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Reactions and Solutions
Chapter Outline
CHEMISTRY CONNECTION: Seeing a Thought
7.1
Writing Chemical Reactions
Combination Reactions
Decomposition Reactions
Replacement Reactions
7.2
Types of Chemical Reactions
Precipitation Reactions
Reactions with Oxygen
Acid-Base Reactions
Oxidation-Reduction Reactions
7.3
Properties of Solutions
General Properties of Liquid Solutions
Solutions and Colloids
Degree of Solubility
Solubility and Equilibrium
Solubility of Gases - Henry's Law
7.4
Concentration of Solutions: Percentage
Weight/Volume Percent
A HUMAN PERSPECTIVE: Scuba Diving: Nitrogen and The Bends
Weight/Weight Percent
7.5
Concentration of Solutions: Moles and Equivalents
Molarity
Dilution
Representation of Concentration of Ions in Solution
7.6
Concentration-Dependent Solution Properties
Vapor Pressure Lowering
Freezing Point Depression and Boiling Point Elevation
Osmotic Pressure
A CLINICAL PERSPECTIVE: Oral Rehydration Therapy
7.7
Water as a Solvent
7.8
Electrolytes in Body Fluids
A HUMAN PERSPECTIVE: An Extraordinary Molecule
A CLINICAL PERSPECTIVE: Hemodialysis
Summary
Key Terms
Questions and Problems
Critical Thinking Problems
Instructional Objectives
Conceptual Objectives
• Recognize the various classes of chemical reactions: precipitation, reactions with oxygen, acidbase, and oxidation-reduction.
• Know what is meant by the terms solution, solute, and solvent.
• Describe the various types of solutions, and give examples of each.
• Know the importance of solution chemistry in chemical processes.
• Recognize the relationship between solubility and equilibrium.
•
•
Describe and explain concentration-dependent solution properties: vapor-pressure lowering,
freezing-point depression, boiling point elevation, and osmotic pressure.
Describe the chemical and physical properties of water that make it a truly unique solvent.
Performance Objectives
• Classify chemical reactions by type: combination, decomposition, or replacement.
• Calculate solution concentration in weight/volume percent and weight/weight percent.
• Calculate solution concentration using molarity.
• Perform dilution calculations.
• Interconvert molar concentration of ions and milliequivalents/liter.
Health Applications
• Understand the relationship of pressure and nitrogen solubility in the blood, which results in a
condition known as the bends.
• Understand the role of hemodialysis in removal of waste products from the blood.
In-Chapter Examples
Example 7.1:
Example 7.2:
Example 7.3:
Example 7.4:
Example 7.5:
Example 7.6:
Example 7.7:
Example 7.8:
Example 7.9:
Example 7.10:
Example 7.11:
Example 7.12:
Example 7.13:
Example 7.14:
Predicting whether precipitation will occur.
Calculating weight/volume percent.
Calculating the weight of solute from a weight/volume percent.
Calculating weight/weight percent.
Calculating molarity from moles.
Calculating molarity from mass.
Calculating volume from molarity.
Calculating molarity after dilution.
Calculating a dilution volume.
Calculating ion concentration (in units of equivalents/liter from a concentration
represented as moles/liter).
Calculating osmolarity.
Calculating osmotic pressure.
Predicting structure from observable properties.
Calculating electrolyte concentrations.
Chapter Overview
Writing Chemical Reactions
Chemical reactions may be classified as combination, decomposition, or replacement. Replacement
reactions are subclassified as either single- or double-replacement.
Types of Chemical Reactions
Reactions that produce products with similar characteristics are often classified as a single group. For
example, the formation of a precipitate denotes precipitation reactions.
Chemical reactions that have a common reactant may be grouped together. Reactions involving
oxygen - combustion reactions - are such a class.
Another approach to the classification of chemical reactions is based on charge transfer. Acid-base
reactions involve proton (H+) transfer and oxidation-reduction reactions involve the transfer of one or
more electrons.
Properties of Solutions
The majority of chemical reactions, and virtually all important organic and biochemical processes,
occur in solution. A solution is composed of one or more solutes dissolved in a solvent. In aqueous
solutions, the solvent is water.
Liquid solutions are clear and transparent with no visible particles of solute. They may be colored or
colorless, depending on the properties of the solute and solvent.
In solutions of electrolytes, the solutes are ionic compounds that have dissociated; the solution
conducts electricity. Solutions of nonelectrolytes are nonconducting.
The degree of solubility depends on the difference between the polarity of solute and solvent ("like
dissolves like"), the temperature, and, for solutions of gases, the pressure.
A saturated solution contains all the solute that can be dissolved at a particular temperature. A
supersaturated solution is an unstable condition that occurs when more than the theoretical amount of
solute is temporarily held in solution. Colloidal suspensions have particle sizes between those of true
solutions and precipitates. A suspension is a heterogeneous mixture that contains particles much larger
than a colloidal suspension. Over time, these particles may settle, forming a second phase.
Henry's law describes the solubility of gases in liquids. At a given temperature the solubility of a gas
is proportional to the partial pressure of the gas.
Concentration of Solutions: Percentage
The amount of solute dissolved in a given amount of solution is the solution concentration. The more
widely used percentage-based concentration units are weight/volume percent and weight/weight percent.
Concentration of Solutions: Moles and Equivalents
Molarity (M) is the number of moles of solute per liter of solution. Dilution calculations require both
concentrations (V1 and V2) to be in the same units.
When discussing solutions of ionic compounds, molarity emphasizes the number of individual ions.
In contrast, equivalents per liter emphasizes charge. One equivalent of an ion is the number of grams of
the ion corresponding to a mole of electrical charges. Changing from molarity to equivalents per liter (or
the reverse) is done using conversion factors.
Concentration-Dependent Solution Properties
Colligative properties are properties that depend on concentration rather than the identity of the
solute. Four colligative properties are: vapor pressure lowering, freezing-point depression, boiling-point
elevation and osmotic pressure. Each has a number of very practical ramifications.
Molality (m), is the number of moles of solute per kilogram of solvent in a solution, and is more
commonly used in calculations involving colligative properties. Osmolarity (osmol) is the molarity of
particles that is used for osmotic pressure () in the equation:  = MRT.
Hypertonic solutions cause the outflow of water from a cell, resulting in crenation (cell collapse).
Hypotonic solutions cause water to flow into the cell, resulting in hemolysis (cell rupture). Isotonic
solutions have identical osmotic pressures.
Water as a Solvent
The role of water in the solution process deserves special attention. Water is often referred to as the
"universal solvent," and it is the principal biological solvent. These characteristics are a direct
consequence of the molecular geometry and structure of water and its ability to undergo hydrogen
bonding.
Electrolytes in Body Fluids
The concentration of cations, anions, and other substances in biological fluids is critical to health. As
a result, the osmolarity of body fluids is carefully regulated by the kidney using the process of dialysis.
Hints for Faster Coverage
Molarity, osmolarity, and dilution are the most important concentration calculations. If desired, the
percentage composition calculations might be minimized or omitted.
Suggested Problem Sets
Chemical Reactions: 25, 27, 29, 31
Concentration of Solutions: Percentage: 33, 35, 37, 39
Concentration of Solutions: Moles and Equivalents: 41, 43, 45, 47, 49
Concentration-Dependent Solution Properties: 51, 53, 55, 57, 59
Water as a Solvent: 61, 63, 65
Electrolytes in Body Fluids: 67, 69, 71
In-Chapter Perspectives
A HUMAN PERSPECTIVE: Scuba Diving - Nitrogen and the Bends. The dependence of nitrogen
solubility in the blood on pressure is the basis for a discussion of a deep-water diver's worse fear, the
bends.
A CLINICAL PERSPECTIVE: Oral Rehydration Therapy. This perspective connects the concept of
osmolarity with the only effective treatment for cholera, oral rehydration therapy.
A HUMAN PERSPECTIVE: An Extraordinary Molecule. The structure-property relationship is best
exemplified by the molecule H2O. The role of water in environmental and biochemical systems is
prescribed by its unique set of physical and chemical properties that result from its size, shape and
electron arrangement.
A CLINICAL PERSPECTIVE: Hemodialysis. This perspective connects chemistry to medicine through
continually-improving technology. In the classroom, the instructor could solicit examples of other such
applications and this could help to promote classroom discussion.
Additional Perspectives
The relationship between concentration, discussed in the chapter, and dosage of medication may be
shown. The importance of minimizing side reactions, many potentially serious, is something which all
students interested in a health related profession must know. Many of the rules governing reactions in
beakers and flasks also apply to those reactions occurring in the human body.
Critical Thinking Problems
1. and 2. These questions emphasize the importance of the number of particles, rather than mass of
particles, in predicting the effect of concentration on colligative properties.
3., 4., 5., and 6. These problems test the ability of a student to design an experiment targeted to solve a
specific problem.
Chapter 7
Reactions and Solutions
Solutions to the Even-Numbered Problems
In-Chapter Questions and Problems
7.2
a.
b.
c.
d.
D
D
DR
C
7.4
a.
NaOH(aq) + NH4Cl(aq)  no reaction
The solubility rules predict that both potential products, sodium chloride (NaCl) and
ammonium hydroxide (NH4OH) are soluble; no precipitation reaction occurs.
b.
2NaOH(aq) + FeCl2(aq) --> 2NaCl(aq) + Fe(OH)2(s)
The solubility rules predict the iron(II) hydroxide is insoluble; a precipitation reaction
occurs.
7.6
grams of solute
milliliters of solution
% (W/V) =
x 100%
Solving for volume,
mL solution
=
g solute
x 100%
% (W/V)
=
10.0 g NaCl
x 100% = 40.0 mL
25.0 % (W/V)
Substituting,
mL solution
7.8
grams of solute
milliliters of solution
% (W/V) =
x 100%
Solving for mass,
g solute
=
Substituting,
g NaOH =
7.10
% (W/V) =
[% (W/V)](mL solution)
100%

103 mL 

(1.0 %)2.0 L x
1 L 

100%
grams of solute
milliliters of solution
1
= 2.0 x 10 g NaOH
x 100%
% (W/V) =
50.0 g Ar
3

10 mL solution 
476 L solution x

1 L solution 

% (W/V) = 10.5 x 10
7.12
-3
x 100%
%
% (W/W) =
grams of solute
grams of solution
% (W/W) =
50.0 g Ar
x 100% = 38.5 %
50.0 g Ar + 80.0 g He
7.14
M AgNO3 =
mol AgNO
L solution
x 100%
3
Solving for moles of AgNO3,
mol AgNO3 = (MAgNO3)(L solution)
mol AgNO3 = (0.500 M)(2.00 L) = 1.00 mol AgNO3
Then, convert mol AgNO3  g AgNO3
1.00 mol AgNO 3 x
7.16
169.9 g AgNO 3
1 mol AgNO 3
= 1.70 x 10 2 g AgNO 3
(M1)(V1) = (M2)(V2)
M1= 0.400 M
V1 = 50.0 mL x (1L/103 mL) = 5.00 x 10–2 L
M2 = 0.200 M
V2 = ? L
Solving for the final volume of 0.200 M sugar, V 2,
(M1)(V1)
(0.400 M)(5.00 x 10 -2 L)
V2 =
=
(M 2)
(0.200 M)
V2 = 1.00 x 10-1 L (or 1.00 x 102 mL) of 0.200 M sugar solution
7.18
C6H12O6 is a non-electrolyte; it does not dissociate.
5.0 x 10 -3 mol C 6H12O6
1 mol particles
x
L solution
1 mol C 6H12O 6
= 5.0 x 10 -3
mol particles
L solution
and,
5.0 x 10
7.20
-3
mol particles
L solution
= 5.0 x 10
-3
osmol
In Question 7.18 we found that the solution was 5.0 x 10 -3 osmol; substituting this value for M
(M represents osmolarity) in the expression:
 = MRT
 = 5.0 x 10 -3
mol particles
0.0821 L  atm
x
x (25 + 273K)
L solution
K  mol
 = 0.12 atm
7.22
Ammonia is a polar substance, as is water. The rule “like dissolves like” predicts that ammonia
would be water soluble. Methane, a nonpolar substance, would not be soluble in water.
7.24
+
40 meq K +
1 eq K+
1 mol charge
1 mol K +
-2 mol K
x 3
x
x
=
4
x
10
1L
1 mol charge
L
10 meq K +
1 eq K +
End-of Chapter Questions and Problems
7.26
a.
The combination of a metal and a non-metal to form a salt, for example,
Cu(s) + Cl2(g)  CuCl2(s)
b.
Reaction of two soluble substances to form an insoluble precipitate, for example,
HCl(aq) + AgNO3(aq)  AgCl(s) + HNO3(aq)
7.28
Any combustion reaction, for example, the combustion of butane,
2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(g)
7.30
a.
C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(g)
b.
6H2O(l) + P4O10(s)  4H3PO4(aq)
c.
PCl5(g) + 4H2O(l)  5HCl(aq) + H3PO4(aq)
a.
4Li(s) + O2(g)  2Li2O(s)
b.
3Ca(s) + N2(g)  Ca3N2(s)
c.
2Al(s) + 3S(s)  Al2S3(s)
7.32
7.34
% (W/V) =
grams of solute
milliliters of solution
a.
x 100%
0.700 g KCl
x 100% = 70.0 % KCl
1.00 mL
% (W/V) =
b.
95.21 g MgCl 2
1 mol MgCl 2
1.00 g MgCl 2 x
% (W/V) =
7.36
% (V/V) =
2.50 x 10 2 mL solution
x 100% = 38.1 % MgCl 2
mL solute
x 100%
mL solution
a.
% (V/V) =
20.0 mL acetic acid
3

10 mL soln 
2.50 L soln x

1 L soln 

x 100%
% (V/V) = 0.800 % acetic acid
b.
Use the density of benzene as a conversion factor to calculate the volume of benzene:
20.0 g benzene x
then
% (V/V) =
7.38
22.8 mL benzene
1.0 x 10 2 mL soln
grams of solute
grams of solution
% (W/W) =
a.
% (W/W) =
b.
1 mL
= 22.8 mL
0.879 g benzene
x 100% = 22.8 % benzene
x 100%
1.00 g KCl
1.00 x 10 2 g solution
x 100% = 1.00 % KCl
Use the density of the potassium chloride solution as a conversion factor to calculate the
volume of the potassium chloride solution.
2
5.00 x 10 mL soln x
Then,
% (W/W) =
1.14 g solution
1 mL solution
50.0 g KCl
5.70 x 10 2 g solution
2
= 5.70 x 10 g soln
x 100% = 8.77 % KCl
7.40
grams of solute
grams of solution
% (W/W) =
x 100%
Solve for g solute,
g solute =
[% (W/W)](g soln)
100%
a.
[5.00% (W/W)](2.50 x 10 2 g soln)
 12.5 g NH 4Cl
100%
g NH 4Cl =
b.
g Na 2CO 3 =
7.42
a.
M KCl =
M KCl
mol KCl
L solution

0.700 g KCl x 1 mol KCl 

74.55 g KCl 
=
= 9.39 M KCl

1L

1.00 mL solution x
3

10 mL solution 
b.
M MgCl2 =
M MgCl2 =
7.44
M =
[3.50% (W/W)](2.50 x 10 2 g soln)
 8.75 g Na 2CO3
100%
mol MgCl 2
L solution
1.00 mol MgCl 2 

1L

2.50 x 10 2 mL solution x
3

10 mL solution 
mol solute
L solution
Solve for mol solute,
mol solute = (M)(L solution)
a.
mol NaBr = (0.100 M)[(2.50 x 10 2 mL soln x (1 L soln/103 mL soln)]
mol NaBr = 2.50 x 10-2 mol NaBr
= 4.00 M MgCl 2
and,
b.
2.50 x 10
-2
102.9 g NaBr
1 mol NaBr
mol NaBr x
= 2.57 g NaBr
mol KOH = (0.200 M)[(2.50 x 102 mL soln x (1 L soln/103 mL soln)]
mol KOH = 5.00 x 10-2 mol KOH
and,
5.00 x 10
7.46
M AgCl =
M AgCl
7.48
-2
mol KOH x
56.11 g KOH
= 2.81 g KOH
1 mol KOH
mol AgCl
L solution

1.58 x 10 -4 g AgCl x 1 mol AgCl 

143.3 g AgCl 
-5
=
= 1.10 x 10 M AgCl

1L

1.00 x 10 2 mL solution x
3

10 mL solution 
(M1)(V1) = (M2)(V2)
M1 = 0.250M
V1 = 50.0 mL x (1 L/103 mL) = 5.00 x 10–2 L
M2 = ?
V2 = 5.00 x 102 mL x (1 L/103 mL) = 5.00 x 10–1 L
Solve for M2,
(M1)(V1)
V2
M2 =
Substitute,
M2 =
7.50
(0.250)(5.00 x 10 -2 )
5.00 x 10
-1
= 2.50 x 10
(M1)(V1) = (M2)(V2)
M1 = 8.00 M
V1 = 6.00 mL x (1 L/103 mL) = 6.00 x 10–3 L
M2 = 0.400 M
V2 = ?
Solve for V2,
Substitute,
V2 =
(M1 )(V1 )
M2
-2
M
V2 =
7.52
7.54
(8.00)(6.00 x 10 -2 )
= 1.20 x 10 -1 L
0.400
Four colligative solution properties are:
•
Vapor pressure lowering. Raoult's Law states that when a solute is added to a solvent, the
vapor pressure of the solvent decreases in proportion to the concentration of the solute.
•
Freezing-point depression. When a nonvolatile solute is added to a solvent, the freezingpoint of the resulting solution decreases.
•
Boiling-point elevation. When a nonvolatile solute is added to a solvent, the boiling point of
the resulting solution increases.
•
Osmosis. Osmosis is the movement of solvent from a dilute solution to a more concentrated
solution through a semipermeable membrane.
A wilted plant regains its "health" when watered due to osmosis. The water, a dilute solution of
ions, moves to a more concentrated region (the dehydrated plant). The resulting osmotic pressure
reverses the drooping phenomenon.
Overview of questions 7.56 - 7.60:
The molar concentration of both solutions is identical. Sodium chloride is an ionic compound
that dissociates in water to produce 2 ions for each ion pair. So, the molarity of particles is
2 x 0.50 M = 1.0 M.
Sucrose is a covalent, nondissociating compound; the number of particles is the same as the
number of molecules. Thus, the molarity of particles is 0.50 M. Armed with this information, we
can now answer the questions.
7.56
The NaCl solution would boil at a higher temperature since it has a larger number of particles/L.
Consequently, the sodium chloride solution would have a higher boiling point. (Remember,
solute particles elevate the boiling point).
7.58
Osmotic pressure results from the movement of solvent from a dilute solution to a more
concentrated solution. (Concentration refers to the number of particles/L.) Since the NaCl
solution has the higher concentration, it must have the lower osmotic pressure.
7.60
Using our definition of osmotic pressure,

= MRT
M must be represented as osmolarity.
M =
0.50 mol sucrose
1 mol particles
x
L
1 mol sucrose
=
0.50 mol particles
L
Now, substituting in our osmotic pressure equation:

= (0.50 mol particles/L) x (0.0821 Latm/Kmol particles) x 298 K = 12 atm
(Note, two significant figures for the answer; the sucrose concentration has two significant
figures)
7.62
The interactive network of water molecules in the liquid state is illustrated in Figure 6.8 of the
textbook. It should be emphasized that this structure is 3-dimensional and extends throughout
the liquid.
7.64
Water’s high boiling point ensures that water’s principal state at normal earth temperatures is
liquid, and the liquid phase is easily contained, stored, and transferred.
7.66
H
Cl–

O
H
The partial (+) end of the water molecule attracts the negative ion, Cl –.
7.68
If the dialysis solution has an elevated potassium ion concentration (perhaps equivalent to the
potassium ion concentration in the blood) there will be no net potassium ion loss due to the lack
of a potassium ion concentration difference between the two fluids. Remember, it is this
concentration difference that is the driving force for dialysis.
7.70
Depressed concentration of potassium ion in blood may lead to heart irregularities and ultimately
to death as a consequence of heart failure.
7.72
Dangerously low concentrations of potassium ion in blood may result from strenuous exercise
that results in excessive perspiration.
1. Classify the following reaction as precipitation, acid-base or
oxidation-reduction:
Ce4+(aq) + Fe2+(aq) ® Ce3+(aq) + Fe3+(aq)
2. Classify the following reaction as precipitation, acid-base or
oxidation-reduction:
H2SO4(aq) + 2KOH(aq) ® K2SO4(aq) + 2H2O(aq)
3. Classify the following reaction as precipitation, acid-base or
oxidation-reduction:
Na2S(aq) + CuSO4(aq) ® Na2SO4(aq) + CuS(s)
4. Complete the products and balance the following equation for an acidbase
reaction:
HCl(aq) + KOH(aq)
5. Complete the products and balance the following equation for a
precipitation reaction:
FeSO4(aq) + NaOH(aq)
6. Write a balanced equation for the reaction between zinc metal and
iron(III) ions to form zinc(II) ions and iron(II) ions (Symbols: zinc =
Zn; iron = Fe).
7. Name the two products formed when octane (C8H18) burns completely in
excess oxygen gas.
8. Classify the following reaction as decomposition, combination, singlereplacement
or double-replacement:
2H2O(l) ® 2H2(g) + O2(g)
9. Classify the following reaction as decomposition, combination, singlereplacement
or double-replacement:
H2SO4(aq) Ba(OH)2(aq) ® BaSO4(s) + 2H2O(g)
10. Classify the following reaction as decomposition, combination, singlereplacement
or double-replacement:
Mg(s) + 2HCl(aq) ® MgCl2(g) + H2(g)
11. Classify the following reaction as decomposition, combination, singlereplacement
or double-replacement:
2Na(s) + Cl2(g) ® 2NaCl(s)
12. What would you call a liquid that displayed the Tyndall effect?
13. Outline a simple method to distinguish between a beaker containing a
true solution and one containing a colloidal suspension.
14. A solution contains 1.65 g of NaOH in a total volume of 150.0 mL. What
is its concentration expressed as % (W/V)?
15. How many grams of glucose are present in 250.0 mL of a 5.00% (W/V)
solution?
16. Define "molarity".
Page 1
17. How many moles of KNO3 are contained in one liter of 0.2 M KNO3
solution?
18. How many moles of KCl are present in 50.0 mL of a 0.552 M solution?
19. How many grams of KCl are present in 250.0 mL of a 0.125 M solution?
[Use formula weight: KCl, 74.55 amu]
20. If 5.20 g of HCl is added to enough distilled water to form 3.00 L of
solution, what is the molarity of the solution? [Use molecular weight:
HCl, 36.46 amu]
21. Calculate the molarity of a solution if 300.0 mL of it contains 16.8 g
of KNO3. [Use formula weight: KNO3, 101.11 amu ]
22. What is the molarity of 50.0 mL of a 0.660 M NaOH solution after it has
been diluted to 450.0 mL?
23. In one sentence, explain what is meant by a colligative property.
24. An aqueous solution is warmed from 20°C to 30°C. Does this change in
temperature affect either the molarity or the molality of the solution?
Explain.
25. How much more will the freezing point of water be lowered by adding one
mole of sodium chloride rather than one mole of glucose, C6H12O6?
26. What type of membranes allow solvent molecules to pass through but do
not allow solute molecules to pass through?
27. What happens when a hypotonic solution is separated from a hypertonic
solution by an osmotic membrane?
28. What term do we use to describe a solution of lower osmolarity compared
to one with a higher osmolarity?
29. Describe what happens when red blood cells are placed in a hypertonic
solution. Give the name of the process which occurs
30. What happens to red blood cells that are placed in an isotonic
solution?
31. Calculate the osmolarity of a 2.0 ´ 10-3 M Na3PO4 solution. Na3PO4 is an
ionic compound and produces an electrolytic solution.
32. Calculate the osmotic pressure of a 6.0 ´ 10-2 M solution of NaCl at
20°C (293K).
33. What liquid is referred to as the "universal solvent"?
34. In what way are dialysis and osmosis similar? In what way are they
different?
35. Name two compounds which can move through the membrane used in
hemodialysis.
Page 2
36. Choose the best classification of the reaction represented by the
following equation:
C6H12O6(s) + 6O2(g) ® 6CO2(g) + 6H2O(l)
A. combustion
B. acid-base
C. precipitation
D. decomposition
E. combination
37. Choose the best classification of the reaction represented by the
following equation:
HNO3(aq) + KOH(aq) ® KNO3(aq) + H2O(l)
A. combustion
B. acid-base
C. precipitation
D. decomposition
E. combination
38. Choose the best classification of the reaction represented by the
following equation:
2Fe3+(aq) + Fe(s) ® 3Fe2+(aq)
A. combustion
B. acid-base
C. precipitation
D. decomposition
E. oxidation-reduction
39. Choose the best classification of the reaction represented by the
following equation:
Pb(NO3)2(aq) + CaCl2(aq) ® PbCl2(s) + Ca(NO3)2(aq)
A. combustion
B. acid-base
C. decomposition
D. precipitation
E. oxidation-reduction
40. Choose the best classification of the reaction represented by the
following equation: MgO(s) + CO2(g) ® MgCO3(s)
A. single-replacement
B. double-replacement
C. combination
D. decomposition
E. precipitation
Page 3
41. Choose the best classification of the reaction represented by the
following equation: CuSO45H2O(s) ® CuSO4(s) + 5H2O(l)
A. single-replacement
B. double-replacement
C. combination
D. decomposition
E. oxidation-reduction
42. Choose the best classification of the reaction represented by the
following equation: Zn(s) + CuSO4(aq) ® ZnSO4(aq) + Cu(s)
A. single-replacement
B. double-replacement
C. combination
D. decomposition
E. acid-base
43. Choose the best classification of the reaction represented by the
following equation: BaCl2(aq) + K2SO4(aq) ® BaSO4(s) + 2KCl(aq)
A. single-replacement
B. double-replacement
C. combination
D. decomposition
E. oxidation-reduction
44. Which one of the following processes cannot separate solute from
solvent in a true solution?
A. chromatography
B. filtration
C. boiling
D. adsorption
E. evaporation
45. What term describes a solution in which the solute concentration
exceeds its equilibrium concentration under the prevailing conditions?
A. hypotonic
B. hypertonic
C. isotonic
D. supersaturated
E. saturated
46. What term describes a solution which is in equilibrium with undissolved
solute?
A. precipitating
B. aqueous
C. saturated
D. unsaturated
E. supersaturated
Page 4
47. The solubility of gases in liquids is highest at
A. low temperature and low pressure
B. low temperature and high pressure
C. high temperature and low pressure
D. high temperature and high pressure
E. high pressure; temperature is immaterial
48. The number of moles of a gas dissolved in a liquid at a given
temperature is proportional to the partial pressure of the gas. What is
this law called?
A. Dalton's Law
B. Henry's Law
C. The Tyndall Law
D. Raoult's Law
E. Boyle's Law
49. Calculate the concentration (% W/V) of NaCl solution that was made by
dissolving 15.0 g of sodium chloride in enough water to make 300.0 mL
of solution.
A. 50.0% (W/V)
B. 0.0500% (W/V)
C. 0.356% (W/V)
D. 35.6% (W/V)
E. 5.00% (W/V)
50. Calculate the mass in grams of NaCl that is present in 500.0 mL of a
0.900% (W/V) solution.
A. 50.0 g B. 0.500 g C. 5.00 g D. 45.0 g E. 4.50 g
51. Assuming that air is a solution containing four molecules of N2 for
every one molecule O2, what is the concentration of O2 in this solution,
expressed as % (W/V)? [Use formula weights: N2, 28.02 amu; O2, 32.00
amu]
A. 22% (W/V)
B. 47% (W/V)
C. 53% (W/V)
D. 78% (W/V)
E. 114% (W/V)
52. Calculate the molarity of 2.00 L of solution that contains 200.0 g of
NaOH. [Use formula weight: NaOH, 40.0 amu]
A. 40.0 M B. 4.00 M C. 0.250 M D. 2.50 M E. 25.0 M
53. How many milliliters of 0.1250 M KCl solution contain 2.330 g of KCl?
[Use formula weight: KCl, 74.55 amu]
A. 20.50 mL B. 26.95 mL C. 25.00 mL D. 1.500 mL E. 250.0 mL
Page 5
54. How many milliliters of 12.0 M HCl are needed to prepare 250.0 mL of
2.50 M HCl solution?
A. 12.0 mL B. 25.0 mL C. 52.1 mL D. 5.21 mL E. 2.50 mL
55. Which one of the following is NOT a colligative property of a solution?
A. vapor pressure lowering
B. density
C. boiling point elevation
D. freezing point depression
E. osmotic pressure
56. What is the law which states that vapor pressure of the solvent
decreases in proportion to the concentration of the solute?
A. Dalton's Law
B. Henry's Law
C. the Tyndall Law
D. Raoult's Law
E. Boyle's Law
57. What concentration term is defined as the number of moles of solute per
kilogram of solvent in a solution?
A. normality
B. osmolarity
C. % (W/V)
D. molarity
E. molality
58. What is the relationship among intracellular fluids, 0.9% (W/V) NaCl
and 5.0% (W/V) glucose?
A. they are saturated
B. they are hypotonic
C. they are hypertonic
D. they are isotonic
E. they are supersaturated
59. If the concentration of Mg2+ in solution is 3.0 ´ 10-3 M, what is its
concentration expressed in meq/L?
A. 6.0 meq/L
B. 3.0 meq/L
C. 1.5 meq/L
D. 6.0 ´ 10-6 meq/L
E. 1.5 ´ 10-6 meq/L
60. T F Sodium, potassium and ammonium compounds are generally insoluble.
61. T F Nitrates and acetates are generally soluble.
62. T F Some double-replacement reactions are also precipitation
reactions.
63. T F The usual products of an acid-base reaction are a salt and water.
Page 6
64. T F Complete combustion of a hydrocarbon compound in oxygen produces
carbon dioxide and hydrogen.
65. T F An alloy, such as brass, is an example of a solution in the solid
state.
66. T F A solution in equilibrium with undissolved solute is said to be
unsaturated.
67. T F One liter of alcohol combined with one liter of water does NOT
produce two liters of solution.
68. T F In normal room lighting, the eye cannot distinguish a true
solution from a colloidal one.
69. T F If a solvent is colorless, all of its true solutions will also be
colorless.
70. T F Change in pressure has very little effect on the solubility of
solids in liquids.
71. T F The solid material that separates from solution when its
solubility is exceeded, is called a precipitate.
72. T F The solubility of solids in water usually increases with
increasing temperature.
73. T F The solubility of gases in liquids increases with increasing
temperature.
74. T F Both molarity and molality are affected by change in temperature.
75. T F An aqueous solution containing a nonvolatile solute will boil
above 100°C.
76. T F An aqueous solution containing a nonvolatile solute will have a
freezing point above 0°C.
77. T F Colligative properties depend only on the concentration of solute
particles, not on their identity.
78. T F The heavier the solute molecule is, the greater the effect on the
freezing point of a solution.
79. T F Osmosis may be defined as the movement of a solvent through a
semipermeable membrane from a region of higher solvent
concentration to one of lower concentration.
80. T F Osmosis is the process that regulates the sodium/potassium ratio
in living cells.
Page 7
Answer Key for Test "chapter7.tst", 8/17/04
No. in
Q-Bank
No. on
Test Correct Answer
7 1 1 oxidation-reduction
7 2 2 acid-base
7 3 3 precipitation
7 4 4 HCl(aq) + KOH(aq) ® KCl(aq) + H2O(l)
7 5 5 FeSO4(aq) + 2NaOH(aq) ® Fe(OH)2(s) + Na2SO4(aq)
7 6 6 Zn(s) + 2Fe3+(aq) ® Zn2+(aq) + 2Fe2+(aq)
7 7 7 carbon dioxide and water
7 8 8 decomposition
7 9 9 double-replacement
7 10 10 single-replacement
7 11 11 combination
7 12 12 colloidal suspension
7 13 13 Direct a narrow beam of light horizontally through the two
beakers. The colloidal suspension will scatter light (the
Tyndall effect) making the beam visible as it passes through;
the true solution will show no scattering.
7 14 14 1.10% (W/V)
7 15 15 12.5 g
7 16 16 It is the concentration of a solution expressed as the number of
moles of solute per liter of solution.
7 17 17 0.2 mol
7 18 18 0.0276 mol
7 19 19 2.33 g
7 20 20 0.0473 M
7 21 21 0.533 M
7 22 22 0.0733 M
7 23 23 It is a solution property which depends on the concentration of
solute particles rather than on their identity.
7 24 24 The volume of the solution is likely to increase, causing a
decrease in the molarity. Since mass is not affected by
temperature, the molality will stay the same.
7 25 25 twice as much
7 26 26 semipermeable
7 27 27 water molecules move from the hypotonic solution to the
hypertonic solution
7 28 28 hypotonic
7 29 29 The cells lose water, by osmosis, to the hypertonic solution,
and they collapse. The process is known as crenation.
7 30 30 they remain unchanged
7 31 31 8.0 ´ 10-3 Osm
7 32 32 2.9 atm
7 33 33 Water
7 34 34
Both involve the selective movement of small molecules through a
membrane, from a solution of high concentration of those molecules to
one of lower concentration. They differ in that osmosis involves only
movement of solvent (water) molecules, whereas in dialysis, solute
molecules can also pass through the membrane.
7 35 35 water, urea
7 36 36 A
7 37 37 B
7 38 38 E
Page 1
Answer Key for Test "chapter7.tst", 8/17/04
No. in
Q-Bank
No. on
Test Correct Answer
7 39 39 D
7 40 40 C
7 41 41 D
7 42 42 A
7 43 43 B
7 44 44 B
7 45 45 D
7 46 46 C
7 47 47 B
7 48 48 B
7 49 49 E
7 50 50 E
7 51 51 A
7 52 52 D
7 53 53 E
7 54 54 C
7 55 55 B
7 56 56 D
7 57 57 E
7 58 58 D
7 59 59 A
7 60 60 F
7 61 61 T
7 62 62 T
7 63 63 T
7 64 64 F
7 65 65 T
7 66 66 F
7 67 67 T
7 68 68 T
7 69 69 F
7 70 70 T
7 71 71 T
7 72 72 T
7 73 73 F
7 74 74 F
7 75 75 T
7 76 76 F
7 77 77 T
7 78 78 F
7 79 79 T
7 80 80 F
Page 2
Chemical and Physical Change:
Energy, Rate, and Equilibrium
Chapter Outline
CHEMISTRY CONNECTION: - The Cost of Energy? More Than You Imagine
8.1
Thermodynamics
The Chemical Reaction and Energy
Exothermic and Endothermic Reactions
Enthalpy
Spontaneous and Nonspontaneous Reactions
Entropy
Free Energy
A HUMAN PERSPECTIVE: Triboluminescence: Sparks in the Dark with Candy
8.2
Experimental Determination of Energy Change in Reactions
8.3
Kinetics
The Chemical Reaction
Activation Energy and the Activated Complex
A CLINICAL PERSPECTIVE: Hot and Cold Packs
Factors that Affect Reaction Rate
Mathematical Representation of Reaction Rate
8.4
Equilibrium
Rate and Reversibility of Reactions
Physical Equilibrium
Chemical Equilibrium
The Generalized Equilibrium-Constant Expression for a Chemical Reaction
LeChatelier's Principle
Summary
Key Terms
Questions and Problems
Critical Thinking Problems
Instructional Objectives
Conceptual Objectives
• Correlate the terms endothermic and exothermic with heat flow between a system and its
surroundings.
• State the meaning of the terms enthalpy, entropy, and free energy and know their implications.
• Describe experiments that yield thermochemical information.
• Describe the concept of reaction rate and the role of kinetics in chemical and physical change.
• Describe the importance of activation energy and the activated complex in determining reaction
rate.
Performance Objectives
• Calculate fuel values based on experimental data.
• Predict the way reactant structure, concentration, temperature, and catalysis affect the rate of a
chemical reaction.
• Write rate equations for elementary processes.
• Recognize and describe equilibrium situations.
• Write equilibrium-constant expressions and use these expressions to calculate equilibrium
constants.
• Use LeChatelier's Principle to predict changes in equilibrium position.
Health Applications
 Relate the concepts of chemical energy and fuel value to diet and obesity.
 Understand the chemical processes involved in the manufacture and use of hot and cold packs.
In-Chapter Examples
Example 8.1:
Example 8.2:
Example 8.3:
Example 8.4:
Example 8.5:
Example 8.6:
Example 8.7:
Example 8.8:
Example 8.9:
Determining whether a process is exothermic or endothermic.
Calculating energy involved in calorimeter reactions.
Calculating energy involved in calorimeter reactions.
Calculating the fuel value of foods.
Writing rate equations.
Writing an equilibrium-constant expression.
Writing an equilibrium constant expression.
Calculating an equilibrium constant.
Predicting changes in equilibrium composition.
Chapter Overview
Thermodynamics
Two laws of thermodynamics, the science involved with energy flow in physical and chemical
change, are of particular importance to us. The first law, the law of conservation of energy, and the
second law, entropy or disorder, can provide us with basic information regarding the spontaneity of
chemical or physical processes as well as the amount of energy absorbed or released by the reaction.
Exothermic reactions release energy; endothermic reactions require energy input. The practical side of the
thermodynamics is presented through discussion of enthalpy, entropy, and free energy.
Experimental Determination of Energy Change in Reactions
A calorimeter measures heat changes (in calories or joules) that occur in chemical reactions. The
specific heat of a substances is the number of calories of heat needed to raise the temperature of one gram
of the substance by one degree Celsius.
The amount of energy per gram of food is its fuel value, commonly reported in units of nutritional
Calories (1 nutritional Calorie = 1 kcal). A bomb calorimeter is useful for measurement of the fuel value
of foods.
Kinetics
Chemical kinetics is the study of the rate of a chemical reaction. Recognizing that chemical reactions
result from effective collisions between potentially reacting particles, we show that factors such as the
structure of the reactant, its concentration, reaction temperature, and the presence or absence of a catalyst
influence reaction rates. The role of the activation energy in chemical processes determines reaction rates,
for we view the formation of the activated complex as an energy barrier to be overcome.
Equilibrium
Not all reactions proceed to completion. Many establish a dynamic equilibrium between products and
reactants, where the rates of the forward and reverse reactions are equal. LeChatelier's Principle states
that if a stress is placed on a system in equilibrium, the system will respond by altering the equilibrium in
such a way as to minimize the stress.
Hints for Faster Coverage
Coverage of the sections discussing thermodynamics and kinetics might be shortened, emphasizing
the experimental factors that affect reaction rate. For example, the section discussing the mathematical
representation of reaction rate may be omitted, without loss of continuity. The derivation of the
equilibrium-constant expression for physical and chemical systems may likewise be omitted.
Suggested Problem Sets
Energy and Thermodynamics: 27, 29, 31, 33, 35, 37, 39
Kinetics: 41, 43, 45, 47, 49, 51
Equilibrium: 53, 55, 57, 59, 61, 63, 65, 67, 69
In-Chapter Perspectives
A HUMAN PERSPECTIVE: Triboluminescence: Sparks in the Dark with Candy. This perspective
presents an example of the scientific method improving our understanding of everyday occurrences.
A CLINICAL PERSPECTIVE: Hot and Cold Packs. Hot and cold packs provide a good opportunity for
the student to see a direct connection between basic science (endothermic and exothermic reactions) and a
commonplace product of technology used in a medical setting.
Additional Perspectives
Energy, rate, and equilibrium have almost unlimited practical applications. The storage and release of
energy in fossil fuels, the use of industrial catalysts to speed up the formation of commercially valuable
products, and the effect of enzymes on biological processes are just a few ideas that can be used to
stimulate student interest and discussion.
Critical Thinking Problems
1.
2.
3.
4.
5.
6.
7.
This problem can be used to emphasize the concept of reversibility; we reach the same equilibrium
point whether we begin with products or reactants. Also, the labels "product" and "reactant" are
somewhat arbitrary when used with equilibrium reactions.
This problem can be used to initiate a discussion of relationships and differences between the
concepts of kinetics and thermodynamics.
Modeling a reaction is a useful way to enhance student understanding of the processes that must
occur during chemical change.
Recognition of the real-life implication of exclusion of the terms for any solids from the equilibriumconstant expression results from the solution of this problem.
This problem allows the student to see LeChatelier's Principle in broader terms.
This problem is a practical, predictive application of LeChatelier's Principle.
This problem may help students understand heat flow in systems.
Chapter 8
Chemical and Physical Change:
Energy, Rate, and Equilibrium
Solutions to the Even-Numbered Problems
In-Chapter Questions and Problems
8.2
8.4
a.
Exothermic. As the dissolution reaction proceeds, the solution gets hotter because heat is
produced.
b.
Exothermic. H˚ is negative, meaning that energy is released by the reaction.
c.
Endothermic. Energy is shown as a reactant in the reaction.
Q = ml x Tl x SHl
Solving for Tl, T l =
Q
ml x SH l
Substituting,
T l =
6.5 x 10 2 cal
(1.00 x 10 2 g)(0.800 cal/g  C)
4.18 J
= 2.9 x 103 J
1 cal
8.6
7.0 x 102 cal x
8.8
Q = mw x Tw x SHw
Solving for mw, mw =
Substituting,
Q
Tw x SH w
= 8.1˚C
mw =
 10 3 cal


3.00 nutritional Cal x 
1 nutritional Cal 
(5.00ÞC)(1.00 cal/g C)
= 6.0 x 102 g
8.10
Net energy is the difference in energy between products and reactants. Activation energy is the
threshold energy that must be overcome in order to cause a chemical reaction.
8.12
The rate of chemical reactions critical to the growth process decreases at the lower temperature
of the refrigerator.
8.14
a.
rate = k[CH4]n[O2]n'
b.
rate = k [NO2]n
8.16
An increase in the rate of the forward reaction favors products. This increases the numerator of
the rate constant expression, hence, the magnitude of the equilibrium constant.
8.18
No. The equilibrium process is dynamic; products and reactants continuously interconvert at the
same rate.
8.20
a.
K eq =
b.
Keq = [Cl2]
8.22
[H2 ][I 2 ]
[HI] 2
The equilibrium favors reactants. A very low value for the equilibrium constant indicates that the
ratio of products to reactants is very low. Therefore reactant concentration is larger than product
concentration.
8.24
K eq =
K eq =
8.26
[H2S]2
[H2 ]2 [S2 ]
[0.80]2
2
-6
[0.22] [1.0 x 10 ]
 1.3 x 10 7
a.
A would increase; C and D would react to produce more B, to compensate for its removal,
and more A as well. the equilibrium shifts to the left.
b.
A would decrease; A and B would react to produce more C, to compensate for its removal.
The equilibrium shifts to the right.
c.
A would increase; excess D would react with C to produce more A and B. The equilibrium
shifts to the left.
d.
A would remain the same; removal of the catalyst would delay the attainment of
equilibrium, but not affect the equilibrium position.
End-of-Chapter Questions and Problems
8.28
a.
Entropy is a measure of randomness or disorder.
b.
The specific heat of water is the number of calories of heat needed to raise the temperature
of one gram of water one degree Celsius.
c.
The fuel value of a substance is the amount of energy, in units of nutritional Calories,
derived from one gram of that substance.
8.30
Entropy is a measure of randomness or disorder.
8.32
a.
Q = mw x Tw x SHw
Q = 2.00 x 102 g H2O x 5.70˚C x (1.00 cal/gH2O˚C) = 1.14 x 103 cal
1.14 x 103 cal lories are derived from 0.0500 moles of the nutrient substance.
1.14 x 10 3 cal
1 mol subst
x
0.500 mol subst
114 g subst
= 2.00 x 10 2 cal/g subst
and
fuel value = (2.00 x 102 cal/g subst) x (1 nutritional Cal/103 cal)
fuel value = 2.00 x 10-1 nutritional Calories/g subst
8.34
2.00 x 10 -1 C
1 kcal 10 3 cal
4.18 J
1 kJ
x
x
x
x
g subst
1C
1 kcal
1 cal
103 J
8.36
= 8.36 x 10 -1 kJ/g subst
a.
Entropy increases. Conversion of a single solid object, the log, into carbon monoxide,
carbon dioxide, and water (all gases) and ashes (disordered solid) increases the disorder of
the substance.
b.
Entropy decreases. A gas (more disordered) is converted to a liquid (less disordered). The
randomness of the water molecules has decreased.
8.38
A deck of cards is arranged in numerical order by suit (an organized system). Simply drop it on
the floor; randomness or disorder has increased. To return the deck of cards to its original
condition required an expenditure of energy to overcome the increase in entropy.
8.40
Energy is released when chemical bonds form. The bond energy is defined as the energy
released when a chemical bond is formed.
8.42
The activation energy is the threshold energy barrier that must be overcome to cause a chemical
reaction to take place.
8.44
Figure 8.10 of the textbook depicts a generalized endothermic reaction. The presence of a
catalyst would lower the activation energy (similar to the effect shown in Figure 8.11 of the
textbook, illustrated for an exothermic reaction) and would leave the initial and final energies
unchanged. Consequently, the change in enthalpy would remain the same.
8.46
An explosion is a very rapid and exothermic reaction. Fuel leaks are dangerous simply because
their reaction with oxygen is so rapid.
An adequate rate of reaction is essential for plant growth, the basis of agriculture. Chlorophyll,
present in the green leaves of plants, catalyzes the process of photosynthesis.
8.48
An increase in the temperature of reactants increases the rate of a reaction for two reasons:
• The speed of reactant molecules increases as the temperature increases. Faster-moving
molecules collide more frequently, increasing the probability of reaction.
•
Reactant molecules have more kinetic energy at higher temperatures; thus, a greater fraction
of all collisions will be effective collisions.
8.50
Rate = k[H2S] [Cl2]
8.52
The catalyst is involved in the formation of the activated complex, an unstable intermediate in
the reaction. When the complex breaks apart to form products, the catalyst is regenerated; it is
free to repeat the process over and over again.
8.54
The rates of the forward and reverse reactions are equal for a reaction at equilibrium.
[HCl] 2
=
[H2S][Cl2 ]
8.56
K eq
8.58
The rate constant is a measure of how rapidly a reaction reaches equilibrium or completion.
The equilibrium constant describes the ratio of product and reactant concentrations at
equilibrium. It is a measure of the completeness of the reaction.
8.60
8.62
8.64
8.66
a.
Equilibrium shifts to the right. Increasing the temperature provides more energy to drive the
endothermic reaction to the right.
b.
Equilibrium shifts to the left. An increase in pressure will favor the direction that favors a
smaller volume: Three moles of product will convert to two moles of reactant.
c.
No change in equilibrium position. The catalyst has no effect on equilibrium position.
a.
a reaction is at equilibrium when no reactants remain -- False; If a reaction has gone to
completion, no reactants remain. During equilibrium, the concentration of products and
reactants remain unchanged.
b.
a reaction at equilibrium is undergoing continual change -- True; The rate at which
reactants form products equal the rate at which products reform reactants.
a.
Adding more carbon has no effect on the equilibrium because carbon is a solid. Solids and
pure liquids are not a part of the equilibrium expression.
b.
Adding more H2 shifts the equilibrium to the right, favoring the formation of methane.
c.
Removing CH4 shifts the equilibrium to the right.
d.
Increasing the temperature shifts the equilibrium to the left. Increasing the temperature
increases heat energy and heat is viewed as a product.
e.
Addition of a catalyst has no effect on the equilibrium position.
Since the total moles of reactants in the gaseous state is equal to the total moles of products in
the gaseous state, an increase in pressure will have no effect on the concentration of NO(g).
[NO]2
[N2 ][O2 ]
8.68
K eq =
8.70
True.
In an endothermic reaction, heat energy is absorbed from the surroundings; heat can be viewed
as a reactant. Adding reactant shifts the equilibrium to the right.