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1 AP Chemistry Review Preparing for the AP Chemistry Exam
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AP Chemistry Review
Preparing
for the
AP Chemistry Exam
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Table of Contents
Title Pg
Top 25 Things to Know before you take the AP Chemistry Exam 3
About the AP Chemistry Exam 5
AP Chemistry Study Guide 8
Types of Reactions 19
AP Subject Review 24
Multiple Choice Practice test 40
Practice Free Response Questions 58
Practice Free Response Scoring Guide 63
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Top 25 Things to know before you take the AP chemistry Exam.
Assume the person grading your exam is an idiot. Make it clear for them to understand your process and grade your
exam with ease. It is not acceptable to lose points because of not showing work/units or messy handwriting. If you
need to cross something out do it like this “The reaction is spontaneous because.” . No need to scribble and make a
mess.
Focus on your weakest areas; it is doubtful you can do/know everything. The AP Chemistry Exam is designed so that
it is impossible to know absolutely everything on it (in case you haven’t noticed).
I might as well place the biggest two at the start – You need to review your incorrect MC from the Practice Exam(s)
and Princeton exam(s) and understand the concepts. It can really increase your MC score.
Know the 6 strong acids HCl, HI, HBr, H2SO4, HClO4, HNO3 and the one weak by formula acetic acid CH3COOH,
everything else is weak. Remember that strong acids/bases don’t make buffers!!! You should be 100% confident
what ionizes and what doesn’t
Know the strong bases: Group 1 hydroxides, Ba(OH)2, Sr(OH)2, Ca(OH)2
Know how to determine which molecule has the largest dipole moment (difference in electronegativity).
Know hybridization on a carbon atom (Remember Double Bonds and Triple bonds don’t count) Lone Pairs do
count!!! Single bonds are sigma, double and triple are pi-bonds.
Solubility Rules are a must. If you still don’t know them get cracking!!!
Know some basic geometry’s of molecules. Do not expect complicated ones like See-Saw etc. Stick with the basics.
If any really hard ones are on there you will prob. need to guess.
Know the bond angles on a Bent Geometry, Trigonal Planer, Trigonal Pyrimidal, and Tetrahedral. Also know why
bond angles shrink as lone pairs are added (b/c if increased repulsion amongst the electrons causing the bond angles
to squeeze)
It would be a safe bet to assume that when a metal by itself it placed in acid you will get H2 gas and some aqueous
salt and the negative ion is the spectator ion. Cu does not react in HCl, but in AP chemistry it is safe to assume that
there is always a reaction. Also groups 1 and group 2 metal by themselves placed in water always give you H2 gas
and a base (which usually ionizes).
Know the carbonate reactions! You always get CO2 on the other side….
Be ready for a complex ion. You should be confident in determining the answer. Remember to “cheat” on the
reaction and determine the product first (using the double it rule) and work backwards to the get the reactants.
Know the units on the rate constant k for kinetics.
Know the kinetics graphs.
Lighter atoms (molar mass) are faster than heavier atoms (molar mass) at the same temperature. Also lighter atoms
“effuse” (leak) out of small hole in a container faster than heavier ones.
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Know the periodic table trends. It would useful to know the E.N values for F =4.0, O=3.5, N=3.0 C=2.5, H=2.1, and
what type of bonds occur as a result.
Know your intermolecular forces, how to I.D. than and what they mean in terms of boiling points. Know about
covalent network bonds and what they mean!
Know what the signs on Delta H, S and G mean and be able to explain each.
Know how to use Ksp
Be sure to look over your Labs, there will a question on the part two this topic.
The Chart. You will need to know it. No doubt. Be Ready
G Ecell Keq
+ -‐ <1
-‐ + >1
0 0 1
OIL RIG and REDCAT. The only time reduction does not occur at the cathode is when it is an electrolysis reaction
in which energy is added. Usually you are given the reduction or the oxidation on this type of diagram/question.
Decide what else to study from the exam memorization guide.
25)The test will be hard, but don’t get frustrated, just keep plugging through, and don’t give up. You might surprise
yourself. You are prepared. Good luck and Congratulations on finishing the course!!!
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About the AP Chemistry Exam
Section I: Multiple-Choice Section: 90 minutes (50% of your grade)
Section I consists of 60 multiple-choice questions, either as discrete questions or question sets, that represent the
knowledge and science practices outlined in the AP Chemistry Curriculum Framework, which students should understand
and be able to apply. Question sets are a new type of question: They provide a stimulus or a set of data and a series of
related questions. Calculators are not permitted on the Multiple Choice section.
Examine each question for a maximum of thirty seconds (on the average, some will take less time allowing more time
for others).
Quickly determine the subject of the question.
By the end of the thirty seconds either:
Mark the correct answer.
Mark a “Y” next to the questions that you know how to work but need more time.
Mark a “N” next to the questions that you don’t have any idea how to work.
Force yourself to move through twenty questions each ten minutes and the full seventy-five questions in forty minutes.
Now make a second pass concentrating on the “Y” questions only. Do not spend any time on the “N” questions. If you
don’t know the correct answer see if some key piece of knowledge will allow you eliminate two or three of the choices.
Complete this pass in forty minutes.
Now make your third pass. Focus only on the “N” questions. Attempt to eliminate at least two choices. If you can, then
make an intelligent guess. If not, guess. You are not penalized for the wrong answer. Any correct choices on this pass are
bonus points. You have only ten minutes, so make it count!
Before time expires, count the number that you have answered. You should answer at least sixty
(60) questions.
Section II: Free-Response Section 90 minutes (50% of your grade)
Section II contains two types of free-response questions (short and long), and each student will have a total of 90 minutes
to complete all of the questions. This section also contains questions pertaining to experimental design, analysis of
authentic lab data and observations to identify patterns or explain phenomena, creating or analyzing atomic and molecular
views to explain observations, articulating and then translating between representations, and following a logical/analytical
pathway to solve a problem.
Beginning with the May 2014 administration of the AP Chemistry Exam, multiple-choice questions will contain four
answer options, rather than five. This change will save students valuable time without altering the rigor of the exam in any
way. A student's total score on the multiple choice section is based on the number of questions answered correctly. Points
are not deducted for incorrect answers or unanswered questions.
What to Bring to the Exam
2-3 Sharpened #2 pencils
Acceptable calculator
A positive attitude
Get a good night’s rest and eat breakfast that morning!
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AP Chemistry Exam Format
Section I
Question Type Number of Questions Timing
Multiple Choice 60 90 minutes
Section II
Long Free Response 3
90 minutes
Short Free Response 4
Calculators are allowed on the free-response section for the first 55 minutes. During that time, students will work
on three required problems. For the last 40 minutes, calculators must be put away as students work on the
remaining free-response questions.
Section II Free Response
Questions require you to apply and explain chemical concepts and solve multiple step problems.
You do not have to answer in essay form and may save time using one of the following methods: bullet format,
chart format or outline format.
Write your answers in the space provided and number your answer clearly.
There is a slight penalty for incorrect sig figs.
Stating a Principle, Law, Theory or stating the name of the Principle law or theory is not an explanation or
justification for an answer. Stating a trend is also not an explanation. State and apply the Principle , Law or
Theory to the specific situation in the questions to explain your answer and use the reason for trends as
explanations for trends not just stating the trend.
Answer everything, no matter what
If you don’t know the answer, substitute in a number to complete the remaining parts of the question. You will be
given credit for following the correct procedure even if the answer may be wrong
Part A (Question 1) Equilibrium
Read all of the question before doing any work. Items later in the problem may provide keys to earlier sections.
Part A is always equilibrium. Determine which type (Gaseous equilibrium, acid/base, buffer, or precipitation). Look for
key words and clues.
Acid/Base: Look for the words acid or base, Ka or Kb, [H+], [OH-], or [H3O+]. Any of these indicate an
acid/base problem.
Buffer: Look for the word buffer. Also, check for a weak acid and its conjugate base.
Precipitation: Look for Ksp or the word solubility.
Gas Equilibrium: Look for (g) on most of the reactants and products.
After determining the type of reaction, write a reaction if one is not provided. Use the general forms given below:
Acid HA + H2O -----> H3O+ or HA -----
> H+ + A-
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Base A-
Precipitation
+ H2O ----->
HA + MA(s) ---
--> M+(aq)
OH-
+ A-(aq)
Write an equilibrium constant expression. Leave out solids and liquids.
Solve the problem. THINK! Put in all of the given quantities in the equilibrium constant expression and solve for the
unknown allowing the units to direct the problem.
Part A (Questions 2 and 3)
Read both problems all the way through before doing any work.
Determine which type of problem each is.
Select the problem you know the most about and solve it. Remember that if you cannot solve an earlier part you may still get
some credit for a later section by showing how you could use the earlier answer in succeeding parts of the problem.
Question 2 & 3 generally cover the following:
Lab procedure
Kinetics
Electrochemistry
Stoichiometry
Thermochemistry
Part B (Question 4)
Consists of three reactions and usually a lab question about the reaction.
Write the reactant in symbol form for all reactions showing each reactant in net ionic form as follows:
Strong Acids, bases, and soluble salts written as ions. Weak acids, bases, and insoluble salts written as
molecules.
Classify the reactions as:
Acid/Base - Look for H or OH or salts which could act as a weak acid or base.
Precipitation - Look for insoluble salts which could form as products according to the solubility rules.
Redox - If it is not acid/base or precipitation, it is probably oxidation-reduction. Check for elements which could
change oxidation states. Pay particular attention to the common oxidizing agents (NO3-, MnO4-, Cr2O72-,
H2O2) and reducing agents (Cl-, Br
-, I
- and elemental metals).
Other - Anything else which doesn’t fit above (usually either organic or complexation).
Remember you score one point for getting the reactants in the correct form and two points for each product. At least get all of
the reactants correct and possibly two or three products.
Part B (Questions 5-7)
Typically includes the following topics:
Bonding/intermolecular forces/hybridization
Electrochemistry
Lewis dot structures/Periodic Trends
Be as specific as possible in your answer. Look for clues in the question as to what is really important.
Answer the question. State exactly what you are asked not what you would like to answer.
Do not simply restate the question.
Remember that you will be getting partial credit. Answer any part about which you have any knowledge.
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AP Chemistry Study Guide
2 • Atoms and Elements
The Development of the Atomic Theory: Define the three theories that Dalton explained in terms of atoms:
o Law of Conservation of Matter o Law of Definite/Constant Proportions o Law of Multiple Proportions
Give examples and solve calculation problems related to each of the three theories. Sketch a cathode ray tube as demonstrated in class and state how J.J. Thomson’s experiments led to the idea that atoms have positive and
negative parts, the negative parts are all the same, and the negative parts (called electrons) have a certain charge/mass ratio. Define cathode rays. State the factors that determine how much a moving charged particle will be deflected by an electric or magnetic field. Explain Millikan’s oil drop experiment & how it added to the atomic theory. Sketch the set-up used by Ernest Rutherford (the gold-foil experiment), show what he observed, and explain how these observations led to the
idea that most of the mass of the atom is concentrated into a tiny, amazingly massive, positively-charged nucleus. Parts of the Atom: State the three particles that make up an atom, their symbol, their charge, their mass, and their location. State the number of protons, neutrons, and electrons in any atom or ion. Explain that isotopes are two atoms with the same atomic number (number of protons) but different mass numbers (number of nucleons—
protons + neutrons).
Represent the nucleus with isotopic notation, such as: 220
86 Rn
Recognize when two nuclei are isotopes of each other.
Molar Mass Calculations: Calculate the isotopic mass of an atom given the resting mass of protons and neutrons. Explain that a mole of any element is actually made up of various isotopes in a constant percentage abundance. Calculate the average atomic mass of an element using the percent abundance and mass of each isotope. Calculate the percent abundance of isotopes given the average atomic mass and isotopic masses of an element. The Families of the Periodic Table: List the common families of the periodic table and recognize to which family any element belongs. Recognize metals, non-metals, and metalloids (semi-metals) on the periodic table. State and define the terms conductivity, malleability, and ductility. State some element facts such as which elements are too radioactive to exist, which is the largest non-radioactive element, which element has
the greatest density, and which element has the highest melting point. Explain how Dmitri Mendeleev put together the periodic table and why we give him credit for the table even though others were working along
the same lines. List the three elements that Mendeleev predicted and where they are located on the periodic table. Nuclear Chemistry: State that Henri Becquerel discovered radioactivity and Marie Curie studied it. List the three “Becquerel rays” (alpha, beta, and gamma) and state why alpha particles were the perfect tool for Ernest Rutherford to study the
structure of atoms. State that the alpha particle is the same as a helium nucleus, a beta particle is a high-speed electron, and a gamma ray is a high-energy form of
light.
3 Chemical Formulas
Formulas Look at a formula and state how many elements and atoms are in that compound. Calculate the molecular mass or molar mass of any compound. State that the mass of a molecule is measured in amu’s and the mass of a mole is measured in grams. Give examples of empirical formulas, molecular formulas, and structural formulas. Identify a formula as empirical, molecular, or structural.
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Ionic Compounds State whether a compound is an ionic compound or a nonmetal compound. Write the formula of an ionic compound given the two ions or its name. Know when to use parentheses. Name an ionic compound given the formula. Determine the charge on an ion from information in an ionic formula.
Nonmetal Compounds (aka Molecular Compounds) Write the formula of a binary nonmetal compound (molecular compound) given its name. Name a binary nonmetal compound (molecular compound) given its formula.
Percent Composition Calculate the percent composition (by mass) for any compound. Calculate the empirical formula from percent composition data. Determine the molecular formula of a compound given its empirical formula and molar mass.
Hydrates Give examples of hydrates and anhydrous compounds. Calculate the formula of a hydrate from dehydration data.
The Mole State the significance of the mole. State the three mole facts for any substance (molar volume, molar mass, Avogadro’s number)
o 1 mole = 22.4 Liters @ STP (gases only) o 1 mole = 6.02 x 1023 particles o (particles = molecules or atoms) o 1 mole = gram molecular mass of chemical
Use dimensional analysis to convert between moles, mass, volume, and number of particles for a chemical. Use density as a conversion factor in mole problems. Use gas density to calculate molar mass.
4 Chemical Equations and Stoichiometry
Chemical Equations Give examples of products and reactants in a chemical equation. State that Antoine Lavoisier introduced the law of conservation of matter. Combustion State that combustion is another name for burning. Write an equation for a combustion reaction given only the fuel that is burned. Correctly label substances in an equation as solid (s) , gas (g), liquid (l), or aqueous (aq) Balancing Equations Balance equations by adding coefficients. Recognize when an equation is balanced. State that the formulas of reactants and products should not be changed in order to balance equations. Stoichiometry Problems
Use the stoichiometric factor ( of the problem) to convert from moles of one substance to moles of a different substance.
(i.e. In the equation: N2 + 3H2 2NH3, 3 mol H2 2 mol NH3) Convert between the quantities of mass, volume, molecules and moles using dimensional analysis (i.e. use 1 mol = 22.4 L, 1 mol = 6.02 x 1023 molecules, and 1 mol = gram molecular mass) Show the units of molar mass as grams/mol or g·mol-1. Limiting Reactant Problems Recognize that a problem with two “given values” is a limiting reactant problem. Determine the limiting reactant and excess reactant in a problem. Solve problems involving Limiting Reactants Calculate how much excess chemical is left over after a reaction. Percent Yield Problems Use stoichiometry to calculate the theoretical yield (mass of a product) in a problem. State that actual yields are usually given in a problem. Use the theoretical yield and actual yield to calculate the percent yield. Chemical Analysis Problems
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Calculate the mass of each element in a given compound given data such as the masses of CO2 and H2O formed in a combustion reaction.
Use mass and mole information to calculate the empirical formula of an unknown substance. Use percent composition to equalize mass and mole information derived from different samples.
5 Reactions in Aqueous Solution____________________________________________________________ Properties of Aqueous Solutions Define solute, solvent, and solution. Give examples. Define electrodes. Give operational and theoretical definitions of electrolytes. Know that soluble ionic compounds and strong acids are strong electrolytes. Ionic compounds of low solubility [e.g. Mg(OH)2]
and weak acids/bases are weak electrolytes. Know that molecular compounds (except acids) are non-electrolytes. Know that alcohols (e.g. CH3OH )are not ionic hydroxides. Bases are usually metallic hydroxides. Know the solubility rules. State whether an ionic compound is soluble in water. Precipitation Reactions Know that ppt reactions are double replacement reactions that produce an insoluble product. Given two ionic compounds in solution, correctly determine the products. (Know your ions). Determine which product(s) is/are precipitates. Use (aq) and (s) symbols correctly. Correctly write the ions in a soluble ionic compound. [e.g. CaCl2(aq) becomes Ca2+ + 2Cl ] Identify spectator ions. Write molecular, detailed ionic, and net ionic equations for a ppt reaction. Acids and Bases Give operational (cabbage juice) and theoretical (ions) definitions of acids and bases. Know that acids increase the H+ ion concentration in an aqueous solution. (Theoretical definition) Memorize the 8 strong acids. Know that acids are molecular compounds that form ions when in aqueous solution. Be able to name acids according to their anion.
[ide hydro__ic acid; ate __ic acid; ite __ous acid; sulfur: add “ur”; phosphorus: add “or”] Know that bases increase the OH ion concentration in an aqueous solution. (Theoretical definition) Memorize the soluble hydroxides (except NH4OH) that are the strong bases. Understand that ammonia(aq), NH3 + H2O NH4
+ + OH forms a weak basic solution.
Know that metal oxides form bases [CaO + H2O Ca(OH)2] while nonmetal oxides form acids [CO2 + H2O H2CO3] Know that acids react with bases to form H2O and a salt. (Neutralization) Write equations for acid-base reactions including NH3 (example on page 199) as the base. Know that strong acids and strong bases are written as ions in the ionic equations. Gas Forming Reactions Recognize the six products that turn into gases. Memorize the gases formed. Organizing Reactions in Aqueous Solution Double Replacement reactions (text calls them exchange reactions) (Fred-Wilma/Barney-Betty reactions) also have the old
fashioned name: metathesis reactions. Know the three examples of double replacement reactions and the “driving force” for each.
Precipitate reactions form an insoluble product. Acid-Base reactions form water (a very weak electrolyte therefore, a very stable product). Gas-forming reactions form a gas.
Know that a driving force is something that keeps the new combinations of ions from reforming the old combinations of ions. Oxidation-Reduction is a fourth type of reaction driven by the transfer of electrons. Oxidation-Reduction Reactions Know that an important type of reaction gets its name from atoms that combine with oxygen. During the refining of iron,
carbon monoxide combines with oxygen (from the iron ore), CO CO2 and is oxidized. Large masses of iron ore (Fe2O3) are reduced to a smaller amount of iron metal.
Understand that since CO helps the iron ore to be reduced, CO is called the reducing agent. Since Fe2O3 causes the C to be oxidized, iron ore is called the oxidizing agent. What ever is oxidized acts as the reducing agent. What ever is reduced acts as the oxidizing agent.
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Know that oxidation-reduction (redox) is driven by the transfer of electrons. Mnemonics to help: GROL (Gain=Reduce / Oxidize=Lose); LeO the lion says GeR (Losing e ’s = Oxidation / Gaining e ’s = Reduction); OIL RIG (Oxidation is Losing e ’s / Reduction is Gaining e ’s)
A redox reaction can be divided into two half-reactions. The oxidation half-reaction has electrons as a product. The r eduction half-reaction has electrons as a reactant. Be able to assign oxidation numbers to any atom in any substance. Learn the rules on page 207.
Recognize redox reactions because oxidation numbers change. (# = oxidation / # = reduction), electrons are gained or lost, or oxygen atoms are gained or lost.
Know several common oxidizing agents and reducing agents and what they turn into. Measuring Concentrations of Compounds in Solution Know the definition of molarity, M, as one way to communicate concentration of solute. Know that the symbol [X] means the concentration of X in moles/Liter. Be able to determine the concentration of ions in an ionic compound. For example, in 0.25 M AlCl3 [AlCl3] = 0.25 M [Al3+] = 0.25 M [Cl ] = 0.75 M Use the molarity formula to calculate moles, mass, volume, or molarity of a solution. Know that Volume x Molarity = moles of solute. Dilution problems use ViMi = VfMf. Describe how to make a solution correctly. Know what a volumetric flask is. Stoichiometry of Reactions in Aqueous Solution Use molarity as another conversion factor to solve stoichiometry problems. Know that titration is a technique called quantitative chemical analysis because you are measuring. It is also called volumetric
analysis (because you are measuring volumes). [Note: qualitative analysis involves no measurements such as using solubility rules to determine the identity of an unknown ionic compound.]
Understand the terms indicator, equivalence point, standardization, and primary standard. [Note: you saw a titration being done in the Measurement video early in the summer. Chloride ion from the Chesapeake Bay was being titrated against silver nitrate to determine the salinity (saltiness) of the water. Yellow K2CrO4 was used as an indicator because it formed the reddish-brown ppt, Ag2CrO4 (which looked pink) when all the chloride ion was used up.]
Know common indicators such as phenolphthalein for titrations with strong bases. Understand that a titration can be done with an acid-base reaction or a redox reaction. In each case, some sort of indicator
must be used to tell when equivalent amounts of reactants have been mixed. 6 • Energy and Chemical Reactions
Driving Forces state that product-favored (spontaneous) reactions tend toward maximum entropy, S, and minimum enthalpy, H.
state the sign of H based on observation of warming or cooling of the surroundings. correctly apply the terms exothermic and endothermic to situations where the surroundings are warming or cooling. draw a PE curve (uphill or downhill) based on information about warming or cooling of the surroundings.
Measuring Heat state the units of heat capacity, specific heat, and molar heat capacity as well as the significance of each. convert between the heat units of calories and Joules. (4.184 J = 1 calorie)
use calorimetry (q=mCT) to calculate heat changes during temperature changes. calculate the heat transferred when two objects, at different temperatures, come into contact.
Energy = Heat and Work state the difference between work and heat energy. state the difference between system and surroundings. recognize the system and the surroundings in a chemical or physical system. calculate the change in internal energy based on changes in heat absorbed by the system and work done by the system.
state that H is a more general (and useful) measure of energy than E and that H = q when a reaction occurs at constant pressure.
Chemical Work = Expanding Gases relate physical work (w=F·d) and chemical work (w=P·V). calculate PV work done by an expanding gas. state that no work is done in a constant volume situation such as a bomb calorimeter.
Calculating H -- Hess’s Law state the definition of a state function. list examples of properties that are and are not state functions.
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write the equation for the heat of formation of a substance. state that the heat of formation of an element under standard conditions has a value of zero. use Hess’s Law to calculate the energy of a chemical or physical change.
Calculating Heat During Phase Changes – Heats of Fusion and Vaporization use heats of vaporization or heats of fusion to calculate heat changes during phase changes. write an equation showing the heat of fusion or heat of vaporization.
7 • Atomic Theory and Bonding
Periodic Trends Atomic Radius Decreases -Up due to the n = # becoming smaller (less shells) and to the right due to the Effective Nuclear Charge (Zeff)
increasing or greater attraction
Ionic Radius
o + ion < neutral < - ion
o + ion has more attraction between nucleus and valence electrons (lose e- gets smaller)
o ion has more repulsion between valence electrons (gain e- gets bigger)
Ionization Energy Increases (Energy required to remove an electron) Up due to the n= # becoming smaller (electrons are closer to the nucleus
therefore will be held more tightly) To the Right due to greater shielding.
Electronegativity ncreases (Ability to attract an electron in a bond – 0 EN for Noble Gases) Same reasons as Ionization Energy
Reactivity
o Alkali Metals : Greater Reactivity going down Group IA due to Ionization Energy.
o Halogens: Greater Reactivity going up Group VIIA due to Electron Affinity
Effective Nuclear Charge (Positive + Charge Affecting the Valence Electrons) Higher Zeff is more attraction on valence electrons
o Zeff = Z – S (where Z = atomic number and S = e- in the inner shell)
K: 19-18 = +1 Br: 35-18 = +17 Bromine has a smaller radius due to greater Zeff
Lewis Dot Structures Count total # of e- Lone atom in center, fill the outside atoms with octets Add left over e- to the center atom If there is less than an octet on the center atom move outer e- to multiple bonds
Stronger Bonds Longer Bonds − < = < ≡ ≡ < = < − Single Bond: σ (1 sigma) Double Bond: σ and π (1 sigma and pi) Triple Bond: σ and 2π (1 sigma and 2 pi) VSEPR Bonding Sites Hybridization Shape Bond Angle Example 1 or 2 sp Linear 180° HF CO2
3 sp2 Trigonal Planar 120° BCl3
Bent < 120° SO2
4 sp3 Tetrahedral 109.5° CH4
Trigonal Pyramidal 107° NH3
Bent 105° H2O
5 sp3d Trigonal Bipyramidal 90°/120° PCl5
Seesaw 90°/<120° SF4
T-Shaped 90° ClF3
Linear 180° XeF2
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6 sp3d2 Octahedral 90° SF6
Square Pyramidal 90° BrF5
Square Planar 90° XeF4
Common Resonance Structures NO2 Dimerization into N2O4
10 • Liquids and Solids
Phase Diagram of Water Triple Point Diagram of Water
Boiling Point is when the Vapor Pressure = Atmospheric Pressure
If the Melting / Freezing Point line is below the temperature of the Triple Point, the substance will sink in its liquid form (solid will be more dense than
the liquid).
Intermolecular Forces Covalent Network Metallic Ionic Covalent:
o Polar - Asymmetric, lone pairs of e- Hydrogen Bonding Dipole-dipole forces
o Nonpolar London Dispersion forces
11 • Solutions
Concentration Calculations
( )
Freezing Point Depression or Boiling Point Elevation Problems
( )
( )
i = Van Hoff’t Factor = # of particles in solution C6H12O6 i = 1 NaCl i = 2 Na2SO4 i = 3 Ca3(PO4)2 i = 5
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Colligative Properties Vapor Pressure is Lowered PA = Ptotal x Mole Fraction Boiling Point Elevation Freezing Point Depression Osmosis = i MRT
Solubility Raoult’s Law – Like Dissolves Like Gases - Henry’s Law – Solubility of a Gas is proportional to Gas Pressure; High Pressure, Low Temperature = Higher Solubility of a Gas Solids and Liquids – Increase Temperature (KE), Increase # of Surface Areas, Add a Catalyst, Add More Solvent will all Increase Solubility
12 • Chemical Kinetics: Rates of Reaction
Reaction Rate
rate = [chemical]/time Common Units: M/s, mol·L-1·s-1
o rate of disappearance of reactant or o rate of appearance of product o use coefficients to change one rate to another
Reaction: 2A + 3B 4C
t
A][21
=
t
B][31
=
t
C][41
watch your signs ([React.] = -[Prod.]) From a graph of [R] vs time
Average rate is the slope of a segment. Instantaneous rate is slope of the tangent. Initial rate is often used.
How to Speed Up a Reaction [Use Collision Theory, Kinetic Molecular Theory] o increase the concentration of reactants
- increase molarity of solutions - increase partial pressure of gases [collision model: more collisions] o more surface area between unlike phases [collision model: more collisions] o increase the temperature [collision model: more & harder collisions] o add a catalyst requires lower energy collision orensures that correct particles collide]
- homogeneous catalyst (used & reformed) - heterogeneous catalyst (surface catalyst)[
Rate Laws
Equation: A + B C Rate = k [A]x[B]y k is the “specific rate constant” Use experimental data to determine x, y, and k. The Rate Law CANNOT be determined from the overall reaction. It MUST be determined experimentally because the rate law reflects only the “rate determining step.” Rate law can be determined from initial rates. Rate Law matches the Molecularity of the Rate Determining Step in the Mechanism
Examples for: 2A + 3B C
Rate Law Rate Determining Step in the mechanism
Rate = k [A][B] A + B X (slow)
Rate = k [A]2 A + A X (slow)
Rate = k [A]2[B]
B + X Y (slow) Each step is usually bimolecular. A third order overall reaction often comes from a fast equilibrium before a slow step.
Rate = k
This could be a mechanism that depends on a catalyst only. The concentrations would not matter.
order of rxn
- first and second order reactions
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- what these look like graphically - how you can graphically tell the order of a reaction
order straight-line plot Slope
0 [A] vs. t -k
1 ln[A] vs. t -k
2 1/[A] vs. t k
Order Rate law Integrated rate law Half life
0 Rate = k [A] = -kt + [A]0 t1/2 = [A]0/2k
1 Rate = k[A] ln[A] = -kt + ln[A]0 t1/2 = 0.693/k
2 Rate = k[A]2 1/[A] = kt + 1/[A]0 t1/2 = 1/k[A]0
Two Important Diagrams
o PE energy profile of a reaction +50
+35
0
–35
–50
a
b
c
d
e
PE
reaction coordinate ∆H of the reaction relates reactant and product PE’s / exo- or endothermic/ downhill, -H, or uphill, +H
activation energy (Ea) = energy barrier
activated complex (at the peak)
whether a reaction is fast or slow depends on the activation energy in the PE profile
PE profile does not change with change in temperature of the reactants?
adding a catalyst lowers the Ea Reaction mechanisms - step-by-step...two particles at a time - example - overall reaction is sum of steps - slowest step is rate-determining step half-life
- relationship to radioactivity (a first order reaction) - the equation
ktA
A
t
o ][
][ln
- the special case of half-life ln(2) = 0.693 = kt½ Arrhenius Equation Finding activation energy: k = Ae-Ea/RT
or ln(k) = - Ea (1/T) + ln(A) R ln(k2/k1) = Ea/R (1/T1 – 1/T2)
13 Chemical Equilibria
aA +bB + . . . rR +sS + . . .
Kc =
ba
sr
[B][A]
[S][R]
and for gases:
16
Kp = b
B
a
A
SR
)(P)(P
)P()(P sr
K > 1 products favored K < 1 reactants favored
3. Excluded: solids; pure liquids; water (in aqueous solutions) because their [ ]’s do not change. Convert from Kc to Kp
Kp = Kc(RT)n
where n = moles of gaseous product – moles of gaseous reactant. Typical question: Given Kc and the starting concentrations of reactants, find concentrations of products at equilibrium.
Example: Kc for acetic acid = 1.8 x 10-5.
What is the equilibrium concentration of [H+] in a 0.100 M solution of the acid?
Equilibrium constant for a reverse reaction = K
1 the value of the forward reaction.
Equilibrium constant for a doubled reaction = K2. When using Hess’s Law:
Koverall = K1 x K2
Le Châtelier’s Principle: effect of changes in concentration, pressure, & temperature. Equilibrium always “shifts” away from what you add. “Stress” means too much or too little: chemical, heat, or room.
If out of equilibrium: Calculate the reaction quotient (Q) similar to the way an equilibrium constant would be found. If:
Q < K forward reaction occurs to reach equilibrium Q > K reverse reaction occurs to reach equilibrium
Problem solving:
Set up problems using the “magic box” (or ICE box) C = “change” or . Example: A B + C
A B C
initial 5.0 M 0 M 0 M
equilibrium
“” row only follows the stoichiometry of the equation.
Learn when to make an approximation (needed for multiple choice questions!) 5% rule usually works when value of K is 103 smaller than value of known concentrations.
Example: A B + C
K = 3.0 x 10-6 if [A] = 5.0M initially; find [C] at equilibrium.
Another easy to solve situation is the perfect squares situation. Example: H2 + I2 2HI K = 3.5 x 102 Calculate [HI] when [H2] = [I2] = 0.10 M
14&15 Acids and Bases, Aqueous Equilibria
Acid-Base Theories Arrhenius: Acids – Have H+ Bases – Have OH-
Bronsted-Lowry: Acids – H+ donors Bases – H+ acceptors
Lewis: Acids – e- pair acceptor Bases – e- pair donors
BF3 + :NH3 BF3NH3
Strong and Weak Acids/Bases Strong Acids HCl, HBr, HI, HNO3, HClO4, H2SO4
Strong Bases Group I and II Hydroxides
Weak Acids All others including HF and HC2H3O2 (can abbreviate them as HA)
Weak Bases All others including NH3 (can abbreviate them as B)
17
pH Calculations pH + pOH = 14
[H+] [OH-] = 1 x 10-14
pH = - log [H+] [H+] = 10^[-pH]
pOH = - log [OH-]
Dilutions or Titrations M1V1 = M2V2
Equivalence Point - when the moles of acid = moles of base
Titration Curves Strong Acid Titrated with a Strong Base (Equivalence Point at pH = 7.00)
Weak Acid Titrated with a Strong Base (Equivalence Point above pH = 7.00)
Weak Base Titrated with a Strong Acid (Equivalence Point below pH = 7.00)
16 Entropy and Free Energy
There are two driving forces for reactions. Reactions tend toward:
minimum Enthalpy, H (heat energy)
H<0, downhill
maximum Entropy, S (randomness)
S +, S>0, uphill
Recognize whether S >0 or < 0.
Entropy increases, S +, S > 0:
from solid to liquid to gas
fewer moles (g) to more moles (g)
simpler molecules to more complex molecules
smaller molecules to longer molecules
ionic solids with strong attractions to ionic solids with weaker attractions
separate solute & solvent to solutions
gas dissolved in water to escaped gas
Product or Reactant favored reactions depend on H, S, and absolute Temp
H S Product-Favored…
+ + at higher temperatures
at lower temperatures
+ at all temperatures
+ never
(reactant-favored at all temps)
Many books use the term “spontaneous” for “product-favored.” A spontaneous reaction does not necessarily mean a fast reaction. The SPEED of a reaction is Kinetics (Ch 15)… we are discussing whether a reaction CAN OCCUR which is Thermodynamics (Ch 6 and Ch 20).
Gibbs Free Energy, G, puts the effects of H, S, and Temperature together.
G = H - TS
G<0, -favored reaction
G>0, G +, reactant-favored reaction
G=0, reaction is at equilibrium Important:
Note that H is usually in kJ/mol
S is usually in J/mol·K
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Convert between K, G, and E using equations given on the AP Exam.
17 Electrochemistry Electrochemistry is all oxidation-reduction chemistry. Leo Ger OIL RIG Oxidation: loss of e ; ox # increases Reduction: gain of e ; ox # decreases
example: Fe2+ + 2e Fe(s) (reduction) In a reaction, the
oxidizing agent gets reduced; the reducing agent gets oxidized.
Balancing redox reactions: half-reaction method. o determine oxidation & reduction o write two separate half-reactions o balance all atoms except H & O o balance O’s (add H2O’s) o balance H’s (add H+’s) o add e ‘s to more positive side o balance e-‘s between half-reactions o combine half-reactions o adjust for basic solution if needed
Electricity can either cause a reaction (electrolysis, electrolytic cell) or can be produced by the reaction (Galvanic cell, electrochemical cell,
Voltaic cell). Electrolysis / Electroplating coulomb (C) = an amount of charge amp = current = charge per second
1 amp · 1 second = 1 Coulomb 1 C / amp·s
Faraday constant, F:
1 mole e- = 96,500 C Electrolysis calculations begin with amp·s Example: How many moles of copper metal can be plated using a 10 amp circuit for 30 s? 10amp x 30s x 1C x 1 mol e- x 1 mol Ag = 1 amp·s 96500C 1 mol e- = 3.1 x 10-3 mole Ag Spontaneous redox reactions (unlike electrolysis/electroplating) can simply occur (as in the ornament lab) or can be separated so the oxidation
and reduction occur in different containers (half-cells). In this way, the electrons must move through an outside wire (this is an electrochemical cell—a battery).
Every atom has a different “potential” to accept electrons… “reduction potential”
Ag+(aq) + e¯ Ag(s) E° = +0.80 v
Cd2+(aq) + 2e¯ Cd(s)
These are measured by comparing every chemical to the same “standard half-cell.”
The reduction with the more positive E value will occur as written; the other reaction will reverse (oxidation). Ex: 2Ag+ + Cd 2Ag + Cd2+
The difference in the E values is the voltage of a cell made using these two reactions. Ex: +0.80 v – (-0.40 v) = 1.20 volts NOTE that you do not multiply the Cd voltage by 2. Comparing every cell to the same standard cell accounts for this. Any change that drives the reaction forward will increase the cell’s voltage. In all electrochemical cells: Oxidation occurs at the Anode
19
Reduction occurs at the Cathode
TYPES OF REACTIONS
SOLUBILITY RULES
SOLUBLE COMPOUNDS EXCEPTIONS
All Group 1 salts None
All ammonium (NH4+) salts None
All NO3−, ClO3
−, ClO4−, and C2H3O2
− salts None
All Cl−, Br−, I− salts Ag+, Hg22+ (mercury (I)), Pb2+
All F− salts Mg2+ Ca2+, Sr2+, Ba2and Pb2+
All salts of SO42− Ca2+, Sr2+, Ba2+, Pb2+, Ag+, Hg2
2+
INSOLUBLE COMPOUNDS EXCEPTIONS
All salts of OH− Group I, NH4+, Ba2+, Sr2+, Ca2+
All salts of S2−, SO32−, CO3
2−, PO43−, CrO4
2− and any other polyatomic not named! Group I and NH4+
Oxides*
* some of these oxides are actually “soluble” because they are basic anhydrides and react with water to form a base:
MO + H2O M(OH). More about this later!
STRONG ACIDS - ionize 100% in water
Type Formula
Hydrogen halides (aq) HCl HBr HI
Oxyacids of halogens HClO3 HBrO3 HIO3
HClO4 HBrO4 HIO4
Sulfuric (1st H+ only!!) H2SO4
Nitric Acid HNO3
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STRONG BASE - dissociate 100% in water. All hydroxides of group I and II except beryllium and magnesium (okay,
Mg(OH)2 tends to decompose but we will neglect that little detail!)
DOUBLE REPLACEMENT ODDITIES How to recognize: Ionic &/or acid with Ionic &/or acid
Decompose: H2CO3 H2O + CO2
Decompose: H2SO3 H2O + SO2
Decompose: NH4OH NH2+ H2O
Aqueous NH3 as a reactant: WRITE NH3 + H2O THINK NH4OH
SINGLE REPLACEMENT ODDITIES How to recognize:
metal plus an ionic compound
Active metal plus water
Active metal plus acid
halogen plus an ionic halide
hydrogen with an ionic compound Higher Oxidation States: “As Snoopy Fell, Huge Cups Cracked”
Diatomics: HOFBrINCl
COMPLEX ION ODDITIES How to recognize: Al or transition metal ion with a ligand such as CN─, OH─, halogen ion, SCN─, NH3, H2O
Metal ion is the Lewis Acid
Ligand has extra pair of electrons and is the Lewis Base
If reactant is Al metal, H2 gas forms
Fe2+
has a coordination # of “6” instead of “4”
Fe3+
only coordinates one SCN─ (ie Fe(SCN)
2+)
Decomposition is like a double replacement with an acid
REDOX RULES How to recognize: Look for words like:
Acidified or acidic
Alkaline or basic
Concentrated or dilute
Oxidizing agents (ie will be reduced) Products formed
MnO41−
(acidic sol’n) Mn2+
MnO41−
(neutral or basic sol’n) MnO2(s)
MnO2 (acidic sol’n) Mn2+
Cr2O72−
(acidic sol’n) Cr3+
HNO3 (concentrated) NO2
HNO3 (dilute) NO
H2SO4 (Concentrated, hot) SO2
21
Na2O2 (peroxide) NaOH
H2O2 (peroxide) H2O (or HOH)
ClO4− (In HClO4 – str acid) Cl
−
Free Halogens (F2, Cl2, Br2, I2) Halide ion
Metal-ic ions (higher oxidation state) Metal-ous ions (lower)
Reducing Agent (ie will be oxidized) Product formed
C2O42−
(oxalate ion) CO2
HCOOH (formic acid) CO2
H2O2 O2
Halide ions (F−, Cl
−, Br
−, I
−) Free halogen
Free metals metal ion*
Metal-ous Metal-ic
SO32−
(sulfite) or SO2 SO42−
(sulfate)
NO2− (nitrite) NO3
− (nitrate)
Free halogens (dilute basic) ex. Cl2 Hypohalite (ClO−)
Free Halogens (conc. Basic) ex. Cl2 Halate ion (ClO3−)
* “As Snoopy Fell, Huge Cups Cracked”
Electrolysis of pure molten 2NaCl 2Na + Cl2
Electrolysis of aqueous NaCl: water is electrolyzed instead
Know the ½ rxns for water electrolysis!
Cathode (reduction): 2H2O(l) + 2e− → H2(g) + 2OH−
(aq)
Anode (oxidation): 4OH−(aq) → O2(g) + 2H2O(l) + 4e−
22
SYNTHESIS/COMBINATION/ADDITION
How to recognize: a. Element + element b. Element + compound c. Compound + compound
Phosphorus exists as the red form (P) or the white form (P4). Sulfur’s most stable form is rings with eight sulfur atoms (S8)
Rules to memorize: a. Sulfur dioxide + metal oxide metal sulfite b. Sulfur trioxide + metal oxide metal sulfate c. Carbon dioxide + metal oxide metal carbonate d. BX3 + NY3 X3BNY3
DECOMPOSITION
How to recognize: Compound heated, often in presence of a catalyst.
Rules to memorize: a. Metal carbonates metal oxides + CO2 b. Metal sulfites metal oxides + SO2 c. Metallic chlorates metallic chlorides + oxygen. d. Ammonium Carbonate ammonia + water + CO2
e. Ammonium nitrate N2 + O2 + H2O (or N2O + H2O)
f. Hydrogen peroxide H2O + O2 g. Carbonic acid SPONTANEOUSLY H2O + CO2 h. Sulfurous acid SPONTANEOUSLY H2O + SO2 i. Sodium hydrogen carbonate sodium carbonate + CO2 + H2O
COMBUSTION
How to recognize: a. adding oxygen to a compound b. Look for words such as “burned”, “undergoes combustion”. NOTE: You need to supply the oxygen because it
will not be explicitly given!
RULES TO MEMORIZE!
element Most common oxygen containing cmpd
C CO in limited oxygen
C CO2 in excess oxygen
S8 SO2 in limited oxygen (Assume if lim or xs not indicated)
S8 SO3 in excess oxygen
N NO in limited oxygen
N NO2 in excess oxygen(Assume if lim or xs not indicated)
P4 P2O5 or P4O10
H H2O
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Metal Metallic oxide
ANHYDRIDES
How to recognize: Anhydride means “without water” so we are taking a substance that is without water and re-hydrating it so to speak
Rules: 1. Metal oxides are “basic anhydrides” 2. Metallic hydrides plus water yield metallic hydroxides and hydrogen gas. 3. Phosphorus halides and phosphorus oxyhalides react with water to produce two acids:
phosphorus oxyacid and a hydrohalic acid (HCl, HBr, HI). 4. Group I and II nitrides react with water to form a base and ammonia
5. Metal Carbides react with water to form metal hydroxides and methane. KEY: OXIDATION NUMBERS DO NOT CHANGE EXCEPT THE HYDRIDES! CAUTION: Sometimes an anhydride will be combined with an acid base reaction! Example Sulfur dioxide is bubbled through a solution of excess strontium hydroxide.
SO2 + H2O + Sr(OH)2 (aq) H2SO3 + Sr(OH)2 (aq) SrSO3 (s) + H2O
SO2 + H2O + Sr2+ + OH− (aq) H2SO3 + Sr(OH)2 (aq) SrSO3 (s) + H2O
ACIDS & BASES & SALT HYDROLYSIS
How to Recognize: a. Look for an acid plus a base. This is really a subset of double replacement with products
typically being a salt and water. b. If the acid is polyprotic, the product may still be an acid. c. Look for the words “equal molar”, “equal volume”, “same moles”, “twice the moles”, “excess”.
You will need to consider whether the acid or base is fully or partially neutralized. BE CAREFUL! Sometimes anhydrides are mixes with acid/base neutralization. CAUTION! Don’t forget to decompose H2CO3 & NH4OH if they form in a hydrolysis reaction!
Strategy d. Follow a similar strategy as double replacement. e. If the words such as “equal molar” are used, then put the mole ratio under the species in the
complete molecular.
Example: Equimolar volumes of phosphoric acid and sodium hydroxide solutions are mixed. H3PO4 (aq) + LiOH(aq) LiH2PO4 (aq) + H2O
24
AP Subject Review
3 Stoichiometry
Percentage Composition
Calculate the percent of each element in the total mass of the compound
(#atoms of the element)(atomic mass of element) x 100 (molar mass of the compound
Determining the empirical formula
Determine the percentage of each element in your compound
Treat % as grams, and convert grams of each element to moles of each element
Find the smallest whole number ratio of atoms
If the ratio is not all whole number, multiply each by an integer so that all elements are in whole number ratio
Determining the molecular formula
Find the empirical formula mass
Divide the known molecular mass by the empirical formula mass, deriving a whole number, n
Multiply the empirical formula by n to derive the molecular formula
Examples:
60 grams propane gas is burned in excess oxygen: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
How much water is produced?
7.321 mg of an organic compound containing carbon, hydrogen, and oxygen was analyzed by
combustion. The amount of carbon dioxide produced was 17.873 mg and the amount of
water produced was 7.316 mg. Determine the empirical formula of the compound.
Sodium metal reacts vigorously with water to produce a solution of sodium hydroxide and
hydrogen gas:
2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)
What mass of hydrogen gas can be produced when 10 grams of sodium is added to 15 grams
of water?
0.1101 gram of an organic compound containing carbon, hydrogen, and oxygen was analyzed
by combustion. The amount of carbon dioxide produced was 0.2503 gram and the amount
of water produced was 0.1025 gram. A determination of the molar mass of the compound
indicated a value of approximately 115 grams/mol. Determine the empirical formula and
the molecular formula of the compound.
4 Chemical Equations and Stoichiometry
Solving Stoichiometry Problems for Reactions in Solution
Identify the species present in the combined solution, and determine what reaction occurs.
Write the balanced net ionic equation for the reaction.
Calculate the moles of reactants.
Determine which reactant is limiting.
Calculate the moles of product(s), as required.
Convert to grams or other units, as required.
Example:
10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate
solution (assume no volume change).
25
Types of Reactions
Precipitation Reactions
A double displacement reaction in which a solid forms and separates from the solution.
Simple Rules for Solubility
Most nitrate (NO3-) salts are soluble.
Most alkali metal (group 1A) salts and NH4+ are soluble.
Most Cl-, Br
-, and I
-salts are soluble (except Ag
+, Pb
2+, Hg2
2+).
Most sulfate salts are soluble (except BaSO4, PbSO4, Hg2SO4, CaSO4).
Most OH- are only slightly soluble (NaOH, KOH are soluble, Ba(OH)2, Ca(OH)2 are marginally soluble).
Most S2-
, CO32-
, CrO42-
, PO43-
salts are only slightly soluble, except for those containing the cations in Rule 2.
Complete Ionic Equation
All substances that are strong electrolytes are represented as ions.
Ag+(aq) + NO3
-(aq) + Na
+(aq) + Cl
-(aq)
AgCl(s) + Na+(aq) + NO3
-(aq)
Net Ionic Equation
Includes only those solution components undergoing a change.
Show only components that actually react.
Acid–Base Reactions (Brønsted–Lowry)
Acid—proton donor
Base—proton acceptor
For a strong acid and base reaction:
H+(aq) + OH
–(aq) H2O(l)
Redox Reactions
Reactions in which one or more electrons are transferred.
Rules for Assigning Oxidation States
Oxidation state of an atom in an element = 0
Oxidation state of monatomic ion = charge of the ion
Oxygen = -2 in covalent compounds (except in peroxides where it = -1)
Hydrogen = +1 in covalent compounds
Fluorine = -1 in compounds
Sum of oxidation states = 0 in compounds
Sum of oxidation states = charge of the ion in ions
Balancing Oxidation–Reduction Reactions by Oxidation States
Write the unbalanced equation.
Determine the oxidation states of all atoms in the reactants and products.
Show electrons gained and lost using “tie lines.”
Use coefficients to equalize the electrons gained and lost.
Balance the rest of the equation by inspection.
Add appropriate states.
Example:
What coefficients are needed to balance the remaining elements?
Zn(s) + 2HCl(aq) Zn2+
(aq) + 2Cl–(aq) + H2(g)
dilute nitric acid is added to crystals of pure calcium oxide.
hydrogen sulfide is bubbled through a solution of silver nitrate.
26
sodium metal is added to water.
a solution of tin(II) chloride is added to a solution of iron(III) sulfate.
phosphorus(V) oxytrichloride is added to water.
solid sodium oxide is added to water.
excess concentrated potassium hydroxide solution is added to a precipitate of zinc hydroxide.
excess concentrated sodium hydroxide solution is added to solid aluminum hydroxide.
5 Gases
Standard conditions
STP = 1 atm (760mmHg or 101kPa) and 273K ALL temperatures must be in Kelvin! (C + 273)
Gas Laws:
Boyles Law: P1V1 = P2V2
Charles Law: V1/T1 = V2/T2
Guy Lussac’s Law: P1/T1 = P2/T2
Combined Gas Law: V1P1/n1T1 = V2P2/n2T2
Ideal Gas Law: PV = nRT volume in liters R = 0.0821 atm L/molK
Daltons Law: Pt = P1 + P2 …
Mole Fraction: mol A/total moles
Root Mean Square: FW
RTuu rms
32
Grahams Law: FW
FW =
2 gas of effusion of Rate
1 gas of effusion of Rate
1
2
Examples:
A sample of hydrogen gas (H2) has a volume of 8.56 L at a temperature of 0ºC and a pressure of
1.5 atm. Calculate the moles of H2 molecules present in this gas sample.
A sample of gas at 15ºC and 1 atm has a volume of 2.58 L. What volume will this gas occupy at
38ºC and 1 atm ?
Sulfur dioxide (SO2), a gas that plays a central role in the formation of acid rain, is found in the
exhaust of automobiles and power plants. Consider a 1.53- L sample of gaseous SO2 at a
pressure of 5.6 x 103 Pa. If the pressure is changed to 1.5 x 10
4 Pa at a constant temperature,
what will be the new volume of the gas ?
Suppose we have a sample of ammonia gas with a volume of 3.5 L at a pressure of 1.68 atm.
The gas is compressed to a volume of 1.35 L at a constant temperature. Use the ideal gas
law to calculate the final pressure
Mixtures of helium and oxygen are used in scuba diving tanks to help prevent “the bends.” For
a particular dive, 46 L He at 25ºC and 1.0 atm and 12 L O2 at 25ºC and 1.0 atm were
27
pumped into a tank with a volume of 5.0 L. Calculate the partial pressure of each gas and
the total pressure in the tank at 25ºC.
Calculate the ratio of the effusion rates of hydrogen gas (H2) and uranium hexafluoride (UF6), a
gas used in the enrichment process to produce fuel for nuclear reactors.
7 • Atomic Theory and Bonding
Know the names and locations of the groups on the periodic table: alkali metals, alkaline earth metals, inner and outer
transition metals, halogens and noble gases and their common charges as ions
Electrons
Be able to write electron configurations for any element
Aufbau Principle Electrons fill their orbitals from lower to higher energy: 1s2s293s3p4s3d4p5s4d5p6s4f5d6p7s
Hund’s Rule An electron will half fill an orbital until all orbitals of that sublevel are occupied
Periodic Trends
Ionization Energy
o Energy required to remove an electron from a gaseous atom or ion. X(g) → X+(g) + e
–
o In general, as we go across a period from left to right, the first ionization energy increases.
Why? Electrons added in the same principal quantum level do not completely shield the
increasing nuclear charge caused by the added protons.
o In general, as we go down a group from top to bottom, the first ionization energy decreases.
Why? The electrons being removed are, on average, farther from the nucleus.
Electron Affinity
o Energy change associated with the addition of an electron to a gaseous atom. X(g) + e– → X
–(g)
o In general as we go across a period from left to right, the electron affinities become more negative.
o In general electron affinity becomes more positive in going down a group.
Atomic Radius
o In general as we go across a period from left to right, the atomic radius decreases.
o Effective nuclear charge increases, therefore the valence electrons are drawn closer to the nucleus,
decreasing the size of the atom.
o In general atomic radius increases in going down a group.
o Cations have lost electrons and are generally smaller than the neutral atom (more nuclear attraction)
o Anions have gained electrons and are larger than the neutral atom (less nuclear attraction)
Examples
Write the electron configuration for Strontium and Sodium using the shorthand notation for the
noble gas cores.
How many unpaired electrons are there in a nitrogen atom?
Arrange the elements S, Ge, P, and Si in order of increasing atomic size.
Arrange the ions Na+, K
+, Cl , and Br in order of increasing size.
Arrange the elements Be, Ca, N, and P in order of increasing ionization energy.
8•Bonding
Types of Chemical Bonds
Ionic Bonding – electrons are transferred
28
Covalent Bonding – electrons are shared equally by nuclei
Polar Covalent Bond Unequal distribution of electrons between atoms in a molecule resulting in a partial positive
and negative region in the molecule
Nonpolar covalent Bond: Equal distribution of the electron cloud
Electronegativity:
increases as you go across the period table and decreases as you go down. Fluorine has an electronegativity of 4
and is the most electronegative element. Noble gases have 0 electronegativity. The greater the e difference, the
more ionic charater the bond has and the stronger it is.
Dipole Moment
Property of a molecule whose charge distribution can be represented by a center of positive charge and a center of
negative charge.
Use an arrow to represent a dipole moment.
Lattice Energy
The change in energy that takes place when separated gaseous ions are packed together to form an ionic solid.
k = proportionality constant
Q1 and Q2 = charges on the ions
r = shortest distance between the centers of the cations and anions
Bond Energy
ΔH = Σn X D(bonds broken) – Σn X D(bonds formed)
Lewis Structure
Shows how valence electrons are arranged among atoms in a molecule.
Octet Rule All atoms should have 8 electrons exept for hydrogen with 2
Steps for Writing Lewis Structures
Sum the valence electrons from all the atoms.
Use a pair of electrons to form a bond between each pair of bound atoms.
Arrange the remaining electrons to satisfy the octet rule (or duet rule for hydrogen).
Resonance occurs when you have a combination of multiple and single bonds:
Formal Charge
Formal charge = (# valence e– on free neutral atom) – (# valence e
– assigned to the atom in the molecule).
VSEPR Model
29
Hybridization
explains why bonds in molecules with different atomic orbitals behave as identical bonds ie. CH4
sigma bond: overlap of two s orbitals or an s and a p orbital or head-to-head p orbitals.
Electron density of a sigma bond is greatest along the axis of the bond.
Pi (π) bonds--come from the sideways overlap of p atomic orbitals; the region above and below the internuclear
axis. NEVER occur without a sigma bond first!
30
HYBRIDIZATION # OF HYBRID
ORBITALS
GEOMETRY
sp 2 Linear
sp2 3 Trigonal planar
sp3 4 Tetrahedral
dsp3 5 Trigonal bipyramidal
d2sp
3 6 Octahedral
Example
Draw a Lewis structure for NH3; CH4; SF6
What molecular shapes are associated with the following electron pairs around the central
atom?
a. 3 bonding pairs and 2 lone pairs
b. 4 bonding pairs and 1 lone pair
Name the shapes and hybridization for the following molecules or polyatomic ions:
a. O3
b. GaH3
Determine whether the following molecules are polar or nonpolar:
a. CCl4
b. XeF4
10 • Liquids and Solids
Intermolecular Forces
London Dispersion forces found in all molecules. It is the only IMF in nonpolar molecules. The larger the
molecule, the stronger the LDFs
Dipole dipole forces found in polar molecules
Hydrogen bonding between H and F,O or N
A polar molecule has polar bonds and is asymmetric
Some Properties of a Liquid
Surface Tension: The resistance to an increase in its surface area (polar molecules). High ST indicates strong
IMF’s.
Capillary Action: Spontaneous rising of a liquid in a narrow tube.
Viscosity: Resistance to flow (molecules with large intermolecular forces). Modeling a liquid is difficult.
Gases have VERY SMALL IMFs and lots of motion. Solids have VERY HIGH IMFs and next to no motion. Liquids
have both strong IMFs and quite a bit of motion.
Types of Solids
Crystalline Solids: highly regular arrangement of their components [often ionic, table salt (NaCl), pyrite (FeS2)].
Amorphous solids: considerable disorder in their structures (glass).
Representation of Components in a Crystalline Solid
Lattice: A 3-dimensional system of points designating the centers of components (atoms, ions, or molecules) that make
up the substance.
(a) network covalent—carbon in diamond form—here each molecule is covalently bonded to each neighboring C
with a tetrahedral arrangement. Graphite on the other hand, make sheets that slide and is MUCH softer! (pictured
later)
(b) ionic salt crystal lattice
(c) ice—notice the “hole” in the hexagonal structure and all the H-bonds. The “hole” is why ice floats—it makes
it less dense than the liquid!
31
Types of Crystalline Solids
Ionic Solid: contains ions at the points of the lattice that describe the structure of the solid (NaCl). VERY high
MP’s. Hard. Ion-Ion Coulombic forces are the strongest of all attractive forces. “IMF” usually implies
covalently bonded substances, but can apply to both types.
Molecular Solid: discrete covalently bonded molecules at each of its lattice points (sucrose, ice).
Atomic Solid: atoms of the substance are located at the lattice points. Carbon—diamond, graphite and the
fullerenes. Boron, and silicon as well.
Know this chart well:
Structure and Bonding in Metals
Metals are characterized by high thermal and electrical conductivity, malleability, and ductility. These properties are
explained by the nondirectional covalent bonding found in metallic crystals.
Electron Sea Model: A regular array of metals in a “sea” of electrons. I A & II A metals pictured at left.
Band (Molecular Orbital) Model: Electrons assumed to travel around metal crystal in MOs formed from valence
atomic orbitals of metal atoms.
Metal alloys: a substance that has a mixture of elements and has metallic properties
substitution alloys—in brass 1/3 of the atoms in the host copper metal have been replaced by zinc atoms. Sterling
silver—93% silver and 7% copper. Pewter—85% tin, 7% copper, 6% bismuth and 2% antimony. Plumber’s
solder—95% tin and 5% antimony.
interstitial alloy—formed when some of the interstices [holes] in the closest packed metal structure are occupied
by small atoms. Steel—carbon is in the holes of an iron crystal. There are many different types of steels, all
depend on the percentage of carbon in the iron crystal.
Network Atomic Solids—a.k.a. Network Covalent
Composed of strong directional covalent bonds that are best viewed as a “giant molecule”. Both diamond and
graphite are network solids. The difference is that diamond bonds with neighbors in a tetrahedral 3-D fashion,
while graphite only has weak bonding in the 3rd
dimension. Network solids are often:
brittle—diamond is the hardest substance on the planet, but when a diamond is “cut” it is actually fractured to
make the facets
do not conduct heat or electricity
carbon, silicon-based
Diamond is hard, colorless and an insulator. It consists of carbon atoms ALL bonded tetrahedrally, therefore sp3
hybridization and 109.5 bond angles.
11 • Solutions
Definitions:
Solution- a homogeneous mixture of two or more substances in a single phase.
solute--component in lesser concentration; dissolvee
solvent--component in greater concentration; dissolver
solubility--maximum amount of material that will dissolve in a given amount of solvent at a given temp. to
produce a stable solution.
Saturated solution- a solution containing the maximum amount of solute that will dissolve under a given set of
conditions.
Unsaturated solution- a solution containing less than the maximum amount of solute that will dissolve under a
given set of conditions. (more solute can dissolve)
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Supersaturated solution- a solution that has been prepared at an elevated temperature and then slowly cooled.
miscible—When two or more liquids mix (ex. Water and food coloring)
immiscible—When two or more liquids DON’T mix.--they usually layer if allowed to set for a while. (ex.
Water and oil)
Units of solution concentration
Molarity (M) = # of moles of solute per liter of solution
Mole fraction () = ratio of the number of moles of a given component to the total number of moles of solution.
Molality (m) = # of moles of solute per kilogram of solvent
Heat of solution (Hsoln) = the enthalpy change associated with
Factors Affecting Solubility
Pressure Effects:
o The solubility of a gas is higher with increased pressure.
o Henry’s Law- the amount of a gas dissolved in a solution is directly proportional to the pressure of the gas
above the solution. P = kC
o P = partial pressure of the gaseous solute k = constant C = concentration of the gas
Temperature Effects:
o The amount of solute that will dissolve usually increases with increasing temperature. Solubility
generally increases with temperature if the solution process is endothermic (Hsoln > 0). Solubility
generally decreases with temperature if the solution process is exothermic (Hsoln < 0). Potassium
hydroxide, sodium hydroxide and sodium sulfate are three compounds that become less soluble as the
temperature rises. This can be explained by LeChatelier’s Principle.
o The solubility of a gas in water always decreases with increasing temperature.
Colligative Properties- properties that depend on the number of dissolved particles
Vapor Pressure Lowering- The presence of a nonvolatile solute lowers the vapor pressure of a solvent. Raoult’s
Law: Psolution = (solvent) (Posolvent)
o Psolution = observed vapor pressure of the solvent in the solution
o solvent = mole fraction of solvent
o Po
solvent = vapor pressure of the pure solvent
An ideal solution is a solution that obeys Raoult’s Law. There is no such thing. In very dilute solutions
o We can find the molecular weight of a solute by using the vapor pressure of a solution.
Boiling Point Elevation
BPE = mki m = molality k = BP constant i= number of ions
Freezing point depression
FPD = mki m = molality k = BP constant i= number of ions
12 • Chemical Kinetics: Rates of Reaction
FACTORS THAT AFFECT REACTION RATES
Nature of the reactants--Some reactant molecules react in a hurry, others react very slowly.
Concentration of reactants--more molecules, more collisions.
Temperature--heat >em up & speed >em up;
Catalysts--accelerate chemical reactions but are not themselves transformed.
Surface area of reactants--exposed surfaces affect speed
Rate = change in concentration of a species per time interval
When writing rate expressions, they can be written in terms of reactants disappearance or products appearance.
* Rate is not constant, it changes with time. Graphing the data of an experiment will show an average rate of reaction.
You can find the instantaneous rate by computing the slope of a straight line tangent to the curve at that time.
reaction rate--expressed as the Δ in concentration of a reagent per unit time or Δ[A]/Δt
Initial rxn rate = k[A]m
[B]n
[C]p
Exponents can be zero, whole numbers or fractions and are determined by experiment
ORDER OF A REACTION
order with respect to a certain reactant is the exponent on its concentration term in the rate expression
order of the reaction is the sum of all the exponents on all the concentration terms in the expression
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Zero order: The change in concentration of reactant has no effect on the rate. These are not very common. General
form of rate equation: Rate = k
First order: Rate is directly proportional to the reactants concentration; doubling [rxt], doubles rate. These are
very common! Nuclear decay reactions usually fit into this category. General form of rate equation: Rate = k [A]
Second order: Rate is quadrupled when [rxt] is doubled and increases by a factor of 9 when [rxt] is tripled etc.
These are common, particularly in gas-phase reactions.
General form of rate equation: Rate = k [A]2
TWO TYPES OF RATE LAWS
o differential rate law--expresses how the rate depends on concentration
o integrated rate law--expresses how the concentrations depend on time
INTEGRATED RATE LAW: CONCENTRATION/TIME RELATIONSHIPS
first order: second order:
ln[A] = -kt + ln[A]o 1/[A] = kt + 1/[A]o
HALF-LIFE AND REACTION RATE FOR FIRST ORDER REACTIONS, t1/2
the time required for one half of one of the reactants to disappear.
T1/2 = 0.693
k
Half life is INDEPENDENT OF ORIGINAL CONCENTRATION for 1st
order!!!
HALF-LIFE AND REACTION RATE FOR SECOND ORDER REACTIONS, t1/2
the time required for one half of one of the reactants to disappear.
k t2 = 1
[A]o
HALF-LIFE AND REACTION RATE FOR ZERO ORDER REACTIONS t2 = [A]o rxn
RATE EXPRESSIONS FOR ELEMENTARY STEPS--the rate expression cannot be predicted from overall
stoichiometry. The rate expression of an elementary step is given by the product of the rate constant and the
concentrations of the reactants in the step.
ELEMENTARY STEP MOLECULARITY RATE EXPRESSION
A products unimolecular rate = k[A]
A + B products bimolecular rate = k[A][B]
A + A products bimolecular rate = k[A]
2
2 A + B products* termolecular* rate = k[A]
2[B]
THE EFFECT OF TEMPERATURE OF REACTION RATE: ARRHENIUS EQUATION
k = reaction rate constant = Ae-E*/RT
o R is the ―energy‖ R or 8.31 x 10-3
kJ/K mol
o A is the frequency factor units of L/(mol s) & depends on the frequency of collisions and the fraction
of these that have the correct geometry--# of effective collisions
o e-E*/RT
is always less than 1 and is the fraction of molecules having the minimum energy required for reaction
CATALYSIS
catalysts are not altered during the reaction--they serve to lower the activation energy and speed up the reaction by offering
a different pathway for the reaction
HETEROGENEOUS CATALYST--
different phase than reactants, usually involves gaseous reactants adsorbed on the surface of a solid catalyst
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HOMOGENEOUS CATALYST—exists in the same phase as the reacting molecules.
13 Chemical Equilibria
THE EQUILIBRIUM EXPRESSION:
A general description of the equilibrium condition proposed by Gudberg and Waage in 1864 is known as the Law of Mass
Action. Equilibrium is temperature dependent, however, it does not change with concentration or pressure.
equilibrium constant expression--for the general reaction
aA + bB cC + dD
Equilibrium constant: K = [C]c[D]
d
[A]a[B]
b
o Pure solids--do not appear in expression—you’ll see this in Ksp problems soon!
o Pure liquids--do not appear in expression—H2O (l) is pure, so leave it out of the calculation
o Water--as a liquid or reactant, does not appear in the expression. (55.5M will not change significantly)
CHANGING STOICHIOMETRIC COEFFICIENTS
o when the stoichiometric coefficients of a balanced equation are multiplied by some factor, the K is raised to the
power of the multiplication factor (Kn). 2x is K squared; 3x is K cubed; etc.
o REVERSING EQUATIONS
o take the reciprocal of K ( 1/K)
o ADDING EQUATIONS
o multiply respective K=s (K1 x K2 x K3 …)
Kc & Kp--NOT INTERCHANGEABLE! Kp = Kc(RT)Δn
where
o Δn is the change in the number of moles of gas going from reactants to products:
o R = universal gas law constant 0.0821 L atm/ mol K
o T = temperature in Kelvin
Kc = Kp if the number of moles of gaseous product = number of moles of gaseous reactant.
THE REACTION QUOTIENT
For use when the system is NOT at equilibrium.
For the general reaction
aA + bB cC + dD
Reaction quotient = Qc = [C]c[D]
d
[A]a[B]
b
Qc has the appearance of K but the concentrations are not necessarily at equilibrium.
1. If Q<K, the system is not at equilibrium:
2. If Q = K, the system is at equilibrium.
3. If Q>K, the system is not at equilibrium:
EXTERNAL FACTORS AFFECTING EQUILIBRIA
Le Chatelier=s Principle: If a stress is applied to a system at equilibrium, the position of the equilibrium will shift in the
direction which reduces the stress.
Temperature—exothermic: heat is a product; endothermic: heat is a reactant.
Adding or removing a reagent--shift tries to reestablish Q.
Pressure--increase favors the side with the least # of gas moles; the converse is also true.
catalysts--NO EFFECT on K; just gets to equilibrium faster!
14&15 Acids and Bases, Aqueous Equilibria
ACID-BASE THEORIES:
ARRHENIUS DEFINITION
acid--donates a hydrogen ion (H+) in water
base--donates a hydroxide ion in water (OH-)
This theory was limited to substances with those "parts"; ammonia is a MAJOR exception!
BRONSTED-LOWRY DEFINITION
acid--donates a proton in water
base--accepts a proton in water
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conjugate acid-base pair--A pair of compounds that differ by the presence of one H+ unit. This idea is critical
when it comes to understanding buffer systems. Pay close attention here!
LEWIS DEFINITION
acid--accepts an electron pair
base--donates an electron pair
This theory explains all traditional acids and bases + a host of coordination compounds and is used widely in
organic chemistry. Uses coordinate covalent bonds
monoprotic--acids donating one H+
(ex. HC2H3O2)
diprotic--acids donating two H+'s (ex. H2C2O4)
polyprotic--acids donating many H+'s (ex. H3PO4)
polyprotic bases--accept more than one H+; anions with -2 and -3 charges (ex. PO4
3- ; HPO4
2-)
ACIDS ONLY DONATE ONE PROTON AT A TIME!!!
RELATIVE STRENGTHS OF ACIDS AND BASES
Strength is determined by the position of the "dissociation Do Not confuse concentration with strength!
STRONG ACIDS:
Hydrohalic acids: HCl, HBr, HI
Nitric: HNO3
Sulfuric: H2SO4
Perchloric: HClO4
STRONG BASES:
Hydroxides OR oxides of IA and IIA metals
THE STRONGER THE ACID THE WEAKER ITS CB, the converse is also true.
WEAK ACIDS AND BASES: are in equilibrium
HA + H2O H3O+ + A
-
Ka = [H3O+][A
-] < 1
[HA]
for weak base reactions:
B + H2O HB+ + OH
-
Keq[H2O]2 = Kw = [H3O
+][OH
-]
Kw = 1.0 x 10-14
( Kw = 1.008 x 10
-14 @ 25 degrees Celsius)
Kw = Ka x Kb (another very beneficial equation)
The pH Scale
Used to designate the [H+] in most aqueous solutions where H
+ is small.
pH = - log [H+]
pOH = - log [OH-]
pH + pOH = 14
Calculating pH of Weak Acid Solutions
Calculating pH of weak acids involves setting up an equilibrium. Always start by writing the equation, setting up the
acid equilibrium expression (Ka), defining initial concentrations, changes, and final concentrations in terms of X,
substituting values and variables into the Ka expression and solving for X. (use the RICE diagram learned in general
equilibrium!)
Determination of the pH of a Mixture of Weak Acids
Only the acid with the largest Ka value will contribute an appreciable [H+]. Determine the pH based on this acid and
ignore any others.
Calculating pH of polyprotic acids
Acids with more than one ionizable hydrogen will ionize in steps. Each dissociation has its own Ka value.
The first dissociation will be the greatest and subsequent dissociations will have much smaller equilibrium constants. As
each H is removed, the remaining acid gets weaker and therefore has a smaller Ka. As the negative charge on the acid
increases it becomes more difficult to remove the positively charged proton.
ACID-BASE PROPERTIES OF SALTS: HYDROLYSIS
Salts are produced from the reaction of an acid and a base. (neutralization)
Salts are not always neutral. Some hydrolyze with water to produce acidic and basic solutions.
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Neutral Salts- Salts that are formed from the cation of a strong base and the anion of a strong acid form neutral
solutions when dissolved in water. A salt such as NaNO3 gives a neutral solution.
Basic Salts- Salts that are formed from the cation of a strong base and the anion of a weak acid form basic
solutions when dissolved in water. The anion hydrolyzes the water molecule to produce hydroxide ions and thus
a basic solution. K2S should be basic since S-2
is the CB of the very weak acid HS-, while K
+ does not hydrolyze
appreciably. S2-
+ H2O OH- + HS
-
Acid Salts- Salts that are formed from the cation of a weak base and the anion of a strong acid form acidic
solutions when dissolved in water. The cation hydrolyzes the water molecule to produce hydronium ions and
thus an acidic solution. NH4Cl should be weakly acidic, since NH4+ hydrolyzes to give an acidic solution, while
Cl- does not hydrolyze. NH4
+ + H2O H3O
+ + NH3
If both the cation and the anion contribute to the pH situation, compare Ka to Kb.
If Kb is larger, basic! The converse is also true.
1. Strong acid + strong base = neutral salt
2. Strong acid + weak base = acidic salt
3. Weak acid + strong base = basic salt
4. Weak acid + weak base = ? ( must look at K values to decide )
THE LEWIS CONCEPT AND COORDINATE BONDS
acid--can accept a pair of electrons to form a coordinate covalent bond
base--can donate a pair of electrons to form a coordinate covalent bond
6 & 16 Thermochemistry
ENERGY AND WORK
E = q(heat) + w(work)
Signs of q
+q if heat absorbed
–q if heat released
Signs of w
+ w if work done on the system (i.e., compression)
-w if work done by the system (i.e., expansion)
When related to gases, work is a function of pressure
w = -PV
NOTE: Energy is a state function. (Work and heat are not.)
ENTHALPY
H is a state function
H = q at constant pressure (i.e. atmospheric pressure)
Enthalpy can be calculated from several sources including:
Coffee-cup calorimetry. q = H @ these conditions.
Bomb calorimetry – weighed reactants are placed inside a steel container and ignited.
Heat capacity – energy required to raise temp. by 1 degree (Joules/ C)
Specific heat capacity (Cp) – same as above but specific to 1 gram of substance
change) re temperatuof (degrees material) of (
ferredheat trans ofquantity heat specific
g
Molar heat capacity -- same as above but specific to one mole of substance
(J/mol K or J/mol C )
Energy (q) released or gained -- q = mCpT
q = quantity of heat ( Joules or calories)
m = mass in grams
ΔT = Tf - Ti (final – initial)
Cp = specific heat capacity ( J/gC)
Specific heat of water (liquid state) = 4.184 J/gC ( or 1.00 cal/g C)
Heat lost by substance = heat gained by water
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Enthalpy of a Reaction
Hrxn = Hf (products) - Hf (reactants)
Hess’s Law
sum up the H’s for the individual reactions to get the overall Hrxn.
First decide how to rearrange equations so reactants and products are on appropriate sides of the arrows.
If equations had to be reversed, reverse the sign of H
If equations had be multiplied to get a correct coefficient, multiply the H by this coefficient since H’s are in
kJ/MOLE (division applies similarly)
Check to ensure that everything cancels out to give you the exact equation you want.
Hint** It is often helpful to begin your work backwards from the answer that you want!
Bond Energies
Energy must be added/absorbed to BREAK bonds (endothermic). Energy is released when bonds are FORMED
(exothermic).
H = sum of the energies required to break old bonds (positive signs) plus the sum of the energies released in the
formation of new bonds (negative signs).
H = bonds broken – bonds formed
H = + reaction is endothermic
H = - reaction is exothermic (favored – nature tends towards lower energy)
ENTHALPY (H)
heat content (exothermic reactions are generally favored)
ENTROPY (S)
disorder of a system (more disorder is favored) Nature tends toward
chaos! Think about your room at the end of the week! Your mom will love this law.
S = - H
T
S = + MORE DISORDER (FAVORED CONDITION)
S = - MORE ORDER
Calculating Entropy
Srxn = S (products) - S (reactants)
FREE ENERGY
G = H - TS
G = G + RT ln (Q)
G = free energy not at standard conditions
G = free energy at standard conditions
R = universal gas constant 8.3145 J/molK
T = temp. in Kelvin
ln = natural log
Q = reaction quotient: Q = [products]
[reactants]
“RatLink”: G = -RTlnK
Grxn = G (products) - G (reactants)
SUMMARY OF FREE ENERGY:
G = + NOT SPONTANEOUS
G = - SPONTANEOUS
Conditions of G:
H S Result
negative positive spontaneous at all temperatures
positive positive spontaneous at high temperatures
negative negative spontaneous at low temperatures
positive negative not spontaneous, ever
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17 Electrochemistry
OIL RIG – oxidation is loss, reduction is gain (of electrons)
Oxidation – the loss of electrons, increase in charge
Reduction – the gain of electrons, reduction of charge
Oxidation number – the assigned charge on an atom
Oxidizing agent (OA) – the species that is reduced and thus causes oxidation
Reducing agent (RA) – the species that is oxidized and thus causes reduction
GALVANIC CELLS
Parts of the voltaic or galvanic cell:
Anode--the electrode where oxidation occurs. After a period of time, the anode may appear to become smaller as
it falls into solution.
Cathode-- the electrode where reduction occurs. After a period of time it may appear larger, due to ions from
solution plating onto it.
inert electrodes—used when a gas is involved OR ion to ion involved such as Fe3+
being reduced to Fe2+
rather
than Fe0. Made of Pt or graphite.
Salt bridge -- a device used to maintain electrical neutrality in a galvanic cell. This may be filled with agar which
contains a neutral salt or it may be replaced with a porous cup.
Electron flow -- always from anode to cathode. (through the wire)
Standard cell notation (line notation) -
anode/solution// cathode solution/ cathode Ex. Zn/Zn2+
(1.0 M) // Cu2+
(1.0M) / Cu
Voltmeter
measures the cell potential (emf) . Usually is measured in volts.
cell potential
Ecell, Emf, or cell—it is a measure of the electromotive force or the “pull” of the electrons as they travel from the
anode to the cathode [more on that later!]
volt (V)
The unit of electrical potential; equal to 1 joule of work per coulomb of charge transferred
voltmeter
measures electrical potential; some energy is lost as heat [resistance] which keeps the voltmeter reading a tad
lower than the actual or calculated voltage. Digital voltmeters have less resistance. If you want to get picky and
eliminate the error introduced by resistance, you attach a variable-external-power source called a potentiometer.
Adjust it so that zero current flows—the accurate voltage is then equal in magnitude but opposite in sign to the
reading on the potentiometer.
STANDARD REDUCTION POTENTIALS
Each half-reaction has a cell potential
Each potential is measured against a standard which is the standard hydrogen electrode [consists of a piece of inert
Platinum that is bathed by hydrogen gas at 1 atm]. The hydrogen electrode is assigned a value of ZERO volts.
standard conditions—1 atm for gases, 1.0M for solutions and 25C for all (298 K)
naught, Ecell, Emf, or cell become Ecello , Emf
o , or cell
o when measurements are taken at standard conditions.
Calculating Standard Cell Potential
Decide which element is oxidized or reduced using the table of reduction potentials. Remember: THE MORE
POSITIVE REDUCTION POTENITAL GETS TO BE REDUCED.
Write both equations AS IS from the chart with their voltages.
Reverse the equation that will be oxidized and change the sign of the voltage [this is now Eoxidation]
Balance the two half reactions **do not multiply voltage values**
Add the two half reactions and the voltages together.
Ecell = Eoxidation + Ereduction means standard conditions: 1atm, 1M, 25C
AN OX – oxidation occurs at the anode (may show mass decrease)
RED CAT – reduction occurs at the cathode (may show mass increase)
FAT CAT – The electrons in a voltaic or galvanic cell ALWAYS flow From the Anode To the CAThode
Ca+hode – the cathode is + in galvanic cells
Salt Bridge – bridge between cells whose purpose is to provide ions to balance the charge. Usually made of a salt
filled agar (KNO3) or a porous cup.
39
EPA--in an electrolytic cell, there is a positive anode.
CELL POTENTIAL, ELECTRICAL WORK & FREE ENERGY
V = work (J)/charge (C)
The work that can be accomplished when electrons are transferred through a wire depends on the “push” or emf which is
IF work flows OUT it is assigned a MINUS sign
When a cell produces a current, the cell potential is positive and the current can be used to do work THEREFORE and
work have opposite signs!
= - w
q therefore -w = q
faraday(F)—the charge on one MOLE of electrons = 96,485 coulombs
q = # moles of electrons x F
For a process carried out at constant temperature and pressure, wmax [neglecting the very small amount of energy
that is lost as friction or heat] is equal to G, therefore….
ΔGo = -nFE
o
G = Gibb’s free energy.
n = number of moles of electrons.
F = Faraday constant 9.6485309 x 104 J/V mol
-Eo implies nonspontaneous.
+Eo implies spontaneous (would be a good battery!)
Strongest Oxidizers are weakest reducers.
As Eo reducing strength .
As Eo oxidizing strength .
DEPENDENCE OF CELL POTENTIAL ON CONCENTRATION
Voltaic cells at NONstandard conditions: LeChatlier’s principle can be applied. An increase in the concentration of a
reactant will favor the forward reaction and the cell potential will increase. The converse is also true!
0.0592
NERNST EQUATION: E = Eo - ---------- log Q @ 25C (298K)
n
As E declines with reactants converting to products, E eventually reaches zero.
Zero potential means reaction is at equilibrium [dead battery]. Also, Q =K AND G = 0 as well.
Advanced Placement Chemistry
40
Practice AP Exam: Multiple Choice Questions
Notes
Note: For all questions involving solutions and/or chemical equations, assume that the system is in pure water at room temperature unless otherwise noted.
Questions 1-4 refer to the following types of energy.
(A) Activation energy
(B) Free energy
(C) Ionization energy
(D) Kinetic energy
(E) Lattice energy
1. The energy required to convert a ground-state atom in the gas phase to a gaseous positive ion
2. The energy change that occurs in the conversion of an ionic solid to widely separated gaseous ions
3. The energy in a chemical or physical change that is available to do useful work
4. The energy required to form the transition state in a chemical reaction
Question 5-8 refer to atoms for which the occupied atomic orbitals shown below.
5. Represents an atom that is chemically unreactive
6. Represents an atom in an excited state
7. Represents an atom that has four valence electrons.
8. Represents an atom of a transition metal.
Question 9-12 refer to aqueous solutions containing 1:1 mole ratios of the following pairs of substances.
Assume all concentrations are 1 M.
41
(A) NH3 and NH4Cl
(B) H3PO4 and NaH2PO4
(C) HCl and NaCl
(D) NaOH and NH3
(E) NH3 and HC2H3O2 (acetic acid)
9. The solution with the lowest pH
10. The most nearly neutral solution
11. A buffer at a pH > 8
12. A buffer at a pH < 6
Questions 13-16 refer to the following descriptions of bonding in different types of solids.
(A) Lattice of positive and negative ions held together by electrostatic forces.
(B) Closely packed lattice with delocalized electrons throughout
(C) Strong single covalent bonds with weak intermolecular forces.
(D) Strong multiple covalent bonds (including bonds.) with weak intermolecular forces
(E) Macromolecules held tgether with strong polar bonds.
13. Cesium chloride, CsCl (s)
14. Gold, Au(s)
15. Carbon dioxide, CO2(s)
16. Methane, CH4(s)
Question 17-18 refer to the following elements.
(A) Lithium
(B) Nickel
(C) Bromine
(D) Uranium
(E) Fluorine
17. Is a gas in its standard state at 298 K
18. Reacts with water to form a strong base
Directions: Each of the questions or incomplete statements below is by five suggested answers or completions.
Select the one that is best in each case and then fill in the corresponding oval on the answer sheet.
19. Which of the following best describes the role of the spark from the spark plug in an automobile engine?
42
(A) The spark decreases the energy of activation for the slow step.
(B) The spark increases the concentration of the volatile reactant.
(C) The spark supplies some of the energy of activation for the combustion reaction.
(D) The spark provides a more favorable activated complex for the combustion reaction.
(E) The spark provides the heat of vaporization for the volatile hydrocarbon.
20. What mass of Au is produced when 0.0500 mol of Au2S3 is reduced completely with excess H2?
(A) 9.85 g
(B) 19.7 g
(C) 24.5 g
(D) 39.4 g
(E) 48.9 g
21. When a solution of sodium chloride is vaporized in a flame, the color of the flame is
(A) blue
(B) yellow
(C) green
(D) violet
(E) White
22. Of the following reaction, which involves the largest decrease in entropy?
(A) CaCO3(s) ---> CaO(s) + CO2(g)
(B) 2 CO(g) + O2(g) ---> 2 CO2
(C) Pb(NO3)3 + 2 KI ---> PbI2 + 2 KNO3
(D) C3H8 + O2 ---> 3 CO2 + 4 H2O
(E) 4 La + 3 O2 ---> 2 La2O3
23. A hot-air balloon, shown right, rises. Which of the following
is the best explanation for this observation?
(A) The pressure on the walls of the balloon increases with
increasing tempearature.
(B) The difference in temperature between the air inside and
outside the ballon produces convection currents.
(C) The cooler air outside the balloon pushes in on the walls of
the ballon.
(D) The rate of diffusion of cooler air is less than that of
warmer air.
(E) The air density inside the ballon is less than that of the
surrounding air.
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24. The safest and most effective emergency procedure to treat an acid splash on skin is to do which of the
following immediately?
(A) Dry the affected area with paper towels
(B) Sprinkle the affected area with powdered Na2SO4(s)
(C) Flush the affected area with water and then with a dilute NaOH solution
(D) Flush the affected area with water and then with a dilute NaHCO3 solution
(E) Flush the affected area with water and then with a dilute vinegar solution
25. The cooling curve for a pure substance as it changes from a
liquid to a solid is shown right. The solid and the liquid coexist at
(A) point Q only
(B) point R only
(C) all points on the curve between Q and S
(D) all points on the curve between R and T
(E) no point on the curve
. . .C10H12O4S(s) + . . O2(g) ---> . . . CO2(g) + . . . SO2(g) + . . . H2O(g)
26. When the equation above is balanced and all coefficients are reduced to their lowest whole-number terms,
the coefficient for O2(g) is?
(A) 6
(B) 7
(C) 12
(D) 14
(E) 28
27. Appropriate uses of a visible-light spectrophtometer include which of the following?
I. Determining the concentration of a solution of Cu(NO3)2
II. Measuring the conductivity of a solution of KMnO4
III. Determining which ions are present in a solution that may contain Na+, Mg2+, Al3+
(A) I only
(B) II only
(C) III only
44
(D) I and II only
(E) I and III only
28. The melting point of MgO is higher than that of NaF. Explanations for this observation include which of the
following?
I. Mg2+ is more positively charged than Na+
II. O2¯ is more negatively charged than F¯
III. The O2¯ ion is smaller than the F¯ ion
(A) II only
(B) I and II only
(C) I and III only
(D) II and III only
(E) I, II, and III
29. The organic compound represented above is an example of
(A) an organic acid
(B) an alcohol
(C) an ether
(D) an aldehyde
(E) a ketone
H2Se(g) + 4 O2F2(g) ---> SeF6(g) + 2 HF(g) + 4 O2(g)
30. Which of the following is true regarding the reaction represented above?
(A) The oxidation number of O does not change.
(B) The oxidation number of H changes from -1 to +1.
(C) The oxidation number of F changes from +1 to -1.
(D) The oxidation number of Se changes from -2 to +6.
(E) It is a disproportionation reaction for F.
31. If the temperature of an aqueous solution of NaCl is increased from 20 °C to 90 °C, which of the following
statements is true?
(A) The density of the solution remains unchanged.
(B) The molarity of the solution remains unchanged.
(C) The molality of the solution remains unchanged.
(D) The mole fraction of solute decreases.
(E) The mole fraction of solute increases.
45
32. Types of hybridization exhibited by the C atoms in propene, CH3CHCH2, include which of the following?
I. sp
II. sp2
III. sp3
(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
33. A 1.0 L sample of an aqueous solution contains 0.10 mol of NaCl and 0.10 mol of CaCl2. What is the
minimum number of moles of AgNO3 that must be added to the solution in order to precipitate all of the Cl¯ as
AgCl(s) ? (Assume that AgCl is insoluble.)
(A) 0.10 mol
(B) 0.20 mol
(C) 0.30 mol
(D) 0.40 mol
(E) 0.60 mol
Questions 34-35 refer to an electrolytic cell that involves the following half-reaction.
AlF63¯ + 3 e¯ ---> Al + 6F¯
34. Which of the following occurs in the reaction?
(A) AlF 63¯ is reduced at the cathode.
(B) Al is oxidized at the anode.
(C) Aluminum is converted from the -3 oxidation state to the 0 oxidation state.
(D) F¯ acts as a reducing agent.
(E) F¯ is reduced at the cathode.
35. As steady current of 10 amperes in passed though an aluminum-production cell for 15 minutes. Which of
the following is the correct expressions for calculating the number of grams of aluminum produced? (1 faraday
= 96,500 coulombs)
46
Experiment Initial [NO]
(mol L¯1
Initial [O2]
(mol L¯1
Initial Rate of
Formation of NO2
(mol L¯1 s¯1)
1 0.10 0.10 2.5 x 10¯4
2 0.20 0.10 5.0 x 10¯4
3 0.20 0.40 8.0 x 10¯3
36. The initial-rate data in the table above were obtained for the reaction represented below. What is the
experimental rate la for the reaction?
(A) rate = k[NO] [O2]
(B) rate = k[NO] [O2]2
(C) rate = k[NO]2 [O2]
(D) rate = k[NO]2 [O2]2
(E) rate = k[NO] / [O2]
Ionization Energies for element X (kJ mol¯1)
First Second Third Fourth Five
580 1815 2740 11600 14800
37. The ionization energies for element X are listed in the table above. On the basis of the data, element X is
most likely to be
47
(A) Na
(B) Mg
(C) Al
(D) Si
(E) P
38. A molecule or an ion is classified as a Lewis acid if it
(A) accepts a proton from water
(B) accepts a pair of electrons to form a bond
(C) donates a pair of electrons to form a bond
(D) donates a proton to water
(E) has resonance Lewis electron-dot structures
39. The phase diagram for a pure substance is shown above. Which point on the diagram corresponds to the
equilibrium between the solid and liquid phases at the normal melting point?
(A) A
(B) B
(C) C
(D) D
(E) E
40. Of the following molecules, which has the largest dipole moment?
48
(A) CO
(B) CO2
(C) O2
(D) HF
(E) F2
2 SO3 (g) <===> 2 SO2 (g) + O2 (g)
41. After the equilibrium represented above is established, some pure O2 (g) is injected into the reaction vessel
at constant temperature. After equilibrium is reestablished, which of the following has a lower value cmpared to
its value at the original equilibrium?
(A) Keq for the reaction
(B) The total pressure in the reaction vessel.
(C) The amount of SO3 (g) in the reaction vessel.
(D) The amount of O2 (g) in the reaction vessel.
(E) The amount of SO2 (g) in the reaction vessel.
. . . Li3N(s) + . . . H2O(l) ---> . . . Li+ (aq) + . . . OH¯(aq) + . . . NH3(g)
42. When the equation above is balanced and all coefficients reduced to lowest whole number terms, the
coefficient for OH¯(aq) is
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6
43. A sample of 61.8 g of H3BO3, a weak acid is dissolved in 1,000 g of water to make a 1.0-molal solution.
Which of the following would be the best procedure to determine to molarity of the solution? (Assume no
additional information is available.)
(A) Titration of the solution with standard acid
(B) Measurement of the pH with a pH meter
(C) Determination of the boiling point of the solution
(D) Measurement of the total volume of the solution
(E) Measurement of the specific heat of the solution
44. A rigid metal tank contains oxygen gas. Which of the following applies to the gas in the tank when
additional oxygen is added at constant temperature?
(A) The volume of the gas increase.
(B) The pressure of the gas decreases.
(C) The average speed of the gas molecules remains th same.
(D) The total number of gas molecules remains the same.
(E) The average distance between the gas molecules increases.
49
45. What is the H+(aq) concentration in 0.05 M HCN (aq) ? (The Ka for HCN is 5.0 x 10¯
10)
(A) 2.5 x 10¯11
(B) 2.5 x 10¯10
(C) 5.0 x 10¯10
(D) 5.0 x 10¯6
(E) 5.0 x 10¯4
46. Which of the following occurs when excess concentrated NH3(aq) is mixed throughly with 0.1 M
Cu(NO3)2(aq) ?
(A) A dark red precipitate forms and settles out.
(B) Separate layers of immiscible liquids form with a blue layer on top.
(C) The color of the solution turns from light blue to dark blue.
(D) Bubbles of ammonia gas form.
(E) The pH of the solution decreases.
47. When hafnium metal is heated in an atmosphere of chlorine gas, the product of the reaction is found to
contain 62.2 percent Hf by mass and 37.4 percent Cl by mass. What is the empirical formula for this
compound?
(A) HfCl
(B) HfCl2
(C) HfCl3
(D) HfCl4
(E) Hf2Cl3
48. If 87.5 percent of a sample of pure 131
I decays in 24 days, what is the half-life of 131
I?
(A) 6 days
(B) 8 days
(C) 12 days
(D) 14 days
(E) 21 days
49. Which of the following techniques is most appropriate for the recovery of solid KNO3 from an aqueous
solution of KNO3?
(A) Paper chromatography
(B) Filtration
(C) Titration
(D) Electrolysis
(E) Evaporation to dryness
50. In the periodic table, as the atomic number increases from 11 to 17, what happens to the atomic radius?
50
(A) It remains constant.
(B) It increases only.
(C) It increases, then decreases.
(D) It decreases only.
(E) It decreases, then increases.
51. Which of the following is a correct interpretation of the results of Rutherford's experiments in which gold
atoms were bombarded with alpha particles?
(A) Atoms have equal numbers of positive and negative charges.
(B) Electrons in atoms are agganged in shells.
(C) Neutrons are at the center of an atom.
(D) Neutrons and protrons in atoms have nearly equal mass.
(E) The positive charge of an atom is concentrated in a small region.
52. Under which of the following sets of conditions could the most O2(g) be dissolved in H2O(l)?
Pressure of O2(g)
Above H2O(l)
(atm)
Temperature
of H2O(l)
°(C)
A) 5.0 80
B) 5.0 20
C) 1.0 80
D) 1.0 20
E) 0.5 20
W(g) + X(g) --> Y(g) + Z(g)
53. Gases W and X react in a closed, rigid vessel to form gases Y and Z according to the equation above. The
initial pressure of W(g) is 1.20 atm and that of X(g) is 1.60 atm. No Y(g) or Z(g) is initially present. The
experiment is carried out at constant temperature. What is the partial pressure of Z(g) when the partial pressure
of W(g) has decreased to 1.0 atm?
A) 0.20 atm
B) 0.40 atm
C) 1.0 atm
D) 1.2 atm
E) 1.4 atm
2NO(g) + O2(g) <===> 2 NO2(g)
H < 0
51
54. Which of the following changes alone would cause a decrease in the value of Keq for the reaction
represented above?
A) Decreasing the temperature
B) Increasing the temperature
C) Decreasing the volume of the reaction vessel
D) Increasing the volume of the reaction vessel
E) Adding a catalyst
10 HI + 2 KMnO4 + 3 H2SO4 --> 5 I2 + 2 MnSO4 + K2SO4 + 8 H2O
55. According to the balanced equation above, how many moles of HI would be necessary to produce 2.5 mol
of I2, starting with 4.0 mol of KMnO4 and 3.0 mol of H2SO4?
A) 20
B) 10
C) 8.0
D) 5.0
E) 2.5
56. A yellow precipitate forms when 0.5 M NaI(aq) is added to a 0.5 M solution of which of the following ions?
A) Pb2+(aq)
B) Zn2+(aq)
C) CrO42¯(aq)
D) SO42¯(aq)
E) OH¯(aq)
M(s) + 3 Ag+(aq) --> 3 Ag(s) + M3+(aq) E = +2.46 V
Ag+(aq) + e¯ --> Ag(s) E = +0.80 V
57. According to the information above, what is the standard reduction potential for the half-reaction M3+
(aq) +
3 e¯ --> M(s)?
A) -1.66 V
B) -0.06 V
C) 0.06 V
D) 1.66 V
E) 3.26 V
58. On a mountaintop, it is observed that water boils at 90°C, not at 100°C as at sea level. This phenomenon
occurs because on the mountaintop the
A) equilibrium water vapor pressure is higher due to the higher atmospheric pressure
B) equilibrium water vapor pressure is lower due to the higher atmospheric pressure
52
C) equilibrium water vapor pressure equals the atmospheric pressure at a lower temperature
D) water molecules have a higher average kinetic energy due to the lower atmospheric pressure
E) water contains a greater concentration of dissolved gases
59. A 40.0 mL sample of 0.25 M KOH is added to 60.0 mL of 0.15 M Ba(OH)2. What is the molar
concentration of OH¯(aq) in the resulting solution? (Assume that the volumes are additive)
A) 0.10 M
B) 0.19 M
C) 0.28 M
D) 0.40 M
E) 0.55 M
NH4NO3(s) --> N2O(g) + 2 H2O(g)
60. A 0.03 mol sample of NH4NO3(s) is placed in a 1 L evacuated flask, which is then sealed and heated. The
NH4NO3(s) decomposes completely according to the balanced equation above. The total pressure in the flask
measured at 400 K is closest to which of the following? ( The value of the gas constant, R, is 0.082 L atm mol¯1
K¯1)
(A) 3 atm
(B) 1 atm
(C) 0.5 atm
(D) 0.1 atm
(E) 0.03 atm
C2H4(g) + 3 O2(g) --> 2 CO2(g) + 2 H2O(g)
61. For the reaction of ethylene represented above, H is - 1,323 kJ. What is the value of H if the combustion
produced liquid water H2O(l), rather than water vapor H2O(g)? (H for the phase change H2O(g) --> H2O(l) is -
44 kJ mol¯1.)
A) -1,235 kJ
B) -1,279 kJ
C) -1,323 kJ
D) -1,367 kJ
E) -1,411 kJ
HC2H3O2(aq) + CN¯(aq) <===> HCN(aq) + C2H3O2¯(aq)
62. The reaction represented above has an equilibrium constant equal to 3.7 x 104. Which of the following can
be concluded from this information?
A) CN¯(aq) is a stronger base than C2H3O2¯(aq)
B) HCN(aq) is a stronger acid than HC2H3O2(aq)
C) The conjugate base of CN¯(aq) is C2H3O2¯(aq)
53
D) The equilibrium constant will increase with an increase in temperature.
E) The pH of a solution containing equimolar amounts of CN¯(aq) and HC2H3O2(aq) is 7.0.
63. The graph above shows the results of a study of the reaction of X with a large excess of Y to yield Z. The
concentrations of X and Y were measured over a period of time. According to the results, which of the
following can be concluded about the rate of law for the reaction under the conditions studied?
A) It is zero order in [X].
B) It is first order in [X].
C) It is second order in [X].
D) It is the first order in [Y].
E) The overall order of the reaction is 2.
64. Equal numbers of moles of He(g), Ar(g), and Ne(g) are placed in a glass vessel at room temperature. If the
vessel has a pinhole-sized leak, which of the following will be true regarding the relative values of the partial
pressures of the gases remaining in the vessel after some of the gas mixture has effused?
A) PHe < PNe < PAr
B) PHe < PAr < PNe
C) PNe < PAr < PHe
D) PAr < PHe < PNe
E) PHe = PAr = PNe
54
65. Which of the following compounds is NOT appreciably soluble in water but is soluble in dilute hydrochloric
acid?
A) Mg(OH)2(s)
B) (NH4)2CO3(s)
C) CuSO4(s)
D) (NH4)2SO4(s)
E) Sr(NO3)2(s)
66. When solid ammonium chloride, NH4Cl(s) is added to water at 25 °C, it dissolves and the temperature of
the solution decreases. Which of the following is true for the values of H and S for the dissolving process?
H S
A) Postive Positive
B) Positive Negative
C) Positive Equal to zero
D) Negative Positive
E) Negative Negative
67. What is the molar solubility in water of Ag2CrO4? (The Ksp for Ag2CrO4 is 8 x 10¯12
.)
A) 8 x 10¯12 M
B) 2 x 10¯12 M
C) (4 x 10¯12 M)1/2
D) (4 x 10¯12 M)1/3
E) (2 x 10¯12 M)1/3
68. In which of the following processes are covalent bonds broken?
A) I2(s) --> I2(g)
B) CO2(s) --> CO2(g)
C) NaCl(s) --> NaCl(l)
D) C(diamond) --> C(g)
E) Fe(s) --> Fe(l)
69. What is the final concentration of barium ions, [Ba2+
], in solution when 100. mL of 0.10 M BaCl2(aq) is
mixed with 100. mL of 0.050 M H2SO4(aq)?
A) 0.00 M
B) 0.012 M
C) 0.025 M
55
D) 0.075 M
E) 0.10 M
70. When 100 mL of 1.0 M Na3PO4 is mixed with 100 mL of 1.0 M AgNO3, a yellow precipitate forms and
[Ag+] becomes negligibly small. Which of the following is a correct listing of the ions remaining in solution in
order of increasing concentration?
A) [PO43¯] < [NO3¯] < [Na+]
B) [PO43¯] < [Na+] < [NO3¯]
C) [NO3¯] < [PO43¯] < [Na+]
D) [Na+] < [NO3¯] < [PO43¯]
E) [Na+] < [PO43¯] < [NO3¯]
71. In a qualitative ananlysis for the presence of Pb2+
, Fe2+
, and Cu2+
ions in a aqueous solution, which of the
following will allow the separation of Pb2+
from the other ions at room temperature?
A) Adding dilute Na2S(aq) solution
B) Adding dilute HCl(aq) solution
C) Adding dilute NaOH(aq) solution
D) Adding dilute NH3(aq) solution
E) Adding dilute HNO3(aq) solution
72. After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student
reported a value of 38 percent. The correct value for the percentage of water in the hydrate is 51 percent. Which
of the following is the most likely explanation for this difference?
A) Strong initial heating caused some of the hydrate sample to spatter out of the crucible.
B) The dehydrated sample absorbed moisture after heating.
C) The amount of the hydrate sample used was too small.
D) The crucible was not heated to constant mass before use.
E) Excess heating caused the dehydrated sample to decompose.
73. The volume of distilled water that should be added to 10.0 mL of 6.00 M HCl(aq) in order to prepare a
0.500 M HCl(aq) solution is approximately
A) 50.0 mL
B) 60.0 mL
C) 100. mL
D) 110. mL
E) 120. mL
74. Which of the following gases deviates most from ideal behavior?
A) SO2
B) Ne
C) CH4
56
D) N2
E) H2
75. Which of the following pairs of liquids forms the solution that is most ideal (most closely follows Raoult's
law)?
A) C8H18(l) and H2O(l)
B) CH3CH2CH2OH(l) and H2O(l)
C) CH3CH2CH2OH(l) and C8H18(l)
D) C6H14(l) and C8H18(l)
E) H2SO4(l) and H2O(l)
Practice Test Answers
Key Percent
Correct Key
Percent
Correct Key
Percent
Correct
1. C 74 26. C 74 51. E 59
2. E 48 27. A 16 52. B 31
3. B 42 28. B 53 53. A 46
4. A 71 29. E 26 54. B 35
5. D 75 30. D 75 55. D 69
6. A 78 31. C 21 56. A 32
7. C 55 32. D 38 57. A 44
8. E 83 33. C 54 58. C 39
9. C 52 34. A 42 59. C 28
10. E 41 35. C 43 60. A 22
11. A 33 36. B 52 61. E 31
12. B 35 37. C 35 62. A 31
13. A 60 38. B 38 63. B 34
57
14. B 64 39. C 71 64. A 65
15. D 65 40. D 54 65. A 35
16. C 61 41. E 65 66. A 31
17. E 70 42. C 71 67. E 25
18. A 67 43. D 60 68. D 54
19. C 73 44. C 54 69. C 31
20. B 55 45. D 40 70. A 34
21. B 33 46. C 22 71. B 32
22. E 68 47. C 50 72. B 25
23. E 67 48. B 42 73. D 33
24. D 20 49. E 53 74. A 48
25. C 57 50. D 62 75. D 33
Advanced Placement Chemistry
58
Practice Free Response Questions
1) The acid ionization constant, Ka, for propanoic acid, C2H5COOH, is 1.3 x 10¯5.
(a) Calculate the hydrogen ion concentration, [H+], in a 0.20-molar solution of propanoic acid.
(b) Calculate the percentage of propanoic acid molecules that are ionized in the solution in (a).
(c) What is the ratio of the concentration of propanoate ion, C2H5COO¯, to that of propanoic acid in a buffer
solution with a pH of 5.20 ?
(d) In a 100-milliliter sample of a different buffer solution, the propanoic acid concentration is 0.50-molar and
the sodium propanoate concentration is 0.50-molar. To this buffer solution, 0.0040 mole of solid NaOH is
added. Calculate the pH of the resulting solution.
2) The molecular formula of a hydrocarbon is to be determined by analyzing its combustion products and
investigating its colligative properties.
(a) The hydrocarbon burns completely, producing 7.2 grams of water and 7.2 liters of CO2 at standard
conditions.
(b) Calculate the mass in grams of O2 required for the complete combustion of the sample of the hydrocarbon
described in (a).
(c) The hydrocarbon dissolves readily in CHCl3. The freezing point of a solution prepared by mixing 100.
grams of CHCl3 and 0.600 gram of the hydrocarbon is -64.0 °C. The molal freezing-point depression constant
of CHCl3 is 4.68 °C / molal and its normal freezing point is -63.5 °C. Calculate the molecular weight of the
hydrocarbon.
(d) What is the molecular formula of the hydrocarbon?
3)
2 ClO2(g) + F2(g) ---> 2 ClO2F(g)
The following results were obtained when the reaction represented above was studied at 25 °C.
(a) Write the rate law expression for the reaction above.
59
(b) Calculate the numerical value of the rate constant and specify the units.
(c) In experiment 2, what is the initial rate of decrease of [F2]?
(d) Which of the following reaction mechanisms is consistent with the rate law developed in (a)? Justify your
choice.
I.
ClO2 + F2 <---> ClO2F2 (fast)
ClO2F2 ---> ClO2F + F (slow)
ClO2 + F ---> ClO2F (fast)
II.
F2 ---> 2 F (slow)
2 (ClO2 + F ---> ClO2F) (fast)
4) Give the formulas to show the reactants and the products for FIVE of the following chemical reactions. Each
of the reactions occurs in aqueous solution unless otherwise indicated. Represent substances in solution as ions
if the substance is extensively ionized. Omit formulas for any ions or molecules that are unchanged by the
reaction. In all cases a reaction occurs. You need not balance.
Example: A strip of magnesium is added to a solution of silver nitrate.
Mg + Ag+ ---> Mg
2+ + Ag
(a) Solid aluminum oxide is added to a solution of sodium hydroxide.
(b) Solid calcium oxide is heated in the presence of sulfur trioxide gas.
(c) Equal volumes of 0.1-molar sulfuric acid and 0.1-molar potassium hydroxide are mixed.
(d) Calcium metal is heated strongly in nitrogen gas.
(e) Solid copper(II) sulfide is heated strongly in oxygen gas.
(f) A concentrated solution of hydrochloric acid is added to powdered manganese dioxide and gently heated.
(g) A concentrated solution of ammonia is added to a solution of zinc iodide.
(h) A solution of copper(II) sulfate is added to a solution of barium hydroxide.
60
5) BCl3(g) + NH3(g) <---> Cl3BNH3(s)
The reaction represented above is a reversible reaction.
(a) Predict the sign of the entropy change, S, as the reaction proceeds to the right. Explain your prediction.
(b) If the reaction spontaneously proceeds to the right, predict the sign of the enthalpy change, H. Explain your
prediction.
(c) The direction in which the reaction spontaneously proceeds changes as the temperature is increased above a
specific temperature. Explain.
(d) What is the value of the equilibrium constant at the temperature referred to in (c); that is, the specific
temperature at which the direction of the spontaneous reaction changes? Explain.
6)
An experiment is to be performed to determine the molecular mass of a volatile liquid by the vapor density
method. The equipment shown above is to be used for the experiment. A barometer is also available.
(a) What data are needed to calculate the molecular mass of the liquid?
(b) What procedures are needed to obtain these data?
(c) List the calculations necessary to determine the molecular mass.
61
(d) If the volatile liquid contains nonvolatile impurities, how would the calculated value of the molecular mass
be affected? Explain your reasoning.
7) Explain each of the following.
(a) When an aqueous solution of NaCl is electrolyzed, Cl2(g) is produced at the anode, but no Na(s) is produced
at the cathode.
(b) The mass of Fe(s) produced when 1 faraday is used to reduce a solution of FeSO4 is 1.5 times the mass of
Fe(s) produced when 1 faraday is used to reduce a solution of FeCl3.
Zn + Pb2+ (1-molar) ---> Zn2+ (1-molar) + Pb
(c) The cell that utilized the reaction above has a higher potential when [Zn2+
] is decreased and [Pb2+
] held
constant, but a lower potential when [Pb2+
] is decreased and [Zn2+
] is held constant.
(d) The cell that utilizes the reaction given in (c) has the same cell potential as another cell in which [Zn2+
] and
[Pb2+
] are each 0.1-molar.
8) Experimental data provide the basis for interpreting differences in properties of substances.
Account for the differences in properties given in Tables 1 and 2 above in terms of the differences in structure
and bonding in each of the following pairs.
(a) MgCl2 and SiCl4
(b) MgCl2 and MgF2
(c) F2 and Br2
(d) F2 and N2
9) Explain each of the following in terms of nuclear models.
(a) The mass of an atom of 4He is less than the sum of the masses of 2 protons, 2 neutrons, and 2 electrons.
62
(b) Alpha radiation penetrates a much shorter distance into a piece of material than does beta radiation of the
same energy.
(c) Products from a nuclear fission of a uranium atom such as 90
Sr and 137
Ce are highly radioactive and decay by
emission of beta particles.
(d) Nuclear fusion requires large amounts of energy to get started, whereas nuclear fission can occur
spontaneously, although both processes release energy.
Copyright © 1991 by College Entrance Examination Board and
Educational Testing Service. All rights reserved.
Advanced Placement Chemistry
63
Practice Free Response Scoring Guide
Notes
[square root] applies to the numbers enclosed in parenthesis immediately following All simplifying assumptions are justified within 5%. One point deduction for a significant figure or math error, applied only once per problem. No credit earned for numerical answer without justification.
1)
a) three points
Ka = ( [H+] [C3H5O2¯] ) ÷ [HC3H5O2]
1.3 x 10¯5 = x
2 ÷ 0.20
x = [H+] = 1.6 x 10¯
3
b) one point
% dissoc. = [H+] ÷ [HC3H5O2]
= 1.6 x 10¯3 ÷ 0.20 = 0.80%
c) two points
[H+] = antilog (- 5.20) = 6.3 x 10¯
6
1.3 x 10¯5 = (6.3 x 10¯
6) x ([C3H5O2¯] ÷ [HC3H5O2]
[C3H5O2¯] ÷ [HC3H5O2] = 1.3 x 10¯5 ÷ 6.3 x 10¯
6 = 2.1
An alternate solution for (c) based on the Henderson-Hasselbalch equation.
pH = pKa + log ([base] ÷ [acid])
5.20 = 4.89 + log ([C3H5O2¯] ÷ [HC3H5O2])
log ([C3H5O2¯] ÷ [HC3H5O2]) = 0.31
[C3H5O2¯] ÷ [HC3H5O2] = 2.0
64
d) six points
0.10 L x 0.35 mol/L = 0.035 mol HC3H5O2
0.10 L x 0.50 mol/L = 0.050 mol C3H5O2¯
0.035 mol - 0.004 mol = 0.031 mol HC3H5O2
0.050 mol + 0.004 mol = 0.054 mol C3H5O2¯
1.3 x 10¯5 = [H
+] x [(0.054 mol/0.1 L) ÷ (0.031 mol/0.1 L)]
Can use 0.54 and 0.31 instead.
[H+] = 7.5 x 10¯
6
pH = 5.13
An alternate solution for (d) based on the Henderson-Hasselbalch equation.
use [ ]s or moles of HC3H5O2 and C3H5O2¯
pH = pKa + log (0.054 / 0.031)
= 4.89 + 0.24 = 5.13
2)
a) three points
7.2 g H2O ÷ 18.0 g/mol = 0.40 mol H2O
0.40 mol H2O x (2 mol H / 1 mol H2O) = 0.80 mol H
7.2 L CO2 ÷ 22.4 L/mol = 0.32 mol CO2
0.32 mol CO2 x (1 mol C / 1 mol CO2) = 0.32 mol C
OR
n = PV ÷ RT = [(1 atm) (7.2 L)] ÷ [(0.0821 L atm mol¯1 K
1) (273 K)] = 0.32 mol CO2
0.80 mol H ÷ 0.32 = 2.5
0.32 mol C ÷ 0.32 = 1
65
2.5 x 2 = 5 mol H
1 x 2 = 2 mol C
empirical formula = C2H5
b) two points
mol O2 for combustion = mol CO2 + 1/2 mol H2O = 0.32 + 0.20 = 0.52 mol O2
0.52 mol O2 x 32 g/mol = 17 g O2
alternate approach for mol O2 from balanced equation
C2H5 + 13/4 O2 ---> 2 CO2 + 5/2 H2O
other ratio examples:
1, 6.5 ---> 4, 5
0.25, 1.625 ---> 1, 1.25
mol O2 = 0.40 mol H2O x (13/4 mol O2 / 5/2 mol H2O) = 0.52 mol O2
Note: starting moles of C2H5 = 0.16 mol C2H5
c) three points
MM stands for molar mass.
T = (Kf (g/MM)) / kg of solvent
0.5 °:C = ((4.68 °:C kg mol¯:1) x (0.60 g / MM)) / 0.1 kg
MM = (4.68 x 0.60) / (0.5 x 0.1) = 56 or 6 x 101
an alternate solution for (c)
molality = 0.5 °:C / (4.68 0.5 °:C/m) = 0.107 m
mol solute =( 0.107 mol / kg solvent) x 0.100 kg solvent = 0.0107 mol
MM = 0.60 g / 0.0107 mol = 56 or 6 x 101
d) one point
66
(56 g/mol of cmpd) / (29 g/mol of empirical formula) = 1.9 empirical formula per mol
OR
6 x 101 / 29 = 2.1
empirical formula times 2 equals molecular formula = C4H10
3)
a) four points
rate = k [ClO2] [F2]
one point - rate equation form, k
one point - F2 order
two points - ClO2 order
b) two points
k = rate / ([ClO2] [F2])
= 2.4 x 10¯3 mol L¯
1 sec¯
1 / ((0.010 mol/L) (0.10 mol/L))
= 2.4 L mol¯1 sec¯
1
one point - value consistent with equation in (a)
one point - units consistent with equation in (a)
c) one point
2 ClO2 + F2 ---> 2 ClO2F
- d[F2] / dt = 1/2 (d[ClO2F] / dt)
= 1/2 (9.6 x 10¯3)
= 4.8 x 10¯3 mol L¯
1 sec¯
1
d) two points
67
mechanism I
defense:
slow step is first order
three equations add to proper stoichiometry
Note: if ClO2 order in rate equation of part (a) is zero, mechanism II must be chosen to obtain credit.
4)
a) Al2O3 + OH¯ ---> Al(OH)4¯
OR
Al2O3 + H2O ---> Al(OH)3
(Personal note by John Park: I think the H2O in the second equation above comes from the fact that the NaOH
concentration was not specified in the original problem. For example, suppose [OH¯] were 10¯12
M. Then the
second equation becomes the predominate one.)
b) CaO + SO3 ---> CaSO4
c) H+ + OH¯ ---> H2O
d) Ca + N2 ---> Ca3N2
e) CuS + O2 ---> Cu + SO2 (also CuO, Cu2O)
f) H+ + Cl¯ + MnO2 ---> Mn
2+ + Cl2 + H2O (one pt. for either redox product, two pts. for all three products)
g) Zn2+
+ NH3 ---> Zn(NH3)42+
OR
Zn2+
+ NH3 + H2O ---> Zn(OH)2 + NH4+
h) Cu2+
+ SO42¯ + Ba
2+ + OH¯ ---> Cu(OH)2 + BaSO4
A rare double precipitation.
Partial credit was allowed for some alternate solutions, e.g.
Cu2+
+ OH¯ ---> Cu(OH)2
Ba2+
+ SO42¯ ---> BaSO4
68
5)
a) two points
S will be negative. The system becomes more ordered as two gases form a solid.
b) two points
H must be negative. For the reaction to be spontaneous, G must be negative, so H must be more negative
than -TS is positive.
c) two points
As T increases, -TS increases. Since S is negative, the positive -TS term will eventually exceed H (which
is negative), making G positive. (In the absence of this, G = H - TS and general discussion of the effect of
T and S gets 1 point.)
d) two points
The equilibrium constant is 1. The system is at equilibrium at this temperature with an equal tendancy to go in
either direction.
OR
G = 0 at equilibrium so K = 1 in G = -RT ln K
(In the absence of these, G = -RT ln K gets 1 point).
The above concludes the AP scoring standards published in 1991. The following is simply alternate ways of
answering which the AP readers may or may not have given full credit to.
a) The amount of entropy goes down, S is negative.
b) G = H - TS. If S is negative, then H must also be negative to get a negative G.
c) Let us say G is positive when H is positive and S is positive. As T goes up - TS becomes more negative
until it makes G (which equals H - TS) become negative.
d) At the temperature when the direction changes, the rate forward = the rate reverse. Since K = kf / kr, this
equals 1.
69
6)
a) two points
Mass of vaporized liquid (or liquid or substance)
two of three in parts a or b
atmosperic pressure
volume of flask
temperature of vapor (water)
b) three points
Procedure for:
mass - the mass difference between flask + air and flask + vaporized liquid volume - volume of flask by filling with H2O and then using graduated cylinder for measuring.
Flask containing liquid is heated until liquid disappears
c) one point
mass ÷ mole where mole is determined from PV = nRT
d) two points
Molar mass is too high
because the non-volatile inpurity contribute additional mass (but insignificant volume).
7)
a) two points
Cl¯ is more easily oxidized than H2O
H2O is more easily oxidized than Na+
no 2nd pt is awarded for H+ ---> H2 unless H2O is implied.
no 2nd pt for Na(s) + H2O ---> Na + OH¯ unless H2 is indicated.
70
b) two points
Fe2+
requires 2 Faraday / mol Fe (s) or 1Faraday ---> 1/ 2 mol Fe(s)
Fe3+
require 3 Faraday / mol Fe (s) or 1 Faraday ---> 1/3 mol Fe (s)
for equal numbers of Faraday (1/2 : 1/3 as 1.5 : 1) (Or inverse relationship is clear)
no 2nd point unless flow of e¯ to mass (moles) is clear and logically correct
c) two points
Le Châtlier's argument if [Zn
2+ ] goes down; reaction shifts right, i.e. cell potential goes up
if [Pb2+
] goes down; reaction shifts left, i.e. cell potential goes down
OR
Nernst Equation argument E = E° - RT ln Q with Q = [Zn
2+ ] / [Pb
2+ ]
if [Zn2+
] goes down Q < 1, therefore E > E°
if [Pb2+
] goes down Q > 1, therefore E < E°
reasoning must indicate correct usage of equation
d) two points
[Zn2+
] / [Pb2+
] does not change; regardless of values; i.e. E=E°
OR
[Zn2+
] / [Pb2+
] = 1 so ln Q = 0; i.e. E=E°
no pt is awarded for just stating concentrations are equal
ratio or proportion concept is required for 2nd point
8)
a) three points
71
MgCl2 is ionic and SiCl4 is covalent. The elctrostatic, interionic forces in MgCl2 are much stronger than the
intermolecular (dispersion) forces in SiCl4 and lead to a higher melting point. Molten MgCl2 contains mobile
ions that conduct electricity whereas molten SiCl4 is molecular, not ionic, and has no conductivity.
b) two points
MgF2 has a higher melting point than MgCl2 because the smaller F¯ ions and smaller interionic distances in MgF2 cause
stronger forces and higher melting point.
c) one point
The bond length in Br2 is larger than in F2 because the Br atom is larger than the F atom.
d) two points
The bond length in N2 is less than in F2 because the N-N bond is triple and the F-F is single. Triple bonds are
stonger and therefore shorter than single bonds.
9)
a) two points
When nucleons are combined in nuclei, some of their mass is converted to energy (binding energy) which is
released and stabilizes the nucleus. (Key concepts: mass defect; binding energy)
b) two points
Alpha particles have a greater mass than beta particles. Thus their speed (penetrating potential) is less.
(Alternate explanation could be based on charge.)
c) two points
The neutron/proton ratio in Sr-90 and Cs-137 is too large and they emit beta particles (converting neutrons to
protons) to lower this ratio.
72
d) two points
Large amounts of energy are neded to initiate fusion reactions in order to overcome the repulsive forces
between the positively charged nuclei. Large amounts of energy are not required to cause large nuclei to split.
Copyright © 1991 by College Entrance Examination Board and
Educational Testing Service. All rights reserved.
