AP Chemistry:
Study Guide
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product.
Key Exam Details
The AP® Chemistry exam is a 3-hour 15-minute, end-of-course test comprised of 60 multiple-
choice questions, for which you will have 1 hour and 30 minutes (this counts for 50% of your
score) and 7 free-response questions, for which you will have 1 hour and 45 minutes (this counts
for 50% of your score).
The exam covers the following course content categories:
• Atomic Structure and Properties: 7–9% of test questions
• Molecular and Ionic CompoundStructure and Properties:7–9% of test questions
• Intermolecular Forces and Properties: 18–22% of test questions
• Chemical Reactions: 7–9% of test questions
• Kinetics: 7–9% of test questions
• Thermodynamics: 7–9% of test questions
• Equilibrium: 7–9% of test questions
• Acids and Bases: 11–15% of test questions
• Applications of Thermodynamics: 7–9% of test questions
This guide will offer an overview of the main tested subjects, along with sample AP multiple-
choice questions that look like the questions you will see on test day.
Atomic Structure and Properties
Around 7‒9% of questions on your AP Chemistry exam will cover Atomic Structure and
Properties.
The physical world is made of matter, which is any substance that has mass and occupies space.
Atoms are the fundamental unit of matter, and the smallest unit that retains all the properties
of an element. Molecules are a group of atoms that are bonded together to form a chemical
compound. This section will go into detail about the structure and properties of atoms.
Moles and Molar Mass
The international standard unit of measure for the number of molecules in a substance is a mole.
A mole is equal to Avogadro’s number, or 6.022 ×10
23
, which is standardized to the number of
atoms that are present in 12 grams of Carbon-12. A mole of a substance is always the same
number of particles, regardless of what the substance is (e.g., hydrogen atoms, water particles,
1
electrons). You can think of this in the same way that a dozen always means 12, regardless of
whether it refers to eggs or days. The molar mass of substance, also called the molecular weight
or molecular mass, is the total mass of one mole of that substance, expressed as grams/mole.
Molar mass is used to convert the mass of a substance to the number of molecules present using
the following conversion:
# of moles of substance = (# of moles of substance) (Avogadro’s number)
This relationship is used to determine the number of molecules present given the mass
of a pure chemical using the atomic mass on the periodic table. To determine the number of
molecules present in a compound, calculate the number of moles present using mass in grams
and the molar mass. This number is then multiplied by Avogadro’s number to determine the
number of molecules.
Mass Spectroscopy
Scientists can measure the abundance of different atoms in a sample using mass spectroscopy.
A mass spectrometer separates molecules in a sample based on their charge and weight. To do
this, the sample is first charged by bombarding it with electrons. Magnetic fields then separate
ions by charge, and the relative abundance of ions in a sample are read by a detector. Ions of
different masses need different strengths of magnetic fields to reach the detector. Results are
plotted showing the mass-to-charge ratio on the x-axis (m/z) and the relative abundance of each
atom type on the y-axis. You may be asked on the AP exam to determine the relative abundance
of different isotopes of an element based on mass spec results.
Elemental Composition of Pure Substances
Pure substances are made of a single type of substance that has consistent characteristics and
cannot be broken down further through physical processes. A pure substance that is made of a
single type of atom is called an element. A pure substance that is made of only one type of
molecule is called a compound. According to the law of definite proportions, a pure chemical
compound broken down into elements always contains elements of a fixed ratio, independent of
where and how it was created. For example, pure water will always contain the same ratio of
hydrogen and oxygen, regardless of where it is found.
2
Composition of Mixtures
A mixture is made of more than one type of element or compound. The components in a mixture
can have different proportions. For example, you can make a 20% saline solution out of 20%
sodium chloride and 80% water, or you can make a 5% saline solution out of 5% sodium chloride
and 95% water; yet both are still saline.
When you make a mixture, no chemical reactions take place, so you could still
theoretically recover the individual components back to their pure forms. In the case of saline,
you could evaporate off the water and recover both the sodium chloride and the water. Mixtures
can be homogenous, meaning all parts of the mixture are identical to other parts due to even
distribution of compounds, or mixtures can be heterogenous, as in there is a non-uniform
distribution of compounds.
Atomic Structure and Electron Configuration
Atoms are made of smaller subatomic particles including positively charged protons, negatively
charged electrons, and uncharged neutrons. Protons and neutrons together occupy the tightly
packed nucleus of an atom, while electrons orbit outside of the nucleus in electron shells.
Together, subatomic particles determine the identity, mass, and charge of an atom.
Protons and neutrons have the same approximate mass (1.67 × 10
-27
kg), while electrons
contribute a negligible amount of mass (roughly 1/1,800 the mass of the protons and neutrons).
Thus, protons and neutronsare included in estimatingmass of an element and electrons are not.
The mass number is a whole number equal to the number of protons and neutrons of an element.
The mass of a single proton or neutron is 1 atomic mass unit (amu). The number of protons in
the nucleus is called the atomic number. The atomic symbol of an element is represented as:
Where X is the element symbol, A is the mass number, and Z is the atomic number. The
atomic number defines which element an atom is. For example, carbon has 6 protons; the
addition of a proton to a carbon atom would make it change to nitrogen. On the other hand, the
number of neutrons in an individual atom of an element can vary, changing the mass number of
an atom. Atoms with the same number of protons but different number of neutrons are called
isotopes of an element. To determine the number of neutrons in an atom, subtract the atomic
number from the mass number.
Atomic mass is the average mass number of all the atoms of that element. Atomic mass
on the periodic table can give you an idea of the proportion of isotopes present. For example,
lithium (L) has an atomic number of 3 and an atomic mass of 6.941, indicating that many lithium
atoms on Earth have more than the 3 neutrons in their nucleus.
3
Atomic charge is determined by the relative number of electrons and protons of an atom.
While the number of protons of an atom is stable, the number of electrons varies. Atoms with a
non-zero charge are called ions. An atom with an equal number of protons and electrons has a
neutral charge. Since electrons are negatively charged, the addition of an electron gives a
negative charge to an atom, creating an anion. The loss of an electron gives a positive charge to
an atom, which is called a cation.
In current atomic models, electrons occupy space outside of the nucleus at discreet
distances and energy levels called shells. The closer the shell is to the nucleus, the lower the
relative energy level. Within each shell, there are subshells that have slightly different energy
levels. Electron shells are numbered (n = 1, 2, 3 and so on), the number of the shell describes the
number of subshells it holds. Subshells (l) are denoted by the letters and corresponding numbers:
0=s, 1=p, 2=d, 3=f, 4=g, 5=h in order of increasing energy and electron holding capacity. The space
within a subshell that an electron has the highest probability of being is called an orbital. Each
orbital can hold a maximum of two electrons.
You can calculate the number of electrons possible in a subshell or shell easily by using
the shell number or the subshell letter. The number of orbitals in a subshell (l) is calculated by
the equation 2l + 1. In other words, subshell d has 5 possible orbitals (2 × 2 + 1=5), and since each
orbital can hold 2 electrons, d has an electron holding capacity of 10. Each shell has n
2
orbitals
and can hold 2n
2
electrons. For example, the first shell has 1 orbital (s) and can hold 2 electrons
(2 × 1
2
=2) and the second shell has 4 orbitals (1 in s and 3 in p) and can hold 8 electrons (2 × 2
2
=8).
Electrons fill the lowest energy orbitals first. This is called the Aufbau principle. To figure
out what shells fill first, follow the diagram showing the orbital diagonal rule. Remember that in
some cases, orbitals in shells with a higher number can fill before those of a lower number; for
instance, 4s has a lower energy than 3d, and will fill first. You may be asked on the exam to show
the electronic configuration for an atom. In this notation, each subshell is written out in order of
4
increasing energy and the number of electrons in each subshell is written in superscript. For
example, neutral carbon has 6 electrons. The electronic configuration for thiswould be 1s
2
2s
2
2p
2
,
meaning there are 2 electrons in subshell 1s, 2 electrons in subshell 2s, and2 electrons in subshell
2p. To figure out the configuration on your own, you can follow the orbital diagram to map out
which shells will be filled first.
According to Hund’s rule, electrons fill all orbitals of equal energy with one electron
before pairing electrons. That means that for carbon, the two electrons in the 2p subshell would
not occupy the same orbital. To draw this out, you can represent each orbital in a subshell using
the following diagram:
5
Here, each orbital is shown as a line and each electron for iron is shown as an arrow. You
can fill in each electron for the atom in order of increasing energy. In the case of iron, the highest
energy subshell 3d is not full and only has 6 of 10 possible electrons. Thus, according to Hund’s
rule, 4 of the 5 orbitals will have one electron and 1 orbital will have two electrons. The direction
of an arrow in this diagram indicates the spin of an electron, either spin up or spin down.
According to Pauli’s Exclusion Principle, two electrons of the same spin cannot occupy the same
orbital; thus, a single orbital can only have one spin up and one spin down electron.
When subshells are not filled, unpaired electrons are present and can interact with
magnetic fields, making the atom paramagnetic. When all subshells are filled, the element does
not interact with magnetic fields and is called diamagnetic. The outer shell is called the valence
shell, which will be discussed in further detail.
The energy that is needed to remove an electron from an atom is called ionization energy.
Coulomb’s law is used to calculate ionization energy, or the energy needed to move an electron
from one energy shell to another. According to Coulomb’s law, the force of attraction or repulsion
(F) between two charges is proportional to the product of charges (Q) divided by the square of
the distance between them (r).
From this equation, you’ll find that the attractive force between protons and electrons is
highly dependent on their distance from the nucleus. In other words, it will take more energy to
remove electrons that are closer to the nucleus than those that are farther away. Be prepared to
use this calculation to estimate relative ionization energies on the AP exam.
Photoelectron Spectroscopy
Photoelectron spectroscopy (PES) is used to determine how many electrons an atom has and
where they reside. This technique relies on ionization energy. In photoelectron microscopy, high
energy radiation is focused onto a substance and the kinetic energy of electrons and relative
abundance of electrons at each energy is detected. This information can be used to calculate an
electron’s binding energy, the energy needed to remove an electron from a subshell of an atom.
To remove an electron from an atom, the energy applied must be greater than the binding
energy of the electron. The closer an electron is to the nucleus of an atom, the stronger the
attraction to the nucleus; thus, electrons closer to the nucleus have greater bindingenergies. In
a PES spectra, high energy peaks are made by electrons emitted from orbitals closer to the
nucleus of the atom, and low energy peaks are made from electrons in or near the valance shell.
The heights of the peaks can be used to calculate relative abundance, which is then used to
6
determine the identity of an atom. In the free response exam section, you may be asked to
identify elements and electron configurations based on PES spectra.
Periodic Trends
Besides organizing elements based on atomic numbers, the periodic table provides an
organized structure that helps define categories of elements. The rows of the periodic table are
called periods, which define the highest energy level an electron of that element occupies at rest.
The columns of the periodic table are called families that share valence electron configurations.
Physical properties of elements are shared in periods and families, with families generally
having stronger effects than periods. For example, when you move right to left and top to bottom
on the periodic table elements become more metallic. These physical properties are due to the
electron configurations. The atomic radius gets larger as you move down a family or from left to
right across a period. This is because there are more electron shells as you move down the period
and the diameter of these shells contracts when the valence shell becomes occupied.
7
On the other hand, the ionization energy increases as you move from left to right and
from bottom to top. This is due to the higher stability of electron shells closer to the nucleus and
the higher magnetic pull of nuclei with more positively charged protons on negatively charged
electrons. In the free response section, you may be asked to explain the basis for periodictrends.
You should be familiar with the basic families on the periodic table and their features:
Alkali metals (1): These are shiny, soft, and very reactive. All have their outer electron in an s-
orbital that reacts easily. They are never found uncombined in nature, as they are highly reactive.
They react violently with water, causing an explosion. These often lose one electron to take on a
+1 charge.
Alkaline earth metals (2): These are shiny, silver-white metals that are semi-reactive. The
outermost s orbital is full, so they can form cations with a charge of +2. These are known to react
with halogens and form ionic halides, and all but Beryllium react with water to form a hydroxide
and hydrogen gas. These often lose two electrons to take on a +2 charge.
Transition metals: These have a partially filled d sub-shell, so they can form cations with an
incomplete valence shell. These are less reactive, so are often found uncombined in nature. These
can conduct electricity, form magnets, and make compounds with color.
Chalcogens (16): These have 6 electrons in their outer shell, so often form anions with a -2
charge. These form compounds through covalent bonding.
Halogens (17): These have 7 electrons in their valence shell, so they tend to form anions in water
with a -1 charge. These tend to form compounds with covalent bonds or are found as highly
reactive diatomic molecules. When combined with hydrogen, these form strong acids.
Noble gasses (18): These are virtually unreactive, colorless, odorless, gases. These have a full
octet of valence electrons, so are found uncompounded in nature and are difficult to get to form
reactions.
Valence Electrons and Ionic Compounds
When the valence shell is not full, electrons in this shell can interact with other atoms by forming
ionic bonds. An ionic bond forms when an atom donates a valence electron to another atom,
typically a metal and a non-metal. The atom that loses an electron becomes a positively charged
cation, and the atom that gains an electron becomes a negatively charged anion.
The opposite charges of the atoms attract them together, causing an ionic bond to form.
Ionic bonds help to increase the stability of the atoms involved by having the optimal number of
electrons in the valence shell. The first five elements in the periodic table are most stable when
8
they have 2 electrons in their 1s orbital; this is called the duet rule. Thus, in ionic compounds,
these elements aim to gain or lose enough electrons to have two total. For most elements with
more than 5 protons, the optimal configuration is to have eight electrons in the valence shell;
this is called the octet rule.
In ioniccompounds these atoms will gain or lose electrons,to have a total of eight in their
valence shell. As mentioned previously, columns on the periodic table are organized by the
number of valence shell electrons. Thus, elements in the same column tend to have similar
charges as ionic compounds and will tend to form analogous compounds (e.g., NaCl and KCl).
Outside Reading
• For more information on atomic structure and mass:
o https://courses.lumenlearning.com/boundless-chemistry/chapter/the-
structure-of-the-atom/
o https://dlc.dcccd.edu/biology1-2/atoms-and-elements
• For more information on mass spectroscopy:
o https://www.chemguide.co.uk/analysis/masspec/howitworks.html
• For more information on photoelectron microscopy:
o https://www.khanacademy.org/science/chemistry/electronic-structure-of-
atoms/electron-configurations-jay-sal/a/photoelectron-spectroscopy
• For more detailed information periodic trends:
o https://courses.lumenlearning.com/boundless-chemistry/chapter/periodic-
trends/
• For further reading on ionic compounds:
o https://courses.lumenlearning.com/introchem/chapter/ionic-bonds/
9
Atomic Structure and Properties Questions
Water has a molecular mass of 18 and has one atom of oxygen to two atoms of hydrogen. Which
of the following choices is the procedure to calculate the mass ratio of hydrogen to oxygen in
water?
A.
2
18
= 0.111
B.
2
16
= 0.125
C.
16
18
= 0.888
D.
16
2
= 8.000
Explanation:
The correct answer is B. Water (molecular mass 18) has one atom of oxygen (atomic mass 16) to
two atoms of hydrogen (atomic mass 1). The mass ratio of hydrogen to oxygen in water is
2
16
= 0.125
.
A 100-g sample of CaCO
3
(FW = 100 g/mol) was heated at 500°C for 12 hours. Afterwards, the
remaining solid was found to have a weight of only 56 g. Which balanced chemical equation best
describes the reaction that took place?
A. CaCO
3(s)
+ heat → CaCO
(s)
+ O
2(g)
B. 2CaCO
3(s)
+ heat → 2CaC
(s)
+ 3O
2(g)
C. CaCO
3(s)
+ heat → CaO
(s)
+ CO
2(g)
D. CaCO
3(s)
+ heat → 2CaO
2(s)
+ CO
(g)
Explanation:
The correct answer is C. The molecular weight of CaCO
3
is 100 g/mol, meaning one mole of
material was heated. The remaining solid must also represent one mole of material and have a
molecular weight of 56 g. CaCO has as molecular weight of 68 g/mol.
10
Which of the following electron configurations is isoelectronic to Cl
–
?
A. 1s
2
2s
2
2p
6
3s
2
3p
3
B. 1s
2
2s
2
2p
6
3s
2
3p
4
C. 1s
2
2s
2
2p
6
3s
2
3p
5
D. 1s
2
2s
2
2p
6
3s
2
3p
6
Explanation:
The correct answer is C. Isoelectronic means having the same electrons. Chlorine has 17
electrons and the chloride ion has one more (18). This answer has 18 electrons.
11
Molecular and Ionic Compound Structure
and Properties
Around 7‒9% of the questions on your AP exam will cover Molecular and Ionic Compound
Structure and Properties.
Types of Chemical Bonds
Chemical bonds are the attractive forces between atoms that hold them together, including
sharing electrons and other electrostatic forces. These forces can be intramolecular, as in they
occur within a molecule, or intermolecular, as in occurring between molecules. Intramolecular
forces are stronger than intermolecular forces. The main types of intramolecular binding forces
are summarized in the following table.
Bond
Description
Example
Covalent
The sharing of electrons between a pair of atoms, also called a
molecular bond. Covalent bonds can result in atoms having complete
valence shells, stabilizing the atoms. These are usually gasses, liquids,
or soft solids and include most biological molecules. The first covalent
bond is called a sigma bond () and additional bonds are called pi
bonds (). The sigma bond is the most stable.
CO
2
Ionic
Ionic bonds describe the electrostatic attraction between cations and
anions or with charged polyatomic ions. Ionic bonds involve the
transfer of electrons from a metal to a non-metal, resulting in full
valence shells for both atoms. Ionic bonds tend to be easily separated
by external forces, like the addition of polar solvents, causing
substances to break apart into anions and cations. Solid, ionically
bonded structures form ordered and strong lattices that are non-
conductive and have high melting and boiling points. This group
includes salts.
NaCl
Metallic
Metallic bonds are the sharing of free electrons in a metallic
structure, composed of charged cations. In metallic bonds, valence
electrons are only loosely bound to the nucleus and form a sort of
cloud around metallic cations, allowing them to be shared. This
results in a highly stable configuration of positively charged cations in
a sea of negatively charged electrons. Metallic bonds account for
certain features of metals like malleability, conductivity, and strength.
Copper
wire,
aluminum
foil
12
One way to predict whether a bond is ionic or covalent is to look at the electronegativity
ofthe atoms involved. Electronegativity describeshow much an atom attracts bondingelectrons.
The scale runs from 4 (fluorine) at the high end to 0.7 (cesium and francium) at the low end. On
the periodic table, electronegativity increases from left to right and bottom to top (excluding
noble gasses).
Another way to think of electronegativity is as a quantitative way to consider how factors
like the atomic structure, valence shells, and Coulomb’s law affect attraction between ions. To
determine bond type, remember that when the difference between electronegativity is high
(greater than 1.7), ionic bonds form. When the difference is low (1.7 and lower), covalent bonds
form.
Covalent bonds formed between atoms with similar electronegativities (<0.5) are non-
polar covalent bonds. This means that electrons are shared equally across the atoms. A polar
covalent bond forms when there is a larger difference in electronegativity across atoms. The
atom with the greater charge pulls electrons closer to it, forming a dipole. In a dipole, electrons
spend more time at one end of the molecule, giving that side a slightly more negative charge (-
) and the other end a slightly positive charge (+). This can result in molecules that are charged,
or polar, and can interact with other molecules through these charges.
The size of a dipole is measured by the dipole moment, . When the difference between
the dipole is very high, the molecule at the + end of the bond essentially transfers its electrons
to the more electronegative atom, forming an ionic bond. The direction of a dipole is ind icated
in a bond as an arrow from the less electronegative atom to the more electronegative atom.
Intramolecular Force and Potential Energy
During covalent bond formation, unstable atoms with incomplete valence shells share electrons
to increase their stability. Attractive forces between the nuclei and electron clouds of unstable
atoms attract each other to form a bond. This is counteracted by repulsive forces between each
atom’s nuclei and between each atom’s electron clouds. This balance between repulsive and
attractive intramolecular forces can be shown in a potential energy diagram.
13
In this diagram, when the two molecules are far apart there is no potential energy, or
stored energy. The closer the molecules get to each other, the lower the potential energy due to
attractive forces between the molecules (remember Coulomb’s law). The distance where this
potential energy is lowest is the average length of the covalent bond that forms between them.
In other words, covalent bond lengths are optimal when the potential energy of the bond is
lowest. You’ll notice in the diagram that when the molecules get even closer together, the
potential energy rapidly shoots back up. This is due to repulsive intramolecular forces between
the atoms. In the case of double and triple bonds, the bond length becomes shorter and the
strength of the bond becomes higher; thus, you can expect to see shorter lengths between nuclei
and higher bond energies.
Structure of Ionic Solids
Ionic solids form in a manner that reduces the repulsive forces between similarly charged ions
and increases the attractive force between oppositely charged ions. This forms a chemical lattice
with an orderly arrangement. The lattice energy (E) is the energy needed to separate 1 mol of an
ionic solid into its gas phase. This is proportional to the product of the electrical charges of the
ions (Q) divided by the sum of the radii of the atoms (r):
14
Note that when the ions are oppositely charged, the lattice energy is negative. This
indicates energy will be released when the lattice forms due to attraction between the molecules.
The more negative the lattice energy is, the stronger the ionic bonds will be. These strong bonds
holding ionic solids together explains some of the features of ionic solids such as crystal
structures and brittle properties.
Structure of Metals and Alloys
Metallic solids are formed by an array of metallic cations surrounded by a sea of freely moving
shared electrons. The loose shared electron sea helps to understand some of the features of pure
metals, such as conductance, malleability, and ductility. When metals are mixed with other
metals or non-metals, alloys are formed. When two metals of similar size are mixed, a
substitutional alloy is formed. The percentages of metals in a substitutional alloy can be varied
to alter the properties of the alloy, such as increasingstrength and conductivity while decreasing
corrosion. When a pure metal is mixed with a much smaller metal or non-metal, an interstitial
alloy is formed. Interstitial alloys are stronger, harder, and less ductile than their pure host
metals.
Lewis Diagrams
Lewis dot diagrams are a quick way to represent the valence electrons of an atom, where each
dot around a molecule represents a valence electron. A covalent bond is represented by dots
between atoms, or by dashes corresponding to the number of covalent bonds. According to the
octet rule, most atoms are considered stable when this valence shell is occupied by eight atoms;
thus, the optimal configuration is to have 2 electrons on each of the 4 sides or 4 total bonds.
Hydrogen atoms only need two electrons in their valence shells.
15
When drawing a dot diagram for a compound, first arrange the ions so they form a
symmetrical structure. Remember that hydrogen is always on the outside of the structure, never
central. Then calculate the total number of valence electrons the molecule has vs. the number of
molecules needed (generally 8 for every molecule except hydrogen). The difference between the
number of electrons the molecule has and the number needed tells you how many electrons are
shared. Then divide shared electrons by two to find out how many are between elements in this
molecule. You then draw the bonds and complete the structure by adding the remaining
electrons so that each atom has the correct number of valence electrons.
Lewis dot diagrams do not handle unpaired electrons well or show the difference
between the strength of bonds; these cases are better described by models that take into
account 3-D structure.
Resonance and Formal Charge
Resonance structures are not adequately represented by dot diagrams. These are instance
where two forms of a molecule are possible with the same connectivity but with differences in
the distribution of electrons. For instance, ozone can exist as O=O−O or O−O=O. In reality, these
would exist as partial bonds, but for the purpose ofdemonstration would be written as follows:
O=O−O O−O=O.
You may come across instances where there are multiple possible Lewis structures. To
determine which is the most appropriate, you can calculate the formal charge of the atoms in
the chemical compound:
The most favorable Lewis structures will have small formal charges on each element:
more electronegative elements will have lowerformal charges, like charges will be far apart, and
unlike charges will be near. The total formal charge for the molecule is the sum of the formal
charges of the atoms.
Note that you may be asked on the exam to compare the relative contribution of different
resonance structures possible for a molecule. In that case, the resonant structures thatare more
common are those that satisfy the octet (or duet) rule for all atoms that have a formal charge on
each atom closest to zero, and where more electronegative elements have lower formal charges.
16
VSEPR and Bond Hybridization
Valence shell repulsion theory, or VSEPR, asserts that electron pairs repel each other and
unbonded electrons are even more repulsive to bonded pairs. This causes molecules to take on
distinct shapes depending on the number of bonds and hybridized orbitals. For example, a
molecule with two sigma bonds will take on a linear shape so that both bonds will be as far apart
as possible. A molecule with four bonds will take on a tetrahedral shapeto optimize distance.
You will need to estimate molecular shapefor the exam using VSEPR. To determine the shape
of a chemical compound using VSEPR:
1. Draw out the Lewis dot diagram.
2. Count the number of electron-dense regions. These include each element in the
compound as well as each pair of electrons.
3. Mentally arrange these electron dense regions to allow equal space between them; this
gives the overall shape of the structure
4. Take into account electronic geometry, due to the number of lone electron pairs, as well
as molecular geometry, due to the number of bonds.
17
Another way to represent molecular geometry is through Valence bond theory. In this
theory, covalent bonds form through the mixing of orbitals to form hybrid orbitals. The number
of hybrid orbitals formed is equal to the number of contributing orbitals. The types of orbitals
that contribute determine the hybrid orbital type (e.g., s and p orbitals hybridizing means that
you will have two sp orbitals).
18
Outside Reading
• For more information on bonds, see:
o https://www.chemguide.co.uk/atoms/bondingmenu.html#top
o https://www.visionlearning.com/en/library/Chemistry/1/Chemical-
Bonding/55
• For more details on VSEPR theory:
o https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_
Chemistry_-
_The_Central_Science_(Brown_et_al.)/09._Molecular_Geometry_and_B
onding_Theories/9.2%3A_The_VSEPR_Model
• For an in-depth explanation of hybrid orbitals, see:
o https://chem.libretexts.org/LibreTexts/University_of_Kentucky/UK%3A_
General_Chemistry/08%3A_Advanced_Theories_of_Covalent_Bonding/
8.2%3A_Hybrid_Atomic_Orbitals
19
Sample Molecular and Ionic Compound Structure and Properties
Questions
When two chlorine atoms are brought together, they form a Cl–Cl bond and energy is released.
The potential energy for this system as a function of Cl–Cl distance is shown above. According to
this graph, what amount of energy is required to break the Cl–Cl bond?
A. 17 eV
B. 8 eV
C. 0 eV
D. −8 eV
Explanation:
The correct answer is B. The right side of the graph asymptotically approaches 8 eV, which is the
energy required to separate the two Cl atoms infinitely far apart.
20
Which ofthe followinghydrocarbonsis expectedto have single bondsbetween allcarbonatoms?
A. propane (C
3
H
8
)
B. butene (C
4
H
8
)
C. hexene (C
6
H
10
)
D. octyne (C
8
H
14
)
Explanation:
The correct answer is A. All of the bonds between the carbon atoms are the same (single bonds).
The ending –ene signifies there are double bonds between carbon atoms along with single bonds
and the ending –yne signifies there are triple bonds, which can be seen by drawing the Lewis
diagram.
Which of the following statements best explains why LiBr is expected to boil at a higher
temperature than SF
4
?
A. LiBr contains ionic bonds and SF
4
has dipole-dipole forces
B. The density of SF
4
is greater than that of LiBr
C. SF
4
has a higher molar mass than LiBr
D. LiBr is linear and SF
4
is not
Explanation:
The correct answer is A. LiBr forms an ionic bond, because Li is a very active metal and Br is a
very electronegative nonmetal. SF
4
is a polar molecule with dipole-dipole forces between the
molecules. Generally, species with ionic bonds have higher boiling points than species with
dipole-dipoleforces.
21
Intermolecular Forces and Properties
Around 18‒22% of the questions on your AP exam will cover Intermolecular Forces and
Properties.
Intermolecular Forces
Intermolecular forces describe the attractive and repulsive forces between molecules.
Intermolecular forces are not chemical bonds; rather, they are the forces between bonded
molecules and other molecules or ions. These include a variety of forces including hydrogen
bonds, dispersion forces, and dipole interactions.
London dispersion forces are temporary and weak forces that occur due to Coulombic
interactions. These occur when electrons from two atoms are positioned to create temporary
dipoles. Dispersion forces are larger for molecules with more electrons and increase when there
is more contact area between molecules. All molecules have dispersion forces. Although they are
individually small, they can add up across a molecule to be large. One temporary dipole can
induce other dipoles to occur nearby. The collective inductioncauses polar substances to freeze
when temperatures are lowered, as in the case of water turning into ice.
Dipole-Dipole interactions are interactions between two polar molecules. Here, the
partial positive side of one molecule is attracted to the partial negative side of anothermolecule.
Dipoles can also induce temporary dipoles in non-polar molecules. These dipole-induced dipole
interactions can increase the solubility of non-polar molecules in water.
Hydrogen bonds are attractive forces between an electronegative atom and hydrogen
atom that is bound to an electronegative atom (N, O, or F). The most common example of a
hydrogen bond is between two water molecules. This isan example of a dipole-dipole interaction,
where the hydrogen end of one water molecule (+) interacts with the oxygen from another
molecule (-).
Solids, Liquids, and Gases
Matter exists in three primary states, or phases, on Earth: solid, liquid, or gas. States of matter
can change due to temperature or pressure. Solids and liquids are relatively incompressible.
Solids have a definite shape while liquids will take the form of whatever container they are put
in over time. Gasses, on the other hand, are compressible and will fill any shape that they are
held in. While molecules in solids and liquids have constant contactwith neighboringmolecules,
22
gas molecules do not. This means that properties of solids and liquids are dependent on
intermolecular forces, while gasses are not.
There are two forms of microscopic solids: amorphous solids and crystalline sol ids.
Amorphous solids are solids that lack a definite organization. These solids do not have a definite
melting point; rather, they decrease in viscosity when the temperature increases. Amorphous
solids include substances like glass and rubber.
Crystalline solids have crystal lattice structures with an organized microscopic
appearance. These include ionic, metallic, molecular, and atomic solids as well as covalent
network solids. Covalent network solids are crystal lattices where molecules are joined by
covalent bonds, thus forming one giant molecule. Diamonds and graphite are examples of crystal
lattices. You should be familiar with the properties of different types of solids and how they relate
to intermolecular properties for the AP exam.
Molecules in liquidsform very brief interactionswith their neighbors, thusintermolecular
forces are momentary; however, liquids do have three unique and notable macroscopic
properties. Molecules within a liquid are very attracted to each other due to cohesive forces. This
means that molecules atthe surface of a liquid are constantly drawn towards the liquid molecules
below. The result of this is a flattening at the air-liquid interface, noticeable in a pool of liquid as
well as in droplet formation. The cohesive forces also cause liquids to resist increasing surface
area at the air-liquid interface—this is called surface tension.
Viscosity is the resistance of liquids to flow. This is caused by the strength of
intermolecular forces between liquid molecules. The stronger the intermolecular forces, the
more viscous the liquid. Finally, when liquids are placed in narrow chambers, like tubes, another
property, called capillary action, becomes apparent. Capillary action is a rising of a liquid in a
tube against the force of gravity. This occurs due to the attractive forces between the liquid and
the solid walls of the tube, called adhesive forces. The stronger the attraction between liquid and
solid, the higher the rise will be.
Ideal Gas Law
Gases have no fixed volume or shape. The molecules within gasses are able to freely expand, to
fill any space. Due to the large spacing of molecules in gasses, they tend to behave in similar ways
regardless of type of gas. Thus, a set of equations can be used to describe the relat ionships
between pressure (P), volume (V), temperature (T), and number of molecules (n) of a gas.
Equations that describe how these properties relate, called ideal gas laws, are described in the
following table.
23
Law
Description
Equations
Boyle’s Law
Pressure is inversely related to volume when
temperature and number of molecules are
constant.
For example, imagine a fixed amount of gas
trapped in a sealed container at a constant
temperature. If you then increase the volume (V
1
)
of the container to V
2
, the molecules will have
more room to move around within the container,
increasing the pressure (P
1
) to P
2
.
&
or
P
1
V
1
= P
2
V
2
Charles’
Volume is proportional to temperature when
Law
pressure and number of molecules of gas are held
constant.
If pressure and number of molecules are held
constant, a rise in temperature (T
1
) to T
2
will cause
a proportional rise in volume of the container (V
1
)
to V
2
.
or
Guy-
Pressure is proportional to temperature when
Lussac’s
volume and number of molecules of gas are held
or
Law
constant.
If you now hold the volume and n constant, and
increase the temperature (T
1
to T
2
), the pressure
will rise proportionally from P
1
to P
2
.
Avogadro’s
Volume is proportional to the number of gas
Law
molecules, when temperature and pressure are
constant.
If the container is filled with more molecules (n
1
to
n
2
), while the temperature and pressure are held
constant, the volume of the container will increase
(V
1
to V
2
) to accommodate the additional particles.
or
Ideal Gas
Law
A combination of the three gas laws that describes
how the properties of a perfect theoretical gas
relate to each other. R is the empirically
determined gas constant (R = 8.3145 ). For an
ideal gas,
PV = nRT
or
Dalton’s
The total pressure of gasses in a mixture is equal to
Law of
the sum of the pressures of the individual gasses.
Partial
The partial pressure of a gas is equal to the total
Pressures
pressure of the mixture multiplied by the mole
fraction (X) of the gas.
where
24
As the number of molecules in a substance (n) can be quite large, it is easier to define
them based on moles. As discussed above, moles are related to mass using the equation:
For gasses, the number of moles is calculated based on volume rather than weight, based
on a similar equation:
In other words, the molar volume is equal to the volume that one mole occupies in set
temperature and pressure conditions. For ideal gasses the standard molar volume is 22.4 L. This
is the volume one mole of an ideal gas occupies at 273.15 K and 1 atm of pressure. Given the
equation and the ideal gas laws, the molar volume can be calculated for any temperature and
pressure:
Ideal gas equations ignore effects between molecules that real gasses have but are a good
approximation of relationships between properties at high temperatures and low pressures. In
conditions where intermolecular forces are higher, such as low temperatures and high pressures
or when describing molecules with high intermolecular forces, the ideal gas laws lead to errors.
Kinetic Molecular Theory
Kinetic molecular theory (KMT) describes how gas particles behave in order to explain the overall
properties of gasses. In KMT, gasses are made primarily of empty spaces that contain molecules
that are constantly moving independently of the neighboring molecules. Here are the important
points of KMT:
• Gas is made of tiny molecules whose total volume is much smaller than the volume of the
gas. Put another way, the distance between molecules in a gas is much greater than the
size of individual molecules.
• The molecules in a gas have no attractive forces between them, or with the container that
holds them.
• Individual molecules are constantly moving in a random direction until they collide with
another molecule or with the container. These collisions are completely elastic, as in no
change in energy occurs due to a collision. In other words, if one molecule slows down
due to a collision, the other molecule will speed up an equal amount so there is no net
loss of energy.
25
• The average kinetic energy of the gas is solely determined based on the absolute
temperature. All gasses have the same average kinetic energy at a given temperature.
Kinetic energy (E
k
) is dependent on mass (m) and velocity (v) of a molecule:
Molecules of a gas have an average kinetic energy, but individual molecules can have different
velocities. When the numberof moleculesat each velocity are plotted, they fit alonga Boltzmann
distribution. This distribution represents several important properties of gasses. For example,
the peak of the distribution corresponds to the most probable velocity of a molecule. The
distribution of velocities changes according to temperature for a given molecule, where higher
temperatures have broader distributions of velocities with higher average velocities.
Solutions and Mixtures
Solutions are homogenous mixtures of substances that are not physically bonded. In solutions,
solutes are dissolved into solvents. Solutions can involve different states of matter dissolving into
another. For instance, sugar can be dissolved into water, carbon dioxidecan dissolve into a drink
to make soda, and metal alloys are solid metal solutions.
Concentration is expressed in several ways. Some common expressions are:
For the AP exam, be sure that you can use these equations to calculate the number of
particles in a solution, molarity, and volume. Also, be prepared to represent these graphically by
showing the relative abundance of particles in a solution and how different particles may interact.
26
Separation of Solutions and Mixtures Chromatography
Components of solutions can be physically separated using evaporation, distillation , or
chromatography.
In evaporation, aqueous or volatile solvents are evaporated or gently boiled away to
recover the solute—this works in the case of recovering sodium chloride from salt water.
In distillation, different vapor pressures and boiling points are used to separate solutes
from a solution. In this method, the solution is placed in a round bottom flask equipped with a
condenser and a receiving vessel. The flask is heated to the boiling point of the more volatile
component in the mixture. It then vaporizes and is collected in the condenser, where it returns
to liquid phase and is collected in the receiving vessel.
In chromatography, liquid mixtures are separated by taking advantage of the differences
between the intermolecular forces of components in liquid (mobile phase) with absorbent paper
or silica gel (stationary phase). The molecules in the solution will interact differently with the
solvent than they will with a solid, polar substance; thus, when the stationary phase is placed
into the mobile phase, the components will travel at different speeds along the stationary phase.
Solubility
Factors that affect miscibility, the ability of a solute to be soluble in a solution, are solute-solvent
interactions, concentration, temperature, and pressure. The attraction between a solute and a
solvent can determine the solubility. In general, like molecules attract like so polar substances
dissolve in other polar solutes andnon-polar substances dissolve in non-polar solutes. The effect
of concentration is described by the common-ion effect, wherein adding more of an ion to a
solute where that ion already exists decreases the solubility of additional ions. When the solution
is saturated, the solid solute is at equilibrium with the dissolved solute, and additional solute
cannot be dissolved.
Temperature affects solutes differently depending on whether the solvation reactions are
endothermic (
H
solvation
>0) or exothermic (
H
solvation
<0).In endothermicreactions, the solubility
of solids into liquids increases when temperature increases. In exothermic reactions, the
solubility of solids into liquids decreases when temperature increases. For gases dissolved in
liquids, increases in temperature decrease solubility; this is why sodas tend to get flat when
heated.
Pressure can affect the solubility of gasses in liquids according to Henry’s law, which
states that when temperature is constant, the amount of gas that can dissolve in a liquid (C) is
proportional to the partial pressure of the gas (p):
p=k
h
C
In this equation, C is the molar concentration of gas in the solution expressed in mol/L, and k
h
is
Henry’s law constant, expressed in L atm/mol, which depends on the gas molecule.
27
Spectroscopy and Electromagnetic Spectrum
Atoms or molecules can absorb or emit energy and depending on the amount of energy absorbed
or emitted, different effects can occur. These effects depend on what frequency on the
electromagnetic spectrum the energy absorbed or emitted lies. The electromagnetic spectrum
describes the range of energy frequencies that exist and how they relate to wavelength and
energy level.
For example, when an atom is energized by the absorption of UV light, electrons in the
atom’s shell undergo an energy level transition to a higher orbital. When the electrons return to
a restingstate, they will emit photons of light that correspond to the change in energy level. This
is the basis of spectroscopy, which can be used to analyze the unknown chemicals in a sample.
When atoms absorb energy at levels lower than the visible spectrum, in the infrared range, the
energy absorbed is not enough to induce a change in electron energy levels; rather, it can cause
the chemical bonds in a molecule to vibrate at specific frequencies. This is used in infrared
spectroscopy to analyze the bonds in molecules.
Photoelectric Effect
When light shines on a metal, electrons are sometimes emitted; this is called the photoelectric
effect. This occurs because the light energy absorbed by atoms causes atoms to shift energy
28
states. If the energy is high enough, this can cause the electron to break away and escape the
atom. Different types of metals require different frequencies of light to eject electrons,
depending on that metal’s unique threshold frequency. Furthermore, the higher the frequency
of light the higher the kinetic energy of emitted electrons.
Beer-Lambert Law
Beer-Lambert law describes the relationship between the amount of light absorbed by a material
that the light travels through. Absorbance (A) is equal to the product of the concentration of
solution (c), length of the light path (b), and the molar absorptivity ():
Molar absorptivity describes how well a given molecule or ion absorbs light of a specific
wavelength. The higher the absorptivity, the more light is absorbed in solution. Similarly, the
higher the concentration of the solution (c) or the longer the path that light travels through the
solution (b), the more light will interact with molecules of solute, creating more chances for light
to be absorbed. In UV-spectroscopy, this equation can be used to measure the concentration of
solutes in a solution.
29
Outside Reading
• For an explanation of kinetic energy in gasses:
o https://opentextbc.ca/introductorychemistry/chapter/kinetic-
molecular-theory-of-gases-2/
• For an explanation of how KMT relates to gas laws, see:
o https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical
_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_T
heoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matt
er/Properties_of_Gases/Kinetic_Theory_of_Gases/Basics_of_Kinetic_
Molecular_Theory
• For information on how KMT relates to liquids and solids:
o https://study.com/academy/lesson/the-kinetic-molecular-theory-
properties-of-solids-and-liquids.html
• For more on Beer-Lambert law:
o https://chem.libretexts.org/Ancillary_Materials/Reference/Organic_Ch
emistry_Glossary/Beer-Lambert_Law
30
Sample Intermolecular Forces and Properties Questions
The figure above shows the Boltzmann probability distributions of the velocities of molecules in
a container at four different temperatures. Based on the figure and what you know about
molecular interactions, what conclusions can you draw about the effect of temperature on
reaction rate?
A. Lower temperatures increase the rate of reaction because a greater percentage of molecules
travel at lower velocity, which improves the likelihoodof productive collisions.
B. Lower temperatures increase the rate of reaction because more molecules will be above the
minimum energy needed to react.
C. Higher temperatures increase the reaction rate because the average kinetic energy of the
molecules decreases.
D. Higher temperatures increase the rate of reaction because a larger population of molecules
possesses the minimum energy required for the reaction to proceed.
Explanation:
The correct answer is D. Increased temperature leads to a higher average velocity at which
molecules travel. Molecules traveling at higher velocity possess greater kinetic energy which
improves the likelihood that a collision will lead to a reaction.
31
Arrange the following molecules in order of increasing boiling point:
propane (CH
3
CH
2
CH
3
), ethanol (CH
3
CH
2
OH), and diethyl ether (CH
3
OCH
3
)
A. diethyl ether < ethanol < propane
B. propane < diethyl ether < ethanol
C. diethyl ether < propane < ethanol
D. propane < ethanol < diethyl ether
Explanation:
The correct answer is B. Ethanol (b.p. 78 °C) interacts through hydrogen bonding, which is a
strong type of dipole–dipole attraction. Diethylether (b.p. 35 °C) cannot hydrogen bond, butdoes
associate through dipole–dipole forces. Propane (b.p. −42 °C) can only associate through weak
London dispersion forces.
Which of the following pairs are most likely to be miscible, or soluble in each other?
A. CH
3
CH
2
OH and CH
3
CH
2
CH
3
B. H
2
O
2
and CH
4
C. NH
3
and C
2
H
6
D. CH
3
OCH
3
and CH
3
CHO
Explanation:
The correct answer is D. A solution is a homogeneous mixture. The choices include polar and
nonpolar molecules. To form a solution, both molecules haveto be either polar or nonpolar. Both
substances are polar.
32
Chemical Reactions
Around 7‒9% of questions on your exam will cover Chemical Reactions.
Physical and Chemical Changes
Thus far, we have discussed physical changes in chemicals—that is, changes in the state of
chemicals or the creation or separation of solutions. Chemical reactions, on the other hand,
transform substances into new substances, usually with different chemical compositions. Typical
signs that a chemical reaction has occurred include changes like the production of heat, light, or
precipitate, or changes in the color of a substance.
Net Ionic Equations
Chemical reactions are often written as linear equations with reactants on the left, products on
the right, and a single or double arrow between, signifying the direction of the reaction. When
balancing equations, remember:
• The Law of Conservation of Mass must be followed; thus, the number and types of atoms
present in the reactants must also be present in the products. Make sure the number of
each element in the products and the reactions are equal.
• The physical state of molecules in the reaction are denoted by subscripts as gas (g),
aqueous (aq), and solid (s).
• Molecular equations show chemical formulas for reactants and products, but do not take
into account ionizationin aqueous solutions.
• Complete ionic equations denote molecules as ions in aqueous solution when
appropriate.
• Net ionic equations remove spectator ions from the equation and only show molecules
that participate in the reaction.
For example, consider the ionizationof calcium chloride salt (CaCl
2 (s)
) in water:
In this reaction, CaCl
2 (s)
is ionized from solid to individual aqueous ions in water. The
reactants consist of 1 part calcium ion, 2 parts chloride ions, and 1 part water.
33
To balance this equation, the reactants also must include these proportions. In this
reaction, water participates as a spectator; thus, the net ionic equation would be:
Another way to represent reactionsis through the particle model. In this model showing propane
combustion, each atom in the reaction is represented as a particle. To balance the reaction, make
sure that all particles that are present on the reactant side are also present on the product side.
Stoichiometry
Stoichiometry deals with the quantitative relationship between elements in a reaction. Simply
stated, it uses ratios of ions to calculate the amounts of reactants and products in an equation.
The factor label-method is often used to convert ratios to molar quantities. This method uses
conversion factors to convert one quantity to another quantity. This concept can be used to
balance a chemical reaction, determine how many moles are present in a reaction, and even
determine conversions for units of measure.
For example, we know that 1 foot = 12 inches but we would like to solve how many inches
are in 0.6 feet. We can use multiplication and division to solve for this. By remembering that the
units must cancel out to give the final unit, inches, we can solve this problem:
34
This same concept can be used to for molar conversions. By Avogadro’s number, we know
that there are 6.02 10
23
atoms in every mole; thus, if we want to identify the number of atoms
in 0.4 moles of a substance:
Similarly, we can use this concept to covert moles of substance to mass of substance and
vice versa. For example, if we would like to find the mass of 5 moles of potassium, we can turn
to the periodictable to find that the atomicweight of potassium is 39.098 g/mol. Thus:
This relationship can be used to convert moles to volume. For gasses at STP, one can refer
to the standard molar volume of an ideal gas, 22.4 L for every mole. Thus, if we have 15 L of any
gas at STP, we can convert to moles of substance:
Solutions are often described in terms of molarity (M = moles/liter). Thus, if we would like
to find the number of moles in 2.5 L of a 0.5 M solution:
The ratios of molecules in a chemical reaction can be used in a similar way to find the
amount of substance involved in a reaction. For these kinds of problems, first make sure the
equations are balanced. Then you can treat ratios of molecules in the reaction to relate given
information to the substance needed in the answer.
Introduction to Titration
Titration is a technique used to find the concentration of a solution based on the known
concentration of another solution, called the titrant. Titration is based on a chemical reaction
between the two solutions that is able to come to completion, which is often marked by a
measurable change like the appearance of a color or a precipitate.
The point at which the titrant neutralizes the solution is called the equivalence point.
Titration curves are used to monitor the progress of the reaction by plotting the titrant added
on the x-axis compared to the monitored change, such as change in pH or color.
35
Note that free response questions about titration often appear on the AP exam. Be sure that you
are able to estimate the equivalence point and the concentration of the solution based on
chemical equations and stoichiometry, as well as features of the titration curve.
Types of Chemical Reactions
Types of reactions are classified by the processes that produced the reaction. For the purposes
of the AP exam, it is useful to be able to identify certain types of chemical reactions. Acid-base
reactions involve the transfer of protons between chemicals. Oxidation-reduction reactions
involve the movement of electrons between chemicals. Precipitation reactions involve the
formation of an insoluble product, called a precipitate.
Introduction to Acid-Base Reactions
Acids and bases are defined by several theories. Arrhenius theory states that substances that
ionize to produce protons (H
+
) in water are acids and substances that produce hydroxide (OH
-
) in
water are bases. This theory limits acid and base definitions to those that take place in water;
thus, the Brønsted-Lowry theory expands on this, stating that substances that donate protons
are acids, and those that accept protons are bases. In Brønsted-Lowry equations, the molecule
that receives the proton is called a conjugate acid, while the molecule that is left after donating
36
the hydrogen ion is a conjugate base. Acid-base reactions will be covered in more detail in later
sections.
Oxidation-Reduction (Redox) Reactions
Oxidation-reduction (redox) reactions are reactions involving a transfer of electrons between
reactants. In redox reactions, there is always an oxidized species, whose electrons are lost, and
reduced species, who gain electrons. This can be remembered using the mnemonic “LEO says
GER” (Loss of Electron is Oxidation, Gain of Electron is Reduction).
Oxidation state or number is the number of electrons that an atom gains, loses, or uses in a
reaction and can be a positive or negative integer or 0. An increase in oxidation number reflects
oxidation and a decrease reflects reduction. Several rules can be used to assign oxidation
numbers:
• Free atoms have an oxidation state of 0
• Hydrogen has an oxidation state of +1 (except for hydride)
• Oxygen has an oxidation state of -2 (except for peroxide)
• In a compound or ion, the oxidation state is equal to the total charge
• Fluorine, bromine, and chlorine have an oxidation state of -1, except when bound to a
lighter halogen, oxygen, or nitrogen
• Group 1 metals are +1, group 2 are +2
Redox reactions occur in various forms:
• Combination reactions:
• Decomposition reactions:
• Replacement reactions:
• Combustion with oxygen:
Redox reactions are balanced using two reactions called half-reactions:
To write net ionic equations for redox reactions, both the number of atoms and the
charge are balanced on both sides of the reaction. The method to do this involves sequential
steps:
1. Write out the unbalanced equation.
2. Identify and assign oxidation numbers for each atom.
3. Separate out the half reactions, one with the oxidized atoms and one with reduced
atoms. Include the number of electrons lost and gained on each side.
37
4. Multiply coefficients so that the gains and losses of electrons are balanced in the half
reactions.
a. In acidic or aqueous solutions, first balance atoms that are not H or O, then
balance charge using H
+
and OH
-
. Then balance oxygen using H
2
O.
5. Recombine the half reactions.
6. Balance any elements whose oxidation state did not change.
7. Double-checkyour work by confirmingthat all elements and charges are balanced.
Outside Reading
• For additional information on balancing equations, see:
o http://www.occc.edu/kmbailey/Chem1115Tutorials/Net_Ionic_Eqns.h
tm
• For examples of how to balance redox reactions, see:
o https://www.periodni.com/balancing_redox_equations.php
• For more information on stoichiometery and balancing equations, see:
o https://www.khanacademy.org/science/chemistry/chemical-
reactions-stoichiome/stoichiometry-ideal/a/stoichiometry
o https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supple
mental_Modules_(Inorganic_Chemistry)/Chemical_Reactions/Stoichio
metry_and_Balancing_Reactions
• For more information on titrations, see:
o https://sites.google.com/a/ramapocentral.net/ap-chemistry/chapter-
15-applications-of-aqueous-equilibria/ph
38
Sample Chemical Reactions Questions
What is the final concentration of anions when a solution is made by adding 300.0 mL of 4.0 M
MgCl
2
to 200.0 mL of 3.0 MNaNO
3
?
A. 0.6 M
B. 3.0 M
C. 5.0 M
D. 6.0 M
Explanation:
The correct answer is D. Anions are negative ions, and the solutions in the question will contain
Cl
−
and NO
3
−
ions in solution. Remember that MgCl
2
will produce 2 moles of Cl
−
and NaNO
3
will
produce 1 mole of NO
3
−
. Calculate the moles of each anion that will be present in the solution,
and divide by the volume of the solution in liters to calculate the molarity.
(0.3000 L)
4.0 mol MgCl
2
1 L
æ
è
ç
ö
ø
÷
(2) = 2.4 mol MgCl
2
(0.2000 L)
3.0 mol NaNO
3
1 L
æ
è
ç
ö
ø
÷
= 0.60 mol NaNO
3
3.0 mol anions
0.500 L
= 6.0 M Total Anions
When 2.0 g K
2
CO
3(s)
was added to a beaker of 75 mL 3 M HNO
3(aq)
, vigorous bubbling occurred.
Which statement best explains this phenomenon?
A. Water vapor is released.
B. The solution was poured too fast.
C. K
2
CO
3(s)
reacts violently when added to water
D. Carbon dioxideis released by the reaction.
39
Explanation:
The correct answer is D. This is a gas-evolution reaction. When carbonates and bicarbonates
(CO²
3
−
and HCO
3
−
) react with acids, the gas product is CO
2
. The reaction is K
2
CO
3(s)
+ 2HNO
3(aq)
→
CO
2(g)
+ H
2
O
(l)
+ 2KNO
3(aq)
. There is not enough energy released to boil the water.
What type of reaction best describes this process?
Zn
0
+ Cu
2+
→ Zn
2+
+ Cu
0
A. Synthesis reaction
B. Decomposition reaction
C. Redox reaction
D. Neutralization reaction
Explanation:
The correct answer is C. Redox (or oxidation–reduction) reactions involve the transfer of
electrons from one reagent to the other. In this case, Zn
0
transfers two electrons to Cu
2+
to form
Zn
2+
and Cu
0
.
40
Kinetics
Around 7‒9% of questions on the AP exam will cover Kinetics.
Reaction Rate
Kinetics is the study of reaction rates, the change in concentration over a given period of time:
Reaction rate can also be expressed in terms of the reaction constituents. Consider the reaction:
The reaction rate can be expressed as any of the following:
Note that as reactants decrease over time, [reactants] will be negative; thus, a negative
sign is added to the equation to generate a positive overall reaction rate. To account for
differences in molar quantities due to the stoichiometry of the reaction, the coefficients are
divided for B and C in this reaction.
The rate constant is determined by the Arrhenius equation:
As shown by the equation, the rate constant is dependent on the temperature that the
reaction takes place, as an increase in temperature increases the kinetic energy of a reaction up
to a certain point. An increase in kinetic energy increases the chances of molecules colliding and
thus reacting with each other, in turn increasing the reaction rate. This rate follows an
exponential scale, meaning an increase in 10 K doubles the reaction rate.
41
Introduction to Rate Law
Rate law describes the relationshipbetween reaction rate and the concentrationsofreactants.
Consider the reaction:
The rate law is expressed as:
Where k is the experimentally determined rate constant for the reaction at a given
temperature and s and t are the reaction orders for the A and B, respectively. The reaction order
indicates the effect of a reactant on the overall reaction. For instance, if doubling a reactant
results in doublingthe rate, then the reaction order for that molecule will be 1 ([2]
1
= 2).
If doubling a reactant causes a four-fold increase in rate, then the reaction order for that
molecule is 2 ([2]
2
= 4). The overall reaction order is equal to the sum of the individual reaction
orders of reactants:
reaction order = s + t
In practical terms, these equationsmean that the concentration of reactant A decreases
by a factor of s and the concentration of reactant B decreases by a factor of t. For example, given
the reaction:
In this reaction, both reactants are decreasing by a factor of 1; thus, the reaction order with
respect to A = 1 and the reaction order with respect to B = 1. In cases where the reaction rate is
not affected by the concentration of reactants, the reaction rate = k and the reaction order is 0.
Concentration Changes over Time
Using rate equations, one can calculate the reaction order for reactants based on their
concentrations in the reaction and the effect on reaction rate, or rate of formation of C.
42
Once the value of the reaction orders is known, the appropriate equations can be used to
determine reaction properties at specific times (t) in a reaction:
Zero Order
First Order
Second Order
Rate law
Integrated rate law
Plot to determine k
[A] vs t
ln[A] vs t
vs t
Relationship of k
slope of plot
Half-life
Half-life is the time it takes for half the reactants to convert to products. In radioactive decay,
half-life demonstrates first-order kinetics and can range from millisecondsto billions of years. To
determine the rate of decay () given a half-life (
we use the equation:
Given the and the initial amount of the sample (
it is possible to find the amount of a
radioactive sample remaining (
after a given amount of time (t), with the equation:
Since half-life is constant for a given isotope, it can be used to estimate the age of an item, as is
used in carbon-14 dating using the above equations.
Elementary Reactions and Reaction Mechanisms
Elementary reactions are the simplest type of reaction, involving only a single step. These are
defined by the number of reactants. For example, a unimolecular reaction involves a single
reactant. These can include isomerization of a reactant or decomposition reactions, like the
decompositionof hydrogen peroxide to water and oxygen:
In bimolecular reactions, two molecules react to form a product. These are the most common
types of reactions that you’ll come across. Reactions that involve the collision of three or more
43
molecules at a time are extremely rare; more often, reactions that involve more than two
reactants take place in a stepwise fashion and form intermediates between steps that do not
show up in the final equation.
For example, take the reaction:
The reaction takes place in two steps that, when added together, result in the overall reaction
given above.
In these types of multi-step reactions, the reaction mechanisms are built on several
elementary reaction steps; the components include reactants, intermediates, products, and
catalysts. The reaction intermediates (E in the equation above) do not show up in the final
products. Identifying the reaction intermediates can help you identify the mechanism of the
reaction.
The slowest step in a multi-step reaction is referred to as the rate-determining (or rate-
limiting) step. The rate law for the whole reaction would thenbe the rate of the rate-determining
step. In multi-step reactions where there is no rate limiting step, the rate law is estimated through
a steady-state approximation. The assumption for this is that there should be an intermediate
that is consumed at the same rate that it is created; thus, its concentration remains constant
through the reaction—this is called reachingsteady state. Be sure to practice identifyingthe rate
law for steady state reactions as well.
Collision Model
For reactions to occur, reactant molecules must collide together with enough energy to break
chemical bonds and in a proper orientation to form new ones. There are several factors that alter
how often these events happen, and thus affect the reaction rate including concentration of
reactants, surface area of reactants, the complexity of reactant molecules, temperature, and
presence of a catalyst.
Reaction Energy Profile
Reaction energy diagrams show the relationship between reaction energy (y-axis) and the phase
of the reaction (x-axis). By looking at the diagram for a reaction where A converts to B via a
transitional state of C, you should be able to quantify the activation energy (the difference in
energy between A and B) as well as the energy change between reactants and products.
44
In multi-step reactions, the energy profile appears with multiple peaks, which correlate to the
activation energy for each elementary reaction.
45
Catalysis
Activation energy can be modulated by the presence of a catalyst. Catalysts are molecules that
lower the activation energy without actually being consumed by a reaction. In biology, catalysts
are generally enzymes that put molecules in optimal configurations for reactions to proceed
efficiently. When a catalyst is added to a reaction, the reaction rate becomes:
Where
is the amount the activation is energy is lowered by the addition of the
catalyst.
46
Outside Reading
• For additional information on kinetics, see:
o https://chem.libretexts.org/Bookshelves/Physical_a
nd_Theoretical_Chemistry_Textbook_Maps/Supple
mental_Modules_(Physical_and_Theoretical_Chemi
stry)/Kinetics/Reaction_Rates/Reaction_Rate
o https://courses.lumenlearning.com/boundless-
chemistry/chapter/reaction-mechanisms/
Sample Kinetics Questions
Which of the following actions will decrease the rate of the reaction?
A. Lowering the temperature
B. Increasing the pressure
C. Adding a catalyst
D. Increasing [N
2
] (at constant pressure)
Explanation:
The correct answer is A. Lowering the temperature decreasing the kinetic energy of the gas
molecules, meaning fewer have the minimum energy necessary to react. Increasing the pressure
will increase the rate at which collisions occur and will accelerate the rate of reaction. Catalysts
lower the energy of activation and increase the rate of reaction. Industrially, heterogeneous iron
catalysts are used to accelerate this reaction. Finally, increasing the concentration of [N
2
] will
shift the equilibrium towards the right, but it will not affect the rate of reaction.
The reaction of nitrogen dioxide with carbon monoxide to produce nitric oxide and carbon
dioxide is a multi-step reaction. Experimental work has determined that the rate law for the
equation follows the equation: rate = k[NO
2
]². Which elementary reaction is most likely the rate-
determining step (RDS)?
NO
2
+ CO → NO + CO
2
47
A. 2NO
2
→ NO
3
+ NO
B. CONO
2
→ NO + CO
2
C. NO
2
+ CO → NO + CO
2
D. NO
2
→ NO + O
Explanation:
The correct answer is A. The overall rate law reflects the rate equation that corresponds to the
rate-determining step. Here, since the rate law is second order in [NO
2
] (rate = k[NO
2
]²), the RDS
must involve 2 molecules of NO
2
and no other species.
If the half-life of a radioactive element is 6 minutes, which of the following was the initial mass
of a sample with 40.0 grams left after 36 minutes?
A. 80 grams
B. 160 grams
C. 2,560 grams
D. 5,120 grams
Explanation:
The correct answer is C. After 36 minutes, six half-life cycles have occurred:
1
6
® 6®
1
64
.
Unknown mass ×
1
64
= 40 grams unknown mass = 2,560 grams
48
Thermodynamics
About 7‒9% of the questions on your exam will cover Thermodynamics.
Endothermic and Exothermic Processes
Thermodynamics deals with the relationship between heat and other forms of energy. During a
chemical reaction, temperature changes indicate that there has been a change in energy. These
changes in energy are due to phase changes or changes in bonds.
Exothermic reactions release energy into their surroundingsin the form of heat.
49
Endothermic reactions absorb energy from their surroundings, resulting in a drop in
temperature.
The potential energy of a reaction represented in an energy diagram tells you whether
the reaction is endothermic or exothermic. If potential energy is decreased in the products
relative to the reactants, this represents a release of energy from the reaction to the environment
indicating an exothermic reaction. If the potential energy of products is increased relative to
reactants, this means energy is absorbed from the environment during the reaction, indicating
an endothermic reaction. Thedifference in energy between reactants and products is theamount
of energy absorbed or released.
Heat Transfer and Thermal Equilibrium
When two substances of different temperatures come into contact with each other, they will
exchange thermal energy. The point when the two substances have uniform temperatures is
thermal equilibrium. At thermal equilibrium, there is no net exchange of heat energy. The
process of reaching thermal equilibrium can be explained through a kinetic model. The molecules
in each substance have their own kinetic energy, with the warmer substance having greater
kinetic energy than the cooler substance.
When the molecules of the substances come into contact with each other, the molecules
begin to collide. In doing so, the molecules from the warmer substance transfer energy as heat
to the molecules of the cooler substance. These molecular collisions and energy transfers
50
continue, eventually causing both substances to have the same average kinetic energy at thermal
equilibrium.
Heat Capacity and Calorimetry
The first law of thermodynamics states that energy is conserved in chemical and physical
reactions. Thus, in a closed system the change in energy is equal to the energy supplied minus
the work done on its surroundings. This means that heat is not created or destroyed, but rather
transferred between systems. Energy changes in a system occur in the form of heat, phase
transitions, and chemical reactions.
Warmer systems transfer energy as heat to cooler systems. The amount of heat
transferred (q) between the systems is expressed using the following equation:
where m is the mass, c is the heat capacity (the amount of heat needed to change the
temperature by 1), and
T is change in temperature. It is important to note that the
temperatures of different substances of the same mass with the same amount of energy
transferred may not have equivalent temperature changes. This is due to differences in the heat
capacity (c) of the substances. Heat capacity is expressed as specific heat capacity or molar heat
capacity, the amount of heatneededto change the temperature of 1gram or 1 mole ofsubstance
by 1 degree, respectively.
Calorimetry is used to measure heat in a reaction by measuring how much heat is
absorbed or produced by a reaction. Coffee cup calorimeters measure the heat given off in the
form of water according to the equation:
Bomb calorimeters, on the other hand, use materials other than water to measure the
change in heat. These rely on changes in heat capacity, the amount of heat needed to change
the temperature by 1. Bomb calorimeters follow the equation:
where C is the heat capacity of the calorimeter in J/C.
Energy of Phase Changes
Energy transfer must occur for phase changes to take place. The energy of a system increases to
go from solid, to liquid, to gas. An equivalent amount of energy is released during a phase change
in the reverse direction (gas to liquid to solid). The heat of fusion (H
f
) is the amount of energy
needed for a unit of substance to change from solid to liquid. This same amount of energy is
released when liquid turns into a solid. The heat of vaporization (H
v
) is the energy needed for
51
a substance to go from liquid to gas. Again, the same amount of energy is released when a
substance condenses from gas to liquid. The heat (q) required for a phase change is dependent
on the number or moles of substance (n). For example, the amount of energy needed for a
substance to go from solid to liquid is determined by the equation:
Enthalpy of Reaction
Enthalpy refers to the energy of a system, usually expressed as heat. It is equal to the internal
energy of a system, plus the product of pressure and volume of the system. For the sake of the
AP exam, pressure and volume is held constant, so enthalpy can be thought of as equivalent to
the energy of the system.
The enthalpy change of reaction (H) is equivalent to the amount of energy released
(negative H) or absorbed (positive H). The enthalpy of reaction is the difference between
enthalpy of products vs. reactants.
This is expressed in kJ/mol, and under standard conditions (constant pressure and
volume), is related to the number of moles of reactants. Thus, the enthalpy of reaction is scalable
based on the amount of reactants in an equation. For example, in the equation:
The overall enthalpy of reaction would be –500 kJ, as 500 kJ of energy was released. If
you double the reactants (4 moles of A and 2 moles of B), the enthalpy or reaction becomes
–1,000 kJ.
Bond enthalpy is the amount of energy needed to break one mole of a bond. Bond
breaking is always endothermic, as it requires absorbing energy. Bond making, on the other hand,
is always an exothermic process, as it requires releasing energy. The amount of energy needed
in a reaction is estimated by adding up the average energies of each bond in the reaction.
Remember that the sign for bond formation is negative, as there is a release of energy.
For example, in the above reaction, we can see that 500 kJ is released by the formation
of A
2
B. Given that this chemical reaction does not break any bonds, and only forms 2 A-B bonds,
52
Enthalpy of Formation
The standard heat of formation (H
f
) is the change in enthalpy that occurs when one mole of a
substance is formed from its pure elements under standard conditions. The H
f
for a reaction is
equal to the sum of H
f
for products minus the sum of H
f
for reactants:
In other words, given the reaction between chemical A and B to form C:
The enthalpy of the formation would be:
Hess’s Law
In complicated reactions with multiple states, the heat of the reaction is measured by the sum of
the heats of individual reactions. This is called Hess’s Law. Remember that when the reaction
moves in the opposite direction, the heat of reaction is the same in magnitude, but the sign
changes (i.e., it is endothermicin one direction and exothermicin the other).
This concept can be extended to measure the heat of state changes as the product of the
final state minus the initial state. Heat of vaporization, for example, is the heat required for a
liquid to turn into a vapor:
53
Outside Reading
• For more information on energy in phase changes, see:
o https://chem.libretexts.org/Courses/can/health/07%3A_En
ergy_and_Chemical_Processes/7.4%3A_Phase_Changes
• For more on reaction enthalpy, see:
o https://chem.libretexts.org/Bookshelves/General_Chemistr
y/Map%3A_Chemistry_-
_The_Central_Science_(Brown_et_al.)/05._Thermochemist
ry/5.4%3A_Enthalpy_of_Reaction
Sample Thermodynamics Questions
Which of the following is a true statement about an endothermic reaction?
A. It releases heat.
B. Increasing the temperature shifts the equilibrium towards the products.
C. It is thermodynamically favored at all temperatures.
D. The use of a catalyst slows the reaction rate.
Explanation:
The correct answer is B. In an endothermic reaction, heat can be thought of as a reagent in the
chemical equation. By increasing the temperature (adding heat), the equilibrium is shifted
towards formation of products. Exothermic reactions release heat, while endothermic reactions
absorb heat from their surroundings. For a reaction to be spontaneous, the free energy (ΔG = ΔH
– TΔS) must be negative. In an endothermic reaction ΔH > 0, so in order to be spontaneous, ΔS
must be positive and the reaction must be done at a high enough temperature such that TΔS >
ΔH. Finally, the use of a catalyst will always accelerate the rate of reaction.
54
Species
ΔH
f
0
(kJ/mol)
Al
2
O
3
−1,670
Fe
2
O
3
−826
SO
2
−297
TiO
2
−945
MnO
−385
MnO
2
−520
The standard enthalpies of formation of several oxides are listedin the table above. Based on the
data in the table, which metal oxide has the strongest average metal-oxygen bond?
A. Al
2
O
3
B. SO
2
C. TiO
2
D. MnO
Explanation:
The correct answer is C. The strength of the metal–oxygen bond is proportional to the energy
released in its formation. TiO
2
has an enthalpy of −945 kJ/mol. This is divided over 2 Ti–O bonds,
giving an average bond enthalpy of −473 kJ/mol.
55
A 100-gram aluminum block is heated to 100°C and placed in an insulated vessel containing 100
grams of ice-cold water (0°C). After the system is allowed to reach thermal equilibrium, the
temperature of the water was recorded as 20°C. What is the magnitude of the specific heat
capacity of water compared to the specific heat capacity of aluminum?
A. The specific heat capacity of water is 4 times less than the specific heat capacity of aluminum.
B. The specific heat capacity of water is 5 times less than the specific heat capacity of aluminum.
C. The specific heat capacity of water is 4 times greater than the specific heat capacity of
aluminum.
D. The specific heat capacity of water is 5 times greater than the specific heat capacity of
aluminum.
Explanation:
The correct answer is C. The energy lost from the Al block must be equal to the energy gained by
the water. The energy lost by the Al block is equal to the change in temperature multiplied by
the heat capacity (C) of Al: (ΔT × C
Al
). Likewise, the energy gained by the water is: (ΔT × C
H2
O).
Setting these equations equal we have:
(80 °C × C
Al
) = (20 °C × C
H2
O)
Rearranging, we have:
The heat capacity of water is 4 times greater than that of Al.
56
Equilibrium
Around 7‒9% of questions on your AP exam will cover Equilibrium.
Introduction to Equilibrium
Most reactions do not proceed to completion; rather, while the reactants make products a
reverse reaction occurs wherein the products convert back to reactants. This sort of reaction is
represented with a chemical equation with both forward and reverse arrows:
As the reactions proceed, the reaction rates become equal, called chemical equilibrium.
When a reaction has reached chemical equilibrium, chemical concentrations in the reaction (A,
B, C, and D) are steady. This state is also referred to as dynamic equilibrium. In this state,
reactions continue to occur to generate reactants and products, but there is no net change in the
ratios. On the contrary, static equilibrium refers to a state where the reaction is completed and
there is no conversion occurring between products and reactants.
To identify if a reaction is moving in the forward direction (reactants converting to
products) or the reverse direction (products converting to reactants), one can look at the rate of
reaction for each reaction. If the reaction rate of the forward reaction is greater than reverse,
then that is the preferred reaction and the conversion to products will be favored. If the reaction
rate of the reverse reaction is higher, then there will be a greater conversion of products to
reactants.
Reaction Quotient and Equilibrium Constant
The equilibrium constant (K
c
or K
eq
) is the ratio of product concentrations divided by reactant
concentrations, each raised to their respective stoichiometric coefficients. For the equation:
When calculating the equilibrium constant, remember that pure solids and pure liquids
are generally not included in the calculation. To get the correct K
c
, make sure the equation is first
balanced, and that the coefficients are reduced to the lowest integers. In processes that have
multiple steps, K
c
is equal to the product of the K
c
of each step. K
c
is constant for a reaction at set
temperature and pressure. If these variables change, then the K
c
will change as well.
57
Reactions that have not reached equilibrium have a similar expression, called a reaction
quotient (Q). When a reaction has reached equilibrium, Q is equal to K
C.
Comparison of K
C
and Q
can be used to determine whether a reaction is at equilibrium. For example, if Q is less than K
c
,
the reaction has not yet reached equilibrium and will favor products. If Q is higher than K
c
, the
reaction will favor reactants. K
c
can also tell us whether reactants or products are favored at
equilibrium. A K
c
of more than 1 favors products while a K
c
less than 1 favors reactants. The
magnitude of the number tells how strongly the reactants or products are favored. For example,
a K
c
of 1.5 has slightly more products than reactants. A K
c
of 1,000 would be mostly products and
very little reactants.
For gasses, the equilibrium constant can be expressed in either mol/L (K
c
) or in pressure
units (K
p
). The K
p
calculation only takes into account molecules that are in the gas phase and is
calculated using a similar equation as K
c
, where pressures are substituted for concentrations. To
convert between K
c
and K
p,
a modified version of the ideal gas law is used:
Calculating Equilibrium Concentrations
Equilibrium constant is used to calculate the concentration of products or reactants. For the AP
exam, use a RICE table to calculate unknown concentrations. Remember that RICE stands for
Reaction, Initial concentrations, Change, Equilibrium concentrations, which is the format of the
table. Imagine you are given the formula:
To fill in a RICE table, first fill in reaction components, then fill in the concentrations of
initial reaction components given, then calculate the expected change in reaction based on the
stochiometric equation. This will then allow you to fill in the equilibrium concentrations based
on the initial concentrations and change. For example, if you know that at the beginning of the
reaction molarities of A and B are both 1 and C is not present then at equilibrium the molarity of
C is 0.5, and you can then fill in the table accordingly:
Reaction
aA
bB
cC
Initial
Concentrations (M)
1
1
0
Change
–ax
–bx
+cx
Equilibrium
Concentrations (M)
1–ax
1–bx
0.5 = 0+cx
58
The concentrations of reactants are both known to be 1 M while c is not present, so a
molarity of 0 is a given. As reactants are converted to product in the reaction, the changes in
reactants are both negative and the change in product is positive. The stochiometric coefficients
determine the magnitude of the reaction. The unknown change in reaction is denoted by the
term x. The equilibrium concentrations are then determined by writing out the sums of initial
concentration and the change. As the equilibrium concentration of C is given, you can then solve
for x to find the equilibrium concentrations of the reactants in this case it would be:
Note that on the free response section, you may be asked to determine the equilibrium
concentrations and then determine the equilibrium constantbased on these values.
Introduction to Le Châtelier’s Principle
LeChâtelier’s principle asserts that when a change in pressure, temperature, or concentration of
molecules (products or reactants) are applied to a system at equilibrium, the reaction will shift
to re-establish equilibrium. In other words, when there is a change in the system that disrupts
the balance between Q and K, then the reaction will shift to bring these back to equality by
changing concentrations or partial pressures.
• Pressure increases favor fewer moles of gas. For example, increasing pressure in a
reaction that involves gas dissolving into liquid causes more of the gas to stay dissolved
in liquid and lowering pressure favors gas to be released from the liquid. Think of how
opening a bottle of soda releases pressure and dissolved carbon dioxide quickly bubbles
out of solution.
• Temperature increases cause the reaction to shift to use the heat. To determine which
way the reaction will move, you will need to know whether the reaction is exothermic or
endothermic. Adding heat to an exothermic reaction will shift towards the reactants to
balance the temperature change. Adding heat to an endothermic reaction will shift
towards the products. Note that changing temperature changes the K of a reaction, not
Q.
• Changing the concentration of products or reactants causes a compensatory change to
the opposite side of the reaction. Adding products increases reactants and adding
reactants increases products. Note that a change in concentration causes a change in Q
but not K.
• Adding catalysts does nothing to a system at equilibrium. The catalyst only serves to
change the rate of the reaction, not to change the balance between reactants and
products.
59
Introduction to Solubility Equilibria
The solubility product constant (K
sp
) refers to the equilibrium constant for a solid thatis dissolved
in a solute. This represents the amount of solid that is able to dissolve. The higher the K
sp
, the
more soluble the substance. When calculating K
sp
, remember that solids are not taken into
account and neither is solute, so for the dissolution reaction:
A solution is considered saturated when Q = K
sp
. When Q is less than K
sp
, a solid will
dissolve in solute, if Q is greater than K
sp
, the reaction is saturated and any ionic solid will
precipitate.
Common-Ion Effect
The common ion effect occurs when an ion is shared between two equilibria. For example,
imagine mixing a solution of KCl and NaCl; Cl
-
is shared between two ions. The solubility is
measured to account for the different concentration of each ion in solution. Thus, the solubility
constant would be calculated as:
If KCl is added to a saturated solution of NaCl that was in equilibrium:
The extra chloride ions would decrease the solubility of NaCl in solution, and thus shift
equilibrium to favor the production of reactants.
pH and Solubility
When one of the molecules in a reaction is a weak acid or base, the pH of the solution can affect
solubility. Decreasing the pH acidifies the environment and increases the solubility of weak bases,
while increasing the pH reduces the solubility. The effect is the opposite in the presence of weak
acids. You will not be asked to calculate solubility based on pH for the AP exam, but you may be
asked to identify qualitative changes of pH based on LeChâtelier’s principle.
60
Free Energy of Dissolution
When a molecule, like salt, dissolves in solution there is a change in energy (
due to changes
in bonds and phases. This is called the energy of dissolution. When some salts, like sodium
chloride, dissolve in water the overall change in temperature is negative. When others, like
magnesium chloride, dissolve there is an increase in temperature. This can be explained due to
factors like breakingof intermolecularbonds, interaction between solvent and dissolved sol ute,
and changes in the structure of solvent around the solute. The sum of these changes can
theoretically be used to estimate the energy of dissolution; however, for the sake of the exam
you will only be required to explain these changes qualitatively.
Outside Reading
• For additional information on equilibrium:
o https://chem.libretexts.org/Bookshelves/General_Che
mistry/Map%3A_Principles_of_Modern_Chemistry_(Ox
toby_et_al.)/UNIT_4%3A_EQUILIBRIUM_IN_CHEMICAL
_REACTIONS
• For additional information on RICE tables, see:
o https://www.ausetute.com.au/ricetable.html
• For more information on free energy of dissolution:
o https://chem.libretexts.org/Bookshelves/General_Che
mistry/Book%3A_CLUE_(Cooper_and_Klymkowsky)/6%
3A_Solutions/6.4%3A_Gibbs_Energy_and_Solubility
61
Sample Equilibrium Questions
After the system has reached equilibrium, if a quantity of NH
3
is rapidly introduced to the system,
which diagram best describes how pressure will change over time (arrow indicates time gas was
injected)?
A.
B.
C.
D.
62
Explanation:
The correct answer is A. Introducing NH
3
to the mixture will cause an initial rise in pressure and
move the system out of equilibrium. To move back into equilibrium, some of the added ammonia
will be converted back to N
2
and H
2
, leading to a further increase in pressure. This is because 3
moles of H
2
and 1 mole of N
2
are produced for every 2 moles of NH
3
.
The following reaction is in equilibrium at pressure P and temperature T. Which of the following
changes to the reaction conditions would favor the product?
CO
(g)
+ 2H
2(g)
→ CH
3
OH
(g)
ΔH = −128.1 kJ/mol
A. Decrease P while keeping T constant.
B. Decrease T while keeping P constant.
C. Remove some hydrogen gas from the container without changing P and T.
D. The reaction is always in equilibrium.
Explanation:
The correct answer is B. The reaction is exothermic according to the negative heat of reaction,
ΔH, so decreasing the temperature will force the reaction to move in the direction to release
more heat, towards the product side.
Which of the following is NOT considered a biologically relevant, reversible chemical reaction?
A. Hemoglobin transports oxygen by binding it in the lungs and then releasing it in cells
B. Sugar and O
2
are converted into H
2
O and CO
2
, which is expelled from the body
C. Nicotine from a cigarette binds receptors in the brain, giving a temporary high
D. The synthesis of ATP from ADP to store energy and the decomposition of ATP to ADP to provide
energy for chemical reactions
Explanation:
The correct answer is B. The oxidation of sugar to CO
2
is an irreversible reaction. This process is
highly exergonic, a common trait of irreversible reactions. Choice A is incorrect because the
binding and release of O
2
by hemoglobin is a reversible process; if the reaction were irreversible,
O
2
would bind permanently and could not be used in biological processes. Choice C is incorrect
63
because the binding and release of O
2
by hemoglobin is a reversible process; if the reaction were
irreversible, O
2
would bind permanently and could not be used in biological processes. Choice D
is incorrect because ATP is a higher energy molecule that transfers chemical energy to other
biological devices by decomposing to ADP; ADP is regenerated in living systems back to ATP and
the process is repeated.
64
Acids and Bases
About 11‒15% of all questions on your exam will cover Acids and Bases.
Introduction to Acids and Bases
Acids and bases dissociate in water in equilibrium reactions. The strength of an acid or base is
related to the equilibrium constant of a reaction (K
a
and K
b
, respectively), where higher
equilibrium constantsindicate strongeracids or bases and more dissociation. The strength of an
acid or base is expressed as pH or pOH respectively. pH represents the concentration of
hydronium ions (H
3
O
+
, often used interchangeably with hydrogen ions H
+
) in solution by the
equation:
pOH is related to the concentration of hydroxide ions (OH
-
) in solution, calculated by the
equation:
Water is amphoteric, meaning it can act as either an acid or base. Pure water autoionizes
according to the equation:
The equilibrium constantfor this reaction is written as:
Remember that the equilibrium constant is dependent on temperature, thus the K
w
will
change if the temperature deviates from 25° C. In neutral water, the concentrations of hydronium
and hydroxide are equal; thus, pH and pOH are equal at 7.
The pH scale is logarithmic, and ranges from 0 (strong acids) to 14 (strong bases). At 25°C,
7 is a neutral pH. In aqueous solutions at 25°C, you can convert between pH and pOH using the
K
w
or pK
w
equations.
pH and pOH of Strong Acids and Bases
Strong acids dissociate completely in water. This means that the concentration of hydronium or
hydroxide ionsis equalto the initialconcentration of the strongacid or base. Thus, the pHor pOH
can easily be calculated. For example, the strong acid HCl:
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For the exam, it is important to be able to spot common strongacids (HCl, HBr, HI, HNO
3
,
HClO
4
, H
2
SO
4
). Strong bases to remember are hydroxidesof alkali earthmetals (NaOH, KOH, LiOH,
CsOH, and RbOH), as well as alkaline earth metal hydroxides (Sr(OH)
2
, Be(OH)
2,
Ca(OH)
2
).
A way to determine the strength of an acid or base is to consider the structure of the
molecule. For example, strong acids tend to be composed of protons bound to a highly
electronegative ion. This causes the electrons to concentrate around the electronegative atom,
thus making the hydrogen ion more easily donated. Another factor is the strength of the bond.
The stronger the H-A bond, the less likely the H
+
ion will be donated, making for a weaker acid.
Weak Acid and Base Equilibria
Weak acids and bases only partially dissociate in water with equilibrium constants below 1.
Another way to express the strength of an acid is through pK
a
:
Low pK
a
indicates a strong acid, high pK
a
indicates a weak acid. As pH for Brønsted-Lowry
acids are calculated using the dissociation of H
3
O
+
, we can then use the equilibrium constant to
calculate the pH of a weak acid or base in solution given the molarity. To calculate pH of a weak
acid (HA), first write out the balanced equation for dissociation in water:
Note that since water is ignored in calculating equilibrium constants the equilibrium
constant equation for this acid will be:
If we are given the initial molarity, (Y), we can assume for any amount of weak acid that
dissociates (X), the same amount of proton and weak base will form; thus, we can calculate:
HA
(aq)
A
-
+
Before equilibrium:
Y
0
0
After equilibrium:
Y – X
X
X
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Then the equilibrium constantbecomes:
As we know this is a weak acid, and thus there is a tiny amount of dissociation compared
to the initial acid, we can disregard the change in [HA] and simplify the equation further:
We then solve for X, as this is the concentration of dissociated protons:
We can then put this into the equation for pH to estimate pH of the weak acid:
The same process can be used for bases by calculatingthe concentration of OH
-
.
Acid-Base Reactions
In acid-base reactions, the pH of the resulting solution is dependent on the strength and
concentration of acids and bases involved.
In a strong acid-strong base reaction, the reaction will proceed according to the equation:
Thus, the pH of the solution is calculated using the concentration of excess acid or base.
In a weak acid-strong base reaction, the equation is represented as:
If the strong base is in excess, the pH is determined by the moles of excess hydroxide ions
in the solution. If the weak acid is in excess, then a buffer is created and the Henderson -
Hasselbalch equation is used to determine pH (see section on buffers). If acid and base are
equimolar, then the equilibrium concentrations are used to calculate the pH according to the
written equilibrium reaction.
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Strong acid-weak base reactions follow a similar logic as weak acid-strong base reactions.
If the weak base is in excess, a buffer is created, and the Henderson-Hasselbalch equation is used.
If the strong acid is in excess, then the moles of excess acid are used to determine pH. Again, if
acid and base are equimolar, thanthe equilibrium reactionconcentrationsare used to determine
the pH.
In weak acid-weak base reactions, the reaction will reach equilibrium, which is
represented by the equation:
Acid-Base Titrations
A titration curve for an acid-base reaction plots the pH against the amount of titrant added. At
the equivalence point, the number of moles of titrant is equal to the number of moles of analyte.
The concentration and volume of titrant is then used to determine the concentration of analyte.
In strong acid-strong base reactions, the equivalence point is a neutral pH (pH = 7 at 25ºC). This
point is identified on the titration curve as the point where there is a sharp rise or drop in pH.
For titrations involving weak acids or weak bases, this relationship is more complicated
because of the larger buffering capacity. It is useful to identify the half-equivalence point. At this
point there are equal concentrations of the conjugate acid-base pair. When conjugate acid-base
pairs have equal concentrations, pH is equal to pK
a
; thus, the concentration of titrant added at
this point can be used to determine concentration.
This shows two titration curves with a base as titrant and acid as analyte. Both curves have an equivalence point of 40.
The grey curve shows a strong acid-base reaction with a sharp rise in pH. The red dotted curve shows a weak acid
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reaction with a long buffering region. In this case, the half-equivalence point, at 20, is used to determine species
concentration.
For polyprotic acids (acids with more than one proton to donate), titration curves can be
used to determine the number of protons present in the analyte. In this case, multiple
equivalence points can be identified in the titration curve, each corresponding to the number of
acidic protons present. You will not be asked to determine the concentration of species from
titration curves of polyprotic acids, but you may be asked to qualitatively describe the major
species present at points along the curve.
pH and pK
a
The pK
a
of a molecule can be thought of as the pH at which the molecule will accept or donate a
proton. While pHis dependent on the concentration of a solution, the pK
a
is a more stable value
that is unaffected by concentration. The relationship between pH of a solution and pK
a
of the
acid in solution can be used to determine the protonation state of acid or base in the solution.
When pH < pK
a
, the acid form has a higher concentration than the base form. When pH > pK
a
,
the base form has a higher concentration than the acid form.
Properties of Buffers
Buffers are solutions containing large amounts of a conjugate acid base or a weak pair. Buffers
resist changes in pH when small amounts of acid or base are added by creating water. The
conjugate acid reacts with added base while the conjugate base reacts with added acid. The pH
of a buffer can be approximated usingthe Henderson-Hasselbalch equation for an acid (HA) and
its conjugate base (A
-
):
Maximum buffering capacity is reached when pH=pK
a
.
Buffering capacity can be increased by increasing the concentrations of buffer
components but keeping the ratios constant. By changing the ratios of conjugate acid or base,
the bufferingcapacity changesfor base or acid. For instance, increasingthe amountof conjugate
acid relative to conjugate base increases the buffering capacity for acid compared to base.
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Outside Reading
• For further reading on acid base structure and strength, see:
o https://chem.libretexts.org/Bookshelves/General_Chemistry/
Map%3A_General_Chemistry_(Petrucci_et_al.)/16%3A_Acids
_and_Bases/16.8%3A_Molecular_Structure_and_Acid-
Base_Behavior
• For more reading on pH and acids and bases:
o https://www.khanacademy.org/science/chemistry/acids-and-
bases-topic#acids-and-bases
Sample Acids and Bases Questions
What is the expected result when Ba(OH)
2
and CH
3
COOH are combined in stoichiometric
amounts in aqueous solution?
Ba(OH)
2(aq)
+ 2CH
3
COOH
(aq)
→
A. solution of pH > 7
B. solution of pH = 7
C. solution of pH < 7
D. no reaction occurs
Explanation:
The correct answer is A. Recognize that this is an acid-base reaction—CH
3
COOH is acetic acid, a
weak acid, which can be written as HC
2
H
3
O
2
. Barium hydroxide is a strong base. When the
reaction is balanced the acid and base form a solution of water and a salt, which remains
dissolved in solution:
Ba(OH)
2(aq)
+ 2CH
3
COOH
(aq)
→ 2H
2
O
(l)
+ Ba(C
2
H
3
O
2
)
2(aq)
The salt of a weak acid and a strong base is a basic salt, which will, through hydrolysis, make the
solution at the equivalence point basic.
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Which of the following aqueous solutions will be basic and act as a buffer? Assume the solutions
contain 1:1 mole ratios and all concentrations are 0.10 M.
A. methylamine (CH
3
NH
2
) and methylammonium chloride (CH
3
NH
3
Cl)
B. hydrofluoric acid (HF) and sodium fluoride (NaF)
C. potassium hydroxide (KOH) and hydrobromic acid (HBr)
D. sodium hydroxide (NaOH) and ammonia (NH3)
Explanation:
The correct answer is A. A basic buffer is a solution made from a weak base and its conjugate
acid. Adding a strong acid or base to a buffer solution does not change the pH significantly; these
solutions are used when a constant pH is desired, in this case a pH > 7.
A stock solution of an acid has a concentration of 0.15 M at STP. What is the pH of the acid?
A. 0.15
B. 12.5
C. 0.8
D. –0.15
Explanation:
The correct answer is C. The concentration of the acid is .15 M. The pH of an acid is the negative
logarithm of the hydrogen-ion concentration. Thus,
pH log H
+
=−
, i.e.,
pH log 0.15 0.8= − =
.
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Applications of Thermodynamics
Finally, 7‒9% of the question on your exam will cover Applications of Thermodynamics.
Introduction to Entropy
Entropy is a measure of the amount of energy in a system that is not available to do work. This
can also be thought of as the amount of disorder in a system. The second law of thermodynamics
states that the universe moves in the direction of increased entropy over time. This can also be
thought of as meaningthat spontaneousprocesses move in the direction of increased entropy.
As matter in a system becomes more dispersed, the entropy increases. For example, when
the state of matter changes from solid to liquid to gas, molecules become more dispersed. In this
case, the entropy of the system increases as there is less available energy to do work. If you then
take the gas phase and increase the volume of the container while keeping temperature
constant, the entropy will further increase as the molecules move further apart. If you instead
condense the gas to a liquid, the entropy of that system will decrease. Entropy also increases
when energy becomes more dispersed. Kinetic energy distribution increases when the
temperature increases; thus, if you increase the temperature of a system containing gas, kinetic
energy becomes more dispersed, resulting in an increase in entropy.
The change in entropy of the system is calculated as:
Gibbs Free Energy and Thermodynamic Favorability
Gibbs free energy (G) is used to determine the thermodynamic favorability of a process. The
Gibbs free energy equation takes into account changes in enthalpy (), changes in entropy (),
and temperature (T):
Reactions are favorable when . Thermodynamically favorable reactions are also
called spontaneousreactions. At standard state, Gibbs free energy is given the symbol ; this
is most often the form seen on the AP exam. From the equation above, you should be able to
predict whether a reaction is favored given values for enthalpy, entropy, and temperature.
Gibbs free energy of formation (
is the change in free energy needed to convert 1
mole of a substance from its reactants at standardstate.
is defined by the Gibbs free energy
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equation, but is generally given in tables as the numbers are standard. Given energy of formation,
the Gibbs free energy of reaction (
of a substance is calculated as the difference
between the free energy of the substance and its reactants. This describes the amount of heat
energy available to do work in a reaction.
In this equation, the sign of
tells us which direction the reaction must proceed
to reach equilibrium ( , and the size of
tells us how far the reaction is from
equilibrium.
Note that the overall favorability of the reaction depends on the sum of the free energy
of products being greater than the sum of the free energy of reactants. This means that a
thermodynamically unfavorable reaction can be made favorable by coupling it with other
reactions that are more favorable. For instance, this can occur when the individu al reactions
share common intermediates. Another way to make a reaction more thermodynamically
favorable is to increase energy through an external source, such as electrical energy or a sunlight.
Thermodynamic and Kinetic Control
It is possible for a reaction to be thermodynamically favorable but either occur very slowly or not
at all. These reactions are under kinetic control. The most common reason for kinetic control of
a reaction is that the activation energy is too high for the thermodynamically favorable products
to form. Generally, increasing the temperature of a reaction will overcome the activation barrier
and form the more thermodynamically favorable product. In this case, the system is under
thermodynamic control.
Free Energy and Equilibrium
Both equilibrium constant and Gibbs free energy can be used to determine whether a reaction is
thermodynamically favored. Remember that this is the case when or when .
Similarly, the systems are considered at equilibrium when or . Thus, it is
reasonable that the free energy and equilibrium constant are related (where R is the universal
gas constant):
You can estimate these constants in relation to each other by considering the above
equations. When is close to zero, you can predict that K will be near 1.
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Galvanic (voltaic) and Electrolyte Cells
There are two types of electrochemical cells: galvanic and electrolytic. In both cases, remember
that oxidation occurs at the anode and reduction occurs at the cathode. The anodes and
cathodes in both cases are electrodes made of metal. A salt bridge is usually present, which acts
as a source of ions that balance building charges.
Galvanic cells, also called voltaic cells, derive energy from thermodynamically favored
redox reactions. Galvanic cells are usually made of two half cells connected by a salt bridge. In
the example in the following figure, the two half cells are composed of zinc (anode) and copper
(cathode) connected by a wire and a salt bridge. When connected, energy flows through the wire
from anode to cathode. Energy is harnessed to power devices through the wire connecting the
anode and cathode.
Electrolytic cells use a power source to force non-spontaneous reactions to occur,
converting electrical energy into chemical energy. In electrolytic cells, electrodes are placed to a
liquid containingelectrolytes. A battery acts as an electron supply to the negative cathode, which
reduces the electrolytes. A positive anode oxidizes and passes the electrons through the circuit.
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Cell Potential and Free Energy
A cell’s potential voltage (E
cell
) is determined by the tendency for the half-reactions to occur
spontaneously. To calculate the voltage of a cell, break the equation up into half reactions and
then add the reduction potential of the oxidation and reduction reactions based on standard
reduction potentials (remember that oxidation should be negative). In calculating cell voltage,
the coefficients of ions in the reaction are not taken into account. For non-standard conditions,
the Nernst equation can be used to calculate cell voltage:
The Nernst equation, which describes the free energy of an electrochemical cell, is also
related to free energy equations given the equation:
If a reaction is thermodynamically favored, the cell will have a positive cell voltage (E°). If
the reaction is thermodynamically unfavored, the cell will have a negative cell voltage and require
external potential for the reaction to occur.
Under non-standardconditions, cell potentialvaries dependingon concentrations of cell
components. The cell potential in this case is the driving force for equilibrium. The farther the
reaction is from equilibrium, the greater the cell potential.
The relationship between charge, current, and time is expressed in the equation:
Given this relationship andFaraday’slaw(1 mol electrons = 96,486 C), quantitative measures can
also be calculated for electrolytic cells.
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Outside Reading
• For further reading on entropy and the second law of thermodynamics,
see:
o https://chem.libretexts.org/Bookshelves/Physical_and_Theoreti
cal_Chemistry_Textbook_Maps/Supplemental_Modules_(Physic
al_and_Theoretical_Chemistry)/Thermodynamics/The_Four_La
ws_of_Thermodynamics/Second_Law_of_Thermodynamics
• For further reading on electrolytic cells, see:
o https://courses.lumenlearning.com/boundless-
chemistry/chapter/electrochemical-cells/
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Sample Applications of Thermodynamics Questions
Which of the following must be true for a reaction that is thermodynamically favored at both low
and high temperatures?
A. ΔH is negative and ΔS is negative.
B. ΔH is negative and ΔS is positive.
C. ΔH is positive and ΔS is negative.
D. ΔH is zero and ΔS is zero
Explanation:
The correct answer is B. It is not necessary to know the reaction or the values of ΔH and ΔS to
answer this question. For a reaction to be spontaneous, ΔG must be negative, which is
determined from the equation ΔG = ΔH − TΔS. Therefore, if ΔS is positive, then −TΔS will be
negative for all temperatures (remember that T is in Kelvin, which is always a positive value). If
ΔH is negative and ΔS is positive, then ΔG will be negative for all temperatures and thus
spontaneous.
Use the following image to answer the two questionsthat follow.
In the voltaic cell, which of the following is considered the anode? Assume the standard reduction
half-cell potential for iron is −0.44 V and the standardreduction half-cellpotential for zincis −0.76
V.
A. iron metal strip
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B. zinc metal strip
C. Fe
2+
solution
D. Zn
2+
solution
Explanation:
The correct answer is B. You must determine several things before answering thisquestion. First,
you must compare the reduction potentials for the half reactions. Fe
2+
+ 2e
−
→ Fe
(s)
has a
reduction potential E
0
= −0.44 V and Zn
2+
+ 2e
−
→ Zn
(s)
has a reduction potential E
0
= −0.76 V. The
half reaction with the more positive E
0
has a greater tendency to reduce, and thus the other half
reaction will oxidize. Therefore, the iron (Fe) side will reduce (making it the cathode, choice A),
accepting electrons and creating a positively charged cathode. The zinc side (Zn) will oxidize,
giving up electrons and creating a negatively charged anode.
In the previous diagram, what would happento the voltage if 100ml of 2.0 M Zn(NO
3
)
2
are added
to the beaker on the left?
A. The voltage decreases.
B. The voltage increases.
C. The voltage does not change.
D. The voltage becomes zero.
Explanation:
The correct answer is A. The reaction is Zn
(s)
+ Fe
2+
(aq)
→ Zn
2+
(aq)
+ Fe
(s)
. Zn(NO
3
)
2
added to the left
beaker increases the Zn
2+
concentration on the left side and “pushes” the reaction towards the
left; this implies a lower voltage.
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