Cl2]1/2(k−1+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(k2k−2[Cl][Cl2])[CO]Rate=k3(k2k−2[Cl2])(k1[Cl2]1/2(k−1+k2[Cl2]))[CO]Rate=k3(k2k−2)(k1k−1)[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+CO→CCl2+O slowO+Cl2→Cl2O fastCl2O+CCl2→COCl2+Cl2 fast

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:

12Cl2⇌Cl fastCl+CO⇌COCl fastCOCl+Cl2→COCl2+Cl slowCl⇌12Cl2 fast

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,

Rateforward reaction 1=k1[Cl2]1/2Ratereverse reaction 2=k−1[Cl]Rateforward reaction 2=k2[Cl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rateforward reaction 2k1[Cl2]1/2=k−1[Cl]+k2[Cl][Cl2]=(k−1+k2[Cl2])[Cl][Cl]=k1[Cl2]1/2(k−1+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,

Rateforward reaction 2=k2[Cl][CO]Ratereverse reaction 2=k−2[COCl]Rate reaction 3=k3[COCl][Cl2]

Now,

Rateforward reaction 2=Ratereverse reaction 2+Rate reaction 3k2[Cl][CO]=k−2[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.

Rateforward reaction 2≅Ratereverse reaction 2k2[Cl][CO]=k−2[COCl][COCl]=k2k−2[Cl][CO]

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

Rate=k3[COCl][Cl2]Rate=k3(k2k−2[Cl][CO])[Cl2]Rate=k3(k2k−2)(k1[Cl2]1/2(k−1+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(k2k−2)(k1[Cl2]1/2(k−1+k2[Cl2]))[CO][Cl2]Rate=k3(k2k−2)(k1k−1)[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+CO⇌COCl+Cl fastCOCl +Cl2→COCl2+Cl slowCl+Cl→Cl2 fast

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,

Rateforward reaction 1=k1[Cl2][CO]Ratereverse reaction 1=k−1[COCl][Cl]Rate reaction 2=k2[COCl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rate reaction 2k1[Cl2][CO]=k−1[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.

Rateforward reaction 2≅Ratereverse reaction 2k1[Cl2][CO]=k−1[COCl][Cl][COCl]=k1k−1[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,

Rateforward reaction 1=k1[Cl2][CO]Ratereverse reaction 1=k−1[COCl][Cl]Rate reaction 2=k2[COCl][Cl2]Rate reaction 3=k3[Cl]2

Now,

Rateforward reaction 1+Rate reaction 2=Ratereverse reaction 1+Rate reaction 3k1[Cl2][CO]+k2[COCl][Cl2]=k−1[COCl][Cl]+k3[Cl]2

This cannot be solved without any assumption.

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

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!

Students have asked these similar questions

First image: I have to explain why the molecule C is never formed in those conditions. Second image: I have to propose a synthesis for the lactone A

First image: I have to explain why the molecule C is never formed in these conditions Second image: I have to propose a synthesis for the lactone A

Help fix my arrows please

Answer 2

Question

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

Explanation of Solution

(a) The accepted mechanism:

12Cl2⇌Cl fastCl+Cl2⇌Cl3 fastCl3+CO→COCl2+Cl slowCl⇌12Cl2 fast

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,

Rateforward reaction 2=k2[Cl][Cl2]Ratereverse reaction 2=k−2[Cl3]Ratereaction 3=k3[Cl3][CO]

Now,

Rateforward reaction 2=Ratereverse reaction 2+Ratereaction 3k2[Cl][Cl2]=k−2[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.

Rateforward reaction 2≅Ratereverse reaction 2k2[Cl][Cl2]=k−2[Cl3][Cl3]=k2k−2[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,

Rateforward reaction 1=k1[Cl2]1/2Ratereverse reaction 2=k−1[Cl]Rateforward reaction 2=k2[Cl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rateforward reaction 2k1[Cl2]1/2=k−1[Cl]+k2[Cl][Cl2]=(k−1+k2[Cl2])[Cl][Cl]=k1[Cl2]1/2(k−1+k2[Cl2])

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

Rate=k3(k2k−2[Cl][Cl2])[CO]Rate=k3(k2k−2[Cl2])(k1[Cl2]1/2(k−1+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(k2k−2[Cl][Cl2])[CO]Rate=k3(k2k−2[Cl2])(k1[Cl2]1/2(k−1+k2[Cl2]))[CO]Rate=k3(k2k−2)(k1k−1)[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+CO→CCl2+O slowO+Cl2→Cl2O fastCl2O+CCl2→COCl2+Cl2 fast

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:

12Cl2⇌Cl fastCl+CO⇌COCl fastCOCl+Cl2→COCl2+Cl slowCl⇌12Cl2 fast

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,

Rateforward reaction 1=k1[Cl2]1/2Ratereverse reaction 2=k−1[Cl]Rateforward reaction 2=k2[Cl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rateforward reaction 2k1[Cl2]1/2=k−1[Cl]+k2[Cl][Cl2]=(k−1+k2[Cl2])[Cl][Cl]=k1[Cl2]1/2(k−1+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,

Rateforward reaction 2=k2[Cl][CO]Ratereverse reaction 2=k−2[COCl]Rate reaction 3=k3[COCl][Cl2]

Now,

Rateforward reaction 2=Ratereverse reaction 2+Rate reaction 3k2[Cl][CO]=k−2[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.

Rateforward reaction 2≅Ratereverse reaction 2k2[Cl][CO]=k−2[COCl][COCl]=k2k−2[Cl][CO]

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

Rate=k3[COCl][Cl2]Rate=k3(k2k−2[Cl][CO])[Cl2]Rate=k3(k2k−2)(k1[Cl2]1/2(k−1+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(k2k−2)(k1[Cl2]1/2(k−1+k2[Cl2]))[CO][Cl2]Rate=k3(k2k−2)(k1k−1)[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+CO⇌COCl+Cl fastCOCl +Cl2→COCl2+Cl slowCl+Cl→Cl2 fast

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,

Rateforward reaction 1=k1[Cl2][CO]Ratereverse reaction 1=k−1[COCl][Cl]Rate reaction 2=k2[COCl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rate reaction 2k1[Cl2][CO]=k−1[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.

Rateforward reaction 2≅Ratereverse reaction 2k1[Cl2][CO]=k−1[COCl][Cl][COCl]=k1k−1[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,

Rateforward reaction 1=k1[Cl2][CO]Ratereverse reaction 1=k−1[COCl][Cl]Rate reaction 2=k2[COCl][Cl2]Rate reaction 3=k3[Cl]2

Now,

Rateforward reaction 1+Rate reaction 2=Ratereverse reaction 1+Rate reaction 3k1[Cl2][CO]+k2[COCl][Cl2]=k−1[COCl][Cl]+k3[Cl]2

This cannot be solved without any assumption.

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

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!

Answer 3

Question

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

Explanation of Solution

(a) The accepted mechanism:

12Cl2⇌Cl fastCl+Cl2⇌Cl3 fastCl3+CO→COCl2+Cl slowCl⇌12Cl2 fast

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,

Rateforward reaction 2=k2[Cl][Cl2]Ratereverse reaction 2=k−2[Cl3]Ratereaction 3=k3[Cl3][CO]

Now,

Rateforward reaction 2=Ratereverse reaction 2+Ratereaction 3k2[Cl][Cl2]=k−2[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.

Rateforward reaction 2≅Ratereverse reaction 2k2[Cl][Cl2]=k−2[Cl3][Cl3]=k2k−2[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,

Rateforward reaction 1=k1[Cl2]1/2Ratereverse reaction 2=k−1[Cl]Rateforward reaction 2=k2[Cl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rateforward reaction 2k1[Cl2]1/2=k−1[Cl]+k2[Cl][Cl2]=(k−1+k2[Cl2])[Cl][Cl]=k1[Cl2]1/2(k−1+k2[Cl2])

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

Rate=k3(k2k−2[Cl][Cl2])[CO]Rate=k3(k2k−2[Cl2])(k1[Cl2]1/2(k−1+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(k2k−2[Cl][Cl2])[CO]Rate=k3(k2k−2[Cl2])(k1[Cl2]1/2(k−1+k2[Cl2]))[CO]Rate=k3(k2k−2)(k1k−1)[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+CO→CCl2+O slowO+Cl2→Cl2O fastCl2O+CCl2→COCl2+Cl2 fast

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:

12Cl2⇌Cl fastCl+CO⇌COCl fastCOCl+Cl2→COCl2+Cl slowCl⇌12Cl2 fast

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,

Rateforward reaction 1=k1[Cl2]1/2Ratereverse reaction 2=k−1[Cl]Rateforward reaction 2=k2[Cl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rateforward reaction 2k1[Cl2]1/2=k−1[Cl]+k2[Cl][Cl2]=(k−1+k2[Cl2])[Cl][Cl]=k1[Cl2]1/2(k−1+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,

Rateforward reaction 2=k2[Cl][CO]Ratereverse reaction 2=k−2[COCl]Rate reaction 3=k3[COCl][Cl2]

Now,

Rateforward reaction 2=Ratereverse reaction 2+Rate reaction 3k2[Cl][CO]=k−2[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.

Rateforward reaction 2≅Ratereverse reaction 2k2[Cl][CO]=k−2[COCl][COCl]=k2k−2[Cl][CO]

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

Rate=k3[COCl][Cl2]Rate=k3(k2k−2[Cl][CO])[Cl2]Rate=k3(k2k−2)(k1[Cl2]1/2(k−1+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(k2k−2)(k1[Cl2]1/2(k−1+k2[Cl2]))[CO][Cl2]Rate=k3(k2k−2)(k1k−1)[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+CO⇌COCl+Cl fastCOCl +Cl2→COCl2+Cl slowCl+Cl→Cl2 fast

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,

Rateforward reaction 1=k1[Cl2][CO]Ratereverse reaction 1=k−1[COCl][Cl]Rate reaction 2=k2[COCl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rate reaction 2k1[Cl2][CO]=k−1[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.

Rateforward reaction 2≅Ratereverse reaction 2k1[Cl2][CO]=k−1[COCl][Cl][COCl]=k1k−1[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,

Rateforward reaction 1=k1[Cl2][CO]Ratereverse reaction 1=k−1[COCl][Cl]Rate reaction 2=k2[COCl][Cl2]Rate reaction 3=k3[Cl]2

Now,

Rateforward reaction 1+Rate reaction 2=Ratereverse reaction 1+Rate reaction 3k1[Cl2][CO]+k2[COCl][Cl2]=k−1[COCl][Cl]+k3[Cl]2

This cannot be solved without any assumption.

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

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!

Answer 4

Question

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

Explanation of Solution

(a) The accepted mechanism:

12Cl2⇌Cl fastCl+Cl2⇌Cl3 fastCl3+CO→COCl2+Cl slowCl⇌12Cl2 fast

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,

Rateforward reaction 2=k2[Cl][Cl2]Ratereverse reaction 2=k−2[Cl3]Ratereaction 3=k3[Cl3][CO]

Now,

Rateforward reaction 2=Ratereverse reaction 2+Ratereaction 3k2[Cl][Cl2]=k−2[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.

Rateforward reaction 2≅Ratereverse reaction 2k2[Cl][Cl2]=k−2[Cl3][Cl3]=k2k−2[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,

Rateforward reaction 1=k1[Cl2]1/2Ratereverse reaction 2=k−1[Cl]Rateforward reaction 2=k2[Cl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rateforward reaction 2k1[Cl2]1/2=k−1[Cl]+k2[Cl][Cl2]=(k−1+k2[Cl2])[Cl][Cl]=k1[Cl2]1/2(k−1+k2[Cl2])

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

Rate=k3(k2k−2[Cl][Cl2])[CO]Rate=k3(k2k−2[Cl2])(k1[Cl2]1/2(k−1+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(k2k−2[Cl][Cl2])[CO]Rate=k3(k2k−2[Cl2])(k1[Cl2]1/2(k−1+k2[Cl2]))[CO]Rate=k3(k2k−2)(k1k−1)[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+CO→CCl2+O slowO+Cl2→Cl2O fastCl2O+CCl2→COCl2+Cl2 fast

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:

12Cl2⇌Cl fastCl+CO⇌COCl fastCOCl+Cl2→COCl2+Cl slowCl⇌12Cl2 fast

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,

Rateforward reaction 1=k1[Cl2]1/2Ratereverse reaction 2=k−1[Cl]Rateforward reaction 2=k2[Cl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rateforward reaction 2k1[Cl2]1/2=k−1[Cl]+k2[Cl][Cl2]=(k−1+k2[Cl2])[Cl][Cl]=k1[Cl2]1/2(k−1+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,

Rateforward reaction 2=k2[Cl][CO]Ratereverse reaction 2=k−2[COCl]Rate reaction 3=k3[COCl][Cl2]

Now,

Rateforward reaction 2=Ratereverse reaction 2+Rate reaction 3k2[Cl][CO]=k−2[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.

Rateforward reaction 2≅Ratereverse reaction 2k2[Cl][CO]=k−2[COCl][COCl]=k2k−2[Cl][CO]

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

Rate=k3[COCl][Cl2]Rate=k3(k2k−2[Cl][CO])[Cl2]Rate=k3(k2k−2)(k1[Cl2]1/2(k−1+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(k2k−2)(k1[Cl2]1/2(k−1+k2[Cl2]))[CO][Cl2]Rate=k3(k2k−2)(k1k−1)[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+CO⇌COCl+Cl fastCOCl +Cl2→COCl2+Cl slowCl+Cl→Cl2 fast

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,

Rateforward reaction 1=k1[Cl2][CO]Ratereverse reaction 1=k−1[COCl][Cl]Rate reaction 2=k2[COCl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rate reaction 2k1[Cl2][CO]=k−1[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.

Rateforward reaction 2≅Ratereverse reaction 2k1[Cl2][CO]=k−1[COCl][Cl][COCl]=k1k−1[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,

Rateforward reaction 1=k1[Cl2][CO]Ratereverse reaction 1=k−1[COCl][Cl]Rate reaction 2=k2[COCl][Cl2]Rate reaction 3=k3[Cl]2

Now,

Rateforward reaction 1+Rate reaction 2=Ratereverse reaction 1+Rate reaction 3k1[Cl2][CO]+k2[COCl][Cl2]=k−1[COCl][Cl]+k3[Cl]2

This cannot be solved without any assumption.

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

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!

Answer 5

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

Explanation of Solution

(a) The accepted mechanism:

12Cl2⇌Cl fastCl+Cl2⇌Cl3 fastCl3+CO→COCl2+Cl slowCl⇌12Cl2 fast

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,

Rateforward reaction 2=k2[Cl][Cl2]Ratereverse reaction 2=k−2[Cl3]Ratereaction 3=k3[Cl3][CO]

Now,

Rateforward reaction 2=Ratereverse reaction 2+Ratereaction 3k2[Cl][Cl2]=k−2[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.

Rateforward reaction 2≅Ratereverse reaction 2k2[Cl][Cl2]=k−2[Cl3][Cl3]=k2k−2[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,

Rateforward reaction 1=k1[Cl2]1/2Ratereverse reaction 2=k−1[Cl]Rateforward reaction 2=k2[Cl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rateforward reaction 2k1[Cl2]1/2=k−1[Cl]+k2[Cl][Cl2]=(k−1+k2[Cl2])[Cl][Cl]=k1[Cl2]1/2(k−1+k2[Cl2])

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

Rate=k3(k2k−2[Cl][Cl2])[CO]Rate=k3(k2k−2[Cl2])(k1[Cl2]1/2(k−1+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(k2k−2[Cl][Cl2])[CO]Rate=k3(k2k−2[Cl2])(k1[Cl2]1/2(k−1+k2[Cl2]))[CO]Rate=k3(k2k−2)(k1k−1)[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+CO→CCl2+O slowO+Cl2→Cl2O fastCl2O+CCl2→COCl2+Cl2 fast

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:

12Cl2⇌Cl fastCl+CO⇌COCl fastCOCl+Cl2→COCl2+Cl slowCl⇌12Cl2 fast

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,

Rateforward reaction 1=k1[Cl2]1/2Ratereverse reaction 2=k−1[Cl]Rateforward reaction 2=k2[Cl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rateforward reaction 2k1[Cl2]1/2=k−1[Cl]+k2[Cl][Cl2]=(k−1+k2[Cl2])[Cl][Cl]=k1[Cl2]1/2(k−1+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,

Rateforward reaction 2=k2[Cl][CO]Ratereverse reaction 2=k−2[COCl]Rate reaction 3=k3[COCl][Cl2]

Now,

Rateforward reaction 2=Ratereverse reaction 2+Rate reaction 3k2[Cl][CO]=k−2[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.

Rateforward reaction 2≅Ratereverse reaction 2k2[Cl][CO]=k−2[COCl][COCl]=k2k−2[Cl][CO]

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

Rate=k3[COCl][Cl2]Rate=k3(k2k−2[Cl][CO])[Cl2]Rate=k3(k2k−2)(k1[Cl2]1/2(k−1+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(k2k−2)(k1[Cl2]1/2(k−1+k2[Cl2]))[CO][Cl2]Rate=k3(k2k−2)(k1k−1)[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+CO⇌COCl+Cl fastCOCl +Cl2→COCl2+Cl slowCl+Cl→Cl2 fast

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,

Rateforward reaction 1=k1[Cl2][CO]Ratereverse reaction 1=k−1[COCl][Cl]Rate reaction 2=k2[COCl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rate reaction 2k1[Cl2][CO]=k−1[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.

Rateforward reaction 2≅Ratereverse reaction 2k1[Cl2][CO]=k−1[COCl][Cl][COCl]=k1k−1[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,

Rateforward reaction 1=k1[Cl2][CO]Ratereverse reaction 1=k−1[COCl][Cl]Rate reaction 2=k2[COCl][Cl2]Rate reaction 3=k3[Cl]2

Now,

Rateforward reaction 1+Rate reaction 2=Ratereverse reaction 1+Rate reaction 3k1[Cl2][CO]+k2[COCl][Cl2]=k−1[COCl][Cl]+k3[Cl]2

This cannot be solved without any assumption.

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

Answer 6

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

Explanation of Solution

(a) The accepted mechanism:

12Cl2⇌Cl fastCl+Cl2⇌Cl3 fastCl3+CO→COCl2+Cl slowCl⇌12Cl2 fast

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,

Rateforward reaction 2=k2[Cl][Cl2]Ratereverse reaction 2=k−2[Cl3]Ratereaction 3=k3[Cl3][CO]

Now,

Rateforward reaction 2=Ratereverse reaction 2+Ratereaction 3k2[Cl][Cl2]=k−2[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.

Rateforward reaction 2≅Ratereverse reaction 2k2[Cl][Cl2]=k−2[Cl3][Cl3]=k2k−2[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,

Rateforward reaction 1=k1[Cl2]1/2Ratereverse reaction 2=k−1[Cl]Rateforward reaction 2=k2[Cl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rateforward reaction 2k1[Cl2]1/2=k−1[Cl]+k2[Cl][Cl2]=(k−1+k2[Cl2])[Cl][Cl]=k1[Cl2]1/2(k−1+k2[Cl2])

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

Rate=k3(k2k−2[Cl][Cl2])[CO]Rate=k3(k2k−2[Cl2])(k1[Cl2]1/2(k−1+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(k2k−2[Cl][Cl2])[CO]Rate=k3(k2k−2[Cl2])(k1[Cl2]1/2(k−1+k2[Cl2]))[CO]Rate=k3(k2k−2)(k1k−1)[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+CO→CCl2+O slowO+Cl2→Cl2O fastCl2O+CCl2→COCl2+Cl2 fast

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:

12Cl2⇌Cl fastCl+CO⇌COCl fastCOCl+Cl2→COCl2+Cl slowCl⇌12Cl2 fast

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,

Rateforward reaction 1=k1[Cl2]1/2Ratereverse reaction 2=k−1[Cl]Rateforward reaction 2=k2[Cl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rateforward reaction 2k1[Cl2]1/2=k−1[Cl]+k2[Cl][Cl2]=(k−1+k2[Cl2])[Cl][Cl]=k1[Cl2]1/2(k−1+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,

Rateforward reaction 2=k2[Cl][CO]Ratereverse reaction 2=k−2[COCl]Rate reaction 3=k3[COCl][Cl2]

Now,

Rateforward reaction 2=Ratereverse reaction 2+Rate reaction 3k2[Cl][CO]=k−2[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.

Rateforward reaction 2≅Ratereverse reaction 2k2[Cl][CO]=k−2[COCl][COCl]=k2k−2[Cl][CO]

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

Rate=k3[COCl][Cl2]Rate=k3(k2k−2[Cl][CO])[Cl2]Rate=k3(k2k−2)(k1[Cl2]1/2(k−1+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(k2k−2)(k1[Cl2]1/2(k−1+k2[Cl2]))[CO][Cl2]Rate=k3(k2k−2)(k1k−1)[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+CO⇌COCl+Cl fastCOCl +Cl2→COCl2+Cl slowCl+Cl→Cl2 fast

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,

Rateforward reaction 1=k1[Cl2][CO]Ratereverse reaction 1=k−1[COCl][Cl]Rate reaction 2=k2[COCl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rate reaction 2k1[Cl2][CO]=k−1[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.

Rateforward reaction 2≅Ratereverse reaction 2k1[Cl2][CO]=k−1[COCl][Cl][COCl]=k1k−1[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,

Rateforward reaction 1=k1[Cl2][CO]Ratereverse reaction 1=k−1[COCl][Cl]Rate reaction 2=k2[COCl][Cl2]Rate reaction 3=k3[Cl]2

Now,

Rateforward reaction 1+Rate reaction 2=Ratereverse reaction 1+Rate reaction 3k1[Cl2][CO]+k2[COCl][Cl2]=k−1[COCl][Cl]+k3[Cl]2

This cannot be solved without any assumption.

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

Answer 7

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.

Answer 8

Expert Solution & Answer

Explanation of Solution

(a) The accepted mechanism:

12Cl2⇌Cl fastCl+Cl2⇌Cl3 fastCl3+CO→COCl2+Cl slowCl⇌12Cl2 fast

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,

Rateforward reaction 2=k2[Cl][Cl2]Ratereverse reaction 2=k−2[Cl3]Ratereaction 3=k3[Cl3][CO]

Now,

Rateforward reaction 2=Ratereverse reaction 2+Ratereaction 3k2[Cl][Cl2]=k−2[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.

Rateforward reaction 2≅Ratereverse reaction 2k2[Cl][Cl2]=k−2[Cl3][Cl3]=k2k−2[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,

Rateforward reaction 1=k1[Cl2]1/2Ratereverse reaction 2=k−1[Cl]Rateforward reaction 2=k2[Cl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rateforward reaction 2k1[Cl2]1/2=k−1[Cl]+k2[Cl][Cl2]=(k−1+k2[Cl2])[Cl][Cl]=k1[Cl2]1/2(k−1+k2[Cl2])

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

Rate=k3(k2k−2[Cl][Cl2])[CO]Rate=k3(k2k−2[Cl2])(k1[Cl2]1/2(k−1+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(k2k−2[Cl][Cl2])[CO]Rate=k3(k2k−2[Cl2])(k1[Cl2]1/2(k−1+k2[Cl2]))[CO]Rate=k3(k2k−2)(k1k−1)[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+CO→CCl2+O slowO+Cl2→Cl2O fastCl2O+CCl2→COCl2+Cl2 fast

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:

12Cl2⇌Cl fastCl+CO⇌COCl fastCOCl+Cl2→COCl2+Cl slowCl⇌12Cl2 fast

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,

Rateforward reaction 1=k1[Cl2]1/2Ratereverse reaction 2=k−1[Cl]Rateforward reaction 2=k2[Cl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rateforward reaction 2k1[Cl2]1/2=k−1[Cl]+k2[Cl][Cl2]=(k−1+k2[Cl2])[Cl][Cl]=k1[Cl2]1/2(k−1+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,

Rateforward reaction 2=k2[Cl][CO]Ratereverse reaction 2=k−2[COCl]Rate reaction 3=k3[COCl][Cl2]

Now,

Rateforward reaction 2=Ratereverse reaction 2+Rate reaction 3k2[Cl][CO]=k−2[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.

Rateforward reaction 2≅Ratereverse reaction 2k2[Cl][CO]=k−2[COCl][COCl]=k2k−2[Cl][CO]

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

Rate=k3[COCl][Cl2]Rate=k3(k2k−2[Cl][CO])[Cl2]Rate=k3(k2k−2)(k1[Cl2]1/2(k−1+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(k2k−2)(k1[Cl2]1/2(k−1+k2[Cl2]))[CO][Cl2]Rate=k3(k2k−2)(k1k−1)[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+CO⇌COCl+Cl fastCOCl +Cl2→COCl2+Cl slowCl+Cl→Cl2 fast

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,

Rateforward reaction 1=k1[Cl2][CO]Ratereverse reaction 1=k−1[COCl][Cl]Rate reaction 2=k2[COCl][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Rate reaction 2k1[Cl2][CO]=k−1[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.

Rateforward reaction 2≅Ratereverse reaction 2k1[Cl2][CO]=k−1[COCl][Cl][COCl]=k1k−1[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,

Rateforward reaction 1=k1[Cl2][CO]Ratereverse reaction 1=k−1[COCl][Cl]Rate reaction 2=k2[COCl][Cl2]Rate reaction 3=k3[Cl]2