The following rate constants were obtained in an experiment in which the decomposition of gaseous N 2 O ; was studied as a function of temperature. The products were NO, and NO,. Temperature (K) 3.5 x 10_i 298 2.2 x 10"4 308 6.8 X IO-4 318 3.1 x 10 1 328 Determine E t for this reaction in kj/mol.

Question

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Answer 1

Textbook Question

Chapter 11, Problem 11.57PAE

The following rate constants were obtained in an experiment in which the decomposition of gaseous N₂O_; was studied as a function of temperature. The products were NO, and NO,.

	Temperature (K)
3.5 x 10_i	298
2.2 x 10"4	308
6.8 X IO-4	318
3.1 x 10 1	328

Determine E_tfor this reaction in kj/mol.

Expert Solution & Answer

Interpretation Introduction

Interpretation:Ea should be determined in kJ/mol for the reaction of decomposition of N2O5 to NO2 and NO3 when the details of a study of that reaction as a function of temperature are given.

Concept introduction:

Rate of a reaction can be explained using either growth of products or reduction of reactants. Both give the same rate, but our concern is different.

Every reaction has activation energy. To overcome this energy sometimes we should provide energy from outside. But for some reactions, we do not have to supply energy. Ambient temperature is enough that reactions. Those reactions are called spontaneous reactions.

Answer to Problem 11.57PAE

Solution:

Ea = 140.277 kJmol−1

Given:

Chemical reaction

N2O5→NO2+ NO3

k/s−1	Temperature/K
3.5×10−5	298
2.2×10−4	308
6.8×10−4	318
3.1×10−3	328

Explanation of Solution

N2O5→NO2+ NO3

The rate of the equation for the reaction can be written as follows.

R= −k [Ν2Ο5]

The only equation relating activation energy and rate constant is Arrhenius equation which is given below. The frequency factor doesn’t depend on the temperature.

Bundle: Chemistry for Engineering Students, 3rd, Loose-Leaf + OWLv2 with QuickPrep 24-Months Printed Access Card, Chapter 11, Problem 11.57PAE

It can be written as

lnk=lnA−EaRT

Therefore, at two different temperatures at T1 and T2 ;

ln k1= ln A – E a R T 1 →1ln k2= ln A – E a R T 2 →2

When equation 1 is subtracted from equation 2,

ln (k2/k1) = EaR (1T1– 1T2)

Formula used:

ln (k2/k1) = EaR (1T1– 1T2)

Calculation:

ln ( k 2 / k 1 ) = E a R ( 1 T 1 – 1 T 2 ) Substitution of valuesln ( 2.2× 10 −4 /3.5× 10 −5 ) = E a 8.314 ( 1 298 − 1 308 )Ea = 140.277 kJmol −1

Conclusion

Ea = 140.277 kJmol−1

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Answer 2

Textbook Question

Chapter 11, Problem 11.57PAE

The following rate constants were obtained in an experiment in which the decomposition of gaseous N₂O_; was studied as a function of temperature. The products were NO, and NO,.

	Temperature (K)
3.5 x 10_i	298
2.2 x 10"4	308
6.8 X IO-4	318
3.1 x 10 1	328

Determine E_tfor this reaction in kj/mol.

Expert Solution & Answer

Interpretation Introduction

Interpretation:Ea should be determined in kJ/mol for the reaction of decomposition of N2O5 to NO2 and NO3 when the details of a study of that reaction as a function of temperature are given.

Concept introduction:

Rate of a reaction can be explained using either growth of products or reduction of reactants. Both give the same rate, but our concern is different.

Every reaction has activation energy. To overcome this energy sometimes we should provide energy from outside. But for some reactions, we do not have to supply energy. Ambient temperature is enough that reactions. Those reactions are called spontaneous reactions.

Answer to Problem 11.57PAE

Solution:

Ea = 140.277 kJmol−1

Given:

Chemical reaction

N2O5→NO2+ NO3

k/s−1	Temperature/K
3.5×10−5	298
2.2×10−4	308
6.8×10−4	318
3.1×10−3	328

Explanation of Solution

N2O5→NO2+ NO3

The rate of the equation for the reaction can be written as follows.

R= −k [Ν2Ο5]

The only equation relating activation energy and rate constant is Arrhenius equation which is given below. The frequency factor doesn’t depend on the temperature.

Bundle: Chemistry for Engineering Students, 3rd, Loose-Leaf + OWLv2 with QuickPrep 24-Months Printed Access Card, Chapter 11, Problem 11.57PAE

It can be written as

lnk=lnA−EaRT

Therefore, at two different temperatures at T1 and T2 ;

ln k1= ln A – E a R T 1 →1ln k2= ln A – E a R T 2 →2

When equation 1 is subtracted from equation 2,

ln (k2/k1) = EaR (1T1– 1T2)

Formula used:

ln (k2/k1) = EaR (1T1– 1T2)

Calculation:

ln ( k 2 / k 1 ) = E a R ( 1 T 1 – 1 T 2 ) Substitution of valuesln ( 2.2× 10 −4 /3.5× 10 −5 ) = E a 8.314 ( 1 298 − 1 308 )Ea = 140.277 kJmol −1

Conclusion

Ea = 140.277 kJmol−1

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

Answer 3

Textbook Question

Chapter 11, Problem 11.57PAE

The following rate constants were obtained in an experiment in which the decomposition of gaseous N₂O_; was studied as a function of temperature. The products were NO, and NO,.

	Temperature (K)
3.5 x 10_i	298
2.2 x 10"4	308
6.8 X IO-4	318
3.1 x 10 1	328

Determine E_tfor this reaction in kj/mol.

Expert Solution & Answer

Interpretation Introduction

Interpretation:Ea should be determined in kJ/mol for the reaction of decomposition of N2O5 to NO2 and NO3 when the details of a study of that reaction as a function of temperature are given.

Concept introduction:

Rate of a reaction can be explained using either growth of products or reduction of reactants. Both give the same rate, but our concern is different.

Every reaction has activation energy. To overcome this energy sometimes we should provide energy from outside. But for some reactions, we do not have to supply energy. Ambient temperature is enough that reactions. Those reactions are called spontaneous reactions.

Answer to Problem 11.57PAE

Solution:

Ea = 140.277 kJmol−1

Given:

Chemical reaction

N2O5→NO2+ NO3

k/s−1	Temperature/K
3.5×10−5	298
2.2×10−4	308
6.8×10−4	318
3.1×10−3	328

Explanation of Solution

N2O5→NO2+ NO3

The rate of the equation for the reaction can be written as follows.

R= −k [Ν2Ο5]

The only equation relating activation energy and rate constant is Arrhenius equation which is given below. The frequency factor doesn’t depend on the temperature.

Bundle: Chemistry for Engineering Students, 3rd, Loose-Leaf + OWLv2 with QuickPrep 24-Months Printed Access Card, Chapter 11, Problem 11.57PAE

It can be written as

lnk=lnA−EaRT

Therefore, at two different temperatures at T1 and T2 ;

ln k1= ln A – E a R T 1 →1ln k2= ln A – E a R T 2 →2

When equation 1 is subtracted from equation 2,

ln (k2/k1) = EaR (1T1– 1T2)

Formula used:

ln (k2/k1) = EaR (1T1– 1T2)

Calculation:

ln ( k 2 / k 1 ) = E a R ( 1 T 1 – 1 T 2 ) Substitution of valuesln ( 2.2× 10 −4 /3.5× 10 −5 ) = E a 8.314 ( 1 298 − 1 308 )Ea = 140.277 kJmol −1

Conclusion

Ea = 140.277 kJmol−1

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Answer 4

Textbook Question

Chapter 11, Problem 11.57PAE

The following rate constants were obtained in an experiment in which the decomposition of gaseous N₂O_; was studied as a function of temperature. The products were NO, and NO,.

	Temperature (K)
3.5 x 10_i	298
2.2 x 10"4	308
6.8 X IO-4	318
3.1 x 10 1	328

Determine E_tfor this reaction in kj/mol.

Expert Solution & Answer

Interpretation Introduction

Interpretation:Ea should be determined in kJ/mol for the reaction of decomposition of N2O5 to NO2 and NO3 when the details of a study of that reaction as a function of temperature are given.

Concept introduction:

Rate of a reaction can be explained using either growth of products or reduction of reactants. Both give the same rate, but our concern is different.

Every reaction has activation energy. To overcome this energy sometimes we should provide energy from outside. But for some reactions, we do not have to supply energy. Ambient temperature is enough that reactions. Those reactions are called spontaneous reactions.

Answer to Problem 11.57PAE

Solution:

Ea = 140.277 kJmol−1

Given:

Chemical reaction

N2O5→NO2+ NO3

k/s−1	Temperature/K
3.5×10−5	298
2.2×10−4	308
6.8×10−4	318
3.1×10−3	328

Explanation of Solution

N2O5→NO2+ NO3

The rate of the equation for the reaction can be written as follows.

R= −k [Ν2Ο5]

The only equation relating activation energy and rate constant is Arrhenius equation which is given below. The frequency factor doesn’t depend on the temperature.

Bundle: Chemistry for Engineering Students, 3rd, Loose-Leaf + OWLv2 with QuickPrep 24-Months Printed Access Card, Chapter 11, Problem 11.57PAE

It can be written as

lnk=lnA−EaRT

Therefore, at two different temperatures at T1 and T2 ;

ln k1= ln A – E a R T 1 →1ln k2= ln A – E a R T 2 →2

When equation 1 is subtracted from equation 2,

ln (k2/k1) = EaR (1T1– 1T2)

Formula used:

ln (k2/k1) = EaR (1T1– 1T2)

Calculation:

ln ( k 2 / k 1 ) = E a R ( 1 T 1 – 1 T 2 ) Substitution of valuesln ( 2.2× 10 −4 /3.5× 10 −5 ) = E a 8.314 ( 1 298 − 1 308 )Ea = 140.277 kJmol −1

Conclusion

Ea = 140.277 kJmol−1

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

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

Interpretation Introduction

Interpretation:Ea should be determined in kJ/mol for the reaction of decomposition of N2O5 to NO2 and NO3 when the details of a study of that reaction as a function of temperature are given.

Concept introduction:

Rate of a reaction can be explained using either growth of products or reduction of reactants. Both give the same rate, but our concern is different.

Every reaction has activation energy. To overcome this energy sometimes we should provide energy from outside. But for some reactions, we do not have to supply energy. Ambient temperature is enough that reactions. Those reactions are called spontaneous reactions.

Answer to Problem 11.57PAE

Solution:

Ea = 140.277 kJmol−1

Given:

Chemical reaction

N2O5→NO2+ NO3

k/s−1	Temperature/K
3.5×10−5	298
2.2×10−4	308
6.8×10−4	318
3.1×10−3	328

Explanation of Solution

N2O5→NO2+ NO3

The rate of the equation for the reaction can be written as follows.

R= −k [Ν2Ο5]

The only equation relating activation energy and rate constant is Arrhenius equation which is given below. The frequency factor doesn’t depend on the temperature.

Bundle: Chemistry for Engineering Students, 3rd, Loose-Leaf + OWLv2 with QuickPrep 24-Months Printed Access Card, Chapter 11, Problem 11.57PAE

It can be written as

lnk=lnA−EaRT

Therefore, at two different temperatures at T1 and T2 ;

ln k1= ln A – E a R T 1 →1ln k2= ln A – E a R T 2 →2

When equation 1 is subtracted from equation 2,

ln (k2/k1) = EaR (1T1– 1T2)

Formula used:

ln (k2/k1) = EaR (1T1– 1T2)

Calculation:

ln ( k 2 / k 1 ) = E a R ( 1 T 1 – 1 T 2 ) Substitution of valuesln ( 2.2× 10 −4 /3.5× 10 −5 ) = E a 8.314 ( 1 298 − 1 308 )Ea = 140.277 kJmol −1

Conclusion

Ea = 140.277 kJmol−1

Answer 8

Expert Solution & Answer

Interpretation Introduction

Interpretation:Ea should be determined in kJ/mol for the reaction of decomposition of N2O5 to NO2 and NO3 when the details of a study of that reaction as a function of temperature are given.

Concept introduction:

Rate of a reaction can be explained using either growth of products or reduction of reactants. Both give the same rate, but our concern is different.

Every reaction has activation energy. To overcome this energy sometimes we should provide energy from outside. But for some reactions, we do not have to supply energy. Ambient temperature is enough that reactions. Those reactions are called spontaneous reactions.

Answer to Problem 11.57PAE

Solution:

Ea = 140.277 kJmol−1

Given:

Chemical reaction

N2O5→NO2+ NO3

k/s−1	Temperature/K
3.5×10−5	298
2.2×10−4	308
6.8×10−4	318
3.1×10−3	328

Explanation of Solution

N2O5→NO2+ NO3

The rate of the equation for the reaction can be written as follows.

R= −k [Ν2Ο5]

The only equation relating activation energy and rate constant is Arrhenius equation which is given below. The frequency factor doesn’t depend on the temperature.

Bundle: Chemistry for Engineering Students, 3rd, Loose-Leaf + OWLv2 with QuickPrep 24-Months Printed Access Card, Chapter 11, Problem 11.57PAE

It can be written as

lnk=lnA−EaRT

Therefore, at two different temperatures at T1 and T2 ;

ln k1= ln A – E a R T 1 →1ln k2= ln A – E a R T 2 →2

When equation 1 is subtracted from equation 2,

ln (k2/k1) = EaR (1T1– 1T2)

Formula used:

ln (k2/k1) = EaR (1T1– 1T2)

Calculation:

ln ( k 2 / k 1 ) = E a R ( 1 T 1 – 1 T 2 ) Substitution of valuesln ( 2.2× 10 −4 /3.5× 10 −5 ) = E a 8.314 ( 1 298 − 1 308 )Ea = 140.277 kJmol −1

Conclusion

Ea = 140.277 kJmol−1

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Interpretation Introduction

Interpretation:Ea should be determined in kJ/mol for the reaction of decomposition of N2O5 to NO2 and NO3 when the details of a study of that reaction as a function of temperature are given.

Concept introduction:

Rate of a reaction can be explained using either growth of products or reduction of reactants. Both give the same rate, but our concern is different.

Every reaction has activation energy. To overcome this energy sometimes we should provide energy from outside. But for some reactions, we do not have to supply energy. Ambient temperature is enough that reactions. Those reactions are called spontaneous reactions.

Answer 12

Answer to Problem 11.57PAE

Solution:

Ea = 140.277 kJmol−1

Given:

Chemical reaction

N2O5→NO2+ NO3

k/s−1	Temperature/K
3.5×10−5	298
2.2×10−4	308
6.8×10−4	318
3.1×10−3	328

Answer 13

Explanation of Solution

N2O5→NO2+ NO3

The rate of the equation for the reaction can be written as follows.

R= −k [Ν2Ο5]

The only equation relating activation energy and rate constant is Arrhenius equation which is given below. The frequency factor doesn’t depend on the temperature.

Bundle: Chemistry for Engineering Students, 3rd, Loose-Leaf + OWLv2 with QuickPrep 24-Months Printed Access Card, Chapter 11, Problem 11.57PAE

It can be written as

lnk=lnA−EaRT

Therefore, at two different temperatures at T1 and T2 ;

ln k1= ln A – E a R T 1 →1ln k2= ln A – E a R T 2 →2