General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 14, Problem 14.107SP

(a)

Interpretation Introduction

Interpretation:

The rate constant of the given reaction has to be calculated.

Concept introduction:

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life.

  • Half-life of a reaction is represented by the symbol as t12
  • Half-life is discovered by Ernest Rutherford's in 1907 from the original term half-life period.
  • The half-life period is then shortened as half-life in early 1950s.

Arrhenius equation:

  • Arrhenius equation is a formula that represents the temperature dependence of reaction rates
  • The Arrhenius equation has to be represented as follows

k=AeEa/RT

  • Ea represents the activation energy and it’s unit is kJ/mol
  • R represents the universal gas constant and it has the value of 8.314 J/K.mol
  • T represents the absolute temperature
  • A represents the frequency factor or collision frequency
  • e represents the base of natural logarithm
  •  Arrhenius equation equation was proposed by Svante Arrhenius in 1889.
  • The activation energy for the given decomposition can be calculated from the modified Arrhenius equation.

lnk1k2=EaR[1T21T1]

Reaction: Substances which are mutually involved each other in a chemical process and changed into different substances.

(a)

Expert Solution
Check Mark

Answer to Problem 14.107SP

The rate constant of the given reaction is k=0.0350min1

Explanation of Solution

We can easily determine the rate constant of a reaction if half-life value is given by using half-life formula.

k=0.693t12

k=0.69319.8min

k=0.0350min1

(b)

Interpretation Introduction

Interpretation:

The activation energy for the decomposition of benzoyl peroxide has to be calculated.

Concept introduction:

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life.

  • Half-life of a reaction is represented by the symbol as t12
  • Half-life is discovered by Ernest Rutherford's in 1907 from the original term half-life period.
  • The half-life period is then shortened as half-life in early 1950s.

Arrhenius equation:

  • Arrhenius equation is a formula that represents the temperature dependence of reaction rates
  • The Arrhenius equation has to be represented as follows

k=AeEa/RT

  • Ea represents the activation energy and it’s unit is kJ/mol
  • R represents the universal gas constant and it has the value of 8.314 J/K.mol
  • T represents the absolute temperature
  • A represents the frequency factor or collision frequency
  • e represents the base of natural logarithm
  •  Arrhenius equation equation was proposed by Svante Arrhenius in 1889.
  • The activation energy for the given decomposition can be calculated from the modified Arrhenius equation.

lnk1k2=EaR[1T21T1]

Reaction: Substances which are mutually involved each other in a chemical process and changed into different substances.

(b)

Expert Solution
Check Mark

Answer to Problem 14.107SP

The activation energy for the decomposition of benzoyl peroxide is Ea=1.1×105J/mol = 110kJ/mol

Explanation of Solution

The activation energy can be calculated by using modified Arrhenius equation and it can represented as follows

lnk1k2=EaR[1T21T1]

Where, k1 value is determined in step (a) k1=0.0350min1

Now, the given rate constant at 70°C is k2=1.58×103min1

Substitute the given values in the above said equation to get activation energy for the

ln0.0350min-11.58×10-3min-1=Ea(8.314J/K.mol)[1343K-1373K]

Ea=1.1×105J/mol = 110kJ/mol

(c)

Interpretation Introduction

Interpretation:

The reactant, product, and intermediate form the given elementary steps have to be written.

Concept introduction:

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life.

  • Half-life of a reaction is represented by the symbol as t12
  • Half-life is discovered by Ernest Rutherford's in 1907 from the original term half-life period.
  • The half-life period is then shortened as half-life in early 1950s.

Arrhenius equation:

  • Arrhenius equation is a formula that represents the temperature dependence of reaction rates
  • The Arrhenius equation has to be represented as follows

k=AeEa/RT

  • Ea represents the activation energy and it’s unit is kJ/mol
  • R represents the universal gas constant and it has the value of 8.314 J/K.mol
  • T represents the absolute temperature
  • A represents the frequency factor or collision frequency
  • e represents the base of natural logarithm
  •  Arrhenius equation equation was proposed by Svante Arrhenius in 1889.
  • The activation energy for the given decomposition can be calculated from the modified Arrhenius equation.

lnk1k2=EaR[1T21T1]

Reaction: Substances which are mutually involved each other in a chemical process and changed into different substances.

(c)

Expert Solution
Check Mark

Answer to Problem 14.107SP

The reactant, product, and intermediate form the given elementary steps are written.

Explanation of Solution

The given steps are all elementary steps, the rate law for the given steps can be deduced simply as follows.

Initiation:           rate = ki[R2]Propagation:      rate = kp[M][M1]Termination:       rate = kt[M'][M'']

  • Reactants are: ethylene monomers
  • Product is: polyethylene
  • Intermediate are: M’, M”, and so on

 (R-species also qualifies as an intermediate)

(d)

Interpretation Introduction

Interpretation:

What condition would favour the growth of long, high-molar-mass polyethylenes has to be explained.

Concept introduction:

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life.

  • Half-life of a reaction is represented by the symbol as t12
  • Half-life is discovered by Ernest Rutherford's in 1907 from the original term half-life period.
  • The half-life period is then shortened as half-life in early 1950s.

Arrhenius equation:

  • Arrhenius equation is a formula that represents the temperature dependence of reaction rates
  • The Arrhenius equation has to be represented as follows

k=AeEa/RT

  • Ea represents the activation energy and it’s unit is kJ/mol
  • R represents the universal gas constant and it has the value of 8.314 J/K.mol
  • T represents the absolute temperature
  • A represents the frequency factor or collision frequency
  • e represents the base of natural logarithm
  •  Arrhenius equation equation was proposed by Svante Arrhenius in 1889.
  • The activation energy for the given decomposition can be calculated from the modified Arrhenius equation.

lnk1k2=EaR[1T21T1]

Reaction: Substances which are mutually involved each other in a chemical process and changed into different substances.

(d)

Expert Solution
Check Mark

Answer to Problem 14.107SP

What condition would favour the growth of long, high-molar-mass polyethylenes is explained.

Explanation of Solution

  • A high rate of propagations and a low rate of termination will favour the growth of long polymers.
  • The rate law of propagation depends on the concentration of ethylene monomer, when we increase the concentration of ethylene the rate of the propagation also increases.
  • The rate law of termination shows that the low concentration of the radical fragment M’ or M” and is lead to slower the rate of termination.  Which is accomplished by taking a low concentration of the initiator, R2

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Chapter 14 Solutions

General Chemistry

Ch. 14.4 - Practice Exercise The first-order rate constant...Ch. 14.4 - Review of Concepts (a) What can you deduce about...Ch. 14.5 - Practice Exercise The reaction between NO2 and CO...Ch. 14.5 - Prob. 1RCCh. 14.6 - Prob. 1RCCh. 14 - Prob. 14.1QPCh. 14 - 15.2 Explain the difference between physical...Ch. 14 - Prob. 14.3QPCh. 14 - Prob. 14.4QPCh. 14 - Prob. 14.5QPCh. 14 - 14.6 Consider the reaction Suppose that at a...Ch. 14 - Prob. 14.7QPCh. 14 - 14.8 What are the units for the rate constants of...Ch. 14 - Prob. 14.9QPCh. 14 - Prob. 14.10QPCh. 14 - Prob. 14.11QPCh. 14 - Prob. 14.13QPCh. 14 - Prob. 14.14QPCh. 14 - Prob. 14.15QPCh. 14 - Prob. 14.16QPCh. 14 - Prob. 14.17QPCh. 14 - Prob. 14.18QPCh. 14 - Prob. 14.19QPCh. 14 - Prob. 14.20QPCh. 14 - 14.21 What is the half-life of a compound if 75...Ch. 14 - 14.22 The thermal decomposition of phosphine (PH3)...Ch. 14 - Prob. 14.23QPCh. 14 - Prob. 14.24QPCh. 14 - 14.25 Consider the first-order reaction A → B...Ch. 14 - Prob. 14.26QPCh. 14 - 14.27 Define activation energy. What role does...Ch. 14 - Prob. 14.28QPCh. 14 - Prob. 14.29QPCh. 14 - 14.30 As we know, methane burns readily in oxygen...Ch. 14 - Prob. 14.31QPCh. 14 - Prob. 14.32QPCh. 14 - Prob. 14.33QPCh. 14 - Prob. 14.34QPCh. 14 - Prob. 14.35QPCh. 14 - Prob. 14.36QPCh. 14 - Prob. 14.37QPCh. 14 - 14.38 The rate at which tree crickets chirp is 2.0...Ch. 14 - 14.39 The diagram here describes the initial state...Ch. 14 - Prob. 14.40QPCh. 14 - Prob. 14.41QPCh. 14 - Prob. 14.42QPCh. 14 - 14.43 Explain why termolecular reactions are...Ch. 14 - 14.44 What is the rate-determining step of a...Ch. 14 - Prob. 14.45QPCh. 14 - Prob. 14.46QPCh. 14 - Prob. 14.47QPCh. 14 - Prob. 14.48QPCh. 14 - Prob. 14.49QPCh. 14 - Prob. 14.50QPCh. 14 - Prob. 14.51QPCh. 14 - Prob. 14.52QPCh. 14 - Prob. 14.53QPCh. 14 - Prob. 14.54QPCh. 14 - Prob. 14.55QPCh. 14 - Prob. 14.56QPCh. 14 - Prob. 14.57QPCh. 14 - Prob. 14.58QPCh. 14 - Prob. 14.59QPCh. 14 - Prob. 14.60QPCh. 14 - Prob. 14.61QPCh. 14 - Prob. 14.62QPCh. 14 - Prob. 14.63QPCh. 14 - Prob. 14.64QPCh. 14 - Prob. 14.65QPCh. 14 - 14.66 The decomposition of N2O to N2 and O2 is a...Ch. 14 - Prob. 14.67QPCh. 14 - Prob. 14.68QPCh. 14 - 14.69 Consider the zero-order reaction a → B....Ch. 14 - Prob. 14.70QPCh. 14 - Prob. 14.72QPCh. 14 - Prob. 14.73QPCh. 14 - Prob. 14.74QPCh. 14 - Prob. 14.75QPCh. 14 - Prob. 14.76QPCh. 14 - Prob. 14.77QPCh. 14 - Prob. 14.78QPCh. 14 - Prob. 14.79QPCh. 14 - Prob. 14.80QPCh. 14 - Prob. 14.81QPCh. 14 - Prob. 14.82QPCh. 14 - 14.83 When a mixture of methane and bromine is...Ch. 14 - 14.84 Consider this elementary step: (a)...Ch. 14 - Prob. 14.85QPCh. 14 - Prob. 14.86QPCh. 14 - 14.87 In recent years ozone in the stratosphere...Ch. 14 - Prob. 14.88QPCh. 14 - Prob. 14.90QPCh. 14 - Prob. 14.91QPCh. 14 - Prob. 14.92QPCh. 14 - Prob. 14.93QPCh. 14 - Prob. 14.94QPCh. 14 - Prob. 14.95QPCh. 14 - Prob. 14.96QPCh. 14 - Prob. 14.97QPCh. 14 - Prob. 14.98QPCh. 14 - Prob. 14.100QPCh. 14 - Prob. 14.101QPCh. 14 - 14.102 Consider the potential energy profiles...Ch. 14 - Prob. 14.103QPCh. 14 - Prob. 14.104QPCh. 14 - 14.105 The activation energy for the...Ch. 14 - Prob. 14.106QPCh. 14 - Prob. 14.107SPCh. 14 - Prob. 14.108SPCh. 14 - Prob. 14.109SPCh. 14 - Prob. 14.110SPCh. 14 - Prob. 14.111SPCh. 14 - Prob. 14.112SPCh. 14 - Prob. 14.113SPCh. 14 - Prob. 14.114SPCh. 14 - Prob. 14.115SPCh. 14 - 14.116 To prevent brain damage, a drastic medical...Ch. 14 - Prob. 14.117SPCh. 14 - Prob. 14.118SP
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