The rate law for the given mechanism has to be determined. The reaction: 2 N O ( g ) + C l 2 ( g ) → 2 N O C l ( g ) The accepted mechanism: N O + C l 2 ⇌ N O C l 2 f a s t N O C l 2 + N O → 2 N O C l s l o w 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 71QRT

(a)

Interpretation Introduction

Interpretation:

The rate law for the given mechanism has to be determined.

The reaction:

2NO(g)+Cl2(g)→2NOCl(g)

The accepted mechanism:

NO+Cl2⇌NOCl2 fastNOCl2+NO→2NOCl slow

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.

(a)

Expert Solution

Explanation of Solution

The reaction:

2NO(g)+Cl2(g)→2NOCl(g)

The accepted mechanism:

NO+Cl2⇌k−1k1NOCl2 fastNOCl2+NO→k22NOCl slow

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

Rate=k2[NOCl2][NO]

Applying the steady-state approximation on the intermediate (NOCl2), the concentration of the intermediate 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.

NOCl2 is created in the forward reaction of step 1 and destroyed in the reverse reaction of step 1 and step 2. Thus,

Rateforward reaction 1=k1[NO][Cl2]Ratereverse reaction 1=k−1[NOCl2]Ratereaction 2=k2[NOCl2][NO]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Ratereaction 2k1[NO][Cl2]=k−1[NOCl2]+k2[NOCl2][NO]

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 1≅Ratereverse reaction 1k1[NO][Cl2]=k−1[NOCl2][NOCl2]=k1k−1[NO][Cl2]

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

Rate=k2(k1k−1[NO][Cl2])[NO]Rate=k'[NO]2[Cl2] [∵k'=k2k1k−1]

Therefore, the rate law of the reaction is Rate=k'[NO]2[Cl2] .

(b)

Interpretation Introduction

Interpretation:

Another mechanism that agrees with the same rate law has to be suggested.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Another mechanism that agrees with the same rate law is given below.

The imaginary mechanism:

2NO2⇌k−1k1N2O2 FastN2O2 +Cl2→k22NOCl2 Slow

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

Rate=k2[N2O2][Cl2]

Applying the steady-state approximation on the intermediate (N2O2), the concentration of the intermediate 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.

N2O2 is created in the forward reaction of step 1 and destroyed in the reverse reaction of step 1 and step 2. Thus,

Rateforward reaction 1=k1[NO2]2Ratereverse reaction 1=k−1[N2O2]Ratereaction 2=k2[N2O2][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Ratereaction 2k1[NO]2=k−1[N2O2]+k2[N2O2][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 1≅Ratereverse reaction 1k1[NO]2=k−1[N2O2][N2O2]=k1k−1[NO]2

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

Rate=k2(k1k−1[NO]2)[Cl2]Rate=k'[NO]2[Cl2] [∵k'=k2k1k−1]

Therefore, the rate law of the reaction is Rate=k'[NO]2[Cl2] .

(c)

Interpretation Introduction

Interpretation:

Another mechanism that does not agree with the same rate law has to be suggested.

(c)

Expert Solution

Explanation of Solution

Another mechanism that does not agree with the same rate law is given below.

The imaginary mechanism:

NO+Cl2⇌k−1k1NOCl+Cl SlowNO +Cl→k2NOCl Fast

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

Rate=k2[NO][Cl2]

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

4. a) Give a suitable rationale for the following cyclization, stating the type of process involved (e.g. 9-endo-dig), clearly showing the mechanistic details at each step. H CO₂Me 1) NaOMe 2) H3O® CO₂Me

2. Platinum and other group 10 metals often act as solid phase hydrogenation catalysts for unsaturated hydrocarbons such as propylene, CH3CHCH2. In order for the reaction to be catalyzed the propylene molecules must first adsorb onto the surface. In order to completely cover the surface of a piece of platinum that has an area of 1.50 cm² with propylene, a total of 3.45 x 10¹7 molecules are needed. Determine the mass of the propylene molecules that have been absorbed onto the platinum surface.

Chem 141, Dr. Haefner 2. (a) Many main group oxides form acidic solutions when added to water. For example solid tetraphosphorous decaoxide reacts with water to produce phosphoric acid. Write a balanced chemical equation for this reaction. (b) Calcium phosphate reacts with silicon dioxide and carbon graphite at elevated temperatures to produce white phosphorous (P4) as a gas along with calcium silicate (Silcate ion is SiO3²-) and carbon monoxide. Write a balanced chemical equation for this reaction.

Answer 2

Question

Chapter 11, Problem 71QRT

(a)

Interpretation Introduction

Interpretation:

The rate law for the given mechanism has to be determined.

The reaction:

2NO(g)+Cl2(g)→2NOCl(g)

The accepted mechanism:

NO+Cl2⇌NOCl2 fastNOCl2+NO→2NOCl slow

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.

(a)

Expert Solution

Explanation of Solution

The reaction:

2NO(g)+Cl2(g)→2NOCl(g)

The accepted mechanism:

NO+Cl2⇌k−1k1NOCl2 fastNOCl2+NO→k22NOCl slow

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

Rate=k2[NOCl2][NO]

Applying the steady-state approximation on the intermediate (NOCl2), the concentration of the intermediate 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.

NOCl2 is created in the forward reaction of step 1 and destroyed in the reverse reaction of step 1 and step 2. Thus,

Rateforward reaction 1=k1[NO][Cl2]Ratereverse reaction 1=k−1[NOCl2]Ratereaction 2=k2[NOCl2][NO]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Ratereaction 2k1[NO][Cl2]=k−1[NOCl2]+k2[NOCl2][NO]

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 1≅Ratereverse reaction 1k1[NO][Cl2]=k−1[NOCl2][NOCl2]=k1k−1[NO][Cl2]

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

Rate=k2(k1k−1[NO][Cl2])[NO]Rate=k'[NO]2[Cl2] [∵k'=k2k1k−1]

Therefore, the rate law of the reaction is Rate=k'[NO]2[Cl2] .

(b)

Interpretation Introduction

Interpretation:

Another mechanism that agrees with the same rate law has to be suggested.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Another mechanism that agrees with the same rate law is given below.

The imaginary mechanism:

2NO2⇌k−1k1N2O2 FastN2O2 +Cl2→k22NOCl2 Slow

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

Rate=k2[N2O2][Cl2]

Applying the steady-state approximation on the intermediate (N2O2), the concentration of the intermediate 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.

N2O2 is created in the forward reaction of step 1 and destroyed in the reverse reaction of step 1 and step 2. Thus,

Rateforward reaction 1=k1[NO2]2Ratereverse reaction 1=k−1[N2O2]Ratereaction 2=k2[N2O2][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Ratereaction 2k1[NO]2=k−1[N2O2]+k2[N2O2][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 1≅Ratereverse reaction 1k1[NO]2=k−1[N2O2][N2O2]=k1k−1[NO]2

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

Rate=k2(k1k−1[NO]2)[Cl2]Rate=k'[NO]2[Cl2] [∵k'=k2k1k−1]

Therefore, the rate law of the reaction is Rate=k'[NO]2[Cl2] .

(c)

Interpretation Introduction

Interpretation:

Another mechanism that does not agree with the same rate law has to be suggested.

(c)

Expert Solution

Explanation of Solution

Another mechanism that does not agree with the same rate law is given below.

The imaginary mechanism:

NO+Cl2⇌k−1k1NOCl+Cl SlowNO +Cl→k2NOCl Fast

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

Rate=k2[NO][Cl2]

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 71QRT

(a)

Interpretation Introduction

Interpretation:

The rate law for the given mechanism has to be determined.

The reaction:

2NO(g)+Cl2(g)→2NOCl(g)

The accepted mechanism:

NO+Cl2⇌NOCl2 fastNOCl2+NO→2NOCl slow

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.

(a)

Expert Solution

Explanation of Solution

The reaction:

2NO(g)+Cl2(g)→2NOCl(g)

The accepted mechanism:

NO+Cl2⇌k−1k1NOCl2 fastNOCl2+NO→k22NOCl slow

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

Rate=k2[NOCl2][NO]

Applying the steady-state approximation on the intermediate (NOCl2), the concentration of the intermediate 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.

NOCl2 is created in the forward reaction of step 1 and destroyed in the reverse reaction of step 1 and step 2. Thus,

Rateforward reaction 1=k1[NO][Cl2]Ratereverse reaction 1=k−1[NOCl2]Ratereaction 2=k2[NOCl2][NO]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Ratereaction 2k1[NO][Cl2]=k−1[NOCl2]+k2[NOCl2][NO]

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 1≅Ratereverse reaction 1k1[NO][Cl2]=k−1[NOCl2][NOCl2]=k1k−1[NO][Cl2]

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

Rate=k2(k1k−1[NO][Cl2])[NO]Rate=k'[NO]2[Cl2] [∵k'=k2k1k−1]

Therefore, the rate law of the reaction is Rate=k'[NO]2[Cl2] .

(b)

Interpretation Introduction

Interpretation:

Another mechanism that agrees with the same rate law has to be suggested.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Another mechanism that agrees with the same rate law is given below.

The imaginary mechanism:

2NO2⇌k−1k1N2O2 FastN2O2 +Cl2→k22NOCl2 Slow

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

Rate=k2[N2O2][Cl2]

Applying the steady-state approximation on the intermediate (N2O2), the concentration of the intermediate 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.

N2O2 is created in the forward reaction of step 1 and destroyed in the reverse reaction of step 1 and step 2. Thus,

Rateforward reaction 1=k1[NO2]2Ratereverse reaction 1=k−1[N2O2]Ratereaction 2=k2[N2O2][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Ratereaction 2k1[NO]2=k−1[N2O2]+k2[N2O2][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 1≅Ratereverse reaction 1k1[NO]2=k−1[N2O2][N2O2]=k1k−1[NO]2

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

Rate=k2(k1k−1[NO]2)[Cl2]Rate=k'[NO]2[Cl2] [∵k'=k2k1k−1]

Therefore, the rate law of the reaction is Rate=k'[NO]2[Cl2] .

(c)

Interpretation Introduction

Interpretation:

Another mechanism that does not agree with the same rate law has to be suggested.

(c)

Expert Solution

Explanation of Solution

Another mechanism that does not agree with the same rate law is given below.

The imaginary mechanism:

NO+Cl2⇌k−1k1NOCl+Cl SlowNO +Cl→k2NOCl Fast

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

Rate=k2[NO][Cl2]

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 71QRT

(a)

Interpretation Introduction

Interpretation:

The rate law for the given mechanism has to be determined.

The reaction:

2NO(g)+Cl2(g)→2NOCl(g)

The accepted mechanism:

NO+Cl2⇌NOCl2 fastNOCl2+NO→2NOCl slow

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.

(a)

Expert Solution

Explanation of Solution

The reaction:

2NO(g)+Cl2(g)→2NOCl(g)

The accepted mechanism:

NO+Cl2⇌k−1k1NOCl2 fastNOCl2+NO→k22NOCl slow

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

Rate=k2[NOCl2][NO]

Applying the steady-state approximation on the intermediate (NOCl2), the concentration of the intermediate 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.

NOCl2 is created in the forward reaction of step 1 and destroyed in the reverse reaction of step 1 and step 2. Thus,

Rateforward reaction 1=k1[NO][Cl2]Ratereverse reaction 1=k−1[NOCl2]Ratereaction 2=k2[NOCl2][NO]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Ratereaction 2k1[NO][Cl2]=k−1[NOCl2]+k2[NOCl2][NO]

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 1≅Ratereverse reaction 1k1[NO][Cl2]=k−1[NOCl2][NOCl2]=k1k−1[NO][Cl2]

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

Rate=k2(k1k−1[NO][Cl2])[NO]Rate=k'[NO]2[Cl2] [∵k'=k2k1k−1]

Therefore, the rate law of the reaction is Rate=k'[NO]2[Cl2] .

(b)

Interpretation Introduction

Interpretation:

Another mechanism that agrees with the same rate law has to be suggested.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Another mechanism that agrees with the same rate law is given below.

The imaginary mechanism:

2NO2⇌k−1k1N2O2 FastN2O2 +Cl2→k22NOCl2 Slow

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

Rate=k2[N2O2][Cl2]

Applying the steady-state approximation on the intermediate (N2O2), the concentration of the intermediate 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.

N2O2 is created in the forward reaction of step 1 and destroyed in the reverse reaction of step 1 and step 2. Thus,

Rateforward reaction 1=k1[NO2]2Ratereverse reaction 1=k−1[N2O2]Ratereaction 2=k2[N2O2][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Ratereaction 2k1[NO]2=k−1[N2O2]+k2[N2O2][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 1≅Ratereverse reaction 1k1[NO]2=k−1[N2O2][N2O2]=k1k−1[NO]2

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

Rate=k2(k1k−1[NO]2)[Cl2]Rate=k'[NO]2[Cl2] [∵k'=k2k1k−1]

Therefore, the rate law of the reaction is Rate=k'[NO]2[Cl2] .

(c)

Interpretation Introduction

Interpretation:

Another mechanism that does not agree with the same rate law has to be suggested.

(c)

Expert Solution

Explanation of Solution

Another mechanism that does not agree with the same rate law is given below.

The imaginary mechanism:

NO+Cl2⇌k−1k1NOCl+Cl SlowNO +Cl→k2NOCl Fast

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

Rate=k2[NO][Cl2]

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 71QRT

(a)

Interpretation Introduction

Interpretation:

The rate law for the given mechanism has to be determined.

The reaction:

2NO(g)+Cl2(g)→2NOCl(g)

The accepted mechanism:

NO+Cl2⇌NOCl2 fastNOCl2+NO→2NOCl slow

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.

(a)

Expert Solution

Explanation of Solution

The reaction:

2NO(g)+Cl2(g)→2NOCl(g)

The accepted mechanism:

NO+Cl2⇌k−1k1NOCl2 fastNOCl2+NO→k22NOCl slow

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

Rate=k2[NOCl2][NO]

Applying the steady-state approximation on the intermediate (NOCl2), the concentration of the intermediate 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.

NOCl2 is created in the forward reaction of step 1 and destroyed in the reverse reaction of step 1 and step 2. Thus,

Rateforward reaction 1=k1[NO][Cl2]Ratereverse reaction 1=k−1[NOCl2]Ratereaction 2=k2[NOCl2][NO]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Ratereaction 2k1[NO][Cl2]=k−1[NOCl2]+k2[NOCl2][NO]

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 1≅Ratereverse reaction 1k1[NO][Cl2]=k−1[NOCl2][NOCl2]=k1k−1[NO][Cl2]

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

Rate=k2(k1k−1[NO][Cl2])[NO]Rate=k'[NO]2[Cl2] [∵k'=k2k1k−1]

Therefore, the rate law of the reaction is Rate=k'[NO]2[Cl2] .

(b)

Interpretation Introduction

Interpretation:

Another mechanism that agrees with the same rate law has to be suggested.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Another mechanism that agrees with the same rate law is given below.

The imaginary mechanism:

2NO2⇌k−1k1N2O2 FastN2O2 +Cl2→k22NOCl2 Slow

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

Rate=k2[N2O2][Cl2]

Applying the steady-state approximation on the intermediate (N2O2), the concentration of the intermediate 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.

N2O2 is created in the forward reaction of step 1 and destroyed in the reverse reaction of step 1 and step 2. Thus,

Rateforward reaction 1=k1[NO2]2Ratereverse reaction 1=k−1[N2O2]Ratereaction 2=k2[N2O2][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Ratereaction 2k1[NO]2=k−1[N2O2]+k2[N2O2][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 1≅Ratereverse reaction 1k1[NO]2=k−1[N2O2][N2O2]=k1k−1[NO]2

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

Rate=k2(k1k−1[NO]2)[Cl2]Rate=k'[NO]2[Cl2] [∵k'=k2k1k−1]

Therefore, the rate law of the reaction is Rate=k'[NO]2[Cl2] .

(c)

Interpretation Introduction

Interpretation:

Another mechanism that does not agree with the same rate law has to be suggested.

(c)

Expert Solution

Explanation of Solution

Another mechanism that does not agree with the same rate law is given below.

The imaginary mechanism:

NO+Cl2⇌k−1k1NOCl+Cl SlowNO +Cl→k2NOCl Fast

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

Rate=k2[NO][Cl2]

Answer 6

Chapter 11, Problem 71QRT

(a)

Interpretation Introduction

Interpretation:

The rate law for the given mechanism has to be determined.

The reaction:

2NO(g)+Cl2(g)→2NOCl(g)

The accepted mechanism:

NO+Cl2⇌NOCl2 fastNOCl2+NO→2NOCl slow

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.

(a)

Expert Solution

Explanation of Solution

The reaction:

2NO(g)+Cl2(g)→2NOCl(g)

The accepted mechanism:

NO+Cl2⇌k−1k1NOCl2 fastNOCl2+NO→k22NOCl slow

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

Rate=k2[NOCl2][NO]

Applying the steady-state approximation on the intermediate (NOCl2), the concentration of the intermediate 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.

NOCl2 is created in the forward reaction of step 1 and destroyed in the reverse reaction of step 1 and step 2. Thus,

Rateforward reaction 1=k1[NO][Cl2]Ratereverse reaction 1=k−1[NOCl2]Ratereaction 2=k2[NOCl2][NO]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Ratereaction 2k1[NO][Cl2]=k−1[NOCl2]+k2[NOCl2][NO]

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 1≅Ratereverse reaction 1k1[NO][Cl2]=k−1[NOCl2][NOCl2]=k1k−1[NO][Cl2]

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

Rate=k2(k1k−1[NO][Cl2])[NO]Rate=k'[NO]2[Cl2] [∵k'=k2k1k−1]

Therefore, the rate law of the reaction is Rate=k'[NO]2[Cl2] .

Answer 7

Chapter 11, Problem 71QRT

(a)

Interpretation Introduction

Interpretation:

The rate law for the given mechanism has to be determined.

The reaction:

2NO(g)+Cl2(g)→2NOCl(g)

The accepted mechanism:

NO+Cl2⇌NOCl2 fastNOCl2+NO→2NOCl slow

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

(a)

Expert Solution

Explanation of Solution

The reaction:

2NO(g)+Cl2(g)→2NOCl(g)

The accepted mechanism:

NO+Cl2⇌k−1k1NOCl2 fastNOCl2+NO→k22NOCl slow

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

Rate=k2[NOCl2][NO]

Applying the steady-state approximation on the intermediate (NOCl2), the concentration of the intermediate 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.

NOCl2 is created in the forward reaction of step 1 and destroyed in the reverse reaction of step 1 and step 2. Thus,

Rateforward reaction 1=k1[NO][Cl2]Ratereverse reaction 1=k−1[NOCl2]Ratereaction 2=k2[NOCl2][NO]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Ratereaction 2k1[NO][Cl2]=k−1[NOCl2]+k2[NOCl2][NO]

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 1≅Ratereverse reaction 1k1[NO][Cl2]=k−1[NOCl2][NOCl2]=k1k−1[NO][Cl2]

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

Rate=k2(k1k−1[NO][Cl2])[NO]Rate=k'[NO]2[Cl2] [∵k'=k2k1k−1]

Therefore, the rate law of the reaction is Rate=k'[NO]2[Cl2] .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

Another mechanism that agrees with the same rate law has to be suggested.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Another mechanism that agrees with the same rate law is given below.

The imaginary mechanism:

2NO2⇌k−1k1N2O2 FastN2O2 +Cl2→k22NOCl2 Slow

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

Rate=k2[N2O2][Cl2]

Applying the steady-state approximation on the intermediate (N2O2), the concentration of the intermediate 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.

N2O2 is created in the forward reaction of step 1 and destroyed in the reverse reaction of step 1 and step 2. Thus,

Rateforward reaction 1=k1[NO2]2Ratereverse reaction 1=k−1[N2O2]Ratereaction 2=k2[N2O2][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Ratereaction 2k1[NO]2=k−1[N2O2]+k2[N2O2][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 1≅Ratereverse reaction 1k1[NO]2=k−1[N2O2][N2O2]=k1k−1[NO]2

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

Rate=k2(k1k−1[NO]2)[Cl2]Rate=k'[NO]2[Cl2] [∵k'=k2k1k−1]

Therefore, the rate law of the reaction is Rate=k'[NO]2[Cl2] .

Answer 13

(b)

Interpretation Introduction

Interpretation:

Another mechanism that agrees with the same rate law has to be suggested.

Concept Introduction:

Refer to part (a).

Answer 14

(b)

Expert Solution

Explanation of Solution

Another mechanism that agrees with the same rate law is given below.

The imaginary mechanism:

2NO2⇌k−1k1N2O2 FastN2O2 +Cl2→k22NOCl2 Slow

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

Rate=k2[N2O2][Cl2]

Applying the steady-state approximation on the intermediate (N2O2), the concentration of the intermediate 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.

N2O2 is created in the forward reaction of step 1 and destroyed in the reverse reaction of step 1 and step 2. Thus,

Rateforward reaction 1=k1[NO2]2Ratereverse reaction 1=k−1[N2O2]Ratereaction 2=k2[N2O2][Cl2]

Now,

Rateforward reaction 1=Ratereverse reaction 1+Ratereaction 2k1[NO]2=k−1[N2O2]+k2[N2O2][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 1≅Ratereverse reaction 1k1[NO]2=k−1[N2O2][N2O2]=k1k−1[NO]2

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

Rate=k2(k1k−1[NO]2)[Cl2]Rate=k'[NO]2[Cl2] [∵k'=k2k1k−1]

Therefore, the rate law of the reaction is Rate=k'[NO]2[Cl2] .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

Another mechanism that does not agree with the same rate law has to be suggested.

(c)

Expert Solution

Explanation of Solution

Another mechanism that does not agree with the same rate law is given below.

The imaginary mechanism:

NO+Cl2⇌k−1k1NOCl+Cl SlowNO +Cl→k2NOCl Fast

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

Rate=k2[NO][Cl2]

Answer 19

(c)

Interpretation Introduction

Interpretation:

Another mechanism that does not agree with the same rate law has to be suggested.

Answer 20

(c)

Expert Solution

Explanation of Solution

Another mechanism that does not agree with the same rate law is given below.

The imaginary mechanism:

NO+Cl2⇌k−1k1NOCl+Cl SlowNO +Cl→k2NOCl Fast

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

Rate=k2[NO][Cl2]

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

Ch. 11.1 - For the reaction of crystal violet with NaOH(aq),...Ch. 11.1 - (a) From data in Table 11.1, calculate the rate of...Ch. 11.1 - For the reaction 4NO2(g)+O2(g)2N2O5(g) (a) express...Ch. 11.1 - Instantaneous rates for the reaction of hydroxide...Ch. 11.1 - Prob. 11.3CE Ch. 11.2 - Prob. 11.4E Ch. 11.2 - Prob. 11.3PSP Ch. 11.2 - Prob. 11.5E Ch. 11.3 - Prob. 11.4PSP Ch. 11.3 - Prob. 11.5PSP

Videos

Explanation of Solution

Interpretation:

Another mechanism that agrees with the same rate law has to be suggested.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Interpretation:

Another mechanism that does not agree with the same rate law has to be suggested.

Explanation of Solution

Want to see more full solutions like this?

Chapter 11 Solutions

Videos

Explanation of Solution

Interpretation: Another mechanism that agrees with the same rate law has to be suggested. Concept Introduction: Refer to part (a).

Explanation of Solution

Interpretation: Another mechanism that does not agree with the same rate law has to be suggested.

Explanation of Solution

Want to see more full solutions like this?

Chapter 11 Solutions

Interpretation:

Another mechanism that agrees with the same rate law has to be suggested.

Concept Introduction:

Refer to part (a).

Interpretation:

Another mechanism that does not agree with the same rate law has to be suggested.