General Chemistry 2

(Q1)

Christian Mangao Foronda

 Table of Contents

Module 1: Intermolecular Forces and Liquids and Solids

Introduction 1
Learning Outcomes 1
Lesson 1. The Kinetic Molecular Theory of Liquids and Solids 2
Lesson 2. Intermolecular forces of attraction 3
Lesson 3. Properties of Liquids 7
Lesson 4. The Structure and Properties of Water 9
Lesson 5. Crystals and Amorphous Solids 11
Lesson 6. Phase Diagrams 13
Lesson 7. Heating and Cooling Curve 15
Assessment Tasks 17
Summary 19
References 19

Module 2: Physical Properties of Solutions

Introduction 20
Learning Outcomes 20
Lesson 1. Types of Solutions 21
Lesson 2. Concentration Units 21
Lesson 3. Stoichiometric Calculations in Solutions 26
Lesson 4. Colligative Properties of Nonelectrolyte and Electrolyte Solution 29
Lesson 5. Calculating Molar Mass using Colligative Properties 34
Assessment Tasks 36
Summary 40
References 40

Module 3: Thermochemistry
Introduction 41
Learning Outcomes 41
Lesson 1. The First Law of Thermodynamics 42
Lesson 2. Enthalpy 43
Lesson 3. Hess Law 46
Assessment Tasks 48
Summary 50
References 50
Module 4: Chemical Kinetics
Introduction 51
Learning Outcomes 51
Lesson 1. Factors Affecting Rate of Reaction 52
Lesson 2. Reaction Order 54
Lesson 3. Collision Theory 56
Lesson 4. Catalysts 59
Assessment Tasks 62
Summary 63
References 64
Course Code: Gen Chem 2

Subject Description: A study of the composition, structure, and properties of

matter; quantitative principles, kinetics, and energetics; and fundamental concepts
of organic chemistry

Content Standards:
The learners demonstrate the understanding of:

1. the properties of liquids and solids to the nature of forces between

particles;
2. phase changes in terms of accompanying changes in energy and
forces between particles;
3. properties of solutions, solubility, and the stoichiometry of solutions;
4. energy changes in chemical reactions;
5. the rate of reaction and the various factors that influence and the
collision theory;
6. spontaneous change, entropy, and free energy;
7. chemical equilibrium and Le Chatelier`s principle;
8. acid base equilibrium and its application to the pH of solutions and
the use of buffer solutions
9. solubility equilibria and its application; and
10. redox reactions as applied to galvanic and electrolytic cells

Performance Standards:

The learners should be able to:

1. design a simple investigation to determine the effect on boiling point

or freezing point when a solid is dissolved in water;
2. prepare a poster on a specific application of one of the following; and
a. Acid-base equilibrium
b. Electrochemistry
3. include in the poster the concepts, principles, and chemical reactions
involved, and diagrams of processes and relevant materials.
Subject Requirements:

 Assessment Tasks
 Written Works -25%
 Performance Task -45%
 Quarterly Exams -30%
Quarterly Grade 100%
Final Grade = (First Quarter Grade + Second Quarter Grade)/2
 MODULE 1
INTERMOLECULAR FORCES AND LIQUIDS AND
SOLIDS

Introduction

Humans has been living in our planet for millions of years. As those years passed by,
we became more immersed on Earth’s surface as well as its atmosphere. Our atmosphere is
composed of different mixtures of gases. We became familiar with the behavior of liquids and
solids since they are visible. Liquids such as water is a necessity for all of us since it is used
in our daily lives.

The movements of the molecule are more restricted in liquid than in gases, while in
solids, the atoms are tightly packed together. In this module, we will examine the structure of
liquids and solids and discuss some of the fundamental properties of the two state of matter.
Aside from that, we will also discuss the transition of states among gases, liquids, and solids
(Chang and Goldsby, 2016).

Learning Outcomes

At the end of this module, students should be able to:

1. use kinetic molecular model to explain properties of liquids and solids;

2. describe and differentiate the types of intermolecular forces;
3. describe the following properties of liquids, and explain the effect of
intermolecular forces on these properties: surface tension, viscosity, vapor
pressure, boiling point, and molar heat of vaporization;
4. explain the properties of water with its molecular structure and intermolecular
forces;
5. describe the difference in structure of crystalline and amorphous solids;
6. interpret the phase diagram of water and carbon dioxide; and

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 7. determine the heating and cooling curve of a substance.

Lesson 1. The Kinetic Molecular Theory of Liquids and Solids

We all know that the distance between the molecules of gases is very great. Since
there is a great distance between the molecules — gases are readily compressed and has a
low density under normal conditions (Chang and Goldsby, 2016). The behaviors of gases
were more explained through the Kinetic Molecular Theory. The story is different in solids and
liquids. The distance between the molecules is their primary difference that is why solids are
less compressible than liquids. Table 1.1 below summarizes the differences between the three
states by focusing on the four characteristics — shape/volume, density, compressibility, and
motion of molecules.

Table 1.1 Characteristic properties of gases, liquids, and solids

State of Matter Volume/Shape Density Compressibility Motion of
Molecules
Gas “Assumes the Low Very Very free motion
volume and compressible
shape of its
container”
Liquid “Has a definite High Only slightly Slide past one
volume but compressible another freely
assumes the
shape of its
container”
Solid “Has definite High Virtually Vibrate about
volume and compressible fixed positions
shape”
Source: Chang and Goldsby (2016, p. 46)

 In a liquid, there is a stronger attraction of particles than in gases because particles

are in contact. Kinetic energy allows the particles of liquids to move freely and past
one another. Since there is only a little space between particles, liquids resist an
applied external force and compressed slightly than gases. They flow and diffuse
slower than gases (Silberberg, 2007).

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  In a solid, there is a strong attraction that it dominates its movement and particles only
jiggles in place. Since the position of the particles are fixed, they have a specific shape
and does not flow. The particles of solids are closely packed together, therefore they
are difficult to compress (Silberberg, 2007).

Phase Changes are referred by chemists as a “homogenous part of the system in

contact with other parts of the system but separated from them by well defined-boundary.
Thus, our glass of ice water contains both the solid phase and the liquid phase” as stated by
Chang and Goldsby (2016).

Lesson 2. Intermolecular Forces of Attraction

Intermolecular forces are “attractive forces between molecules. It is the force that is
needed to surpass for the phase change to proceed”. Intramolecular forces hold atoms
together. Intramolecular force stabilizes molecules while intermolecular forces are the one
responsible for the different properties of matter such as melting point and boiling point (Chang
and Goldsby, 2016).

Generally, intermolecular forces are much weaker than intramolecular forces. The
energy needed to break the bonds in the molecules of a liquid is greater than the energy to
evaporate a liquid. 41 kJ (kilojoules) of energy is needed to vaporize 1 mole of water and 930
kJ of energy to break the bonds between hydrogen and oxygen in 1 mole of water (Chang
and Goldsby, 2016).

To understand the different properties of matter, it is necessary that we know the

different types of intermolecular forces. Dipole-dipole, dipole-induced dipole, and dispersion
forces makes up the van Der Waals forces, named after the Dutch physicist Johannes van
Der Waals. Ions and dipoles were attracted to one another because of electrostatic forces
called ion-dipole forces. Ion-dipole forces are not van Der Waals force. Hydrogen bonding is
a special type of dipole-dipole forces and is particularly strong. Depending on the nature of
chemical bonds, types of elements present, and phase of the substance, more than one type
of interaction can contribute to the total attraction between molecules (Chang and Goldsby,
2016).

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 Dipole – dipole forces – it is an attractive force between polar molecules, that is
molecules that possess dipole moments. Their origin is electrostatic; thus, they can
only be understood in terms of Coulomb’s law. The larger the dipole moment, the
greater the force. Polar molecules in liquids is not held as rigidly as in a solid molecule,
but they align in a way that they can maximize the attractive interaction. Example of a
dipole – dipole interaction is in the iodine monochloride compound (Chang and
Goldsby, 2016).

Figure 1.1 Dipole – dipole interaction between iodine monochloride molecules

Source: Dipole-Dipole Forces (2000)

The symbol - means a partial negative charge due to cloud of electrons and + means
a lower number of electrons. This means that the chlorine atom has more electrons
than the iodine atom. Due to their partial charges, the partial positive charge (+) of
iodine monochloride molecule is attracted to the partial negative side (-) of iodine
monochloride molecule forming a dipole – dipole force of attraction. The dash lines
represent the dipole-dipole force.

HBr, H2O, and HCl (see structures) are examples of compounds that also exhibits
dipole-dipole interaction

 Ion – dipole forces – attraction between “cation or anion and a polar molecule to each
other. The strength of this interaction depends on the charge and size of the ion and
on the magnitude of dipole moment and size of the molecule”. Since cations are
smaller, charges are more concentrated compared to anions, that is why cations
interacts more strongly than anions with dipoles with same magnitude. Hydration of
salt is an example of ion dipole force interaction (Chang and Goldsby, 2016).

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 Figure 1.2 Hydration of sodium chloride
Source: Intermolecular Forces: Chapter 7 (n.d.)
Figure 1.2 shows what happens when water is incorporated in the structure of sodium
chloride. The partial positively charged side of water is attracted to negatively charged
chloride anion and the partial negatively charged side of water is attracted to positively
charged sodium cation, thus, they form an ion – dipole force of attraction.
 Dispersion force – or London dispersion forces, named after Fritz London is a type of
interaction between non-polar molecules. It also exists in all particles, polar or non-
polar, ionic or covalent. It also contributes to the overall energy of attraction of all
substances (Ebbing and Gammon, 2009).

Figure 1.3 Illustration of London dispersion forces in a neon molecule

Source: Ebbing and Gammon (2009, p. 438)

Since electrons in an atom (in our example, Neon atom) are constantly moving, there
are an instance where there are more electrons in one side of neon atom thus that
side forms a partial negative charge (-) and side with lesser electrons form a partial
positive charge (+). If this neon atom is near another neon atom, the electrons on that
side are repelled forming an instantaneous dipole which gives an attraction force.

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 Dispersion forces are stronger in large molecules such as I 2 and Xe, and weaker in
smaller molecules such as H2 and He.
“What is the relationship between the molar mass of molecule and the strength of
dispersion forces? What if the molecules have the same molar mass?”

 Hydrogen Bonding – “is a special type of dipole-dipole interaction between the

hydrogen atom in a polar bond, which is special type dipole-dipole interaction between
the hydrogen atom in a polar bond, such as N-H, O-H, or F-H, and an electronegative
O, N, or F atom (Chang and Goldsby, 2016). The interaction is written as”
A — H  :B or A — H  :A

A and B represents F, O, and N. Take a look at Figure 1.4 below.

Figure 1.4 Hydrogen bonding of H—F, H—O, and H—N

Source: Silberberg (2007, p. 361)

Figure 1.5 Hydrogen bonding in water, ammonia and hydrogen fluoride.

Source: Chang and Goldsby (2016, p. 472)

The solid lines in the molecular structure of water, ammonia, and hydrogen fluoride in
Figure 1.5 represents the covalent bonds and the dotted lines represent the hydrogen bonds.
NOTE: Ion – dipole bond is the strongest intermolecular force and the weakest intermolecular
force is the dispersion force (London Dispersion force). The order of increasing strength of
intermolecular force is: London dispersion force < dipole – dipole forces < hydrogen bonding

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< ion – dipole forces. Note that this order is not always true since structure and the molar
mass should be also accounted.

Lesson 3. Properties of Liquids

Due to intermolecular forces of attraction, different properties and structural features

of liquids arise. In this lesson, we will discuss these phenomena associated to liquids. These
are: surface tension, viscosity, vapor pressure, boiling point, and molar heat of vaporization.
 Surface Tension – “The surface tension is the amount of energy required to stretch or
increase the surface of a liquid by a unit area. Liquids that have a strong intermolecular
force of attraction also have a high surface tension”. One of surface tension is capillary
action or capillarity. It is the rising of liquid through a narrow space against the pull of
gravity. Another example is cohesion and adhesion. Cohesion is the intermolecular
attraction between like molecules while adhesion is attraction between unlike
molecules (Chang and Goldsby, 2016; Silberberg, 2007; Ebbing and Gammon, 2009).

Figure 1.6 Explaining Surface tension

Source: Ebbing and Gammon (2009, p. 434)

A molecule in the surface experiences a net force toward the interior of the liquid while
the molecule is equally attracted in all sides thus there is no net interior forces.
 Viscosity – “is a measure of fluid’s resistance to flow. The higher the viscosity, the
more the slowly the liquid flows. Liquids that have stronger intermolecular forces of
attraction have higher viscosities” (Chang and Goldsby, 2016). That is why water has
higher viscosity than any other liquids because of its ability to form hydrogen bond.
The unit of viscosity is newton-seconds per meter square.

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  Vapor pressure – “the measure of the pressure (force per unit area) exerted by a gas
above a liquid in a container” (Chang and Goldsby, 2016). Stronger intermolecular
forces of attraction mean a lower vapor pressure of liquid. Weak intermolecular forces
of attraction mean a stronger vapor pressure.
 Boiling point – “is the temperature in which the vapor pressure of the liquid is equal to
the pressure exerted on the liquid (atmospheric pressure, unless the vessel containing
the liquid is closed)” (Chang and Goldsby, 2016). Boiling happens when the vapor
pressure is equal to the atmospheric pressure and stable bubbles form within the
liquid. The higher the intermolecular force of attraction, the higher the boiling point of
the compound (Ebbing and Gammon, 2009).

Figure 1.7 Boiling of a liquid

Source: Ebbing and Gammon (2009, p. 11.7)

 Molar heat of vaporization – is the heat needed to vaporize a liquid and is denoted as
ΔHvap. The stronger the intermolecular force, the higher the molar heat of vaporization.
At 100 oC, the heat of vaporization of water is 40.7 kJ per mole (Silberberg, 2007).

Lesson 4. The Structure and Properties of Water

Earth’ s surface is composed of 75% water, making it the most common substance.
Water has some unique and unusual properties of any substance and these properties were
very vital on the existence of life on this planet. Water`s structure is composed of two atoms
of hydrogen around the electronegative atom oxygen and a two lone-pair, thus its structure is
bent and highly polar. This arrangement of atoms is important since it allows each molecules

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of water to engage in four hydrogen bonding sites. Due to this fundamental atomic and
molecular properties, water has a unique and remarkable macroscopic behavior (Silberberg,
2007).

Figure 1.8 Hydrogen bonding ability of water

Source: Silberberg (2007, p. 367)

Since there are two O — H bonds and two lone pairs, one molecule of water can engage
as many as four hydrogen bonds to surrounding molecules, which is arranged
tetrahedrally.

 Solvent properties of water – Water is called the “universal solvent” because it can
almost dissolve any kinds of substances. Water’s polarity and unique hydrogen
bonding ability makes it a great solvent. Ionic compounds can be dissolved by water
through ion – dipole force that can separate cations and anions. Through hydrogen
bonding, water can dissolve many polar nonionic substances like ethanol (CH 3CH2OH)
and glucose (C6H12O6). London dispersion forces and the induced dipole – dipole
interaction is the responsible why water can dissolve, in a limited extent, some
nonpolar gases in the atmosphere. Due to water`s unique solvent properties, it formed
the complex solutions we know as oceans, rivers, lakes, and even the cellular fluids
(Silberberg, 2007).
 Thermal properties of water – Water has a high specific heat capacity (heat needed to
raise the temperature of 1 kilogram of a substance by 1 Kelvin) than almost any other
liquids. This is due to water`s many hydrogen bonding attachment sites. The same
scenario can be also applied to water`s high heat of vaporization (Silberberg, 2007).

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 Surface properties of water – The hydrogen bonding of water is the responsible for its
high surface tension and high capillarity. Water has the highest surface tension of any
liquid except for some molten salts and metals. Thanks to water`s high surface tension,
surface aquatic plants stay rested in the surface giving shelter and foods to many
fishes and microorganisms. During dry seasons, land plants can survive through
capillary action by absorbing deep underground water (Silberberg, 2007).

Figure 1.9 Water strider in water`s surface

Source: Chang and Goldsby (2016, p. 473)
Due to water`s high surface tension, insect such as water strider can “walk” on water.

 The density of solid and liquid water – Due to water`s four hydrogen bond attachment
sites, it forms a tetrahedral pattern. When the tetrahedral pattern is continued through
many molecules in a fixed array, it leads to a hexagonal open structure of ice (solid
water). This symmetrical pattern can be seen in snowflakes. The large spaces within
ice give the solid state a lower density than the liquid state (Silberberg, 2007).

Figure 1.10 Ice in water (left) and benzene in water (right)

Source: Chang and Goldsby (2016, p. 476)

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Lesson 5. Crystals and Amorphous Solids
Solids can be divided into two categories: crystals and amorphous.

Crystals

Crystals have a well-defined shape because their particle — atoms, molecules, or ions
— occur in an orderly and specific arrangement. The arrangement of the particles is such that
the net attractive intermolecular force is at maximum. The reason why crystals are stable are
due to ionic forces, covalent bonds, van Der Waals forces, hydrogen bonding or combination
of these forces (Chang and Goldsby, 2016). The figures below are examples of crystals. Ice
is also a type of crystal.

Figure 1.11 The beauty of crystalline solids (A) Pyrite, (B), Beryl, (C) Barite (left) on calcite (right)
Source: Silberberg (2007, p. 369)

“The unit cell of crystal is the smallest boxlike unit (each box having faces that are
parallelograms) from which you can imagine constructing crystal by stacking the units three
dimensions”. There are seven shapes possible for a unit cell, which give rise the seven crystal
systems used to classify crystal. The seven basic shapes are: cubic, tetragonal, orthorhombic,
monoclinic, hexagonal, rhombohedral, triclinic (Ebbing and Gammon, 2007).

“Crystal lattice as the geometric arrangement of lattice points of a crystal, in which we

choose one lattice point at the same location within each of the basic units of crystal. Take
note that crystal lattice and crystal structure is not the same” (Ebbing and Gammon, 2007).

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Types of Crystal

The types of forces that hold together the crystals dictate the structure and properties,
such as melting point, density and hardness of crystals. We can classify them as ionic,
covalent, molecular, and metallic (Ebbing and Gammon, 2009). Table 1.2 summarizes the
properties of the four types of crystals.

Table 1.2 Types of crystals and general properties

Force(s) Holding the

Types of Crystal General Properties Examples
Units Together
“Hard, brittle, high
melting point, poor NaCl, LiF, MgO,
Ionic Electrostatic attraction
conductor of heat and CaCO3
electricity”
“Hard, high melting
C (diamond), SiO2
Covalent Covalent bond point, poor conductor
(quartz)
of heat and electricity”
“Soft, low melting
Ar, CO2, I2, H2O,
Molecular Intermolecular forces point, poor conductor
sucrose
of heat and electricity”
“Soft to hard, low to
All metallic elements;
high melting point,
Metallic Metallic bonds for example, Na, Mg,
good conductor of
Fe, Cu
heat and electricity”
Source: Chang and Goldsby (2016, p. 491)

Amorphous Solids

Amorphous solids lack regular three-dimensional arrangements of atoms which

means they are non-crystalline. They are formed rapidly (they are cooled quickly), the
molecules or the atom don`t have enough time to align themselves and will be locked in that
position thus forming an irregular pattern. Examples of amorphous solids are charcoal, rubber
and glass. Glasses are formed by mixing molten silicon dioxide, as its chief component, with

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compounds such as sodium oxide, boron oxide with transition metals for colors and other
properties (Chang and Goldsby, 2016).

Figure 1.12 Two-dimensional representation of (a) crystalline quartz and (b) non-crystalline quartz
Source: Chang and Goldsby (2016, p. 493)

Lesson 6. Phase Diagrams

Scientists use phase diagrams to represent the relationship between solids, liquids,
and vapor phases. It summarizes the conditions at which a substance exists as a solid, liquid,
or gas (Chang and Goldsby, 2016). We will discuss here the phase diagram of water and
carbon dioxide.

Figure 1.13 Phase Diagram of (A) CO2 and (B) H2O

Source: Silberberg (2007, p. 357)

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 Figure 1.13 shows the phase diagram of water and carbon dioxide. The graph is
divided into three regions and each region represents a pure phase. The lines that separate
two regions indicates conditions where two phases can exist in equilibrium. For example, in
the phase diagram of water, the line that separates the liquid and solid phase means that solid
water (ice) and liquid water is in equilibrium. The triple point is the condition in which three all
phases can be equilibrium with one another. This means that at this point, water and carbon
dioxide is subliming and depositing, freezing and melting, and condensing and evaporating at
the same time! The triple point of water is at 0.01 oC at 0.006 atm (atmospheric pressure)
while in carbon dioxide is at -57.0 oC at 5.1 atm. Take note that the triple point of carbon
dioxide is beyond our atmospheric pressure which means it is impossible to melt carbon
dioxide at 1 atm. When you heated a carbon dioxide at -78 oC at 1 atm, the solid carbon
dioxide (called dry ice) doesn’t melt but instead sublimes. This is the reason why carbon
dioxide is useful as a refrigerant. At critical point, the phase boundary between liquid and gas
disappears due to extreme heat and pressure. Since more liquid vaporizes, the density will
also increase. The density of liquid and the vapor becomes closer until they are at equal at
the critical temperature (Tc ) and critical pressure (Pc ), that is where the phase boundary
disappear. “Since the average kinetic energy of the molecules is so high, that the vapor cannot
be condensed no matter what pressure is applied (Chang and Goldsby, 2016; Ebbing and
Gammon, 2009; Silberberg, 2007)

The normal boiling point of water is at 1 atm at 100 oC. Raising the pressure above 1
atm will increase the boiling point and decrease the melting point of water. A decrease in
pressure will lower the boiling point and raise the melting point. The only difference in the
phase diagram of water and carbon dioxide is that the slope of the boundary between liquid
and solid of carbon dioxide is positive while in water it is negative (Chang and Goldsby, 2016).

“What do you think is the importance of a negative slope of the phase boundary of solid and
liquid in the phase diagram of water?”

Lesson 7. Heating Curve and Cooling Curve

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 Figure 1.14 Typical Heating Curve of a Substance
Source: Chang and Goldsby (2016, p. 501)

Figure 1.14 shows the typical heating curve of a substance. A heating curve is a line
graph presentation of phase changes of a substance from solid to liquid or liquid to gas. It
shows at what point of time and temperature does a phase coexist with each other in
equilibrium (Chang and Goldsby, 2016). Take a look with line segment AB, this is the
temperature in which both the substance`s solid and liquid state are in equilibrium. This means
that the substance is in both solid and liquid state at the same time! This is also what we call
the melting point of the substance. The line segment BC is the point (temperature and time)
where the substance is completely in liquid state. Line segment CD is the point where liquid
and vapor are in equilibrium or coexist with each other. This means that water and vapor
coexist with other at the same time! This is also the boiling point of the substance. Take note
also that line segment AB is shorter than line segment CD. It means that it takes more time to
boil the substance than to melt it.

When you removed the heat from a substance, you will able to graph the cooling curve
of that substance. The cooling curve will be the reverse of the heating curve of that substance.

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Assessment Task

TASK NO. 1 (WRITTEN WORK)

Direction: Answer the following questions correctly. Use the following structures of
compounds as your basis.

Source: Chang and Goldsby (2016, p. 67) Source: sciencing.com

Source: Chang and Goldsby (2016, p. 67)

1. Which among the four compounds above exhibits a London Dispersion Forces?
(5 points)
2. Which among the four compounds above may exhibit a dipole – dipole force of
attraction? (4 points)
3. Which among the four compounds above may participate in a hydrogen
bonding? (4 points)
4. Which among the four compounds above may participate in an ion – dipole
interaction? (3 points)
5. Except calcium chloride, which among the three remaining compounds has the
highest boiling point? Explain your answer. How about the lowest? (5 points)
6. Which among the following compounds may form a hydrogen bond with water?
Draw the structure of those compounds with a hydrogen bond formed with
water? (10 points)

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TASK NO 2. (PERFORMANCE TASK)
A. Direction: Identify if the following solids are ionic crystal, covalent crystal, molecular
crystal, metallic crystal or amorphous solid. (10 points)
1. CaCO3 6. NaOH pellets
2. ice 7. zinc wires
3. fructose 8. solid krypton
4. silver 9. cotton candy
5. rubber tires 10. SiC

B. Direction: Graph the heating curve of ethanol using the information given below.
Check off each number ass you add additional information so that none is missed.
Background information on ethanol: Boiling point = 60 oC
Melting point = -105 oC
Starting temperature = -120 oC

1. After two minutes, frozen cold ethanol starts to melt. It takes two minutes to melt
completely.
2. After eight more minutes, it begins to boil. It boils for six minutes.
3. Heat is added for two more minutes until ethanol reaches 80 oC.
4. Label “Melting” where this takes place.
5. Label “Vaporization” where this takes place.
6. Label “Phase Change” where a phase change occurs.
7. Indicate where ethanol is only a SOLID, only LIQUID, and only a GAS.
8. Of these three phases, label which phase has weakest IMF (intermolecular
force), medium IMF, and the strongest IMF.

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 Summary

 The intermolecular forces of attraction are the reasons why we have phases of matter.
 Dispersion force or London dispersion force exists in all substances.
 Dipole – dipole force exists in a polar molecule.
 Ion – dipole force is the attraction between a cation or anion and a polar molecule.
 Hydrogen bonding is a special type of dipole – dipole force. It is the attraction between
the most electronegative elements (F, O, N) and hydrogen.
 Surface tension, viscosity, vapor pressure, boiling point, and molar heat of
vaporization are some of the properties of matter that depends on the strength of
intermolecular force.
 Water is a unique substance due to its structure`s ability to participate in a hydrogen
bonding in four sites. It is the only the substance that can exist in all phases in our
planet.
 Solids can be categorized into two: crystalline and amorphous.
 Crystals can be ionic, covalent, molecular, or metallic.
 Phase diagrams shows the relationship between solid, liquid, and gas.
 Heating curve is a line graph presentation of solid state changing to liquid or liquid
changing to gas.

References

 Chang, R. and Goldsby, K.A. (2016). Chemistry. (12th Edition). Mc-Graw Hill Education
 Ebbing D. D. and Gammon S.D. (2009). General Chemistry. (9th Edition). Houghton
Mifflin Company
 [Untitled Illustration of Sodium and Chlorine Anions]. Intermolecular forces and
solutions. shorturl.at/dorsE
 Purdue University. (n.d.). Dipole – dipole forces [Online Image].
https://www.chem.purdue.edu/gchelp/liquids/dipdip.html
 Silberberg M. S. (2007). Principles of General Chemistry. (1st Edition). Mc-Graw Hill
Companies, Inc.

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 MODULE 2
PHYSICAL PROPERTIES OF SOLUTIONS

Introduction

Most reaction does not take place in their pure state (pure liquid, solid, or gas), but
among ions or molecules dissolved in water or other solvents (Chang and Goldsby, 2016).
That is one practical reason why chemists or scientists alike prepare solutions. Aside from
that, solutions are very useful in our life. For instance, the air we breathe is a solution! The
coffee and soft drinks you drink, the bleaching solution that you use in washing your clothes,
and the metal alloys in cars and jeeps you ride is all a solution.

In this module, we will examine the properties of solution, determine how to express
solutions in different concentration units and discuss the different colligative properties of
solutions.

Learning Outcomes

At the end of this module, students should be able to:

1. determine different types of solutions;

2. use different ways of expressing concentration of solutions: percent by mass,
mole fraction, molarity, molality, percent by volume, and ppm;
3. perform stoichiometric calculations for reactions in solutions;
4. describe the effect of concentration on the colligative properties of solution;
5. differentiate the colligative properties of nonelectrolyte solutions and electrolyte
solution;
6. calculate the boiling point elevation and freezing point depression from the
concentration of a solute in a solution; and
7. calculate the molar mass from colligative property data.

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Lesson 1. Types of Solution

Solutions are homogenous mixtures of two or more substances. Solutions can be

characterized by their capacity to dissolve a solute (Chang and Goldsby, 2016); Silberberg,
2007; Ebbing and Gammon, 2009).

 Unsaturated – “contains less solute than it has the capacity to dissolve”

 Saturated – “contains maximum amount of a solute that will dissolve in a given at a
specific temperature”
 Supersaturated – “contains more solute than is present in saturated solution”

We can also distinguish six types of solution, depending on the original states of the solution
components. Take a look at Table 2.1.

Table 2.1 Types of Solutions

State of Resulting
Component 1 Component 2 Examples
Solution

Gas Gas Gas Air

Gas Liquid Liquid Soda water

Gas Solid Solid H2 gas in palladium

Liquid Liquid Liquid ethanol in water

Solid Liquid Liquid NaCl in water

Solid Solid Solid Brass (Cu/Zn)

Source: Chang and Goldsby (2016, p. 519)

Lesson 2. Concentration Units

“Concentration is the amount of solute present in a given amount of solution”.

Chemists use these different expression of concentration units, each with advantage and
limitations (Chang and Goldsby, 2016, p. 522). There are six common units of concentration:

20
  percent by mass
 percent by volume
 mole fraction
 molarity
 molality
 parts per million

Percent by mass (% by mass)

“Percent by mass is also called percent by weight or weight percent, is the ratio of the
mass of a solute to the mass of the solution multiplied by 100 percent”. This concentration
unit is unitless because it is the ratio of two similar quantities (Chang and Goldsby, 2016, p.
522). The formula is shown below:

Sample Problem

A sample of 0.892 g of potassium chloride (KCl) is dissolved in 54.6 g of water. What

is the percent by mass of KCl in the solution?

Given: mass of solute = 0.892 g KCl mass of solvent = 54.8 g water

Unknown: % by mass
mass of solute
Solution: percent by mass of KCl = mass of solute +mass of solvent or mass of solution x 100%
0.892 g
= 0.892 g + 54.6 g x 100%

= 1.61 %

Percent by volume

This often use in expressing the concentration of liquids and gases and often termed
as percent volume or %v/v (Silberberg, 2007).
volume of solute
Volume percent = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 X 100%

21
Sample Problem

What is the %v/v of 40 mL of ethanol in a 240 mL solution?

Given: volume of solute = 40 mL ethanol volume of solution = 240 mL solution

Unknown: % v/v
volume of solute
Solution: %v/v = X 100%
volume of solution
40 mL ethanol
= 240 mL solution X 100%

= 17%

Mole Fraction (X)

“It is the ratio of the number of moles of one component of a mixture to the total number
of moles of all components. It is denoted by Greek letter chi (X), and with subscript to indicate
the component of interest” (Ebbing and Gammon, 2009). The formula is shown below.
moles of component
Mole fraction of the component = total moles of all components

For example, the mole fraction of NaOH in a sodium hydroxide solution is represented as
XNaOH. This is also a unitless concentration unit.

Sample Problem

Find the mole fraction of ethanol in a 40% by mass ethanol solution.

Given: 40% by mass ethanol MMwater = 18 g/mol

MMethanol = 46 g/mol

Unknown: mole fraction of ethanol (Xethanol)

Solution:

Step 1. When converting concentration units based on the mass or moles of a solute and
solvent, it is useful to assume a certain total mass of solution. Assume there is exactly 100 g
of solution. Because the solution is 40% ethanol (C2H6O), it contains 40 grams of ethanol and
60 g of water.

Step 2. Convert the mass of the components (ethanol and water) to number of moles.
40 g
molethanol = 46 g/mol = 0.87 mol

22
 60 g
molwater = 18 g/mol = 3.33 mol

Step 3. Substitute the values in the formula.

mole ethanol
Xethanol = mole ethanol + mole water
0.87 mol
Xethanol = 0.87 mol + 3.33 mol

Xethanol = 0.21

Try to solve Xwater on your own!

Molarity (M)

“It is the number of moles of solute dissolved in 1 liter of solution” (Chang and Goldsby,
2016, p. 523). The formula of molarity is shown below.
moles of solute
Molarity (M) =
volume of solution in liter

Sample Problem

“What is the molarity of an 85.0 mL ethanol (C 2H6O) solution containing 1.77 g of

ethanol?”

Given: 85. O ml ethanol solution and 1.77 g of ethanol

Unknown: Molarity (M) of the solution

Solution:

Step 1. Convert the 1.77 g of ethanol to mol ethanol using the molar mass as the conversion
factor.
1.77 g
molethanol = 46 g/mol = 0.038 mol

Step 2. Substitute the values in the formula.

0.038 mol of ethanol
M = 85.0 ml of ethanol solution

= 0.00045 M

23
Molality (m)

“It is the number of moles of solute dissolved in 1 kg of solvent” (Chang and Goldsby,
2016, p. 523). The formula is shown below.
moles of solute
Molality (m) = mass of solvent in kg

Sample Problem

Calculate the molality of a sulfuric acid solution containing 24.4 g of sulfuric acid in 198
g of water. The molar mass of sulfuric acid is 98.09 g.

Given: 24.4 g of H2SO4 and 198 g of water

Unknown: m of sulfuric acid solution

Solution:

Step 1. Convert the mass of sulfuric acid to moles of sulfuric acid using the molar mass as the
conversion factor.
24.4 g
moles of H2SO4 = 98.09 g/mol = 0.249

Step 2. Convert mass of water to kilograms, that is —

1kg
198 g of water (1000 g ) = 0.198 kg of water

Step 3. Substitute the values in the formula.

moles of solute
m = mass of solvent in kg

0.249 mol H2SO4

m = 0.198 kg of water

= 1.26 m

Parts per million (ppm)

“It is commonly used as measures of smaller levels of concentration of solutions such

as pollutants in air, water, or body fluids” (The Engineering Toolbox, 2008). 1 ppm is equal to
1mg of substance per kg of solid or 1 mg of substance per kg of liquid.
mg of substance mg of substance
ppm = or
kg of solid L of liquid

24
Sample Problem

What is the concentration of the solution in ppm, if 0.03 grams of NaOH is dissolved
in 1000 mL of solution?

Given: 0.03 g of NaOH and 1000 mL of solution

Unknown: concentration in ppm

Solution:

Step 1. Convert 0.03 g of NaOH to mg and 1000 mL of solution to L.

0.03 g NaOH = 30 mg NaOH

1000 mL solution = 1 L solution

Step 2. Substitute the values in the formula.

mg of substance
ppm = L of liquid

30 𝑚𝑔 𝑁𝑎𝑂𝐻
ppm =
1 L of solution

= 30 ppm

Lesson 3. Stoichiometric Calculations in Solutions

“Stoichiometry is the relationship between the relative quantities of substances taking

part in a reaction or compound formation”. In this lesson, we are going to focus on solution
chemistry such that problems presented will incorporate concentrations. It is important that
correct conversions are observed, thus labeling units is a must (Chang and Goldsby, 2016).
Take a look at Figure 2.1 below.

Figure 2.1 Relationship between the concentrations of reactants and products

Source: (Teaching Guide for Senior High School Gen. Chem 2, 2016, p. 211)

25
 Figure 2.1 shows the relationship between the concentration of reactants and
products. We can use this as our model when dealing with stoichiometric calculations with
solutions.
Sample Problem 1

Nitric acid reacts with sodium hydroxide in solution to give sodium nitrate and water.

How many moles of water are formed when 25.0 mL of 0.100 M HNO 3 reacts with
NaOH?

Step 1 “Always check if the chemical equation is balance. Since the equation above is
balance, we can proceed now step 2”.

Step 2 “Relate the situation the model in Figure 2.1”.

Molarity of moles of moles of

HNO3 HNO3 water

Step 3 “We should first get the number of moles of HNO3. This can be done by multiplying the
molarity of HNO3 to the volume of HNO3. Convert first the volume to proper unit which is liters”.

VHNO3 = 25. 0 mL x 1L / 1000 mL = 0.025 L

Thus:

moles HNO3 = MHNO3 X VHNO3 = 0.100 mol/L X 0.025 L = 0.0025 mol

Step 4 “Convert moles of HNO3 to moles of water by using the mole to mole ratio. According
to the balance chemical reaction, for every 1 mole of HNO 3 reacted, one mole of water is
formed. The mole to mole ratio is 1:1”
mol of H2O
moles H2O = mol HNO3 X ( mol HNO3
)

1 mol of H 2O
= 0.0025 mol HNO3 X ( )
1 mol HNO3

=0.0025 mol of H2O is formed.

26
Sample Problem 2

“What volume of a 0.470 M of HCl reacted with enough Zn metals to produce 1.50 g
ZnCl2 based on the following reaction? The molar mass of ZnCl 2 is 136.29 g/mol.”

Step 1. “Check if the reaction is balance. Since the reaction above is balanced, we can
proceed to step 2.”

Step 2. Relate the problem to the model in Figure 2.1.

moles of ZnCl2 moles of HCl Volume of

HCl

Step 3. According to our model in step 2, we should first get moles of ZnCl 2 by converting the
mass of ZnCl2 to moles of ZnCl2 using the molar mass of ZnCl2 as the conversion factor.
1 mol
mol ZnCl2 = 1.50 g ZnCl2 (136.29 g ZnCl ) = 0.0110 mol ZnCl2
2

Step 4. According to our model in step 2, we should get the moles of HCl. This can be obtained
through multiplying the moles of ZnCl2 to the mole ratio of moles ZnCl2 to moles HCl. Since
there 2 moles of HCl is needed to form 1 mole of ZnCl 2, the mole ratio is 2:1.
2 mol HCl
mol HCl – 0.110 mol ZnCl2 ( ) = 0.0220 mol HCl
1 mol ZnCl2

Step 5. Now that we obtained the moles of HCl needed, we can divide the molarity of HCl to
moles of HCl to obtain the volume of HCl.
0.0220 mol HCl
VHCl = 0.0470 mol = 0.0468 L = 46.8 mL
1L

27
Lesson 4. Colligative Properties of Nonelectrolyte and
Electrolyte Solutions

“Colligative properties are properties that does not depend on the nature of the solute
but instead depend on the number of solute particles in solution (Chang and Goldsby, 2016).
The colligative properties are:”
 vapor pressure lowering
 boiling point elevation
 freezing point depression
“Electrolyte is a substance that dissociates into ions in aqueous solution. Nonelectrolytes do
not dissociate into ions at all” (Silberberg, 2007).

1. Colligative Property of Nonelectrolyte Solution

 Vapor pressure lowering (ΔP)
“The vapor pressure of a solution of a nonvolatile nonelectrolyte is
always lower than the vapor pressure of the pure solvent. This relationship can
be quantitatively expressed by Raoult`s Law.”

Psolvent = Xsolvent X Posolvent

“In a solution containing only one solute, X1 = 1 – X2, where X2 is the

mole fraction of the solute. The equation above can be rearranged as
(Silberberg, 2007, p. 407)”

Po1 – P1 = ΔP = X2Po1

 Boiling point elevation (ΔTb)

“The presence of nonvolatile solute lowers the vapor pressure of the
solution, then it also affects the boiling point of the solution. It is defined as the
boiling point of the solution (Tb) minus the boiling point of the pure solvent (T ob)”
(Chang and Goldsby, 2016, p.536). Like the vapor pressure lowering, the
boiling point elevation is proportional to the concentration of solute particles:
ΔTb = Kbm

28
 Where: ΔTb = boiling point

Kb = molal boiling point elevation constant (oC/m) of solvent

m = molality of the solution

ΔTb is a positive quantity, therefore we will subtract the lower

temperature to the high temperature. We will subtract the solvent Tob to
solution Tb.

ΔTb = Tb - Tob
 Freezing point depression (ΔTf)
“It is defined as the difference between freezing point of the pure
solvent (Tof) and the freezing point of the solution (T f). Mathematically written
as:”
ΔTf = Tof - Tf
Again, ΔTf is proportional to the concentration of the solution.

ΔTf = Kfm
Where:
ΔTf = freezing point depression
Kf = molal freezing-point depression constant (oC/m) of solvent
m = molality of the solution
The qualitative explanation of this phenomenon is due to great disorder in the
system of solution than the solvent, thus more energy needs to be removed in a
solution than a solvent. Therefore, the solution`s freezing point is lower than the
solvent (Ebbing and Gammon, 2009).

Figure 2.2 De-icing of airplanes is based on freezing-point depression

Source: Chang and Goldsby (2016, p.537)

29
Table 2.2 Molal boiling point elevation and freezing point depression constants of several
solvents

Solvent Boiling Point Kb (oC/m) Melting Point Kf (oC/m)

(oC) (oC)
Acetic acid 117.9 3.07 16.6 3.90
Benzene 80.1 2.53 5.455 5.12
Carbon disulfide 46.2 2.34 -111.5 3.83
Carbon 76.5 5.03 -23 30.0
tetrachloride
Chloroform 61.7 3.63 -63.5 4.70
Diethyl ether 34.5 2.02 -116.2 1.79
Ethanol 78.5 1.22 -117.3 1.99
Water 100.0 0.512 0.0 1.86
Source: Silberberg (2007, p. 409)

Sample Problems involving the boiling point elevation and freezing point depression
of solutions

Sample Problem 1

“An aqueous solution is 0.0222 m glucose. What are the boiling point
and freezing point of this solution? Kb and Kf of water are 0.512 oC/m and 1.86
oC/m, respectively” (Ebbing and Gammon, 2009, p. 501).

Given: 0.0222 m glucose Kb = 0.512 oC/m Kf = 1.86 oC/m

Unknown: boiling point and freezing point of the solution

Solution:

Step 1. “Solve first for the boiling point elevation and the freezing point
depression. Substitute the values in the formula”.

ΔTb = Kbm = 0.512 oC/m X 0.0222 m = 0.0114 oC

ΔTf = Kfm = 1.86 oC/m X 0.0222 m = 0.0413 oC

30
Step 2. “To obtain the boiling point of the solution, add the boiling point
elevation to the boiling point of solvent. Rearrange the formula and substitute
the values in the formula.”

ΔTb = Tb - Tob

Rearranged to:Tb = ΔTb + Tob = 0.0114 oC + 100 oC = 100.011 oC

The boiling point of the solution is 100.011 oC.

Step 3. “To obtain the freezing point depression of the solution, subtract the
freezing point depression to the freezing point of solvent. Rearrange the
formula and substitute the values in the formula.”

ΔTf = Tof - Tf
Rearranged to: Tf = Tof - ΔTf = 0.00 oC – 0.0114 oC = - 0.041 oC

The freezing point of the solution is - 0.041 oC.

“NOTE: The boiling point of solution is higher than the boiling point of solvent.
The freezing point of solution is lower than the freezing point of solvent.”

Sample Problem 2

“You add 1.00 kg ethylene glycol (C 2H6O6) antifreeze to your car

radiator, which contains 4450 g of water. What are the boiling and freezing
points of the solution?”

Given: 1.00 kg of ethylene glycol and 4450 g of water

Since the solvent is water, Kb = 0.512 oC and Kf = 1.86 oC/m

Unknown: boiling point and freezing point of the solution

Solution:

Step 1 “In order to compute the boiling and freezing point of the solution, we
need the molal boiling and freezing constant, and the molality. Molality is not
given in the problem but we are given with the mass of solute and mass of
solvent. Convert first the mass of solute into moles using the molar mass of
ethylene glycol as the conversion factor. MM of ethylene glycol is 62.07 g/mol.”

31
 1000 g 1 mol C2H6O2
Moles of C2H6O2 = 1.00 kg C2H6O2 X 1 kg
X 62.07 g C H O = 16.1 mol C2H6O2
2 6 2

Step 2 “Calculate the molality of the solution”.

moles of solute 16.1 mol C2H6O2
m= = = 3.62 m
mass of solvent in kg 4.450 kg

Step 3 “Substitute the values in the formula to calculate boiling point elevation
(ΔTb) and boiling point of solution (Tb).”

ΔTb = Kbm = 0.512 oC/m X 3.62 m = 1.85 oC

Tb = ΔTb + Tob = 1.85 oC + 100 oC = 101.85 oC

Step 4 Substitute the values in the formula to calculate the freezing point
depression (ΔTf) and freezing point of the solution (Tf).

ΔTf = Kfm = 1.86 oC/m X 3.62 m = 6.73 oC

Tf = Tof - ΔTf = 0.00 oC – 6.73 oC = - 6.73 oC

2. Colligative Property of Electrolyte Solution

Colligative properties of electrolytes required a different approach compared to

the colligative property of nonelectrolyte. This is because electrolytes dissociate into
two or more ions in a solution. “For example, each unit of NaCl dissociates into two
ions— Na2+ and Cl-. Thus, the colligative property of a 0.1 m NaCl solution should be
twice as great as those of 0.1 m solution containing nonelectrolyte such as sucrose”
(Chang and Goldsby, 2016, p. 544). That is why we are including a multiplying factor
called the van`t Hoff factor (i), named after the Dutch chemist Jacobus van`t Hoff. “It
is the ratio of the measured value of the colligative property in the electrolyte solution
to the expected value for nonelectrolyte solution:”
measured value for electrolyte solution
i = expected value for nonelectrolyte solution

Thus, i should be always 1 for all nonelectrolyte and for strong electrolytes such
as NaCl and KNO3, i should be 2 and for Na2SO4 and CaCl2, i should be 3. The formula
for colligative properties of strong electrolyte solution are:

For vapor pressure lowering: ΔP = i (X2Po1)

For boiling point elevation: ΔTb =i (Kbm)

32
 For freezing point depression: ΔTf = i (Kfm)

Lesson 5. Calculating Molar Mass Using Colligative Properties

Colligative properties of nonelectrolyte solutions provide a mean for chemists to

calculate the molar mass of the solute. Theoretically, you can use all the colligative properties
to compute for the molar mass, but in practice freezing point depression is the most commonly
used. To do this, scientists experimentally determine the freezing point of the solution, then
calculate the molality of the solution. Knowing the mass of the solute, we can readily determine
its molar mass (Chang and Goldsby, 2016, p. 542). The sequence of solving the molar mass
is given below.
freezing point depression molality number of moles molar mass

Sample Problem

“A 7.85 g sample of a compound with the empirical formula C 5H4 is dissolved in 301 g
of benzene. The freezing point of the solution is 1.50 oC below of that pure benzene. What is
the molar mass and molecular formula of this compound?” (Chang and Goldsby, 2016, p. 542)

Given: 7.85 g sample of C5H4 dissolved in 301 g benzene

ΔTf = 1.05 oC Kf of benzene is 5.12 oC/m (see Table 2.2)

Unknown: molar mass and molecular formula of this compound

Solution:

Step 1. “Calculate first the molality of the solution by manipulating the formula for freezing
point depression. We write the new formula”
ΔTf 1.05 oC
molality = = = 0.205 m
Kf 5.12 oC/m

Step 2. “There is 0.205 mole in 1 kg of solution, the number of moles in 301 g or 0.301 kg of
solvent is”
0.205 mol
0.301 kg X 1 kg
= 0.617 mol

Step 3. “Calculate the molar mass.”

33
 grams of compound 7.85 g
molar mass = moles of compound
= 0.0617 mol = 127 g/ mol

Step 4. “Calculate the molecular formula by comparing the molar mass to the molar mass of
the empirical formula.”
molar mass 127 g/mol
empirical molar mass
= 64 g/mol
≈2

Therefore, the molecular formula is C10H8 (naphthalene).

34
Assessment Task

TASK NO. 1(WRITTEN WORK)

A. Direction: Fill in the blank table below by choosing the appropriate answer in the
box given.

water(liquid) mercury(liquid) nitrogen(gas)

gas in gas liquid in a solid liquid in liquid

acetic acid(liquid) oxygen(gas) silver(solid)

Type Example Primary Solute Solvent

Gas Solutions
gas in gas Air 2 3
Liquid Solutions
4 Vinegar 5 6
Solid Solutions
7 Dental Amalgam 8 9

B. Direction: Problem Solving. Show your solution. Follow proper use of sig.
figures.

1. “A sample of 6.44 g of naphthalene (C10H8) is dissolved in 80.1 g of benzene

(C6H6). Calculate the percent by mass of naphthalene in this solution.” (3 points)

35
2. “Calculate the mole fractions of the solute in a100.0 g of C 2H6O in 100 grams of water.”
(5 points)

3. “A solution is prepared by mixing a 1.00 g of ethanol (C 2H6O) with 100.0 g water to

give a final volume of 101 mL. Calculate the mole fraction of solute and solvent, and the
molality of the solution.” (8 points)

4. “What is the molality of a solution containing 7.78 g of urea [(NH 2)2CO] in 203 g of
water?” (5 points)

5. What volume in mL of a 0.315 M NaOH solution contains 6.22 g of NaOH? (3 points)

36
TASK NO. 2 (PERFORMANCE TASK)

A. Direction: Problem Solving. Show your solution and follow proper use significant
figures.

1. Consider the following equation.

Ca(OH)2(s) + HCl → CaCl2(aq) + H2O(l)

How many liters of 0.100 M HCl is required to completely react with 5.00 g of calcium
hydroxide? (7 points)

2. Using the following equation:

NaOH + H2SO4 H2 + Na2SO4

“How many grams of sodium sulfate will be formed, if you start with 1.25 L
of a 4.0 M solution of sodium hydroxide?” (7 points)

37
3. “How many grams of Ca(OH)2 are needed to neutralize 25.0 mL of 0.100 M HNO 3?”
The reaction proceeds as follows:

HNO3(aq) + Ca(OH)2(s) H2O(l) + Ca(NO)3(aq)

B. Direction: Problem Solving. Show your solution and follow proper significant figures.

1. “A solution is prepared by dissolving 0.843 g of sulfur, S 8, in 100.0 g of acetic acid,

HC2H3O2. Calculate the freezing point and boiling point of solution, K b and Kf of acetic
acid is 3.08 oC/m and 3.59 oC/m respectively.” (10 points)

2. “Butylated hydroxytoluene (BHT) is used as an anti-oxidant in processed foods. A

solution of 2.500 g of BHT in 100.0 g of benzene had a freezing point of 4.880 oC.
What is the molar mass of BHT?” (8 points)

38
 Summary

 Solution is a homogenous mixture of two or more substances and can be gases, liquid,
or solids.
 The different units of concentration are: percent by mass, percent by volume, mole
fraction, molality, molarity, and parts per million.
 Colligative properties do not depend on the nature of solute but rather on the number
of solute particles in solution.
 The colligative properties are: vapor pressure lowering, boiling point elevation, and
freezing point depression.

References

 Chang, R. and Goldsby, K.A. (2016). Chemistry. (12th Edition). Mc-Graw Hill Education
 Ebbing D. D. and Gammon S.D. (2009). General Chemistry. (9th Edition). Houghton Mifflin
Company
 Engineering Toolbox. (2008). Parts per million – ppm. Retrieved July 23, 2020, from
https://www.engineeringtoolbox.com/ppm-d_1039.html
 Silberberg M. S. (2007). Principles of General Chemistry. (1st Edition). Mc-Graw Hill
Companies, Inc.

39
 MODULE 3
THERMOCHEMISTRY

Introduction

Energy content of matter will change if it undergoes physical or chemical change. This
means that almost all chemical reactions may produce or absorb energy in the form of heat.
Heat is the transfer of thermal energy between two bodies that are at different temperature.
When describing energy changes that occur during processes, we described it as “heat
absorbed” or “heat released”. Example of this is during burning of woods where energy is
released as heat and light and when ice melts to water, energy is absorbed (Chang and
Goldsby, 2016, p. 232).

“Thermodynamics is the study of the interconversion of heat and other kinds of energy.
Thermodynamics will be explored in the later module”. Our focus in this module is under the
broad topic of thermodynamics, which is thermochemistry or the study of heat change in a
chemical reaction (Chang and Goldsby, 2016, p. 232)

Learning Outcomes

At the end of this module, students should be able to:

1. explain the first law of thermodynamics;

2. explain enthalpy; and
3. calculate the change in enthalpy of a given reaction using Hess` Law.

40
Lesson 1. First Law of Thermodynamics

“The first law of thermodynamics is based on the law of conservation of energy, which
states that energy can be converted from one form to another, but cannot be created or
destroyed.” We can see every day the application of this law. For example, in cars, gasolines
were burned in the car engine and is transformed as mechanical energy. The burning of oils
in power plants can be transformed as electricity and light energy. Biological processes like
photosynthesis cannot proceed without the light energy from the sun, which can be converted
to chemical energy. Energy is not totally lost but instead transformed or transferred to another.
In chemistry, we are more focus on the energy changes associated with the system (which
may be a flask containing the reactants and products), not with surroundings (Chang and
Goldsby, 2016, pp. 234-236). We can express this mathematically as
ΔU = q + w

or simply say the “change in internal energy (ΔU) of a system, is equal to the sum of heat (q)
exchange between the system and surroundings and work (w) done on (or by) the system.”
The sign conventions of q and w is summarized in Table 3.1 below.

Table 3.1 Sign conventions for work and heat

Process Sign
Work done by the system on the surroundings -
Work done on the system by the surroundings +
Heat absorbed by the system from surrounds
+
(endothermic process)
Heat released to the surroundings from the
-
system (exothermic process)
Source: (Chang and Goldsby, 2016, p. 238)

If a system released energy to the surroundings or does work on the surrounding, we

would expect the internal energy to be negative. In contrast, if we add heat to the system or if
work is done on the system, we would an increase on the internal energy (Chang and Goldsby,
2016). Let us try to solve some problem with the application of the first law of thermodynamics.

41
Sample Problem

“The work done when a gas is compressed in a cylinder is 462 J (joules). During this
process, there is a heat transfer of 128 J from the gas to the surroundings. Calculate the
energy change for this process.” (Chang and Goldsby, 2016, p. 239).

Given: work done on gas is 462 J and heat transfer of gas is 128 to the surroundings.

Unknown: change in internal energy (ΔU)

Solution:

Step 1 We need to know first the sign convention of our w and q. Since compression is a work
done in gas, w sign is +462 J. Heat is released to the surroundings, therefore q sign is -128
J.

Step 2 Use the formula for the change in internal energy and substitute the values.

ΔU = q + w

ΔU = -128 J + (+462 J)

= 334 J

The energy of the gas increased by 334 J.

Lesson 2. Enthalpy

“Enthalpy (H) is defined is the sum of internal energy (U) of the system and the product
of pressure and volume (PV)” (Chang and Goldsby, 2016).

H = U + PV

Since U and PV has energy units, therefore H has also energy units. The SI unit of
energy is joules (J). It is a derived unit composed of three basic units (1 J = 1 kg * m 2/s2).
Calorie (cal) is an “older unit that was defined originally as the quantity of energy needed to

42
raise the temperature of 1 g of water by 1 oC.” The conversion factor of joule to calorie and
vice versa is (Silberberg, 2007)

1 cal = 4.184 J or 1 J = 0.2390 cal

If we know the enthalpy of the substances, therefore you can calculate the change in
enthalpy.

ΔH = ΔU + Δ(PV)

Since almost all of reactions that occur is in constant pressure, then

ΔH = ΔU + PΔV

Enthalpy of Reaction (ΔHrxn)

“For any reaction type of

reactants products

we define the change in enthalpy, called the enthalpy of reaction, ΔHrxn, as the difference
between the enthalpies of products and the enthalpies of reactants.”

ΔH = H(products) – H(reactants)

“Enthalpy of reaction can be positive or negative, depending on the process.

Exothermic (heat released by the system) ΔH is negative. For an endothermic (heat absorbed
by the system), ΔH is positive (Chang and Goldsby, 2016, p. 242). Take a look at the
thermochemical equation (shows the enthalpy changes as well as mass relationships) below:”

H2O(s) H2O(l) ΔH = 6.01 kJ/mol

The thermochemical equation above is the melting of ice which is an example of

endothermic process since our ΔH is positive. The “per mole” unit written in the ΔH means
that is enthalpy changer per mole of the reaction (when 1 mole of ice is converted to 1 mole
of liquid water). Let us look now on another example which is the burning of natural gas.

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = - 890.4 kJ/mol

43
 The thermochemical equation above is an exothermic process since the value of ΔH
is negative also burning gases releases heat to the surroundings. To explain the reaction
above, if 1 mole of CH4 reacts with 2 moles of O2 to produce 1 mole of CO2 and 2 mole of
water, 890.4 kJ of energy is released to the surroundings. Always remember that ΔH refers
to all the reacting species in a thermochemical reaction and not to a particular reactant or
product. Therefore, we can create a conversion factor (Chang and Goldsby, 2016, p. 242-
243).
- 890 .4 kJ - 890 .4 kJ - 890 .4 kJ - 890 .4 kJ
1 mol CH4 2 mol O2 1 mol CH4 1 mol CH4

Let us try to solve some problem involving change in enthalpy.

Sample Problem 1

“Given the thermochemical equation

2SO2(g) + O2(g) 2SO3(g) ΔH = - 198.2 kJ/mol

calculate the heat evolved when 87.9 g of SO 2 (molar mass = 64.07 g/mol) is converted to
SO3” (Chang and Goldsby, 2016, p. 244).

Given: mass of SO2 = 87.9 g and ΔH = - 198.2 kJ/mol

Unknown: heat evolved when SO2 is converted to SO3.

Solution:

“Convert the mass of SO2 first to moles of SO2 and then multiply it to the conversion factor.
- 198.2 kJ
The conversion factor is 2 mol SO . Therefore, the enthalpy for this reaction is given by:”
2

1 mol SO2 - 198.2 kJ

ΔH = 87.9 g SO2 X 64.07 g SO X 2 mol SO2
= -136 kJ
2

The heat released to the surroundings is 136 kJ.

Sample Problem 2

Try to solve this on your own!

Calculate the heat evolved when 266 g of white phosphorus (P 4) burns in air
according to the equation.

P4(s) + 5O2(g) P4O10(s) ΔH = -3013 kJ/mol

44
Lesson 3. Hess` Law

“Hess`s law of heat of summation is defined as the enthalpy change of an overall

process is the sum of the enthalpy changes of its individual steps, proposed by Swiss-Russian
chemist Germain Hess. This is because many compounds cannot be directly synthesized
from their elements. There are some cases where the reaction is too slow or there is a side
reaction that produces substances other than the desired compound. In these cases, Hess`
law can be used to find for ΔH. In other words, if we break down the reaction of interest into
a series of reaction for which ΔH can be measured, we can calculate the overall ΔH for the
overall reaction” (Chang and Goldsby, 2016, p. 255).
In calculating the unknown ΔH, it involves 3 steps (Silberberg, 2007, pg, 193).

 “Identify the target equation, the step whose ΔH is unknown, and note the number of
moles of each reactant and product.”
 “Manipulate the equations with known ΔH values so that the target numbers of moles
of reactants and products are on the correct side. Remember to:
1. Change the sign of ΔH when you reverse an equation
2. Multiply number of moles and ΔH by the same factor”
 “Add the manipulated equations to obtain the target equation. All substances except
those in the target equation must cancel. Add their ΔH values to obtain the unknown
ΔH.”

To better understand this, let us try to solve some problem using Hess` Law.

Sample Problem 1

Let`s see how we apply the Hess`s law in the case of the oxidation of sulfur to sulfur
trioxide, the central process in the industrial production of sulfuric acid and in the formation of
acid rain. When we burn S in excess of O 2, sulfur dioxide is formed not sulfur trioxide (SO 3)
as you can see in Equation 1 below. If we oxidized SO 2, we can produce SO3 (Equation 2)
(Silberberg, 2007, p. 142).

Equation 1. S(s) + O2(g) SO2(g) ΔH1 = -296.8 kJ

Equation 2. 2SO2(g) + O2(g) 2SO3(g) ΔH2 = -198.4 kJ

3
Equation 3. S(s) + 2O2(g) SO3(g) ΔH3 = ?

45
You can see in the equations that you cannot obtain SO 3 just by reacting sulfur with oxygen.
Even if that is not possible, we can still determine overall ΔH of the reaction through the use
of Hess`s Law.

Solution:

Step 1. Identify our target equation. Our target equation is equation 3 since its ΔH3 is unknown.
Note that ΔH1 are values of Equation 1 and ΔH2 are values of Equation 2. We will manipulate
Equations 1 and 2 to add up to Equation 3.

Step 2. Equation 1 and 3 contain the same amount of S, so we leave Equation 1 unchanged.
1
Equation 2 has twice as much as SO 3 as Equation 3, so we multiply by 2, does ΔH2 is also
halved.

Equation 1. S(s) + O2(g) SO2(g) ΔH1 = -296.8 kJ

1
(Equation 2. SO2(g) + 2O2(g) SO3(g) ΔH2 = -99.2 kJ) multiply to 1/2

Step 3. With the targeted amounts of reactants and products now present, we add Equation
1 and to the halved Equation 2 and cancel the terms for both sides:

Equation 1. S(s) + O2(g) SO2(g) ΔH1 = -296.8 kJ

1
Equation 2. SO2(g) + O2(g) SO3(g) ΔH2 = -99.2 kJ
2
3
Equation 3. S(s) + O2(g) SO3(g) ΔH3 = ?
2

Step 4. Add the values of ΔH1 and ΔH2.

ΔH3 = ΔH1 + ΔH2 = -296.8 kJ + (- 99.2 kJ) = -396.0 kJ

Sample Problem 2

“Two gaseous pollutants that form into auto exhaust are CO and NO. An environmental
chemist is studying ways to convert them to less harmful gases through the following
equations” (Silberberg, 2007)”
1
Equation 1 CO(g) + 2O2(g) CO2(g) ΔH1 = -283.0 kJ

Equation 2 N2(g) + O2(g) 2NO(g) ΔH2 = 180.6 kJ

1
Equation 3. CO(g) + NO(g) CO2 + N2(g) ΔH3 = ?
2

46
Calculate the unknown ΔH.

Solution:

Step 1. Identify the target equation. Our target equation is Equation 3.

Step 2. Manipulate the given equations. Equation 1 has the same number of moles of CO and
CO2 as the target, so we leave it as a written. Equation 2 has twice the needed amounts of N 2
and NO, they are on the opposite sides from the target; therefore, we will reverse Equation 2
and multiply it to 1/2. When we reverse an equation, we will also reverse the value of ΔH2.
1 1 1
Equation 2 [ 2NO(g) N2(g) + O2(g) ] ΔH = - (ΔH2) = - (180.6 kJ)
2 2 2
1 1
Or, NO(g) N2(g) + O2(g) ΔH = -90.3 kJ
2 2

Step 3. Add the manipulated equations to obtain the target equation.

1
Equation 1. CO(g) + 2O2(g) CO2(g) ΔH = -283.0 kJ
1 1
Equation 2 (reversed) NO(g) N2(g) + O2(g) ΔH = -90.3 kJ
2 2
1
Target: CO(g) + NO(g) CO2(g) + N2(g) ΔH = -373.3 k
2

Assessment Task
TASK NO. 1 (WRITTEN WORK)

Direction: Problem Solving. Show your solution and follow proper significant figures.

1. “In a reaction, gaseous reactants form a liquid product. The heat absorbed by the
surroundings is 26.0 kcal, and the work done on the system is 15825.8 J. Calculate the
change in internal energy.” (5 points)

2. “The work done to compress a gas is 74 J. As a result, 26 J of heat is given off to the
surroundings. Calculate the change in internal energy.” (5 points)

47
TASK NO. 2 (PERFORMANCE TASK)

Direction: Problem Solving. Show your solution and follow proper use of significant
figures.

1. “The first step in the industrial recovery of zinc from the zinc sulfide ore is roasting, that
is, conversion of ZnS to ZnO by heating:”

2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g) ΔH = -879 kJ/mol

Calculate the heat evolved in kJ, per gram of ZnS roasted.

2. “Determine the amount of heat given off when 1.26 x 10 4 g of NO2 are produced
according to the equation”

2NO2(g) + O2(g) → 2NO2 ΔH = -114.6 kJ/mol

3. “Calculate the overall ΔH for and determine if it is exothermic or endothermic.” (7 points)

1
Ca(s) + O2(g) + CO2(g) → CaCO3(s)
2

Given the following reactions:

1
Ca(s) + 2O2(g) → CaO(s) ΔH = -635.1 kJ

CaCO3(s) → CaO(s) + CO2(g) ΔH = 178.3 kJ

4. “Calculate the overall ΔH and determine if it is exothermic or endothermic. (7 points)”

2NOCl(g) → N2(g) + O2(g) + Cl2(g)

Given the following reactions:

1 1
2
N2(G) + 2O2(g) → NO(g) ΔH = 90.3 kJ
1
NO(g) + 2Cl2(g) → NOCl(g) ΔH = -38.6 kJ

48
Summary

 Thermochemistry is the study of heat change in a reaction.

 The first law of thermodynamics states that energy cannot be created or destroyed but
only transformed to other forms or transferred.
 Enthalpy is the sum of change in internal energy (U) and product of pressure and
volume and is denoted as H.
 Hess’s law of heat summation is defined as the overall enthalpy change equals to
individual enthalpy change of reactions involved. It is an indirect method of measuring
enthalpy change ΔH.

References

 Chang, R. and Goldsby, K.A. (2016). Chemistry. (12th Edition). Mc-Graw Hill Education
 Ebbing D. D. and Gammon S.D. (2009). General Chemistry. (9th Edition). Houghton
Mifflin Company
 Silberberg M. S. (2007). Principles of General Chemistry. (1st Edition). Mc-Graw Hill
Companies, Inc.

49
 MODULE 4
CHEMICAL KINETICS

Introduction

In the previous module, we discussed the changes in the internal energy of the system.
In this module, we are going to tackle how fast does the reaction proceeds. Chemical reactions
occur in varying time for completion depending on the characteristics of the reactants and
products. Some reactions occur in just a fraction of a second, some in just an hour and some
takes a longer time, such as days or years. “Chemical Kinetics is the area of chemistry
concerned with speed, or rates, at which a chemical reaction occurs” (Chang and Goldsby,
2016, p. 563).

It is a very essential knowledge for us to know how long a reaction proceeds. This is
because we can learn in details what is happening in a reaction at the molecular level. We
can also use this knowledge to maximize the potential yields with a given time or if we want
to slow down or stop an unwanted reaction (Silberberg, 2007, p. 499).

Learning Outcomes

At the end of this module, students should be able to:

1. describe how various factors influence rate of reaction,

2. differentiate zero, first, and second-order reactions,
3. explain reactions qualitatively in terms of molecular collisions; and
4. cite and differentiate types of catalysts.

50
Lesson 1. Factors Affecting Reaction Rates

Let us begin the lesson with some key factors influencing the reaction rates. Reaction
rate is the increase or decrease in molar concentration of product of a reaction per unit time
(Chang and Goldsby, 2016). Some reactions are fast or slow, but the rate of any reaction is
affected by the following factors:

1. Concentration

“It is one major factor influencing the rate of a given reaction is reactant
concentration. Since higher concentration means a higher number of molecules
present in the container, the more frequent they collide, and the more often the
reaction occurs. An example of this is a piece of steel wood burns with some difficulty
in air (20% O2) but bursts white flame in pure oxygen. The rate of burning increases
with the concentration of O2. However, some is unaffected by the concentration of
particular reactant, as long as it is present at some concentration” (Ebbing and
Gammon, 2009, p. 525). The reaction rate is proportional to the concentration
reactants:

Rate ∝ collision frequency ∝ concentration

2. Physical State

Physical state can also affect the reaction rate in a way that if the “reactants
are in same phase, such as in aqueous solution, random thermal motion brings them
in contact. While if they are in different phase, contact between particles is small and
needs a vigorous stirring or grinding may be required. The more finely divided a solid
or liquid reactant, the greater the surface area per unit volume, the more contact it
makes with the other reactant, and the faster the reaction occurs. You can see this
concept when burning woods, because wood chips and thin branches burns faster
than thick logs” (Silberberg, 2007, p. 500).

51
 Figure 4.1 Lycopodium powder burns easily because of its high surface area
Source: Ebbing and Gammon (2016, p. 525)

3. Temperature

High temperature usually speed ups the rate of reaction. Since increasing
temperature also increases kinetic energy, thus more collision happens at a given time
(Silberberg, 2007). “Increasing the temperature increases the reaction rate by
increasing the number, and especially, the energy of collisions.”

rate ∝ collision energy ∝ temperature

4. Catalysts

“Catalyst is a substance that increases the rate of reaction without being

consumed in the overall reaction. Catalysts is not consumed in the reaction therefore;
it does not appear in the balanced chemical equation (its presence can be written by
writing its formula over the arrow). A solution of pure hydrogen peroxide, H 2O2 , is
stable but when hydrobromic acid, HBr(aq), is added, H2O2 decomposes rapidly into H2
and O2” (Silberberg, 2007), as shown in Figure 4.2 below.

2 H2O2(aq) HBr(aq) 2H2O(l) + O2(g)

52
 Figure 4.2 Catalytic decomposition of hydrogen
Source: Ebbing and Gammon (2009, p. 525)

Lesson 2. Reaction Order

We can classify a reaction by its order. The reaction order with respect to a given
reactant species equals the exponent of the concentration of that species in the rate law, as
determined experimentally. The rate law expresses the “rate as a function of reactant,
concentrations, product concentrations, and temperature” (Chang and Goldsby, 2016, p. 571).
For the general equation.
aA + bB cC + dD

the rate law takes the form

rate = k[A]x[B]y

where x and y are numbers that must be determined experimentally and cannot be determined
just by looking at the balanced chemical equation. Note that x and y are not equal to
stoichiometric coefficients a and b. The symbol [A] and [B] is read as “concentration of A and
concentration of B”. The exponents x and y relate the concentration of A and B to reaction
rates. When we know the values of x, y, and k, we can calculate for the rate of reaction given
the reaction rates of A and B. Now, how are we going to determine the reaction order? To
determine the overall reaction, we just need to add the exponents x and y. We can say that A
is in xth order, B is in yth order and the overall reaction order is x + y (Chang and Goldsby,
2016; Silberberg, 2007; Ebbing and Gammon, 2009). Let’s take a look the different order of
reaction.

53
First Order Reaction

“A reaction is first order overall if the rate is directly proportional to [A]” (Silberberg,
2007, p. 506).

rate = k[A]

Example is heating of cyclopropane (C3H6)

Figure 4.3 Isomerization of cyclopropane to propylene

Source: (Ebbing and Gammon, 2009, p. 531)

The reaction has the experimentally determined rate law

rate = k[C3H6]

The reaction is first order in cyclopropane and first order overall (Ebbing and Gammon, 2009,
p. 531).

Second Order Reaction

“It is second order reaction overall if the rate is directly proportional to the square of
[A]” (Silberberg, 2007, p. 2007).

rate = k[A]2

Example is the reaction between nitrogen monoxide and ozone

NO(g) + O3(g) NO2(g) + O2(g)

54
 The rate law is experimentally determined as
rate = k[NO][O3]

The reaction is first order with respect to NO and O2. To determine the overall reaction, just
add the exponents of the reactants. The reaction is second order overall (1 + 1) (Ebbing and
Gammon, 2009, p. 532).

Zero Order Reaction

“It is zero order if the rate is not dependent on [A] at all, a common situation in a metal-
catalyzed and biochemical processes” (Silberberg, 2007, p. 506).

rate = k[A]0 = k (1) = k

Example is the reaction of acetone, CH3COCH3, with iodine in acidic solution.

CH3COCH3(aq) + I2(aq) H+ CH3CHO2I(aq) + HI(aq)

The experimentally determine rate law is

rate = k[CH3COCH3][H+]

The reaction order is first order in acetone and H +. It is zero in iodine, that is the rate law
contains the factor [I2]0 = 1. This means that the rate does not depend on the concentration of
I2, as long as some concentration of iodine is present. The overall reaction is second order
(Ebbing and Gammon, 2009, p. 532).

Take note also that although reaction orders are frequently positive whole number, they can
be fractional, negative or zero (Ebbing and Gammon, 2009, p. 532).

Lesson 3. Collision Theory

“Collision theory is a theory that assumes that, for every reaction to occur, reactant
molecules must collide with an energy greater than some minimum value and with the proper
orientation” (Ebbing and Gammon, 2009. p. 545).

55
 In collision theory, the rate constant for a reaction is given as a product of three factors:
1) Z, the collision frequency, 2) f , the fraction of collision energy greater than the activation
energy, and 3) p, the fraction of collision that occur with the reactant molecules properly
oriented” (Chang and Goldsby, 2016; Ebbing and Gammon, 2009; Silberberg, 2007) Thus,
k = Zfp

Collision Frequency

In order for a reaction to occur, there should be a collision between reactant molecules.
The collision frequency of reactant molecules depends on the temperature. To further
understand this, remember the kinetic molecular theory of gases where there is a postulate
that gas molecules frequently collide with each other. Increasing the temperature will also
increase the kinetic energy of molecules, there is increases in the motion of the gas molecules,
therefore more collisions with other gas molecules. This is just same concept with reactant
molecules in a chemical reaction (Ebbing and Gammon, 2009).

However not all collision results to a chemical reaction, since you already know that
the rates of reaction differ greatly. We must consider the first if the particles have enough
energy to collide and if the particles have a proper orientation (Ebbing and Gammon, 2009).

Activation Energy (Ea)

Activation energy or Ea is defined as the minimum energy of collision for two molecules
to react. This means that colliding molecules must have equal or greater kinetic energy than
the activation energy for collision to happen. It serves as energy barrier that needed to be
overcome for the reaction to proceed. Lacking this energy, the molecules remain intact, and
no change or results from collision. The value of E a depends on the particular reaction. “When
molecules collide, they form an activated complex (also called transition state), a temporary
species formed by the reactant molecules as a result of the collision before they form a
product” (Ebbing and Gammon, 2009). Figure 4.4 shows the two different potential energy
profiles for reaction

A+B AB± C+D

56
 Figure 4.4 Potential profiles for (a) exothermic and (b) endothermic reactions
Source: Chang and Goldsby (2016, p. 589)

where AB ± denotes an activated complex formed temporarily by collision between A and B.

If the products are more stable than the reactants, then the reaction will be accompanied by
release of heat [Figure 4.4 (a)]. If the reactant is more stable than the product, the heat will be
absorbed by the reacting mixtures from the surrounding [Figure 4.4 (b)]. These plots show the
potential energy changes as reactants are converted to products (Chang and Goldsby, 2016
p. 589).

Which between the two graphs in Figure 4.4 has a longer reaction time?

Orientation of Molecules

For a reaction to proceed, there should be a collision between molecules of reactants

so that the reacting atoms make contact. Before they collide, there should be enough energy
to surpass the activation energy and there should be a particular molecular orientation
(Silberberg, 2007). Take a look at Figure 4.5, where CO made to react with NO 2 to form carbon
dioxide and nitric oxide.

CO(g) + NO2 CO2(g) + NO(g)

57
 Figure 4.5 Effective orientation of the reaction of CO and NO2 (a) and ineffective orientation of the
reaction of CO and NO2 (b)
Source: Chang and Goldsby (2016, p. 593)

The reaction of CO and NO2 is most favorable when the reacting molecules collide
with each other in the orientation shown in Figure 4.5 (a) and products will be formed. If the
molecules of CO and NO2 collide with each other as shown in Figure 4.5 (b), there is no
reaction and no product will be formed (Chang and Goldsby, 2016, 593).

Lesson 4. Catalysts

“Catalysts are substances that increases the rate of reaction by lowering the activation
energy. It does so by providing an alternative reaction pathway. Catalysts may form an
intermediate with the reactant temporarily, but it will regenerate in subsequent step so it is not
consumed in the reaction” (Chang and Goldsby, 2016, p. 599).

A sample of KClO3 is heated as shown in Figure 4.6.

2KClO3(s) 2KCl(s) + 3O2(g)

58
 Figure 4.6 Heating of KClO3
Source: (Chang and Goldsby, 2016, p. 139)

This reaction is very slow in the absence of catalysts, unless you add a small amount of
manganese (IV) dioxide (MnO2). All of the MnO2 can be recovered at the end of the reaction.

Figure 4.7 Comparison of the activation energy of uncatalyzed reaction (a) and catalyzed reaction (b)
Source: Chang and Goldsby (2016, p. 600)

Figure 4.7 shows the difference in the activation energy of uncatalyzed reaction (a)
and catalyzed reaction (b). Take note that the energies of A and B (reactants) and C and D
(products) are not affected by the catalyst. The only difference is that the activation energy of
Figure 4.7 (b) is lower than of Figure 4.7 (a) because of the presence of catalyst.

59
 Catalyst is also a very useful substance especially in organism in the form of enzymes
to speed up cellular processes. Enzymes are biological catalyst. An enzyme is usually a large
protein molecule that contains one or more active sites where substrate interaction takes
place. An enzyme acts only on a certain molecule, in fact, active sites are only for specific
substrate, much in a same way a key fits a particular lock (Chang and Goldsby, 2016). Figure
4.8 shows the lock-and-key model of substrates.

Figure 4.8 The lock-and-key model of an enzyme’s specificity for substrate molecule
Source: Chang and Goldsby (2016, p. 605)

Homogenous catalyst is a type of catalyst with the same phase as the reacting
species. Acids and bases catalysts are the most common important types of homogenous
catalysis in liquid solutions. Examples are the reaction of ethyl acetate with water to form
acetic acid and ethanol normally occurs too slowly to be measured, but with addition of acid
catalyst, the reaction will be faster (Ebbing and Gammon, 2009).

Heterogenous catalyst is a type of catalyst that is different phase with the reacting
species. Usually, the catalyst is solid and the reactants are liquid or gas. A great example of
this is the Haber synthesis of ammonia (Ebbing and Gammon, 2009).

60
Assessment Task
TASK NO. 1 (WRITTEN WORK)

Direction: Answer the following questions correctly.

1. What is meant by the rate of chemical reaction?

2. “Can you suggest two reactions that are very slow (takes days or longer to complete)
and two reactions that are very fast (reactions that are over in minutes or seconds)?”

3. “Define activation energy. What role does activation energy play in chemical kinetics?”

4. “The burning of methane in oxygen is a highly exothermic reaction. Yet a mixture of

methane and oxygen gas can be kept indefinitely without any apparent change. Explain.”

5.How does a catalyst increase the rate of reaction?

6. Explain the Haber process.

61
TASK NO, 2 (PERFORMANCE TASK)

Direction: Determine the order of each substance and overall orders of the reactions to
which the following rate laws apply:

1. rate = k[NO2]2

2. rate = k

3. rate = k[H2][Br2]1/2

4. rate = k[NO]2[O2]

5. rate = k[NO]2[Cl2]

Summary

 Chemical kinetics is the study on the rate of reaction.

 Higher concentration of reactants means a faster reaction rate.
 Higher surface area of reactants means a faster reaction rate.
 Higher temperature means a faster reaction rate.
 Catalysts speed up the reaction rate by lowering the activation energy.
 The rate law is expressed as rate = [A] x[B]y.
 Rate law is determined experimentally.
 The most common order of reactions are the first order and second order. The rare
one is zero order reaction.
 According to collision theory, molecules should collide with each other with enough
energy and proper orientation in order for the reaction to take place.
 The activation energy is the energy that reactant molecules should overcome for them
to collide with each other.
 The three types of catalyst are enzymes, homogenous catalyst, and heterogenous
catalyst.

62
 References

 Chang, R. and Goldsby, K.A. (2016). Chemistry. (12th Edition). Mc-Graw Hill Education
 Ebbing D. D. and Gammon S.D. (2009). General Chemistry. (9th Edition). Houghton
Mifflin Company
 Silberberg M. S. (2007). Principles of General Chemistry. (1st Edition). Mc-Graw Hill
Companies, Inc.

63