(a) Interpretation: A graph of ln k v/s 1/T is to be constructed for the given data of decomposition of PAN. 0 °C k = 5.6 × 10 − 6 s − 1 10 °C k = 3.2 × 10 − 5 s − 1 20 °C k = 1.6 × 10 − 4 s − 1 30 °C k = 7.6 × 10 − 4 s − 1 Concept introduction: Arrhenius equation relates the rate constant (k) of the reaction to its absolute temperature as- k = A e - E a R T o r l n k = l n A - E a R T Where A − Pre-exponential factor E a - Activation energy T- Absolute temperature in Kelvins R- Universal gas constant

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Answer 1

Question

Chapter 20, Problem 53E

Interpretation Introduction

(a)

Interpretation:

A graph of ln k v/s 1/T is to be constructed for the given data of decomposition of PAN.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

Concept introduction:

Arrhenius equation relates the rate constant (k) of the reaction to its absolute temperature as-

k=Ae - E a RTorlnk=lnA-EaRT

Where

A − Pre-exponential factor

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Interpretation Introduction

(b)

Interpretation:

The activation energy (E_a ) is to be calculated for the given data of decomposition of PAN.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

Concept introduction:

Arrhenius equation relates the rate constant (k) of the reaction to its absolute temperature as-

k=Ae -E a RTorlnk=lnA-EaRT

Where

A − Pre-exponential factor

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Interpretation Introduction

(c)

Interpretation:

The half-life for the decomposition of PAN at 40°C is to be calculated.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

E_a = 149.6 KJ/mol

Concept introduction:

At two different temperatures, Arrhenius equation can be modified as:

lnk2k1=-EaR(1T2-1T1)

Where k − rate constants

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Expert Solution & Answer

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Answer 2

Question

Chapter 20, Problem 53E

Interpretation Introduction

(a)

Interpretation:

A graph of ln k v/s 1/T is to be constructed for the given data of decomposition of PAN.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

Concept introduction:

Arrhenius equation relates the rate constant (k) of the reaction to its absolute temperature as-

k=Ae - E a RTorlnk=lnA-EaRT

Where

A − Pre-exponential factor

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Interpretation Introduction

(b)

Interpretation:

The activation energy (E_a ) is to be calculated for the given data of decomposition of PAN.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

Concept introduction:

Arrhenius equation relates the rate constant (k) of the reaction to its absolute temperature as-

k=Ae -E a RTorlnk=lnA-EaRT

Where

A − Pre-exponential factor

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Interpretation Introduction

(c)

Interpretation:

The half-life for the decomposition of PAN at 40°C is to be calculated.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

E_a = 149.6 KJ/mol

Concept introduction:

At two different temperatures, Arrhenius equation can be modified as:

lnk2k1=-EaR(1T2-1T1)

Where k − rate constants

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 20, Problem 53E

Interpretation Introduction

(a)

Interpretation:

A graph of ln k v/s 1/T is to be constructed for the given data of decomposition of PAN.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

Concept introduction:

Arrhenius equation relates the rate constant (k) of the reaction to its absolute temperature as-

k=Ae - E a RTorlnk=lnA-EaRT

Where

A − Pre-exponential factor

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Interpretation Introduction

(b)

Interpretation:

The activation energy (E_a ) is to be calculated for the given data of decomposition of PAN.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

Concept introduction:

Arrhenius equation relates the rate constant (k) of the reaction to its absolute temperature as-

k=Ae -E a RTorlnk=lnA-EaRT

Where

A − Pre-exponential factor

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Interpretation Introduction

(c)

Interpretation:

The half-life for the decomposition of PAN at 40°C is to be calculated.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

E_a = 149.6 KJ/mol

Concept introduction:

At two different temperatures, Arrhenius equation can be modified as:

lnk2k1=-EaR(1T2-1T1)

Where k − rate constants

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 20, Problem 53E

Interpretation Introduction

(a)

Interpretation:

A graph of ln k v/s 1/T is to be constructed for the given data of decomposition of PAN.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

Concept introduction:

Arrhenius equation relates the rate constant (k) of the reaction to its absolute temperature as-

k=Ae - E a RTorlnk=lnA-EaRT

Where

A − Pre-exponential factor

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Interpretation Introduction

(b)

Interpretation:

The activation energy (E_a ) is to be calculated for the given data of decomposition of PAN.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

Concept introduction:

Arrhenius equation relates the rate constant (k) of the reaction to its absolute temperature as-

k=Ae -E a RTorlnk=lnA-EaRT

Where

A − Pre-exponential factor

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Interpretation Introduction

(c)

Interpretation:

The half-life for the decomposition of PAN at 40°C is to be calculated.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

E_a = 149.6 KJ/mol

Concept introduction:

At two different temperatures, Arrhenius equation can be modified as:

lnk2k1=-EaR(1T2-1T1)

Where k − rate constants

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 20, Problem 53E

Interpretation Introduction

(a)

Interpretation:

A graph of ln k v/s 1/T is to be constructed for the given data of decomposition of PAN.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

Concept introduction:

Arrhenius equation relates the rate constant (k) of the reaction to its absolute temperature as-

k=Ae - E a RTorlnk=lnA-EaRT

Where

A − Pre-exponential factor

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Interpretation Introduction

(b)

Interpretation:

The activation energy (E_a ) is to be calculated for the given data of decomposition of PAN.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

Concept introduction:

Arrhenius equation relates the rate constant (k) of the reaction to its absolute temperature as-

k=Ae -E a RTorlnk=lnA-EaRT

Where

A − Pre-exponential factor

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Interpretation Introduction

(c)

Interpretation:

The half-life for the decomposition of PAN at 40°C is to be calculated.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

E_a = 149.6 KJ/mol

Concept introduction:

At two different temperatures, Arrhenius equation can be modified as:

lnk2k1=-EaR(1T2-1T1)

Where k − rate constants

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Answer 6

Chapter 20, Problem 53E

Interpretation Introduction

(a)

Interpretation:

A graph of ln k v/s 1/T is to be constructed for the given data of decomposition of PAN.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

Concept introduction:

Arrhenius equation relates the rate constant (k) of the reaction to its absolute temperature as-

k=Ae - E a RTorlnk=lnA-EaRT

Where

A − Pre-exponential factor

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Answer 7

Chapter 20, Problem 53E

Interpretation Introduction

(a)

Interpretation:

A graph of ln k v/s 1/T is to be constructed for the given data of decomposition of PAN.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

Concept introduction:

Arrhenius equation relates the rate constant (k) of the reaction to its absolute temperature as-

k=Ae - E a RTorlnk=lnA-EaRT

Where

A − Pre-exponential factor

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Answer 8

Interpretation Introduction

(b)

Interpretation:

The activation energy (E_a ) is to be calculated for the given data of decomposition of PAN.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

Concept introduction:

Arrhenius equation relates the rate constant (k) of the reaction to its absolute temperature as-

k=Ae -E a RTorlnk=lnA-EaRT

Where

A − Pre-exponential factor

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Answer 9

Interpretation Introduction

(b)

Interpretation:

The activation energy (E_a ) is to be calculated for the given data of decomposition of PAN.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

Concept introduction:

Arrhenius equation relates the rate constant (k) of the reaction to its absolute temperature as-

k=Ae -E a RTorlnk=lnA-EaRT

Where

A − Pre-exponential factor

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Answer 10

Interpretation Introduction

(c)

Interpretation:

The half-life for the decomposition of PAN at 40°C is to be calculated.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

E_a = 149.6 KJ/mol

Concept introduction:

At two different temperatures, Arrhenius equation can be modified as:

lnk2k1=-EaR(1T2-1T1)

Where k − rate constants

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Answer 11

Interpretation Introduction

(c)

Interpretation:

The half-life for the decomposition of PAN at 40°C is to be calculated.

0 °C k = 5.6×10−6s−1

10 °C k = 3.2×10−5s−1

20 °C k = 1.6×10−4s−1

30 °C k = 7.6×10−4s−1

E_a = 149.6 KJ/mol

Concept introduction:

At two different temperatures, Arrhenius equation can be modified as:

lnk2k1=-EaR(1T2-1T1)

Where k − rate constants

E_a- Activation energy

T- Absolute temperature in Kelvins

R- Universal gas constant

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

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Check out a sample textbook solution

See solution

Answer 15

Ch. 20 - In the reaction 2A+BC+3D , reactant A is found to...Ch. 20 - From Figure 20-2 estimate the rate of reaction at...Ch. 20 - In the reaction A products, [A] is found to be...Ch. 20 - In the reaction A products, at t = 0. [A]= 0.1565...Ch. 20 - In the reaction A products. 4.40 min after the...Ch. 20 - Refer to Experiment 2 of Table 20.3 and to...Ch. 20 - For the reaction A+2BC ,the rate of reaction is...Ch. 20 - If the rate of reaction (20.3) is 5.7104 M s-1 ,...Ch. 20 - In the reaction A(g)B(g)+C(g) , the totalpressure...Ch. 20 - At 65C , the half-life for the first-order...

Solution Summary: The author explains that a graph of ln k v/s 1/T is to be constructed for the given data of decomposition of PAN.

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Chapter 20 Solutions

0 °C	k = 5.6×10−6s−1
10 °C	k = 3.2×10−5s−1
20 °C	k = 1.6×10−4s−1
30 °C	k = 7.6×10−4s−1