EBK CHEMICAL PRINCIPLES
EBK CHEMICAL PRINCIPLES
8th Edition
ISBN: 8220101425812
Author: DECOSTE
Publisher: Cengage Learning US
Question
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Chapter 17, Problem 26E

(a)

Interpretation Introduction

Interpretation:

The sign and magnitude of ΔH1,ΔH2,ΔH3 and ΔHsoln for the solute-solvent combinations needs to be determined.

  EBK CHEMICAL PRINCIPLES, Chapter 17, Problem 26E , additional homework tip  1

Concept Introduction :

The intermolecular forces in a solute can be broken down by new interactions from the solution as each solute particle will be surrounded by solvent particles in a solution. This is possible when there is disruption between the solute -solute and solvent-solvent interaction.

(a)

Expert Solution
Check Mark

Answer to Problem 26E

The values for ΔH1=Small;ΔH2=Large; ΔH3=Small

Therefore ΔHsoln=Large positive and hence no solution will form

Explanation of Solution

The process of formation of solution takes place in 3 main steps

  1. The solutes separating into individual components require energy making it an endothermic reaction.
  2. The intermolecular forces in solvent must be such that it can make space for solute which requires energy making it an endothermic reaction.
  3. To allow the solvent and solute molecules to interact which absorbs energy making it an exothermic reaction.

The formation of solution involves enthalpy changes which is depicted as

  ΔHsoln=ΔH1+ΔH2+ΔH3

    ΔH1ΔH2ΔH3ΔHsolnResult
    Non polar solute, Non polar solventSmallSmallSmallSmallSolution will form
    Non-polar solute, polar solventSmallLargeSmallLarge, positiveNo-solution will form
    Polar solute, Non polar solventLargeSmallSmallLarge negativeNo solution will form
    Polar solute-polar solvent LargeLargeLarge, negativesmallSolution will form

In this reaction, acetone reacts with water and as water is polar in nature, the values of each enthalpy changes are depicted as

    ΔH1ΔH2ΔH3ΔHsolnOutcome
    Polar solute, Non polar solventLargeSmallSmallLarge negativeNo solution will form

(b)

Interpretation Introduction

Interpretation:

The sign and magnitude of ΔH1,ΔH2,ΔH3 and ΔHsoln for the solute-solvent combinations needs to be determined.

  EBK CHEMICAL PRINCIPLES, Chapter 17, Problem 26E , additional homework tip  2

Concept Introduction :

The intermolecular forces in a solute can be broken down by new interactions from the solution as each solute particle will be surrounded by solvent particles in a solution. This is possible when there is disruption between the solute -solute and solvent-solvent interaction.

(b)

Expert Solution
Check Mark

Answer to Problem 26E

The values for ΔH1=Large;ΔH2=Large; ΔH3=Large (negative)

Therefore ΔHsoln=Small and hence solution will form

Explanation of Solution

The process of formation of solution takes place in 3 main steps

  1. The solutes separating into individual components require energy making it an endothermic reaction.
  2. The intermolecular forces in solvent must be such that it can make space for solute which requires energy making it an endothermic reaction.
  3. To allow the solvent and solute molecules to interact which absorbs energy making it an exothermic reaction.

The formation of solution involves enthalpy changes which is depicted as

  ΔHsoln=ΔH1+ΔH2+ΔH3

    ΔH1ΔH2ΔH3ΔHsolnResult
    Non polar solute, Non polar solventSmallSmallSmallSmallSolution will form
    Non-polar solute, polar solventSmallLargeSmallLarge, positiveNo-solution will form
    Polar solute, Non polar solventLargeSmallSmallLarge negativeNo solution will form
    Polar solute-polar solvent LargeLargeLarge, negativesmallSolution will form

In this reaction, ethanol reacts with water and as water is polar in nature, the values of each enthalpy changes are depicted as

    ΔH1ΔH2ΔH3ΔHsolnOutcome
    Polar solute-polar solvent LargeLargeLarge, negativesmallSolution will form

(c)

Interpretation Introduction

Interpretation:

The sign and magnitude of ΔH1,ΔH2,ΔH3 and ΔHsoln for the solute-solvent combinations needs to be determined.

  EBK CHEMICAL PRINCIPLES, Chapter 17, Problem 26E , additional homework tip  3EBK CHEMICAL PRINCIPLES, Chapter 17, Problem 26E , additional homework tip  4

Concept Introduction :

The intermolecular forces in a solute can be broken down by new interactions from the solution as each solute particle will be surrounded by solvent particles in a solution. This is possible when there is disruption between the solute -solute and solvent-solvent interaction.

(c)

Expert Solution
Check Mark

Answer to Problem 26E

The values for ΔH1=Small;ΔH2=Small; ΔH3=Small

Therefore ΔHsoln=Small and hence solution will form

Explanation of Solution

The process of formation of solution takes place in 3 main steps

  1. The solutes separating into individual components require energy making it an endothermic reaction.
  2. The intermolecular forces in solvent must be such that it can make space for solute which requires energy making it an endothermic reaction.
  3. To allow the solvent and solute molecules to interact which absorbs energy making it an exothermic reaction.

The formation of solution involves enthalpy changes which is depicted as:

  ΔHsoln=ΔH1+ΔH2+ΔH3

    ΔH1ΔH2ΔH3ΔHsolnResult
    Non polar solute, Non polar solventSmallSmallSmallSmallSolution will form
    Non-polar solute, polar solventSmallLargeSmallLarge, positiveNo-solution will form
    Polar solute, Non polar solventLargeSmallSmallLarge negativeNo solution will form
    Polar solute-polar solvent LargeLargeLarge, negativesmallSolution will form

In this reaction, heptane reacts with hexane and as both is non-polar in nature; the values of each enthalpy changes are depicted as:

    ΔH1ΔH2ΔH3ΔHsolnOutcome
    Non polar solute, Non polar solventSmallSmallSmallSmallSolution will form

(d)

Interpretation Introduction

Interpretation:

The sign and magnitude of ΔH1,ΔH2,ΔH3 and ΔHsoln for the solute-solvent combinations needs to be determined.

  EBK CHEMICAL PRINCIPLES, Chapter 17, Problem 26E , additional homework tip  5EBK CHEMICAL PRINCIPLES, Chapter 17, Problem 26E , additional homework tip  6

Concept Introduction :

The intermolecular forces in a solute can be broken down by new interactions from the solution as each solute particle will be surrounded by solvent particles in a solution. This can happen when there is disruption between the solute-solute and solvent-solvent interaction.

(d)

Expert Solution
Check Mark

Answer to Problem 26E

The values for ΔH1=Small;ΔH2=Large; ΔH3=Large positive

Therefore ΔHsoln=Large positive  and hence solution will not form

Explanation of Solution

The process of formation of solution takes place in 3 main steps

  1. The solutes separating into individual components require energy making it an endothermic reaction.
  2. The intermolecular forces in solvent must be such that it can make space for solute which requires energy making it an endothermic reaction.
  3. To allow the solvent and solute molecules to interact which absorbs energy making it an exothermic reaction.

The formation of solution involves enthalpy changes which are depicted as:

  ΔHsoln=ΔH1+ΔH2+ΔH3

    ΔH1ΔH2ΔH3ΔHsolnResult
    Non polar solute, Non polar solventSmallSmallSmallSmallSolution will form
    Non-polar solute, polar solventSmallLargeSmallLarge, positiveNo-solution will form
    Polar solute, Non polar solventLargeSmallSmallLarge negativeNo solution will form
    Polar solute-polar solvent LargeLargeLarge, negativesmallSolution will form

In this reaction, heptane reacts with water wherein heptane is non-polar and water is polar in nature. The values of each enthalpy changes are depicted as:

    ΔH1ΔH2ΔH3ΔHsolnResult
    Non-polar solute, polar solventSmallLargeSmallLarge, positiveSolution will not form.

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Chapter 17 Solutions

EBK CHEMICAL PRINCIPLES

Ch. 17 - Prob. 11DQCh. 17 - Prob. 12ECh. 17 - Prob. 13ECh. 17 - Prob. 14ECh. 17 - Prob. 15ECh. 17 - Prob. 16ECh. 17 - Prob. 17ECh. 17 - Prob. 18ECh. 17 - Prob. 19ECh. 17 - Prob. 20ECh. 17 - Prob. 21ECh. 17 - Prob. 22ECh. 17 - Prob. 23ECh. 17 - Prob. 24ECh. 17 - Prob. 25ECh. 17 - Prob. 26ECh. 17 - Prob. 27ECh. 17 - Prob. 28ECh. 17 - Prob. 29ECh. 17 - Prob. 30ECh. 17 - Prob. 31ECh. 17 - Prob. 32ECh. 17 - Prob. 33ECh. 17 - Prob. 34ECh. 17 - Prob. 35ECh. 17 - Prob. 36ECh. 17 - Prob. 37ECh. 17 - Prob. 38ECh. 17 - Prob. 39ECh. 17 - Prob. 40ECh. 17 - Rationalize the temperature dependence of the...Ch. 17 - Prob. 42ECh. 17 - Prob. 43ECh. 17 - Prob. 44ECh. 17 - Prob. 45ECh. 17 - Prob. 46ECh. 17 - Prob. 47ECh. 17 - Prob. 48ECh. 17 - Prob. 49ECh. 17 - Prob. 50ECh. 17 - Prob. 51ECh. 17 - Prob. 52ECh. 17 - Prob. 53ECh. 17 - Prob. 54ECh. 17 - Prob. 55ECh. 17 - Prob. 56ECh. 17 - The following plot shows the vapor pressure of...Ch. 17 - Prob. 58ECh. 17 - Prob. 59ECh. 17 - Prob. 60ECh. 17 - Prob. 61ECh. 17 - Prob. 62ECh. 17 - Prob. 63ECh. 17 - Prob. 64ECh. 17 - Prob. 65ECh. 17 - Prob. 66ECh. 17 - Prob. 67ECh. 17 - An aqueous solution of 10.00 g of catalase, an...Ch. 17 - Prob. 69ECh. 17 - What volume of ethylene glycol (C2H6O2) , a...Ch. 17 - Prob. 71ECh. 17 - Erythrocytes are red blood cells containing...Ch. 17 - Prob. 73ECh. 17 - Prob. 74ECh. 17 - Prob. 75ECh. 17 - Prob. 76ECh. 17 - Prob. 77ECh. 17 - Prob. 78ECh. 17 - Prob. 79ECh. 17 - Prob. 80ECh. 17 - Consider the following solutions: 0.010 m Na3PO4...Ch. 17 - From the following: pure water solution of...Ch. 17 - Prob. 83ECh. 17 - Prob. 84ECh. 17 - Prob. 85ECh. 17 - Prob. 86ECh. 17 - Prob. 87ECh. 17 - Prob. 88ECh. 17 - Prob. 89ECh. 17 - Prob. 90ECh. 17 - Prob. 91ECh. 17 - Prob. 92ECh. 17 - Prob. 93AECh. 17 - Prob. 94AECh. 17 - Prob. 95AECh. 17 - Prob. 96AECh. 17 - The term proof is defined as twice the percent by...Ch. 17 - Prob. 98AECh. 17 - Prob. 99AECh. 17 - Prob. 100AECh. 17 - Prob. 101AECh. 17 - Prob. 102AECh. 17 - Prob. 103AECh. 17 - Prob. 104AECh. 17 - Prob. 105AECh. 17 - Prob. 106AECh. 17 - Prob. 107AECh. 17 - Prob. 108AECh. 17 - Prob. 109AECh. 17 - Prob. 110AECh. 17 - Prob. 111AECh. 17 - Prob. 112AECh. 17 - Prob. 113AECh. 17 - Prob. 114AECh. 17 - Formic acid (HCO2H) is a monoprotic acid that...Ch. 17 - Prob. 116AECh. 17 - Prob. 117AECh. 17 - Prob. 118AECh. 17 - Prob. 119AECh. 17 - Prob. 120AECh. 17 - Prob. 121AECh. 17 - Prob. 122AECh. 17 - Prob. 123AECh. 17 - Prob. 124AECh. 17 - Prob. 125AECh. 17 - Prob. 126AECh. 17 - Prob. 127CPCh. 17 - Prob. 128CPCh. 17 - Prob. 129CPCh. 17 - Plants that thrive in salt water must have...Ch. 17 - Prob. 131CPCh. 17 - Prob. 132CPCh. 17 - Prob. 133CPCh. 17 - Prob. 134CPCh. 17 - Prob. 135CPCh. 17 - Prob. 136CP
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