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Presentation on theme: "SAT II CHEM PREP PPT Mrs. Gupta"— Presentation transcript:

Modified from Mark Rosengarten’s Powerpoint

2 Setup of the SAT II Chem Exam
85 total questions, 1 hour (about 42 s/question) - All multiple choice, 1/4th point taken off for every incorrect answer - if you can narrow down to two choices, then guess otherwise leave blank - scoring scale from

3 What to Bring to the Exam
#2 pencil, eraser No calculators allowed (brush up on your basic math skills) Your brain. Please don’t leave it at home.:)

4 How To Prepare DO NOT CRAM. Get your studying done with by the night before. Get a good night’s sleep and have breakfast the morning of the exam. Actively participate in any and all review classes and activities offered by your teacher.

5 Matter 1) Properties of Phases 2) Types of Matter 3) Phase Changes

6 Properties of Phases Solids: Crystal lattice (regular geometric pattern), vibration motion only Liquids: particles flow past each other but are still attracted to each other. Gases: particles are small and far apart, they travel in a straight line until they hit something, they bounce off without losing any energy, they are so far apart from each other that they have effectively no attractive forces and their speed is directly proportional to the Kelvin temperature (Kinetic-Molecular Theory, Ideal Gas Theory)

7 Solids The positive and negative ions alternate in the
ionic crystal lattice of NaCl.

8 Liquids When heated, the ions move faster and eventually
separate from each other to form a liquid. The ions are loosely held together by the oppositely charged ions, but the ions are moving too fast for the crystal lattice to stay together.

9 Gases Since all gas molecules spread out
the same way, equal volumes of gas under equal conditions of temperature and pressure will contain equal numbers of molecules of gas L of any gas at STP (1.00 atm and 273K) will contain one mole (6.02 X 1023) gas molecules. Since there is space between gas molecules, gases are affected by changes in pressure.

10 Types of Matter Substances (Homogeneous) Mixtures
Elements (cannot be decomposed by chemical change): Al, Ne, O, Br, H Compounds (can be decomposed by chemical change): NaCl, Cu(ClO3)2, KBr, H2O, C2H6 Mixtures Homogeneous: Solutions (solvent + solute) Heterogeneous: soil, Italian dressing, etc.

11 Elements A sample of lead atoms (Pb). All atoms in the sample consist of lead, so the substance is homogeneous. A sample of chlorine atoms (Cl). All atoms in the sample consist of chlorine, so the substance is homogeneous.

12 Compounds Lead has two charges listed, +2 and +4. This is a sample of lead (II) chloride (PbCl2). Two or more elements bonded in a whole-number ratio is a COMPOUND. This compound is formed from the +4 version of lead. This is lead (IV) chloride (PbCl4). Notice how both samples of lead compounds have consistent composition throughout? Compounds are homogeneous!

13 Mixtures A mixture of lead atoms and chlorine atoms. They exist in no particular ratio and are not chemically combined with each other. They can be separated by physical means. A mixture of PbCl2 and PbCl4 formula units. Again, they are in no particular ratio to each other and can be separated without chemical change.

14 The Atom 1) Nucleons 2) Isotopes 3) Natural Radioactivity 4) Half-Life 5) Nuclear Power 6) Electron Configuation 7) Development of the Atomic Model

15 Nucleons Protons: +1 each, determines identity of element, mass of 1 amu, determined using atomic number, nuclear charge Neutrons: no charge, determines identity of isotope of an element, 1 amu, determined using mass number - atomic number (amu = atomic mass unit) 3216S and 3316S are both isotopes of S S-32 has 16 protons and 16 neutrons S-33 has 16 protons and 17 neutrons All atoms of S have a nuclear charge of +16 due to the 16 protons.

16 Isotopes Atoms of the same element MUST contain the same number of protons. Atoms of the same element can vary in their numbers of neutrons, therefore many different atomic masses can exist for any one element. These are called isotopes. The atomic mass on the Periodic Table is the weight-average atomic mass, taking into account the different isotope masses and their relative abundance. Rounding off the atomic mass on the Periodic Table will tell you what the most common isotope of that element is.

17 Weight-Average Atomic Mass
WAM = ((% A of A/100) X Mass of A) + ((% A of B/100) X Mass of B) + … What is the WAM of an element if its isotope masses and abundances are: X-200: Mass = amu, % abundance = 20.0 % X-204: Mass = amu, % abundance = 80.0% amu = atomic mass unit (1.66 × kilograms/amu)

18 Most Common Isotope The weight-average atomic mass of Zinc is amu. What is the most common isotope of Zinc? Zn-65! What are the most common isotopes of: C, H and O? FACT: one atomic mass unit (1.66 × kilograms) is defined as 1/12 of the mass of an atom of C-12.

19 Natural Radioactivity
Alpha Decay Beta Decay Positron Decay Gamma Decay Charges of Decay Particles Natural decay starts with a parent nuclide that ejects a decay particle to form a daughter nuclide which is more stable than the parent nuclide was.

20 Alpha Decay The nucleus ejects two protons and two neutrons. The atomic mass decreases by 4, the atomic number decreases by 2. 23892U 

21 Beta Decay A neutron decays into a proton and an electron. The electron is ejected from the nucleus as a beta particle. The atomic mass remains the same, but the atomic number increases by 1. 146C 

22 Positron Decay A proton is converted into a neutron and a positron. The positron is ejected by the nucleus. The mass remains the same, but the atomic number decreases by 1. 5326Fe 

23 Gamma Decay The nucleus has energy levels just like electrons, but the involve a lot more energy. When the nucleus becomes more stable, a gamma ray may be released. This is a photon of high-energy light, and has no mass or charge. The atomic mass and number do not change with gamma. Gamma may occur by itself, or in conjunction with any other decay type.

24 Charges of Decay Particles

25 Half-Life Half life is the time it takes for half of the nuclei in a radioactive sample to undergo decay. Problem Types: Going forwards in time Going backwards in time Radioactive Dating

26 Going Forwards in Time How many grams of a 10.0 gram sample of I-131 (half-life of 8 days) will remain in 24 days? #HL = t/T = 24/8 = 3 Cut 10.0g in half 3 times: 5.00, 2.50, 1.25g

27 Going Backwards in Time
How many grams of a 10.0 gram sample of I-131 (half-life of 8 days) would there have been 24 days ago? #HL = t/T = 24/8 = 3 Double 10.0g 3 times: 20.0, 40.0, 80.0 g

28 Radioactive Dating A sample of an ancient scroll contains 50% of the original steady-state concentration of C-14. How old is the scroll? 50% = 1 HL 1 HL X 5730 y/HL = 5730y

29 Nuclear Power Artificial Transmutation Particle Accelerators
Nuclear Fission Nuclear Fusion

30 Artificial Transmutation
4020Ca + _____ > 4019K + 11H 9642Mo + 21H > 10n + _____ Nuclide + Bullet --> New Element + Fragment(s) The masses and atomic numbers must add up to be the same on both sides of the arrow.

31 Nuclear Fission 23592U n  9236Kr Ba n + energy The three neutrons given off can be reabsorbed by other U-235 nuclei to continue fission as a chain reaction A tiny bit of mass is lost (mass defect) and converted into a huge amount of energy.

32 Chain Reaction

33 Nuclear Fusion 21H + 21H  42He + energy
Two small, positively-charged nuclei smash together at high temperatures and pressures to form one larger nucleus. A small bit of mass is destroyed and converted into a huge amount of energy, more than even fission.

34 Electron Configuration
Basic Configuration Valence Electrons Electron-Dot (Lewis Dot) Diagrams Excited vs. Ground State Rules for Electron Filling Para and Diamagnetic Lewis Structures and Hybridization

35 Basic Configuration The number of electrons is determined from the atomic number. Look up the basic configuration below the atomic number on the periodic table. (PEL: principal energy level = shell) He: 2 (2 e- in the 1st PEL) Na: (2 e- in the 1st PEL, 8 in the 2nd and 1 in the 3rd) Br: (2 e- in the 1st PEL, 8 in the 2nd, 18 in the 3rd and 7 in the 4th)

36 Valence Electrons The valence electrons are responsible for all chemical bonding. The valence electrons are the electrons in the outermost PEL (shell). He: 2 (2 valence electrons) Na: (1 valence electron) Br: (7 valence electrons) The maximum number of valence electrons an atom can have is EIGHT, called a STABLE OCTET.

37 Electron-Dot Diagrams
The number of dots equals the number of valence electrons. The number of unpaired valence electrons in a nonmetal tells you how many covalent bonds that atom can form with other nonmetals or how many electrons it wants to gain from metals to form an ion. The number of valence electrons in a metal tells you how many electrons the metal will lose to nonmetals to form an ion. Caution: May not work with transition metals. EXAMPLE DOT DIAGRAMS (c) 2006, Mark Rosengarten

38 Example Dot Diagrams Carbon can also have this dot diagram, which it
has when it forms organic compounds.

39 Excited vs. Ground State
Configurations on the Periodic Table are ground state configurations. If electrons are given energy, they rise to higher energy levels (excited state). If the total number of electrons matches in the configuration, but the configuration doesn’t match, the atom is in the excited state. Na (ground, on table): 2-8-1 Example of excited states: 2-7-2, , 2-6-3

40 Ways to Represent Electron Configuration
Expanded Electron Configuration Condensed Electron Configurations Orbital Notation Electron Dot Structure Write the above four electron configurations for Zinc, Zinc ion and Cu ion.

41 Electron Configuration of Ions
-Group configurations: s block ns1-2, p block ns2 np 1-6, d block ns0-2 (n-1) d 1-10, f block ns 0-2 (n-1) d 1 (n-2) f 1-14 - Remember that outmost electrons are lost first (which means that it will always be s or p electrons lost, never d or f). Ex. Sc+ or Sc3+ electron configuration would be:

42 Rules for Electron Filling
- Afbau’s Principle: Electrons tend to occupy the lowest energy orbitals first. - Hund’s Rule: Pairing of e in the degenerate orbitals does not take place till every orbital has one e. - Pauli’s Exclusion Principle: No two electrons can have all four same quantum numbers.

43 Diamagnetism, Paramagnetism
Diamagnetism: does not show magnetic properties in external magnetic field. No unpaired electrons. Paramagnetism: shows magnetic properties in external magnetic field. Has unpaired electrons. Best way to predict dia or paramagnetism is by drawing orbital diagrams.

44 Writing Lewis Structures
Lewis structures are used to depict bonding pairs and lone pairs of electron in the molecule. Step 1 Total number of valence electrons in the system: Sum the number of valence electrons on all the atoms . Add the total negative charge if you have an anion. Subtract the charge if you have a cation. Example: CO32-  Step 2 Number of electrons if each atom is to be happy: Atoms in our example will need 8 e (octet rule) or 2 e ( hydrogen). So, for the ex. Step 3 Calculate number of bonds in the system: Covalent bonds are made by sharing of e. You need 32 and you have 24. You are 8 e deficient. If you make 4 bonds ( with 2 e per bond) , you will make up the deficiency. Therefore, # of bonds= ( e in step 2- e in step 1)/2 =(32-24)/2= 4 bonds Step 4 Draw the structure: The central atom is C ( usually the atom with least electro negativity will be in the center). The oxygens surround it . Because there are four bonds and only three atoms, there will be one double bond. Step 5  Double check your answer by counting total number of electrons.  © 2009, Prentice-Hall, Inc.

45 Formal Charges: Writing Lewis Structures
Then assign formal charges. For each atom, count the electrons in lone pairs and half the electrons it shares with other atoms. Subtract that from the number of valence electrons for that atom: the difference is its formal charge. © 2009, Prentice-Hall, Inc.

46 Writing Lewis Structures
The best Lewis structure… …is the one with the fewest charges. …puts a negative charge on the most electronegative atom. © 2009, Prentice-Hall, Inc.

47 Resonance One Lewis structure cannot accurately depict a molecule like ozone. We use multiple structures, resonance structures, to describe the molecule. © 2009, Prentice-Hall, Inc.

48 Resonance Just as green is a synthesis of blue and yellow…
…ozone is a synthesis of these two resonance structures. © 2009, Prentice-Hall, Inc.

49 Molecular Shapes The shape of a molecule plays an important role in its reactivity. By noting the number of bonding and nonbonding electron pairs we can easily predict the shape of the molecule. © 2009, Prentice-Hall, Inc.

50 What Determines the Shape of a Molecule?
Simply put, electron pairs, whether they be bonding or nonbonding, repel each other. By assuming the electron pairs are placed as far as possible from each other, we can predict the shape of the molecule. © 2009, Prentice-Hall, Inc.

51 Electron Domains We can refer to the electron pairs as electron domains. In a double or triple bond, all electrons shared between those two atoms are on the same side of the central atom; therefore, they count as one electron domain. The central atom in this molecule, A, has four electron domains. © 2009, Prentice-Hall, Inc.

52 Valence Shell Electron Pair Repulsion Theory (VSEPR)
“The best arrangement of a given number of electron domains is the one that minimizes the repulsions among them.” © 2009, Prentice-Hall, Inc.

53 Electron-Domain Geometries
These are the electron-domain geometries for two through six electron domains around a central atom. © 2009, Prentice-Hall, Inc.



56 Hybridization Refers to mixing of orbitals.
Atomic orbitals of central atom undergo change to accommodate incoming atoms. Hybridization could be sp, sp2, sp3, sp3d and sp3d2. How do you tell the hybridization on the central atom?

57 9.1 – 9.2: V.S.E.P.R. Valence-shell electron-pair repulsion theory
Because e- pairs repel, molecular shape adjusts so the valence e- pairs are as far apart as possible around the central atom. Electron domains: areas of valence e- density around the central atom; result in different molecular shapes Includes bonding e- pairs and nonbonding e- pairs A single, double, or triple bond counts as one domain Summary of LmABn (Tables ): L = lone or non-bonding pairs A = central atom B = bonded atoms Bond angles notation used here: < xº means ~2-3º less than predicted << xº means ~4-6º less than predicted

58 Tables 9.1 - 9.3 # of e- domains & # and type of hybrid orbitals
e- domain geometry Formula Molecular geometry Predicted bond angle(s) Example (Lewis structure with molecular shape) 2 Two sp hybrid orbitals Linear AB2 180º BeF2 CO2 |X X |B B A A

59 Three sp2 hybrid orbitals
3 Three sp2 hybrid orbitals Trigonal planar AB3 120º BF3 Cl-C-Cl << 120º Cl2CO LAB2 Bent < 120º NO21- |B B A |X X A |B : B A

60 Example: CH4 H | H—C—H Molecular shape = tetrahedral
Bond angle = 109.5º 109.5º

61 Four sp3 hybrid orbitals
4 Four sp3 hybrid orbitals or Tetrahedral AB4 109.5º CH4 LAB3 Trigonal pyramidal < 109.5º Ex: NH3 = 107º NH3 L2AB2 Bent <<109.5º Ex: H2O = 104.5º H2O B A A X X : A B X X A : A B

62 PCl5 : :Cl: :Cl: \ / :Cl—P—Cl: | :Cl:
\ / :Cl—P—Cl: | :Cl: : Molecular shape = trigonal bipyramidal Bond angles equatorial = 120º axial = 90º 90º 120º

63 Five sp3d hybrid orbitals
5 Five sp3d hybrid orbitals Trigonal bipyramidal AB5 Equatorial = 120º Axial = 90º PCl5 LAB4 Seesaw Equatorial < 120º Axial < 90º SF4 B A | A X | X X : B - A - B B

64 Five sp3d hybrid orbitals
5 Five sp3d hybrid orbitals Trigonal bipyramidal L2AB3 T-shaped Axial << 90º ClF3 L3AB2 Linear Axial = 180º XeF2 B : A | X A | : A B |

65 Six sp3d2 hybrid orbitals
6 Six sp3d2 hybrid orbitals or Octahedral AB6 90º SF6 LAB5 Square pyramidal < 90º BrF5 A X | A B | X X B B A B | .. B B A X | X X

66 Six sp3d2 hybrid orbitals
6 Six sp3d2 hybrid orbitals or L2AB4 Square planar 90º XeF4 L3AB3 T-shaped <90º KrCl31- A .. | .. A B | .. B B B A .. | .. A .. | B B .. B .. B

67 What Is Light? Light is formed when electrons drop from the excited state to the ground state. The lines on a bright-line spectrum come from specific energy level drops and are unique to each element. Ex. Emission and Absorption Spectra ( line spectra)

68 EXAMPLE SPECTRUM This is the bright-line spectrum of hydrogen. The top
numbers represent the energy level transition change that produces the light with that color and the bottom number is the wavelength of the light (in nanometers, or 10-9 m). No other element has the same bright-line spectrum as hydrogen, so these spectra can be used to identify elements or mixtures of elements.

69 Development of the Atomic Model
Thompson Model Rutherford Gold Foil Experiment and Model Bohr Model Quantum-Mechanical Model

70 Thompson Model The atom is a positively charged diffuse mass with negatively charged electrons stuck in it.

71 Rutherford Model The atom is made of a small, dense, positively charged nucleus with electrons at a distance, the vast majority of the volume of the atom is empty space. Alpha particles shot at a thin sheet of gold foil: most go through (empty space). Some deflect or bounce off (small + charged nucleus).

72 Bohr Model Electrons orbit around the nucleus in energy levels (shells). Atomic bright-line spectra was the clue.

73 Quantum-Mechanical Model
Electron energy levels are wave functions. Electrons are found in orbitals, regions of space where an electron is most likely to be found. You can’t know both where the electron is and where it is going at the same time. Electrons buzz around the nucleus like gnats buzzing around your head.

74 Orbital Quantum Numbers
Symbol Name Description Meaning Equations n Principle Q.N. Energy level (i.e. Bohr’s theory) Shell number n = 1, 2, 3, 4, 5, 6, 7 n = 1, 2, 3, … l Angular Momentum Q.N. General probability plot (“shape” of the orbitals) Subshell number l = 0, 1, 2, 3 l = 0 means “s” l = 1 means “p” l = 2 means “d” l = 3 means “f” l = 0, 1, 2, …, n – 1 Ex: If n = 1, l can only be 0; if n = 2, l can be 0 or 1.

75 Symbol Name Description Meaning Equations ml ms
Magnetic Q.N. 3-D orientation of the orbital s has 1 p has 3 d has 5 f has 7 ml = -l, -l +1, …, 0, l, …, +l There are (2l + 1) values.   ms Spin Q.N. Spin of the electron Parallel or antiparallel to field ms = +½ or * s, p, d, and f come from the words sharp, principal, diffuse, and fundamental.

76 Permissible Quantum Numbers
(4, 1, 2, +½) (5, 2, 0, 0) (2, 2, 1, +½) Not permissible; if l = 1, ml = 1, 0, or –1 (p orbitals only have 3 subshells) Not permissible; ms = +½ or –½ Not permissible; if n = 2, l = 0 or 1 (there is no 2d orbital)

77 Phase Changes Phase Change Types Phase Change Diagrams
Heat of Phase Change Evaporation

78 Phase Change Types

79 Phase Change Diagrams AB: Solid Phase BC: Melting (S + L)
CD: Liquid Phase DE: Boiling (L + G) EF: Gas Phase Notice how temperature remains constant during a phase change? That’s because the PE is changing, not the KE.

80 Heat of Phase Change How many joules would it take to melt 100. g of H2O (s) at 0oC? q=mHf = (100. g)(334 J/g) = J How many joules would it take to boil 100. g of H2O (l) at 100oC? q=mHv = (100.g)(2260 J/g) = J

81 Evaporation When the surface molecules of a gas travel upwards at a great enough speed to escape. The pressure a vapor exerts when sealed in a container at equilibrium is called vapor pressure, and can be found on Table H. When the liquid is heated, its vapor pressure increases. When the liquid’s vapor pressure equals the pressure exerted on it by the outside atmosphere, the liquid can boil. If the pressure exerted on a liquid increases, the boiling point of the liquid increases (pressure cooker). If the pressure decreases, the boiling point of the liquid decreases (special cooking directions for high elevations).

82 Reference Table H: Vapor Pressure of Four Liquids
(c) 2006, Mark Rosengarten

83 Phase diagrams: CO2 Lines: 2 phases exist in equilibrium
Triple point: all 3 phases exist together in equilibrium (X on graph) Critical point, or critical temperature & pressure: highest T and P at which a liquid can exist (Z on graph) Temp (ºC) For most substances, inc P will cause a gas to condense (or deposit), a liquid to freeze, and a solid to become more dense (to a limit.)

84 Phase diagrams: H2O For H2O, inc P will cause ice to melt.

85 The Periodic Table Metals Nonmetals Metalloids Chemistry of Groups
Electronegativity Ionization Energy

86 Metals Have luster, are malleable and ductile, good conductors of heat and electricity Lose electrons to nonmetal atoms to form positively charged ions in ionic bonds Large atomic radii compared to nonmetal atoms Low electronegativity and ionization energy Left side of the periodic table (except H)

87 Nonmetals Are dull and brittle, poor conductors
Gain electrons from metal atoms to form negatively charged ions in ionic bonds Share unpaired valence electrons with other nonmetal atoms to form covalent bonds and molecules Small atomic radii compared to metal atoms High electronegativity and ionization energy Right side of the periodic table (except Group 18)

88 Metalloids Found lying on the jagged line between metals and nonmetals flatly touching the line (except Al and Po). Share properties of metals and nonmetals (Si is shiny like a metal, brittle like a nonmetal and is a semiconductor).

89 Chemistry of Groups Group 1: Alkali Metals
Group 2: Alkaline Earth Metals Groups 3-11: Transition Elements Group 17: Halogens Group 18: Noble Gases Diatomic Molecules

90 Group 1: Alkali Metals Most active metals, only found in compounds in nature React violently with water to form hydrogen gas and a strong base: 2 Na (s) + H2O (l)  2 NaOH (aq) + H2 (g) 1 valence electron Form +1 ion by losing that valence electron Form oxides like Na2O, Li2O, K2O

91 Group 2: Alkaline Earth Metals
Very active metals, only found in compounds in nature React strongly with water to form hydrogen gas and a base: Ca (s) + 2 H2O (l)  Ca(OH)2 (aq) + H2 (g) 2 valence electrons Form +2 ion by losing those valence electrons Form oxides like CaO, MgO, BaO

92 Groups 3-11: Transition Metals
Many can form different possible charges of ions If there is more than one ion listed, give the charge as a Roman numeral after the name Cu+1 = copper (I) Cu+2 = copper (II) Compounds containing these metals can be colored.

93 Group 17: Halogens Most reactive nonmetals
React violently with metal atoms to form halide compounds: 2 Na + Cl2  2 NaCl Only found in compounds in nature Have 7 valence electrons Gain 1 valence electron from a metal to form -1 ions Share 1 valence electron with another nonmetal atom to form one covalent bond.

94 Group 18: Noble Gases Are completely nonreactive since they have eight valence electrons, making a stable octet. Kr and Xe can be forced, in the laboratory, to give up some valence electrons to react with fluorine. Since noble gases do not naturally bond to any other elements, one atom of noble gas is considered to be a molecule of noble gas. This is called a monatomic molecule. Ne represents an atom of Ne and a molecule of Ne.

95 Diatomic Molecules Br, I, N, Cl, H, O and F are so reactive that they exist in a more chemically stable state when they covalently bond with another atom of their own element to make two-atom, or diatomic molecules. Br2, I2, N2, Cl2, H2, O2 and F2 The decomposition of water: 2 H2O  2 H2 + O2

96 Electronegativity An atom’s attraction to electrons in a chemical bond. F has the highest, at 4.0 Fr has the lowest, at 0.7 If two atoms that are different in EN (END) from each other by 1.7 or more collide and bond (like a metal atom and a nonmetal atom), the one with the higher electronegativity will pull the valence electrons away from the atom with the lower electronegativity to form a (-) ion. The atom that was stripped of its valence electrons forms a (+) ion. If the two atoms have an END of less than 1.7, they will share their unpaired valence electrons…covalent bond!

97 Ionization Energy The energy required to remove the most loosely held valence electron from an atom in the gas phase. High electronegativity means high ionization energy because if an atom is more attracted to electrons, it will take more energy to remove those electrons. Metals have low ionization energy. They lose electrons easily to form (+) charged ions. Nonmetals have high ionization energy but high electronegativity. They gain electrons easily to form (-) charged ions when reacted with metals, or share unpaired valence electrons with other nonmetal atoms.

98 Ions Ions are charged particles formed by the gain or loss of electrons. Metals lose electrons (oxidation) to form (+) charged cations. Nonmetals gain electrons (reduction) to form (-) charged anions. Atoms will gain or lose electrons in such a way that they end up with 8 valence electrons (stable octet). The exceptions to this are H, Li, Be and B, which are not large enough to support 8 valence electrons. They must be satisfied with 2 (Li, Be, B) or 0 (H).

99 Metal Ions (Cations) Na: 2-8-1 Na+1: 2-8 Ca: 2-8-8-2 Ca+2: 2-8-8
Note that when the atom loses its valence electron, the next lower PEL becomes the valence PEL. Notice how the dot diagrams for metal ions lack dots! Place brackets around the element symbol and put the charge on the upper right outside!

100 Nonmetal Ions (Anions)
Note how the ions all have 8 valence electrons. Also note the gained electrons as red dots. Nonmetal ion dot diagrams show 8 dots, with brackets around the dot diagram and the charge of the ion written to the upper right side outside the brackets. F: 2-7 F-1: 2-8 O: 2-6 O-2: 2-8 N: 2-5 N-3: 2-8

101 Chemical Bonding Intermolecular Bonding: Ionic, Covalent, Metallic and Covalent Network Bonds Intermolecular Bonding: H bond, dipole-dipole interaction, LDFs

102 Ionic Bonding If two atoms that are different in EN (END) from each other by 1.7 or more collide and bond (like a metal atom and a nonmetal atom), the one with the higher electronegativity will pull the valence electrons away from the atom with the lower electronegativity to form a (-) ion. The atom that was stripped of its valence electrons forms a (+) ion. The oppositely charged ions attract to form the bond. It is a surface bond that can be broken by melting or dissolving in water. Ionic bonding forms ionic crystal lattices, not molecules.

103 Example of Ionic Bonding

104 Covalent Bonding If two nonmetal atoms have an END of 1.7 or less, they will share their unpaired valence electrons to form a covalent bond. A particle made of covalently bonded nonmetal atoms is called a molecule. If the END is between 0 and 0.4, the sharing of electrons is equal, so there are no charged ends. This is NONPOLAR covalent bonding. If the END is between 0.5 and 1.7, the sharing of electrons is unequal. The atom with the higher EN will be d- and the one with the lower EN will be d+ charged. This is a POLAR covalent bonding. (d means “partial”)

105 Examples of Covalent Bonding

106 Sigma and Pi bonds Sigma (s) bond: Pi (p) bond:
Covalent bond that results from axial overlap of orbitals between atoms in a molecule Lie directly on internuclear axis “Single” bonds, could form between s-s orbital or s-p orbital or p-p orbital by axial overlapping Ex: F2 Pi (p) bond: Covalent bond that results from side-by-side overlap of orbitals between atoms in a molecule. Are “above & below” and “left & right” of the internuclear axis and therefore have less total orbital overlap, so they are weaker than s bonds. Forms between two p orbitals (py or pz) Make up the 2nd and 3rd bonds in double & triple bonds. Ex: O2 N2

107 Metallic Bonding Metal atoms of the same element bond with each other by sharing valence electrons that they lose to each other. This is a lot like an atomic game of “hot potato”, where metal kernals (the atom inside the valence electrons) sit in a crystal lattice, passing valence electrons back and forth between each other). Since electrons can be forced to travel in a certain direction within the metal, metals are very good at conducting electricity in all phases.

108 Types of Compounds Ionic: made of metal and nonmetal ions. Form an ionic crystal lattice when in the solid phase. Ions separate when melted or dissolved in water, allowing electrical conduction. Examples: NaCl, K2O, CaBr2 Molecular: made of nonmetal atoms bonded to form a distinct particle called a molecule. Bonds do not break upon melting or dissolving, so molecular substances do not conduct electricity. EXCEPTION: Acids [H+A- (aq)] ionize in water to form H3O+ and A-, so they do conduct. Network: made up of nonmetal atoms bonded in a seemingly endless matrix of covalent bonds with no distinguishable molecules. Very high m.p., don’t conduct.

109 Ionic Compounds (c) 2006, Mark Rosengarten

110 Molecular Compounds (c) 2006, Mark Rosengarten

111 Network Solids Network solids are made of nonmetal atoms covalently bonded together to form large crystal lattices. No individual molecules can be distinguished. Examples include C (diamond) and SiO2 (quartz). Corundum (Al2O3) also forms these, even though Al is considered a metal. Network solids are among the hardest materials known. They have extremely high melting points and do not conduct electricity.

112 Intermolecular Forces
The attractions between molecules are not nearly as strong as the intramolecular attractions that hold compounds together. © 2009, Prentice-Hall, Inc.

113 Intermolecular Forces
They are, however, strong enough to control physical properties such as boiling and melting points, vapor pressures, and viscosities. © 2009, Prentice-Hall, Inc.

114 Intermolecular Forces
These intermolecular forces as a group are referred to as van der Waals forces. © 2009, Prentice-Hall, Inc.

115 van der Waals Forces Dipole-dipole interactions Hydrogen bonding
London dispersion forces © 2009, Prentice-Hall, Inc.

116 Ion-Dipole Interactions
Ion-dipole interactions (a fourth type of force), are important in solutions of ions. The strength of these forces are what make it possible for ionic substances to dissolve in polar solvents. © 2009, Prentice-Hall, Inc.

117 Dipole-Dipole Interactions
Molecules that have permanent dipoles are attracted to each other. The positive end of one is attracted to the negative end of the other and vice-versa. These forces are only important when the molecules are close to each other. © 2009, Prentice-Hall, Inc.

118 Dipole-Dipole Interactions
The more polar the molecule, the higher is its boiling point. © 2009, Prentice-Hall, Inc.

119 London Dispersion Forces
While the electrons in the 1s orbital of helium would repel each other (and, therefore, tend to stay far away from each other), it does happen that they occasionally wind up on the same side of the atom. © 2009, Prentice-Hall, Inc.

120 London Dispersion Forces
At that instant, then, the helium atom is polar, with an excess of electrons on the left side and a shortage on the right side. © 2009, Prentice-Hall, Inc.

121 London Dispersion Forces
Another helium nearby, then, would have a dipole induced in it, as the electrons on the left side of helium atom 2 repel the electrons in the cloud on helium atom 1. © 2009, Prentice-Hall, Inc.

122 London Dispersion Forces
London dispersion forces, or dispersion forces, are attractions between an instantaneous dipole and an induced dipole. © 2009, Prentice-Hall, Inc.

123 London Dispersion Forces
These forces are present in all molecules, whether they are polar or nonpolar. The tendency of an electron cloud to distort in this way is called polarizability. © 2009, Prentice-Hall, Inc.

124 Factors Affecting London Forces
The shape of the molecule affects the strength of dispersion forces: long, skinny molecules (like n-pentane tend to have stronger dispersion forces than short, fat ones (like neopentane). This is due to the increased surface area in n-pentane. © 2009, Prentice-Hall, Inc.

125 Factors Affecting London Forces
The strength of dispersion forces tends to increase with increased molecular weight. Larger atoms have larger electron clouds which are easier to polarize. © 2009, Prentice-Hall, Inc.

126 Which Have a Greater Effect
Which Have a Greater Effect? Dipole-Dipole Interactions or Dispersion Forces If two molecules are of comparable size and shape, dipole-dipole interactions will likely the dominating force. If one molecule is much larger than another, dispersion forces will likely determine its physical properties. © 2009, Prentice-Hall, Inc.

127 How Do We Explain This? The nonpolar series (SnH4 to CH4) follow the expected trend. The polar series follows the trend from H2Te through H2S, but water is quite an anomaly. © 2009, Prentice-Hall, Inc.

128 Hydrogen Bonding The dipole-dipole interactions experienced when H is bonded to N, O, or F are unusually strong. We call these interactions hydrogen bonds. © 2009, Prentice-Hall, Inc.

129 Hydrogen Bonding Hydrogen bonding arises in part from the high electronegativity of nitrogen, oxygen, and fluorine. Also, when hydrogen is bonded to one of those very electronegative elements, the hydrogen nucleus is exposed. © 2009, Prentice-Hall, Inc.

130 Attractive Forces Molecules have partially charged ends. The d+ end of one molecule attracts to the d- end of another molecule. Ions are charged (+) or (-). Positively charged ions attract other to form ionic bonds, a type of attractive force. Since partially charged ends result in weaker attractions than fully charged ends, ionic compounds generally have much higher melting points than molecular compounds. Determining Polarity of Molecules Hydrogen Bond Attractions

131 Determining Polarity of Molecules
(c) 2006, Mark Rosengarten

132 Hydrogen Bond Attractions
A hydrogen bond attraction is a very strong attractive force between the H end of one polar molecule and the N, O or F end of another polar molecule. This attraction is so strong that water is a liquid at a temperature where most compounds that are much heavier than water (like propane, C3H8) are gases. This also gives water its surface tension and its ability to form a meniscus in a narrow glass tube.

133 Summarizing Intermolecular Forces
© 2009, Prentice-Hall, Inc.

134 Compounds 1) Types of Compounds 2) Formula Writing 3) Formula Naming 4) Empirical Formulas 5) Molecular Formulas 6) Types of Chemical Reactions 7) Balancing Chemical Reactions 8) Attractive Forces

135 Formula Writing The charge of the (+) ion and the charge of the (-) ion must cancel out to make the formula. Use subscripts to indicate how many atoms of each element there are in the compound, no subscript if there is only one atom of that element. Na+1 and Cl-1 = NaCl Ca+2 and Br-1 = CaBr2 Al+3 and O-2 = Al2O3 Zn+2 and PO4-3 = Zn3(PO4)2 Try these problems!

136 Formulas to Write Ba+2 and N-3 NH4+1 and SO4-2 Li+1 and S-2
Cu+2 and NO3-1 Al+3 and CO3-2 Fe+3 and Cl-1 Pb+4 and O-2 Pb+2 and O-2

137 Formula Naming Compounds are named from the elements or polyatomic ions that form them. KCl = potassium chloride Na2SO4 = sodium sulfate (NH4)2S = ammonium sulfide AgNO3 = silver nitrate Notice all the metals listed here only have one charge listed? So what do you do if a metal has more than one charge listed? Take a peek!

138 The Stock System CrCl2 = chromium (II) chloride Try
CrCl3 = chromium (III) chloride Co(NO3)2 and CrCl6 = chromium (VI) chloride Co(NO3)3 FeO = iron (II) oxide MnS = manganese (II) sulfide Fe2O3 = iron (III) oxide MnS2 = manganese (IV) sulfide The Roman numeral is the charge of the metal ion!

139 Math of Chemistry 1) Formula Mass 2) Percent Composition 3) Mole Problems 4) Gas Laws 5) Neutralization 6) Concentration 7) Significant Figures and Rounding 8) Metric Conversions 9) Calorimetry

140 Formula Mass Gram Formula Mass = sum of atomic masses of all elements in the compound Round given atomic masses to the nearest tenth H2O: (2 X 1.0) + (1 X 16.0) = 18.0 grams/mole Na2SO4: (2 X 23.0)+(1 X 32.1)+(4 X 16.0) = g/mole Now you try: BaBr2 CaSO4 Al2(CO3)3

141 Percent Composition The mass of part is the number of atoms of that element in the compound. The mass of whole is the formula mass of the compound. Don’t forget to take atomic mass to the nearest tenth! This is a problem for you to try.

142 Practice Percent Composition Problem
What is the percent by mass of each element in Li2SO4?

143 Mole Problems Grams <=> Moles Molecular Formula Stoichiometry

144 Grams <=> Moles How many grams will 3.00 moles of NaOH (40.0 g/mol) weigh? 3.00 moles X 40.0 g/mol = 120. g How many moles of NaOH (40.0 g/mol) are represented by 10.0 grams? (10.0 g) / (40.0 g/mol) = mol

145 Molecular Formula Molecular Formula = (Molecular Mass/Empirical Mass) X Empirical Formula What is the molecular formula of a compound with an empirical formula of CH2 and a molecular mass of 70.0 grams/mole? 1) Find the Empirical Formula Mass: CH2 = 14.0 2) Divide the MM/EM: 70.0/14.0 = 5 3) Multiply the molecular formula by the result: 5 (CH2) = C5H10

146 Stoichiometry Moles of Target = Moles of Given X (Coefficent of Target/Coefficient of given) Given the balanced equation N2 + 3 H2  2 NH3, How many moles of H2 need to be completely reacted with N2 to yield 20.0 moles of NH3? 20.0 moles NH3 X (3 H2 / 2 NH3) = 30.0 moles H2

147 Limiting Reactant controls the amount of product formed.
CO(g) + 2H2 (g)  Ch3OH If 500 mol of CO react with 750 mol of H2, which is the limiting reactant? Use either given amount to calculate required amount of other. Compare calculated amount to amount given b. How many moles of excess reactant remain unchanged? H2 125 mol CO

148 Percent yield= (actual yield/ theoretical yield)*100
Theoretical yield is the maximum amount of product that can be produced from a given amount of reactant Actual yield is the measured amount of a product obtained from a reaction Theoretical yield= g SnF2 Actual yield = g SnF2 Percent yield = g SnF2 117.5 g SnF2 *100

149 Determining empirical formula from combustion data
When a compound containing C,H and O undergoes combustion, it forms CO2 and H2O. Then from the mass of CO2 and H2O, we can calculate the mass of C and Hand then find the mass of O by subtracting the sum of masses of C and H from total g present of that substance. From the mass of C,H and O, we can calculate the moles of C,H and O.Then the smallest whole number ratios of these moles will give the empirical formula. Ex. A g sample of the unknown produced g of CO2 and g of H2O. Determine the empirical formula of the compound. Ans. C7H6O2

150 Empirical Formulas Ionic formulas: represent the simplest whole number mole ratio of elements in a compound. Ca3N2 means a 3:2 ratio of Ca ions to N ions in the compound. Many molecular formulas can be simplified to empirical formulas Ethane (C2H6) can be simplified to CH3. This is the empirical formula…the ratio of C to H in the molecule. All ionic compounds have empirical formulas.

151 Molecular Formulas The count of the actual number of atoms of each element in a molecule. H2O: a molecule made of two H atoms and one O atom covalently bonded together. C2H6O: A molecule made of two C atoms, six H atoms and one O atom covalently bonded together. Molecular formulas are whole-number multiples of empirical formulas: H2O = 1 X (H2O) C8H16 = 8 X (CH2) Calculating Molecular Formulas

152 Calculating Empirical Formulas
One can calculate the empirical formula from the percent composition. © 2009, Prentice-Hall, Inc.

153 Calculating Empirical Formulas
The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA. © 2009, Prentice-Hall, Inc.

154 Calculating Empirical Formulas
Assuming g of para-aminobenzoic acid, C: g x = mol C H: g x = 5.09 mol H N: g x = mol N O: g x = mol O 1 mol 12.01 g 14.01 g 1.01 g 16.00 g © 2009, Prentice-Hall, Inc.

155 Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number of moles: C: =  7 H: =  7 N: = 1.000 O: =  2 5.105 mol mol 5.09 mol 1.458 mol © 2009, Prentice-Hall, Inc.

156 Calculating Empirical Formulas
These are the subscripts for the empirical formula: C7H7NO2 © 2009, Prentice-Hall, Inc.

157 Combustion Analysis Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this. C is determined from the mass of CO2 produced. H is determined from the mass of H2O produced. O is determined by difference after the C and H have been determined. © 2009, Prentice-Hall, Inc.

158 Stoichiometric Calculations
The coefficients in the balanced equation give the ratio of moles of reactants and products. © 2009, Prentice-Hall, Inc.

159 Stoichiometric Calculations
Starting with the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant). © 2009, Prentice-Hall, Inc.

160 Stoichiometric Calculations
C6H12O6 + 6 O2  6 CO2 + 6 H2O Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6… use the coefficients to find the moles of H2O… and then turn the moles of water to grams. © 2009, Prentice-Hall, Inc.

161 Molecular Formula Actual ratio of atoms in a compound. Ex. H2O, C6H12O6 To determine the molecular formula, divide the molar mass by empirical formula mass. This will give the number of empirical formula units (n) in actual molecule. n= Molar Mass/ Empirical Formula Mass Ex. Determine the empirical and molecular formula of each of the following: Ethylene glycol, the substance used as antifreeze has % C, 9.70 % H and % O , mm= g Caffeine, a stimulant in coffee has the following percent composition: 49.50 % C, 5.15% H, % N and % O , molar mass= g

162 Types of Chemical Reactions
Redox Reactions: driven by the loss (oxidation) and gain (reduction) of electrons. Any species that does not change charge is called the spectator ion. Synthesis Decomposition Single Replacement Ion Exchange Reaction: driven by the formation of an insoluble precipitate. The ions that remain dissolved throughout are the spectator ions. Double Replacement

163 Synthesis Two elements combine to form a compound 2 Na + O2  Na2O
Same reaction, with charges added in: 2 Na0 + O20  Na2+1O-2 Na0 is oxidized (loses electrons), is the reducing agent O20 is reduced (gains electrons), is the oxidizing agent Electrons are transferred from the Na0 to the O20. No spectator ions, there are only two elements here.

164 Decomposition A compound breaks down into its original elements.
Na2O  2 Na + O2 Same reaction, with charges added in: Na2+1O-2  2 Na0 + O20 O-2 is oxidized (loses electrons), is the reducing agent Na+1 is reduced (gains electrons), is the oxidizing agent Electrons are transferred from the O-2 to the Na+1. No spectator ions, there are only two elements here.

165 Single Replacement An element replaces the same type of element in a compound. Ca + 2 KCl  CaCl2 + 2 K Same reaction, with charges added in: Ca0 + 2 K+1Cl-1  Ca+2Cl K0 Ca0 is oxidized (loses electrons), is the reducing agent K+1 is reduced (gains electrons), is the oxidizing agent Electrons are transferred from the Ca0 to the K+1. Cl-1 is the spectator ion, since it’s charge doesn’t change.

166 Double Replacement The (+) ion of one compound bonds to the (-) ion of another compound to make an insoluble precipitate. The compounds must both be dissolved in water to break the ionic bonds first. NaCl (aq) + AgNO3 (aq)  NaNO3 (aq) + AgCl (s) The Cl-1 and Ag+1 come together to make the insoluble precipitate, which looks like snow in the test tube. No species change charge, so this is not a redox reaction. Since the Na+1 and NO3-1 ions remain dissolved throughout the reaction, they are the spectator ions. How do identify the precipitate?

167 Identifying the Precipitate
The precipitate is the compound that is insoluble. AgCl is a precipitate because Cl- is a halide. Halides are soluble, except when combined with Ag+ and others.

168 Balancing Chemical Reactions
Balance one element or ion at a time Use a pencil Use coefficients only, never change formulas Revise if necessary The coefficient multiplies everything in the formula by that amount 2 Ca(NO3)2 means that you have 2 Ca, 4 N and 12 O. Examples for you to try!

169 Reactions to Balance ___NaCl  ___Na + ___Cl2 ___Al + ___O2  ___Al2O3
___SO3  ___SO2 + ___O2 ___Ca + ___HNO3  ___Ca(NO3)2 + ___H2 __FeCl3 + __Pb(NO3)2  __Fe(NO3)3 + __PbCl2

170 Writing Net Ionic Equations
Cancel all the spectator ions. Dissociate all dissociable ionic compounds (refer to solubility rules) All gases and liquids NEVER dissociate. Write the net ionic equation.

171 Gases

172 Manometers: measure P of a gas
Closed-end: difference in Hg levels (Dh) shows P of gas in container compared to a vacuum closed

173 2. Open-end: Difference in Hg levels (Dh) shows P of gas in container compared to Patm

174 Gas Laws Make a data table to put the numbers so you can eliminate the words. Make sure that any Celsius temperatures are converted to Kelvin (add 273). Rearrange the equation before substituting in numbers. If you are trying to solve for T2, get it out of the denominator first by cross-multiplying. If one of the variables is constant, then eliminate it. Try these problems!

175 Gas Law Problem 1 A 2.00 L sample of N2 gas at STP is compressed to 4.00 atm at constant temp-erature. What is the new volume of the gas? V2 = P1V1 / P2 = (1.00 atm)(2.00 L) / (4.00 atm) = L

176 Gas Law Problem 2 To what temperature must a L sample of O2 gas at K be heated to raise the volume to L? T2 = V2T1/V1 = (10.00 L)(300.0 K) / (3.000 L) = K

177 Gas Law Problem 3 A 3.00 L sample of NH3 gas at kPa is cooled from K to K and its pressure is reduced to 80.0 kPa. What is the new volume of the gas? V2 = P1V1T2 / P2T1 = (100.0 kPa)(3.00 L)(300. K) / (80.0 kPa)(500. K) = 2.25 L

178 Gay Lussac’s Law of Combining Volumes
When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers. N2 1 volume + 3 H2 3 volumes 2 NH3 2 volumes 1. Students write balanced equation for reaction. Note balancing coefficients are same as volume in Liters.

179 Gay Lussac’s Law of Combining Volumes
When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.

180 Avogadro’s Law Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.

181 22.4 L at STP is known as the molar volume of any gas.
Mole-Mass-Volume Relationships Volume of one mole of any gas at STP = 22.4 L. 22.4 L at STP is known as the molar volume of any gas.

182 atmospheres nT V a P Number of molecules or moles affects other three quatities. Increase #, increase collision rate, increase pressure.

183 Determination of Density Using the Ideal Gas Equation
Density = mass/volume Density varies directly with molar mass and pressure and inversely with Kelvin temp D = MP/ RT

184 Mole fraction (X): Ratio of moles of one component to the total moles in the mixture (dimensionless, similar to a %) Ex: What are the mole fractions of H2 and He in the previous example?

185 Collecting Gases “over Water”
When a gas is bubbled through water, the vapor pressure of the water (partial pressure of the water) must be subtracted from the pressure of the collected gas: PT = Pgas + PH2O ∴ Pgas = PT – PH2O See Appendix B for vapor pressures of water at different temperatures.

186 Graham’s Law of Effusion
The rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses. Rate of effusion of A = MB Rate of effusion of B MA M = molar masses, density of a gas varies directly with molar mass, can replace molar mass with density

187 Neutralization 10.0 mL of 0.20 M HCl is neutralized by 40.0 mL of NaOH. What is the concentration of the NaOH? #H MaVa = #OH MbVb, so Mb = #H MaVa / #OH Vb = (1)(0.20 M)(10.0 mL) / (1) (40.0 mL) = M How many mL of 2.00 M H2SO4 are needed to completely neutralize 30.0 mL of M KOH?

188 Concentration Molarity Parts per Million Percent by Mass
Percent by Volume

189 Molarity What is the molarity of a mL solution of NaOH (FM = 40.0) with 60.0 g of NaOH (aq)? Convert g to moles and mL to L first! M = moles / L = 1.50 moles / L = 3.00 M How many grams of NaOH does it take to make 2.0 L of a M solution of NaOH (aq)? Moles = M X L = M X 2.0 L = moles Convert moles to grams: moles X 40.0 g/mol = 8.00 g

190 Parts Per Million 100.0 grams of water is evaporated and analyzed for lead grams of lead ions are found. What is the concentration of the lead, in parts per million? ppm = ( g) / (100.0 g) X = 1.0 ppm If the legal limit for lead in the water is 3.0 ppm, then the water sample is within the legal limits (it’s OK!)

191 Percent by Mass A 50.0 gram sample of a solution is evaporated and found to contain grams of sodium chloride. What is the percent by mass of sodium chloride in the solution? % Comp = (0.100 g) / (50.0 g) X 100 = 0.200%

192 Percent By Volume Substitute “volume” for “mass” in the above equation. What is the percent by volume of hexane if 20.0 mL of hexane are dissolved in benzene to a total volume of 80.0 mL? % Comp = (20.0 mL) / (80.0 mL) X100 = 25.0%

193 Colligative Properties
Vapor Pressure Lowering B.P. Elevation DTf= m. kf (or D Tb= m. kb) F.P. Depression Osmotic Pressure Colligative properties depend upon # of particles (ions, atoms, molecule= particle) Which will have lowest B.P. 1M NaCl, 1 M C6H12O6 or 1M Na3PO4?

194 How many Sig Figs? Start counting sig figs at the first non-zero.
All digits except place-holding zeroes are sig figs. Measurement # of Sig Figs 0.115 cm 3 cm 2 cm cm cm 5 Measurement # of Sig Figs 234 cm 3 67000 cm 2 _ 45000 cm 4 560. cm cm 5

195 What Precision? A number’s precision is determined by the furthest (smallest) place the number is recorded to. 6000 mL : thousands place 6000. mL : ones place mL : tenths place 5.30 mL : hundredths place 8.7 mL : tenths place mL : thousandths place

196 Rounding with addition and subtraction
Answers are rounded to the least precise place.

197 Rounding with multiplication and division
Answers are rounded to the fewest number of significant figures.

198 Metric Conversions Determine how many powers of ten difference there are between the two units (no prefix = 100) and create a conversion factor. Multiply or divide the given by the conversion factor. How many kg are in 38.2 cg? (38.2 cg) /( cg/kg) = km How many mL in dL? (0.988 dg) X (100 mL/dL) = 98.8 mL

199 Calorimetry This equation can be used to determine any of the variables here. You will not have to solve for C, since we will always assume that the energy transfer is being absorbed by or released by a measured quantity of water, whose specific heat is given above. Solving for q Solving for m Solving for DT

200 Solving for q How many joules are absorbed by grams of water in a calorimeter if the temperature of the water increases from 20.0oC to 50.0oC? q = mCDT = (100.0 g)(4.18 J/goC)(30.0oC) = J

201 Solving for m A sample of water in a calorimeter cup increases from 25oC to 50.oC by the addition of joules of energy. What is the mass of water in the calorimeter cup? q = mCDT, so m = q / CDT = (500.0 J) / (4.18 J/goC)(25oC) = 4.8 g

202 Solving for DT If a 50.0 gram sample of water in a calorimeter cup absorbs joules of energy, how much will the temperature rise by? q = mCDT, so DT = q / mC = ( J)/(50.0 g)(4.18 J/goC) = 4.8oC If the water started at 20.0oC, what will the final temperature be? Since the water ABSORBS the energy, its temperature will INCREASE by the DT: 20.0oC + 4.8oC = 24.8oC

203 Reaction Rate Reactions happen when reacting particles collide with sufficient energy (activation energy) and at the proper angle. Anything that makes more collisions in a given time will make the reaction rate increase. Increasing temperature Increasing concentration (pressure for gases) Increasing surface area (solids) Adding a catalyst makes a reaction go faster by removing steps from the mechanism and lowering the activation energy without getting used up in the process.

204 Heat of Reaction Reactions either absorb PE (endothermic, +DH) or release PE (exothermic, -DH) Exothermic, PEKE, Temp Endothermic, KEPE, Temp Rewriting the equation with heat included: 4 Al(s) + 3 O2(g)  2 Al2O3(s) kJ N2(g) + O2(g) kJ  2 NO(g)

205 5.3: Enthalpy, H Since most reactions occur in containers open to the air, w is often negligible. If a reaction produces a gas, the gas must do work to expand against the atmosphere. This mechanical work of expansion is called PV (pressure-volume) work. Enthalpy (H): change in the heat content (qp) of a reaction at constant pressure H = E + PV H = E + PV (at constant P) H = (qp + w) + (-w) H = qp

206 Sign conventions H > 0 Heat is gained from surroundings + H in endothermic reaction H < 0 Heat is released to surroundings - H in exothermic reaction

207 5.4: Enthalpy of Reaction (Hrxn)
Also called heat of reaction: Enthalpy is an extensive property (depends on amounts of reactants involved). Ex: CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l) Hrxn = kJ Combustion of 1 mol CH4 produces 890. kJ … of 2 mol CH4 → (2)(-890. kJ) = kJ What is the H of the combustion of 100. g CH4?

208 Hreaction = - Hreverse reaction
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l) H = kJ CO2 (g) + 2 H2O (l)  CH4 (g) + 2 O2 (g) H = kJ

209 5.6: Hess’ Law Ex. What is DHrxn of the combustion of propane?
If a rxn is carried out in a series of steps, Hrxn =  (Hsteps) = H1 + H2 + H3 + … Germain Hess ( ) Ex. What is DHrxn of the combustion of propane? C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) 3 C (s) + 4 H2 (g)  C3 H8 (g) H1 = kJ C (s) + O2 (g)  CO2 (g) H2 = kJ H2 (g) + ½ O2 (g)  H2O (l) H3 = kJ C3H8 (g)  3 C (s) + 4 H2 (g) H1 = kJ Born in Switzerland, moved to St Petersburg at age 3 with his family. His father was an artist. Hess was also interested in geology. 3[ ] ( ) 4[ ] ( ) Hrxn = ( ) + 4( ) = kJ

210 5.7: Enthalpy of Formation (Hf)
Formation: a reaction that describes a substance formed from its elements NH4NO3 (s) Standard enthalpy of formation (Hf): forms 1 mole of compound from its elements in their standard state (at 298 K) C2H5OH (l) Hf = kJ Hf of the most stable form of any element equals zero. H2, N2 , O2 , F2 , Cl2 (g) Br2 (l), Hg (l) C (graphite), P4 (s, white), S8 (s), I2 (s) Ex: 2 N2 (g) + 4 H2 (g) + 3 O2 (g)  2 2 C (graphite) + 3 H2 (g) + ½ O2 (g) 

211 Hess’ Law (again) Ex. Combustion of propane:
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) Given: Compound Hrxn (kJ/mol) C3H8 (g) CO2 (g) H2O (l) H2O (g) Hrxn = [3( ) + 4( )] – [1( ) + 5(0)] = kJ

212 5.5: Calorimetry q = C DT q = m c DT Measurement of heat flow
Heat capacity, C: amount of heat required to raise T of an object by 1 K q = C DT Specific heat (or specific heat capacity, c): heat capacity of 1 g of a substance q = m c DT Ex: How much energy is required to heat 40.0 g of iron (c = 0.45 J/(g K) from 0.0ºC to 100.0ºC? q = m c DT = (40.0 g)(0.45 J/(g K))(100.0 – 0.0 ºC) = 1800 J

213 Potential Energy Diagrams
Steps of a reactions: Reactants have a certain amount of PE stored in their bonds (Heat of Reactants) The reactants are given enough energy to collide and react (Activation Energy) The resulting intermediate has the highest energy that the reaction can make (Heat of Activated Complex) The activated complex breaks down and forms the products, which have a certain amount of PE stored in their bonds (Heat of Products) Hproducts - Hreactants = DH EXAMPLES

214 Making a PE Diagram X axis: Reaction Coordinate (time, no units)
Y axis: PE (kJ) Three lines representing energy (Hreactants, Hactivated complex, Hproducts) Two arrows representing energy changes: From Hreactants to Hactivated complex: Activation Energy From Hreactants to Hproducts : DH ENDOTHERMIC PE DIAGRAM EXOTHERMIC PE DIAGRAM

215 Endothermic PE Diagram
If a catalyst is added?

216 Endothermic with Catalyst
The red line represents the catalyzed reaction.

217 Exothermic PE Diagram What does it look like with a catalyst?

218 Exothermic with a Catalyst
The red line represents the catalyzed reaction. Lower A.E. and faster reaction time!

219 19.1: Spontaneous Processes
Reversible reaction: can proceed forward and backward along same path (equilibrium is possible) Ex: H2O freezing & melting at 0ºC Irreversible reaction: cannot proceed forward and backward along same path Ex: ice melting at room temperature Spontaneous reaction: an irreversible reaction that occurs without outside intervention Ex: Gases expand to fill a container, ice melts at room temperature (even though endothermic), salts dissolve in water

220 Entropy Entropy (S): a measure of molecular randomness or disorder
S is a state function: DS = Sfinal - Sinitial + DS = more randomness - DS = less randomness For a reversible process that occurs at constant T: Units: J/K

221 Examples of spontaneous reactions:
Particles are more evenly distributed Particles are no longer in an ordered crystal lattice Ions are not locked in crystal lattice Gases expand to fill a container: Ice melts at room temperature: Salts dissolve in water:

222 19.3: 3rd Law of Thermodynamics
The entropy of a crystalline solid at 0 K is 0. How to predict DS: Sgas > Sliquid > Ssolid Smore gas molecules > Sfewer gas molecules Shigh T > Slow T Ex: Predict the sign of DS for the following: CaCO3 (s) → CaO (s) + CO2 (g) N2 (g) + 3 H2 (g) → 2 NH3 (g) N2 (g) + O2 (g) → 2 NO (g) +, solid to gas -, fewer moles produced ?

223 DG = DH - TDS DG° = DH° - TDS°
19.5: Gibbs free energy, G Represents combination of two forces that drive a reaction: DH (enthalpy) and DS (disorder) Units: kJ/mol DG = DH - TDS DG° = DH° - TDS° (absolute T) Josiah Willard Gibbs ( ) Called “free energy” because DG represents maximum useful work that can be done by the system on its surroundings in a spontaneous reaction. (See p. 708 for more details.)

224 Determining Spontaneity of a Reaction
If DG is: reaction is spontaneous (proceeds in the forward direction Positive Forward reaction is non-spontaneous; the reverse reaction is spontaneous Zero The system is at equilibrium

225 19.6: Free Energy & Temperature
DG depends on enthalpy, entropy, and temperature: DG = DH - TDS DH DS DG and reaction outcome - + Always (-); spontaneous at all T 2 O3 (g) → 3 O2 (g) + - Always +; non-spontaneous at all T 3 O2 (g) → 2 O3 (g) - - Spontaneous at low T; non-spontaneous at high T H2O (l) → H2O (s) + + Spontaneous only at high T ; non-spontaneous at low T H2O (s) → H2O (l)

226 Solubility Curves Solubility: the maximum quantity of solute that can be dissolved in a given quantity of solvent at a given temperature to make a saturated solution. Saturated: a solution containing the maximum quantity of solute that the solvent can hold. The limit of solubility. Supersaturated: the solution is holding more than it can theoretically hold OR there is excess solute which precipitates out. True supersaturation is rare. Unsaturated: There are still solvent molecules available to dissolve more solute, so more can dissolve. How ionic solutes dissolve in water: polar water molecules attach to the ions and tear them off the crystal.

227 Solubility Solubility: go to the temperature and up to the desired line, then across to the Y-axis. This is how many g of solute are needed to make a saturated solution of that solute in 100g of H2O at that particular temperature. At 40oC, the solubility of KNO3 in 100g of water is 64 g. In 200g of water, double that amount. In 50g of water, cut it in half.

228 Supersaturated If 120 g of NaNO3 are added to 100g of water at 30oC:
1) The solution would be SUPERSATURATED, because there is more solute dissolved than the solubility allows 2) The extra 25g would precipitate out 3) If you heated the solution up by 24oC (to 54oC), the excess solute would dissolve.

229 Unsaturated If 80 g of KNO3 are added to 100g of water at 60oC:
1) The solution would be UNSATURATED, because there is less solute dissolved than the solubility allows 2) 26g more can be added to make a saturated solution 3) If you cooled the solution down by 12oC (to 48oC), the solution would become saturated

230 How Ionic Solutes Dissolve in Water
Water solvent molecules attach to the ions (H end to the Cl-, O end to the Na+) Water solvent holds the ions apart and keeps the ions from coming back together

231 Formulas, Naming and Properties of Acids
Arrhenius Definition of Acids: molecules that dissolve in water to produce H3O+ (hydronium) as the only positively charged ion in solution. HCl (g) + H2O (l)  H3O+ (aq) + Cl- Properties of Acids Naming of Acids Formula Writing of Acids

232 Properties of Acids Acids react with metals above H2 on Table J to form H2(g) and a salt. Acids have a pH of less than 7. Dilute solutions of acids taste sour. Acids turn phenolphthalein CLEAR, litmus RED and bromthymol blue YELLOW. Acids neutralize bases. Acids are formed when acid anhydrides (NO2, SO2, CO2) react with water for form acids. This is how acid rain forms from auto and industrial emissions.

233 Naming of Acids (polyatomic ion) -ate +ic acid
Binary Acids (H+ and a nonmetal) hydro (nonmetal) -ide + ic acid HCl (aq) = hydrochloric acid Ternary Acids (H+ and a polyatomic ion) (polyatomic ion) -ate +ic acid HNO3 (aq) = nitric acid (polyatomic ion) -ide +ic acid HCN (aq) = cyanic acid (polyatomic ion) -ite +ous acid HNO2 (aq) = nitrous acid

234 Formula Writing of Acids
Acids formulas get written like any other. Write the H+1 first, then figure out what the negative ion is based on the name. Cancel out the charges to write the formula. Don’t forget the (aq) after it…it’s only an acid if it’s in water! Hydrosulfuric acid: H+1 and S-2 = H2S (aq) Carbonic acid: H+1 and CO3-2 = H2CO3 (aq) Chlorous acid: H+1 and ClO2-1 = HClO2 (aq) Hydrobromic acid: H+1 and Br-1 = HBr (aq) Hydronitric acid: Hypochlorous acid: Perchloric acid:

235 Formulas, Naming and Properties of Bases
Arrhenius Definition of Bases: ionic compounds that dissolve in water to produce OH- (hydroxide) as the only negatively charged ion in solution. NaOH (s)  Na+1 (aq) + OH-1 (aq) Properties of Bases Naming of Bases Formula Writing of Bases

236 Properties of Bases Bases react with fats to form soap and glycerol. This process is called saponification. Bases have a pH of more than 7. Dilute solutions of bases taste bitter. Bases turn phenolphthalein PINK, litmus BLUE and bromthymol blue BLUE. Bases neutralize acids. Bases are formed when alkali metals or alkaline earth metals react with water. The words “alkali” and “alkaline” mean “basic”, as opposed to “acidic”.

237 Naming of Bases Bases are named like any ionic compound, the name of the metal ion first (with a Roman numeral if necessary) followed by “hydroxide”. Fe(OH)2 (aq) = iron (II) hydroxide Fe(OH)3 (aq) = iron (III) hydroxide Al(OH)3 (aq) = aluminum hydroxide NH3 (aq) is the same thing as NH4OH: NH3 + H2O  NH4OH Also called ammonium hydroxide.

238 Formula Writing of Bases
Formula writing of bases is the same as for any ionic formula writing. The charges of the ions have to cancel out. Calcium hydroxide = Ca+2 and OH-1 = Ca(OH)2 (aq) Potassium hydroxide = K+1 and OH-1 = KOH (aq) Lead (II) hydroxide = Pb+2 and OH-1 = Pb(OH)2 (aq) Lead (IV) hydroxide = Pb+4 and OH-1 = Pb(OH)4 (aq) Lithium hydroxide = Copper (II) hydroxide = Magnesium hydroxide =

239 Neutralization H+1 + OH-1  HOH
Acid + Base  Water + Salt (double replacement) HCl (aq) + NaOH (aq)  HOH (l) + NaCl (aq) H2SO4 (aq) + KOH (aq)  2 HOH (l) + K2SO4 (aq) HBr (aq) + LiOH (aq)  H2CrO4 (aq) + NaOH (aq)  HNO3 (aq) + Ca(OH)2 (aq)  H3PO4 (aq) + Mg(OH)2 (aq) 

240 pH A change of 1 in pH is a tenfold increase in acid or base strength.
A pH of 4 is 10 times more acidic than a pH of 5. A pH of 12 is 100 times more basic than a pH of 10.

241 16.2: Dissociation of Water
Autoionization of water: H2O (l) ↔ H+ (aq) + OH- (aq) KW = ion-product constant for water H3O+ (aq) or H+ (aq) = hydronium

242 Indicators At a pH of 2: Methyl Orange = red Bromthymol Blue = yellow
Phenolphthalein = colorless Litmus = red Bromcresol Green = yellow Thymol Blue = yellow Methyl orange is red at a pH of 3.2 and below and yellow at a pH of 4.4 and higher. In between the two numbers, it is an intermediate color that is not listed on this table.

243 Alternate Theories Arrhenius Theory: acids and bases must be in aqueous solution. Alternate Theory: Not necessarily so! Acid: proton (H+1) donor…gives up H+1 in a reaction. Base: proton (H+1) acceptor…gains H+1 in a reaction. HNO3 + H2O  H3O+1 + NO3-1 Since HNO3 lost an H+1 during the reaction, it is an acid. Since H2O gained the H+1 that HNO3 lost, it is a base.

244 16.11: Lewis Acids & Bases Lewis acid: “e- pair acceptor”
Brønsted-Lowry acid = H+ donor Arrhenius acid = produces H+ Lewis base: “e- pair donor” B-L base = H+ acceptor Arrhenius base = produces OH- Ex: NH3 + BF3 → NH3BF3 Lewis base Lewis acid Lewis salt 6 CN- + Fe3+ → Fe(CN)63- Lewis base Lewis acid Coordination compound Gilbert N. Lewis (1875 – 1946) Picture of Gilbert N Lewis (USA) who was the first to isolate D2O.

245 15.1: Chemical Equilibrium
Occurs when opposing reactions are proceeding at the same rate Forward rate = reverse rate of reaction Ex: Vapor pressure: rate of vaporization = rate of condensation Saturated solution: rate of dissociation = rate of crystallization Expressing concentrations: Gases: partial pressures, PX Solutes in liquids: molarity, [X]

246 Reversible Reactions and Rate
Forward rate Reaction Rate Time Equilibrium is established: Forward rate = Backward rate Backward rate When equilibrium is achieved: [A] ≠ [B] and kf/kr = Keq

247 15.2: Law of Mass Action Derived from rate laws by Guldberg and Waage (1864) For a balanced chemical reaction in equilibrium: a A + b B ↔ c C + d D Equilibrium constant expression (Keq): Cato Guldberg Peter Waage ( ) ( ) or But Waage and Guldberg were also related through two marriages; Guldberg married his cousin Bodil Mathea Riddervold, daughter of cabinet minister Hans Riddervold, and the couple had three daughters. Waage married Bodil's sister, Johanne Christiane Tandberg Riddervold by whom he had five children, and after her death in 1869, he became Guldberg's brother-in-law a second time, in 1870, by marrying one of Guldberg's sisters, Mathilde Sofie Guldberg, by whom he had six children. Keq is strictly based on stoichiometry of the reaction (is independent of the mechanism). Units: Keq is considered dimensionless (no units)

248 Relating Kc and Kp Convert [A] into PA:
where Dn = = change in coefficents of products – reactants (gases only!) = (c+d) - (a+b)

249 Magnitude of Keq Since Keq a [products]/[reactants], the magnitude of Keq predicts which reaction direction is favored: If Keq > 1 then [products] > [reactants] and equilibrium “lies to the right” If Keq < 1 then [products] < [reactants] and equilibrium “lies to the left”

250 Relationship Between Q and K
Reaction Quotient (Q): The particular ratio of concentration terms that we write for a particular reaction is called reaction quotient. For a reaction, A B, Q= [B]/[A] At equilibrium, Q= K Reaction Direction: Comparing Q and K Q<K, reaction proceeds to right, until equilibrium is achieved (or Q=K) Q>K, reaction proceeds to left, until Q=K

251 Value of K For the reference rxn, A>B,
For the reverse rxn, B >A, For the reaction, 2A > 2B For the rxn, A > C C > B K(ref)= [B]/[A] K= 1/K(ref) K= K(ref)2 K (overall)= K1 X K2

252 15.3: Types of Equilibria Homogeneous: all components in same phase (usually g or aq) N2 (g) + H2 (g) ↔ NH3 (g) 1 3 2 Fritz Haber (1868 – 1934) German chemist, who received the Nobel Prize in Chemistry in 1918 for his development of synthetic ammonia, important for fertilizers and explosives. He is also credited as the "father of chemical warfare" for his work developing and deploying chlorine and other poison gases during World War I; this role is thought to have provoked his wife to commit suicide. Despite his contributions to the German war effort, Haber was forced to emigrate from Germany in 1933 by the Nazis because of his Jewish background; many of his relatives were killed by the Nazis in concentration camps, gassed by Zyklon B. Though he had converted from Judaism in an effort to become fully accepted, he was forced to emigrate from Germany by the Nazis in 1933 on account of his being Jewish in their eyes. He died in the process of emigration. The Haber process now produces 500 million tons of nitrogen fertilizer per year, mostly in the form of anhydrous ammonia, ammonium nitrate, and urea. 1% of the world's annual energy supply is consumed in the Haber process (Science 297(1654), Sep 2002). That fertilizer is responsible for sustaining 40% of the Earth's population, as well as various deleterious environmental consequences.

253 CaCO3 (s) ↔ CaO (s) + CO2 (g)
Heterogeneous: different phases CaCO3 (s) ↔ CaO (s) + CO2 (g) Definition: What we use: Concentrations of pure solids and pure liquids are not included in Keq expression because their concentrations do not vary, and are “already included” in Keq (see p. 548). Even though the concentrations of the solids or liquids do not appear in the equilibrium expression, the substances must be present to achieve equilibrium.

254 15.4: Calculating Equilibrium Constants
Steps to use “ICE” table: “I” = Tabulate known initial and equilibrium concentrations of all species in equilibrium expression “C” = Determine the concentration change for the species where initial and equilibrium are known Use stoichiometry to calculate concentration changes for all other species involved in equilibrium “E” = Calculate the equilibrium concentrations

255 NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)
Ex: Enough ammonia is dissolved in 5.00 L of water at 25ºC to produce a solution that is M ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that [OH1-] is 4.64 x 10-4 M. Calculate Keq at 25ºC for the reaction: NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)

256 NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)
Initial Change Equilibrium NH3 (aq) H2O (l) NH41+ (aq) OH1- (aq) X M 0 M 0 M X - x + x + x X M 4.64 x 10-4 M 4.64 x 10-4 M x = 4.64 x 10-4 M

257 Equilibrium When the rate of the forward reaction equals the rate of the reverse reaction. (c) 2006, Mark Rosengarten

258 Examples of Equilibrium
Solution Equilibrium: when a solution is saturated, the rate of dissolving equals the rate of precipitating. NaCl (s)  Na+1 (aq) + Cl-1 (aq) Vapor-Liquid Equilibrium: when a liquid is trapped with air in a container, the liquid evaporates until the rate of evaporation equals the rate of condensation. H2O (l)  H2O (g) Phase equilibrium: At the melting point, the rate of solid turning to liquid equals the rate of liquid turning back to solid. H2O (s)  H2O (l)

259 Le Châtelier’s Principle
If a system at equilibrium is stressed, the equilibrium will shift in a direction that relieves that stress. A stress is a factor that affects reaction rate. Since catalysts affect both reaction rates equally, catalysts have no effect on a system already at equilibrium. Equilibrium will shift AWAY from what is added Equilibrium will shift TOWARDS what is removed. This is because the shift will even out the change in reaction rate and bring the system back to equilibrium NEXT

260 Steps to Relieving Stress
1) Equilibrium is subjected to a STRESS. 2) System SHIFTS towards what is removed from the system or away from what is added. The shift results in a CHANGE OF CONCENTRATION for both the products and the reactants. If the shift is towards the products, the concentration of the products will increase and the concentration of the reactants will decrease. If the shift is towards the reactants, the concentration of the reactants will increase and the concentration of the products will decrease. NEXT

261 Examples For the reaction N2(g) + 3H2(g)  2 NH3(g) + heat
Adding N2 will cause the equilibrium to shift RIGHT, resulting in an increase in the concentration of NH3 and a decrease in the concentration of N2 and H2. Removing H2 will cause a shift to the LEFT, resulting in a decrease in the concentration of NH3 and an increase in the concentration of N2 and H2. Increasing the temperature will cause a shift to the LEFT, same results as the one above. Decreasing the pressure will cause a shift to the LEFT, because there is more gas on the left side, and making more gas will bring the pressure back up to its equilibrium amount. Adding a catalyst will have no effect, so no shift will happen.

262 Oxidation Numbers Rules for Assigning Oxidation States
The oxidation state of an atom in an uncombined element is 0. The oxidation state of a monatomic ion is the same as its charge. Oxygen is assigned an oxidation state of –2 in most of its covalent compounds. Important exception: peroxides (compounds containing the O2 2- group), in which each oxygen is assigned an oxidation state of –1) In its covalent compounds with nonmetals, hydrogen is assigned an oxidation state of +1 For a compound, sum total of ON s is zero. For an ionic species (like a polyatomic ion), the sum of the oxidation states must equal the overall charge on that ion.

263 16.6: Weak Acids Weak acids partially ionize in water (equilibrium is somewhere between ions and molecules). HA (aq) ↔ A- (aq) + H+ (aq) Ka = acid-dissociation constant in water Weak acids generally have Ka < 10-3 See Appendix D for full listing of Ka values

264 Dissociation Constants
For a generalized acid dissociation, the equilibrium expression would be This equilibrium constant is called the acid-dissociation constant, Ka. HA (aq) + H2O (l) A- (aq) + H3O+ (aq) [H3O+] [A-] [HA] Kc = © 2009, Prentice-Hall, Inc.

265 Dissociation Constants
The greater the value of Ka, the stronger is the acid. © 2009, Prentice-Hall, Inc.

266 Calculating Ka from the pH
The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is Calculate Ka for formic acid at this temperature. We know that [H3O+] [COO-] [HCOOH] Ka = © 2009, Prentice-Hall, Inc.

267 Calculating Ka from the pH
The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is Calculate Ka for formic acid at this temperature. To calculate Ka, we need the equilibrium concentrations of all three things. We can find [H3O+], which is the same as [HCOO-], from the pH. © 2009, Prentice-Hall, Inc.

268 Calculating Ka from the pH
pH = -log [H3O+] 2.38 = -log [H3O+] -2.38 = log [H3O+] = 10log [H3O+] = [H3O+] 4.2  10-3 = [H3O+] = [HCOO-] © 2009, Prentice-Hall, Inc.

269 Calculating Ka from pH Now we can set up a table… [HCOOH], M [H3O+], M
[HCOO-], M Initially 0.10 Change - 4.2  10-3 + 4.2  10-3 At Equilibrium  10-3 = = 0.10 4.2  10-3 © 2009, Prentice-Hall, Inc.

270 Calculating Ka from pH [4.2  10-3] [4.2  10-3] Ka = [0.10]
= 1.8  10-4 © 2009, Prentice-Hall, Inc.

271 Calculating Percent Ionization
[H3O+]eq [HA]initial Percent Ionization =  100 In this example [H3O+]eq = 4.2  10-3 M [HCOOH]initial = 0.10 M 4.2  10-3 0.10 Percent Ionization =  100 = 4.2% © 2009, Prentice-Hall, Inc.

272 HC2H3O2 (aq) + H2O (l) H3O+ (aq) + C2H3O2- (aq)
Calculating pH from Ka Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25C. HC2H3O2 (aq) + H2O (l) H3O+ (aq) + C2H3O2- (aq) Ka for acetic acid at 25C is 1.8  10-5. © 2009, Prentice-Hall, Inc.

273 Calculating pH from Ka The equilibrium constant expression is
[H3O+] [C2H3O2-] [HC2H3O2] Ka = © 2009, Prentice-Hall, Inc.

274 Calculating pH from Ka We next set up a table… [C2H3O2], M [H3O+], M
Initially 0.30 Change -x +x At Equilibrium x  0.30 x We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored. © 2009, Prentice-Hall, Inc.

275 Calculating pH from Ka (x)2 (0.30) 1.8  10-5 =
Now, (x)2 (0.30) 1.8  10-5 = (1.8  10-5) (0.30) = x2 5.4  10-6 = x2 2.3  10-3 = x © 2009, Prentice-Hall, Inc.

276 Calculating pH from Ka pH = -log [H3O+] pH = -log (2.3  10-3)
© 2009, Prentice-Hall, Inc.

277 Polyprotic Acids… …have more than one acidic proton
If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation. © 2009, Prentice-Hall, Inc.

278 Bases react with water to produce hydroxide ion.
Weak Bases Bases react with water to produce hydroxide ion. © 2009, Prentice-Hall, Inc.

279 Weak Bases The equilibrium constant expression for this reaction is
[HB] [OH-] [B-] Kb = where Kb is the base-dissociation constant. © 2009, Prentice-Hall, Inc.

280 Kb can be used to find [OH-] and, through it, pH.
Weak Bases Kb can be used to find [OH-] and, through it, pH. © 2009, Prentice-Hall, Inc.

281 pH of Basic Solutions What is the pH of a 0.15 M solution of NH3?
NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) [NH4+] [OH-] [NH3] Kb = = 1.8  10-5 © 2009, Prentice-Hall, Inc.

282 pH of Basic Solutions Tabulate the data. [NH3], M [NH4+], M [OH-], M
Initially 0.15 At Equilibrium x  0.15 x © 2009, Prentice-Hall, Inc.

283 pH of Basic Solutions (x)2 (0.15) 1.8  10-5 =
© 2009, Prentice-Hall, Inc.

284 pH of Basic Solutions Therefore, [OH-] = 1.6  10-3 M
pOH = -log (1.6  10-3) pOH = 2.80 pH = pH = 11.20 © 2009, Prentice-Hall, Inc.

285 Ka and Kb Ka and Kb are related in this way: Ka  Kb = Kw
Therefore, if you know one of them, you can calculate the other. © 2009, Prentice-Hall, Inc.

286 Factors Affecting Acid Strength
The more polar the H-X bond and/or the weaker the H-X bond, the more acidic the compound. So acidity increases from left to right across a row and from top to bottom down a group. © 2009, Prentice-Hall, Inc.

287 Factors Affecting Acid Strength
In oxyacids, in which an -OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid. © 2009, Prentice-Hall, Inc.

288 Factors Affecting Acid Strength
For a series of oxyacids, acidity increases with the number of oxygens. © 2009, Prentice-Hall, Inc.

289 Factors Affecting Acid Strength
Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic. © 2009, Prentice-Hall, Inc.

290 16.9: Salt Solutions as Acids & Bases
Hydrolysis: acid/base reaction of ion with water to produce H+ or OH- Anion (A-) = a conjugate base A- (aq) + H2O (l) ↔ HA (aq) + OH- (aq) Cation (B+) = a conjugate acid B+ (aq) + H2O (l) ↔ BOH (aq) + H+ (aq)

291 17.1: Common Ion Effect Addition of a “common ion”: solubility of solids decrease because of Le Châtelier’s principle. Ex: AgCl (s) ↔ Ag+ (aq) + Cl- (aq) Addition of Cl- shifts equilibrium toward solid

292 17.4: Solubility Equilibria
Dissolving & precipitating of salts Solubility rules discussed earlier are generalized qualitative observations of quantitative experiments. Ex: PbCl2 (s) ↔ Pb2+ (aq) + 2 Cl- (aq) Ksp = [Pb2+][Cl-]2 = 1.6 x 10-5 Ksp = solubility-product constant (found in App. D) Recall that both aqueous ions and solid must be present in solution to achieve equilibrium Changes in pH will affect the solubility of salts composed of a weak acid or weak base ion.

293 Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO4 in water: BaSO4(s) Ba2+(aq) + SO42−(aq) © 2009, Prentice-Hall, Inc.

294 Solubility Products The equilibrium constant expression for this equilibrium is Ksp = [Ba2+] [SO42−] where the equilibrium constant, Ksp, is called the solubility product. © 2009, Prentice-Hall, Inc.

295 Solubility Products Ksp is not the same as solubility.
Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M). © 2009, Prentice-Hall, Inc.

296 Calculating Ksp from solubility
1. Calculate Ksp for Ag2CrO4, if its solubility is g/L. (Ans: 6.6 X 10^-5)

297 Calculating Solubility given Ksp
2. Ksp for MgF2 is 6.4 X 10^-9 at 250C. Calculate its solubility in mol/L and g/L. (Ans: 1.2 X 10^-3 M, 7.3 X 10^-2 g/L)

298 Factors Affecting Solubility
The Common-Ion Effect If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease. BaSO4(s) Ba2+(aq) + SO42−(aq) © 2009, Prentice-Hall, Inc.

299 Factors Affecting Solubility
pH If a substance has a basic anion, it will be more soluble in an acidic solution. Substances with acidic cations are more soluble in basic solutions. © 2009, Prentice-Hall, Inc.

300 Factors Affecting Solubility
Amphoterism Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases. Examples of such cations are Al3+, Zn2+, and Sn2+. © 2009, Prentice-Hall, Inc.

301 17.2: Buffers: Solutions that resist drastic changes in pH upon additions of small amounts of acid or base. Consist of a weak acid and its conjugate base (usually in salt form) Ex: acetic acid and sodium acetate: HC2H3O2 + NaC2H3O2 Or consist of a weak base and its conjugate acid (usually in salt form) Ex: ammonia and ammonium chloride: NH3 + NH4Cl

302 Buffers Buffers are solutions of a weak conjugate acid-base pair.
They are particularly resistant to pH changes, even when strong acid or base is added. © 2009, Prentice-Hall, Inc.

303 Buffers If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water. © 2009, Prentice-Hall, Inc.

304 Buffers Similarly, if acid is added, the F− reacts with it to form HF and water. © 2009, Prentice-Hall, Inc.

305 Titration In this technique a known concentration of base (or acid) is slowly added to a solution of acid (or base). © 2009, Prentice-Hall, Inc.

306 Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base. © 2009, Prentice-Hall, Inc.

307 Titration of a Strong Acid with a Strong Base
From the start of the titration to near the equivalence point, the pH goes up slowly. © 2009, Prentice-Hall, Inc.

308 Titration of a Strong Acid with a Strong Base
Just before (and after) the equivalence point, the pH increases rapidly. © 2009, Prentice-Hall, Inc.

309 Titration of a Strong Acid with a Strong Base
At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid. © 2009, Prentice-Hall, Inc.

310 Titration of a Strong Acid with a Strong Base
As more base is added, the increase in pH again levels off. © 2009, Prentice-Hall, Inc.

311 Titration of a Weak Acid with a Strong Base
Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. At the equivalence point the pH is >7. Phenolphthalein is commonly used as an indicator in these titrations. © 2009, Prentice-Hall, Inc.

312 Practice Problem on Titration:
If 7.3 mL of 1.25 M HNO3 is required to neutralize mL of a potassium hydroxide solution, what is the molarity of the potassium hydroxide? 0.044 M KOH

313 Titration of a Weak Base and Strong Acid
14 pH 7 Volume of HCl added (mL) Half Equivalence Point , pH= pKa pka or pkb of weak acid or base in a buffer should be clsoe to the desired pH of the buffer solution.

314 Redox: Reduction occurs when an atom gains one or more electrons. Ex:
Oxidation occurs when an atom or ion loses one or more electrons. LEO goes GER Copper metal reacts with silver nitrate to form silver metal and copper nitrate: Cu + 2 Ag(NO3)  2 Ag + Cu(NO3)2.

315 Identifying OX, RD, SI Species
Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20 Oxidation = loss of electrons. The species becomes more positive in charge. For example, Ca0  Ca+2, so Ca0 is the species that is oxidized. Reduction = gain of electrons. The species becomes more negative in charge. For example, H+1  H0, so the H+1 is the species that is reduced. Spectator Ion = no change in charge. The species does not gain or lose any electrons. For example, Cl-1  Cl-1, so the Cl-1 is the spectator ion.

316 Oxidizing Agent and Reducing Agent:
Oxidizing agent gets reduced itself and reducing agent gets oxidized itself, so a strong oxidizing agent should have a great tendency to accept e and a strong reducing agent should be willing to lose e easily. What are strong oxidizing agents- metals or non metals? Why? Which is the strongest oxidizing agent and which is the strongest reducing agent?

317 Agents Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20
Since Ca0 is being oxidized and H+1 is being reduced, the electrons must be going from the Ca0 to the H+1. Since Ca0 would not lose electrons (be oxidized) if H+1 weren’t there to gain them, H+1 is the cause, or agent, of Ca0’s oxidation. H+1 is the oxidizing agent. Since H+1 would not gain electrons (be reduced) if Ca0 weren’t there to lose them, Ca0 is the cause, or agent, of H+1’s reduction. Ca0 is the reducing agent.

318 Steps for Balancing a Redox Reaction: Half Reaction Method
In half reaction method, oxidation and reduction half- reactions are written and balanced separately before combining them into a balanced redox reaction. It is a good method for balancing redox reactions because this method can be used both for reactions carried out in acidic and basic medium .

319 Steps for Balancing Redox Reaction Using Half Reaction Method IN ACIDIC MEDIUM:
Step 1: Write unbalanced equation in ionic form. Step 2: Write separate half reactions for the oxidation and reduction processes. (Use Oxidation Numbers for identifying oxidation and reduction reactions) Step 3: Balance atoms in the half reactions First, balance all atoms except H and O Balance O by adding H2O Balance H by adding H+ Step 4: Balance Charges on each half reaction, by adding electrons. Step 5: Multiply each half reaction by an appropriate number to make the number of electrons equal in both half reactions. Step 6: Add two half reactions and simplify where possible by canceling species appearing in both sides. Step 7: Check equation for same number of atoms and charges on both sides.

320 Writing Half-Reactions
Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20 Oxidation: Ca0  Ca+2 + 2e- Reduction: 2H+1 + 2e-  H20 The two electrons lost by Ca0 are gained by the two H+1 (each H+1 picks up an electron). PRACTICE SOME!

321 Practice Half-Reactions
Don’t forget to determine the charge of each species first! 4 Li + O2  2 Li2O Oxidation Half-Reaction: Reduction Half-Reaction: Zn + Na2SO4  ZnSO4 + 2 Na

322 Steps for Balancing Redox Reaction Using Half Reaction Method IN BASIC MEDIUM:
For balancing redox reactions in basic solutions, all the steps are the same as acidic medium balancing, except you add one more step to it. The H+ ions can then be “neutralized” by adding an equal number of OH- ions to both sides of the equation. Ex.

323 Standard Cell Potential
Just as the water tends to flow from a higher level to a lower level, electrons also move from a higher “potential” to a lower potential. This potential difference is called the electromotive force (EMF) of cell and is written as Ecell. The standard for measuring the cell potentials is called a SHE (Standard Hydrogen Electrode). Description of SHE (Standard Hydrogen Electrode) Reaction 2H+(aq, 1M)+ 2e - H2(g, 101kPa) E0= 0.00 V

324 Standard Reduction Potentials
Many different half cells can be paired with the SHE and the standard reduction potentials for each half cell is obtained. Check the table for values of reduction potential for various substances: Would substances with high reduction potential be strong oxidizing agents or strong reducing agents? Why?


326 Activity Series For metals, the higher up the chart the element is, the more likely it is to be oxidized. This is because metals like to lose electrons, and the more active a metallic element is, the more easily it can lose them. For nonmetals, the higher up the chart the element is, the more likely it is to be reduced. This is because nonmetals like to gain electrons, and the more active a nonmetallic element is, the more easily it can gain them.

327 Metal Activity 3 K0 + Fe+3Cl-13 REACTION Metallic elements start out with a charge of ZERO, so they can only be oxidized to form (+) ions. The higher of two metals MUST undergo oxidation in the reaction, or no reaction will happen. The reaction 3 K + FeCl3  3 KCl + Fe WILL happen, because K is being oxidized, and that is what Table J says should happen. The reaction Fe + 3 KCl  FeCl3 + 3 K will NOT happen. Fe0 + 3 K+1Cl-1 NO REACTION

328 Voltaic Cells (Galvanic Cells)
A voltaic cell converts chemical energy from a spontaneous redox reaction into electrical energy. Ex: Cu and Zn voltaic cell (More positive reduction potential is the cathode) Key Words: Cathode Anode Salt Bridge How a Voltaic Cell Works: An Ox, Red Cat Representing Electrochemical Cells

329 Voltaic Cells Produce electrical current using a spontaneous redox reaction Used to make batteries! Materials needed: two beakers, piece of the metals (anode, - electrode and cathode + electrode), solution of each metal, porous material (salt bridge), solution of a salt that does not contain either metal in the reaction, wire and a load to make use of the generated current! Use Reference Table J to determine the metals to use Higher = (-) anode (lower reduction potential) Lower = (+) cathode (higher reduction potential)

330 Making Voltaic Cells

331 Electrolytic Cells Use electricity to force a nonspontaneous redox reaction to take place. Uses for Electrolytic Cells: Decomposition of Alkali Metal Compounds Decomposition of Water into Hydrogen and Oxygen Electroplating Differences between Voltaic and Electrolytic Cells: ANODE: Voltaic (-) Electrolytic (+) CATHODE: Voltaic (+) Electrolytic (-) Voltaic: 2 half-cells, a salt bridge and a load Electrolytic: 1 cell, no salt bridge, IS the load

332 Decomposing Alkali Metal Compounds
2 NaCl  2 Na + Cl2 The Na+1 is reduced at the (-) cathode, picking up an e- from the battery The Cl-1 is oxidized at the (+) anode, the e- being pulled off by the battery (DC)

333 Decomposing Water 2 H2O  2 H2 + O2
The H+ is reduced at the (-) cathode, yielding H2 (g), which is trapped in the tube. The O-2 is oxidized at the (+) anode, yielding O2 (g), which is trapped in the tube.

334 Electroplating The Ag0 is oxidized to Ag+1 when the (+) end of the battery strips its electrons off. The Ag+1 migrates through the solution towards the (-) charged cathode (ring), where it picks up an electron from the battery and forms Ag0, which coats on to the ring.

335 Spontaneity of Redox Reactions:
E0 = E0red( reduction process-cathode) – E0red (oxidation process-anode) A positive value of E0 indicates a spontaneous process and a negative value of E0 indicates a nonspontaneous value. Steps for Predicting Spontaneity of Redox Reactions First write the reaction as oxidation and reduction half reactions. Then plug standard reduction potential values in the equation given above. Check for the spontaneity by a positive or a negative value of E0 Ex:

336 Hydrocarbons Molecules made of Hydrogen and Carbon
Carbon forms four bonds, hydrogen forms one bond Hydrocarbons come in three different homologous series: Alkanes (single bond between C’s, saturated) Alkenes (1 double bond between 2 C’s, unsaturated) Alkynes (1 triple bond between 2 C’s, unsaturated) These are called aliphatic, or open-chain, hydrocarbons. Count the number of carbons and add the appropriate suffix!

337 Alkanes CH4 = methane C2H6 = ethane C3H8 = propane C4H10 = butane
C5H12 = pentane To find the number of hydrogens, double the number of carbons and add 2.

338 Methane Meth-: one carbon -ane: alkane
The simplest organic molecule, also known as natural gas!

339 Ethane Eth-: two carbons -ane: alkane

340 Propane Prop-: three carbons -ane: alkane
Also known as “cylinder gas”, usually stored under pressure and used for gas grills and stoves. It’s also very handy as a fuel for Bunsen burners!

341 Butane But-: four carbons -ane: alkane
Liquefies with moderate pressure, useful for gas lighters. You have probably lit your gas grill with a grill lighter fueled with butane!

342 Pentane Pent-: five carbons -ane: alkane Draw Hexane: Draw Heptane:
Your Turn!!! Draw Hexane: Draw Heptane:

343 Alkenes C2H4 = Ethene C3H6 = Propene C4H8 = Butene C5H10 = Pentene
To find the number of hydrogens, double the number of carbons.

344 Ethene Two carbons, double bonded. Notice how each carbon has four bonds? Two to the other carbon and two to hydrogen atoms. Also called “ethylene”, is used for the production of polyethylene, which is an extensively used plastic. Look for the “PE”, “HDPE” (#2 recycling) or “LDPE” (#4 recycling) on your plastic bags and containers!

345 Propene Three carbons, two of them double bonded. Notice how each carbon has four bonds? If you flipped this molecule so that the double bond was on the right side of the molecule instead of the left, it would still be the same molecule. This is true of all alkenes. Used to make polypropylene (PP, recycling #5), used for dishwasher safe containers and indoor/outdoor carpeting!

346 Butene This is 1-butene, because the double bond is between the 1st and 2nd carbon from the end. The number 1 represents the lowest numbered carbon the double bond is touching. This is 2-butene. The double bond is between the 2nd and 3rd carbon from the end. Always count from the end the double bond is closest to. ISOMERS: Molecules that share the same molecular formula, but have different structural formulas.

347 Pentene This is 1-pentene. The double bond is on the first carbon from the end. This is 2-pentene. The double bond is on the second carbon from the end. This is not another isomer of pentene. This is also 2-pentene, just that the double bond is closer to the right end.

348 Alkynes C2H2 = Ethyne C3H4 = Propyne C4H6 = Butyne C5H8 = Pentyne
To find the number of hydrogens, double the number of carbons and subtract 2.

349 Ethyne Now, try to draw propyne! Any isomers? Let’s see!
Also known as “acetylene”, used by miners by dripping water on CaC2 to light up mining helmets. The “carbide lamps” were attached to miner’s helmets by a clip and had a large reflective mirror that magnified the acetylene flame. Used for welding and cutting applications, as ethyne burns at temperatures over 3000oC!

350 Propyne This is propyne! Nope! No isomers.
OK, now draw butyne. If there are any isomers, draw them too.

351 Butyne Well, here’s 1-butyne! And here’s 2-butyne!
Is there a 3-butyne? Nope! That would be 1-butyne. With four carbons, the double bond can only be between the 1st and 2nd carbon, or between the 2nd and 3rd carbons. Now, try pentyne!

352 Pentyne Naming: Check this link out
1-pentyne 2-pentyne Now, draw all of the possible isomers for hexyne!

353 Isomers Isomers are compounds that have same molecular formula (same number of atoms) but a different structure. There are three types of isomers: Structural Isomers: Same number of atoms, arranged differently. Geometric Isomers (Cis- trans-): Happens in = or triple bonded compounds since these are inflexible bonds. Ex. Optical Isomers ( D- and L-): Need a central atom that is “Chiral” (all four groups attached to it are different). These are non super imposable mirror images. Usually this central atom is C.


355 Substituted Hydrocarbons
Hydrocarbon chains can have three kinds of “dingly-danglies” attached to the chain. If the dingly-dangly is made of anything other than hydrogen and carbon, the molecule ceases to be a hydrocarbon and becomes another type of organic molecule. Alkyl groups Halide groups Other functional groups To name a hydrocarbon with an attached group, determine which carbon (use lowest possible number value) the group is attached to. Use di- for 2 groups, tri- for three.

356 Alkyl Groups

357 Halide Groups

358 Organic Families Each family has a functional group to identify it.
Alcohol (R-OH, hydroxyl group) Organic Acid (R-COOH, primary carboxyl group) Aldehyde (R-CHO, primary carbonyl group) Ketone (R1-CO-R2, secondary carbonyl group) Ether (R1-O-R2) Ester (R1-COO-R2, carboxyl group in the middle) Amine (R-NH2, amine group) Amide (R-CONH2, amide group) These molecules are alkanes with functional groups attached. The name is based on the alkane name.

359 Alcohol On to DI and TRIHYDROXY ALCOHOLS

360 Di and Tri- hydroxy Alcohols

361 Positioning of Functional Group
PRIMARY (1o): the functional group is bonded to a carbon that is on the end of the chain. SECONDARY (2o): The functional group is bonded to a carbon in the middle of the chain. TERTIARY (3o): The functional group is bonded to a carbon that is itself directly bonded to three other carbons.

362 Organic Acid These are weak acids. The H on the right side is the one that ionized in water to form H3O+. The -COOH (carboxyl) functional group is always on a PRIMARY carbon. Can be formed from the oxidation of primary alcohols using a KMnO4 catalyst.

363 Aldehyde Aldehydes have the CO (carbonyl) groups ALWAYS on a PRIMARY carbon. This is the only structural difference between aldehydes and ketones. Formed by the oxidation of primary alcohols with a catalyst. Propanal is formed from the oxidation of 1-propanol using pyridinium chlorochromate (PCC) catalyst.*

364 Ketone Ketones have the CO (carbonyl) groups ALWAYS on a SECONDARY carbon. This is the only structural difference between ketones and aldehydes. Can be formed from the dehydration of secondary alcohols with a catalyst. Propanone is formed from the oxidation of 2-propanol using KMnO4 or PCC catalyst.*

365 Ether Ethers are made of two alkyl groups surrounding one oxygen atom. The ether is named for the alkyl groups on “ether” side of the oxygen. If a three-carbon alkyl group and a four-carbon alkyl group are on either side, the name would be propyl butyl ether. Made with an etherfication reaction.

366 Ester Esters are named for the alcohol and organic acid that reacted by esterification to form the ester. If the alcohol was 1-propanol and the acid was hexanoic acid, the name of the ester would be propyl hexanoate. Esters contain a COO (carboxyl) group in the middle of the molecule, which differentiates them from organic acids.

367 Amine Component of amino acids, and therefore proteins, RNA and DNA…life itself! - Essentially ammonia (NH3) with the hydrogens replaced by one or more hydrocarbon chains, hence the name “amine”!

368 Amide Synthetic Polyamides: nylon, kevlar Natural Polyamide: silk!
For more information on polymers, go here.

369 Organic Reactions Combustion Fermentation Substitution Addition
Dehydration Synthesis Etherification Esterification Saponification Polymerization

370 Combustion Happens when an organic molecule reacts with oxygen gas to form carbon dioxide and water vapor. Also known as “burning”.

371 Substitution Alkane + Halogen  Alkyl Halide + Hydrogen Halide
The halogen atoms substitute for any of the hydrogen atoms in the alkane. This happens one atom at a time. The halide generally replaces an H on the end of the molecule. C2H6 + Cl2  C2H5Cl + HCl The second Cl can then substitute for another H: C2H5Cl + HCl  C2H4Cl2 + H2

372 Addition Alkene + Halogen  Alkyl Halide
The double bond is broken, and the halogen adds at either side of where the double bond was. One isomer possible. (c) 2006, Mark Rosengarten

373 Etherification* Alcohol + Alcohol  Ether + Water
A dehydrating agent (H2SO4) removes H from one alcohol’s OH and removes the OH from the other. The two molecules join where there H and OH were removed. Note: dimethyl ether and diethyl ether are also produced from this reaction, but can be separated out.

374 Esterification Organic Acid + Alcohol  Ester + Water
A dehydrating agent (H2SO4) removes H from the organic acid and removes the OH from the alcohol. The two molecules join where there H and OH were removed.

375 Saponification The process of making soap from glycerol esters (fats).
Glycerol ester + 3 NaOH  soap + glycerol Glyceryl stearate + 3 NaOH  sodium stearate + glycerol The sodium stearate is the soap! It emulsifies grease…surrounds globules with its nonpolar ends, creating micelles with - charge that water can then wash away. Hard water replaces Na+ with Ca+2 and/or other low solubility ions, which forms a precipitate called “soap scum”. Water softeners remove these hardening ions from your tap water, allowing the soap to dissolve normally.

376 Polymerization A polymer is a very long-chain molecule made up of many monomers (unit molecules) joined together. The polymer is named for the monomer that made it. Polystyrene is made of styrene monomer Polybutadiene is made of butadiene monomer Addition Polymers Condensation Polymers Rubber

377 Addition Polymers Joining monomers together by breaking double bonds
Polyvinyl chloride (PVC): vinyl siding, PVC pipes, etc. Vinyl chloride polyvinyl chloride n C2H3Cl  (-C2H3Cl-)-n Polytetrafluoroethene (PTFE, teflon): TFE PTFE n C2F  (-C2F4-)-n

378 Condensation Polymers
Condensation polymerization is just dehydration synthesis, except instead of making one molecule of ether or ester, you make a monster molecule of polyether or polyester.

379 Rubber The process of toughing rubber by cross-linking the polymer strands with sulfur is called...

380 THE END (c) 2006, Mark Rosengarten

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