Zumdahls Chemistry

AP Chemistry A. Allan Chapter 1 Notes - Chemical Foundations 1.1 Chemistry: An Overview A. Reaction of hydrogen and oxygen 1. Two molecules of hydrogen react with one molecule of oxygen to form two molecules of water 2H2 + O2 à 2H2O 2. Decomposition of water 2H2O à 2H2 + O2 B. Problem Solving in Chemistry (and life) 1. Making observations 2. Making a prediction 3. Do experiments to test the prediction 1.2 The Scientific Method A. General Framework 1. Making observations a. Quantitative ( measurement) b. Qualitative (color, phase, shape, etc) 2. Making a prediction 3. Do experiments to test the prediction B. Vocabulary 1. Observation a. Something that is witnessed and can be recorded 2. Theory (Model) a. Tested hypotheses that explains WHY nature behaves in a certain way b. Theories change as more information becomes available 3. Natural Law a. A summary of observed, measurable behavior 1.3 Units of Measurement A. Measurements 1. Number and Scale (units) are both essential "The number without the units is worthless!" B. SI system Important SI Units for Chemistry Mass kilogram kg Length meter m Time second s Temperature Kelvin K Amount of Substance mole mol Volume liter L

1

C. SI Prefixes mega M 1,000,000 kilo k 1,000 hecto h 100 deka da 10

106 103 102 101

nano micro milli centi deci

n Î m c n

0.000000001 0.000001 0.001 0.01 0.1

10-9 10-6 10-3 10-2 10-1

1.4 Uncertainty in Measurement A. Recording Measurements (Significant figures) 1. Record all digits that are certain 2. Record the first digit that is uncertain (all measurements have some degree of uncertainty) 3. Uncertainty in the last number is + 1, unless otherwise noted B. Accuracy 1. The agreement of a particular value with the accepted value C. Precision 1. The degree of agreement among several measurements made in the same way "You can be precise, but not accurate. If you are accurate, you are necessarily precise." D. Errors 1. Random Errors (indeterminate errors) a. Measurements may be high or low b. Causes: 1) Interpretation of the uncertain digit 2) Procedural ineptness 2. Systematic Errors a. Always occur in the same direction b. Caused by poor measurement calibration 1) gun sight set too high/low 2) balance improperly zeroed 3) thermometer improperly marked

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1.5 Significant Figures and Calculations A. Rules for Counting Significant Figures Number Rule Nonzero integers Always significant Leading zeroes Never significant Captive zeroes Always significant Trailing zeroes Significant if after a decimal Exact numbers

Infinite significance

Scientific notation

All digits are significant

Example 6.34 m (3 sig figs) 0.00634 m ( 3 sig figs) 6.0034 (5 sig figs) 63400 (3 sig figs) 0.63400 (5 sig figs) e.g. There is 1 star at the center of our solar system. There is no doubt about the number "1" 6.3400 x 106 (5 sig figs)

B. Multiplication and Division 1. Keep as many sig figs in your answer as are in the piece of data with the least number of sig figs 2.37 cm x 15.67 cm x 7.4 cm = 274.82046 (keep two sig figs) = 2.7 x 102 cm3 C. Addition and Subtraction 1. Keep the same number of decimal places as the least precise measurement in your calculation 34.039 m + 0.24 m + 1.332 m + 12.7 m = 48.311 m (keep one decimal place) = 48.3 m D. Rules for Rounding 1. Round at the end of a series of calculations, NOT after each step 2. Use only the first number to the right of the last sig fig to decide whether or not to round a. Less than 5, the last significant digit is unchanged b. 5 or more, the last significant digit is increased by 1 Note from this section in your book: Detailed solutions and stepwise examples of problems in this text show correct sig figs at each step. Since you will most often do a sequence of calculations and then round to correct sig figs at the end, your answer will often be slightly different (usually only in the last place) than the answer given in the book. 1. Significant figures rules will be observed in all calculations throughout the year in this course. You need never ask, "Do we have to watch our sig figs?" The answer is always "Yes!"

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1.6 Dimensional Analysis A. Examine examples 1. pages 18 - 21 B. Unit Conversions Questions 1. What units am I given? 2. What units must be in my answer? 3. What is conversion factor? Full credit can never be given for working a problem in which you do not do all of the following: 1. Observe significant figures rules 2. Label all steps of your work with the correct units 3. Correctly label and identify your answer 4. Solve the problem in a manner that can be understood by the reader. 1.7 Temperature A. Celsius (°C) and Kelvin (K) 1. Kelvin = Celsius + 273.15 2. Celsius = Kelvin - 273.15 3. Size of the temperature unit (degree) is the same B. Fahrenheit 1. TC = (TF - 32°F)(5°C/9°F) 2. TF = TC x (9°F/5°C) + 32°F 1.8 Density A. Density = mass/volume 1.9 Classification of Matter A. Matter 1. Anything that occupies space and has mass B. States of Matter 1. Solids - rigid, fixed volume and shape 2. Liquids - definite volume, no specific shape 3. Gases - no fixed volume or shape, highly compressible C. Mixtures - Matter of variable composition 1. Heterogeneous mixtures a. Having visibly distinguishable parts 2. Homogeneous mixtures (solutions) a. Having visibly indistinguishable parts D. Components of Mixtures can be Separated by Physical Means 1. Distillation 2. Filtration 3. Chromatography

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E. Pure substances 1. Elements a. Cannot be decomposed into simpler substances by physical or chemical means 2. Compounds a. Constant composition b. Can be broken into simpler substances by chemical means, not by physical means The Organization of Matter (Slightly different than your book) Matter

Mixtures: a) Homogeneous (Solutions) b) Heterogeneous

Pure Substances Compounds s

Elements

Atoms Nucleus Protons

Quarks

Electrons Neutrons

Quarks

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AP Chemistry A. Allan Chapter 2 Notes - Atoms, Molecules and Ions 2.1 The Early History Refer to the Chemistry History Timeline for this chapter 2.2 Fundamental Chemical Laws A. Law of Conservation of Mass 1. "Mass is neither created nor destroyed" 2. Translation: In ordinary chemical reactions, the total mass of the reactants is equal to the total mass of the products B. Law of Definite Proportion 1. "A given compound always contains the same proportions of elements by mass" 2. Translation: Compounds have an unchanging chemical formula C. Law of Multiple Proportions 1. "When two elements form a series of compounds, the ratios of the masses of the second element that combine with one gram of the first element can always be reduced to small whole numbers 2. Translation: Sometimes two elements can come together in more than one way, forming compounds with similar, though not identical formulas 2.3 Dalton's Atomic Theory A. Atomic Theory 1. Each element is made up of tiny particles called atoms 2. The atoms of a given element are identical 3. Chemical compounds are formed when atoms combine with each other. A given compound always has the same relative numbers and types of atoms 4. Chemical reactions involve reorganizations of the atoms. The atoms themselves are not changed in a chemical reaction B. Avogadro's Hypothesis 1. At the same conditions of temperature and pressure, equal volumes of different gases contain the same number of particles. 2.4 Early Experiments to Characterize the Atom A. J.J. Thomson and the Electron 1. Determined the charge to mass ratio of the electron 2. Reasoned that all atoms must contain electrons 3. Reasoned that all atoms must contain positive charges B. Robert Millikan and the Oil Drop 1. Oil drop experiments determined the charge on an electron 2. With charge information, and Thomson's charge/mass ratio, he determined the mass of an electron (9.11 x 10-31 kg)

C. Radioactivity 1. Gamma (È) rays - high energy light 2. Beta (Ã) particles - high speed electrons 3. Alpha (Â) particles - nuclear particle with a 2+ charge D. The Nuclear Atom - Rutherford's Metal Foil Experiment 1. Most alpha particles pass straight through thin metal foil 2. Some particles were greatly deflected ("like a howitzer shell bouncing off of a piece of paper") a. Could not have been deflected by electrons or single protons b. Must have been deflected by a positively charged object of substantial mass 1) Supported concept of a small, central, positive nucleus where most of the atom's mass was concentrated 2) Disproved Thomson's "plum pudding" model 2.5 The Modern View of Atomic Structure: An Introduction A. Nucleus 1. Protons - positively charged 2. Neutrons - no charge 3. Small size, high density a. The mass of all of the cars in the United States in an object that would easily fit in a teaspoon b. A pea with the mass of 250 million tons B. Electrons 1. Negatively charged 2. The source of varying reactivity of different elements 3. Provide most of the atomic volume C. Atomic Number 1. Number of protons D. Mass Number 1. Number of protons + number of neutrons E. Isotopes 1. Atoms with the same number of protons (same element) but different numbers of neutrons (mass numbers) F. Symbols for the Elements 1. Mass Number 23Na Element symbol 11 Atomic Number

2.6 Molecules and Ions A. Chemical Bonding 1. Covalent bonding - Sharing of electrons 2. Ionic bonding - Attraction of oppositely charged ions due to a reaction in which electrons are transferred B. Representing Molecules (Covalently bonded) 1. Chemical formula a. Symbols for atoms and subscripts 1) H20 2) CH4 2. Structural formula a. Bonds represented by lines

Ball and Stick

Space Filling

C. Ions 1. Cations a. Positive ions formed by the loss of electrons 2. Anions a. Negative ions formed by gaining electrons D. Ionic Bonding 1. Bond formed by the attraction between oppositely charged ions 2. Ionic bonding forms ionic solids (salts) 3. Ions can be monatomic (one atom) or polyatomic (more than one atom) 2.7 An Introduction to the Periodic Table A. Organization 1. Horizontal row is called a "period" (or series) 2. Vertical column is called a "group" or "family" a. Group 1A - Alkali metals b. Group 2A - Alkaline earth metals c. Group 7A - Halogens (Gr, "salt makers") d. Group 8A - Noble gases B. Naming Elements 104 and beyond Nil = 0 un = 1 bi = 2 tri = 3 quad = 4 Pent = 5 hex = 6 sept = 7 oct = 8 enn = 9 Element 109 = un (1) nil(0) enn(9) ium = unnilennium

2.8 Naming Simple Compounds A. Ionic Compounds 1. Positive ion is always named first, negative ion second You were given a list of ions to memorize on the first day of class Tips for memorizing the polyatomics: a. Find the "ate" ion (sulfate, for instance) sulfate = SO42b. The "ite" ion always has one less oxygen than the "ate" ion sulfite = SO32c. The prefix "per" (think hyper, meaning "above") is used with the "ate" prefix to indicate one more oxygen than the "ate" ion persulfate = SO52d. The prefix "hypo" (meaning "under" or "below") is used with the "ite" prefix to indicate one less oxygen than the "ite" ion hyposulfite = SO22Examples (Just because you can name it doesn't mean it exists!) Perchlorate Pernitrate ClO4NO4Chlorate Chlorite

ClO3ClO2-

Nitrate Nitrite

NO3NO2-

hypochlorite hyponitrite ClONO2. Metals with more than one oxidation state (transition metals) must have a roman numeral to indicate the oxidation state Fe3+ = iron (III) Mn+2 = manganese (II) B. Binary Covalent Compounds 1. Must contain two elements, BOTH nonmetals a. First element 1) full element name 2) prefix only if there is more than one atom b. Second element 1) named as if it were an anion (-ide suffix) 2) always gets a prefix mono - 1 penta - 5 octa - 8 di - 2 hexa - 6 nona - 9 tri - 3 hepta - 7 deca - 10 tetra - 4 C. Naming Acids 1. Binary Acids (two elements - hydrogen + one other) a. prefix "Hydro" + root of second element + "ic" suffix 2. Oxyacids a. If the acid contains an anion whose name ends in "ate": Use root of anion name and an "ic" ending (H2SO4 = sulfuric acid) b. If the acid contains an anion whose name ends in "ite": Use the root of the anion name and an "ous" ending (H2SO3 = sulfurous acid)

AP Chemistry A. Allan Chapter 3 Notes - Stoichiometry 3.1 Atomic Masses A. C-12, the Relative Standard 1. C-12 is assigned a mass of exactly 12 atomic mass units (amu) 2. Masses of all elements are determined in comparison to the carbon 12 atom (12C) the most common isotope of carbon 3. Comparisons are made using a mass spectrometer B. Atomic Mass (Average atomic mass, atomic weight) 1. Atomic masses are the average of the naturally occurring isotopes of an element 2. Atomic mass does not represent the mass of any actual atom 3. Atomic mass can be used to "weigh out" large numbers of atoms 3.2 The Mole A. Avogadro's number 1. 6.022 x 1023 units = 1 mole 2. Named in honor of Avogadro (he did NOT discover it) B. Measuring moles 1. An element's atomic mass expressed in grams contains 1 mole of atoms of that element a. 12.01 grams of carbon is 1 mole of carbon b. 12 grams of carbon-12 is 1 mole of carbon-12 3.3 Molar Mass A. Molar Mass (Gram molecular weight) 1. The mass in grams of one mole of a compound 2. The sum of the masses of the component atoms in a compound a. Molar mass of ethane (C2H6): Mass of 2 moles of C = 2(12.01 g) Mass of 6 moles of H = 6(1.008 g) 30.07 g

3.4 Percent Composition of Compounds A. Calculating any percentage 1. "The part, divided by the whole, multiplied by 100" B. Percentage Composition 1. 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)

3.5 Determining the Formula of a Compound A. Determining the empirical formula 1. Determine the percentage of each element in your compound 2. Treat % as grams, and convert grams of each element to moles of each element 3. Find the smallest whole number ratio of atoms 4. If the ratio is not all whole number, multiply each by an integer so that all elements are in whole number ratio B. Determining the molecular formula 1. Find the empirical formula mass 2. Divide the known molecular mass by the empirical formula mass, deriving a whole number, n 3. Multiply the empirical formula by n to derive the molecular formula 3.6 Chemical Equations A. Chemical reactions 1. Reactants are listed on the left hand side 2. Products are listed on the right hand side 3. Atoms are neither created nor destroyed a. All atoms present in the reactants must be accounted for among the products, in the same number b. No new atoms may appear in the products that were not present in the reactants B. The Meaning of a Chemical Reaction 1. Physical States a. Solid - (s) b. Liquid - (l) c. Gas - (g) d. Dissolved in water (aqueous solution) - (aq) 2. Relative numbers of reactants and products a. Coefficients give atomic/molecular/mole ratios 3.7 Balancing Chemical Equations A. Determine what reaction is occurring 1. It is sometimes helpful to write this in word form: Hydrogen + oxygen à water B. Write the unbalanced equation 1. Focus on writing correctly atomic and compound formulas H2 + O2 à H2O C. Balance the equation by inspection 1. It is often helpful to work systematically from left to right 2H2 + O2 à 2H2O D. Include phase information 2H2 (g) + O2 (g) à 2H2O (l)

3.8 Stoichiometric Calculations: Amounts of Reactants and Products A. Balance the chemical equation B. Convert grams of reactant or product to moles C. Compare moles of the known to moles of the desired substance 1. A ratio derived from the coefficients in the balanced equation D. Convert from moles back to grams if required 3.9 Calculations Involving a Limiting Reactant A. Concept of limiting reactant (limiting reagent): " I want to make chocolate chip cookies. I look around my kitchen (I have a BIG kitchen!) and find 40 lbs. of butter, two lbs. of salt, 1 gallon of vanilla extract, 80 lbs. of chocolate chips, 200 lbs. of flour, 150 lbs. of sugar, 150 lbs. of brown sugar, ten lbs. of baking soda and TWO eggs. It should be clear that it is the number of eggs that will determine the number of cookies that I can make." 1. The limiting reactant controls the amount of product that can form B. Solving limiting reactant problems 1. Convert grams of reactants to moles 2. Use stoichiometric ratios to determine the limiting reactant 3. Solve as before, beginning the stoichiometric calculation with the grams of the limiting reactant C. Calculating Percent Yield 1. Actual yield - what you got by actually performing the reaction 2. Theoretical yield - what stoichiometric calculation says the reaction SHOULD have produced Actual Yield x 100% Theoretical Yield

= percent yield

AP Chemistry A. Allan Chapter 4 Notes - Types of Chemical Reactions and Solution Chemistry 4.1 Water, the Common Solvent A. Structure of water 1. Oxygen's electronegativity is high (3.5) and hydrogen's is low (2.1) 2. Water is a bent molecule 3. Water is a polar molecule B. Hydration of Ionic Solute Molecules 1. Positive ions attracted to the oxygen end of water 2. Negative ions attracted to the hydrogen end of water C. Hydration of Polar Solute Molecules 1. Negative end of polar solute molecules are attracted to water's hydrogen 2. Positive end of polar solute molecules are attracted to water's oxygen D. "Like Dissolves Like" 1. Polar and ionic compounds dissolve in polar solvents like water 2. Nonpolar compounds like fats dissolve in nonpolar solvents like ____?_____ 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes A. Definition of Electrolytes 1. A substance that when dissolved in water produces a solution that can conduct an electric current B. Strong electrolytes conduct current very efficiently 1. Completely ionized when dissolved in water a. Ionic compounds b. Strong acids (HNO3(aq), H2SO4(aq), HCl(aq)) c. Strong bases (KOH NaOH) C. Weak electrolytes conduct only a small current 1. Slightly ionized in solution a. Weak acids (organic acids - acetic, citric, butyric, malic) b. Weak bases (ammonia) D. Nonelectrolytes conduct no current 1. No ions present in solution a. alcohols, sugars 4.3 The Composition of Solutions A. Molarity 1. Moles of solute per liter of solution M = molarity = moles of solute/liters of solution

B. Concentration of Ions in Solution 1. Ionic compounds dissociate in solution, multiplying the molarity by the number of ions present C. Moles from Concentration 1. Liters of solution x molarity = moles of solute D. Solutions of Known Concentration 1. Standard solution - a solution whose concentration is accurately known 2. Preparation of Standard solutions How much x How strong x What does it weigh? L x mol/L x g/mol = grams required to prepare the standard E. Dilution 1. Dilution of a volume of solution with water does not change the number of moles present 2. Solving dilution problems M1V1 = M2V2 4.4 Types of Chemical Reactions A. Precipitation reactions 1. When two solutions are mixed, an insoluble solid forms B. Acid-Base reactions 1. A soluble hydroxide and a soluble acid react to form water and a salt C. Oxidation-Reduction reactions (redox rxns) 1. Reactions in which one or more electrons are transferred 4.5 Precipitation Reactions A. Dissociation 1. Ionic compounds dissolve in water and the ions separate and move independently AgNO3(aq) + NaCl(aq) à products Ag+ (aq) + NO3-(aq) + Na+(aq) + Cl- à products B. Determination of Products 1. Recombination of ions a. AgNO3 NaCl AgCl NaNO3 2. Elimination of reactants as products a. AgNO3 and NaCl are reactants and can't be products 3. Identifying the precipitate a. "Switch Partners" of reactant pairs to determine the names of the products. b. AgCl and NaNO3 are the products c. AgCl is insoluble, so it is the white precipitate d. If there is no insoluble product, the reaction does not occur a. NaCl(aq) + KNO3(aq) à NaNO3(aq) + KCl(aq) Both products are soluble and all ions remain independent in solution; no reaction occurs Na+(aq) + Cl-(aq) + K+(aq) + NO3-(aq)

Table 4.1 Simple Rules for the Solubility of Salts in Water 1. Most nitrate (NO3-)salts are soluble. 2. Most salts containing the alkali metal ions (Li+, Na+, K+, Cs+, Rb+) and the ammonium (NH4+) ion are soluble. 3. Most chloride, bromide and iodide salts are soluble. Exceptions are salts containing the ions Ag+, Pb2+, and Hg2+. 4. Most sulfate salts are soluble. Notable exceptions are BaSO4, PbSO4, HgSO4 and CaSO4. 5. Most hydroxide salts are only slightly soluble. The important soluble hydroxides are NaOH and KOH. The hydroxides of barium, strontium and calcium are marginally soluble. 6. Most sulfide (S2-), carbonate (CO32-), chromate (CrO42-) and phosphate (PO43-) salts are only slightly soluble. 4.6 Describing Reactions in Solution A. The Molecular Equation 1. Gives the overall reaction stoichiometry, not necessarily the actual forms of reactants and products in solution Na2CO3(aq) + Ca(NO3)2(aq) à 2NaNO3(aq) + CaCO3(s) B. The Complete Ionic Equation 1. Represents as ions all reactants and products that are strong electrolytes 2Na+(aq) + CO32-(aq) + Ca2+(aq) + 2NO3-(aq) à 2Na+(aq) + 2NO3-(aq) + CaCO3(s) C. The Net Ionic Equation 1. Includes only those components that take part in the chemical change 2. Spectators are eliminated Ca2+(aq) + CO32-(aq) à CaCO3(s) 4.7 Stoichiometry of Precipitation Reactions A. Determine what reaction takes place B. Write the balanced net ionic equation for the reaction C. Calculate the moles of reactants D. Determine which reactant is limiting E. Calculate the moles of product or products F. Convert to grams or other units, as required

4.8 Acid-Base Reactions (Neutralization Reactions) A. Definitions 1. Br∅nsted:Acids are proton donors, bases are proton acceptors 2. Arrhenius: Acids produce H+ ions in water, bases produce OH- ions in solution B. Net ionic equation for acid-base reactions 1. H+(aq) + OH-(aq) à H2O(l) 2. The hydroxide ion can be assumed to completely react with even a weak acid in solution C. Stoichiometry Calculations for Acid-Base Reactions 1. List the species present in the combined solution before any reaction occurs; decide what reaction will occur 2. Write the balanced net ionic equation for this reaction 3. Calculate the moles or reactants a. For reactions in solution, use volumes of the original solutions and their molarities 4. Determine the limiting reactant where appropriate 5. Calculate the moles of the required reactant or product 6. Convert to grams or volume of solution as required D. Acid-Base Titrations 1. Vocabulary a. Titrant - Solution of known concentration b. Analyte - Solution of unknown concentration c. Equivalence point - Point at which the amount of titrant added to analyte results in perfect neutralization d. Indicator - a substance added at the beginning of the titration that changes color at the equivalence point e. Endpoint - the point at which the indicator changes color 2. Requirements for a successful titration a. the exact reaction between titrant and analyte must be known b. the reaction must proceed rapidly c. the equivalence point must be marked accurately (select the appropriate indicator) d. the volume of titrant required to reach the equivalence point must be known accurately e. for acid-base titrations, the titrant should be a strong acid or a strong base 4.9 Oxidation-Reduction Reactions (redox) A. Electron transfer (LEO says GER) 1. Gain electrons = reduction 2. Lose electrons = oxidation B. Examples of redox rxns 1. Photosynthesis 2. Combustion of fuels 3. Oxidation of sugars, fats, proteins for energy

Rules for Assigning Oxidation Numbers 1. the oxidation number of the atom of a free element is zero 2. the oxidation number of a monatomic ion equals its charge 3. In compounds, oxygen has an oxidation number of 2, except in peroxides, where it is -1 4. In compounds containing hydrogen, hydrogen has an oxidation number of +1 5. In compounds, fluorine is ALWAYS assigned an oxidation number of -1 6. The sum of the oxidation states for an electrically neutral compound must be zero

Summary Element = 0

Oxygen = -2 Hydrogen = +1 Fluorine = -1

C. Noninteger Oxidation states 1. Fe3O4 - Magnetite a. Oxidation number for each iron averages to +8/3 b. Magnetite contains two Fe3+ ions and one Fe2+ D. Characteristics of Oxidation-Reduction Reactions 1. the oxidized substance: a. loses electrons b. increases oxidation state c. is the reducing agent 2. the reduced substance a. gains electrons b. decreases oxidation state c. is the oxidizing agent 4.10 Balancing Oxidation-Reduction Equations There are no notes for this section. The only way to master the balancing of redox equations is to actually balance them. While there are some minor variations in the processes used for acidic and basic solutions, the skills involved are identical. We will practice balancing numerous redox equations as a class.

AP Chemistry A. Allan Chapter 5 - Gases 5.1 Pressure A. Properties of gases 1. Gases uniformly fill any container 2. Gases are easily compressed 3. Gases mix completely with any other gas 4. Gases exert pressure on their surroundings a. Pressure = force/area B. Measuring barometric pressure 1. The barometer a. Inventor - Evangelista Torricelli (1643) 2. Units a. mm Hg (torr) (1) 760 torr = Standard pressure b. newtons/meter2 = pascal (Pa) (1) 101,325 Pa = Standard pressure c. atmospheres (1) 1 atmosphere = Standard pressure 5.2 The Gas Laws of Boyle, Charles, and Avogadro A. Boyle's Law (Robert Boyle, 1627 - 1691) 1. the product of pressure times volume is a constant, provided the temperature remains the same PV = k a. P is inversely related to V b. The graph of P versus V is hyperbolic c. Volume increases linearly as the pressure decreases (1/P) 2. At constant temperature, Boyle's law can be used to find a new volumes or a new pressure a. P1V1 = k = P2V2 ∴

PV 1

1

= P 2V 2

P =V P V

or

1

2

2

1

3. Boyle's law works best at low pressures 4. Gases that obey Boyle's law are called Ideal gases B. Charles' Law (Jacques Charles, 1746 - 1823) 1. The volume of a gas increase linearly with temperature provided the pressure remains constant a. V = bT V/T = b (1) V1/T1 = b = V2/T2

V T

1 1

=V2

T

2

or

V V

1 2

=

T T

1 2

b. Temperature must be measured in degrees Kelvin (1) K = °C + 273 (2) 0 K is "absolute zero" C. Avogadro's Law (Amedeo Avogadro, 1811) 1. For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles, n a. V = an V/n = a (1) V1/n1 = a = V2/n2 ∴

V =V n n 1

2

1

2

5.3 The Ideal Gas Law A. Derivation from existing laws 1. V = k/P V = bT V = an

 Tn  V = (k )(b )(a )  P 2. Constants k, b, a are combined into universal gas constant, R

V =

nRT P

or

R = 0.0826

PV = nRT L ∗ atm K ∗ mol

B. Limitations of the Ideal Gas Law 1. Works well at low pressures and high temperatures 2. Most gases do not behave ideally above 1 atm pressure 3. Does not work well near the condensation conditions of a gas C. Solving for new volumes, temp or pressure (n remaining constant) 1. Combined law (from general chem) 2.

PV T 1

1

1

=n

R

=

PV T 2

2

2

or

PV T

 P1  T 2  = V 2 V 1     P 2  T 1 

1

1

1

=

PV T 2

2

2

5.4 Gas Stiochiometry A. Standard temperature and pressure (STP) 1. 0 °C, 273 K 2. 760 torr, 1 atm B. Molar volume 1. One mole of an ideal gas occupies 22.42 liters of volume at STP C. Things to remember

Density =

mass volume

grams of subs tan ce molar mass

n=

5.5 Dalton's Law of Partial Pressures (John Dalton, 1803) A. Statement of law 1. "For a mixture of gases in a container, the total pressure exerted is the sum of the pressures each gas would exert if it were alone" 2. It is the total number of moles of particles that is important, not the identity or composition of the gas particles B. Derivation 1. PTOTAL = P1 + P 2 + P3 + ... =

nRT V 1

2.

P

3.

P

4.

PTOTAL =

5.

1

TOTAL

=

P

nRT V 1

2

+

=

n RT n RT P= V V 2

3

3

n RT V 2



+

(n + n + n + ...) R T V 1

2

3

R T = PTOTAL nTOTAL   V



. . .

n RT V 3

   

   

C. Mole Fraction 1. The ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture 2. For an ideal gas, the mole fraction (x):

x

1

=

n

1

n

TOTAL

=

P P

1

TOTAL

5.6 The Kinetic Molecular Theory of Gases (KMT) A. Postulates of the KMT Related to Ideal Gases 1. The particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be zero 2. The particles are in constant motion. Collisions of the particles with the walls of the container cause pressure 3. Assume that the particles exert no forces on each other. 4. The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas B. Explaining Observed Behavior with KMT 1. P and V (T = constant) a. As V is decreased, P increases: V decrease causes a decrease in the surface area. Since P is force/area, the decrease in V causes the area to decrease, increasing the P 2. P and T (V = constant) a. As T increase, P increases The increase in T causes an increase in average kinetic energy. Molecules moving faster collide with the walls of the container more frequently, and with greater force 3. V and T (P = constant) a. As T increases, V also increases Increased T creates more frequent, more forceful collisions. V must increase proportionally to increase the surface area, and maintain P 4. V and n (T and P constant) a. As n increases, V must increase Increasing the number of particles increases the number of collisions. This can be balanced by an increase in V to maintain constant P 5. Dalton's law of partial pressures a. P is independent of the type of gas molecule KMT states that particles are independent, and V is assumed to be zero. The identity of the molecule is therefore unimportant C. Root Mean Square Velocity 1. Velocity of a gas is dependent on mass and temperature. 2. Velocity of gases is determined as an average a. M = mass of one mole of gas particles in kg b. R = 8.3145 J/K•mol (1) joule = kg•m2/s2

u

rms

=

3RT M

D. Mean Free Path 1. Average distance a molecule travels between collisions a. 1 x 10-7 m for O2 at STP

5.7 Effusion and Diffusion A. Effusion 1. Movement of a gas through a small opening into an evacuated container (vacuum) 2. Graham's law of effusion

Rate of effusion for gas 1 Rate of effusion for gas 2

=

M M

2 1

B. Diffusion 1. The mixing of gases 2. Diffusion is complicated to describe theoretically and mathematically 5.8 Real Gases and van der Waals Equation (Johannes van der Waals, 1873) A. Volume 1. Real gas molecules do have volume 2. Volume available is not 100% of the container volume a. n = number of moles b. b = is an empirical constant, derived from experimental results Ideal Real

P

=

n RT V

P= '

n RT V −n b

B. Pressure 1. Molecules of real gases do experience attractive forces a. a = proportionality constant determined by observation of the gas

n = − P P a  V '

obs

   

2

C. Combining to derive van der Waal's eqn

n n R T −a  P = − V V nb  obs

2

   

and then rearranging…    Pobs + a  

n  V 

    × (V − n b ) = n R T    2

5.9 Chemistry in the Atmosphere A. Composition of the Troposphere Composition of dry air (sea level Component Mole Fraction Nitrogen 0.78084 Oxygen 0.20948 Argon 0.00934 Carbon dioxide 0.000345 Neon 0.00001818 Helium 0.00000524 Methane 0.00000168 B. Photochemical Smog - the problem of nitrogen oxides (NOx) 1. Auto exhaust contains small amounts of NO, which is quickly oxidized 2NO(g) + O2(g) à 2NO2(g) 2. Radiant energy causes NO2 to decompose NO2(g) à NO(g) + O(g) 3. Free oxygen atoms combine with oxygen molecules to form ozone O(g) + O2(g) à O3(g) 4. Ozone may absorb light energy and decompose to excited oxygen atoms and excited oxygen molecules O3(g) à O2* + O* 5. Excited oxygen atoms react with water to form the hydroxyl radical O* + H2O à 2OH 6. Hydroxyl can react with NO2 to form nitric acid OH + NO2 à HNO3 C. Coal and acid rain 1. Most coal, especially cheap coal, contains sulfer S (in coal) + O2 (g) à SO2 (g) 2. Sulfur dioxide is oxidized in air 2SO2 (g) + O2(g) à 2SO3 (g) 3. Acid rain forms at the SO3 combines with water in the air SO3 (g) + H2O (l) à H2SO4 (aq)

AP Chemistry A. Allan Chapter Six Notes - Thermochemistry 6.1 The Nature of Energy A. Definition 1. Energy is the capacity to do work (or to produce heat*) a. Work is a force acting over a distance (moving an object) b. *Heat is actually a form of energy. (1) chemicals may store potential energy in their bonds that can be released as heat energy B. Law of Conservation of Energy 1. Energy can be converted from one form to another, but cannot be created or destroyed a. Potential energy (1) energy due to position or composition b. Kinetic energy (1) energy due to the motion of an object 1 2 (2) KE = 2 m v C. Heat and Temperature 1. Temperature reflects random motion of particles in a substance 2. Temperature indicates the direction in which heat energy will flow 3. Heat is a measure of energy content 4. Heat is what is transferred during a temperature change D. State Functions 1. A property of a system that depends only on its present state. 2. State functions do not depend on what has happened in the system, or what might happen in the system in the future 3. State functions are independent of the pathway taken to get to that state. Example: a liter of water behind a dam has the same potential energy for work regardless of whether it flowed downhill to the dam, or was taken uphill to the dam in a bucket. The potential energy is a state function dependent only on the current position of the water, not on how the water got there. E. Chemical Energy 1. Exothermic reactions a. Reactions that give off energy as they progress b. Some of the potential energy stored in the chemical bonds is converted to thermal energy (random KE) through heat c. Products are generally more stable (stronger bonds) than reactants 2. Endothermic reactions a. Reactions in which energy is absorbed from the surroundings b. Energy flows into the system to increase the potential energy of the system

c. Products are generally less stable (weaker bonds) than the reactants F. Thermodynamics 1. System Energy ∆E = q + w a. q = heat (1) q is positive in endothermic reactions (2) q is negative in exothermic reactions b. w = work (1) w is negative if the system does work (2) w is positive if work is done on the system 2. Work done by gases w = − P∆V a. by a gas (through expansion) (1) ∆V is positive (2) w is negative b. to a gas (by compression) (1) ∆V is negative (2) w is positive 6.2 Enthalpy and Calorimetry A. Enthalpy

H

=

E + PV

1. In systems at constant pressure, where the only work is PV, the change in enthalpy is due only to energy flow as heat (∆H = heat of rxn) ∆H = H products − H reac tan ts a. ∆H is negative for exothermic rxns b. ∆H is positive for endothermic rxns

B. Calorimetry - science of measuring heat 1. Heat capacity (C) a. ratio of heat absorbed to increase in temperature

C

=

heat absorbed Temperature increase

2. Specific Heat Capacity a. Energy required to raise the temp of 1 gram of a substance by 1 °C 3. Molar heat capacity a. Energy required to raise the temp of 1 mole of a substance by 1 °C C. Constant Pressure Calorimetry (solutions) 1. Calculating Heat of Rxn, ∆H a. ∆H = specific heat capacity x mass of sol'n x increase in temp ∆H = s x m x ∆T 2. Heat of rxn is an extensive property - dependent on the amount of substance a. ∆H α moles of reactant D. Constant Volume Calorimetry 1. Volume of bomb calorimeter cannot change, so no work is done 2. The heat capacity of the calorimeter must be known, generally in kJ/°C

2.

∆E = q + w

,

w =0

∴

∆E = q

6.3 Hess's Law A. Statement of Hess's Law 1. In going from a particular set of reactants to a particular set of products, the change in enthalpy (∆H) is the same whether the reaction takes place in one step or in a series of steps One step: N2 (g) + 2O2 (g) à 2NO2 (g)

∆H1 = 68kJ

Two step N2 (g) + O2 (g) à 2NO (g) 2NO (g) + O2 (g) à 2NO2 (g) N2 (g) + 2O2 (g) à 2NO2 (g)

∆H2 = 180kJ ∆H3 = -112kJ ∆H2 + ∆H3 = 68kJ

B. Characteristics of Enthalpy Changes 1. If a reaction is reversed the sign on ∆H is reversed N2 (g) + 2O2 (g) à 2NO2 (g) ∆H = 68kJ 2NO2 (g) à N2 (g) + 2O2 (g) ∆H = - 68kJ 2. The magnitude of ∆H is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ∆H is multiplied by the same integer C. Using Hess's Law 1. Work backward from the final reaction 2. Reverse reactions as needed, being sure to also reverse ∆H 3. Remember that identical substances found on both sides of the summed equation cancel each other 6.4 Standard Enthalpies of Formation A. Standard State 1. For a compound a. Gaseous state (1) pressure of 1 atm b. Pure liquid or solid (1) standard state is the pure liquid or solid c. Substance in solution (1) concentration of 1 M 2. For an element a. the form in which the element exists at 1 atm and 25°C B. Standard Enthalpy of Formation (∆Hf°) 1. The change in enthalpy that accompanies the formation of one mole of a compound from its elements with all elements in their standard state C. Calculating enthalpy change 1. When a rxn is reversed, the magnitude of ∆H remains the same, but its sign changes 2. When the balanced eqn for a rxn is multiplied by an integer, the value of ∆H must be multiplied by the same integer 3. The change in enthalpy for a rxn can be calculated from the enthalpies of formation of the reactants and products ° ° ° ∆H reaction = ∑ n p ∆H f ( products ) − ∑ nr ∆H f (reac tan ts ) 4. Elements in their standard states are not included a. For elements in their standard state, ∆Hf° = 0

6.5 Present Sources of Energy A. Fossil Fuels 1. Energy derived from these fuels was initially captured from solar energy by photosynthesis 2. Combustion of fossil fuels always produces H2O and CO2 B. Petroleum and Natural Gas 1. Petroleum a. Thick, dark liquid composed of hydrocarbon 2. Natural gas a. Methane, with smaller amounts of ethane, propane and butane Some Common Hydrocarbons CH4 Methane C2H6 Ethane C3H8 Propane C4H10 Butane C5H12 Pentane C6H14 Hexane C7H16 Heptane C8H18 Octane C. Petroleum Refining 1. Original refining isolated kerosene (gasoline was a waste product) 2. Tetraethyl lead added as an "anti-knock" agent Petroleum Fraction C5 - C10 C10 - C18 C15 - C25 > C25

Major Uses Gasoline Kerosene, Jet fuel Diesel fuel, Heating oil, lubricating oil Asphalt

D. Coal 1. Four stages of Coal 2. Carbon content increases over time 3. Value of coal is proportional to carbon content Mass Percent of Each Element Type of Coal C H O N Lignite 71 4 23 1 Subbituminous 77 5 16 1 Bituminous 80 6 8 1 Anthracite 92 3 3 1

S 1 1 5 1

E. CO2 and Earth's Climate 1. CO2 is a by-product of cellular respiration 2. CO2 is a by-product of burning fossil fuels 3. CO2 is a greenhouse gas 4. Atmospheric CO2 increased 16% from 1880 to 1980 5. Long-term climatic change seems imminent but is difficult to predict

6.6 New Energy Sources A. Coal Conversion 1. Gasification a. Reduce length of hydrocarbon molecules to create liquid or gaseous fuels b. Produce Syngas (CO and H2) 2. Coal Slurry a. Coal dust suspended in water used as a heavy fuel oil replacement 3. Coal limitations a. Mining of coal has a serious environmental impact B. Hydrogen as a fuel 1. Freeing hydrogen from compounds requires substantial energy a. CH4 (g) + H2O (g) à 3H2 (g) + CO (g) ∆H = 206 kJ b. H2O (l) + H2 (g) + 1/2 O2 (g) ∆H = 286 kJ 2. Hydrogen is difficult to transport a. Hydrogen in contact with metal produces free hydrogen atoms b. Hydrogen attempts penetrate the metal and make it brittle 3. Hydrogen is not dense a. The fuel equivalent of 20 gallons of gasoline occupies a volume of 238,000 liters. b. Liquid hydrogen is stored under great pressure and is potentially explosive C. Other Energy Alternatives 1. Shale a. Must be heated to extract fuel molecules, and produces immense amounts of waste rock 2. Ethanol from fermentation a. Mixture of ethanol and gasoline - gasohol b. Ethanol is renewable 3. Methanol 4. Seed oils a. Renewable

AP Chemistry A. Allan Chapter 7 Notes - Atomic Structure and Periodicity 7.1 Electromagnetic Radiation A. Types of EM Radiation (wavelengths in meters) 10-12

10-10

gamma

xrays

10-8

4 to 7x10

-7

UV

visible

10-4 IR

10-2 micro

102

1

104

Radio waves FM

short

AM

Wavelength increases Frequency decreases Energy decreases Speed is constant = 2.9979 x 108 m/sec B. Properties of EM Waves 1. Wavelength (λ) a. Distance between two consecutive peaks or troughs in a wave b. Measured in meters (SI system) 2. Frequency (ν) a. Number of waves that pass a given point per second b. Measured in hertz (sec-1) 3. Speed ( c ) a. Measured in meters/sec

4. Relationship of properties a. λν =c

7.2 The Nature of Matter A. Max Planck and Quantum Theory 1. Energy is gained or lost in whole number multiples of the quantity hv Frequency = v Planck's constant = h = 6.626 x 10-34 J•S ∆E = nhv 2. Energy is transferred to matter in packets of energy, each called a quantum B. Einstein and the Particle Nature of Matter 1. EM radiation is a stream of particles - "photons"

E

photon

= hv =

hc λ

2. Energy and mass are inter-related 2 E = mc C. de Broglie and the Dual Nature of Light 1. Light travels through space as a wave 2. Light transmits energy as a particle 3. Particle's have wavelength, exhibited by diffraction patterns

λ

=

h mv

a. large particles have very short wavelengths b. All matter exhibits both particle and wave properties 7.3 The Atomic Spectrum of Hydrogen A. Continuous spectra 1. Contains all wavelengths of light B. Bright line spectra 410nm

434nm

486nm

656nm

1. Excited electrons in an atom return to lower energy states 2. Energy is emitted in the form of a photon of definite wavelength 3. Definite change in energy corresponds to: a. Definite frequency b. Definite wavelength

∆E = hv

=

hc λ

4. Only certain energies are possible within any atom

7.4 Bohr Model (Neils Bohr, 1913) A. Quantum Model 1. The electron moves around the nucleus only in certain allowed circular orbits 2. Bright line spectra confirms that only certain energies exist in the atom, and atom emits photons with definite wavelengths when the electron returns to a lower energy state 3. Energy levels available to the electron in the hydrogen atom  2 −18 E = − 2.178 x10 J  Z2  n  n = an integer Z = nuclear charge J = energy in joules

B. Calculating the energy of the emitted photon 1. Calculate electron energy in outer level 2. Calculate electron energy in inner level 3. Calculate the change in energy (∆E) ∆E = energy of final state - energy of initial state 4. Use the equation:

λ

=

hc ∆E

to calculate the wavelength of the emitted photon C. Energy Change in Hydrogen atoms 1. Calculate energy change between any two energy levels   −18  1 − 1  = − x ∆E 2.178 10 J  2 2   n final ninitial  D. Shortcomings of the Bohr Model 1. Bohr's model does not work for atoms other than hydrogen 2. Electron's do not move in circular orbits 7.5 The Quantum Mechanical Model of the Atom A. The Electron as a standing wave 1. Standing waves do not propagate through space 2. Standing waves are fixed at both ends 3. Only certain size orbits can contain whole numbers of half wave lengths a. fits the observation of fixed energy quantities - "quanta" B. The Schrödinger Equation (Erwin Schrödinger) 1. For the motion of one particle, in the along the x axis in space:

h d ψ +V ψ − 8 π m dx 2

2

2

2

=

Eψ

2. Solution of the equation has demonstrated that E (energy) must occur in integer multiples 3. Your book has done you the favor of greatly simplifying this and presents the general equation: ∧

Hψ

= Eψ

∧

H = a set of mathematical functions called an "operator" ψ = a wave function. Specific wave functions are called "orbitals" 4. Orbitals 1. Orbitals are not circular orbits for electron 2. Orbitals are areas of probability for locating electrons

C. Heisenberg Uncertainty Principle (Werner Heisenberg) 1. "There is a fundamental limitation on how precisely we can know both the position and momentum of a particle at a given time"

∆x ⋅ ∆(mv)

≥

h 4π

∆x = uncertainty in the particle's position ∆(mv) = uncertainty in the particle's momentum 2. The more accurately we know the position of any particle, the less accurately we can know its momentum, and vice-versa D. Physical Meaning of a Wave Function 1. Square of the absolute value of the wave function gives a probability distribution.

ψ

2

2. Electron density map indicates the most probable distance from the nucleus 3. Wave functions and probability maps do not describe: a. How an electron arrived at its location b. Where the electron will go next c. When the electron will be in a particular location 7.6 Quantum Numbers A. Principal Quantum Number (n) 1. Integral values: 1, 2, 3, …. 2. Indicates probable distance from the nucleus a. Higher numbers = greater distance b. Greater distance = less tightly bound = higher energy B. Angular Momentum Quantum (l) (this was called the "orbital quantum number" in your general chem book) 1. Integral values from 0 to n - 1 for each principal quantum number n 2. Indicates the shape of the atomic orbitals Table 7.1 Angular momentum quantum numbers and corresponding atomic orbital numbers Value of l 0 1 2 3 4 Letter used s p d f g C. Magnetic Quantum Number (ml) 1. Integral values from l to -l, including zero 2. Magnetic quantum number relates to the orientation of the orbital in space relative to the other orbitals D. Spin Quantum Number 1. Covered in section 7.8

Table 7.2 Quantum numbers for the first four levels of orbitals in the hydrogen atom Orbital # of n l ml designation orbitals 1 0 1s 0 1 2

0 1

2s 2p

0 -1, 0, 1

1 3

3

0 1 2

3s 3p 3d

0 -1, 0, 1 -2, -1, 0, 1, 2

1 3 5

4

0 1 2 3

4s 4p 4d 4f

0 -1, 0, 1 -2, -1, 0, 1, 2 -3, -2, -1, 0, 1, 2, 3

1 3 5 7

7.7 Orbital Shapes and Energies A. Size of orbitals 1. Defined as the surface that contains 90% of the total electron probability 2. Orbitals of the same shape (s, for instance) grow larger as n increases B. s Orbitals 1. Spherical shape 2. Nodes (s orbitals of n=2 or greater) a. Internal regions of zero probability C. p Orbitals 1. Two lobes each 2. Occur in levels n=2 and greater 3. Each orbital lies along an axis (2px, 2py, 2pz)

D. d Orbitals 1. Occur in levels n=3 and greater 2. Two fundamental shapes a. Four orbitals with four lobes each, centered in the plane indicated in the orbital label dxz dyz dxy dx2- y2

b. Fifth orbital is uniquely shaped - two lobes along the z axis and a belt centered in the xy plane dz2

E. f Orbitals 1. Occur in levels n=4 and greater 2. Highly complex shapes 3. Not involved in bonding in most compounds F. Orbital Energies 1. All orbitals with the same value of n have the same energy a. "degenerate orbitals" (hydrogen only!) 2. The lowest energy state is called the "ground state" 3. When the atom absorbs energy, electrons may move to higher energy orbitals - "excited state" 7.8 Electron Spin and the Pauli Principle A. Electronic Spin Quantum Number 1. An orbital can hold only two electrons, and they must have opposite spins 2. Spin can have two values, +1/2 and -1/2 B. Pauli Exclusion Principle (Wolfgang Pauli) 1. "In a given atom no two electrons can have the same set of four quantum numbers" 7.9 Polyelectronic Atoms A. Internal Atomic energies 1. Kinetic energy of moving electrons 2. Potential energy of attraction between nucleus and electrons 3. Potential energy of repulsion between electrons B. The Electron Correlation Problem 1. Electron pathways are not known, so electron replusive forces cannot be calculated exactly

2. We approximate the average repulsions of all other electrons

C. Screening or Shielding 1. Electrons are attracted to the nucleus 2. Electrons are repulsed by other electrons 3. Electrons would be bound more tightly if other electrons weren't present D. Variations in energy within the same quantum level 1. Atoms other than hydrogen have variations in energy for orbitals having the same principal quantum number 2. Electrons fill orbitals of the same n value in preferential order Ens < Enp < End < Enf 3. Electron density profiles show that s electrons penetrate to the nucleus more than other orbital types a. Closer proximity to the nucleus = lower energy 7.10 The History of the Periodic Table No notes. Read this section. Notice especially Table 7.3 and 7.4. The periodic table is as useful for predicting properties of undiscovered elements today as it was in Mendeleev's time 7.11 The Aufbau Principle and the Periodic Table A. The Aufbau Principle 1. "As protons are added one by one to the nucleus to build up elements, electrons are similarly added to these hydrogen-like orbitals B. Hund's Rule 1. "The lowest energy configuration for an atom is the one having the maximum number of unpaired electrons allowed by the Pauli principle in a particular set of degenerate orbitals ***Note: We will review configuration and orbital notation in class. This was covered extensively in General Chem, and will come back to you quickly C. Period Table Vocabulary 1. Valence Electrons a. Electrons in the outermost principal quantum level of an atom b. Elements in the same group (vertical column) have the same valence electron configuration 2. Transition metals a. What we have called the "d" block 3. Lanthanide and Actinide Series a. The sets of 14 elements following lanthanum and actinium b. What we have called the "f" block 4. Main-group, or Representative Elements a. Groups 1A through 8A b. Configurations are consistent 5. Metalloids (semi-metals)

a. Found along the border between metals and nonmetals b. Exhibit properties of metals and nonmetals

7.12 Periodic Trends in Atomic Properties A. Ionization Energy - the energy required to remove an electron from an atom 1. Ionization energy increases for successive electrons 2. Ionization energy tends to increase across a period a. electrons in the same quantum level do not shield as effectively as electrons in inner levels b. irregularities at half filled and filled sublevels due to extra repulsion of electrons paired in orbitals, making them easier to remove 3. Ionization energy decreases with increasing atomic number within a group a. electrons farther from the nucleus are easier to remove B. Electron Affinity - the energy change associated with the addition of an electron 1. Affinity tends to increase across a period 2. Affinity tends to decrease as you go down in a period a. electrons farther from the nucleus experience less nuclear attraction b. Some irregularities due to repulsive forces in the relatively small p orbitals C. Atomic Radius 1. Determination of radius a. half of the distance between radii in a covalently bonded diatomic molecule - "covalent atomic radii" 2r

2. Periodic Trends a. Radius decreases across a period (1) increased effective nuclear charge due to decreased shielding b. Radius increases down a group (1) addition of principal quantum levels 7.13 The Properties of a Group: the Alkali Metals A. Easily lose valence electron (Reducing agents) 1. React with halogens to form salts 2. React violently with water a. Lithium is not the most reactive because the heat of reaction is insufficient to melt lithium and expose all of its surface area B. Large hydration energy 1. Positive ionic charge makes ions attractive to polar water molecules C. Radius and Ionization energy follow expected trends

See Table 7.8 for other properties of the Alkali metal group

AP Chemistry A. Allan Chapter 8 Notes - Bonding: General Concepts 8.1 Types of Chemical Bonds A. Ionic Bonding 1. Electrons are transferred 2. Metals react with nonmetals 3. Ions paired have lower energy (greater stability) than separated ions B. Coulomb's Law Q Q  −19  1 2 1. = ⋅ 2 . 31 x 10 J nm E    r  a. E = energy in joules b. Q1 and Q2 are numerical ion charges c. r = distance between ion center in nanometers d. negative sign indicates an attractive force C. Bond Length (covalent) 1. Distance at which the system energy is at a minimum 2. Forces at work a. Attractive forces (proton - electron) b. Repulsive forces (electron - electron, proton - proton) 3. Energy is given off (bond energy) when two atoms achieve greater stability together than apart D. Covalent Bonds 1. Electrons are shared by nuclei 2. Pure covalent (non-polar covalent) a. Electrons are shared evenly 3. Polar covalent bonds a. Electrons are shared unequally b. Atoms end up with fractional charges (1) δ+ or δ8.2 Electronegativity A. Electronegativity 1. The ability of an atom in a molecule to attract shared electrons to itself B. Electronegativity Trends 1. Electronegativity generally increases across a period (why?) 2. Electronegativity generally decrease within a family (why?) C. Characterizing bonds 1. Greater electronegativity difference between two elements means less covalent character and greater ionic character 2. We will not use the subtraction of electronegativities to determine ionic character. This text uses a practical definition to identify ionic compounds: Any compound that conducts an electric current when melted is an ionic compound.

8.3 Bond Polarity and Dipole Moments A. Dipolar Molecules 1. Molecules with a somewhat negative end and a somewhat positive end (a dipole moment) 2. Molecules with preferential orientation in an electric field + + +

3. All diatomic molecules with a polar covalent bond are dipolar B. Molecules with Polar Bonds but no Dipole Moment 1. Linear, radial or tetrahedral symmetry of charge distribution a. CO2 - linear b. CCl4 - tetrahedral 2. See table 8.2 in your text 8.4 Ions: Electron Configurations and Sizes A. Bonding and Noble Gas Electron Configurations 1. Ionic bonds a. Electrons are transferred until each species attains a noble gas electron configuration 2. Covalent bonds a. Electrons are shared in order to complete the valence configurations of both atoms B. Predicting Formulas of Ionic Compounds 1. Placement of elements on the periodic table suggests how many electrons are lost or gained to achieve a noble-gas configuration a. Group I loses one electron, Group II loses two, Group VI gains two, Group VII gains one…. 2. Formulas for compounds are balanced so that the total positive ionic charge is equal to the total negative ionic charge +3

−2

2

3

Al O

Total positive = +6 Total negative = -6 C. Sizes of Ions 1. Anions are larger than the parent atom 2. Cations are smaller than the parent atom 3. Ion size increases within a family 4. Isoelectronic ions a. Ions with the same number of electrons b. Size decreases as the nuclear charge Z increases

8.5 Formation of Binary Ionic Compounds A. Lattice Energy 1. The change in energy that takes place when separated gaseous ions are packed together to form an ionic solid M+ (g) + X- (g) à MX (s) 2. Energy change is exothermic (negative sign) Example: Formation of lithium fluoride Process Description Li(s) à Li(g) Sublimation energy

Energy Change (kJ) 161

Li(g) à Li+(g) + e-

Ionization energy

520

1/2F2 à F(g)

Bond energy (1/2 mole)

77

F(g) + e- à F-(g)

Electron affinity

-328

Li+(g) + F-(g) à LiF(s)

Lattice energy

-1047

Li(s) + 1/2F2(g) à LiF(s)

∆H

-617

3. The formation of ionic compounds is endothermic until the formation of the lattice 4. The lattice formed by alkali metals and halogens (1:1 ratio) is cubic except for cesium salts B. Lattice Energy Calculations Q Q  1. Lattice Energy = k  1 2     r  a. k = a proportionality constant dependent on the solid structure and the electron configuration b. Q1 and Q2 are charges on the ions c. r = shortest distance between centers of the cations and the anions 2. Lattice energy increases as the ionic charge increases and the distance between anions and cations decreases 8.6 Partial Ionic Character of Covalent Bonds A. Calculating Percent Ionic Character  measured dipole moment of X − Y   x100% Percent ionic character =  + −   calculated dipole moment of X Y  B. Ionic vs. Covalent 1. Ionic compounds generally have greater than 50% ionic character 2. Ionic compounds generally have electronegativity differences greater than 1.6 3. Percent ionic character is difficult to calculate for compounds containing polyatomic ions

8.7 The Covalent Chemical Bond: A Model A. Strengths of the Bond Model 1. Associates quantities of energy with the formation of bonds between elements 2. Allows the drawing of structures showing the spatial relationship between atoms in a molecule 3. Provides a visual tool to understanding chemical structure B. Weaknesses of the Bond Model 1. Bonds are not actual physical structures 2. Bonds can not adequately explain some phenomena a. resonance 8.8 Covalent Bond Energies and Chemical Reactions A. Average Bond Energies Process CH4(g) à CH3(g) + H(g) CH3(g) à CH2(g) + H(g) CH2(g) à CH(g) + H(g) CH(g) à C(g) + H(g)

Energy Required (kJ/mol) 435 453 425 339 Total 1652 Average 413

B. Multiple Bonds 1. Single bonds - 1 pair of shared electrons 2. Double bonds - 2 pairs of shared electrons 3. Triple bonds - 3 pairs of shared electrons Multiple Bonds, Average Energy (kJ/mole) C=C 614 N=O 607 839 N=N 418 C≡C O=O 495 941 N≡N C=O 745 891 C≡N 1072 C=N 615 C≡O 4. As the number of shared electrons increases, the bond length shortens (see table 8.5) C. Bond Energy and Enthalpy (using bond energy to calculate approximate energies for rxns) 1. ∆H = sum of the energies required to break old bonds(endothermic) + sum of the energies released in forming new bonds (exothermic) 2. ∆H = ∑ D ( Bonds broken) − ∑ D ( Bonds formed ) a. D always has a positive sign

8.9 The Localized Electron Bonding Model A. Lone electron pairs 1. Electrons localized on an atom (unshared) B. Bonding electron pairs 1. Electrons found in the space between atoms (shared pairs) C. Localized Electron Model 1. "A molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms D. Derivations of the Localized Model 1. Valence electron arrangement using Lewis structures 2. Prediction of molecular geometry using VSEPR (valence shell electron pair repulsion) 3. Description of the type of atomic orbitals used to share or hold lone pairs of electrons 8.10 Lewis Structures A. Electrons and Stability 1. "the most important requirement for the formation of a stable compound is that the atoms achieve noble gas configurations 2. Duet rule a. Hydrogen, lithium, beryllium, and boron form stable molecules when they share two electrons (helium configuration) 3. Octet Rule a. Elements carbon and beyond form stable molecules when they are surrounded by eight electrons B. Writing Lewis Structures 1. Rules a. Add up the TOTAL number of valence electrons from all atoms b. Use a pair of electrons to form a bond between each pair of bound atoms. Lines instead of dots are used to indicate each pair of bonding electrons c. Arrange the remaining atoms to satisfy the duet rule for hydrogen and the octet rule for the second row elements

8.11 Exceptions to the Octet Rule A. Boron Trifluoride 1. Note that boron only has six electrons around it 2. BF3 is electron deficient and acts as a Lewis acid (electron pair acceptor) 3. Boron often forms molecules that obey the octet rule B. Sulfur Hexafluoride 1. Note that sulfur has 12 electrons around it, exeeding the octet rule 2. Sulfur hexafluoride is very stable 3. SF6 fills the 3s and 3p orbitals with 8 of the valence electrons, and places the other 4 in the higher energy 3d orbital C. More About the Octet Rule 1. Second row elements C, N, O and F should always obey the octet rule 2. B and Be (second row) often have fewer then eight electrons around them, and form electron deficient, highly reactive molecules 3. Second row elements never exceed the octet rule 4. Third row and heavier elements often satisfy (or exceed) the octet rule 5. Satisfy the octet rule first. If extra electrons remain, place them on elements having available d orbitals a. When necessary to exceed the octet rule for one of several third row elements, assume that the extra electrons be placed on the central atom 8.12 Resonance A. Nitrate ion 1. Experiments show that all N-O bonds are equal 2. A single Lewis structure cannot represent the nitrate ion 3. A resonance structure is drawn by writing the three variant structures, connected by a double-headed arrow B. Resonance 1. When more than one valid Lewis structure can be written for a particular molecule 2. The actual structure is an average of the depicted resonance structures C. Odd Electron Molecules 1. Molecules in which there is not an even number of electrons 2. Does not fit localized electron model D. Formal Charge 1. Number of valence electrons on the free atom minus Number of valence electrons assigned to the atom in the molecule a. Lone pair (unshared) electrons belong completely to the atom in question b. Shared electrons are divided equally between the sharing atoms

2. The sum of the formal charges of all atoms in a given molecule or ion must equal the overall charge on that species a. If the charge on an ion is -2, the sum of the formal charges must be -2 E. Using Formal Charge to Evaluate Lewis Structures 1. If nonequivalent Lewis structures exist for a species, those with the formal charges closest to zero, and with negative formal charges on the most electronegative atoms are considered the best candidates 2. Only experimental evidence can conclusively determine the correct bonding situation in a molecule 8.13 Molecular Structure: The VSEPR Model A. Valence Shell Electron Pair Repulsion (VSEPR) 1. The structure around a given atom is determined principally by minimizing electron-pair repulsions 2. Non-bonding and bonding electron pairs will be as far apart as possible Arrangement of Electron Pairs Around an Atom Yielding Minimum Repulsion # of Electron Shape Arrangement of Electron Pairs Pairs 2 Linear

3

Trigonal Planar

4

Tetrahedral

5

Trigonal bipyramidal

6

Octahedral

B. Effect of Unshared Electron Pairs 1. The ideal tetrahedral angle is 109.5° Comparison of Tetrahedral Bond Angles Compound Structure Angle between Hydrogens Methane 109.5°

Ammonia

107°

Water

104.5°

2. Lone (unshared) electron pairs require more room than bonding pairs (they have greater repulsive forces) and tend to compress the angles between bonding pairs 3. Lone pairs do not cause distortion when bond angles are 120° or greater C. VSEPR and Multiple Bonds 1. For the VSEPR model, multiple bonds count as one effective electron pair 2. When a molecule exhibits resonance, ANY of the resonance structures can be used to predict the molecular structure using the VSEPR model D. Molecules Containing No Single Central Atom 1. Apply the principal of distancing shared and unshared electron pairs 2. Look at real 3-dimensional, rotatable models to develop predictive skills E. How Well Does VSEPR Work? 1. For non-ionic compounds, VSEPR works in most cases

Chapter 9 - Covalent Bonding: Orbitals 9.1 Hybridization and the Localized Electron Model A. Hybridization 1. The mixing of two or more atomic orbitals of similar energies on the same atom to produce new orbitals of equal energies B. Hybrid Orbitals 1. Orbitals of equal energy produced by the combination of two or more orbitals on the same atom C. Evidence for hybridization of carbon - Methane and sp3 1. Four bonds of equal length and strength

Carbon's isolated configuration

Carbon's hybridized configuration

2. Four effective pairs of electrons surround the carbon 3. "Whenever a set of equivalent tetrahedral atomic orbitals is required by an atom, this model assumes that the atom adopts a set of sp3 orbitals; the atom becomes hybridized" D. sp2 hybridization 1. Trigonal planar structure, 120° angle, in ethene (ethylene) rules out sp3 hybridization 2. sp2 hybridization creates 3 identical orbitals of intermediate energy and length and leaves one unhybridized p orbital 3. 3 effective pairs of electrons surround the carbon (double bond treated as one effective pair)

1

4. Sigma bonds (σ bond) a. Bond in which the electron pair is shared in an area centered on a line running between the atoms b. Lobes of bonding orbital point toward each other c. All bonds in methane are sigma bonds 5. Pi bonds (π bonds) a. Electron pair above and below the σ bond b. Created by overlapping of nonhybridized 2p orbitals on each carbon 6. Double bonds a. Double bonds always consist of one σ bond and one π bond C. sp Hybridization

1. Each carbon has two hybrid orbitals and two unhybridized 2p orbitals

2

2. Carbon dioxide a. Oxygens have 3 effective pairs of electrons (sp2 hybrids) (1) 1 double bond, two lone pairs b. Carbons have 2 effective pairs (2 double bonds)

Notice that the sp2 orbitals on the two oxygens are at 90° angles, as are the π bond between carbon and oxygen D. dsp3 Hybridization 1. Five effective pairs around a central 2. Trigonal bypyramidal shape 3. PCl5 is an example

atom

E. d2sp3 Hybridization 1. Six effective pairs around a central atom 2. Octahedral structure 3. SF6 is an example Question: Why doesn't carbon undergo dsp3 or d 2sp3 hybridization, while phosphorous and sulfur do undergo this type of hybridization? Atomic Orbitals s, p s, p, p s, p, p, p s, p, p, p, d s, p, p, p, d, d

Type of hybridization sp sp2 sp3 dsp3

# of hybrid orbitals 2 3 4 5

d2sp3

6

3

Geometry Linear Trigonal-planar Tetrahedral Trigonal bipyramidal Octagonal

# of Effective pairs 2 3 4 5 6

9.2 The Molecular Orbital Model A. Shortcomings of the Localized Electron Model 1. Electrons are not actually localized 2. Does not deal effectively with molecules containing unpaired electrons 3. Gives no direct information about bond energies B. Molecular Orbitals 1. Can hold two electrons with opposite spins 2. Square of the orbital's wave function indicates electron probability C. The Hydrogen Molecule (H2 ) 1. Two possible bonding orbitals, shapes determine by Ψ2

2. Bonding takes place in MO1 in which electrons achieve lower energy (greater stability), with electrons between the two nuclei 3. Both orbitals are in line with the nuclei, so they are σ molecular orbitals 4. Higher energy orbital is designated as antibonding (*). 5. Electron configuration of H2 can be written as σ1s 2 D. Bond Order 1. Bond order is the difference between the number of bonding electrons and the number of antibonding electrons, divided by two 2. Larger bond order = a. greater bond strength b. greater bond energy c. shorter bond length 9.3 Bonding in Homonuclear Diatomic Molecules A. In order to participate in molecular orbitals, atomic orbitals must overlap in space B. Larger bond order is favored A. When molecular orbitals are formed from p orbitals, σ orbitals are favored over π orbitals (σ interactions are stronger than π interactions) 1. Electrons are closer to the nucleus = lower energy

4

Energy

D. Paramagnetism 1. Magnetism can be induced in some nonmagnetic materials when in the presence of a magnetic field a. Paramagnetism causes the substance to be attracted into the inducing magnetic field (1) associated with unpaired electrons b. Diamagnetism causes the substance to be repelled from the inducing magnetic field (1) associated with paired electrons Figure 9.39 B2

C2

N2

σ2p*

_______

_______

_______

σ2p*

π 2p*

___ ___

___ ___

___ ___

π 2p*

___ ___

___ ___

σ2p

_______

_______

_______

π 2p

___ ___

___ ___

π 2p

___ ___

___ ___

___ ___

σ2p

_______

_______

σ2s *

_______

_______

_______

σ2s *

_______

_______

σ2s

_______

_______

_______

σ2s

_______

_______

Para 1 290

Dia 2 620

Dia 3 942

Para 2 495

Dia 1 154

159

131

110

121

143

Magnetism Bond Order Observed bond dissocation energy (kJ/mol) Observed bond length (pm)

O2 _______

F2 _______

One can measure magnetic properties FIRST, and use the results (dia- or para-) to determine the energy order of the molecular orbitals

5

9.4 Bonding in Heteronuclear Diatomic Molecules A. Similar, but not identical atoms 1. Use molecular orbital diagrams for homonuclear molecules B. Significantly different atoms 1. Each molecule must be examined individually 2. There is no universally accepted molecular orbital energy order 9.5 Combining the Localized Electron and Molecular Orbital Models A. Resonance 1. Attempt to draw localized electrons in a structure in which electrons are not localized

2. σ bonds can be described using localized electron model 3. π bonds (delocalized) must be described using the molecular orbital model B. Benzene 1. σ bonds (C - H and C - C) are sp2 hybridized a. Localized model 2. π bonds are a result of remaining p orbitals above and below the plane of the benzene ring

6

Chapter 10 - Liquids and Solids 10.1 Intermolecular Forces A. Dipole-Dipole Forces 1. Attraction between molecules with dipole moments a. Maximizes (+) ----- ( - ) interactions b. Minimizes (+) ----- ( + ) and (-) ----- ( - ) interactions 2. About 1% of strength of ionic bonds a. Unimportant in gas phase due to distance between molecules B. Hydrogen Bonding 1. Special dipole-dipole attraction a. Hydrogen covalently bonded to highly electronegative elements (N, O, F) has a higher than normal δ+ charge 2. Bond strength is higher than other dipole-dipole attractions 3. Important in the bonding of molecules such as water and DNA C. London Dispersion Forces 1. Instantaneous dipoles a. Random movement of electrons can create a momentary nonsymmetrical distribution of charge even in nonpolar molecules b. Instantaneous dipoles can induce a short-lived dipole in a neighboring molecule 2. London dispersion forces exist between all molecules, but are the weakest forces of attraction 3. "Polarizability" increases with the number of electrons in a molecule a. CCl4 experiences greater London forces than CH4 10.2 The Liquid State A. Surface Tension 1. The resistance of a liquid to an increase in its surface area 2. High intermolecular forces greater high surface tension B. Capillary Action 1. Cohesive forces between liquid molecules 2. Adhesive forces between polar liquid molecules and polar bonds in the material making up the container a. Water's adhesive forces are greater than its cohesive forces, thus the increase in surface area (concave meniscus) b. Oxygen in glass is attracted to hydrogen in water C. Viscosity 1. Measure of a liquid's resistance to flow a. Viscosity increases with intermolecular forces b. Viscosity increase with molecular size D. Structural Model for Liquids 1. Strong intermolecular forces (like solids) 2. Considerable molecular motion (like gases)

1

10.3 An Introduction to Structures and Types of Solids A. Types of solids 1. Crystalline solids a. Highly regular arrangement of components b. Components organized in a three-dimensional lattice (1) Smallest repeating unit of the lattice is a unit cell 2. Amorphous solids a. Components "frozen in place" and lacking orderly arrangement (1) glass (2) plastic B. X-ray Analysis of Solids 1. Structure of crystalline solids can be determined by xray diffraction a. waves in parallel beams that are "in phase" produce constructive interference b. waves in parallel beams that are "out of phase" produce destructive interference

θ = angle of incidence and reflection d = distance between atoms n = integer λ = wavelength 2. Bragg's Law a.

xy + yz = nλ ∴

and

xy + yz = 2d sin θ

nλ = 2d sin θ

3. Diffractometer a. computerized device to rotate crystal samples in an x-ray field, gather incidence and reflection data, and construct models of crystal structure

2

C. Types of Crystalline Solids 1. Ionic solids a. Ions occupy lattice points b. Sodium chloride is an example 2. Molecular solids a. Discrete covalent molecules occupy lattice points b. Ice and sucrose are examples 3. Atomic Solids (pure elements) a. Metallic solids b. Network solids c. Group 8A solids 10.4 Structure and Bonding in Metals A. Closest Packing 1. Arrangement of metallic atoms in the tightest pattern possible a. Each atom has twelve nearest neighbors (1) six in same layer (2) three above (3) three below 2. Hexagonal closest packed structure (a, b, a, b, a …) a. Each layer is identical to the layer two below it 3. Cubic closest packed structure (a, b, c, a, b, c, …) a. Each layer is identical to the layer three below it 4. Not all metals crystallize in closest packing B. Bonding Models for Metals 1. Electron sea model a. Metal cations b. Mobile sea of valence electrons that conduct heat and electricity 2. Band Model (Molecular Orbital Model) a. Electrons travel within molecular orbitals formed by the valence orbitals of the metallic atoms b. Molecular orbitals occupied by conducting electrons are called conduction bands C. Metal Alloys 1. Alloys are substances that contain a mixture of elements and have metallic properties 2. Substitutional alloys a. Host metal atoms are replaced in the lattice by other atoms of similar size (1) Brass, sterling silver, pewter 3. Interstitial alloys a. Holes in the closest packed metallic structure are filled by small atoms b. Carbon is often added to iron to produce steels with higher than normal hardness ***Ignore the references to steel frame bikes in Table 14.4. Everyone knows that REAL racing bikes are made out of TITANIUM! Don't believe me? See page 950! 3

10.5 Carbon and Silicon: Network Atomic Solids A. Network Solids 1. Atomic solids with strong directional covalent bonds B. Diamond 1. Large gap in energy between occupied orbitals and unoccupied orbitals makes diamond a poor conductor of electricity 2. Bonds formed by overlap of hybridized sp3 orbitals C. Graphite 1. Very strong σ bonds between carbons in fused rings using sp2 hybridization 2. π molecular orbitals provide weaker π bonding between layers a. Delocalized electrons make graphite an electrical conductor D. Silica 1. Quartz (SiO2) is actually based on interconnected SiO4 tetrahedra 2. Glass is produced by quickly cooling silica, producing an amorphous solid a. Other compounds added to molten silica prior to cooling produce glass with varying properties (see table 10.5) E. Ceramics 1. Made from clays containing silicates (silicon-oxygen anions) 2. Heterogeneous a. crystals of silicates suspended in a glassy cement 3. Remarkable range of uses F. Semiconductors 1. Silicon a. Energy gap between occupied and unoccupied orbitals is smaller than in diamond, allowing some electrons to be conducted b. Conductivity increases with temperature 2. n-type semiconductors a. "Dope" silicon with elements such as arsenic (1) extra valence electron increases conductivity 3. p-type semiconductors a. "Dope" silicon with elements such as boron (1) shortage of valence electrons creates a "hole" that electrons fill as they move (conduct) 10.6 Molecular Solids A. Examples 1. Water 2. Dry ice 3. P4 4. S8 B. Bonding 1. London dispersion forces 2. Dipole-dipole (including hydrogen bonding) a. Only in polar molecules

4

10.7 Ionic Solids A. Closest Packing 1. Large particles (usually anions) arrange in closest packing 2. Smaller particles (usually cations) fit into holes between anions a. not all holes need be filled 3. Maximize (+) ---- (-) attraction 4. Minimize (+) ---- (+) and (-) ---- (-) repulsion B. Types of holes 1. Trigonal holes - three spheres in the same layer 2. Tetrahedral holes - a sphere sits in the dimple of three spheres in the same layer 3. Octahedral holes - two layers of three spheres in adjoining layers 10.8 Vapor Pressure and Changes of State A. Vaporization (Evaporation) 1. The escape of molecules of a liquid from the surface to form a gas 2. Vaporization is always endothermic a. Heat of vaporization (Enthalpy of vaporization, ∆Hvap) is the energy required to vaporize one mole of a liquid at 1 atm B. Vapor Pressure 1. Pressure of the vapor present at equilibrium (also called equilibrium vapor pressure) ***remember vapor pressure corrections made in gas law problems involving water displacement 2. Variation in Vapor Pressure a. Liquids with high intermolecular attraction have relatively low vapor pressures b. Liquids with low intermolecular attraction have relatively high vapor pressures (the are "volatile") c. Vapor pressure increases with temperature 3. Clausius-Clapeyron equation  PTvap ln  1vap  PT  2

 ∆H vap =  R 

 1 1  −   T2 T1 

C. Sublimation 1. A process in which a substance goes directly from the solid to the gaseous state 2. Reasons for sublimation a. Solids have vapor pressure, but it is normally very low b. Solids with little intermolecular attraction may have substantial vapor pressures and be able to sublime at room conditions

5

D. Changes of State 1. Heating curve

2.

3.

4.

5.

6.

a. Note that temperature remains constant during a phase change b. Chemical bonds are not being broken during phase changes Heat of fusion (Enthalpy of fusion, ∆Hfus ) a. Energy required to convert a mole of solid substance to a mole of liquid substance Normal melting point a. The temperature at which the solid and liquid states have the same vapor pressure under conditions where the total pressure is 1 atm Normal boiling point a. The temperature at which the vapor pressure of the liquid is exactly 1 atmosphere Supercooling a. Rapid cooling of a liquid may allow it to exist as a liquid at temperatures below its normal melting point (1) quick temperature change does not allow time for molecules to become organized as they must be to become solids (2) When crystallization does begin, it occurs rapidly Superheating a. Rapid heating of a liquid may allow it to exist as a liquid at temperatures above the normal boiling point (1) not enough high energy molecules accumulate in one place to form bubbles (2) when bubbles do form, they tend to be very large (3) superheating can be avoided by adding boiling chips

6

10.9 Phase Diagrams A. Diagram for a closed system

1. Triple Point a. Solid and liquid have identical vapor pressure b. All three phases exist together in equilibrium ***Note that the "great outdoors" does not constitute a closed system. Seeing snow, water, and water vapor on a day at the slopes does not constitute the triple point, since the system is not closed and is not at equilibrium 2. Critical temperature a. The temperature above which the substance cannot exist as a liquid, regardless of how great the pressure 3. Critical pressure a. The pressure required to produce liquefaction at the critical temperature 4. Critical point a. Point defined by the critical temperature and critical pressure b. For water: 374°C and 218 atm ***I sincerely apologize about the references to ice skating on page 486 and 487. Ignore the references and think HOCKEY instead! Table 10.10 Boiling Point of Water at Various Locations Location Feet above sea level Patm (torr) Boiling Poing (°C) Top of Mt. Everest, Tibet 29,028 240 70 Top of Mt. Denali, Alaska 20,320 340 79 Top of Mt. Whitney, California 14,494 430 85 Leadville, Colorado 10,150 510 89 Top of Mt. Washington, N.H. 6,293 590 93 Boulder, Colorado 5,430 610 94 Madison, Wisconsin 900 730 99 New York City, New York 10 760 100 Death Valley, California -282 770 100.3

7

Chapter 11 – Properties of Solutions 11.1 Solution Composition A. Molarity 1.

Molarity( M ) =

moles solute liters of solution

B. Mass Percent    mass of solute  1. Mass percent =  × 100  mass of solution    C. Mole Fraction Mole fraction of componentA = x A =

1.

nA nA + nB

D. Molality 1.

moles of solute

Molality =

ki log ram of solvent E. Normality 1. 2. 3.

Normality =

equivalent s

liter of solution Equivalents of acids and bases a. Mass that donates or accepts a mole of protons Equivalents of oxidizing and reducing agents a. Mass that provides or accepts a mole of electrons

11.2 The Energies of Solution Formation A. “Like Dissolves Like” 1. Polar molecules and ionic compounds tend to dissolve in polar solvents 2. Nonpolar molecules dissolve in nonpolar compounds B. Steps of Solution Formation 1. Breaking up the solute into individual components (expanding the solute) a. Endothermic 2. Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent) a. Endothermic 3. Allowing the solute and solvent to interact to form the solution a. Often exothermic C. Enthalpy (Heat) of Solution 1. ∆H sol 'n = ∆H step1 + ∆H step2 + ∆H step3 1. ∆Ηsol’n may have a positive sign or a negative sign 3. Enthalpy (Heat) of Hydration, ∆Ηhyd a. Include ∆Η of step 2 and step 3

1

D. Factors Favoring the Solution Process 1. Negative value for ∆Ηsol’n 2. Increase in entropy 3. For positive values of ∆Ηsol’n it is the increase in entropy that outweighs the increase in energy and causes the solution process of occur 11.3 Factors Affecting Solubility A. Structure Effects 1. Polar (hydrophilic) dissolves in polar a. Water soluble vitamins 2. Nonpolar (hydrophobic) in nonpolar a. Fat soluble vitamins B. Pressure Effects 1. Henry’s Law a. The amount of a gas dissolved in a solution is directly proportional to the pressure of the gas above the solution b. P = kC (1) P = partial pressure of the gaseous solute above the solution (2) C = concentration of the dissolved gas (3) k = constant characteristic of a particular solution c. Henry’s Law works best for dilute solutions of gases that do not dissociate or react with the solvent C. Temperature Effects 1. Solids a. Increases in temperature always cause dissolving to occur more rapidly b. Increases in temperature usually increases solubility (the amount that can be dissolved) 2. Gases a. Solubility of gases always decreases with increasing temperature 11.4 The Vapor Pressure of Solutions A. Nonvolatile Solutes 1. Nonvolatile electrolytes lower the vapor pressure of a solute a. Nonvolatile molecules do not enter the vapor phase b. Fewer molecules are available to enter the vapor phase B. Raoult's Law 0 1. Pso ln = X solvent Psolvent where Xsolvent is the mole fraction of the solvent in the solution a. a linear equation (y = mx) 2. Molar mass of a solute can be calculated from experimental results for vapor pressure lowering C. Ionic solutes 1. Dissociation of ionic compounds has nearly two, three or more times the vapor pressure lowering of nonionic (nonelectrolyte) solutes.

2

D. Non-ideal Solutions 1. Liquid-liquid solutions in which both components are volatile 2. Modified Raoult's Law: PTOTAL = PA + PB = X A PA0 + X B PB0 a. P0 is the vapor pressure of the pure solvent b. PA and PB are the partial pressures E. Ideal Solutions 1. Liquid-liquid solution that obeys Raoult's law a. No solution is perfectly ideal, though some are close 2. Negative deviations from Raoult's law (lower than predicted vapor pressure for the solution) a. Solute and solvent are similar, with strong forces of attraction b. ∆Hsol'n is large and negative 3. Positive deviations from Raoult's law (higher than predicted vapor pressure for the solution) a. Solute and solvent are dissimiliar, with only weak forces of attraction b. Particles easily escape attractions in solution to enter the vapor phase 11.5 Boiling Point Elevation and Freezing-Point Depression A. Colligative Properties 1. Properties dependent on the number of solute particles but not on their identity a. Boiling-Point elevation b. Freezing-Point depression c. Osmotic Pressure B. Boiling-Point Elevation 1. Nonvolatile solutes elevate the boiling point of the solvent ∆T = K b msolute a. ∆Τ is the boiling point elevation b. Kb is the molal boiling point elevation constant of the solvent c. m solute is the molality of the solute in the solution C. Freezing-Point Depression 1. Solutes depress the freezing point of the solvent ∆T = K f msolute a. ∆Τ is the freezing point depression b. Kf is the molal freezing point depression constant of the solvent c. m solute is the molality of the solute in the solution

3

11.6 Osmotic Pressure A. Osmosis 1. The flow of solvent molecules into a solution through a semipermeable membrane

B. Osmotic Pressure 1. The pressure necessary to keep water from flowing across a semipermeable membrane 2. Osmotic pressure can be used to characterize solutions and determine molar masses π = MRT a. π is osmotic pressure in atmospheres b. M is the molarity of the solution c. R is the gas law constant d. T is the Kelvin temperature C. Dialysis 1. Transfer of solvent molecules as well as small solute molecules and ions D. Isotonic Solutions 1. Solutions that have the same osmotic pressure E. Osmotic Pressure and Living Cells 1. Crenation a. Cells placed in a hypertonic solution lose water to the solution, and shrink 2. Hemolysis a. Cells placed in a hypotonic solution gain water from the solution and swell, possibly bursting F. Reverse Osmosis 1. External pressure applied to a solution can cause water to leave the solution a. Concentrates impurities (such as salt) in the remaining solution b. Pure solvent (such as water) is recovered on the other side of the semipermeable membrane

4

11.7 Colligative Properties of Electrolyte Solutions A. van't Hoff factor i: 1.

i=

moles of

particles in solution

moles of solute dissolved 2. For ionic compounds, the expected value of i is an integer greater than 1 a. NaCl, i = 2 b. BaCl2, i = 3 c. Al2(SO4)3, i = 5 B. Actual values of i 1. Values of i are less than expected due to ion pairing (clustering) 2. Refer to table11.6 3. Values of i are closer to the expected the more dilute the solution becomes C. Incorporation of van't Hoff factor in Problem Solving 1. Boiling-elevation and freezing-point depression a. ∆T = imK 2. Osmotic pressure a. π = iMRT 11.8 Colloids A. Colloidal Dispersions (Colloids) 1. Tiny particles suspended in some medium 2. Particles range in size from 1 to 1000 nm. B. Tyndall Effect 1. Scattering of light by particles a. Light passes through a solution b. Light is scattered in a colloid Table 11.7

Types of Colloids

Examples Fog, aerosol sprays Smoke, airborne bacteria Whipped cream, soap suds Milk, mayonnaise Paint, clays, gelatin Marshmallow, polystyrene foam Butter, cheese Ruby glass

Dispersing Medium Gas Gas Liquid Liquid Liquid Solid Solid Solid

5

Dispersed Substance Liquid Solid Gas Liquid Solid Gas Liquid Solid

Colloid Type Aerosol Aerosol Foam Emulsion Sol Solid foam Solid emulsion Solid sol

Chapter 12 - Chemical Kinetics 12.1 Reaction Rates A. Chemical kinetics 1. Study of the speed with which reactants are converted to products B. Reaction Rate 1. The change in concentration of a reactant or product per unit of time A at time t 2 − concentrat ion of A at time t1 ∆[ A] = t 2 − t1 ∆t a. Rates decrease with time b. It is customary to express reaction rates as positive values c. Instantaneous rate can be determined by finding the slope of a line tangent to a point representing a particular time C. Decomposition of NO2 2NO2(g) à 2NO(g) + O2(g) Rate =

concentration of

Rate of consumption of NO2 = rate of production of NO = 2(rate of production of O2) −

∆[NO2 ] ∆t

=

∆[ NO] ∆t 1

=

 ∆[O2 ]  2   ∆t 

12.2 Rate Laws: An Introduction A. Reversibility of reactions 1. For the reaction 2NO2 (g) à 2NO(g) + O2(g) The reverse reaction 2NO(g) + O2 (g) à 2NO2(g) may also take place 2. The reverse reaction effects the rate of change in concentrations a. ∆[A] depends on the difference in the rates of the forward and reverse reactions B. Rate Law (Ignoring reverse reaction) 1. Rate = k[NO2]n a. k is a proportionality constant called the rate constant n is the order of the reactant (an integer, including zero, or a fraction) (1) k and n must be determined experimentally 2. Concentrations of products do not appear in the rate law C. Types of Rate Laws 1. Differential Rate Law (Rate Law) a. Expresses how the rate of a reaction depends upon concentration Rate = k[NO2]n 2. Integrated Rate Law a. Expresses how the concentration of a species (reactant or product) in a reaction depend on time ∆[NO2 ] ∆[ NO]  ∆[O2 ]  = = 2  ∆t ∆t  ∆t  3. Choice of rate law depends on what data is easiest to collect −

***Why study rate law? Re-read the paragraph at the bottom of page 556. 12.3 Determining the Form of the Rate Law A. Determining the value of n (Order of the reactant) 1. Example [A] Rate (mol/L ⋅s) 1.00 M 8.4 x 10-3 0.50 M 2.1 x 10-3 a. Doubling the concentration of species A quadruples the rate of the reaction. Therefore, the reaction is of second order with respect to A Rate = k[A]2 B. Method of Initial Rates 1. Initial rate a. Instantaneous rate just after t = 0 2. Experimental method a. Vary initial concentration of reactant(s) b. Determine initial rate for each concentration c. Examine relationship between rate and initial concentration

2

3. Example: Table 12-4 Experiment

1 2 3

Intial Rates from Three Experiments for the Reaction: NH4 +(aq) + NO2-(aq) à N2(g) + 2H2 O(l) Initial Initial Concentration Concentration Initial rate + Of NH4 Of NO2 (mol/L⋅s) 0.100 M 0.0050 M 1.35 x 10-7 0.100 M 0.010 M 2.70 x 10-7 0.200 M 0.010 M 5.40 x 10-7

a. In experiment 1 and 2, the concentration of NH4+ is held constant b. In experiment 2 and 3, the concentration of NO2- is held constant c. Basic rate law for the reaction: Rate = k[NH4+]n [NO2-]m 4. Calculations a. Experiment 1 and 2 (1) Rate doubles when concentration of NO2- doubles (2) m = 1 (first order) b. Experiment 2 and 3 (1) Rate doubles when concentration of NH4+ doubles (2) n = 1 (first order) 5. Overall reaction order a. Overall reaction order is the sum of m and n (1) m +n = 2 so overall the reaction is second order Rate = k[NH4+][NO2-] 6. Calculate k, the rate constant a. Rate is known b. Both concentrations are known c. Exponents are known 1.35 x 10-7 mol/L⋅s = k(0.100 M)(0.0050 M) k = 1.35 x 10-7 mol/L ⋅s (0.100 M)(0.0050 M) 12.4 The Integrated Rate Law A. Integrated First-Order Rate Law (single reactant)  [A]  1. ln[A] = -kt + ln[A]0 or ln  0  = kt  [A]  a. [A] is concentration of reactant A at time = t b. [A]0 is concentration of reactant A at t = 0 c. Equation is linear (y = mx + b) 2. Half-life of a First-Order Reaction a. Half life of a reaction is the time required for a reactant to reach half its original concentration [ A]0 à ln  [ A]0  = kt à ln (2 ) = kt  [A]  ln( 2) ln  0  = kt à [A] = à t1 / 2 = 1/ 2 1/ 2   2 k  [A]   [A]0 / 2  3

B. Integrated Second-Order Rate Law (single reactant) 1 1 1. = kt + [ A] [ A]0 a. Graph of 1/[A] versus t is a straight line with slope k 2. Half-life of a Second-Order Reaction (derivation on page 567) 1 a. t1 / 2 = k [A]0 b. For a second-order reaction each successive half-life is double the preceding one C. Zero Order reactions 1. Rate is constant; it does not change with changing concentration 2. Zero order sometimes happens with catalysis 3. Integrated rate law for zero-order reactions a. [A] = −kt + [ A]0 4. Half-life of a zero-order reaction [ A]0 a. t1 / 2 = 2k D. Integrated Rate Laws for Reactions with More than One Reactant 1. Examine rate with one reactant in very low concentration and the others much higher Rate = k[A]n[B]m[C]p a. If [B]> >[A] and [C] > >[A] then [B] and [C] do not change as greatly relative to [A] so… Rate = k'[A]n 2. Pseudo-first-order rate law (or zero-order , or second-order) a. Simplification yields a rate law of a particular order 12.5 Rate Laws: A Summary No Notes - This is review. Rate law problems are a virtual certainty on the AP test. Bookmark this section and review it thoroughly prior to the exam! 12.6 Reaction Mechanisms A. Reaction Mechanism 1. A series of elementary steps that must satisfy two requirements a. The sum of the elementary steps must give the overall balanced equation for the reaction b. The mechanism must agree with the experimentally determined rate law B. Intermediates 1. A species that is neither a reactant nor a product, but that is formed and consumed during a chemical reaction C. Elementary steps 1. Reactions whose rate law can be written from their molecularity (balanced eqn for the step) 4

D. Molecularity 1. The number of species that must collide to produce the reaction indicated by that step a. Unimolecular step - a reaction involving one molecule b. Bimolecular step - reaction involving the collisions of two species c. Termolecular step - reaction involving the collisions of three species Table 12.7 Examples of Elementary Steps Elementary Step Molecularity A à products Unimolecular A + A à products Bimolecular (2A à products) A + B à products Bimolecular A + A + B à products Termolecular (2A + B à products) A + B + C à products Termolecular

Rate Law Rate = k[A] Rate = k[A]2 Rate = k[A][B] Rate = k[A]2[B] Rate = k[A][B][C]

E. Rate-Determining Step 1. The slowest step in a reaction determines the rate of the reaction F. Example 1. Overall rxn: NO2(g) + CO(g) à NO(g) + CO2 (g) 2. Rate law determined experimentally: Rate = k[NO2]2 3. Elementary Steps k1 NO2(g) + NO2(g) → NO3(g) + NO(g) k2 NO3(g) + CO(g) → NO2 (g) + CO2(g) 4. Which step is rate determining? a. IF step 1 , Rate = k[NO2]2 (1) this agrees with experimental results b. IF step 2 , Rate = k[NO3][CO] (1) this does not agree with experimental results 12.7 A Model for Chemical Kinetics A. Collision Model 1. Molecules must collide in order to react a. They must collide with sufficient energy b. They must collide with correct orientation B. Activation Energy (Ea) 1. The energy required to convert atoms or molecules into the activated complex (transition state) 2. The minimum energy required for an effective collision C. The Arrhenius Equation 1. Form #1: k = zpe − E a / RT a. z is collision frequency b. p is the stearic factor representing the fraction of collisions with correct orientation 5

2. Form #2: k = Ae − E a / RT a. z and p are combined to make A, the frequency factor (frequency of effective collisions) 3. Taking the natural log of both sides: ln( k ) = ln( Ae − E a / RT ) ln( k ) = ln( A) + ln( e − E a / RT ) ln( k ) = −

Ea R

 1  + ln( A)   T 

a. Linear equation y = ln(k) slope = -Ea/R

x = 1/T

intercept = ln(A)

12.8 Catalysis A. Catalysts 1. A substance that speeds up a reaction without being consumed itself B. Effects of Catalysis 1. Catalysts lower activation energy but do not change ∆E for the reaction 2. Catalysts provide alternate reaction pathways 3. Catalysis results in a higher percentage of effective collisions C. Heterogeneous Catalysis (usually gaseous reactants on a solid surface) 1. Four steps a. Adsorption and activation of the reactants (1) Adsorption refers to the collection of one substance on the surface of another substance b. Migration of the adsorbed reactants on the surface c. Reaction of the adsorbed substances d. Escape, or desorption, of the products 2. Uses a. Conversion of alkenes to alkanes b. Conversion of CO and NO in auto exhaust to CO2 and N2 D. Homogeneous Catalysis 1. Reactants and catalysts are in the same phase 2. Example - The decomposition of ozone a. Freon, CCl2F2(g) light  → CClF 2(g) + Cl(g) b. Chlorine catalyzes the decomposition of ozone Cl(g) + O3 (g) à ClO(g) + O2(g) O(g) + ClO(g) à Cl(g) + O2 (g) O(g) + O3(g) à 2O2(g) c. Notice that the Cl(g) is not used up, a qualification for it to be catalytic. This is also why small amounts of chlorofluorocarbons such as freon can do immense damage to the ozone layer

6

Chapter 13 - Chemical Equilibrium Intro A. Chemical Equilibrium 1. The state where the concentrations of all reactants and products remain constant with time 2. All reactions carried out in a closed vessel will reach equilibrium a. If little product is formed, equilibrium lies far to the left b. If little reactant remains, equilibrium lies far to the right 13.1 The Equilibrium Condition A. Static Equilibrium does not occur in chemical systems 1. No reaction is taking place 2. All product molecules will remain product 3. All unused reactant molecules will remain unreacted B. Dynamic Equilibrium 1. Reactions continue to take place 2. Reactant molecules continue to be converted to product 3. Product continues to be converted to reactant (reverse reaction) 4. Forward and reverse reactions take place at the same rate at equilibrium C. Causes of Equilibrium 1. Beginning of reaction a. Only reactant molecules exist, so only reactant molecules may collide 2. Middle a. As product concentration increases, collisions may take place that lead to the reverse reaction 3. At equilibrium a. Rates of forward and reverse reactions are identical D. Example - The Haber Process 1. N2(g) + 3H2(g) ¾ 2NH3(g) a. hydrogen is consumed at 3x the rate of nitrogen b. ammonia is formed at 2x the rate at which nitrogen is consumed

1

13.2 The Equilibrium Constant A. The Law of Mass Action 1. For the balanced equation: jA + kB ¾ lC + mD

(j, k, l, m) are coefficients

The law of mass action is represented by the equilibrium expression: [ C] l [ D]m K= [ A] j [ B] k a. K is a constant called the equilibrium constant (1) K varies depending on temperature and upon the coefficients of the balanced equation b. [X] represents concentration of chemical species at equilibrium 2. For the reverse reaction lC + mD ¾ jA + kB The law of mass action is represented by the equilibrium expression: [ A] j [ B] k 1 K' = = l m [ C] [ D ] K B. Equilibrium Position 1. A set of equilibrium concentrations 2. There is only one value of K for a reaction at a given temperature, but an infinite number of possibilities for equilibrium positions C. Summary of the Equilibrium Expression 1. The equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse 2. When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression raised to the nth power Knew = (Koriginal )n 3. For a particular reaction at a given temperature, the value of K is constant regardless of the amounts of gases that are mixed together (homogeneous rxn system) 13.3 Equilibrium Expressions Involving Pressure A. Derivation of Equilibrium Gas Laws n 1. PV = nRT therefore… P =   RT V  n/V is the molar concentration of the gas, represented by C P = CRT 2. For the reaction N2(g) + 3H2(g) ¾ 2NH3(g) In terms of concentrations K c =

C NH 3 2 (C N 2 )(C H 2 3 )

In terms of partial pressures K p =

2

PNH 3 2 ( PN 2 )( PH 2 3 )

B. Relationship between K and Kp 1. K p = K ( RT ) ∆ n a. ∆n = (l + m) - (j + k) the difference in the sums of the coefficients for the gaseous products and reactants 13.4 Heterogeneous Equilibria A. The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present B. If pure solids or liquids are involved in a chemical reaction, their concentrations are not included in the equilibrium expression for the reaction C. Pure liquids are not the same as solutions, whose concentration can change 13.5 Applications of the Equilibrium Constant A. Extent of a Reaction 1. Reactions with large equilibrium constants (K>>1) go essentially to completion a. Equilibrium position is far to the right a. Generally large, negative ∆E 2. Reactions with small equilibrium constants (K