Блог Анкара

In This Chapter

^ Connecting chemistry you can’t see to chemistry you can see

^ Knowing how to predict the outcome of common chemical reactions

^ Predicting properties from position on the periodic table

There are a bunch of elements. These elements assemble into a ridiculously large variety of compounds. Then, just to make things more complicated, the compounds react with one another in different ways. To reduce the headache-inducing complexity of chemistry, it helps to know the lay of the land, to be familiar with a few details and themes that pop up over and over. Knowing these patterns helps you to develop a sense for whether certain compounds will react and what products they might generate. Also, sometimes things happen during a chemical reaction that are visible to the naked eye — solid particles or bubbles appear as if from nowhere — and these events give important clues as to the microscopic details of the reaction. Getting a feel for these kinds of things is what some people call "chemical intuition." In this chapter, you gain some chemical intuition so you can make predictions about how reactions will turn out.

Using Your Senses: Qualitative Aspects of Common Reaction Types

Seen from the outside, chemistry can seem weird. Mix two clear solutions and you never know what might happen. Maybe a solid (called a Precipitate) Will form, settling to the bottom of the beaker. Maybe gas will bubble out of the mixture. Maybe the final solution will be green. Maybe nothing will happen. How can you predict? Knowing which compounds you’re mixing is a good start. Armed with that information, you might be able to identify a likely reaction. The sections that follow list some common themes and patterns that might prove helpful on the AP chemistry exam.

Precipitation reactions

Precipitation occurs when an insoluble compound forms during the course of a reaction. Usually, the insoluble product is an ionic compound, formed between metal and nonmetal components. Being able to predict when precipitates form requires you to know which compounds are soluble and which are insoluble. Here are some helpful guidelines to help you sniff out which compounds are likely to precipitate and which are likely to stay in solution.

Precipitation reactions can occur when ionic compounds react, typically in Double displacement Reactions. In double displacement reactions, the cation components of ionic compounds swap anionic partners. So, if you are presented with two ionic reactant compounds,

AB(ag) + CD(a<7) — ? perform the following steps to complete the reaction and predict the products.

1. Swap the anion and cation partners to predict the reaction equation:

AB(aq) + CD(aq) — CB(?) + AD(?)

2. Next, Inspect the products to see if either or both are insoluble. Here are some basic rules for judging whether an ionic compound is soluble.

Compounds that contain the following ions are Soluble:

•NH4+

• Group IA (column 1) cations (that is, alkali metal ions) •NCV

• ClO3-, ClO2-

•CH3COO" (acetate ion, which can also be written as C2H3O2")

Compounds that contain the following ions are Usually soluble: •Cl-, Br- and I – (except When paired with Ag+, Pb2+, and Hg22+) •SO42- (except When paired with Ag+, Pb2+, Hg22+,Sr2+, Ca2+, and Ba2+)

Compounds that contain the following ions are Usually insoluble:

•OH- (that is, hydroxides, Except When paired with Group IA cations, Sr2+, Ca2+, and Ba2+)

•CO32-, PO43-, SO32-, and S2- (except When paired with Group I cations or NH4+)

3. After you determine whether you have a product that is insoluble, Rewrite the completed reaction equation as a Net ionic equation. Net ionic equations list only those components of the reaction that change state during the reaction. Ions in aqueous solution are listed separately, as ions, and not as part of a compound. Ions that appear on both the reactant and product sides (these are called Spectator ions) Are omitted. Thus,

Na2CO3(a<7) + Ba(NO3)2(aq) — ?

Undergoes a double displacement reaction, yielding:

4. Na2CO3(ag) + Ba(NO3)2(aq) — 2NaNO3(?) + BaCO3(?)Once you have the completed reaction and have assigned phases, Write a Total Ionic equation. The solubility rules state that the sodium nitrate pair is soluble, but when barium ion and carbonate ion combine, a solid precipitate forms. The resulting Total Ionic equation is

2Na+(ag) + CO32-(aq) + Ba2+(aq) + 2NO3-(aq) — 2Na+(aq) + 2NO3-(aq) + BaCO3(s)

5. Then, remove the spectator ions (Na+ and NO3-), leaving the Net Ionic equation:

CO32-(a< ) + Ba2+(a< ) — BaCO3(s)

Aside from precipitation reactions, you should be familiar with the basic patterns and common players in other reaction types, including gas-releasing reactions, reactions of hydrides and oxides, combustion reactions, reactions involving colored compounds, and acid-base reactions.

Gas-releasing reactions

Gas-releasing reactions often include decomposition or the reaction of metals/metal hydrides with water.

Gas-releasing decomposition can result from several processes:

Heating: Adding heat to some compounds can cause them to decompose (break apart) into several products, some of which are gases. For example

•CaCO3(s) — CaO(s) + CO2(g)

• 2KNO3(s) — 2KNO2(s) + O2(g)

• 2KClO3(s) — 2KCl(s) + 3O2(g)

•2H2O2(l) — 2H2O(l) + O2(g)

Electrolysis:. Molten ionic compounds can be decomposed when an electric current passes through them, releasing gas in the process:

2Na+(l) + 2Cl-(l) — 2Na(l) + Cl2(g)

Or, shown as two redox half-reactions

2Na+(l) + 2E – — 2Na(l) (reduction)

2Cl-(l) — Cl2(g) + 2e" (oxidation)

Unstable acids: Some weak acids are unstable and can decompose, especially if they aren’t in solution while heat is added.

H2SO3(s) — H2O(l) + SO2(g)

H2CO3(s) — H2O(l) + CO2(g)

Metals and metal hydride reactions

Metals and metal hydrides can release hydrogen gas when they react with water. These reactions result in a basic solution as hydrogen is taken from water leaving excess hydroxide ions:

LiH(s) + H2O(l) — Li+(aq) + OH-(aq) + H2(g) CaH2(s) + 2H2O(l) — Ca2+(aq) + 2OH-(aq) + 2H2(g) 2Na(s) + 2H2O(l) — 2Na+(aq) + 2OH-(aq) + H2(g)

Combustion reactions

Complete combustion occurs when hydrocarbons react with oxygen, forming carbon dioxide and water.

2C2H6(g) + 7O2(g) — 4CO2(g) + 6H2O(g) (ethane combusts) C2H4(g) + 3O2(g) — 2CO2(g) + 2H2O(g) (ethene combusts) 2C2H2(g) + 5O2(g) — 4CO2(g) + 2H2O(g) (ethyne combusts)

Oxide reactions

The pattern of an oxide reaction depends on whether it is a metal oxide or nonmetal oxide that reacts.

Metal oxides often react with water to produce basic solutions.

Li2O(s) + H2O(l) — 2Li+(a< ) + 2OH-(a< ) Nonmetal oxides often react with water to produce acidic solutions.

N2O5(s) + H2O(l) — 2NO3-(a< ) + 2H+(a< )

Colored compounds

Knowing something about which compounds are colored and what colors those compounds are can help you identify an unknown compound.

Salts and oxides that contain transition metals are often colored.

Ni-containing compounds are often green. Mn-containing compounds are often pink/purple. Colored soluble salts

• Fe-containing soluble salts are red and/or brown.

• Cu-containing soluble salts are blue and/or green.

• Co-containing soluble salts are blue. Colored insoluble salts

• Precipitates containing CrO42- (chrmate) are yellow.

• Precipitates containing Cr2O72- (dichromate) are orange.

• Precipitates containing OH – (hydroxide) are white.

• Precipitate of silver chloride (AgCl) is white.

• Precipitate of silver bromide (AgBr) is pale yellow.

Acids and bases

Acid-base reactions are very common and confuse many students. Key to understanding these reactions is being able to identify the acid and the base, and understanding how strong acids and bases behave differently from weak acids and bases. Acids and bases have predictable properties and reactions:

A strong acid is a weak base. A strong base is a weak acid.

The conjugate acid of a weak base is a weak acid. The conjugate base of a weak acid is a weak base.

Strong acids react with strong bases to form neutral salts and water. Strong acids react with weak bases to form acidic solutions. Strong bases react with weak acids to form basic solutions.

Surveying the Table: Predicting

Properties and Reactivity

There are two basic ways to look at the periodic table. You can pick a property and observe the trends in that property as you move left and right or up and down the table. Or, you can pick a row or column of the table and observe changes or the lack of changes in various properties as you move along the members of the column or row.

Reading trends across the table

The key periodic property is electronegativity, the tendency of an atomic nucleus to draw electrons toward itself within a bond. Electronegativity tends to increase to the right and upward within the main body of the periodic table, as shown in Figure 23-1. Electronegativity represents the result of a balance between an unbound atom’s ionization energy (tendency to lose an electron) and its electron affinity, the ability of an unbound atom to attract an electron. Ionization energy is the energy it takes to remove an electron from a gas-phase atom.

Trends in electronegativity, ionization energy (Figure 23-2), and electron affinity underlie other periodic trends, such as atomic radius (Figure 23-3), and metallic character (Figure 23-4). Atomic radius and metallic character are also largely determined by electronegativity, but both follow a pattern that is the mirror image of electronegativity, as each increases downward and to the left within the periodic table.

Increasing Ionization Energy

Figure 23-2-

Lonization energy increases upward and to the right within the periodic table.

IA

Low Energy

_ High Energy

VIIIA

Increasing Size

Figure 23-3-

Atomic radius increases downward and to the left within the periodic table.

IA

1 2 3 4 5 6 7

Large Radii

Small Radii

VIIIA

Increasing Metallic Character

Figure 23-4-

Metalli c c ha r a c te r increases downward and to the left within the periodic table.

IA

Least Metallic

VIIIA

Most Metallic

Finding patterns in columns and rows

The trends described in the previous section help explain why the elements within a vertical group of the periodic table tend to have properties similar to one another, whereas elements across a horizontal period display incremental changes in their properties, with a consistent trend.

Some groups hang together more closely than others. Within the periodic table, the two leftmost and two rightmost groups show the greatest kinship, keeping tightly to the characteristic properties of their respective vertical families. So, you can feel most confident making predictions based on the group similarity of leftmost columns (columns 1 and 2) and two columns on the right (columns 16 and 17). In addition, the elements in group VIIIA (column 18), the noble gases, have closely identifiable properties to one another. Note that two methods are used to identify columns in the periodic table. The older one uses A and B columns, the newer one has been adopted by the International Union of Pure and Applied Chemistry (IUPAC) and the American Chemical Society. Both will be used here.

Alkali metals (Group IA or 1)

• Elemental alkali metals are extremely reactive, and their cations are extremely unreactive.

• Alkali metals are the least electronegative, easily giving up electrons to form cations.

• In compounds, alkali metals have an oxidation number of +1.

• Though listed with Group IA, Hydrogen is not a metal And has some very different properties:

• In compounds with nonmetals, hydrogen assumes an oxidation number of +1.

• In compounds with metals, hydrogen assumes "hydride" form, with an oxidation number of -1.

Alkaline earth metals (Group IIA or 2)

• Alkaline earth metals are very reactive, and their cations are very unreactive.

• Alkaline earth metals have very low electronegativity.

• In compounds, alkaline earth metals have an oxidation number of +2. VIA or 16

• These elements are reactive because they have six electrons in their valence shell, not the stable, eight-electron valence shell of a noble gas.

• Ions of this group tend to have -2 charge. Halogens (Group VIIA or 17)

• Halogens are very electronegative, with the most electronegative at the top.

• In compounds, halogens usually have an oxidation number of -1.

• Halogens range in reactivity (though none are unreactive), with the most reactive at the top.

• Halogens form ionic compounds with metals and molecular compounds with nonmetals.

Noble gases (Group VIIIA or 18)

• Noble gases are the least reactive of the elements, due to their filled valence shells.

The 5th Wave By Rich Tennant

I Kenny didn’t need ike distraction, but it vras I Just his luck that the Zorlocks’ invasion oЈ "Earth J Viould begin JV3ST as he opened his | !__AP Chemistry booklet. \

In this part. . .

Y the time you make it to this part, chances are you’ve had your fill of Theory. It’s time for Practice. However you feel about chemistry, you definitely want to turn the AP chemistry exam into a showcase for your skill and mastery. When you stride out of the exam hall, you want to be wearing a smile that says, "Nice try, guys, but I was ready for you." For that tasty moment to occur, you need practice. This part gives you that practice. Herein, you’ll find two full-length practice tests. In case you don’t know Every Answer, we’ve got answer keys with complete explanations. Earn your test-day smile here.

In This Chapter

^ Finding your way around atoms and the periodic table

^ Figuring out ions and electron configuration

^ Illuminating the info you need on isotopes

^ Getting up to speed on decay

^ Understanding the excited state of electrons

For hundreds of years, scientists have operated under the idea that all matter is made up of smaller building blocks called Atoms. So small, in fact, that until the invention of the electron microscope, the only way to find out anything about these tiny, mysterious particles was to design a very clever experiment. These experiments allowed chemists to get around the fact that they couldn’t exactly corner a single atom in a back alley somewhere and study it alone — they had to study the properties of whole gangs of atoms and try to guess what individual ones might be like. Through some remarkable cleverness and some incredibly lucky shots in the dark, chemists now understand a great deal about the atom, and guess what? You need to, too! Because we don’t want to leave you hanging, in this chapter we introduce you to everything you need to know about atoms, their properties, and what they are made of.

Picking Apart Atoms and the Periodic Table

The following sections begin by giving you the basics on how atoms are structured so that you can not only get to know their components, but also, you can easily figure out how atoms help to classify elements. Even better, we give you the ultimate road map for the periodic table, showing you how the periodic table can give you some valuable info that helps you predict certain properties of elements.

Getting a gander at what makes up an atom

For now, picture an atom as a microscopic building block. Atoms come in all shapes and sizes, and you can build larger structures out of them. However, like building blocks, atoms are extremely hard to break. In fact, there is so much energy stored inside an atom that breaking one in half is the process that drives a nuclear explosion.

All substances consist of the 120 unique varieties of atoms, each of which is made up of a combination of three types of Subatomic particles:

Protons: Protons have equal and opposite charges to electrons and have very nearly the same mass as neutrons.

Electrons: Electrons have equal and opposite charges to protons, and electrons are much lighter than protons and neutrons.

^ Neutrons: Neutrons are neutral and have the same mass as protons. We summarize the must-know information about the three subatomic particles in Table 3-1.

Atoms always have an equal number of protons and electrons, which makes them overall electrically neutral. Many atoms, however, actually prefer to have an unbalanced number of protons and electrons, which leaves them with an overall charge. We discuss these charged atoms, called Ions, Even further in the section "Exercising Electrons: Ions and Electron Configuration," later in this chapter.

The atom can still safely be called the smallest possible unit of an element because after you break an atom of an element into its subatomic particles, it loses the basic properties that make that element unique.

So what does all of this mean for the structure of an atom? What does an atom actually look like? It took scientists a very long time to figure it out through clever experimentation and tricky math, and over time a succession of models grew closer and closer to an accurate description:

^ The Thompson model, Also called the "Plum Pudding" model, pictured discrete, negatively charged electrons evenly distributed through a positively charged medium that composed the rest of the atom. The electrons were like plums in a positive pudding.

Iir The Rutherford model Modified the Thompson model by making clear that most of the volume of the atom is empty space, with a large amount of charge concentrated at the center of the atom.

^ The Bohr model Built on the Rutherford model by describing the compact, central charge as a nucleus composed of distinct proton and neutron particles. The positive charge of the nucleus derived from the protons. Bohr envisioned electrons as discrete particles that orbited the nucleus along distinct paths, like planets in orbit around the sun.

I The Quantum Mechanical model Modified the Bohr model, pointing out that electrons do not orbit the nucleus like planets around the sun. Instead, they occupy their orbitals in a cloudlike manner; one can only describe their location in terms of probability, with some "dense" regions having a very high probability of having an electron and other regions having lower probability.

Classifying elements with mass and atomic numbers

Two very important numbers associated with an atom are the Atomic number And the Mass number. Chemists tend to memorize these numbers like sports fans memorize baseball stats, but clever chemistry students do not need to resort to memorization when they have the all-important periodic table at their disposal. Here are the basics about atomic numbers and mass numbers:

I The atomic number is the number of protons in the nucleus of an atom. Atomic numbers identify elements, because the number of protons is what gives an element its unique identity. Changing the number of protons (and therefore the atomic number) changes the identity of the element. Atomic numbers are listed as a subscript (on the bottom) to the left side of an element’s chemical symbol.

Notice that the atoms of the periodic table are lined up according to their atomic number. Atomic numbers increase by one each time you move to the right in the periodic table, continuing their march, row by row, from the top of the table to the bottom.

I The mass number is the sum of the protons and neutrons in the nucleus of an atom. Subtracting the atomic number from the mass number gives you the number of neutrons in the nucleus of an atom. Different atoms of the same element can have different numbers of neutrons, and we call those different atoms Isotopes. Mass numbers are listed as a superscript (on the top) to the left side of an element’s chemical symbol.

The mass number gives the mass of an individual atom in atomic mass units, amu, where 1 amu = 1.66 x 10-27 kg.

Why, you may wonder, don’t we care about electrons? Remember that an electron has only Hs36 of the mass of a proton or neutron. So, though electrons are incredibly important from the standpoint of charge, they make little contribution to mass. Atomic masses in amu are never whole numbers for a variety of reasons, including the fact that most elements exist in nature as a combination of different isotopes. The percent of each isotope relative to the whole is called its Natural abundance. The nonwhole number atomic masses you see listed on the periodic table are the average of all isotopes of a given element, in which each isotope is weighted by its natural abundance within the average.

In order to specify the atomic and mass numbers of an element, chemists will typically write that element in the form

AX

Where Z is the atomic number, A is the mass number, and X is the chemical symbol for that element.

Grouping elements within the periodic table

The whole purpose of the periodic table, aside from providing interior decoration for chemistry classrooms, is to organize elements to help predict and explain their properties. Notice in Figure 3-1 that the atoms of the periodic table are lined up according to their atomic numbers, as if their teacher has just taken a roll call to make sure they are all in their proper places. Here are some characterisitics of how the periodic table is organized:

I Atomic numbers increase by one each time you move to the right in the periodic table and when they run out of space in one row, they move down one and continue increasing to the right again.

I The rows are called Periods.

The columns are called Groups.

The elements within any group have very similar properties. The properties of the elements emerge mostly from their different numbers of protons and electrons, and from the arrangement of their electrons. As you move across any period, you pass over a series of elements whose properties change in a predictable way. Here are the groups you can find on the periodic table:

I Group IA (alkali metals): Starting on the far left of the periodic table is group IA

(notice the group label atop each column), also known as the alkali metals. These elements, the most metallic of all elements, are very reactive, and are never found naturally in a pure state, but always in a bonded state.

Group IIA (alkaline earth metals): Just like the alkali metals, the alkaline earth metals are highly reactive, and not found free in nature.

I Group B. . . mostly (transition metals): The large central block of the periodic table is occupied by the transition metals. Transition metals have properties that vary from extremely metallic, at the left side, to far less metallic, on the right side. The rightmost boundary of the transition metals is shaped like a staircase, shown in bold in the figure, and the staircase separates the elements further:

• Metals: Elements to the left of the staircase (except hydrogen) are metals. Metals tend to be solid, shiny, malleable (easily shaped), and ductile (easily drawn out into wire). Metals also conduct electricity and heat and easily give up electrons.

• Nonmetals: Elements to the right of the staircase are nonmetals, which have properties opposite to those of metals: They also tend to be solids, but are generally dull, hard, and do not conduct electricity.

• Metalloids Or Semi-metals: Elements bordering the staircase are called metalloids or semi-metals because they have properties between metals and nonmetals: They are somewhat ductile and a little shiny and can conduct electricity, but not as well as a metal.

I Group VIIIA (noble gases): The most extreme nonmetals are the noble gases which are inert, or extremely unreactive.

Group VIIA (halogens): One column to the left of the noble gases, is another key family of nonmetals, the halogens. In nature, the reactive halogens tend to bond with metals to form salts, like sodium chloride, NaCl.

Be aware that some periodic tables use a different system for labeling the groups, in which each column is simply numbered from 1 to 18, left to right.

Rummaging through reactivity in the periodic table

The properties of elements change as a function of the numbers of protons and electrons in the element. Increasing numbers of protons increases the positive charge of the nucleus, which contributes to Electronegativity, The pull exerted by the nucleus of one atom on electrons in a bond with a second atom. Increasing numbers of electrons changes the reactivity of the element in predictable ways, based on how those electrons fill up successive energy levels.

DpfREft The outermost electrons (in the highest energy level, which is discussed later) are called

Valence electrons. Valence electrons are the ones most important in determining whether an element is reactive or unreactive and are the electrons involved in bonding. Because much of chemistry is about the making and breaking of bonds, valence electrons are the most important particles for chemistry. The details of how electrons fill up the various energy levels are covered later in this chapter, in the section "Exercising Electrons: Ions and Electron Configuration." For now, understand that atoms are most stable when their valence shells (or clouds) are completely filled with electrons.

Chemistry happens as a result of atoms attempting to fill the valence shells. Alkali metals are especially reactive because they need only give up one electron to have a completely filled valence shell. Halogens are especially reactive because they need only acquire one electron to have a completely filled valence shell. The elements within a group tend to have the same number of valence electrons, and for that reason tend to have similar chemical properties. Elements in the A groups have the same number of valence electrons as the roman numeral of their group. For example, magnesium in Group IIA has two valence electrons.

Rounding up the atomic radius in the periodic table

In addition to reactivity, another property that varies across the table is atomic radius, or the geometric size (not the mass) of the atoms. As you move down a column or to the right along a row on the periodic table, elements have both more protons and more electrons. As you move down the table the added electrons occupy discretely higher energy levels. So, atomic radius tends to decrease as you move to the right because the increasing positive charge of the nuclei pulls inwardly on the electrons of that energy level. As you move down the table, atomic radius tends to increase because, even though you are adding positively charged protons to the nucleus, you are now adding electrons to higher and higher energy levels, which are farther and farther from the nucleus and therefore feel a much lower attraction to the nucleus.

PERIODIC TABLE OF THE ELEMENTS

1

IA

18 0

§ Note: Elements 113, 115, and 117 are not known at this time, but are included in the table to show their expected positions.

Exercising Electrons: Ions and Electron Configuration

The outermost shell of an element, which is the highest energy level, is the Valence shell. Elements are so insistent about having filled valence shells that they’ll actively gain or lose valence electrons to do so. Atoms that gain or lose electrons in this way are called Ions. When atoms become ions, they lose the one-to-one balance between their protons and electrons, and therefore acquire an overall charge.

IU Atoms that lose electrons (like metals) acquire positive charge, becoming Cations, Such as Na+ or Mg2+. You can remember this with the simple phrase "cats have paws (pos +)," so cations are positively charged.

I Atoms that gain electrons (like nonmetals) acquire negative charge, becoming Anions, Such as Cl – or O2-.

In the following sections we show you how to figure out which elements gain or lose electrons as well as how to diagram and write out electron configurations.

Knowing which elements gain or lose electrons

You can discover what kind of ion many elements will form simply by looking at their position on the periodic table. With the exception of row one (hydrogen and helium), all elements are most stable with a full shell of eight valence electrons, known as an Octet. Atoms tend to take the shortest path to a complete octet, whether that means ditching a few electrons to achieve a full octet at a lower energy level, or grabbing extra electrons to complete the octet at their current energy level. In general, metals tend to lose electrons, and nonmetals tend to gain electrons. However, a partner atom is usually required for these processes to happen.

You can predict just how many electrons an atom will gain or lose to become an ion (but things do get unpredictable in the transition metal region and then get more predictable with the nonmetals). Elements tend to lose or gain as many electrons as necessary to have valence shells resembling the elements in Group VIIIA, the noble gases:

Group IA (alkali metal) elements lose one electron. I Group IIA (alkaline earth metal) elements lose two electrons. I Group VIIA (halogen) elements gain one electron. I Group VIA elements tend to gain two electrons. I Group VA elements tend to gain three electrons.

Figuring out electron configuration

An awful lot of detail goes into determining just how many electrons an atom has and just which energy levels those electrons occupy. Several different schemes exist for annotating all of this important information, but the Electron configuration Is a type of shorthand that captures much of the pertinent information. Each electron gets a symbol code that indicates the type of energy and shape (called an electron orbital) that it will have in the atom.

Here’s how electron configuration works:

Each numbered row of the periodic table corresponds to a different Principal energy level, With higher numbers indicating higher energy. Remember higher energy means farther from the nucleus.

Within each energy level, electrons can occupy different Subshells (or clouds). Different types of subshells have slightly different energy levels.

Each subshell has a certain number of Orbitals.

• Each orbital can hold up to two electrons, but electrons won’t double up within an orbital unless no other open orbitals exists within the same subshell.

• Electrons fill up orbitals from the lowest energies to the highest, to try to keep everything in the lowest energy state, like everything else in chemistry.

The four types of subshells or clouds, named S, p, d, And f fill with electrons in the following ways:

Row 1 consists of a single 1s subshell. A single electron in this subshell corresponds to the electron configuration of hydrogen, written as 1s1. The superscript written after the symbol for the subshell indicates how many electrons occupy that subshell. Filling the subshell with two electrons, 1s2, corresponds to the electron configuration of helium. Each higher principal energy level contains its own S Subshell (2s, 3s, and so on), and these subshells are the first to fill within those levels.

In addition to S Subshells, principal energy levels 2 and higher contain P Subshells. There are three P Orbitals at each level, accommodating a maximum of six electrons. Because the three P Orbitals (also known as Px, py, And pz) have equal energy in an isolated atom they are each filled with a single electron before any receives a second electron. The elements in rows 2 and 3 on the periodic table contain only S And P Subshells. The P Subshells of each energy level are filled only after the S Subshell is filled.

Rows 4 and higher on the periodic table include D Orbitals, of which there are five at each principal energy level, accommodating a maximum of 10 electrons. D Subshell electrons are a major feature of the transition metals. These are also equal in energy unless the atom is involved in bonding.

Rows 5 and higher include F Orbitals, numbering seven at each level, accommodating a maximum of 14 electrons. F Subshell electrons are a hallmark of the Lanthanides And the Actinides, The two rows of elements detached from the rest of the periodic table, and set aside at the bottom.

After you get to row 4, the exact order in which you fill the energy levels can get a bit confusing. To keep things straight, it’s useful to refer to the Aufbau filling diagram, shown in Figure 3-2. Start at the bottom of the diagram and work your way up from the lowest arrows to the highest. For example, always start by filling 1 S, Then fill 2S, Then 2P, Then 3S, Then 4S, Then 3D, And so on.

Figure 3-2:

The Aufbau filling diagram.

Noting exceptions to the Aufbau fitting diagram

Sadly, there are a few exceptions to the tidy picture presented by the Aufbau filling diagram. Copper, chromium, and palladium are notable examples. These exceptional electron configurations arise from situations where electrons get transferred from their proper, Aufbau-filled subshells to create half-filled or fully filled sets of D Subshells; these half – and fully filled states are slightly more stable than the states produced by strict Aufbau-based filling.

Two conditions typically lead to exceptional electron configurations:

U Successive orbital energies must lie close together, as is the case with 3d and 4s orbitals, for example.

U Shifting electrons between these energetically similar orbitals must result in a half-filled or fully filled set of identical orbitals, an energetically happy state of affairs.

Here are a few examples:

U Strictly by the rules, chromium should have the following electron configuration: [Ar]3d*4s2.

U Because shifting a single electron from 4s to the energetically similar 3d level half-fills the 3D Set, the actual configuration of chromium is [Ar]3D54S1.

U For similar reasons, the configuration of copper is not the expected [Ar]3d°4s2, but instead is [Ar]3D104S1.

Writing out electron configuration

To come up with a written electron configuration you

1. Determine how many electrons the atom in question actually has.

2. You assign those electrons to subshells, one electron at a time, from the lowest energy subshells to the highest.

In a given type of subshell (like a 2p or 3d subshell, for example) you only place two electrons within the same subshell when there is no other choice. For example, only at oxygen (1s22s22p4) would electrons begin to double up in the 2p subshells.

3. Write out the configuration based on the assignment of electrons.

From lowest subshell to highest, you write the subshell and add the number of electrons assigned to it in superscript, like oxygen, 1 S22S22P4.

For example, suppose you want to find the electron configuration of carbon:

1. Determine the amount of electrons. Carbon has six electrons, the same as the number of its protons as described by its atomic number, and it’s in row 2 of the periodic table.

2. Assign electrons to subshells.

First, the S Subshell of level 1 is filled. Then, the S Subshell of level 2 is filled.

These subshells each accept two electrons, leaving two more with which to fill the P Subshells of level 2.

Each of the remaining electrons would occupy a separate P Subshell.

3. Write out the configuration: 1s22s22p2.

Ions have different electron configurations than their parent atoms because ions are created by gaining or losing electrons. Atoms tend to gain or lose electrons so they can achieve full valence octets, like those of the noble gases. Guess what? Many of the resulting ions have precisely the same electron configurations as those noble gases. So, by forming the Br-anion, bromine achieves the same electron configuration as the noble gas, krypton. This is called being Isoelectronic. You can use this concept to condense electron configurations, which can get a bit long to write. For example, [Ne]3s23p3 is the condensed configuration for phosphorous. The expanded electron configuration for phosphorous is 1s22s22p63s23p3. The condensed form simply means that the atom’s electron configuration is just like that of neon (Ne), with additional electrons filled into subshells 3S And 3P As annotated. Warning: Don’t try to get too creative and write something like [Be]2p4 for oxygen. Only noble gases Can be used for the base element in the condensed form! On the AP test be careful — if the problem asks for the complete formula, the condensed version does not receive credit.

Locating electrons with quantum numbers

Chemists use a set of four numbers called collectively the Quantum numbers To describe the state of any particular electron within an electron configuration of an atom. No two electrons in the same atom can share the same set of quantum numbers. In other words, there is no such thing as identical electron twins. They Must Differ from one another in at least one of their quantum numbers. Note that chemists invented this scheme to describe electrons in an atom. There are not "slots" to fill in the atom.

U The Principal quantum number, Which indicates the principal energy level of an electron, is denoted with the variable N And can be any integer number greater than or equal to 1. Electrons that share the same principal quantum number are said to be in the same Shell.

U The Azimuthal quantum number, L, which denotes the subshell which an electron occupies, has only four values: 0, 1, 2, or 3 corresponding to S, p, d, And F Subshells respectively. Any particular N Value, or Shell, Has subshells ranging from L = 0 to L = n-1. The third shell, for example, has subshells with L Values 0, 1, and 2. In other words, the third shell has S, p, And D Subshells.

U The Magnetic quantum number, Ml, builds off of the azimuthal quantum number and describes the three-dimensional geometry of each subshell. The magnetic quantum number describes the number of Orbitals In each subshell, and each orbital can be occupied by at most two electrons. The magnetic quantum number has values ranging from -l To L For each subshell. In other words, a P Subshell (which has an azimuthal quantum number of L = 1) has ml values of -1, 0, and 1, while a D Subshell with an azimuthal quantum number of L = 2 has mL Values of -2, -1, 0, 1, and 2. This rule means that a P Subshell has three orbitals in which electrons may exist, while a D Subshell has five.

U The spin quantum number, S, Has only two values: —2 and K. Each orbital shares the same N, l, And mL Numbers by definition and, because no two electrons can share the same set of quantum numbers, the spin quantum number must be where the two electrons in an orbital differ. Two electrons can be in one orbital as long as one has a spin of —2 and the other has a spin of K. Counting up the ml and S Quantum numbers leads you directly to the number of total electrons that can fit in each subshell. If an S Orbital has an L Value of 0, for example, this means that it has a single mL Value of 0. Because two electrons (one with each type of spin) can fit in this mL = 0 lobe, an S Orbital will always fit two and only two electrons no matter what its principle quantum number is. Similarly, a P Subshell has three mL Values (-1, 0, and 1) and each of those can have two electrons of opposing spin so a P Subshell can have up to six electrons.

Isolating Info on Isotopes

Isotopes Are atoms of the same element, but they have different atomic masses, which must translate to different numbers of neutrons. If that’s all there is to them, then why do chemists get so titillated by isotopes? A neutron is, after all, a Neutral Particle, so you wouldn’t think adding one would change much about an atom. Indeed, much of the time, adding a neutron does not alter the properties of an element significantly, but merely makes it slightly heavier.

Occasionally, however, adding another sinister little neutron can push an atom over to the dark side of radioactivity. If an atom has just the right number of neutrons, it becomes unstable, or radioactive, which means that it will break down into another element at a predicable rate. These radioactive isotopes are called radioisotopes for short. The more unstable the added neutron makes the atom, the faster it tends to decay into another.

Although the word "radioactivity" conjures up images of three-headed frogs and fish with ten eyes, the truth is that it doesn’t deserve its bad rep. Many radioactive elements are harmless. Some radioactive isotopes are very useful, such as those used in medical imaging.

Regardless of how you feel about radioactivity, we give you some useful calculations related to radioisotopes and istopes that you definitely need to check out in the sections that follow.

Calculating the remainder of a radioisotope

Science and medicine are stuffed with useful, friendly applications for radioisotopes. Many of these applications center on the predictable decay rates of various radioisotopes. These predictable rates are characterized by Half lives. The half life of a radioisotope is simply the amount of time it takes for exactly half of a sample of that isotope to decay into daughter nuclei. For example, if a scientist knows that a sample originally contained 42mg of a certain radioisotope and measures 21mg of that isotope in the sample four days later, then the half life of that radioisotope is four days. The half lives of radioisotopes range from seconds to billions of years.

Radioactive dating Is the process scientists use to date samples based on the amount of radioisotope remaining. The most famous form of radioactive dating is carbon-14 dating, which has been used to date human remains and other organic artifacts. However, radio-isotopes have also been used to date the Earth, the solar system, and even the universe.

Table 3-2 lists some of the more useful radioisotopes, along with their half lives and decay modes, which are described in detail later in the chapter.

Consider the element carbon, for example, which has three naturally occurring isotopes. carbon-12 (or carbon with six protons and six neutrons), written as:

Is boring old run-of-the-mill carbon, and it accounts for 99 percent of all of the carbon out there. carbon-13 (or carbon with six protons and Seven Neutrons), written as:

Is a slightly more exotic, though equally dull, isotope, which makes up most of the remaining 1 percent of carbon atoms. Taking on an extra neutron makes carbon-13 slightly heavier than carbon-12, but does little else to change it. However, even this minor change has some very real scientific consequences. Scientists compare the ratio of carbon-12 to carbon-13 in meteorites to help them determine where that meteorite came from. These ratios have been especially useful for identifying Martian meteorites. Earth is significantly more massive than Mars, and therefore, has a stronger magnetic field, allowing it to hold onto its atmosphere. Mars, on the other hand, is too small to hold onto the upper part of its atmosphere. carbon-12 is lighter than carbon-13, so it floats up to the upper atmosphere of Mars, where the solar wind comes along like the big bad wolf and blows it away. This leaves Mars with a higher percentage of carbon-13 than you would find on Earth. So, minerals on Mars that take carbon from the atmosphere and turn it into rock end up with a little more carbon-13 and a little less carbon-12 than similar minerals on earth. When a meteor from Mars lands on Earth, scientists can verify its origin by testing the ratio of carbon-12 to carbon-13.

Carbon-14, the most exotic and interesting isotope of carbon, has been very important in the process of radioactive dating. Carbon is one of the building blocks of organic matter, including the human body. Only one out of every trillion or so carbon atoms is the carbon-14 (or carbon with six protons and eghtneutrons) radioiso-tope, which looks like

You have many, many trillions of trillions of trillions of carbon atoms in your body, which means that you contain trillions of atoms of radioactive carbon! Now before you go checking the mirror to see if you have sprouted a third eye, rest assured that this radioactive carbon will not harm you in any way. In fact, it is what allows scientists to determine the age of fossils.

As an example, take Matilda the Mammoth, who met her untimely end 4,500 years ago. While Matilda was alive, the carbon in her body was constantly being replenished, so she was always made of about 99 percent carbon-12, 1 percent carbon-13, and 0.0000000001 percent carbon-14. However, when poor Matilda kicked the bucket, the biological processes that were replacing the carbon in her body came to an abrupt end. With their supply of carbon-14 cut off, Matilda’s bones slowly lost their carbon-14 as it broke down through radioactive decay into nitrogen. Paleontologists digging up Matilda 4,500 years later will run straight to their friendly neighborhood chemist, Dr. Isotopian, and ask him to tell them how much carbon-14 is left in Matilda. Because carbon-14 breaks down at a very predictable rate, Dr. Isotopian is able to guess to within a few hundred years exactly how long ago Matilda kicked the bucket.

To calculate the remaining amount of a radioisotope, use the following formula where A0 Is the amount of the radioisotope that existed originally, T Is the amount of time the sample has had to decay and T Is the half life:

A = A0 x(0.5F

Calculating the average mass of an element

That number beneath each element in the periodic table is calculated using percentages such as the carbon abundances mentioned earlier, which allow scientists to give an Average Atomic mass for the isotope. Certain elements, such as chlorine, have several commonly occurring isotopes, so their average atomic masses are rarely close to whole integers. Many elements, however, such as carbon, have one very commonly occurring isotope and several rare isotopes, resulting in an average atomic mass that is very close to the mass of the most common isotope.

To calculate an average atomic mass you

1. Make a list of the masses of each isotope, noting the percentage, in decimal form, at which each isotope is found in nature, a quantity called the Relative abundance Of the isotope.

2. Multiply the mass and the relative abundance together.

3. Add up the results of each calculation in step 2 to get the average atomic mass.

On the AP chemistry exam, you will be expected to know how to calculate this weighted average or to determine at which isotope is the most common based on that weighted average.

The most commonly occurring isotope Often Has a mass number closest to the average atomic mass. Be careful when the average atomic mass differs from the nearest whole number by more than about 0.1, however. Such numbers can lead you to false conclusions regarding the most abundant isotope. Chlorine, for example, has an average atomic mass of 35.5amu. This may lead you to conclude that the isotopes

7 Cl

36 Cl

Both have roughly 50 percent abundances because the average atomic mass is almost exactly halfway between 35 and 36. In reality, however

35 Cl

17

Exists with roughly 75 percent abundance, while

3177 Cl

17

Exists with roughly 25 percent abundance.

Is about as common as a Chemistry teacher without chalk on the butt of his pants.

Running into Radioactivity

Atomic numbers are like name tags, identifying elements by telling you the number of protons in the nucleus of that element. Adding or removing a proton to the nucleus of an atom changes its atomic number (and thus its identity), and atoms are very fond of their identities, so adding or removing protons from an atom is usually impossible.

However, you Can Change atomic numbers:

U Certain large elements can be split in half, and certain small elements can be smashed together in processes called Nuclear fission And Nuclear fusion, Respectively. This splitting or joining of atoms releases a tremendous amount of energy and is not something that you should try at home, even if you do have a nuclear reactor in your basement.

U The only other way for an atom to change its atomic number (and therefore its identity) is to "decay" into something else through a radioactive process, and this happens about as often as chemists remember to brush their hair in the mornings. There are three key ways in which an element can decay, and these are described following.

Many elements in the periodic table have unstable versions called Radioisotopes. These radioisotopes decay into other, generally more stable elements at periodic intervals in a process called Radioactive decay. There are many radioisotopes in existence; however not all radioisotopes are created equal. Radioisotopes break down through three separate decay processes. Yep, you guessed it: We’re gonna tell you all about ‘em.

Alpha decay

The first decay process, called Alpha decay, Involves the emission of an alpha particle by the nucleus of an unstable atom. Emitting an alpha particle, which is nothing more exotic than the nucleus of a helium atom (two protons and two neutrons), causes the atomic number of the daughter nucleus to decrease by two and the mass number to decrease by four. The following pattern shows the decay of the parent nucleus X into a daughter nucleus Y and an alpha particle.

AX — AB-42Y + 2He

Beta decay

The second type of decay, called Beta decay, Comes in three forms:

U Beta plus decay: A proton in the nucleus is converted into a neutron, a positron (e+), and a tiny weakly interacting particle called a neutrino (u), resulting in the atomic number decreasing by one. The mass number, however, Does not change, Even though the actual mass does change very slightly. A proton becoming a neutron, after all, should have absolutely no effect on the mass number because both protons and neutrons are Nucleons, Or particles contained in the nucleus, and converting one to the other does not change the overall number of particles in the nucleus.

Beta plus decay follows the form of the equation below, with the atomic number being decreased by one and the mass number remaining the same.

BX

AY

U Beta minus decay: A neutron is converted into a proton, emitting in this case an electron and an antineutrino (u). Again, the mass number remains the same throughout the decay because the number of nucleons remains the same; however the atomic number Increases By one. In case you haven’t guessed, this is the reverse of beta plus decay.

Beta minus decay follows the form of the equation below, with the atomic number being increased by one and the mass number remaining the same.

AX— *Y+E-+v

U Electron capture: The final form of beta decay occurs when an inner electron is captured by an atomic proton, converting it into a neutron and emitting a neutrino in the process.

Electron capture follows the form of the equation below, with the atomic number being decreased by one and the mass number remaining the same.

BX

AY

B-1

BB9

All three of these processes involve the emission or capture of an electron or a positron and result in a change in the atomic number of the daughter atom.

Gamma decay

The last form of decay, termed Gamma decay, Involves the emission of a high-energy form of light, called a gamma ray, by an excited nucleus. Although it does not change the atomic number or the mass number of the daughter nucleus, gamma decay often accompanies alpha or beta decay and it is a means for the nucleus of a daughter atom to reach its lowest possible energy state. The general form of gamma decay is shown below, where

Represents the excited state of the parent nucleus, and the greek letter gamma (y) represents the gamma ray.

The forces that hold atomic nuclei together are extremely powerful, and the energy required to join two light nuclei together or split one heavy nucleus in two is tremendous. The only place where nuclear reactions happen in nature is in the very center of stars like our sun, where extreme temperatures cause atoms of hydrogen, helium, and other light elements to smash together and join into one. This extremely energetic process, called Nuclear fusion, Is what ultimately causes the sun to shine and provides the outward pressure required to support the sun against gravitational collapse.

Nuclear fission, on the other hand, occurs only rarely in nature. At a site in central Africa called Oklo a natural nuclear reactor has been found to have occurred many, many, years ago. Humans first harnessed the tremendous power released when splitting an atom through fission during the United States’s Manhattan Project, which led to the development of the first atomic bomb. Fission has since been used for more benign purposes in nuclear power plants, which produce energy much more efficiently and cleanly than traditional fossil fuel-burning power plants.

Special circumstances can drive electrons to configurations other than their ideal Ground state (or lowest energy) configurations. You will need to know the causes and consequences of electrons being driven from their ground states and the equations that will allow you to qualitatively analyze those transitions, all of which are laid out for you in this chapter.

Electrons can be driven to higher energy levels than they would otherwise occupy by absorbing energy in the form of heat, light, or electric current. An electron can only absorb an amount of energy exactly equivalent to the difference in energy between the state it is already occupying and another nearby energy state. Electrons in these higher-than-expected energy levels are said to be in an Excited state. Excited electrons quickly release amounts of energy equivalent to the difference between their excited state and a lower energy state by emitting a Photon, Or particle of light, with an equivalent energy. The Wavelength (which in the realm of visible light is equivalent to color) of this photon characterizes the energy difference between the excited and lower energy state. It is only by measuring these that chemists know about the energy structure of atoms. There isn’t any way to learn about the inside of an atom by just looking at a ground state, stable one.

A B

X*

A B

Getting Electrons Excited

Excited state

An electron that should be in the 3s subshell, for example, may be kicked up into the 4f Sub-shell, where it wants to remain just about as badly as you want to remain on the couch during your Great Aunt Maybell’s slideshow of her summer at the lake. In order to get back to the 3s subshell where it belongs, it must emit an amount of energy exactly equal to the energy difference between the 4f And the 3s subshells. This energy difference corresponds to a very specific wavelength, or color, of light because of the relationship between the energy and wavelength of a photon of light.

Converting energies to Wavelengths

Max Planck developed the equation

E = hf-

To show the relationship between the wavelength and the energy of a photon. However, this equation can also be written as E = hu

This alternative equation is based on another convenient relationship between the properties of light, specifically the relationship between wavelength and frequency:

C = Au

Here’s an explanation of the variables in those two equations: Energy (E) Is expressed in Joules (J).

H Is a constant called Planck’s constant equal to 6.63 x 10-34 m2 x kg/s.

Frequency (u) is expressed in reciprocal seconds (1/s or s-1) also known as Hertz (1 Hz = 1 s-1).

C is the speed of light, Specifically 3.0 x 108 m/s. Wavelength (A) is expressed in meters.

Niels Bohr, in order to complete his model of the atom, used Planck’s ideas to explain why the orbits of electrons in the hydrogen atom are Quantized, Or occur only at certain distances from the nucleus. Planck had proposed that the energy of light particles (photons) Came in Quanta, Or discrete values. Bohr found that the energy levels of the electron orbits were also quantized and followed the formula

E n

-2.178 x 10-

Joules

Where n is the principal quantum number. You will be expected to recognize and use this formula for predicting energy levels in the hydrogen atom on the AP exam, but make sure that you note that it applies to the Hydrogen atom alone And will give you an incorrect answer if you apply it to any other element.

Emission spectra

Although it sounds like the next blockbluster sci-fi movie, Planck’s formula E= hU also helps explain the phenomenon of atomic Emission spectra, Which are characterized by bright lines of light of very specific colors. Chemists had realized long before Bohr and Planck came along that if you used a prism or spectrograph to disperse the light emitted when a pure solid element is burned or a pure gaseous element has an electronic current passed through it, then every individual element would emit a characteristic sequence of bright-colored lines called its Spectrum. No two excited elements emitted identical spectra, so spectra became a definitive method for identifying elements.

An element’s characteristic spectrum contains lines of specific colors because every element has its own unique set of allowed energies and therefore its own unique set of energy transitions. Because excited electrons must emit photons in order to reach the ground state and photon energies correspond directly to wavelengths, or colors, the appearance of every atomic spectrum is unique in the color and brightness of each line. The brighter a spectral line is, the more likely that transition is to occur. Most elements have hundreds of allowed transitions and emit light from infrared to X-ray wavelengths. The specific wavelength of light emitted when an electron drops to a lower energy level can be found by taking the energy difference between the excited and ground states and translating it into a wavelength using Planck’s formula and substituting this energy change (A) for E as in

AE = M

A

In This Chapter

^ Trying your hand at 75 multiple-choice questions ^ Taking on six free-response questions

N the AP exam, you will be given a list of formulas and constants. You may use your Cheat Sheet (which mimics the AP formula list) and a periodic table for this exam, but

Do not use your calculator until the section specifically says that you may. Make sure to

Time yourself to get a sense of your pacing.

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Multiple-Choice Questions (90 Minutes)

CALCULATORS MAY NOT BE USED

Questions 1 through 3 refer to the following reaction types:

(A) Combustion

(B) Single Replacement

(C) Double Replacement

(D) Combination

(E) Acid/Base

1. Characteristic products of this reaction type are carbon dioxide and water.

2. The activity series of metals determines whether or not these reactions will take place.

3. Titrations take place by way of this type of reaction.

Questions 4 through 8 refer to the following elements:

(A) Na

(B) Mg

(C) Al

(D) S

(E) Cl

4. Is the heaviest metal

5. Is the most electronegative

6. Has the largest first ionization energy

Has the largest jump between second and third ionization energies

8. Has the largest atomic radius

Questions 9 through 11 refer to the following atomic models:

(A) Thompson Model

(B) Bohr Model

(C) Rutherford Model

(D) Quantum Mechanical Model

(E) Boyle Model

9. This method was developed via the famous gold foil experiment.

10. This model represents our best modern understanding of the actual structure of the atom.

11. Only one type of charge came in particle form in this atomic model.

Questions 12 through 14 refer to the following gas laws:

(A) Boyle’s Law

(B) Charles’ Law

(C) Graham’s Law

(D) The Combined Gas Law

(E) The Ideal Gas Law

12. To use this law, you must assume that temperature is constant.

13. This law can be used to derive many of the other gas laws.

14. This law described the phenomenon of effusion.

15. Which of the following species is not planar?

(A) CO2

(B) Cl2

(C) BF3

(D) CH4

(E) NO3-

16. A compound has a formula X2O. Which of the following species could be X?

(A) Be

(B) Mg

(C) Fe

(D) Li

(E) Sr

17. Zn(s) + 2AgNO3(aq) M 2Ag(s) + Zn(NO3)2(aq)

According to the reaction represented above, about how many grams of Zn must go into this reaction to produce 1.0mol of Ag?

(A) 17g

(B) 25g

(C) 33g

(D) 65g

(E) 130g

18. A 0.25M solution has an [H+] of 4.2 x 10-6. What is its pH?

(A) 5.00

(B) 5.37

(C) 6.00

(D) 6.27

(E) 7.00

19. Which of the following bonds is the most polar?

(A) C-H

(B) C-N

(C) C-O

(D) C-F

(E) N-O

20. Which of the following is a correct representation of the electron configuration for molybdenum?

(A) 1s22s22p63s23p64s23d104p65s24d4

(B) [Ar]5s24d4

(C) [Ar] 5s14d5

(D) Kr] 5s14d5

(E) [Kr] 5s25d4

21. Which of the following Ka Values would be most suitable to buffer a solution with a pOH of 4.30?

(A) 6.4 x 10-3

(B) 8.2 x 10-4

(C) 5.1 x 10-5

(D) 1.8 x 10-9

(E) 2.0 x 10-10

22. Which of the following organic compounds contains 2 pi bonds?

(A) CH3CHCHCHCH3

(B) CH3CH2COOH

(C) CH3CH2CH2CH2OH

(D) CH3CHCHCH2CH3

(E) Benzene

23. Which of the following does NOT favor the formation of products in the reaction below?

C(s) + H2O(g) — CO(g) + H2(g)

(A) Increasing the concentration of H2O

(B) Removing H2 as it is formed

(C) Increasing the pressure

(D) Increasing the volume of the reaction container

(E) Adding a catalyst

24. What are the possible values of the quantum number L For an atom with a principal quantum number of 3?

(A) 0, 1, 2

(B) -2, -1, 0, 1, 2

(C) 0, 1, 2, 3

(D) -3, -2, -1, 0, 1, 2, 3

(E) -34 and >2

25. Which of the following is NOT a property of an ideal gas?

(A) Its molecules occupy no volume.

(B) Its particles do not interact with one another.

(C) Its molecules are in constant random motion.

(D) Collisions between molecules are completely inelastic.

(E) There are no intermolecular forces acting between molecules.

26. Compound A combines with chlorine as ACl2. Which of the following is likely to be its electron configuration?

(A) 1s22s22p63s1

(B) 1s22s22p63s2

(C) 1s22s1

(D) 1s22s22p6

(E) 1s22s22p63s23p1

27. Which of the following substances contains both ionic and covalent bonds?

(A) CH3CH2F

(B) NaCl

(C) BF3

(D) CH3COOH

(E) NH4Cl

Figure 29-1:

Titration

14 13 12 11 10 9 8

^ 7

6 5 4 3 2 1 0

14

13

12

11

10

9

8

7 ^

6 5 4 3 2 1 0

10

20

30

40

50

60

Curve.

0

28. Which of the following matches the titration curve shown in Figure 29-1?

(A) A strong acid is titrated into a weak base.

(B) A strong acid is titrated into a strong base.

(C) A strong base is titrated into a weak acid.

(D) A strong base is titrated into a strong acid.

(E) A weak base is titrated into a weak acid.

29. Miscible liquids are best separated from one another by

(A) filtration.

(B) condensation.

(C) centrifugation.

(D) distillation.

(E) precipitation.

30. HNO2(aq) + OH-(aq) M NO2-(aq) + H2O(l)

In the reaction above, which species acts as the Lewis base?

(A) HNO2

(B) OH-

(C) NO2-

(D) H2O

(E) This is not an acid-base reaction.

31. One mole of a pure hydrocarbon undergoes complete combustion, creating six moles of water and five moles of carbon dioxide. Which of the following gives the empirical formula for that hydrocarbon?

(A) C2H6

(B) C3H8

(C) C4H10

(D) C5H12

(E) C6H14

32. How many valence electrons does phosphorous have?

(A) 3

(B) 4

(C) 5

(D) 6

(E) 7

33. The molecule PCl5 has what type of geometry according to VSEPR theory?

(A) Linear

(B) Trigonal Planar

(C) Trigonal bipyramidal

(D) Tetrahedral

(E) Octahedral

34. 4Ag(s) + O2(g) + 4H+(aq) — 4Ag+(aq) + 2H2O(l)

O2 (g) + 4H+(aq) + 4e – — 2H2O(l) E red = 1.23V

Ag+(aq) + e — Ag(s)

0.80V

What is the EMF of the voltaic cell driven by the above reaction?

(A) -0.43V

(B) 0.43V

(C) 0.80V

(D) 1.23V

(E) 2.03V

35. Which of the following ions will have the largest atomic radius according to periodic table trends?

(A) N3-

(B) O2-

(C) F-

(D) Na+

(E) Mg2+

36. What is the electron configuration of O2-?

(A) 1s22s22p2

(B) 1s22s22p4

(C) 1s22s22p6

(D) 1s22s23p2

(E) 1s22s23p6

37. The net ionic equation for the reaction when solutions of calcium chloride and sodium carbonate are mixed is

(A) CaCl2(aq) + Na2CO3(aq) — CaCO3(s) + 2NaCl(aq).

(B) CaCl2(aq) + Na2CO3(aq) — CaCO3(aq) + 2NaCl(aq).

(C) Ca2+(aq) + 2Cl-(aq) + Na+(aq) + CO32-(aq) — CaCO3(s) + 2Na+(aq) + 2Cl-(aq).

(D) Ca2+(aq) + 2Cl-(aq) + Na+(aq) + CO32-(aq) — Ca2+(aq) +CO32-(aq) + 2Na+(aq) + 2Cl-(aq).

(E) Ca2+(aq) + CO32-(aq) — CaCO3(s).

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38. An oxide of sulfur is found to contain 40% sulfur. What is its identity?

(A) SO

(B) SO2

(C) SO3

(D) S6O2

(E) S7O2

39. Mg(s) + 2HCl(aq) M MgCl2(aq) + H2(g)

According to the reaction above, how many grams of HCl are needed to produce 11.2L of hydrogen gas at STP?

(A) 18.3g

(B) 36.5g

(C) 38.2g

(D) 46.9g

(E) 73.0g

Figure 29-2:

Phase diagram.

Temperature

40. Which of the arrows on the phase diagram in Figure 29-2 represents the process of sublimation?

(A) Arrow 1

(B) Arrow 2

(C) Arrow 3

(D) Arrow 4

(E) Arrow 5

Questions 41 through 42 refer to the following unbalanced half-reaction

MnO4- + H+ — Mn2+

41. How many electrons will appear on the left-hand side in the final, balanced half-reaction?

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6

42. How many moles of water will need to appear on the righthand side of the equation to balance out the H and O on the left-hand side?

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6

43. Compared to light with low energy, high energy light has

(A) a shorter wavelength.

(B) a smaller frequency.

(C) a larger amplitude.

(D) a faster speed.

(E) X-rays.

44. Which of the following salts forms an acidic solution when dissolved in water?

45. Which of the following functional groups is characteristic of an amine?

(A) -CHO

(B) -COOH

(C) -OH

(D) -NH2

(E) -NH3

46. If you dilute 20mL of a 1M NaOH stock solution with 30mL of water, what is the molarity of your diluted solution?

(A) 0.3M

(B) 0.4M

(C) 0.5M

(D) 0.6M

(E) 0.7M

47. The data in the table above were obtained for the reaction A + B — C. Which of the following is the rate law for the reaction?

48. When aqueous barium chloride and potassium sulfate are mixed together, what is the identity of the solid precipitate that forms?

(A) Ba(SO4)2

(B) Ba2SO4

(C) BaSO4

(D) KCl

(E) K2Cl

49. Arrange these species in order of decreasing oxidation number:

Fluoride, hydrogen bonded to a nonmetal, oxygen not in a peroxide, magnesium cation

(A) Magnesium < oxygen< hydrogen < fluoride

(B) Fluoride < magnesium < hydrogen < oxygen

(C) Magnesium < hydrogen < oxygen < fluoride

(D) Magnesium < hydrogen < fluoride < oxygen

(E) Hydrogen < oxygen < fluoride < magnesium

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50. Which of the following polyatomic ions has the greatest amount of negative charge?

(A) Nitrate

(B) Sulfate

(C) Phosphate

(D) Permanganate

(E) Ammonium

51. What formula could be expected for a binary compound of potassium and sulfur?

52. What is the pKa of a substance with a Kbof 1 x 10-6?

(A) 5

(B) 6

(C) 7

(D) 8

(E) 9

53. Li+(aq) + e – M Li(s)E° = -3.04 Zn2+(aq) + 2e – — Zn(s)E = -0.76 F2(g) + 2e – — 2P(aq)E = 2.87

Based on the standard electrode potentials given above, which of the following is the strongest oxidizing agent?

(A) Li+

(B) Li

(C) Zn

(D) F2

(E) F-

54. Which of the following represents a reaction that can occur at the cathode of an electrolytic cell?

(A) Cu2+(aq) + 2e – — Cu(s)

(B) Sn2+(aq) + 2e – — Sn(s)

(C) Au3+(aq) + 3e – — Au(s)

(D) Fe3+(aq) + e – — Fe2+(aq)

(E) Br2(Q + 2e – — 2Br-(aq)

55. Which of the following organic compounds is the most soluble in water?

(A) 2-fluoropentane

(B) Benzene

(C) Ethene

(D) Propanol

(E) Pentane

56. Which of the following pairs of compounds are not a conjugate acid/base pair?

(A) H3O+ and H2O

(B) H2O and OH-

(C) NH3 and NH2-

(D) CH3COOH and CH3COO-

(E) H3PO4 and PO43-

57. The half-life of 125I is about 60 days. A company in the United States has ordered that a sample of this radioisotope be shipped to their lab from a lab in China. The sample is sent by boat on a trip that takes 120 days.

If the lab needs to receive 1.5g of the radioisotope, how much should be sent from the lab in China?

(A) 1.0g

(B) 1.5g

(C) 1.75g

(D) 3.0g

(E) 6.0g

58. Which of the following compounds cannot form hydrogen bonds?

59. What is the theoretical yield of water for a reaction between 6.00g of H2 and 160.g O2?

(A) 53.4g

(B) 98.0g

(C) 108g

(D) 120.g

(E) 134g

60. At constant temperature, if the volume of a 65. gas is halved, its pressure will be

(A) halved.

(B) remain the same.

(C) doubled.

(D) tripled.

(E) quadrupled.

61. The hybridization of the carbon atoms in

Ethene can be described as 66.

(A) sp.

(B) sp2.

(C) sp3.

(D) sp3d.

(E) sp3d2.

62. Which of the following operations should

Have a final answer containing four signifi – 67. cant figures?

(A) 12.2 x 13.51

(B) (62.315)2

(C) 0.023 + 1.311

(D) 1.010 – 11.623

(E) 64.5 * 3.2

63. Which of the following elements is not 68. isoelectronic with the others?

(A) S2-

(B) Cl-

(C) Ar

(D) K+

(E) Mg2+

64. If 149g of KCl is dissolved in 400.g of water,

What is the new freezing point of the water 69. (Kf = 1.86°C/m)

(A) 255K

(B) 282K

(C) -0.930°C

(D) 9.30°C

(E) 109°C

10.0mL of 1.00M NaOH will create a solution of pH = 7 when mixed with which of the following (assume the solution is 1.00M)?

(A) 3.33mL HNO3

(B) 3.33mL CH3COOH

(C) 5.00mL HCl

(D) 5.00mL H2SO4

(E) 30.0mL H3PO4

How many lone electron pairs are in the molecule H-Cl?

(A) one

(B) two

(C) three

(D) four

(E) six

Which of the following substances is insoluble in water?

(A) K2S

(B) AgCl

(C) (NH4)3PO4

(D) Sr(OH)2

(E) Li2SO4

How many moles of nitric acid are in 5.00L of nitric acid solution with a molarity of 0.250?

(A) 0.250

(B) 1.25

(C) 2.50

(D) 4.75

(E) 5.25

An element undergoes alpha decay to form Lead-206. What is the identity of that element?

(A) Mercury-206

(B) Thalium-207

(C) Lead-209

(D) Bismuth-209

(E) Polonium-210

70. A 2.00L container holding oxygen at 1.50atm pressure is joined to a 3.00L container holding nitrogen at 0.500atm. What will

The partial pressure of oxygen be when the gases mix and equilibrate in the new 5.00L volume?

(A) 0.300atm

(B) 0.450atm

(C) 0.600atm

(D) 1.25atm

(E) 2.00atm

71. What is the molarity of a solution of NaOH if 3.00mL of the solution must be titrated with 5.00mL of 0.750M H2SO4 to reach the equivalence point?

72. What is the [H+] of a solution with a pOH

73. A reaction is exothermic. Which of the following must also be true?

(I) The reaction is spontaneous.

(II) AH is negative.

(III) Reactants have lower energy than products.

(A) I only

(B) II only

(C) I and II

(D) I, II, and III

(E) None of the above

74. Which of the following is not a likely decay product for the decay of iodine-131?

(A) Antimony-127

(B) Tellurium-131

(C) Iodine-131

(D) Xenon-131

(E) Cesium-135

75. Polonium-218 undergoes an alpha decay followed by a beta minus decay. What is the end product of this reaction?

(A) Thallium-214

(B) Lead-214

(C) Polonium-218

(D) Bismuth-214

(E) Astatine-218

STOP

DO NOT TURN THE PAGE UNTIL TOLD TO DO SO. DO NOT RETURN TO A PREVIOUS TEST.

Free-Response Questions

Part A (55 minutes)

YOU MAY USE YOUR CALCULATOR

CLEARLY SHOW ALL STEPS YOU TAKE TO ARRIVE AT YOUR ANSWER. It is to your advantage to do this, because you will receive partial credit for partially correct responses. Make sure to pay attention to significant figures.

Answer questions 1, 2, and 3. The score weighting for each question is 20 percent.

C6H5COOH(aq) + H2O(l) — H3O+(aq) + C6H5COO-(aq) Ka = 6.17 x 10-5

1. Benzoic acid acid dissociates in water according to the reaction above.

(a) Write the equilibrium constant expression for the reaction.

(b) Calculate the molar concentration of C6H5COO – in a 0.0100M benzoic acid solution.

(c) Write a balanced neutralization reaction equation for the reaction of benzoic acid and calcium hydroxide.

(d) After adding 10.0mL of 5.00 x 10-6 M Ca(OH)2 to 90.0mL of an unknown concentration of benzoic acid, the pH of the solution is 5.26. Calculate each of the following:

(i) The [H+] of the solution after the addition of Ca(OH)2.

(ii) The [OH-] of the solution after the addition of Ca(OH)2.

(iii) The initial molarity of the benzoic acid solution before the addition of Ca(OH)2.

(e) State whether the solution at the equivalence point of this titration is acidic, basic, or neutral. Explain your reasoning.

2. Reaction A: N2(g) + 3H2(g) — 2NH3(g) AH = -92kJ AS = -198 J/K

Reaction B: N2F4(g) — 2NF2(g) AH = 85kJ AS = 198 J/K

(a) Identify each reaction as exothermic or endothermic. Explain your reasoning.

(b) In each reaction, which side is more ordered: the reactants or the products? Explain your reasoning.

(c) Calculate the Gibbs free energy change for each reaction and identify it as spontaneous or nonspontaneous under standard conditions. Explain your reasoning.

(d) How will the Gibbs free energy for each of these reactions change with increasing temperature?

(e) What is the value of the equilibrium constant for reaction A at 298K?

3. A voltaic cell is constructed based on the redox reaction between cadmium and tin (both to their 2+ ions).

(a) Write the half-reactions that occur and identify them as occurring at the anode or the cathode.

(b) Calculate the value of the standard cell potential.

(c) Sketch and label a diagram of the cell, indicating the anode, cathode, salt bridge, and voltmeter.

(d) Draw an arrow on your diagram showing the direction that electrons flow through the wire.

Write a balanced net ionic equation for the spontaneous cell reaction.

Calculate the value of the standard free energy change for the reaction.

(e)

(f)

DO NOT TURN THE PAGE UNTIL TOLD TO DO SO. DO NOT RETURN TO A PREVIOUS TEST.

Part B (40 minutes)

CALCULATORS MAY NOT BE USED

Answer question 4 below. The score weighting for this question is 10 percent.

4. For each of the following three reactions, in part (i) write a balanced equation for the reaction and in part (ii) answer the question about the reaction. In part (i), coefficients should be in terms of lowest whole numbers. Assume that solutions are aqueous unless otherwise indicated. Represent substances in solution as ions if the substances are extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction.

(a) Solid lead carbonate is added to an aqueous solution of ammonium bromide.

(i) Balanced equation:

(ii) Which product(s) is soluble, if any? Which product(s) is insoluble, if any?

(b) Calcium metal is exposed to oxygen gas.

(i) Balanced reaction:

(ii) What is the oxidation number of oxygen before and after the reaction?

(c) Hexane undergoes complete combustion in the presence of oxygen gas.

(i) Balanced reaction:

(ii) How would the products be different if hexane were combusted in air instead?

Answer questions 5 and 6 below. The score weighting for each question is 15 percent.

5. Answer each of the following questions.

(a) Draw a complete Lewis electron-dot diagram for carbon dioxide and for

PO43-.

(b) On the basis of your Lewis diagram from part (a), identify the hybridization of the central carbon atom in CO2 and the central phosphorous atom in PO43-.

(c) When carbon dioxide dissolves in water, a small fraction (at equilibrium) of the carbon dioxide reacts with water to form carbonic acid. Write out a complete, balanced equation for this reaction and identify the Lewis acid and the Lewis base in the reaction.

(d) Is CO2 polar? Explain.

(e) What is the O-P-O bond angle in PO43-?

6. Use chemical and physical principles to account for each of the following.

(a) An opened soda can becomes "flat" (loses its carbonation) over time.

(b) Most medicines designed to relieve heartburn and calm the stomach contain hydroxide.

(c) A pure elemental gas emits distinct colors of light when viewed through a prism and no two are alike.

(d) The water molecule has a tetrahedral electron pair arrangement but bent geometry.

Answers to Practice Test 1

Multiple choice

1. (A). Carbon dioxide and water are the characteristic products of a combustion reaction.

2. (B). The activity series of metals determines whether or not single replacement reactions will occur.

3. (E). Titrations are used in acid/base reactions.

4. (C). Sodium, magnesium, and aluminum are all metals, but aluminum has the greatest mass.

5. (E). Chlorine is the farthest right on the periodic table and is therefore the most electronegative.

6. (E). Chlorine has the highest first ionization energy because it has the strongest pull on its electrons. (Chlorine wants to gain electrons, not lose them.)

7. (B). All of the alkaline earth metals have extremely large jumps between their second and third ionization energies because after the second ionization, they achieve the electron configuration of a noble gas and are reluctant to relinquish it.

8. (A). Atomic radius increases to the left across a period and down in a family. All of these elements are in the same row so only the first trend applies. Sodium has the largest radius because it is the farthest left.

9. (C). Rutherford discovered the nucleus through his famous gold foil experiment, which led to the development of the Rutherford model of the atom.

10. (D). Our modern understanding of the atom is based on the Quantum Mechanical model.

11. (A). The Thompson model of the atom, often referred to as the "plum pudding model," consisted of negatively charged particles in a diffuse soup of positive charge.

12. (A). Boyle’s law assumes constant temperature.

13. (E). The ideal gas law can be used to derive many of the other gas laws.

14. (C). Graham’s law describes the phenomenon of effusion.

15. (D). CH4 has a tetrahedral geometry and is therefore not planar.

16. (D). Oxygen has a 2- charge as an ion, so it will form a salt in a 1:2 ratio with any alkali metal, of which lithium is the only one on the list.

17. (C). Convert 1mol of Ag to grams of Zn using the mole ratios in the balanced equation to get your answer.

LOOmolAg x 1molZn x^i. = 32.7GZn 1 2molAg 1molZn

18. (B). pH does not depend on compound molarity, only on [H+]. You do not have a calculator to take its logarithm, but you don’t need one because the answers don’t require you to be nit-picky. You know that a solution with an [H+] concentration that has a coefficient greater than 1 will have a pH value between its coefficient (neglecting the negative sign) and its coefficient minus 1. In this case, 5.37 is the only number between 6 and 5.

19. (D). Fluorine is the most electronegative element on the periodic table and therefore pulls most heavily on the electrons of a bound carbon atom. Nitrogen and oxygen, lying so closely together on the periodic table, are quite close in electronegativity and therefore form a nonpolar bond.

20. (D). Molybdenum is one of the transition metals with an exceptional electron configuration. In order to achieve greater stability, it will steal an electron from its 5s orbital in order to obtain a half-full 4d orbital.

21. (E). The best buffers are those whose pKa values are close to the desired pH. You’re given the pOH here, so you need to subtract it from 14 to get a desired pH of 9.7. Here again, you cannot take a logarithm without a calculator but you know that you have a pH with a value between 9 and 10, which means that your desired exponent will be -10. Choice E is the only one that fits the bill.

22. (A). If you draw out the Lewis structures (see Figure 29-3) of the compounds given, you will find that choice A is the only one with two double bonds.

Figure 29-3: H H

A compound .. ‘ n „ n ’ ..

. . . H — C — C = C = C — C — H

Containing I i i i I

2pi bonds. H H H H H

23. (C). Increasing the pressure in the reaction vessel will favor the side of the reaction with fewer moles of gas, which is the reactants side. Adding a catalyst promotes the formation of both products and reactants, so although the equilibrium concentration of product isn’t altered by the catalyst, adding it nonetheless promotes product formation prior to attaining equilibrium.

24. (A). The quantum number l has values ranging from 0 to n-1. Since n is 3 in this case, you’re left with three possible values: 0, 1, and 2.

25. (D). Collisions between the particles of an ideal gas are completely Elastic, Meaning that no energy is lost in the process of colliding.

26. (B). A compound that combines with chlorine in a 1:2 ratio is likely to be an alkaline earth metal, meaning its electron configuration should end with a full s orbital. B is the only choice that fits this criterion, as it is the electron configuration for magnesium.

27. (E). Ammonium is a polyatomic ion held together by covalent bonds, but its bond with chlorine is ionic.

28. (B). This is a steep curve that moves from low pH to high pH, with an equivalence point of 7. It must be a strong acid titrated with a strong base.

29. (D). Distillation is the best process for separating miscible liquids.

30. (B). A Lewis base donates a pair of electrons to a coordinate covalent bond. In this case, that pair of electrons comes from one of the lone pairs on the oxygen of OH-, so it is the Lewis base.

31. (D). The information provided should lead you to the formula C5H12 + 8O2 — 6H2O + 5CO2 for the complete combustion of pentane.

32. (C). Phosphorous has 5 valence electrons.

33. (C). Phosphorous pentachloride has trigonal bipyramidal geometry.

34. (B). You calculate EMF, or E cell, by subtracting the reduction potential of the anode from that of the cathode. Oxidation happens at the anode, so the reaction Ag+(aq) + e – — Ag(s) must be the anode reaction. This leads you to an EMF of 1.23 – 0.80 = 0.43V.

35. (A). All of the ions listed have the same number of electrons, but nitrogen has the smallest number of protons, which means less positive charge to pull in the negatively charged electron cloud.

36. (C). O2- has the regular electron configuration of oxygen plus an additional 2 electrons, giving it the electron configuration of neon.

37. (E). The complete reaction is answer choice A, however the net ionic reaction takes account of dissociation and eliminates spectators, which is answer choice E. CaCO3 is also a carbonate and all carbonates are insoluble except for those containing ammonium or alkali metals so that product must be a solid, which choice E also shows.

38. (C). The easiest way to approach this problem is to use the answer choices and do rough calculations of the percent composition of sulfur in each one. Choice A gives you 32 4 48, which is too large. Choice B is also too large and choices D and E are too small. Choice C, which gives you a percent composition of 32480 = 0.4, is the correct answer.

39. (B). Convert liters of H2 to grams of HCl using the mole ratios in the balanced equation and the conversion 22.4L per mole of gas at STP.

11.2LH2 1molH2 2molHCl 36.5GHCl „„ c rr„,

-i— x, r2 x "„ x -—°! = 36.5GHCl

1 22.4L 1MolH2 1MolHCl

40. (A). Arrow 1 shows a transition from the solid to the gas regions on the phase diagram, which is the process of sublimation.

41. (D). Both questions 41 and 42 refer to the balanced half-reaction. Balance the excess oxygen on the lefthand side by adding 4mol H2O to the right. Then balance the hydrogen on the left with the hydrogen atoms in the water on the right by adding the coefficient 8. Finally, balance charge by adding five electrons to the lefthand side, giving you a total charge of +2 on both sides, leaving you with. MnO4- + 8H+ +5e – — Mn2+ + 4H2O.

42. (C). According to the equation you created to answer question 41, 4mol of water are needed.

43. (A). The higher the energy of light, the shorter its wavelength. Your spectroscopy equation E = hc 4 X can be used to verify this inverse relationship.

44. (E). NH4I is the only solution that will form a strong acid (HI) and a weak base NH4OH when dissolved in water, leaving it with a low pH. All of the others will form both a strong acid and a strong base when they react with water.

45. (D). Primary amines have the functional group – NH2.

46. (B). Use the formula M1V1 = M2V2to Solve for M2. Be careful though! V2 Is the final diluted volume, which in this case is 50mL.

M = 1M x 0.02 L = 0.4M 2 0.05L

47. (E). When comparing experiments 1 and 2, the reaction rate does not change, indicating A is zero order. While comparing experiments 2 and 3, the rate doubles while B doubles indicating a first-order reaction.

48. (C). The balanced double replacement reaction between barium chloride and potassium sulfate is BaCl2(aq) + K2SO4(aq) — BaSO4(s) + 2KCl(aq). You know that the sulfate is the precipitate because barium is one of the elements whose presence makes a sulfate insoluble according to your solubility rules.

49. (D). The magnesium cation has an oxidation number of +2, hydrogen is +1 when bonded to a nonmetal, the fluoride ion is -1, and oxygen is always -2 except in peroxides.

50. (C). This question tests purely your ability to memorize the polyatomic ions and their charges. Phosphate has the greatest with a -3 charge.

51. (B). Potassium forms a +1 ion while sulfur forms a -2 ion, so they will combine in a 2:1 ratio.

52. (D). Use Kw and the given Kb to solve for Ka.

Since the coefficient on KA is 1, the pKA is simply the exponent (neglecting the negative sign).

53. (D). Oxidizing agents have positive values of E.

54. (B). This reaction requires the use of your chart of standard reduction potentials (which you will be given on the exam). Recall that reduction happens at the cathode of an electrolytic cell and choice B is the only answer with a negative reduction potential, so it is the only choice that is a reduction reaction.

55. (D). Remember the concept "like dissolves like." The intermolecular forces in propanol are the most similar to those of water.

56. (E). All of these pairs involve the loss of hydrogen ions, but choice E involves the loss of more than just a single hydrogen, so it is not a conjugate acid/base pair.

57. (E). 120 days amounts to two half-lives of the radioisotope, so they must send 6.0g total.

58. (B). O-H, N-H, and F-H bonds can all participate in hydrogen bonding, but I-H bonds cannot.

59. (A). The balanced reaction here is 2H2 + O2 — 2H2O. Determine the limiting reagent by converting grams of hydrogen to grams of oxygen using the mole ratios from the balanced equation.

6.00GH2 1mol02 16.0G.„„ „

-r-^- x-z-Tft – X -Sr = 48.0 GO 2

1 2molH 2 1molO2

This reveals that H2 is the limiting reagent so now you need to convert grams of H2 to grams of H2O through a similar calculation.

6.00gH2 1molH2 2molH2O 18.0g 1AO, r „ 1 2.02 2molH2 1molH2O ^2

60. (C). According to Boyle’s law, pressure and volume are inversely proportional, so halving one should double the other.

61. (B). Both of the carbon atoms in ethene are sp2 hybridized.

62. (C). Follow the rules for addition and subtraction; the answer contains the same number of decimal places as the least number of decimal places of the measurements used in the calculation. So, the calculation in answer C yields 1.334, with four significant figures.

63. (E). All of the elements listed have the same electron configuration as (are isoelectronic with) argon except for Mg2+.

64. (A). Begin by calculating the molality of the solution by converting grams of KCl to moles and g of solvent to kg:

149GKCl X1 MolKCl X1000 G =5.00M

Notice that KCl has a van’t Hoff factor of 2; in other words, 1 mole of KCl dissolves into 2 moles of particles, K+ and Cl-. So, 5.00 molal KCl contributes a Solute particle molality Of 10.0. Multiply the particle molality by the KF to get a freezing point depression of 18.6 degrees. This means that the new freezing point of the water is 0.00°C-18.6°C = -18.6°C or 255K.

65. (D). 10.0mL of 1.00M NaOH will be neutralized by 10.0mL of a monoprotic acid, 0.500mL of a diprotic acid, or 3.33mL of a triprotic acid. D is the only answer choice that fits.

66. (C). There are three nonbonding pairs of electrons in the molecule HCl, which can be seen by drawing its Lewis dot structure (see Figure 29-4).

Figure 29-4: Lewis dot H — C*! structure. "

67. (B). All chlorides, bromides, and iodides are soluble except those containing Ag, Pb, and Hg.

68. (B). Multiply the molarity by the volume to get the number of moles present. 5.00 x 0.250 = 1.25.

69. (E). When an element undergoes alpha decay, it loses 4 from its mass number and 2 from its atomic number, so the progenitor of Lead-206 must be polonium-210.

70. (B). The question requires you to use Boyle’s Law, P1 V1 = P2V2. You’re asked to find a final pressure, so rearrange the gas law to the form P2 = (Pi Vf) / V2. Substituting in known values gives you P2 = (1.50atm x 2.00L) / 5.00L = 0.600atm.

71. (A). Convert the molarity of acid to the molarity of the base, taking account of equivalents and volumes.

0.750molH2SO, 5 00X10 3L 2molNaOH 1 „>„ .T

-71-x —-:-x– x ——— = 2.50mNaO1

1L 1 1moH2SO4 3.00 x 10 3 L

72. (D). A solution with a pOH of 4.50 has a pH of 9.50, so the exponent on its H+ concentration must be -10.

73. (B). Although most exothermic reactions are spontaneous, all are not. All exothermic reactions have a negative AH and the energy of their products is always lower than that of their reactants.

74. (E). Iodine-131 could undergo alpha decay to create antimony-127, beta plus decay to create tellurium-131, gamma decay to create iodine-131(gamma decay does not change the identity of an atom), or beta minus decay to create xenon-131. Cesium-135 is the odd man out.

75. (D). An alpha decay of polonium-218 will change it into lead-214. Beta minus decay will not change the mass number, but it will increase the atomic number by one leaving bismuth-214.

Free Response

1. (a) The equilibrium constant is constructed by taking the concentration of all of the products raised to the powers of their coefficients divided by the concentrations of all of the reactants (except for water) raised to the power of their coefficients. Because the reaction is an acid dissociation, the Keq in this case is also the Ka.

K = K =[C 6H 5C°° J Br ] Eq A \_CH 3COOH J

(b) 7.85 x 10-4M The given molarity is equivalent to [C6H5COOH]. When C6H5COOH dissociates in water, it creates 1 H3O+ molecule for every C6H5COO – molecule, so those two concentrations are the same, and we can say that they both equal x. The final concentration of C6H5COOH is 0.0100M – x, which we can safely approximate to be 0.0100M, leaving you with the equation

2

KA = 0 0100 = 6.17 x 10 5

Solve the equation for X To get your answer.

(c) 2 C6H5COOH + Ca(OH)2 — 2H2O + Ca(C6H5COO-)2.

(d) This is clearly a titration problem.

(i) 5.50 x 10-6M. To calculate the [H+] you simply need to solve the equation 10-pH = [H+].

(ii) 1.82 x 10-9M You know that [H+] = 5.50 x 10-6M. Further, you know that in aqueous solution, [H+]x[OH-] = 10-14. So, you can substitute in your known value for hydrogen ion concentration and solve for the hydroxide ion concentration.

(iii) 1.66 x 10-6M benzoic acid. The road to this answer is a bit long and winding, and has to do with understanding what happens when you titrate a weak acid with a strong base. Because the base is strong and the acid is weak, essentially all added base titrates away hydrogen ions from the weak acid — initally, at least. This 1-for-1 pattern holds through the "buffer region" of the weak acid, about 1 to 1.5 units above its pKa. The pKa of benzoic acid is – log(6.17 x 10-5) = 4.21, so the final pH of 5.26 is within the buffer region. So, all added base acts to neutralize hydrogen ion from the weak acid. The total moles of base added in this case are:

(5.00 x 10-6Af) x (10.0 x 10-3L) x (2 OH – / Ca(OH)2) = 1.00 x 10-7 moles OH-

Because you are in the buffer region, you know that the original number of moles of weak acid equals the final number of moles Plus The 1.00 x 10-7 moles that have been neutralized. That’s fine, but what are the final moles of weak acid?

To find the final moles of weak acid, use your answers from part (b) and part (d-i). You know the Ka = 6.17 x 10-5. Further, you know the final concentrations of hydrogen ion and benzoate ion, [H+] = [C6H5COO-] = 5.50 x 10-6M. Plug these three known values into the equation for Ka, and you can solve for the final benzoic acid concentration, [C6H5COOH] = 4.90 x 10-7M. Multiply this value by the 0.100L final volume (10.0mL + 90.0mL) to obtain the final moles of benzoic acid, 4.90 x 10-8 moles. Now, add this final amount to the 1.00 x 10-7 moles that were neutralized, and you get the initial moles of benzoic acid, 1.49 x 10-7 moles. These moles were initially present in 90.0mL of solution. Calculate the initial molarity of benzoic acid by dividing 1.49 x 10-7 moles by 0.0900L to obtain 1.66 x 10-6M. Phew.

(e) This is the titration of a weak acid with a strong base, so the solution will be basic at the equivalence point. The reason for this is that reaching the equivalence point involved the formation of a salt, calcium benzoate in this case, or Ca(C6H5COO-)2. The benzoate anion of this salt is slightly basic, so the equivalence point solution will be slightly basic.

2. (a) Reaction A is exothermic because AH/<0 and Reaction B is endothermic because AH/>0.

(b) Reaction A has a negative entropy change (AS), which means that the products are more ordered than the reactants. This is consistent with the fact that Reaction A takes 4 moles of gas and converts it to 2 moles of gas. Reaction B has a positive entropy change, which means the reactants are more ordered than the products. This is consistent with the fact that Reaction B takes 1 mole of gas and converts it to 2 moles of gas. This also makes sense qualitatively because Reaction A is a combination reaction and Reaction B is a decomposition reaction.

(c) AG = -33kJ for Reaction A and AG = 26kJ for Reaction B. The Gibbs free energy change is governed by the equation AG =AH/ – TAS. Here the temperature must be in K, and because they tell us that these reactions take place under standard conditions, we know that the temperature is 25°C or 298K. Plug the given values of AH/ and AS into the equation for each reaction to get your answers. Reaction A is spontaneous because AG <0 and Reaction B is nonspontaneous because AG >0.

(d) As temperature increases in the equation AG = AH/ – TAS, the AG for reaction A will become nonspontaneous, because the negative entropy change makes the term -TASA positive quantity. Reaction B will become spontaneous.

(e) Keq= 1.6 X 10"6.To answer this question, you must use the expression AG = AG + 2.303R71ogQ. At equilibrium, AG=0 and Q Is the equilibrium constant, so the expression becomes Keq = 10(AG/(2.303R7). Plug in your known values, remembering to convert AG into joules (form kilojoules), and you calculate Keq = 1.6 x 10-6.

3. (a) Your two half-reactions must be. Cd2+ + 2e – — Cd and Sn2+ + 2e – — Sn. You must look up their standard reduction potentials to determine which occurs at the anode and which at the cathode. Cd2+ + 2e – — Cd has an E of -0.40 and Sn2+ + 2e – — Sn has an E of -0.14. The E for Sn2+ is the higher of the two so it is reduced at the cathode. This means that Cd must be oxidized at the anode in the reaction Cd — Cd2+ + 2e-.

(b) The standard cell potential is calculated using the equation E cell = E red(cathode) – E red (anode). Plug in the values you looked up in the table to get E cell = -0.14V – (-0.40V) = 0.26V.

(c) Be sure to label the anode, cathode, salt bridge, voltmeter, and the wire connecting the anode and cathode (see Figure 29-5).

Voltaic Cell

Sn(s) (cathode) • electrode

Figure 29-5:

Anode, cathode, salt bridge, voltmeter, and connecting wire.

Voltmeter

Salt Bridge

Sn2+ (aq)

Half-cell 1

E- flow

4*

Switch

Cd(s) (anode) ele c trode

Cd2+ (aq)

Half-cell 2

(d) Electrons flow from anode to cathode.

(e) The net ionic equation for this cell is Sn2+(ag) + Cd(s) — Cd2+(aq) + Sn(s).

(f) Use the expression AG = -nFE", Where N Is the number of moles of electrons exchanged in the net ionic equation (2 in this case) and F Is Faraday’s constant (96,500 C per mol e-). This gives you a AG of -2 x 96,500 x 0.26 = -50.2kJ

4. (a) (i) Your balanced reaction should be NH4Br(aq) + PbCO3(s) — (NH4)2CO3(aq) + PbBr2(s). It is a relatively simple double replacement reaction as long as you have memorized your solubility rules and charges of the polyatomic ions. Eliminating spectators (NH4+), you should be left with Br-(aq) + PbCO3(s) — CO32-(aq) + PbBr2(s).

(ii) Carbonates are generally insoluble except for those containing ammonium and alkali metals so (NH4)2CO3 is soluble. Bromides are generally soluble except for those containing Ag, Pb, or Hg, so PbBr2 is insoluble.

(b) (i) This is a simple combination reaction of the form 2Ca(s) + O2(g) — 2CaO(s).

(ii) Pure elements have oxidation numbers of zero and oxygen has an oxidation number of -2 in the compound because it assumes its usual -2 charge.

(c) (i) The complete combustion of any hydrocarbon in the presence of oxygen alone will lead to the production of water and carbon dioxide. This reaction follows the equation 2C6H14 + 19O2 — 14H2O + 12CO2.

(ii) Nitrogen is also present in air and takes place in a combustion reaction, leading to the addition of N2 among the reactants and NO2 among the products.

5. See Figure 29-6.

(a) The carbon in CO2 is sp hybridized while the phosphorous in PO43- is sp3 hybridized.

(b) The balanced reaction has the form CO2 + H2O — H2CO3. In this reaction, the oxygen on the H2O donates the lone pair to form a coordinate covalent bond with CO2, so it is the Lewis base and CO2 is the Lewis acid.

(c) No. The carbon-oxygen bonds of CO2 are polar because O is more electronegative than C, and so the two oxygen atoms in CO2 pull the electrons in their covalent bonds with carbon closer to them, leaving the oxygens with a slight negative charge and the carbon with a slight positive charge. However, the linear geometry of CO2 opposes these polar bonds such that they cancel out and the molecule as a whole is nonpolar.

(d) PO43- is tetrahedral, so the O-P-O bond angles are 109.5 degrees.

Figure 29-6:

Lewis Structures.

IO = C = O!

I

• P_ •

‘,or • ,o’,

99 • n • • •

6. (a) Carbon dioxide has a higher solubility in water at a higher pressure. Opening a can of soda lowers the pressure in the air in the headspace to 1atm and CO2 begins to escape the solution.

(b) Heartburn and stomach upset are caused by an overproduction of stomach acid. Medications designed to relieve these symptoms contain hydroxide in order to fuel a neutralization of the stomach acid.

(c) No two atoms have the same available electron energy levels and since the colors emitted by an excited elemental gas depend on the energy difference between those levels, no two elements have the same emission spectrum.

(d) The two lone pairs on the oxygen atom of H2O repel the hydrogen bonds, bending them downward slightly and changing the bond angles.

In This Chapter

^ Packing the tools you need ^ Following a study schedule ^ Building your confidence ^ Relieving stress

Reparing yourself for the AP exam involves several things, and they are all equally " important. First you should familiarize yourself with AP test (in case you haven’t done so already, check out Chapter 1). In this chapter, we take you a step further by explaining what you need to take with you to the test as well as how to study and mentally prepare for the test.

Taking Your Tools to the Test

Although you’re not packing to go to a remote island in Fiji (albeit, a great visualization skill to use right before the test! See "Practicing visualization" at the end of this chapter), we do want you to put the same amount of thought into what you need to bring and not bring to the test site on test day. Pay special attention to the lists in the following sections. Being prepared in advance will minimize your stress level and assure that you have what you need.

Packing what you need

It’s not advisable to ask the test proctor to wait 20 minutes while you run home to get your forgotten calculator. You definitely need to be sure you have the permitted tools for the test on test day. The night before the test try to pack the following list of items:

Pencils with erasers (#2): Used for the multiple-choice test Black pens (with nonfelt tip): For the free-response questions Your Social Security number: This number identifies you

Your school code: Your school counselor or AP chemistry teacher can give you your school code. The AP folks will give you one if you are home-schooled

Photo I. D.: Just in case they suspect that you paid your neighbor to take the test for you

A watch: You will need to pay close attention to the time

A scientific calculator: For use on the free-response section

Snacks and drinks: Quiet snacks, like soft granola bars, are a better choice than loud snacks like cookies. First, you won’t disrupt others, and second, you won’t come crashing down after a sugar high. Water is a better choice than sugar-filled soda for the same reason.

Appropriate clothing: You don’t know whether there will be an arctic freeze or global warming at the test site, so be prepared for both. Layering your clothes works for either instance.

Tissues: No one wants to be distracted by sleeve wipes or sniffing. Bring something to wipe your nose!

Knowing what not to bring

Ft\NG/ The following list needs just as much attention as the list of what you can bring (see the section above, "Packing what you need"). The below items are some things you definitely Don’t Want to bring. You run the risk of being thrown out of the test site for bringing some of these items, so read closely:

Scratch paper: The AP will let you use portions of the exam booklets to take notes. Notes, books, dictionaries, highlighters, or cheat sheets: Leave it all at home, folks!

Electronic devices like MP3 players, cell phones, beepers, and watches that beep:

No one wants to get interrupted by your Beyonce ringtone.

Your smart, nerdy friend to take the test for you: For some reason they don’t like that substitution.

Energy drinks: Unless you want to come crashing down and have a brain melt, leave them in the fridge.

Needing Special Attention

Normal is so last year. If you require special accommodations to take the AP exam, you are not alone. Not everyone takes the test under the same conditions. You may have special circumstances that require that the test be administered to you in a different way. Here is a brief list of special instances that might warrant an adjustment to the AP chemistry exam:

Learning disabilities: If you have a diagnosed learning disability, you may be able to get special accommodations. You may have extended time, large print, a reader, and frequent breaks, among other things, but you must specifically request learning disability accommodations on your application form. The accommodations you request will depend on your specific, diagnosed learning disability. You should make sure that your school has a SSD (Services for Students with Disabilities) Coordinator’s Form on file with the College Board. You must fill out this form and send it to the College Board. Allow seven weeks for the pros at College Board to review your request.

To get special testing

• You must have been formally diagnosed with a learning disability by a professional.

• You must have a current, individualized educational plan at school.

• In most cases, the evaluation and diagnostic testing should have taken place within five years of the request for accommodations.

• You must also describe the comprehensive testing and techniques used to arrive at the diagnosis, including test results with subtest scores.

• Your best bet is to log on to the College Board Web site (Www. collegeboard. com/ssd/student/index. html) To see all up-to-the-minute requirements for accommodations.

Physical disabilities: If you have a physical disability, you may be allowed to take a test in a special format — in Braille, in large print, or on an audio cassette or CD. Follow the same instructions detailed above to request accommodations, and if you have further questions contact the College Board directly for more information.

Religious obligations: If your religion prohibits you from taking a test on a specific day, you may test on an alternate date. Again, the College Board folks can guide you in the right direction for alternate dates.

Military duty: If you’re an active military person you don’t need to complete the normal registration form. Instead, ask your Educational Services Officer about the testing through DANTES (Defense Activity for Non Traditional Educational Support).

Using Long-Term Strategies for Training and Survival

When you decide to take the AP chemistry exam you’re basically committed to getting hitched. Think of the test day as your wedding day and the months leading up to it your engagement. You can consider the day after the test your divorce date, if you will, but for all intents and purposes, you are pretty darn committed to nurturing your AP perpetual partner until then.

You need to plan about ten months in advance for this test. You cannot study for this test like you might a regular test in an ordinary class. In other words, you can’t cram — it just plain doesn’t work for the AP tests. The AP people are a smart group — they know how to ask questions in such a way as to eliminate (or not "qualify") the students who didn’t plan in advance, and simply stayed up all night with a few cans of Red Bull as support. About now, you may be wondering what you should be doing for those ten months. Well, don’t despair: We’ve outlined the entire ten months for you in the following sections. Consider us your AP wedding planners!

Early planning

The early-planning phase takes you from September to October. In the following list, we’ve given you some solid ideas on what to do during each month as you plan for your upcoming exam:

September: As soon as you are able, enroll in an AP chemistry class. If your school doesn’t offer one, speak with your counselor and see if there is a community college chemistry course you could take. If an AP course is not available to you, get your hands on a college level chemistry textbook, but keep in mind that nothing comes close to a passionate teacher teaching you the concepts face-to-face.

I November: Because you’ve been studying chemistry for a few months, you should be ready to take a diagnostic test. You can use one of the sample tests in this book or you can go to Www. collegeboard. com And find some there. Taking such tests gives you an early indicator of your strengths and weaknesses, which can help guide your studying

In the months to come. Your first diagnostic score can also help to identify what concepts you need to pay more attention to in this book.

I December: Begin reading this book from the beginning. You will find that being in the classroom for the last four months has provided you a wealth of information. You should already be familiar with many of the concepts covered in the first third of this book.

Midyear planning

You lay the groundwork for taking the exam in the earlier stages of planning to discover your strengths and weaknesses. Although you’ve been honing your knowledge and skills the last four months, you still need to continue to prepare from January to May:

I January through March: Continue reviewing this book and take another practice test in early March. You still have two months before your wedding day, so your score on this test will help you pinpoint further concepts from this book that you might still need to review or re-review.

I April and May: Take another practice test in early May. Make this your last practice test. Whatever you do, Do not Take another practice test the week of the real test. Like a long-distance runner who has been preparing all year for the big race, she doesn’t do the race right before the race!

Up-to-the-minute planning

Last-minute planning does not equal cramming! For ten months, you should’ve been preparing for your exam, and the last few days and minutes count as well. Make the best use of your final time before the test by following this schedule:

I The night before the test: Relax the night before the test. You might be tempted to pull an all-nighter, but our experience has proven that if you don’t know the material the night before the test, you aren’t going to learn it the night before the test. Pamper your brain instead by eating a good dinner, watching a movie, reading a good book. If you want to glance at this book one last time, that’s cool, but just don’t turn it into a cram-fest.

I The morning of the test: Listen to your mama and eat breakfast, please. We know you may be nervous and your appetite might be slim, but your brain needs the energy. If you’re too nervous, or Heaven forbid, you’ve gotten up late, consider bringing some healthy breakfast items for the car ride. It can actually help keep your mind less stressed.

I Test day: Take the test.

June: Wait patiently for your score. . . very patiently. Did we mention to wait Patiently?

Feeling Confident: It’s All in Your Head

You have been studying and preparing for this test for a long time. You know more than you think you do. You might not remember every single detail of every single concept, but you know at least something about something. If you come to the test feeling completely insecure, please remind yourself that you did the work, you studied, and you will be fine.

Anticipating the outcome

Like we have been saying throughout this chapter, the AP exam tests core concepts and material. There are no surprises. After finishing this book, you will have been exposed to all the material you need to excel on this test. Anticipate the best possible outcome.

Remembering: This is a test… This is only a test

Listen, your score on the AP test is not going to change your life. It’s not a do-or-die type of experience. Your life’s success is not tied into the outcome of this test. Treat this test as you would treat any other test you have taken (you’ve taken hundreds, right?). And don’t worry, you don’t know the AP folks. They are not going to snicker as you walk by and say under their breath, "There goes the one that didn’t know what Dalton’s Law of Partial Pressure was." Although the AP guys don’t know you, we realize that your peers do. And it’s inevitable that "score gossip" will creep into your conversations with your friends. Comparing scores is not advisable, as it creates stress and underlying feelings about your chemistry capabilities. Remembering that this is only a test and not a life-defining moment will help you stay balanced and keep it all in perspective.

Surviving During the Test with These Four Stress-Busters

It’s completely normal to feel some fear the day of the test. Everyone does. Research has shown us that a little anxiety is actually good for the brain. A touch of anxiety helps the brain stay ultra-focused and attentive. A little anxiety tells the brain, "Hey, brain, if I’m feeling this way, whatever I’m doing must be important, so I better stay alert." However, excess anxiety and panic actually have the opposite effect. High levels of fear make the brain go into flight-or-fight mode. Your brain tells your body to conserve your higher-order thinking skills in an effort to send extra physical strength to the body in case of emergency. So, if you’re feeling too much anxiety during the test, here are some stress-busters that will calm those nerves.

Counting to four

Breathing is grossly underrated. Take a deep breath until your belly expands, hold it in four counts, and then expel the air for four counts. Be sure not to exhale like you’re in the middle of an aerobic workout so as not to disturb the other test-takers. Oh, and some breath mints wouldn’t hurt either.

Try not to take shallow breaths, which can cause you to become even more anxious because your body is deprived of oxygen. Hyperventilating or sucking on your inhaler during the exam is grossly overrated.

Stretching

Several stretches can help relieve stress and make your body feel more relaxed and comfortable. Do the practices in the list discretely so you don’t disrupt others around you — in other words, refrain from doing the downward-dog position from your yoga class:

I Rotate your head around to stretch out and relax your neck muscles. (We suggest keeping your eyes closed so the proctor doesn’t think you are trying to cheat.)

I Hunch and roll your shoulders to help relax your back and spine. You’ll be sitting for quite some time, so maintaining good posture is crucial.

I Shake out your hands like you have writer’s cramp. Imagine that all your tension and stress is going out through your fingertips.

I Extend and push out your legs like you’re pushing something away with your heels.

I With your legs stretched out in front of you, point your toes back toward your knees and hold that position for a count of three.

Right before the exam or during a short break, practice creative visualization. Close your eyes and imagine yourself in the test room cheerfully looking at questions that you know the answer to, filling in the answers, finishing early, and double-checking your work. Picture yourself leaving the exam room full of energy, and then getting your score and rejoicing. Think of how proud of you your parents are (if thinking about them stresses you out leave this one alone). Imagine not having to take early chemistry in college. The goal is to associate the AP exam with feelings of joy.

Any time you feel yourself starting to panic or think negative thoughts, make a conscious decision to

1. Say to yourself, "Stop, self! Don’t dwell on anything negative."

2. Drop that negative thought like a hot Bunsen burner.

3. Then switch over to a positive train of thought.

For example, suppose you catch yourself thinking, "Why didn’t I pay more attention to chemical equilibrium?" Change that script to, "I’ve got most of this right, maybe I’ll get this right, too. No sense worrying now, overall I’m rocking my world!"

Practicing visualization

Don’t do this visualization exercise during the test. You’ll just waste time and lose concentration.

Stop, drop, and, change your mind

In This Chapter

^ Getting comfy with the difference between kinetics and thermodynamics ^ Using rate laws to describe reaction kinetics

^ Using state functions and heat to describe reaction thermodynamics Keeping tabs on heat flow by using calorimetry and Hess’s Law

Ost people don’t like waiting. And nobody likes waiting for nothing. Research has tentatively concluded that chemists are people, too. It follows that chemists don’t like to wait, and if they must wait, they’d prefer to get something for their trouble.

To address these concerns, chemists study things such as

Kinetics Is concerned with the rates of things, such as chemical reactions. Kinetics answers the question, "How much product will I get in the next minute?"

Thermodynamics Is concerned with the relationship between energy and chemical systems. Thermodynamics answers questions like, "How far will my reaction go?" and "How hot will my beaker get?"

Kinetics and thermodynamics are separate. There’s no simple connection between how long it takes for a reaction to proceed and how productive a reaction can be. In other words, chemists have good days and bad days, like everyone else. At least they have a little bit of science to help them make sense of these things. In this chapter, you get an overview of this science. Don’t wait. Read on.

Getting There Rapidly or Slowly: Kinetics and Reaction Rates

So, you’ve got this beaker, and a reaction is going on inside of it. Is the reaction a fast one or a slow one? How fast or how slow? How can you tell? These are questions about rates. You can measure a reaction rate by measuring how fast a reactant disappears or by measuring how fast a product appears. If the reaction occurs in solution, the molar concentration of reactant or product changes over time, so rates are often expressed in units of molarity per second (M s"1).

For the following reaction, A + B — C

You can measure the reaction rate by measuring the decrease in the concentration of either reactant A (or B) or the increase in the concentration of product C over time. So, by measuring changes in concentration over a defined change in time, you can measure Average rate, The rate of the reation averaged over a specific time period:

Average rate = – A[A]/At = A[C]/At

Note that A means "average change in." Here, the expression for average rate is given in terms of reactant A or in terms of product C. The expression in terms of A includes a minus sign because that reactant disappears in the course of the reaction. No sign in the C expression means "+" as [C] increases with time. Note that if the reaction were reversed so that [C] decreased and [A] increased over time, the signs of [A] and [C] would still be opposite, but reversed.

As the period of time gets smaller (that is, as At approaches zero), the average rate approaches the Instantaneous rate, The rate of the reaction at a given instant:

Instantaneous rate = -d[A]/dt = D[C]/dt

In these kinds of equations, if you have done calculus you’ll smile down upon those less-fortunate mortals, but "d" Is only math-speak for a change in the amount of something at any given moment. If you plot the concentration of product against the reaction time, for example, you might get a curve like the one shown in Figure 21-1. Each tiny point on the graph corresponds to a slope that is an instantaneous value of D[product]/dt.

Reactions usually occur most quickly at the beginning of a reaction, when the concentration of products is the lowest and the concentration of reactants is highest. Chemists usually measure "initial rates" under these conditions and use these rates to characterize the reactions.

Time

The stoichiometry of a reaction helps determine the relative rates at which reactant and product concentrations change. For example, in the reaction

A + 2B — C

The following is true:

Instantaneous rate = – d[A]/dt = -1/2d[B]/dt = d[C]/dt

Because two moles of reactant B are consumed for every mole of reactant A, moles of B disappear at twice the rate of moles of A.

Measuring and Calculating with Rates

How can chemists follow the concentration of reactants and products as a reaction proceeds, anyway? One common method is by using Absorption spectroscopy, Which measures the concentration of substances in a sample by their ability to absorb electromagnetic radiation. If the substance you want to measure absorbs green light, you shine green light through the sample and measure how much green light passes through to the other side — less transmitted light means more absorbed light, which means a greater concentration of sample. Beer’s Law Is a useful equation that expresses this idea:

A = Abc

Where A Is the light absorbed by the sample, A Is the molar absorptivity of the substance (a property that varies from one substance to another), and B Is the length of the path traveled by the radiation through the sample. To make this easy in practice, solutions are often put into flat sided cells, so the length of the path is easily measured. The variable C Is the concentration of the substance.

Equations that relate the rate of a reaction to the concentration of some species (reactant or product) in solution are called Rate laws. The exact form a rate law assumes depends on the reaction involved. Countless research studies have described the intricacies of rate laws in chemical reactions. Here, we focus on rate laws for only the simplest reactions.

In general, rate laws take the form

Reaction rate = K [reactant A]"[reactant B]n

The rate law shown describes a reaction whose rate depends on the concentration of two reactants, A and B. Other rate laws for other reactions may include factors for greater or fewer reactants. In this equation, K Is the Rate constant, A number that must be experimentally measured for different reactions and reaction conditions (that is, different temperatures, solvents, etc.). The exponents " And N Are called Reaction orders, And must also be measured for different reactions. A reaction order reflects the impact of a change in concentration in overall rate. If M > N, Then a change in the concentration of A affects the rate more than does ^jjjABEfl Changing the concentration of B. The sum, M + N, Is the overall reaction order.

J Some simple kinds of reaction rate laws crop up frequently, so they’re worth your notice:

Zero-order reactions: Rates for these reactions don’t depend on the concentration of any species, but simply proceed at a characteristic rate:

Rate = K

First-order reactions: Rates for these reactions depend linearly on the concentration of a single species:

Rate = K[A]

Second-order reactions: Rates for these reactions may depend on the concentration of one or two species (or some intermediate combination):

Rate = k[A]2

Or

Rate = k[A][B]

Figure 21-2 illustrates how changing the reaction order changes the progress of a reaction; the more the rate of a reaction depends on the concentration of a reactant, the more severely the reaction slows down as that reactant is used up.

Figure 21-2:

Disappearance of reactant in zero-order, first-order, and second-order reactions.

Zero Order First Order Second Order

Time

Because the initial concentration of reactants may vary from one reaction to another, chemists often find it useful to employ Integrated rate equations, Which describe the reactant concentration at time T ([A]t) with respect to the initial concentration of a reactant, usually designated [A]0.

Another useful mathematical tool is Half-life, The time it takes for the concentration of a reac-tant to drop to one half of its initital value. This value is usually given the symbol T1/2.

Figure 21-3 summarizes the rate laws, integrated rate laws, rate constant units, and half-life equations for the three most common reaction types. Other types are not expected in AP.

Figure 21-3:

Summary of expressions for the kinetics of zero-, first-, and second-order reactions.

Changing Rates: Factors That After Kinetics

Despite the impression given by their choice in clothing, chemists are finicky, tinkering types. They often want to change reaction rates to suit their own needs. What can affect rates, and why? Temperature, concentration, and catalysts influence rate. In this section, we describe how these factors can change reaction rates, and we show how those changes are expressed mathematically in the Arrhenius equation, an important equation that relates reaction rate to the activation energy of a chemical reaction. Finally, we give a brief summary of how reaction rates are used to help figure out chemical reaction mechanisms, the nitty-gritty details about how exactly chemical reactions occur.

Seeing rate changes in terms of collisions

The effects of these changes can be understood in light of the Collision model, In which chemical reactions occur when reactatnts slam into one another with sufficient energy and in just the right way, As shown in Figure 21-4, if the collision isn’t "just right" the reactants will bounce away and no new products are formed.

A)

Figure 21-4:

The collision model of nonproductive and productive chemical reactions.

B)

9 «9

Here is a summary of the major factors that affect reaction rates:

Reaction rates tend to increase with temperature. This trend results from the fact that reactants must collide with one another to have the chance to react. If reactants collide with the right orientation and with enough energy, the reaction can occur. So, the greater the number of collisions, and the greater the energy of those collisions, the more actual reacting takes place. An increase in temperature corresponds to an increase in the average kinetic energy of the particles in a reacting mixture — the particles move faster, colliding more frequently and with greater energy. So an increase in temperature provides a "double whammy" effect — increased number of colissions (per second) and increased average energy per collision. Make sure you include both of these in your AP exam answer.

Ii Increasing concentration or surface area tends to increase reaction rate. The reason for this trend also has to do with collisions. Higher concentrations mean there are more reactant particles to undergo more collisions per second and have a greater chance of reacting. Increasing the concentration of reactants may mean dissolving more of those reactants in solution. Some reactants aren’t completely dissolved, but come in larger, undissolved particles. In these cases, smaller particles lead to faster reaction. Smaller particles have a greater fraction of the reactant molecules (like Compound A) on the surface where they can more easily react with molecules of Compound B, making a greater portion of the particles available for reaction.

Catalysts increase reaction rates. Catalysts don’t themselves undergo overall chemical change, and they don’t alter the amount of product a reaction can eventually produce (the Yield). However, they do enhance the probablility of a reaction occurring. Most catalysts form temporary complexes with one or both reactant molecules, holding them in more appropriate orientation for reaction. Then the product is released, leaving the catalyst to "fight another day" or engage more reactants over and over. Thus catalysts are often only needed in very small amounts to increase the rate of a reaction. Chemists (who like to define things, if you didn’t realize that by now), separate catalysts into two types. Homogenous catalysts Are of the same phase (solid, liquid, gas) as the reacting molecules, and Heterogeneous catalysts Are in a different phase. Catalysts operate mostly in the way described, but all of those ways have to do with decreasing Activation energy, The hill in chemical potential energy that reactants must climb to reach a Transition state, The highest-energy state along a reaction pathway. Lower activation energies mean faster reactions. Figure 21-5 shows a Reaction progress diagram, The energetic pathway reactants must traverse to become products. The figure also shows how a catalyst affects reactions, by lowering the activation energy without altering the energies of the reactants or products. If you think this diagram looks much like a roller coaster, you are right! In that case, the energy is gravitational potential energy, rather than chemical potential energy — but the effect is the same. Too little kinetic energy or too big a hill — no ride or no reaction.

Transition state

Figure 21-5:

A reaction progress diagram highlighting the effect of a c atalyst on activation energy.

Reaction Progress

Relating rate to activation energy

Concentration, temperature, and the the activation energy, Ea, Help determine the rate of a reaction. The influence of concentration is clearly evident in the rate laws shown in Figure 21-3. But where is the influence of temperature and activation energy? These two factors help determine the rate constant, K, As shown in the Arrhenius equation:

In the Arrhenius (yes the same Arrhenius who worked out what acids are doing) equation, K Is calculated by raising the natural logarithm (base number e, approximately 2.718) to a rational exponent. The exponent, – Ea / RT, Is the negative of the activation energy divided by the product of the gas constant, R, And the temperature, T. This whole power is multiplied by a prefactor (sometimes called the frequency factor), A. This factor has the same units as K And depends on the reaction order. It is hard to calculate theoretically, though it relates to the effectiveness of the collisions in "going over the hill," and is often an experimental measurement.

The Arrhenius equation can be rearranged into different forms that are more or less convenient for different purposes. If you know a variety of values for K And T, And would like to determine EA, then you may want to use this form:

E, , Ln K = —+ ln A

The previous equation corresponds to a linear plot of ln K Versus 1/T. The slope of the line is equal to -EA / RT And the y-intercept equals ln A.

If you know the rate constant at one temperature and would like to predict it at another temperature, then you may want to use this form:

Ln, =—fT K2 R

T, T

Using rates to determine mechanisms

Measuring reaction rates can help you figure out Mechanism, The molecular details by which a reaction takes place. Chemical reactions can be broken down into a series of Elementary steps, Simple events involving only one or two molecules that occur with characteristic rates. Each elementary step has a certain Molecularity, The number of molecular bits that must collide within that step. Unimolecular steps require one molecule, bimolecular steps require two molecules, and termolecular steps require three molecules. Termolecular steps are rare, as are collisions at one intersection of three cars (except in the movies).

Most reaction mechanisms consist of several elementary steps. Only a few of these steps consume or produce chemical players that you see in the formal reaction equation. The other players, the ones on the path between the formal reactants and products, are called Intermediates. Much chemical research on reaction mechanisms is concerned with establishing the important elementary steps and detecting and defining intermediates.

Each elementary step in a multireaction has its own rate law. The slowest step acts as a bottleneck and determines the overall rate of the reaction. For this reason, the slow step is called the Rate determining step. If an intermediate step is the slow step, chemists often assume that the previous, faster steps come to equilibrium during all the time afforded to them by the bottleneck.

Whatever the details, two things are always true:

I For a mechanism to be true, the rate law proposed by that mechanism must agree with the observed rate law.

I Mechanisms can never be proved correct directly by kinetics, but incorrect mechanisms can be disproved.

Getting There at All: Thermodynamics and State Functions

Figure 21-6 shows a reaction progress diagram like the one shown in Figure 21-5, but highlights the difference in energy between reactants and products. This difference is completely independent of activation energy, which is discussed in the previous section. Although activation energy controls the rate of a reaction, the difference in energy between reactants and products determines the extent of a reaction — how much reactant will have converted to product when the reaction is complete.

Transition state

Reaction Progress

A reaction that has produced as much product as it ever will is said to be at Equilibrium. We discuss equilibrium at length in Chapter 15. You may want to refer to that chapter for a quick refresher on that important concept, particularly on how equilibrium is quantified by the equilibrium constant, KEq. Here, we are primarily concerned with the Thermodynamics Of equilibrium — how the equilibrium position of a reaction relates to energy changes as a reaction proceeds.

^jjjMj|E# The difference in energy between reactants and products is built into the rules that determine the Keq for a reaction. The particular form of energy important in this relationship is called Free energy, G. You may sometimes see this called the Gibbs free energy (hence the G), in honor of one of the earliest American chemists, Josiah Willard Gibbs. According to Gibbs every chemical possesses a quantity of "free energy" G Depending on what it is and what the conditions (phase, temperature, pressure) are. While it is extremely difficult to find out what this is, changes in this value can easily be tracked by smart chemists. The relationship between free energy and equilibrium is

AG = – RT ln Keq

Or

K = e _AG /RT

In these equations, R Is the gas constant and T Is the temperature in Kelvin degrees. Favorable (spontaneous) reactions possess negative values for AG And unfavorable (nonspontaneous) reactions possess positive values for AG. Energy must be added to drive an unfavorable reaction forward. If the AG for a set of reaction conditions is 0, the reaction is at equilibrium. But be careful! These values only apply at a specific set of conditions. A reaction that is non-spontaneous under one set of conditions may be spontaneous if those conditions change. For

Example, most people think of hydrogen and oxygen reacting pretty spontaneously to form water, but heat water to thousands of degrees and it will split into hydrogen and oxygen!

Free energy is an example of a state function, a property of a system that depends only on its characteristics in the given moment. State functions don’t care about how you got here, only about who you are. But they change as conditions change! Your state function is different in bed at home as compared to sitting at your desk in school. In addition to free energy, other state functions include Enthalpy, H, And Entropy, S. At constant pressure, enthalpy equals the the thermal energy content of a system. Entropy is a measure of the energy of disorder in a system.

Energy is a pretty abstract concept. Before looking further into how energy governs chemical reactions, it might be useful to remind yourself of three things that are always true — the laws of thermodynamics:

First Law: In any process, the total energy of the universe remains constant. Remember, chemists divide the universe into two parts, the system and the surroundings. Universe = system + surroundings. The system and the surroundings can exchange energy; the universe cannot.

Second Law: In any spontaneous process, the overall entropy of the universe increases. Things that happen on their own do so with an increase in disorder. If you focus only on the system, the increased disorder may not be apparent, as it may occur in the surroundings. In fact, complex chemical reactions (such as the biochemical ones in our bodies) often increase order at the expense of the disorder in the surroundings.

Third Law: As temperature approaches absolute zero, the entropy of a system approaches a constant minimum. If the system is a pure, perfectly ordered crystalline solid, then that minimum of entropy is also zero.

Chemical reactions are processes that occur in the universe. So, chemical reactions obey the laws of thermodynamics. If you encounter a chemical reaction that fails to obey these laws, patent it. You’ll probably find some investors willing to lose money on it.

Two other concepts follow closely on the heels of the thermodynamic laws: Reversibility And Irreversibility. In a reversible process, a system can go back to its original state by reversing its steps Along the same path It originally took. When a reaction is at equilibrium, mass (in the form of atoms) moves between reactant and product states reversibly.

In an irreversible process, the system may be able to return to its original state, but must do so along a different path. Spontaneous processes are irreversible, as they irrevocably increase the entropy of the universe. Never fear, irreversible processes and their properties do not show up on AP chemistry exams.

Adding Up the Energies: Free Energy, Enthalpy, and Entropy

Three different kinds of energy are important for determining whether or not a particular change occurs, such as whether or not a chemical reaction goes forward. These three kinds of energy are

Gibbs free energy, G, The portion of a system’s energy that is available to do work

Enthalpy, H, The thermal energy content of a system at constant pressure

Entropy, S, A measure of the portion of a system’s energy that is unavailable to do work; entropy is related to disorder

By tracking changes in these three energies between different states (like the reactant and product states of a chemical reaction), you can determine whether a change will occur on its own, or whether that change needs to be driven forward by the addition of energy. The change in Gibbs free energy, AG, Is negative for spontaneous processes, ones that occur on their own. The change in Gibbs free energy can be seen in terms of changes in enthalpy and entropy in the Gibbs equation:

AG = AH - TAS (where T Is temperature in Kelvin)

The Gibbs equation reveals that spontaneity arises from the interplay of enthalpy and entropy, two state functions that underlie free energy. A lot of AP questions revolve around which is the winning term on the righthand side, so get all your ducks (sorry, arguments) in a row here. Negative changes in enthalpy are spontaneous, as are positive changes in entropy. Note that the effect of entropy changes on the free energy (which are generally 1,000 times smaller than enthalpy changes) is magnified by the temperature. The upshot of this equation is that chemical reactions can be driven forward by spontaneous changes in enthalpy, entropy, or both. If the changes in neither enthalpy nor entropy are spontaneous, then the reaction will not occur spontaneously. These patterns are summarized in Figure 21-7.

+

Spontaneous Reactions:

TAS

Favorable AH overcomes unfavorable TAS

The 5th Wave By Rich Tennant

In this part. . .

Studying chemistry without spending time in a lab is a bit like reading up on the history of skydiving from the confines of an armchair in your basement. Something is missing from the experience. For this reason, AP chemistry courses include extensive labwork as part of the standard curriculum. Although the specific experiments may vary from one high school course to another, they all conform to the same set of guidelines and principles, and most of them are drawn from an official set of recommended experiments put forth by the AP chemistry Powers That Be. In this part, we give you a brief review of common laboratory equipment and techniques, and a complete survey of the Official List of AP chemistry experiments.

In This Chapter

^ Summarizing what you need to remember

^ Testing your knowledge with some practice questions

^ Finding explanations for the answers

Chapter 11 describes the different kinds of interactions between particles within different phases of matter and how changes in temperature and pressure can cause matter to move between phases. Chapter 11 also explores solutions — what they are, how to measure their concentration, and how their behavior changes with added solute. Just to be certain you remember the key points of Chapter 11, we’ve included those important concepts for you to review in this chapter before you take a stab at answering questions about solids, liquids, and solutions.

Getting a Grip on What Not to Forget

Here is a collection of the most important points to remember for the AP exam about how particles behave in different states of matter, how samples move between states, and how solute particles interact with solvent within solutions.

Kinetic theory

Just as kinetic theory can explain gas behavior, the following list shows how kinetic theory can also help explain the properties and behavior of the Condensed states — liquids and solids, in case you haven’t read Chapter 11:

Particles within liquids are close together, but can slide past one another (that is, Translate), So that liquids are fluid, meaning liquids have Short-range order.

Particles within a solid cannot translate past one another, but vibrate about an average position. So solids have Long-range order.

In condensed states, interparticle attractions and repulsions are critical in determining the properties of the substance.

Solid states

Solid states include several categories with distinct characteristics:

The particles within Ionic solids Are strongly attracted to one another, and these solids have high melting points. Lattice energy Measures the strength of the ion-ion interactions within these solids.

Molecular solids Have weaker attractions between particles and typically have lower melting points.

Covalent (network) solids Are bound together by multiple covalent bonds and tend to be exceptionally strong.

I Metallic solids Consist of a lattice of positively charged atoms within a sea of mobile electrons coming from valence shells, which accounts for the high electrical conductivity of metals.

I Particles within Crystalline solids Pack into highly ordered arrangements of a repeating Unit cell And tend to have well-defined melting temperatures. Particles within Amorphous solids Exhibit far less long-range order and tend to melt over broad temperature ranges.

Condensed states

Within condensed states, a collection of "weak" forces (that is generally, but not always, weaker than covalent or ionic bonds) can play important roles:

I Hydrogen bonds Can occur when two electronegative atoms (like oxygen, nitrogen, or fluorine) essentially "share" an electropositive hydrogen atom. Formally, the hydrogen is covalently bound to one of the electronegative atoms.

Dipole-dipole Forces can occur between the partially charged parts of polar molecules. Opposite partial charges attract and same-sign partial charges repel. Dipoles can also interact with ions according the the same basic concept.

I London dispersion forces Are attractive forces that are generally weaker than hydrogen bonds, dipole-dipole, and ion-dipole forces at long distances. London forces occur between Induced dipoles, Partial charges that flicker into and out of existence as electron clouds interact with one another.

Phase diagrams

Phase diagrams Reveal the effects of temperature and pressure on the Phase (physical state) of a substance. The following list describes how to interpret the diagrams:

I A field of the phase diagram represents boundaries between phases. Crossing a line indicates a Phase change, Such as melting, freezing, boiling, condensing, subliming, or desubliming (depositing).

I The Triple point Represents a substance-specific combination of temperature and pressure at which the solid, liquid, and gas phases of a substance coexist.

At very high temperatures and/or very low pressures, gases can move into Plasma phase, In which ionized particles intermingle with free electrons.

I At temperatures and pressures above a substance-specific condition called the Critical point, The distinction between liquid and gas phase disappears. A substance in this condition is called a Supercritical fluid.

I Heating curves Show how the temperature of a substance varies as heat is added to it. These curves tend to exhibit a staircase-pattern in which temperature increases as heat is added at a constant rate, and then levels off during phase changes because heat must be added simply to shift the substance from one phase to another.

Solutions

Solutions Are mixtures in which different components are completely and evenly mixed down to the level of individual molecules, ions, or atoms.

The major component of a solution is the Solvent. Minor components are Solutes.

I Substances Dissolve If and when solvent-solute attractions dominate over solvent-solvent and solute-solute attractions.

I Saturated Solutions are those that hold the maximum amount of a dissolved solute. Unsaturated Solutions can accommodate further solute additions.

I The Solubility Of a solute is the concentration of the substance required to produce a saturated solution. So, substances with large solubility can be dissolved to a large degree, while smaller amounts of substances with low solubility will dissolve.

I Solubility can vary in the following ways, depending on temperature, pressure, and the physical/chemical properties (especially Polarity) Of the solute and solvent:

• Increasing temperature increases the solubility of most (but not all) solid solutes in liquid solutions.

• Increasing temperature decreases the solubility of gases in liquid solutions.

• Increasing pressure increases the solubility of gases in liquid solutions, as expressed by Henry’s law: S1 / P1 = S2 / P2.

• Polar solutes dissolve best in polar solvents. Nonpolar solutes dissolve best in nonpolar solvents. In other words, "like dissolves like."

Ideal solutions have some important points to remember:

Ideal solutions obey Raoult’s law. This law states that in the vapor over a liquid solution, each solution component should have a partial pressure in proportion to its mole fraction within the solution. Raoult’s law implies that adding a solute to a solvent should reduce the partial pressure of the solvent within the vapor over the solution.

I In Ideal solutions (those which are dilute and in which all intermolecular forces are about the same strength), adding solute particles has predictable effects on Colligative properties Like boiling point, freezing point, and osmotic pressure. Most AP questions involve only nonvolatile solutes — those that have essentially no vapor pressure of their own to add to that of the solvent. Thus only the solvent properties are affected.

• Adding solute particles raises the boiling point of a solvent in a phenomenon called Boiling point elevation: ATb = Kb X m.

• Adding solute particles lowers the freezing point of a solvent in a phenomenon called Freezing point depression: ATf = Kf X m.

• Adding solute particles increases the Osmotic pressure, N, of a solution.

Many real-life solutions are nonideal solutions, especially at high concentrations or when one kind of intermolecular force dominates over others in solution. (Fortunately, AP examinations never ask about these.) We give you examples of some nonideal solution effects in the following list:

I Strong solute-solvent interactions lead to vapor pressures lower than those predicted by Raoult’s law.

I Weak solute-solvent interactions lead to vapor pressures higher than those predicted by Raoult’s law.

I The activity coefficient, y, is used to correct for nonideal solution effects on solute concentration: Activity = yx Concentration.

More concentrated solutions can be diluted with solvent to yield less-concentrated solutions, as expressed by the Dilution equation: C1 x V1 = C2 x V2.

Reporting the concentration of a solute

Different methods for reporting the concentration of a solute include molarity, molality, percent solution, mole fraction, and partial pressure. The following list reminds you how those methods are expressed:

Ii* Molarity: M = moles solute / liters solution Ii* Molality: M = moles solute particles / kilograms solution

I Percent solution Can refer to mass percent, volume percent, or mass/volume percent:

• Mass % = 100% x (mass solute / total mass solution)

• Volume % = 100% x (volume solute / total volume solution)

• Mass/Volume % = 100% x (grams solute / 100 milliliters solution) I Mole fraction: % = moles component / total moles of all components I Partial pressure Of A, PA = mole fraction of A x total pressure

Testing Your Knowledge

No matter how we summarize it, clearly you need to know a lot about solids, liquids, gases, and solutions. Test your mastery of the material in Chapter 11 by giving these questions a try.

Questions 1 through 5 refer to the phase diagram below.

Temperature

1. Which points correspond to a melting/freezing equilibrium?

(A) 1 and 5

(B) 1 and 3

(C) 2 and 4

(D) 6 and 7

(E) 7 and 8

2. Which point(s) correspond(s) to homogenous phase(s)?

3. Which point corresponds to a sublimation/deposition equilibrium?

(A) 1

(B) 3

(C) 5

(D) 6

(E) 8

4. Which point corresponds to the critical point?

(A) 1

(B) 2

(C) 5

(D) 7

(E) 9

5. Which point corresponds to the triple point?

(A) 1

(B) 2

(C) 5

(D) 7

(E) 9

6. How much water must be added to 150mL of 0.500M KCl to make 0.150M KCl?

(A) 45mL

(B) 150.mL

(C) 350.mL

(D) 500.mL

(E) 650.mL

7. The Ksp Of lead (II) chloride is 1.7 x 10-5. What is the molar concentration of a chloride ion in 1.0L of a saturated PbCl2 solution?

(A) 5.7 x 10-6

(B) 1.1 x 10-5

(C) 1.6 x 10-2

(D) 2.6 x 10-2

(E) 3.2 x 10-2

8. You can prepare 0.75 molal NaCl by dissolving 15g NaCl in what amount of water?

(A) 0.40kg

(B) 0.34kg

(C) 0.27kg

(D) 0.20kg

(E) 0.26kg

9. If 1.6 x 10-3 moles of an ideal gas are dissolved in 2L of a saturated aqueous solution at 1atm pressure, what will be the molar concentration of the gas under 2.5atm pressure?

(A) 1.6 x 10-3

(B) 0.8 x 10-3

(C) 4.0 x 10-3

(D) 2.0 x 10-3

(E) 3.2 x 10-3

Questions 10 through 14 refer to the following set of choices:

(A) Particles vibrate about average positions.

(B) Particles are ordered and occur within a sea of mobile electrons.

(C) Particles are ionized, disordered and highly energetic.

(D) Particles diffuse rapidly and can dissolve many solutes.

(E) Particles do not translate but lack long-range order.

10. Supercritical fluid

11. Metallic solid

12. Amorphous solid

13. Plasma

14. Crystalline solid

15. One mole of glucose, C6H12O6, dissolved in 1.00kg water results in a solution that boils at 100.52 °C and freezes at -1.86 °C. If 300. grams of ribose, C5H10O5, are dissolved in 4.00kg water, what will be the boiling and freezing points of the resulting solution?

(A) 100.26 °C, -0.93 °C

(B) 100.13 °C, -0.47 °C

(C) 100.52 °C, -1.86 °C

(D) 101.04 °C, -3.72 °C

(E) 101.56 °C, -5.58 °C

17. Ethanol, CH3CH2OH, engages in which of the following types of intermolecular interactions?

I. Hydrogen bonding

II. Dipole-dipole interactions

III. London forces

(A) I only

(B) II only

(C) III only

(D) I and II only

(E) I, II, and III

Questions 18 through 22 refer to the following diagram and the accompanying set of choices with regard to the heat being transferred:

Heat Added

(A) Liquid phase

(B) Increase in average kinetic energy of particles

(C) Decrease in average kinetic energy of particles

(D) Heat of fusion

(E) Heat of vaporization

18. Segment 2 corresponds to this phase or phase change.

19. Moving left to right on segment 1 represents this.

20. Moving right to left on segment 5 represents this.

21. Segment 4 corresponds to this phase or phase change.

22. Segment 3 corresponds to this phase or phase change..

Checking Your Work

Dissolve any lingering uncertainties you have about your answers. Check them here. The following diagram refers to items explained in the answers to questions 1 through 5:

Supercritical fluid

Liquid

Critical point

Gas

Tempe rature

1. (B). The melting/freezing equilibrium occurs at the boundary line between solid and liquid phases. Although different substances have differently shaped phase diagrams, it is typically the case that low temperature-high pressure conditions favor solid phase; high temperature-low pressure conditions favor gas phase; and liquid phases occur at intermediate temperatures and pressures.

2. (B). Homogenous phases are those that are uniform throughout and not in equilibrium between multiple phases. So, on a phase diagram, homogenous phases are found in the open spaces between phase boundary lines.

3. (E). Sublimation is direct movement from solid to gas phase, and deposition is the reverse of that process. So, a sublimation/deposition equilibrium occurs at the boundary line between solid and gas phases.

4. (C). The critical point for a substance is the unque combination of pressure and temperature beyond which liquid and gas phases cease to be distinct. Any combination of temperature and pressure that is simultaneously above both the critical temperature and critical pressure causes the substance to become a supercritical fluid.

5. (D). The triple point for a substance is a unique combination of temperature and pressure at which all three major phases (solid, liquid, and gas) are in simultaneous equilibrium. The triple point occurs at the convergence of the three phase boundary lines.

6. (C). Attack this problem by using the dilution equation: C1 x V1 = C2 X V2.

(0.500M) x (150mL) = (0.150M) x V2

Solving for V2 gives you 500mL. So, the final solution should have a volume of 500mL. Because the initial solution had a volume of 150mL, you must add 350mL of water to achieve the desired final volume.

7. (E). Solving this question requires you to understand the dissolution reaction of lead (II) chloride and the definition of the solubility product constant, Ksp.

PbCl2(s) + H2O(Q Pb2+(aq) + 2Cl-(ag)

Ksp = 1.7 x 10-5 = [Pb2+][Cl-]2

Notice from the dissolution reaction equation that for every mole of PbCl2 that dissolves, you get one mole of Pb2+ cation and Two Moles of Cl – anions. Using this fact, we can substitute into the equation for the Ksp:

1.7 x 10-5 = [x][2x]2

1.7 x 10-5 = 4×3

Solving for x, you get 1.6 x 10-2. However, the concentration of Cl – is 2x, so the answer is 3.2 x 10-2.

8. (B). To answer this question, you must know the definition of molality, M = (moles solute) / (kilograms solvent). Next, you must find the moles of NaCl solute by converting from the given mass of 15 grams. Because the molar mass of NaCl is 58.45 g mol-1, you have 0.257mol NaCl. Next, substitute into the definition of molality:

0.75m = (0.257mol NaCl) / (X kilograms water)

Solving for X, you get 0.34kg water.

9. (D). This question requires you to apply a result of Henry’s law, stating that the solubility of an ideal gas in a liquid solvent is directly proportional to the pressure: S1 / P1= S2 / P2.

Before substituting into the proportion, however, you should calculate the initial molar solubility:

Molarity = (1.6 x 10-3mol) / (2L) = 0.80 x 10-3 M Substituting into the proportion, you get (0.80 x 10-3 M) / (1atm) = S2 / (2.5atm) Solving for S2 gives you 2.0 x 10-3 M.

10. (D). Supercritical fluids occur in high temperature-high pressure conditions beyond the thermodynamic critical point of a substance. Under these conditions, the substance has properties of both a gas (high diffusivity) and a liquid (ability to dissolve solute).

11. (B). The atoms of a metallic solid are packed within an ordered lattice, but are so electropositive that electrons move freely about the lattice, which accounts for the high electrical conductivity of metals.

12. (E). In an amorphous solid, particles — often polymers — are packed tightly enough that the substance isn’t fluid, but lack the extreme order and structural regularity of a crystalline solid.

13. (C). Provided enough energy, the particles of a gas (often, a superheated gas) can ionize to create plasma, a state of matter that is diffuse and fluid like gas, but has unusual properties due to its ionic nature.

14. (A). Within a crystalline solid, particles are tightly packed within highly ordered, regularly repeating structural units called unit cells. So little translational freedom is available to the particles within a crystalline solid that all they can usually do is to vibrate in place. Such is the price of perfection.

15. (A). This question tests your knowledge of two colligative properties, boiling point elevation (ATb) and freezing point depression, (A7f). You are required to know that colligative properties, which apply to ideal solutions, depend solely on the number of solute particles in solution, and not on the physical or chemical properties of those particles. The number of solute particles is measured by molality, M. The effects of added solute on boiling and freezing points depend on the solvent. By giving you data on the boiling point and freezing point of a

1 M Glucose solution, the question essentuially gives you the boiling point and freezing point proportionality constants, Kb = 0.52 °C m"1 and Kf = -1.86 °C m"1. Next, you must determine the molality of the ribose solution. To do so, you must calculate the molar mass of ribose, 150 g mol-1. Because 300g ribose are dissolved, there are 2.0 moles ribose in the solution.

Molality = (2.0mol ribose) / (4.0kg solvent) = 0.50M

Next, substitute into the equations for boiling point elevation and freezing point depression:

ATb = (0.52 °C m"1) x (0.50m) = 0.26 °C

A7f = (°1.86 °C m"1) x (0.50m) = -0.93 °C

Adding these quantites to the standard boiling point and freezing point of water gives TB = 100.26 °C and TF = -0.93 °C.

16. (D). This question requires to understand that boiling point, as a colligative property, depends on the number of moles of solute Particles In solution, not on the number of moles of solute compound. For example, one mole of dissolved KCl contributes 2 moles of particles (one mole K+ and one mole Cl-). Taking this factor into account, it is clear that in 1.0M NaNO3, the NaNO3 contributes the fewest particles to solution (one mole Na+ and one mole of the polyatomic ion NO3-), and therefore elevates the boiling point the least.

17. (E). With its hydroxyl functional group, ethanol is quite capable of forming hydrogen bonds, which require a hydrogen covalently bonded to an electronegative atom (like oxygen) and also require a second electronegative atom (like the oxygen of a second ethanol molecule). Hydrogen bonds are themselves a kind of dipole-dipole interaction, employing the strong dipoles set up by the electronegative atoms. Finally, all condensed phase molecules engage in London dispersion interactions. So, ethanol participates in all three kinds of inter-molecular interactions.

18. (D). The heat added within this segment clearly does not go toward rasing the temperature of the substance. Instead, the added heat goes toward rearranging particles at a given temperature from solid phase into liquid phase. We know this as it is the first of the phase changes to be shown (moving left to right on the diagram).The heat required to accomplish this phase change is called the heat of fusion, AHfus.

19. (B). Heat added in this segment clearly goes toward increasing the temperature of the substance. Temperature is a measure of the average kinetic energy of the particles within a sample.

20. (C). Moving from right to left within the diagram corresponds to heat transferred out of the system, and doing so within this segement clearly results in a decrease in the temperature of the substance. Falling temperature corresponds to a decrease in the average kinetic energy of the particles in a system.

21. (E). The heat added within this segment clearly does not go toward raising the temperature of the substance. Instead the added heat goes toward rearranging particles at a given temperature from liquid phase to gas phase (the second phase change shown in the diagram). The heat required to accomplish this phase change is called the heat of vaporization, AHvap.

22. (A). This segment lies beween the plateaus corresponding to fusion (melting) and vaporization (boiling). The phase that sits between melting and boiling is liquid. Note that the average kinetic energy of the particles is increasing during this segment.

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In This Chapter

^ Reviewing important points on bonding

^ Breaking yourself in with some practice questions on bonding ^ Finding out what you did right. . . and not so right

Chapter 5 introduced you to chemical bonds, the attractive forces that bring atoms together to form compounds. Atoms come together to form different kinds of bonds and a multitude of different compounds. Yet, for all the complication, the basic rules of the game are pretty straightforward. Review them here and test your mastery of the material with the practice questions.

Brushing Up on Bonding Forget-Me-Nots

We’ve included a helpful summary of the points from Chapter 5 that you need to remember. So, do some bonding of your own with the following list:

Compounds Are collections of atoms held together by bonds.

• Ionic bonds Arise from electrostatic attraction between ions of opposite charge (between anions and cations).

• Covalent bonds Arise when atoms share electrons.

• Metallic bonding Involves a sea of electrons that moves freely through a lattice of metal cations.

Ionic bonds tend to form between a metal and a nonmetal. Covalent bonds tend to form between nonmetals.

Inorganic compounds are named in a systematic way (see the figure in Chapter 5). Naming and recognizing compounds will be much easier if you are familiar with the common polyatomic ions (see the table in Chapter 5).

Solid ionic compounds consist of a highly ordered lattice of ions. The strength of an ionic bond within a lattice is expressed by Lattice energy. Large, positive lattice energies correspond to strong ionic bonds.

Other factors being equal, ionic bonds tend to be stronger between ions with more charge, and between smaller ions.

Electrons within covalent bonds distribute with greater density in the region between the two atomic nuclei.

Usually, each atom of a covalent bond contributes one electron per bond. When one atom contributes both electrons, we say the bond is a Coordinate covalent bond.

The strength of covalent bonds is described by Bond enthalpy. Large, positive bond enthalpies correspond to strong covalent bonds.

Polar bonds Are marked by unequal sharing of electrons. Electronegativity Is the tendency of an atom to draw electrons to itself.

• Bonds between atoms with extreme differences in electronegativity are ionic.

• Bonds between atoms with significant differences in electronegativity are polar covalent.

• Bonds between atoms with insignificant differences in electronegativity are nonpolar covalent.

The polarity of a bond is described by the Bond dipole, U. = Qd, where Q is the magnitude of the separated charge and D Is the distance of separation.

Bond dipoles sum within molecules to produce Molecular dipoles. Dipole-dipole and ion-dipole interactions are major contributors to the properties of different compounds, especially boiling points and melting points.

Testing Your Knowledge

Try your hand at these practice questions on bonding. Do yourself a favor by not looking back in the text to get the questions correct on the first try. Do your best, and then check your answers afterward so that you can discover what you need to review.

1. Which of the following are compounds that might reasonably form from combining iron and oxygen?

(I) Fe2O3

(II) Fe3O2

(III) FeO

(A) I only

(B) II only

(C) III only

(D) I and II only

(E) I and III only

2. Which of the following is least likely to be a stable compound?

(A) CF4

(B) S2O

(C) PCl3

(D) SiO3

(E) NO

3. Which of the following compounds is least likely to form?

(A) Na2Cr7O2

(B) KC2H3O2

(C) Li2CN

(D) Rb2C2O4

(E) HNO2

4. Which of the following is correctly named?

(A) CsCl; cesium (I) chloride

(B) Fe2O3; iron (II) oxide

(C) CBr4; carbon quatrobromide

(D) NO2; dioxygen mononitride

(E) MnO2; manganese (IV) oxide

5. Which empirical formula best fits the following percent composition? 50.00% carbon, 5.59% oxygen, 44.41% hydrogen

Questions 6 through 10 refer to the following types of bonds:

(A) Ionic bonds

(B) Polar covalent bonds

(C) Nonpolar covalent bonds

(D) Coordinate covalent bonds

(E) Metallic bonds

6. Form in compounds like N2 and O2

7. Tend to support electrical conductivity

8. Tend to form strong, brittle compounds

9. Tend to form from an atom containing a lone pair of electrons 10. Form in compounds like CO and H2S

Checking Your Work

You’ve done your best. Now check your work. Make sure to read the explanations thoroughly for any questions you got wrong — or for any you got right by guessing.

1. (E). Both Fe2O3 (iron (III) oxide) and FeO (iron (II) oxide) can form, the former from the Fe3+ cation and the latter from the Fe2+ cation. Fe3O2 is a bad choice because the overall charge of this compound is not neutral; even using Fe2+, there is an excess of positive charge because each oxygen can contribute only -2 charge.

2. (D). As shown, SiO3 is least likely to be stable. Although the SiO32- (silicate) anion exists, the compound shown in the question is neutral. In order to fill its valence shell and remain neutral, silicon requires four covalent bonds.

3. (C). Li2CN is least likely to occur because the "ion math" doesn’t work. Lithium forms cations with +1 charge (Li+) and the cyanide ion has -1 charge (CN-). So, a more likely version of this compound is LiCN.

4. (E). If a manganese ion with +4 charge is used (Mn4+), the compound can form; that ion is specified by the Roman numeral IV. As described in the answer for question 1, Fe2O3 is correct as long as the Fe3+ ion is used. But the Roman numeral II in the name specifies the Fe2+ ion. Answer A is incorrect because the alkali metal cesium forms only ions with +1 charge, so no Roman numeral is used in its name. The correct names for the compounds in choices C and D are carbon tetrabromide and nitrogen dioxide, respectively.

5. (C). Percent composition refers to percent weight. So, a 100.g sample of the mystery compound would contain 50.00g carbon, 44.41g oxygen, and 5.59g hydrogen. To convert these masses to relative numbers of each type of atom, divide each mass by the molar mass of that atom:

50.00g carbon / (12.01g mol-1) = 4.163 mol carbon 5.59g hydrogen / (1.01g mol-1) = 5.53 mol hydrogen 44.41g oxygen / (16.00g mol-1) = 2.776 mol oxygen

Next, take the smallest mol value of the set (2.776 mol in this case) and divide all the mol values by that number. You get: 1.5 mol carbon, 1 mol oxygen, and 2.0 mol hydrogen. You can’t build compounds with fractions of an atom, so multiply all results by the smallest number that will produce a set of whole numbers. In this case, multiplying all values by 2 does the trick, yielding: 3 mol carbon, 2 mol oxygen, and 4 mol hydrogen. Finally, assemble the atom symbols and the whole numbers into an empirical formula: C3H4O2.

6. (C). Covalent bonds between identical atoms are nonpolar because each atom has identical electronegativity.

7. (E). The sea of mobile electrons that is the defining feature of metallic bonding supports electrical conductivity; the mobile electrons can move in response to applied voltage.

8. (A). Ionic bonds are strong, but the extreme order of ionic bonding lends itself to fracturing.

9. (D). Coordinate covalent bonds form when one atom contributes both electrons to the bond; lone pairs are perfect for this task, because they include two electrons not otherwise involved in bonding.

10. (B). Polar covalent bonds form between the atoms in compounds like these because the atoms have significantly different electronegativities — but not so different that ionic bonds form instead.