Gen Chem Review Test Prep

General Chemistry

MCAT

FOUNDATION REVIEWCOMPANION BOOKLET

®

MM3258A

GENERAL CHEMISTRY 1

Foundation Review

ATOMIC STRUCTURE AND THE PERIODIC TABLE

Atomic Structure

Atomic Number (Z)

number of protons in the nucleus

Mass Number (A)

number of protons and neutrons: AZX

Isotopes

atoms with the same atomic number but different mass numbers

1.

FOUNDATION REVIEW

• An atom consists of a nucleus and surrounding electrons.• There are two types of particles in the nucleus: protons

(positively charged) and neutrons (neutral). Both can alsobe referred to as nucleons.

• The atomic number defines an element. Atoms of the sameelement always have the same number of protons. Con-versely, if two atoms have the same number of protons,they belong to the same element.

• Elements on the periodic table are shown in the order ofincreasing atomic number.

• In a neutral atom, the number of protons equals the num-ber of electrons.

• The number of neutrons is therefore equal to the mass number minusthe atomic number.

• We represent a nucleus by a nuclide symbol: AZ X, where X is the symbolfor the chemical element.

• Atoms or nuclei with the same atomic number but different mass num-bers are isotopes. Since isotopes have the same atomic number, theyare the same element.

• E.g. Carbon-12, carbon-13, and carbon-14 have six protons. Carbon-12has six neutrons, carbon-13 has seven, and carbon-14 has eight. Thenuclide symbols are 126 C, 136 C, and 146 C, respectively.

• To report the mass of an individual atom, we oftenuse the unit amu (atomic mass unit).

• 1 amu = �112�

the mass of a carbon-12 atom. In otherwords, a carbon-12 atom weighs 12 amu.

• Different isotopes of the same element have different atomic masses, because they have differentnumbers of neutrons. The atomic mass (or atomic weight) reported for an element in the PeriodicTable is a weighted average of the masses of all the isotopes that exist for that element. Chlorine,for example, has two naturally occurring isotopes: Cl-35 and Cl-37. The atomic mass of chlorine is35.5. (The Cl-35 isotope is more abundant, hence the weighted average is closer to 35 than it is to 37.)

• For convenience, we can also think of a proton and a neutron as each having a mass of 1 amu, whilean electron has negligible mass.

• The atomic mass of an element is also numerically equal to its mass in g/mol. A carbon-12 atomhas an atomic mass of 12 amu. One mole of carbon-12 therefore weighs 12 g. Similarly, one mole ofnaturally-occurring chlorine has a mass of 35.5 g, numerically equal to its atomic weight.

Atomic Mass/Weight

calculated mass of an atom (average of the masses of each isotope of an element); units are a.m.u.(atomic mass units)

the mass in grams of 1 mole of the element (�m

g

ol�)

Mole

the amount of a substance that contains the same number of particles as there are atoms in 12 g of 12C

1 mole corresponds to 6.022 × 1023 particles.

How many kilograms of helium are in a blimp that holds 1.3 × 109 mol of helium?

2.

MCAT: GENERAL CHEMISTRY

• Just like a pair corresponds to two, a dozen corresponds to 12, and a scorecorresponds to 20, a mole corresponds to a number. Specifically, a mole corre-sponds to about 6.022 × 1023, a number known as Avogadro’s number.

• Where does this number come from? It is the number of atoms in 12 g of 12C.In other words, 1 mole of a substance corresponds to an amount that containsthe same number of particles as there are atoms in 12 g of carbon-12.

A: He weighs 4.0 g/mol. 1.3 × 109 mol of He therefore weighs1.3 x 109 x 4.0 = 5.2 × 109 g, or 5.2 × 106 kg.

Models of the Atom

Heisenberg Uncertainty Principle

states that it is impossible to determine both the position and the momentum of an electron withcomplete accuracy at the same time

Quantum Numbers

Principal Quantum Number (n)

energy shell; can have integral values of 1, 2, 3, etc.

nucleus

Bohr Model Quantum MechanicalModel

"microscopicsolar system"

"electron clouds"

3.

FOUNDATION REVIEW

• Rutherford’s scattering experiment demonstrates that the atom consists of a dense, positively chargednucleus surrounded by negatively charged electrons.

• The absorption and emission spectrum of atoms also suggest that an electron can exist only in certain fixedenergy states; the energy of an electron is “quantized.”

• According to the Bohr model, electrons revolve around the nucleus in orbits. The fact that only certain energyvalues are allowed means that only certain orbit sizes are allowed. Absorption or emission occurs when anelectron moves from one energy level (one orbit) to another.

• The Bohr model is not successful in explaining the data on atoms with more than one electron.• The quantum mechanical model does away with the notion of prescribed circular orbits of electrons. Instead,

the position of an electron is described by the probability of its being at different points in space. The func-tion describing the probability of finding an electron at different points in space is known as an orbital.

• The more precisely we know the position, the larger theuncertainty in momentum, and vice versa.

• Any electron can be completely described by 4 quantum numbers: n, l,ml, and ms. In other words, the four quantum numbers determine theproperties of an electron: its energy, angular momentum, orbital, etc.

• The larger the value, the higher the energy level, and the fartheraway it is from the nucleus in general.

Azimuthal (angular) Quantum Number (�)

shape of subshell; integral values from 0 → n–1

Magnetic Quantum Number (m�)

spatial orientation of orbital; integer values from –� → +�

�

4.


s p d f

• labels the subshell• Can take on integer

values from 0 → n–1 for each value of n. The n = 1 shell contains only one subshell, the onewith l = 0; the n = 2 shell contains two subshells withl values of 0 and 1; etc.

• Each subshell consists of one or more orbitals. The number of orbitalsin a subshell is (2l + 1).

s subshell: one orbital (2l + 1 = 1)

p subshell: three orbitals (2l + 1 = 3)

d subshell: five orbitals (2l + 1 = 5)

f subshell: seven orbitals (2l + 1 = 7)

• s orbitals are spherical in shape and p orbitals are dumb-bell shaped.

• In most circumstances, orbitals belonging to the same subshell aredegenerate; i.e., they are at the same energy level.

• specifies the particular orbital within a sub-shell occupied by an electron

• integer value ranges from –l to l.

Spin quantum number (mS)

any electron can have only one of two values: +1/2 or –1/2

Pauli Exclusion Principle

states that no two electrons in an atom can have the same four quantum numbers

What are the possible quantum numbers for an electron in third energy level (n = 3)?

How many electrons can the third energy level hold?

Electron Configuration and Orbital Filling

Orbital Filling

filling follows diagonal arrows. The order of electron orbital filling is thus 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d, etc.

What is the electron configuration of Cl?

What is the electron configuration of Cl–?

1s 2s 2p3s 3p 3d 4s 4p 4d 4f5s 5p 5d 5f6s 6p 6d 6f7s 7p 7d 7f

n = 3 � m� ms

5.

FOUNDATION REVIEW

• Regardless of the shell, subshell, or orbital, any electron can have only one of two values

for the spin quantum number; the two spin orientations are designated +�21� and – �2

1�.

• Electrons in the same orbital have the same n, l, and ml quantum numbers. They musttherefore have different spin quantum numbers. As a consequence, each orbital canhold only two electrons maximum.

A: There are nine orbitals total. Each orbital can holdtwo electrons. Therefore the n = 3 shell can holdup to 18 electrons.

012

0–1, 0, 1

–2, –1, 0, 1, 2

+ �21�, – �2

1�

+ �21�, – �2

1�

+ �21�, – �2

1�

• The order in which subshellsare filled can be determinedusing the diagram shown inthe Lesson Book. We follow adiagonal arrow until we reachits tip, and then start overat the next arrow. Therefore,order of filling is: 1s, 2s, 2p,3s, 3p, 4s, 3d, 4p, 5s, 4d, anso on.

A: For neutral chlorine, the number of electrons is equal to the number of protons (atomic #),therefore we have 17 electrons. Keeping in mind that each orbital can hold 2 electrons, andthat the p subshell has three orbitals, the electron configuration is: 1s22s22p63s23p5

A: Now Cl has gained 1 electron, and has 18 electrons, therefore,1s22s22p63s23p6

Hund’s rule

Electrons add to unfilled degenerate orbitals (orbitals of the same energy) one at a time with their spinsparallel before they add to half-filled orbitals with their spins antiparallel.

What is the ground state electron configuration for nitrogen (N)?

Ene

rgy

1s

2s

2px y z

6.


• So far we have only dealt with the order of filling subshells. But what about individual orbitals? In other words,when we want to put two electrons in the p subshell, do we put them in the same p orbital or in different ones?

• Hund’s Rule states that within a given subshell, orbitals are half-filled so that they each have one electron, allwith parallel spins, before any orbital is fully occupied with two electrons of opposite spins. In other words, elec-trons would tend to avoid pairing as much as possible.

A: Nitrogen has an atomicnumber of 7, thus itselectron configuration is1s22s22p3. In this case,we want to be able tosay exactly how the 2psubshell is filled.According to Hund’srule, the 1s and 2sorbitals will fill com-pletely, while the threeorbitals of the 2p sub-shell will each containone electron, all withparallel spins.

1 1 1

1?

1?

• The ionization energy is the energy required to completely remove an electron from an atom or ion inthe gas phase.

• The trend in ionization energy follows closely that of electronegativity, since both have to do with anatom’s attraction for electrons.

• Ionization energies are low when the ionization process results in a noble gas (closed shell) configura-tion. For example: the alkali metals (elements in the first column) have low ionization energiesbecause removing an electron would bring them to a full octet, noble gas configuration. The noble

gases have high ionization energies because they want to “hold onto” their fullvalence shell configuration.

• One common shorthand in designating an electron configuration is to write explic-itly only the valence electrons. For example, for the chlorine example we saw earlier,we can write that its electron configuration is [Ne]3s23p5. We use the noble gas asa shorthand to represent the core electrons of the chlorine atom.

• Valence electrons are the electrons in the outermost shell. These are the electronsthat can participate in bonding and determine the chemical properties of the atom.

• Elements in the same column in the periodic table (the same group) have the samenumber of valence electrons. (Helium is an exception.) Elements inthe same group therefore have similar chemical properties, i.e.,they tend to react the same way.

Periodic Trends

Valence Electrons

the electrons in the outermost shell

Atomic Radius

indication of the size of the atom

Electronegativity

a measure of the attraction an atom has for electrons in a chemical bond

Ionization Energy

the energy required to completely remove an electron from a gaseous atom or ion

Electron Affinity

the energy that is released when an electron is added to a gaseous atom or ion; represents the ease withwhich an atom can accept an electron

7.

FOUNDATION REVIEW

• The atomic radius is an indication of the size of an atom. As we go down a group, the principalquantum number increases. The highest energy shell is therefore farther and farther away fromthe nucleus. Hence, atomic radius increases as we go down a group.

• Atomic radius decreases as we move towards the right along the same row (across a period).We will go more into the reason why in the actual Lesson. The largest atoms are therefore in thelower left hand-corner of the periodic table.

• Electronegativity is a measure of the attraction an atom has for electrons in a chemicalbond (how strong an atom holds onto its electrons).

• Electronegativity increases from left to right on the periodic table and increases as we goup a column. Electronegative elements are therefore found in the upper right-hand cornerof the periodic table.

• The energy that is released when an electron is added to a gaseous atom,and it represents the ease with which the atom can accept an electron.

• The stronger the attractive pull of the nucleus for electrons, the greater theelectron affinity will be.

Types of Elements

Metals

Transition metals

Nonmetals

Metalloids

alkali metals alkaline earth metals halogens noble gases

electronegativity atomic radiusionization energy

electronegativityatom

ic radius

ionization energy

8.


• Metals are found on the left portion of the periodic table (exception: hydrogen).• Metals are good conductors of heat and electricity. They are malleable and duc-

tile as solids.• Metals in Group 1A (first column) are known as alkali metals. Metals in Group

2A (second column) are known as alkaline earth metals. They are very reactive.

• Transition metals are “Group B” elements. They have partly filled or filled d orbitals.They form coordination complexes which can be highly colored.

• The inner transition metals are the ones in the lanthanide and actinide series. Theyhave partly filled or filled f orbitals.

• Nonmetals are found in the right portion of the periodic table.• They are brittle in solid state and exhibit no metallic luster; they are

poor conductors of heat and electricity.• Group 7A elements are known as halogens. Group 8A elements are

known as noble or inert gases.• Metalloids, or semimetals, are found along a “ladder” extending

from boron to astatine.• They have characteristics intermediate between those of metals

and nonmetals.

CHEMICAL BONDING

Types of Bonding

Ionic Bonding

occurs between metals and nonmetals; transfer of electrons

Covalent Bonding

occurs between nonmetals; sharing of electrons

polar covalent bonds

nonpolar covalent bonds

Metallic Bonding

occurs in metals (sea of electrons)

9.

FOUNDATION REVIEW

• Occurs between metals and nonmetals.• Atoms have large electronegativity differences which

means that the nonmetals hold onto electrons and pullelectrons away from other atoms and metals give awayelectrons. As a result, a transfer of electrons occurs.

• Formed from the interactions of cations and anions• Example: Na + Cl → Na+Cl– • Occurs between nonmetals

• Covalent bonding involves sharing of electrons. • The number of electron pairs shared between two atoms

determines the bond order of the covalent bond. Singlebonds are the longest and weakest; triple bonds are thestrongest and the shortest.

• Nonpolar covalent bonds: This type ofbonding occurs between atoms thathave the same electronegativity. Thebonding electrons are shared equally.These bonds occur primarily in diatomicmolecules: H2, N2, O2, F2, Cl2, Br2, I2.

• The valence orbitals of constituent atoms interact to formmolecular orbitals that are delocalized over the entire metalliccrystal. Valence electrons therefore can move freely through thecrystal, forming what can be thought of as a “sea of electrons.”The mobility of electrons is what makes metals good conductors.

• This type of bonding occurs between atoms with different electronegativities. Thebonding pair is not shared equally, but pulled more towards the element with thehigher electronegativity. As a result, the more electronegative atom acquires apartial negative charge , δ–, and the less electronegative atom acquires a partialpositive charge, δ+, giving the molecule partially ionic character:

δ+ δ–H–Cl

Intermolecular Forces (in order of decreasing strength)

Ion-Dipole Forces

occur between an ion and a polar molecule

Hydrogen Bonding

specific type of dipole-dipole interaction; occurs between molecules that have hydrogen bound to ahighly electronegative atom such as fluorine, oxygen, or nitrogen

Dipole-Dipole Forces

occur between polar molecules

Dispersion Forces (London Forces)

interactions between transient dipoles; predominant intermolecular attraction among nonpolarmolecules

10.

MCAT: GENERAL CHEMISTRY• Most liquids (except those that are molten

forms of ionic solids) are held together byintermolecular attractions that are gener-ally weaker than ionic and covalent interac-tions. These attractive forces areelectrostatic in nature, and affect theatoms or molecules even though they areneutral because of the asymmetric distribu-tion of charge density.

• These are the interactions between an ion and a polarmolecule. The most common context in which they occur isin the solvation of ionic compounds. Sodium chloride disso-ciates into positive sodium ions and negative chloride ionswhen put in water. Each ion is surrounded by, or solvated by,the polar water molecules.

• When hydrogen is bonded to a highly electronegative atom such as fluorine, oxygen, ornitrogen, the hydrogen atom carries little of the electron density of the covalent bond,most of which is shifted over to the electronegative atom. This positively charged hydro-gen atom interacts with the partial negative charge located on the electronegativeatoms of nearby molecules, causing the two molecules to experience an attraction foreach other. Substances which display hydrogen bonding tend to have unusually high boil-ing points compared with compounds of similar molecular formula that do not partici-pate in hydrogen bonding. Hydrogen bonding is particularly important in the behavior ofwater, alcohols, amines, and carboxylic acids.

• Polar molecules tend to orient themselves such that the positive region of one moleculeis close to the negative region of another molecule. This arrangement is energeticallyfavorable because of the electrostatic attraction between unlike charges. The magnitudeof this kind of interaction increases with increasing polarity of the molecules.

• weakest of all intermolecular forces, their effect is evident in nonpolar molecules,for which no other type of intermolecular attractive forces operate.

• These are short-lived electrostatic interactions arising from random fluctuationsin electron density.

MOLECULAR STRUCTURE

Compounda pure substance that is composed of two or more elements in a fixed proportion

Moleculea combination of two or more atoms held together by covalent bonds

Molecular Weightthe sum of the weights of the atoms that make up the molecule

What is the molecular weight of SOCl2?

Molar Mass

How many moles are in 9.52 g of MgCl2?

11.

FOUNDATION REVIEW

• A compound is a pure substance that is composed of two or more elementsin a fixed proportion. Compounds can be broken down chemically to producetheir constituent elements or other compounds.

• Compounds may be either ionic or covalent.

• A molecule is a combination of two or more atoms held together by covalent bonds. It is the smallest unitof a compound that displays the properties of that compound. Molecules may contain two atoms of thesame element, as in N2 and O2, or may be comprised of two or more different atoms, as in CO2 and SOCl2.

• Ionic compounds are not considered to be made up of molecules because they are vast lattices of ions.

A: 1 S = 1 × 32 amu = 32 amu1O = 1 × 16 amu = 16 amu2Cl = 2 × 35.5 amu = 71 amumolecular weight = 32 + 16 + 71 = 119 amu

• Just as the mass in amu of an atom and the mass in g of 1 mol of atoms arenumerically equivalent, the same is true for molecules.

• The mass of 1 mole of a substance is known as its molar mass, and for covalentcompounds, it is numerically equivalent to the mass of one molecule in amu. SOCl2,for example, has a molar mass of 119 g/mol.

A: (1) Find the molar mass of MgCl2 .1(24.3 g/mol) + 2(35.5 g/mol) = 95.3 g/mol.(2) Solve for the number of moles.

�959.3.5

g2/mg

ol� = 0.10 mol of MgCl2

Empirical and Molecular Formulas

A compound with the empirical formula CH2O has a weight of 180 g/mol. What is themolecular formula?

12.


• The empirical formula gives the simplest whole number ratio of the elementsin the compound. The molecule N2O4 has an empirical formula of NO2, whichcaptures the ratio between the two types of atoms: for every nitrogen atomthere are two oxygen atoms.

• The empirical formula describes a formula unit.• The molecular formula gives the exact number of atoms of each element in a

molecule of the compound, and is a multiple of the empirical formula(including a multiple of 1; i.e. same as the empirical formula).

• If given the empirical formula, we can determine the correct molecular for-mula if we also know the molecular weight.

A: (1) Find the formula weight using the empirical formula:1 × mass of carbon atom + 2 × mass of hydrogen atom + 1 × mass of oxy-gen atom = (1 × 12 + 2 × 1 + 1 × 16) g/mol = 30 g/mol

(2) The actual molecular weight is 6 times this; therefore the molecularformula must be 6 times the empirical formula: C6H12O6.

Percent Composition

The percent composition by mass of an element is the percentage by mass contributed by each elementin a compound.

% composition = × 100%

What is the percent composition of chromium in K2Cr2O7?

What is the empirical formula of a compound which contains 40.9% carbon, 4.58%hydrogen, 54.52% oxygen, and has a molecular weight of 264 g/mol?

mass of X in formula��formula weight of compound

13.

FOUNDATION REVIEW

• The percent composition by mass of anelement is the percentage by mass con-tributed by each element in a compound.To determine the percent composition ofan element X in a compound, the followingformula is used:

• The percent composition of an element maybe determined using either the empirical ormolecular formula.

A: The formula weight of K2Cr2O7 is:2(39 g/mol) + 2(52 g/mol) + 7(16 g/mol) = 294 g/mol

% composition of Cr = �22×9

54

2� × 100% = 35.4 %

A: (1) In problems like this one, we start by assuming that we have a sample thatweighs 100 g total. This allows us to work with concrete weights rather than themore abstract percentages. In this sample, there are 40.9 g of carbon, 4.58 g ofhydrogen, and 54.52 g of oxygen.

(2) Next, we convert grams to moles by dividing the weight of each element by itsmolar atomic mass:

# mol C = �124g0/.m9

ol� = 3.41 mol

# mol H = �14

g.5/8mo

gl� = 4.58 mol

# mol O = �1564g.5/m2

ogl� = 3.41 mol

(3) Find the simplest whole number ratio of the elements by dividing the numberof moles by the smallest number obtained in the previous step.

C: �33..44

11� = 1 H: �43

..5481� = 1.33 O: �33

.

.44

11� = 1

(4) Convert the numbers obtained into whole numbers (multiplying them by aninteger value). In this case, we want to turn 1.33 into an integer; the smallestnumber we can multiply it by to make it an integer is 3: 1.33 × 3 = 4

The empirical formula is therefore 3 × C1H1.33O1 = C3H4O3. Note that we do notneed to use the molecular weight at all if we are only interested in the empiricalformula. If we are asked to find the molecular formula, then we would need to usethe molecular weight.

Lewis Dot Structures

Octet Rule

states that an atom will form bonds in order to have 8 valence electrons (“stable octet”)

can be violated by hydrogen, boron, beryllium, and the elements from period 3 on down

When drawing Lewis dot structures, generally the least electronegative element assumes the central position.

14.


• Lewis structures are used to depict covalent compounds and ions. The “octet rule” statesthat an atom will form bonds in order to have 8 valence electrons.

• This rule can be violated. Hydrogen, for example, can only hold two valence electrons. Ele-ments which typically form incomplete octets include beryllium, as in BeCl2, and boron, as inBF3. Elements in period three and beyond can hold more than eight valence electrons.

• Steps to follow in drawing a Lewis Dot Structure:

(1) Write the skeletal structure of the compound (in general, the least electronegative atomis the central atom). Hydrogen and the halogens (F, Cl, Br, and I) usually occupy the endposition: H C N

(2) Add the valence electrons of the atoms: HCN has a total of 10 valance electrons.

(3) Draw single bonds between the central atom and the atoms surrounding it: H-C-N

(4) Complete the octets of all atoms around the central atom:

H:C:N:

(5) Place any extra electrons on the central atom. If the central atom has less than anoctet, try to write double or triple bonds between the central and surrounding atoms usingthe nonbonding, unshared lone electron pairs: H—C N

::

Formal Charge

used as an aid in determining alternative Lewis structures—the structure with the smallest formalcharge is the preferred one

determined using Lewis dot structures and the formula:

Draw the Lewis dot structure for methylchloride (CH3Cl) and calculate the formal chargeof carbon.

Resonance

Resonance Structures

Lewis structures reveal that, in some molecules, electrons can be shared among the atoms in several ways.

The true structure is a hybrid of the component Lewis structures.

Delocalization

Resonance hybridization diffuses the charge of a polyatomic ion, stabilizing the structure.

S OO S OO S OO

15.

FOUNDATION REVIEW

• The formal charge of all the atoms in HCN is zero. In general, wewant a Lewis structure that minimizes the formal charges on atoms.

A: We place the carbon atom in the center. It has four atoms around it: threehydrogen and one chlorine. A single bond between carbon and each ofthese will complete the octet of carbon. The remaining electrons are lonepairs on chlorine.

Carbon has eight bonding electrons. It has a valence of 4 and so its formalcharge is zero.

H C

H

Cl

H

• For some molecules, two or more identical Lewis structures can be drawn; these arecalled resonance structures. Resonance structures differ only in the way that elec-trons are distributed; the arrangement of the actual atoms doesn’t change.

• For example, we can draw three equivalent structures for SO2.

(Note that we have not explicitly included the formal charges.)• The true structure is a hybrid of the component Lewis structures.

This often means that electrons are delocalized. (Example: benzene)

• Delocalization of charge leads to the stabilization of a molecule or an ion.

The Valence Shell Electron-Pair Repulsion Theory (VSEPR Theory)

Definition

The three-dimensional arrangement of atoms surrounding a central atom is determined by therepulsion between the bonding and the nonbonding electron pairs in the valence shell of the central atom.

Number ofelectron pairs Example Geometric

Arrangement# of nonbonded electron pairs Shape Angles between

electron pairs

16.


• The theory states that the three-dimensional arrangement of atoms surrounding a central atomis determined by the repulsions between the pairs in the valence shell of the central atom.

• These electron pairs arrange themselves as far apart as possible, minimizing repulsion.

17.


GASES

Units

Pressure

1 atm = 760 mmHg = 760 torr

Volume

expressed in liters (L) or milliliters (mL)

Temperature

T(K) = T(°C) + 273.15

Standard conditions = 298 K (25°C) and 1 atm STP = 273.15 K (0°C) and 1 atm

Ideal Gas Laws

Boyle’s Law

For a given gaseous sample held at a constant temperature (isothermal conditions), the volume of thegas is inversely proportional to its pressure:

P1V1 = P2V2

• The pressure of a gas is the force per unit area that the atoms ormolecules exert on the walls of the container through collision. TheSI unit for pressure is the pascal (Pa), which is equal to 1 newton permeter squared. In chemistry, however, gas pressures are more com-monly expressed in units of atmospheres (atm) or millimeters ofmercury (mmHg or torr), which are related as follows: 1 atm = 760mmHg = 760 torr

• The temperature of a gas is usually given in Kelvin, and its value, also known as theabsolute temperature, is related to the temperature in degrees Celsius by theexpression T(K) = T(¡C) + 273.15.

• Gases are often discussed in terms of standard temperature and pressure (STP),which refer to conditions of 273.15 K (0¡C) and 1 atm.

• To simplify our description of how gases behave, we usually assume that the fol-lowing conditions hold: (1) The actual particles of a gas take up no volume. (2)There are no intermolecular attractions between gas particles. (3) The particlesare in random motion, colliding elastically with one another and with the walls ofthe container. (4) The average kinetic energy of the gas particles is proportionalto the absolute temperature, and is the same for all gases at a given tempera-ture. These are the assumptions of the kinetic molecular theory of gases.

Charles’s Law

For a given gaseous sample held at a constant pressure (isobaric conditions), the volume of a gas isdirectly proportional to its absolute temperature:

=

Avogadro’s Law

For a given gaseous sample held at a constant pressure and temperature, the volume of a gas is directlyproportional to the number of moles of gas present:

=

Ideal Gas Equation

PV = nRT

n2�V2

n1�V1

V2�T2

V1�T1

18.

FOUNDATION REVIEW

• The ideal gas equation contains all the informationof the other three laws.

Dalton’s Law

Mole Fraction

The mole fraction of a gas A in a mixture of gases is defined as:

XA =

Partial Pressure

The partial pressure of a gas is related to its mole fraction and can be determined using the following equation:

PA = PTXA

Dalton’s Law states that the total pressure of a gas mixture is the sum of the pressures of the component gases:

PT = PA + PB + PC +. . .

A vessel contains 0.75 mol of nitrogen, 0.20 mol of hydrogen, and 0.05 mol of fluorine at atotal pressure of 2.5 atm. What is the partial pressure of each gas?

number of moles of A��total number of moles of gases

19.


• When we have a mixture of two or more gases in the same container, eachexerts its own pressure known as partial pressure. Dalton’s law is a quantitativeexpression of the partial pressures of ideal gas mixtures.

• The partial pressure of a gas is related to its molefraction and can be determined using the followingequation:

PA = PT XAwhere PT is the total pressure of the system.

• The total pressure of a gaseous mixture isequal to the sum of the partial pressuresof the individual components:

PT = PA + PB + PC +...

A: (1) First calculate the mole fraction of each gas.

(2) Then calculate the partial pressure:

PA = XAPTPN2 = (2.5 atm)(0.75) = 1.875 atmPH2 = (2.5 atm)(0.20) = 0.50 atmPF2 = (2.5 atm)(0.05) = 0.125 atm

• Each gas behaves independently of the other(s). Therefore, the pressure exerted byeach gas in the mixture will be equal to the pressure that gas would exert if it werethe only one in the container.

X N 2

= 0.75 mol1.00mol

= 0. 75, X H 2= 0.20 mol

1.00mol= 0. 20, X F2

= 0.05 mol1.00mol

= 0.05

GENERAL CHEMISTRY 2

Foundation Review

STOICHIOMETRY, KINETICS AND EQUILIBRIUM

Balancing Equations

Balance the following reaction.

C4H10 (l) + O2 (g) → CO2 (g) + H2O (l)

How many grams of calcium chloride are needed to prepare 72 g of silver chloride accordingto the following equation?

CaCl2 (aq) + 2AgNO3 (aq) → Ca(NO3)2 (aq) + 2AgCl (s)

22.

FOUNDATION REVIEW

• Chemical equations express how much and what type of reactants must be usedto obtain a given quantity of product. From the law of conservation of mass, themass of the reactants in a reaction must be equal to the mass of the products.More specifically, chemical equations must be balanced so that there are the samenumber of atoms of each element in the products as there are in the reactants.

• When balancing an equation, the important thing to realize is that we can only change the number infront of the compound, not the subscripts ( this would change the nature or the identity of the com-pound, and hence the reaction itself).

A. First, balance the carbons: C4H10 + O2 → 4 CO2 + H2O

Second, balance the hydrogens : C4H10 + O2 → 4 CO2 + 5 H2O

Third, balance the oxygens (the element that appears in thegreatest number of species): 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

A: We can verify that the equation is balanced. Then, we need to fig-ure out how many moles of AgCl we want. The molar mass of AgClis 144 g. Therefore:

# mol AgCl = �14472

g/gmol� = 0.5 mol

Based on the equation, for every mole of CaCl2, 2 moles of AgClare formed: Therefore, we can set up a proportion:

=

Therefore, we know that we need 0.25 mol CaCl2. The mass ofCaCl2 needed is 0.25 mol × 110 g/mol = 27.5 g.

x mol CaCl2��0.5 mol AgCl1 mol CaCl2��2 mol AgCl

Limiting Reactant

Given the following unbalanced reaction:

AgNO3 + MgCl2 → AgCl + Mg(NO3)2

If 340g of AgNO3 and 285g of MgCl2 are allowed to react, how many grams of AgCl areproduced?

23.


• When reactants are mixed, they are seldom added in the exact stoichiometricproportions as shown in the balanced equation. Therefore, in most reactions,one of the reactants will be used up first. This reactant is known as the lim-iting reactant (or limiting reagent) because it limits the amount of productthat can be formed in the reaction.

• The reactant that remains after all of the limiting reactant is used up iscalled the excess reactant.

A: We need to first balance the equation. Upon inspection, we notice that there are twonitrate ions on the right hand side, but only one on the left. Similarly, there are two chlo-ride ions on the left, but only one on the right. We can solve both these problems andbalance the equation by putting a 2 in front of the silver nitrate on the left and in frontof silver chloride on the right.

2AgNO3 + MgCl2 → 2AgCl + Mg(NO3)2

Next, we determine how many moles of each reactant we have by dividing the mass ofeach by its molar mass. The molar mass of silver nitrate is 108 + 14 + 16 × 3 = 170g/mol. The molar mass of magnesium chloride is 24.3 + 35.5 × 2 = 95.3 g/mol. Therefore,

mol AgNO3 = �1730

4g0/m

gol� = 2 mol

mol MgCl2 = �952.385

g/gmol� = 3 mol

Since two moles of silver nitrate are needed to react with every mole of magnesium chlo-ride, magnesium chloride is the limiting reagent. Only one mole will react before all thesilver nitrate is used up. There will be two moles of magnesium chloride left over.One mole of magnesium chloride will yield two moles of silver chloride, which has a molarmass of 143.5 g/mol. We will therefore end up with 143.5 × 2 = 287 g of silver chloride.

• Note that these two steps add up to the overall (net) reaction. N2O4, whichdoes not appear in the overall reaction, is called an intermediate.

• The slowest step in a proposed mechanism is called the rate-determiningstep (or the rate-limiting step). It is the bottleneck of the reaction, and

determines the rate by imposing an upper limiton how fast it goes.

Reaction MechanismsThe mechanism of a reaction is the actual series of steps through which a chemical reaction occurs.

Consider the following reaction:

The reaction occurs in two steps:

The slowest, or rate-determining step, determines the overall reaction rate.

NO2 + NO2 N2O4

N2O4 + CO CO2 ++ NO2NO

rate limiting

NO2 + CO + CO2NO

24.

FOUNDATION REVIEW

• This equation seems to imply some sort of encounterbetween a molecule of nitrogen dioxide and a moleculeof carbon monoxide. However, this is not necessarilytrue. The equation just gives us the overall outcome. Infact, experiment suggests that the reaction actuallyoccurs in two steps:

Reaction Rates

Rate Law

For nearly all forward, irreversible reactions, the rate is proportional to the product of theconcentrations of the reactants of the rate determining step.

For the general reaction: a A + b B → c C + d D, the rate law is

rate = k [A]x [B]y,

where k is known as the rate constant and x and y are the orders of reaction.

The orders of the reaction are usually determined experimentally.

Reaction Orders

the sum of the orders of the reaction (x + y)

Given the data below, find the rate law for the following reaction at 300K.

A + B → C + D

Trial [A]initial(M) [B]initial(M) rinitial(M/sec)

1 1.00 1.00 2.0

2 1.00 2.00 8.1

3 2.00 2.00 15.9

25.


• This expression is the rate law for thegeneral reaction above, where k is knownas the rate constant, and is differentfor different reactions and may alsochange depending on the reaction con-ditions. The exponents x and y arecalled the orders of reaction; x is theorder of the reaction with respect to Aand y is the order with respect to B.[A] and [B] are the concentrations ofspecies A and B respectively, and are reported in units of molar, �mli

otelers

�.

• The reaction order generally needs to bedetermined experimentally.

• The overall order of a reaction (or the reac-tion order) is defined as the sum of theexponents, here equal to x + y.

A: a) In trials 1 and 2, the concentration of A is kept constant while the concen-tration of B is doubled. The rate increases by a factor of �2

8..01

�, approximately 4.Therefore, the order with respect to B is 2.

b) In trials 2 and 3, the concentration of B is kept constant while the concen-

tration of A is doubled; the rate is increased by a factor of �185.

.91�

, approxi-

mately 2. Therefore, the order with respect to A is 1.

c) So r = k[A] [B]2, i.e. the order of the reaction with respect to A is 1 andwith respect to B is 2; the overall reaction order is 1 + 2 = 3.

Factors Affecting Reaction Rate

Reactant Concentration

Temperature

Solvent

Catalysts (enzymes)

Catalysts increase the rate of chemical reactions by lowering the activation energy.

26.

FOUNDATION REVIEW

• The greater the concentration of reactants (the more particles perunit volume), the greater will be the number of effective collisions perunit time, and therefore, the reaction rate will generally increase.

• For zero order reactions, however, the reaction rate is not dependenton the concentration of reactants. Increasing the reactant concen-tration will have no effect on the rate.

• The reaction rate will increase as the temperature of the systemincreases. As the temperature increases, the reactant moleculeshave more energy. They thus find it easier to climb the energy barrierto the reaction (the activation energy).

• The reaction rate will increase as the temperature of the systemincreases. As the temperature increases, the reactant moleculeshave more energy. They thus find it easier to climb the energy barrierto the reaction (the activation energy).

• Catalysts increase the rate of chemical reactions by lowering theactivation energy.

• Catalysts are not used up in a chemical reaction (they are recycled).

EquilibriumAt equilibrium (a dynamic state), the forward and reverse reaction rates are equal, but the molarconcentrations of the reactants and products are usually not equal.

Equilibrium Expression

ratio of product to reactant concentrations. Pure liquids and solids do not appear in the expression.

Le Châtelier’s Principle

If a stress is applied to a system at equilibrium, the system will shift in such away as to relieve the applied stress.

aA + +bB cC dD Rf = Rr

27.


• Pure solids and liquids do not appear in the equilib-rium constant expression.

• Keq is a characteristic of a given system at a giventemperature

• If the value of Keq is very large compared to 1, thenat equilibrium, there will much more of the prod-ucts around.

• If a stress is applied to a system at equilibrium, the system will shift in such away as to relieve the applied stress.

• Increasing the concentration of a species will tend to shift the equilibrium awayfrom the species that is added (to use it up).

• An increase in the pressure of a system will shift the equilibrium so as todecrease the number of moles of gas present.

• If the temperature was increased, the endothermic reaction will shift to the rightand the exothermic reaction will shift to the left. If the temperature isdecreased, the endothermic reaction will shift to the left to produce more heat toraise the temperature, and the exothermic reaction will shift to the right, to pro-duce more heat to raise the temperature.

• The equilibrium constant for dissolution is known as the solu-bility product constant, Ksp.

• For the compound iron (III) hydroxide, Fe(OH)3, for example, thereaction is

Fe(OH)3 (s) Fe3+ (aq) + 3OH– (aq)Keq = Ksp = [Fe3+ ][OH–]3 (concentrationsat saturation level)• Notice that the solid compound does

not appear in the expression because it is a pure solid. The higherthe Ksp, the harder it is to reach saturation (the more compoundwill dissolve).

• Like other equilibrium constants, the solubility product constant depends on temperature. Generally,for a particular compound, it increases with temperature; i.e., more of the compound dissolves athigher temperatures.

• Another important quantity is the molar solubility, the maximum number of moles of the compoundthat will dissolve in 1 L of solution. The molar solubility and Ksp are related.

Solubility

Saturated Solutions

a dynamic equilibrium between dissolved and solid solute

Solubility Product Constant (Ksp)

the greater the Ksp, the more soluble the solute in that solvent

Keq = Ksp = [Na+]sat[Cl–]sat

The solubility of Fe(OH)3 in an aqueous solution was determined to be 4.5 × 10 –10 mol/L.What is the value of the Ksp for Fe(OH)3?

28.

FOUNDATION REVIEW

• When we add an ionic compound to a water, it usually dissolves by dissociatinginto its component ions. (Recall ion-dipole forces.) As we add more and more of thecompound, however, we eventually reach a point where the solid no longer dissolves,but instead settles down at the bottom. We say that the solution is saturated.

• When a solution is saturated, a dynamic equilibrium exists between the dissolvedsubstance (the solute) and the solid.

A: If the solubility of the compound is 4.5 × 10–10 mol/L, then atsaturation, the concentration of Fe3+ ions is 4.5 × 10–10 mol/L.The concentration of hydroxide ions, however, is three times asmuch, since for every mole of compound that dissolves, threemoles of hydroxide ions are released.

Ksp = [Fe3+][OH–]3 = [Fe3+] (3[Fe3+])3 = 27[Fe3+]4

Ksp = 27(4.5 × 10 –10)4

Ksp = 1.1 × 10–36

THERMOCHEMISTRY

First Law of Thermodynamics conservation of energy

Heat

Conduction

Radiation

Convection

29.


• The first law of thermodynamics is a statement of conservation of energy.• The energy of a system can be changed either by heat transfer, or by doing

work. ∆E = q – w where q is the heat absorbed by the system from its sur-roundings, and w is the work done by the system on its surroundings. (Amore modern convention is to define w as the work done ON the system, inwhich case the equation is →E = q + w.)

• Heat is energy transfer that occurs as a result of a temperature difference between thesystem and its surroundings; this transfer will occur spontaneously from a warmer sys-tem to a cooler system.

• Heat, being an exchange of energy, is measured in the same units of energy, e.g., calories(cal) or joules (J), although kcal (kilocalorie, equals 1000 cal) or kJ (kilojoule, or 1000 J)is often more convenient. The conversion between calories and joules is done via the rela-tion: 1 cal = 4.184 J; similarly, 1 kcal = 4.184 kJ.

• Conduction is a process in which heat is transferred from one particle to another throughcollisions.

• Metals are good conductors of heat.

• Radiation is the transfer of energy by electromagnetic waves. This typeof heat transfer does not require a material medium; i.e. it can occur ina vacuum. This is how the sun warms the earth.

• Convection refers to the transfer of heat by the bulk movement offluids. For example, warm air rises and cool air falls because of thedifference in density.

State Functions

Entropy (S)

measure of the disorder, or randomness, of a system

Enthalpy (H)

measure of the “heat content” of a system

Exothermic Reaction Endothermic Reaction

Reactants

Products

Pote

ntia

l Ene

rgy

Reaction Coordinate

—∆H Reactants

Products

Pote

ntia

l Ene

rgy

Reaction Coordinate

+∆H

30.

FOUNDATION REVIEW

• State functions are those whose value depends only on the position of the sys-tem, and not on how the system got there. For example, drawing a more physi-cal analogy, elevation is a state function, but distance traveled is not.

• Units of entropy are �temepneerragtyure�, commonly J/K or cal/K. Often, we will specify J/K·mol.

• The Second Law of Thermodynamics states that all spontaneous processesproceeding in an isolated system lead to an increase in entropy.

• The Third Law of Thermodynamics states that the absolute entropy of a purecrystalline substance at absolute zero is zero: this corresponds to a state of“perfect order” because all the atoms in this hypothetical state possess nokinetic energy and do not vibrate at all, thus there is absolutely no random-ness and no disorder in the spatial arrangement of the atoms.

• In terms of phases of matter: gas > liquid > solid in entropy. (The dissolutionof particles in solution increases the entropy.)

• Entropy increases with temperature.

• The change in enthalpy (∆H) of a system is equal tothe heat absorbed or evolved by the system at con-stant pressure.

• A positive ∆H (an increase in enthalpy) corresponds toan endothermic process (absorbs heat), and a nega-tive ∆H (a decrease in enthalpy) corresponds to anexothermic process (releases heat).

Standard Heat of Formation

The enthalpy of formation of a compound, ∆Hf° , is the enthalpy change that would occur if one mole ofthat compound were formed directly from its elements in their standard states.

Standard Heat of Reaction

The standard heat of a reaction, ∆H°rxn, is the hypothetical enthalpy change that would occur if thereaction were carried out under standard conditions.

∆H°rxn = (sum of ∆Hf° of products) – (sum of ∆Hf° of reactants)

Substance ∆H°f (kJ/mol)NH3(g) –46.19NO(g) 90.37H2O(g) –241.8

What is the standard enthalpy change for the following reaction?

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

31.


• The standard enthalpy (or heat) of formation of a compound , ∆H°f , is the enthalpy change thatwould occur if one mole of the compound were formed directly from its constituent elements in theirstandard states. For example, the heat of formation of water is the enthalpy change of the reactionbetween 1 mole of diatomic hydrogen and half a mole of diatomic oxygen.

• The ∆H°f of an element in its standard state is zero.

• The standard heat of a reaction, ∆H°rxn, is the hypothetical enthalpy changethat would occur if the reaction were carried out under standard conditions;i.e. when reactants in their standard states are converted to products understandard conditions (1 atm pressure, and usually 298 K).

A: Sum of ∆H°f of products = (4)(90.37) + (6)(–241.8) = –1089 kJSum of ∆H°f of reactants = (4)(–46.19) + (5)(0) = –185 kJ(–1089) – (–185) = –905 kJ (exothermic)

Hess’s Law

It states that if a reaction can be broken down into a series of steps, the enthalpy change for the overallnet reaction is just the sum of the enthalpies of each step.

Given the following thermochemical equations:

a) C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (l) ∆Ha = –2220.1 kJ

b) C (graphite) + O2 (g) → CO2 (g) ∆Hb = –393.5 kJ

c) H2 (g) + 1/2 O2 (g) → H2O (l) ∆Hc = –285.8 kJ

Calculate ∆H for the reaction:

3 C (graphite) + 4 H2 (g) → C3H8 (g)

Gibbs Free Energy(G)

∆G = ∆H – T ∆S; describes overall spontaneity of a reaction

∆H ∆S Outcome– + spontaneous at all temperatures

+ – nonspontaneous at all temperatures

+ + spontaneous only at high temperatures

– – spontaneous only at low temperatures

32.

FOUNDATION REVIEW

• This is just a consequence of enthalpy being a state function.

A: 3 × (C + O2 → CO2 ) ∆Hb = –393.5 kJ × 3+ 4 × (H2 + �2

1� O2 → H2O) ∆Hc = –285.8 kJ × 4

+ 1 × (3 CO2 + 4 H2O → C3H8 + 5 O2) ∆Ha = +2220.1 kJ

3 C (graphite) + 4 H2 (g) → C3H8 (g) ∆H = –103.6 kJ

• describes overall spontaneity of a reaction • ∆G = ∆H – T ∆S, where T is the absolute

temperature • If ∆G is negative, the reaction is sponta-

neous; if ∆G is positive, the reaction isnot spontaneous and if ∆G is zero, thesystem is in a state of equilibrium.

• Because the temperature is always posi-tive, i.e., in Kelvins, the effects of thesigns of ∆H and ∆S and the effect oftemperature on spontaneity can be sum-marized as follows:

A: First we need to melt the ice. The heat of fusion corresponds to the energy needed to effect thisphase change. 18 g of H2O corresponds to 1 mol, the heat needed to melt the ice is 6 kJ. Thenext step is to raise the temperature f the water from 0¡C to 100¡C. The heat needed for thisprocess is18 × 4.18 ×100 = 7500J = 7.5 kJ.Finally, weneed to vaporize this one mole of water to steam. The enthalpy change of the process is 40kJ. The total amount of heat needed is therefore 6 + 7.5 + 40 = 53.5 kJ.

• During a phase change, the temperature of the sys-tem stays constant. The thermal energy supplied isused to increase the potential energy of the sub-stance, instead of the kinetic energy, which corre-sponds to a change in temperature.

• q = mL• Breaking

bondsrequires energy. Thusmelting, vaporization,and sublimation areendothermic. Formingbonds releases energy.Thus, freezing, condensa-tion, and deposition areexothermic.

PHASE CHANGES

Specific Heat (c) the amount of heat needed to raise the temperature of one mass unit of a substance by 1 degree Celsius

The amount of heat (q) gained or given off by a substance that changes in temperature is given by the equation:

q = mc∆T

Heat of Transformation (L)The amount of heat required to change the phase of one mass unit of a substance.

The amount of heat (q) gained or given off by a substance that changes phase is given by the equation:

q = mL

Heating Curve

How much heat must be added to change 18 g of ice at 0°C to steam at 100°C? (cH20 = 4.18 J/g•K; ∆HfusH20

= 6 kJ/mol; ∆HvapH20= 40 kJ/mol)

0

25

-25

50

75

100

125

(heat of fusion)

(heat of vaporization)

ice

water

steam

water-iceequilibrium

water-steamequilibrium

∆Hfus

∆Hvap

33.

FOUNDATION REVIEW

Phase Diagram

Triple Point

point at which all 3 phases are in equilibrium

Critical Point

point at which the differences between the properties of the liquid and gas phases disappear

pressure

temperature

triple point

solid liquidgas

fusion/melting

freezing

vaporization

condensation

sublimationdeposition

critical point

34.


• Each substance has its own characteristic phase diagram whichdescribes its physical properties.

• The intersection of the three boundaries is called thetriple point. At this temperature and pressure, uniquefor a given substance, all three phases are in equilibrium.

• Beyond the critical point, the substance exists as asupercritical fluid.

GENERAL CHEMISTRY 3

Foundation Review

SOLUTIONS AND COLLIGATIVE PROPERTIES

Components of Solutions

Solute

Solvent

Expressions of Concentration

Molality (m)

If 10 g of NaOH are dissolved in 500 g of water, what is the molality of the solution?

Molarity (M)

If enough water is added to 11 g of CaCl2 to make 100 mL of solution, what is the molarityof the solution?

Parts per million (ppm)

A concentration of 1 ppm means that every million grams of solution contain 1 g of solute.

DilutionMiVi = MfVf

How many mL of water must be added to 65 mL of a 5.5 M solution of NaOH in order toprepare a 1.2 M NaOH solution?

moles solute��liters of solution

moles solute��kilograms of solvent

37.

FOUNDATION REVIEW

• Normally the component present in the smaller amount.

• The dissolving medium, normally the component present in the larger

A: Formula weight of NaOH = 23 + 16 + 1 = 40 g/mol10 g NaOH = 0.25 mol NaOH500 g = 0.5 kgmolality = 0.25 mol solute / 0.5 kg solvent = 0.50 m

A: Formula weight of CaCl2 =40 + 35.5 × 2 = 11111 g CaCl2 = 0.10 mol CaCl2volume of solution = 100 mL = 0.10 Lmolarity = 0.10 mol / 0.10 L = 1.0 M

• 1 ppm = 1 g for every million g of solution• useful in expressing very low concentrations, esp. in environmental chemistry

• A solution is diluted when solvent is added to a solution of high concentrationto produce a solution of lower concentration. Since the number of moles ofsolute is conserved, we can write:

A: The first step is to find the final volume of thesolution:5.5 M × 0.065 L = 1.2 M × VfVf = 5.5 × �0.

10.265� = 0.3 L = 300 mL

The volume of water that needs to be added is therefore (300 – 65) mL = 235 mL

Electrolytes and ConductivityIn the case of aqueous solutions, electrical conductivity is governed by the presence and concentrationof ions in solution; solutes whose solutions are conductive are called electrolytes.

Colligative PropertiesThe presence of solute particles can make the physical properties of a solution different from those ofthe pure solvent.

depend on the number of dissolved particles in the solution but not on their chemical identity or nature

Vapor-Pressure Depression

Vapor pressure is the pressure of the vapor above a fluid due to evaporated molecules.

The vapor pressure of a liquid decreases with the addition of a nonvolatile solute.

38.


• In the case of aqueous solutions, electrical conductivity is governed by the presence and concentra-tion of ions in solution. The movement of these ions in response to an electric field is what makesup a current. Therefore, pure water does not conduct an electrical current well since the concentra-tions of hydronium and hydroxide ions are very low.

• Solutes whose solutions are conductive are called electrolytes. A solute is considered a strong elec-trolyte if it dissociates completely into its constituent ions. Examples of strong electrolytes includeionic compounds, such as NaCl and KI, and molecular compounds with highly polar covalent bondsthat dissociate into ions when dissolved, such as HCl in water.

• A weak electrolyte, on the other hand, ionizes or hydrolyzes incompletely in aqueous solution andonly some of the solute is present in ionic form. Examples of weak electrolytes include acetic acidand other weak acids, ammonia and other weak bases, and HgCl2.

• Many compounds do not ionize at all in aqueous solution, retaining their molecular structure insolution. These compounds are called nonelectrolytes and include many nonpolar gases and organiccompounds, such as oxygen and sugar.

• The presence of solute particles can make the physical properties of a solution differentfrom those of the pure solvent. The more numerous these solute particles are in solution,the more pronounced the changes on the physical properties. Physical properties thatdepend on the number of dissolved particles in the solution but not on their chemicalidentity or nature are known as colligative properties.

• Assuming that the solute itself is nonvolatile, Psoln = XsolnPosolvent

Boiling-Point Elevation

A liquid boils when its vapor pressure is equal to the ambient pressure.

The boiling point of a liquid increases with the addition of solute.

Freezing-Point Depression

The freezing point of a liquid decreases with the addition of solute.

The addition of ethylene glycol (“antifreeze”) to water in car engines protects them fromoverheating in the summer and freezing up in the winter. How?

Osmotic Pressure (π)

Osmotic pressure is the pressure that would have to be applied to a solution to prevent diffusion of puresolvent through a semipermeable membrane into that solution.

π = MRT

39.

FOUNDATION REVIEW

• A liquid boils when its vapor pressure equals the atmospheric pressure. Since,as we have seen above, the vapor pressure of a solution is lower than that of thepure solvent, a higher temperature will be required before its vapor pressureequals atmospheric pressure. In other words, the boiling point of a solution ishigher than that of the pure solvent.

• Solute particles interfere with the process of crystal formation that occursduring freezing; the solute particles lower the temperature at which themolecules can align themselves into a crystalline structure.

• Freezing-point depression is the principle behind spreading salt on ice: thefreezing point of water is lowered by the presence of the salt, and so the icemelts. Antifreeze (mostly ethylene glycol) also operates by the same principle.

REDUCTION AND OXIDATION REACTIONS (REDOX)

REDOX Reactions

Reduction

gain of electrons (RIG)

Oxidation

loss of electrons (OIL)

Oxidizing Agent

gets reduced; oxidizes something else

Reducing Agent

gets oxidized; reduces something else

40.


• Reduction refers to a reaction in which a species gains electrons.

• Oxidation refers to a reaction in which a species gives up or loses electrons.

• Since electrons can neither be created nor destroyed in normal chemical reactions (as opposed tonuclear reactions which will be discussed in the next chapter), an isolated loss or gain of elec-trons cannot occur; in other words, neither oxidation nor reduction can occur all by itself: theyoccur simultaneously in a redox reaction, resulting in net electron transfer between the species.

• The electrons released during oxidation are taken up in the reduction process. The reducedspecies is called an oxidizing agent because it causes something else (the species giving up theelectrons) to be oxidized.

• Similarly, a reducing agent causes another species to be reduced, andis itself oxidized. This is summarized below:

oxidizing agent reducing agentreduced oxidizedgains electrons loses electrons

Oxidation Numbers

Assign oxidation numbers to the following reaction and determine the oxidizing andreducing agent:

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

Assign oxidation numbers to the following reaction:

SnCl2 + PbCl4 → SnCl4 + PbCl2

41.

FOUNDATION REVIEW

• Rules for assigning oxidation numbers:

(1) The oxidation number of an element in its elemental form is zero (N2 , O2, He, etc.)(2) The oxidation state of a monatomic ion is the same as its charge (e.g., the oxida-tion # of Na+ is +1, Cu2+ = 2; Cl– = –1, Fe3+ = 3).(3) In binary compounds (those with two different elements), the element with greaterelectronegativity is assigned a negative oxidation number equal to its charge in simpleionic compounds of the element (e.g. PCl3, Cl has an oxidation state of –1).(4) The sum of the oxidation numbers equals zero for an electrically neutral compoundand equals the overall charge for an ionic species (e.g. PCl3: oxidation number for Cl is –1and therefore, oxidation must be +3 for P).(5) Periodic trends: Alkali metals ( Group IA) = +1; Alkaline earth metals (Group IIA) =+2; Group 3A = +3; Halogens (Group VIIA) = –1, except when combined with an elementof higher electronegativity (HOCl, Cl = +1); oxygen = –2, with a few exceptions; hydrogenhas an oxidation number of +1 when it is bonded to a more electronegative element(most nonmetals) and of –1 when bonded to a less electronegative element (most met-als, NaH and CaH2).

A:

Zn is oxidized and is therefore the reducing agent. H is reduced and istherefore the oxidizing agent.

A:

Z n(s) + 2H+ (aq) Z n2+(aq) + H2(g)0 +1 +2 0

SnC l2 + PbC l4 SnC l4 + PbC l2+2

–1

–2

+2

–4+4

–1

–4

+4

+4

–1

–2+2

+2–1+4

Balancing Redox Reactions

Fill in the blanks as you balance the following REDOX reaction in acid

Half-reactions

a complex REDOX reaction can be separated into its component reactions; elements other than O andH should be balanced

Balancing Half-reactions

the loss and gain of electrons is indicated where appropriate; charge is then balanced with H+(when inacid) or OH– (when in base); elements O and H are then balanced with H2O

Combining Reactions

each half-reaction is multiplied by an appropriate factor so that the number of electrons gained equalsthose lost; the reactions are then combined

__Mn2+__e- + +__H+ + __H2O+ +__MnO4- __e-+__C2O4

2- __CO2

__Mn2++ +___H+ __H2O+ +__MnO4- __C2O4

2- __CO2

[ ] [ ]__ x x ____Mn2+__e- +__H+ + __H2O+__MnO4- __e-+__CO2__C2O4

2-

__Mn2+__e- +__H+ + __H2O+__MnO4- __e-+__C2O4

2- __CO2

__Mn2+__e- +__H+ + __MnO4-

__Mn2+__e- +__H+ + __H2O+__MnO4-

__e-+__C2O42- __CO2

__Mn2+__e- + __MnO4-

Reduction Oxidation

__e-+__CO2__C2O42-

MnO4- + +C2O4

2- Mn2+ CO2acid

42.


• When balancing equations, we must obey the law of conservation of mass,as well as make sure that the gains and losses of electrons are equal. Inother words, when balancing an equation, both the net charge and numberof atoms must be equal on both sides of the equation. The most commonmethods for balancing redox reactions is the half-reaction method inwhich the equation is separated into two half-reactions: the oxidationpart and the reduction part. Each half-reaction is balanced separatelyand they are then added to give a balanced overall reaction.

• Mn has an oxidation state of +7 on the reactant side and an oxidationstate of +2 on the product side. C has an oxidation state of +3 on thereactant side and an oxidation state of +4 on the product side.

• The half-reactions with masses (other than O and H) balanced are therefore:5e– + MnO4

– → Mn2+

C2O42– → 2CO2 + 2e–

(Note that even though the oxidation state of carbonchanges only by one, there are two carbon atoms in C2O4

2–

and so we need two electrons. Also we have balanced themass of carbon.)

• Charge is not balanced for the reduction half-reaction. Since the reaction occurs in acid, we use H+ to balance the charge.8H+ + 5e– + MnO4

– → Mn2+

The masses of O and H need to be balanced with water.8H+ + 5e– + MnO4

– → Mn2+ + 4H2O• The oxidation half-reaction

is already balanced in chargeand in the mass of O.

16H+ + 5C2O42– + 10e– + 2MnO4

– → 2Mn2+ + 10CO2 + 8 H2O + 10e–

Canceling the electrons on either side, we have 16H+ + 5C2O42– +

2MnO4– → 2Mn2+ + 10CO2 + 8 H2O.

Electrochemistry

Galvanic/Voltaic Cells

REDOX half reactions carried out in separate compartments

Negative ∆G and spontaneous; supply energy and are used to do work

SaltBridge

voltmeter

Cu2+

SO42—

Zn2+

SO42—

Anode

(—)

Cathode

(+)

Cu(s)Zn(s)

43.

FOUNDATION REVIEW

• To combine the two half-reactions, we multiply the reduction half-reaction by2 and the oxidation half-reaction by 5. Only then is the number of electronsgained by one species equal to the number of electrons lost by the other.16H+ + 10e– + 2MnO4

– → 2Mn2+ + 8 H2O5C2O4

2– → 10CO2 + 10e–

• An electrochemical cell is a contained system in which a redox reaction occursin conjunction with the passage of electric current.

• There are two types of electrochemical cells, galvanic cells (also known asvoltaic cells), and electrolytic cells. Both kinds of electrochemical cells containtwo electrodes, which are essentially two pieces of metal, that serve as thesites for the oxidation and reduction half-reactions separately. The electrodeat which oxidation occurs is called the anode, and the electrode where reduc-tion occurs is called the cathode. This is true for both galvanic (voltaic) andelectrolytic cells.

• Negative DG; spontaneous.• supply energy and are used to do work.• Oxidation and reduction reactions take place in different cells called half-cells.• Oxidation occurs at the anode: Zn(s) → Zn2+(aq) + 2e–

• Reduction occurs at the cathode: Cu2+(aq) + 2e– → Cu(s) • Salt bridge: If only a wire were pro-

vided for this electron flow, thereaction would soon cease becausean excess negative charge wouldbuild up in the solution surroundingthe cathode and an excess positivecharge would build up in the solu-tion surrounding the anode. Thischarge gradient is dissipated bythe presence of a salt bridge, whichpermits the exchange of cationsand anions. The salt bridge con-tains an inert electrolyte. At thesame time the anions from the saltbridge (e.g., Cl–) diffuse from thesalt bridge of the cell into the

ZnSO4 solution to balance out the charge of the newly created Zn2+ ions, the cations of thesalt bridge (e.g. K+) flow into the CuSO4 solution to balance out the charge of the SO4

2– ionsleft in solution when the Cu2+ ions deposit as copper metal.

• The common dry cell battery and the lead-acid storage battery found in cars are examples ofgalvanic cells.

Electrolytic Cells

Positive ∆G and nonspontaneous; electrical energy is required to induce the reaction

Electrodes

Oxidation happens at the anode (An Ox) and reduction happens at the cathode (Red Cat).

44.


• Positive ∆G and nonspontaneous• Electrical energy is required to induce the reaction.

• Oxidation at the anode; reduction at the cathode. (An Ox, Red Cat)• Electrons flow through the wire from anode to cathode.

Reduction Potentials

The species in a reaction that will be oxidized or reduced can be determined from the reductionpotential of each species, defined as the tendency of a species to acquire electrons and be reduced. Themore positive the potential, the greater the species’ tendency to be reduced.

The Electromotive Force

Standard reduction potentials are used to calculate the standard electromotive force (EMF or E°cell) of areaction, the difference in potential between two half-cells.

EMF = E°red + E°ox

The standard EMF of a galvanic cell is positive, while the standard EMF of an electrolytic cell is negative.

45.

FOUNDATION REVIEW

• The species in a reaction that will be oxidized or reduced can be determined from the reductionpotential of each species, defined as the tendency of a species to acquire electrons and be reduced.The more positive the potential, the greater the species’ tendency to be reduced. A reduction poten-tial is measured in volts and is defined relative to the standard hydrogen electrode, which is arbitrar-ily given a potential of 0.00 volts. Standard reduction potential (E¡) is measured under standardconditions (25¡ C, a 1M concentration for each ion in the reaction, a partial pressure of 1 atm foreach gas in reaction, and metals in their pure state).

Example:

(1) F2(g) + 2e– → 2F–(aq) E¡red = +2.87 V

F2(g) likes to be reduced (it has a large, positive E¡red), and is therefore a strong oxidizing agent.

(2) Li+(aq) + e– → Li(s) E¡red = –3.05 V

Li+ doesn’t like to undergo reduction (it has a large, negative E¡red). But if we flip the equation andtherefore flip the sign of the E¡ value, the equation becomes spontaneous.

Li(s) → Li+(aq) + e– E¡ox = +3.05

Therefore, Li(s) likes to undergo oxidation and is therefore, a strong reducing agent.

• Standard reduction potentials are used to calculate the standard electromotive force (EMF orE¡cell) of an electrochemical cell, the difference in potential between two half-cells. The EMF isdetermined by adding the standard reduction potential of the reduced species and the stan-dard oxidation potential of the oxidized species. When adding standard potentials, do not mul-tiply by the number of moles oxidized or reduced.

• The standard EMF of a galvanic cell is positive, while the standard EMF of an electrolytic cell is negative.

ACIDS AND BASES

Three Different Definitions

Arrhenius

Arrhenius defined an acid as a species that produces H+ (protons) in an aqueous solution, and a base asa species that produces OH– (hydroxide ions) in an aqueous solution.

Brønsted-Lowry Definition

A Brønsted-Lowry acid is a species that donates protons, while a Brønsted-Lowry base is a species thataccepts protons.

Lewis Definition

Lewis defined an acid as an electron-pair acceptor, and a base as an electron-pair donor.

Conjugate Acid-Base Pairs

Dissociation of Water and pH Scale

Autoionization, Kw

H2O can act as either a proton donor or acceptor and can act as a proton donor and acceptor towarditself (amphoteric).

Keq = Kw = [H3O+][OH–] = 1 x 10–14

H2O + +H2O H3O+ OH-

HNO2(aq) + H2O(l) NO2-(aq) + H3O+(l)

ConjugateAcid

ConjugateBase

ConjugateAcid

ConjugateBase

46.


• A Brønsted-Lowry acid is a species that donates protons, while a Brønsted-Lowry base is a species that accepts protons. For example, NH3 and Cl– areboth Brønsted-Lowry bases because they accept protons. However, they can-not be called Arrhenius bases since in aqueous solution they do not dissoci-ate to form OH–

• Brønsted-Lowry acids and bases always occur in pairs,called conjugate acid-base pairs. The two members of aconjugate pair are related by the transfer of a proton.

• The stronger the acid, the weaker its conjugate base andthe weaker the acid, the stronger the conjugate base.

• The equilibrium constant for this reaction is designatedby Kw and has a value of 1 × 10–14 at 25¡C.

pH

general definition: pX = –log X

pH = –log[H+] = log ( )

Acidic: pH < 7.00; Basic: pH > 7 .00; Neutral: pH = 7.00

Strong Acids and Bases

Strong acids include: HClO4, HNO3, H2SO4, HCl, HBr and HI; Strong bases include: NaOH (sodiumhydroxide), KOH (potassium hydroxide), and other soluble hydroxides of Group IA and IIA metals.

1�[H+]

47.

FOUNDATION REVIEW

• Hydrogen ion or proton concentration, [H+], like concentrations ofother particles, can of course be measured in the familiar units ofmolarity, etc. However, it is more generally measured as pH, where:

• where [H+] is its molarity and the logarithm is of base10. (Log x) is the power to which 10 would be raised toobtain the number x. I.e.:

• If [H+] = 1 × 10–3, then pH = 3.

Acidic Solutions: pH value < 7.00relative excess of H+ ionslemon juice, gastric juice

Basic Solutions: pH value >7.00relative excess of OH– ionsbaking soda, bleach,ammonia

Neutral Solutions: pH value = 7.00;water

• Completely dissociate into their component ions in aqueous solution.For example, when NaOH is added to water, it dissociates completely: NaOH(s) + H2O(l) → Na+(aq) + OH–(aq)Hence, in a 1 M solution of NaOH, complete dissociation gives 1 mole of OH– ions per literof solution.

• Strong acids commonly encountered in the laboratory include HClO4 (perchloric acid), HNO3(nitric acid), H2SO4 (sulfuric acid), and HCl (hydrochloric acid), HBr and HI.

• Commonly encountered strong bases include NaOH (sodium hydroxide), KOH (potassiumhydroxide), and other soluble hydroxides of Group IA and IIA metals.

Weak Acids and Bases

do not ionize completely; weak electrolytes

Ka = equilibrium constant for acid ionization

Kb = equilibrium constant for base ionization

Calculating pH for solutions of weak acids

Complete the following chart:

CH3COOH (aq) H+ (aq) + CH3COO– (aq)

48.


• Weak acids and bases are those that only partially dissociate in aqueous solution. A weak monopro-tic acid, HA, in aqueous solution will achieve the following equilibrium after dissociation: HA (aq) +H2O (l) H3O+ (aq) + A– (aq)

• The equilibrium constant for this reaction is known as the acid dissociation constant, Ka. It is ameasure of the degree to which an acid dissociates.

Ka =

• The weaker the acid, the smaller the Ka. Weak acids have values of Ka that are much smaller than 1.• To calculate [H+] in a 2.0 M aqueous solution of acetic acid, first write the equilibrium reaction:

CH3COOH(aq) H+(aq) + CH3COO–(aq)The expression for the acid dissociation constant is:

Ka = 1.8 × 10–5 = [H+][CH3COO–]��[CH3COOH]

[H3O+][A–]��[HA]

• The concentration of CH3COOH at equilibrium is equal to its initial concen-tration, 2.0 M, less the amount dissociated, x. Likewise [H+] = [CH3COO–] =x, since each molecule of CH3COOH dissociates into one H+ ion and oneCH3COO– ion. Thus, the equation can be rewritten as follows:

Ka = �2x.·0x� – x = 1.8 × 10–5

• We can approximate that 2.0 – x × 2.0 since acetic acid is a weak acid, andonly slightly dissociates in water. This simplifies the calculation of x:

Ka = �2x.2

0�= 1.8 × 10–5

x = 6.0 × 10–3 M = [H+]

2.0

–x

2.0 – x

0

x

x

0

x

x

Titrations

the addition of a solution of known concentration and volume to another solution to determine itsunknown concentration

pH

base added

equivalencepoint

pH

base added

equivalencepoint

49.

FOUNDATION REVIEW

• Titration involves in general the adddition of a solution to determine itsconcentration.

• For an acid-base titration, we add base of precisely known concentrationdrop by drop to an acid solution (or the other way round) until the numberof moles of base added equals the number of moles of acid initially pre-sent. The point at which this occurs is the equivalence point.

• At the equivalence point, then, we can write:VaMa = VbMb

where Va is the initial volume of the acid solution, and Ma is its molaritywhich we are trying to determine. Vb is the volume of base we have added(which we monitor as we add it drop by drop), and Mb is the molarity ofthe base (known precisely beforehand).

• We know that the equivalence point is reached either by monitoring the pHof the mixture as we add the base (i.e. by obtaining a titration curve), or byusing a few drops of an indicator that changes color at a pH around theequivalence point.

• The equivalence point need not occur at pH 7. Titrations between strongacids and strong bases have an equivalence point at pH 7, but a titrationbetween a weak acid and a strong base has an equivalence point at pH > 7.

Gen Chem Review Test Prep

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