Presentation on theme: "SAT II CHEM PREP PPT Mrs. Gupta Modified from Mark Rosengartens Powerpoint."— Presentation transcript:
SAT II CHEM PREP PPT Mrs. Gupta Modified from Mark Rosengartens Powerpoint
Setup of the SAT II Chem Exam 85 total questions, 1 hour (about 42 s/question) - All multiple choice, 1/4 th point taken off for every incorrect answer - if you can narrow down to two choices, then guess otherwise leave blank - scoring scale from
What to Bring to the Exam #2 pencil, eraser No calculators allowed (brush up on your basic math skills) Your brain. Please dont leave it at home.:)
How To Prepare DO NOT CRAM. Get your studying done with by the night before. Get a good nights sleep and have breakfast the morning of the exam. Actively participate in any and all review classes and activities offered by your teacher.
Matter 1) Properties of PhasesProperties of Phases 2) Types of MatterTypes of Matter 3) Phase ChangesPhase Changes
Properties of Phases Solids: Crystal lattice (regular geometric pattern), vibration motion only Solids Liquids: particles flow past each other but are still attracted to each other. Liquids 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) Gases
Solids The positive and negative ions alternate in the ionic crystal lattice of NaCl.
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.
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 ) gas molecules. Since there is space between gas molecules, gases are affected by changes in pressure.
Types of Matter Substances (Homogeneous) – Elements (cannot be decomposed by chemical change): Al, Ne, O, Br, H Elements – Compounds (can be decomposed by chemical change): NaCl, Cu(ClO 3 ) 2, KBr, H 2 O, C 2 H 6 Compounds Mixtures – Homogeneous: Solutions (solvent + solute) – Heterogeneous: soil, Italian dressing, etc.
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.
Compounds Lead has two charges listed, +2 and +4. This is a sample of lead (II) chloride (PbCl 2 ). 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 (PbCl 4 ). Notice how both samples of lead compounds have consistent composition throughout? Compounds are homogeneous!
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 PbCl 2 and PbCl 4 formula units. Again, they are in no particular ratio to each other and can be separated without chemical change.
The Atom 1) NucleonsNucleons 2) IsotopesIsotopes 3) Natural RadioactivityNatural Radioactivity 4) Half-LifeHalf-Life 5) Nuclear PowerNuclear Power 6) Electron ConfiguationElectron Configuation 7) Development of the Atomic ModelDevelopment of the Atomic Model
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) S and S 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.
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.weight-average atomic mass Rounding off the atomic mass on the Periodic Table will tell you what the most common isotope of that element is.common isotope of that element
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)
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.
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.
Alpha Decay The nucleus ejects two protons and two neutrons. The atomic mass decreases by 4, the atomic number decreases by U
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 C
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 Fe
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.
Charges of Decay Particles
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 forwards in time – Going backwards in time Going backwards in time – Radioactive Dating Radioactive Dating
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
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
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
Artificial Transmutation Ca + _____ -----> K H Mo H -----> 1 0 n + _____ Nuclide + Bullet --> New Element + Fragment(s) The masses and atomic numbers must add up to be the same on both sides of the arrow.
Nuclear Fission U n Kr 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.
Nuclear Fusion 2 1 H H 4 2 He + 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.
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
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)
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.
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
Example Dot Diagrams Carbon can also have this dot diagram, which it has when it forms organic compounds.
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 doesnt match, the atom is in the excited state. Na (ground, on table): Example of excited states: 2-7-2, , 2-6-3
Ways to Represent Electron Configuration 1.Expanded Electron Configuration 2.Condensed Electron Configurations 3.Orbital Notation 4.Electron Dot Structure Write the above four electron configurations for Zinc, Zinc ion and Cu ion.
Electron Configuration of Ions -Group configurations: s block ns1-2, p block ns 2 np 1-6, d block ns0-2 (n-1) d 1-10, f block ns 0- 2 (n-1) d 1 (n-2) f 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:
Rules for Electron Filling - Afbaus Principle: Electrons tend to occupy the lowest energy orbitals first. - Hunds Rule: Pairing of e in the degenerate orbitals does not take place till every orbital has one e. - Paulis Exclusion Principle: No two electrons can have all four same quantum numbers.
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.
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?
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 L m AB n (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
Tables # 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 AB 2 Linear 180º BeF 2 CO 2 A |X X A |B B
3 Three sp 2 hybrid orbitals Trigonal planar AB 3 Trigonal planar 120º BF 3 Cl-C-Cl << 120º Cl 2 CO LAB 2 Bent < 120º NO 2 1- A |X XX A |B BB A :B
4 Four sp 3 hybrid orbitals or Tetrahedral AB 4 Tetrahedral 109.5º CH 4 LAB 3 Trigonal pyramidal < 109.5º Ex: NH 3 = 107º NH 3 L 2 AB 2 Bent <<109.5º Ex: H 2 O = 104.5º H2OH2O X A X X X X A X XX B A B BB : A B BB : A B B:
5 Five sp 3 d hybrid orbitals Trigonal bipyramidal AB 5 Trigonal bipyramidal Equatorial = 120º Axial = 90º PCl 5 LAB 4 Seesaw Equatorial < 120º Axial < 90º SF 4 X X X A X |X|X B B B A B |B|B : B - A - B B B
5 Five sp 3 d hybrid orbitals Trigonal bipyramidal L 2 AB 3 T-shaped Axial << 90º ClF 3 L 3 AB 2 Linear Axial = 180º XeF 2 X X X A X |X|X B : : A B |B|B : : : A B |B|B
6 Six sp 3 d 2 hybrid orbitals or Octahedral AB 6 Octahedral 90º SF 6 LAB 5 Square pyramidal < 90º BrF 5 X X A X |X|X X X B B A B |B|B B B B B A B |.. B B A X |X|X XX X X
6 Six sp 3 d 2 hybrid orbitals or L 2 AB 4 Square planar 90º XeF 4 or L 3 AB 3 T-shaped <90º KrCl 3 1- A B |B|B BB.. B B A |.. B B B B A.. |.... B A |B|B BB
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)
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 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.
Development of the Atomic Model Thompson Model Rutherford Gold Foil Experiment and Model Bohr Model Quantum-Mechanical Model
Thompson Model The atom is a positively charged diffuse mass with negatively charged electrons stuck in it.
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).
Bohr Model Electrons orbit around the nucleus in energy levels (shells). Atomic bright-line spectra was the clue.
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 cant 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.
Orbital Quantum Numbers SymbolNameDescriptionMeaningEquations n Principle Q.N. Energy level (i.e. Bohrs 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.
SymbolNameDescriptionMeaningEquations mlml Magnetic Q.N. 3-D orientation of the orbital s has 1 p has 3 d has 5 f has 7 m l = -l, -l +1, …, 0, l, …, +l There are (2l + 1) values. msms Spin Q.N. Spin of the electron Parallel or antiparallel to field m s = +½ or -½ * s, p, d, and f come from the words sharp, principal, diffuse, and fundamental.
Permissible Quantum Numbers (4, 1, 2, +½) (5, 2, 0, 0) (2, 2, 1, +½) 76 Not permissible; if l = 1, m l = 1, 0, or –1 (p orbitals only have 3 subshells) Not permissible; m s = +½ or –½ Not permissible; if n = 2, l = 0 or 1 (there is no 2d orbital)
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? Thats because the PE is changing, not the KE.
Heat of Phase Change How many joules would it take to melt 100. g of H 2 O (s) at 0 o C? q=mH f = (100. g)(334 J/g) = J How many joules would it take to boil 100. g of H 2 O (l) at 100 o C? q=mH v = (100.g)(2260 J/g) = J
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.vapor pressure, and can be found on Table H. When the liquid is heated, its vapor pressure increases. When the liquids 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).
Reference Table H: Vapor Pressure of Four Liquids (c) 2006, Mark Rosengarten
83 Phase diagrams: CO 2 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) 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.) Temp (ºC)
84 Phase diagrams: H 2 O For H 2 O, inc P will cause ice to melt.
The Periodic Table Metals Nonmetals Metalloids Chemistry of Groups Electronegativity Ionization Energy
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 Lose electrons to nonmetal atoms to form positively charged ionsionic bonds Large atomic radii compared to nonmetal atoms Low electronegativity and ionization energyelectronegativityionization energy Left side of the periodic table (except H)
Nonmetals Are dull and brittle, poor conductors Gain electrons from metal atoms to form negatively charged ions in ionic bonds Gain electrons from metal atoms to form negatively charged ions Share unpaired valence electrons with other nonmetal atoms to form covalent bonds and moleculescovalent bonds Small atomic radii compared to metal atoms High electronegativity and ionization energyelectronegativityionization energy Right side of the periodic table (except Group 18)
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).
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
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) + H 2 O (l) 2 NaOH (aq) + H 2 (g) 1 valence electron Form +1 ion by losing that valence electron Form oxides like Na 2 O, Li 2 O, K 2 O
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 H 2 O (l) Ca(OH) 2 (aq) + H 2 (g) 2 valence electrons Form +2 ion by losing those valence electrons Form oxides like CaO, MgO, BaO
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.
Group 17: Halogens Most reactive nonmetals React violently with metal atoms to form halide compounds: 2 Na + Cl 2 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.
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.
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. Br 2, I 2, N 2, Cl 2, H 2, O 2 and F 2 The decomposition of water : 2 H 2 O 2 H 2 + O 2
Electronegativity An atoms 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!
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.
Ions Ions are charged particles formed by the gain or loss of electrons. – Metals lose electrons (oxidation) to form (+) charged cations. Metals lose electrons – Nonmetals gain electrons (reduction) to form (-) charged anions. Nonmetals gain electrons 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).
Metal Ions (Cations) Na: Na +1 : 2-8 Ca: Ca +2 : Al: Al +3 : 2-8 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!
Nonmetal Ions (Anions) F: 2-7 F -1 : 2-8 O: 2-6 O -2 : 2-8 N: 2-5 N -3 : 2-8 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.
Chemical Bonding Intermolecular Bonding: Ionic, Covalent, Metallic and Covalent Network Bonds Intermolecular Bonding: H bond, dipole-dipole interaction, LDFs
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.
Example of Ionic Bonding
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 - and the one with the lower EN will be + charged. This is a POLAR covalent bonding. ( means partial)
Examples of Covalent Bonding
Sigma and Pi bonds Sigma ( ) 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: F 2 Pi ( ) 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 bonds. Forms between two p orbitals (py or pz) Make up the 2 nd and 3 rd bonds in double & triple bonds. Ex: O 2 N 2
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.
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, K 2 O, CaBr 2 Ionic 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 H 3 O + and A -, so they do conduct. Molecular Network: made up of nonmetal atoms bonded in a seemingly endless matrix of covalent bonds with no distinguishable molecules. Very high m.p., dont conduct. Network
Ionic Compounds (c) 2006, Mark Rosengarten
Molecular Compounds (c) 2006, Mark Rosengarten
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 SiO 2 (quartz). Corundum (Al 2 O 3 ) 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.
Attractive Forces Molecules have partially charged ends. The + end of one molecule attracts to the - 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
Determining Polarity of Molecules (c) 2006, Mark Rosengarten
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, C 3 H 8 ) are gases. This also gives water its surface tension and its ability to form a meniscus in a narrow glass tube.
Compounds 1) Types of CompoundsTypes of Compounds 2) Formula WritingFormula Writing 3) Formula NamingFormula Naming 4) Empirical FormulasEmpirical Formulas 5) Molecular FormulasMolecular Formulas 6) Types of Chemical ReactionsTypes of Chemical Reactions 7) Balancing Chemical ReactionsBalancing Chemical Reactions 8) Attractive ForcesAttractive Forces
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 = CaBr 2 Al +3 and O -2 = Al 2 O 3 Zn +2 and PO 4 -3 = Zn 3 (PO 4 ) 2 Try these problems!
Formulas to Write Ba +2 and N -3 NH 4 +1 and SO 4 -2 Li +1 and S -2 Cu +2 and NO 3 -1 Al +3 and CO 3 -2 Fe +3 and Cl -1 Pb +4 and O -2 Pb +2 and O -2
Formula Naming Compounds are named from the elements or polyatomic ions that form them. KCl = potassium chloride Na 2 SO 4 = sodium sulfate (NH 4 ) 2 S = ammonium sulfide AgNO 3 = 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!Take a peek!
The Stock System CrCl 2 = chromium (II) chloride Try CrCl 3 = chromium (III) chloride Co(NO 3 ) 2 and CrCl 6 = chromium (VI) chloride Co(NO 3 ) 3 FeO = iron (II) oxideMnS = manganese (II) sulfide Fe 2 O 3 = iron (III) oxideMnS 2 = manganese (IV) sulfide The Roman numeral is the charge of the metal ion!
Math of Chemistry 1) Formula MassFormula Mass 2) Percent CompositionPercent Composition 3) Mole ProblemsMole Problems 4) Gas LawsGas Laws 5) NeutralizationNeutralization 6) ConcentrationConcentration 7) Significant Figures and RoundingSignificant Figures and Rounding 8) Metric ConversionsMetric Conversions 9) CalorimetryCalorimetry
Formula Mass Gram Formula Mass = sum of atomic masses of all elements in the compound Round given atomic masses to the nearest tenth H 2 O: (2 X 1.0) + (1 X 16.0) = 18.0 grams/mole Na 2 SO 4 : (2 X 23.0)+(1 X 32.1)+(4 X 16.0) = g/mole Now you try: – BaBr 2 – CaSO 4 – Al 2 (CO 3 ) 3
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. Dont forget to take atomic mass to the nearest tenth! This is a problem for you to try.This is a problem for you to try
Practice Percent Composition Problem What is the percent by mass of each element in Li 2 SO 4 ?
Mole Problems Grams Moles Molecular Formula Stoichiometry
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
Molecular Formula Molecular Formula = (Molecular Mass/Empirical Mass) X Empirical Formula What is the molecular formula of a compound with an empirical formula of CH 2 and a molecular mass of 70.0 grams/mole? 1) Find the Empirical Formula Mass: CH 2 = ) Divide the MM/EM: 70.0/14.0 = 5 3) Multiply the molecular formula by the result: 5 (CH 2 ) = C 5 H 10
Stoichiometry Moles of Target = Moles of Given X (Coefficent of Target/Coefficient of given) Given the balanced equation N H 2 2 NH 3, How many moles of H 2 need to be completely reacted with N 2 to yield 20.0 moles of NH 3 ? 20.0 moles NH 3 X (3 H 2 / 2 NH 3 ) = 30.0 moles H 2
Limiting Reactant controls the amount of product formed. CO(g) + 2H 2 (g) Ch 3 OH a.If 500 mol of CO react with 750 mol of H 2, which is the limiting reactant? 1.Use either given amount to calculate required amount of other. 2.Compare calculated amount to amount given b. How many moles of excess reactant remain unchanged? 147 H2H2 125 mol CO
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 148 Theoretical yield= g SnF 2 Actual yield = g SnF 2 Percent yield = g SnF g SnF 2 *100
Determining empirical formula from combustion data When a compound containing C,H and O undergoes combustion, it forms CO 2 and H 2 O. Then from the mass of CO 2 and H 2 O, 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 CO 2 and g of H 2 O. Determine the empirical formula of the compound. Ans. C 7 H 6 O2
Empirical Formulas Ionic formulas: represent the simplest whole number mole ratio of elements in a compound. Ca 3 N 2 means a 3:2 ratio of Ca ions to N ions in the compound. Many molecular formulas can be simplified to empirical formulas – Ethane (C 2 H 6 ) can be simplified to CH 3. This is the empirical formula…the ratio of C to H in the molecule. All ionic compounds have empirical formulas.
Molecular Formulas The count of the actual number of atoms of each element in a molecule. H 2 O: a molecule made of two H atoms and one O atom covalently bonded together. C 2 H 6 O: 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: – H 2 O = 1 X (H 2 O) – C 8 H 16 = 8 X (CH 2 ) Calculating Molecular Formulas
Molecular Formula Actual ratio of atoms in a compound. Ex. H 2 O, C 6 H 12 O 6 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: 1.Ethylene glycol, the substance used as antifreeze has % C, 9.70 % H and % O, mm= g 2.Caffeine, a stimulant in coffee has the following percent composition: % C, 5.15% H, % N and % O, molar mass= g
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 Synthesis – Decomposition Decomposition – Single Replacement 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 Double Replacement
Synthesis Two elements combine to form a compound 2 Na + O 2 Na 2 O Same reaction, with charges added in: – 2 Na 0 + O 2 0 Na 2 +1 O -2 Na 0 is oxidized (loses electrons), is the reducing agent O 2 0 is reduced (gains electrons), is the oxidizing agent Electrons are transferred from the Na 0 to the O 2 0. No spectator ions, there are only two elements here.
Decomposition A compound breaks down into its original elements. Na 2 O 2 Na + O 2 Same reaction, with charges added in: – Na 2 +1 O -2 2 Na 0 + O 2 0 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.
Single Replacement An element replaces the same type of element in a compound. Ca + 2 KCl CaCl K Same reaction, with charges added in: – Ca K +1 Cl -1 Ca +2 Cl K 0 Ca 0 is oxidized (loses electrons), is the reducing agent K +1 is reduced (gains electrons), is the oxidizing agent Electrons are transferred from the Ca 0 to the K +1. Cl -1 is the spectator ion, since its charge doesnt change.
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) + AgNO 3 (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 NO 3 -1 ions remain dissolved throughout the reaction, they are the spectator ions. How do identify the precipitate?
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.
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(NO 3 ) 2 means that you have 2 Ca, 4 N and 12 O. Examples for you to try!
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.
Manometers: measure P of a gas 1.Closed-end: difference in Hg levels ( h) shows P of gas in container compared to a vacuum 2.http://www.chm.davidson.edu/ChemistryApplets/GasLaws/Pressure.html 172 closed
2. Open-end: Difference in Hg levels ( h) shows P of gas in container compared to P atm 173
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 T 2, get it out of the denominator first by cross-multiplying. If one of the variables is constant, then eliminate it. Try these problems!
Gas Law Problem 1 A 2.00 L sample of N 2 gas at STP is compressed to 4.00 atm at constant temp- erature. What is the new volume of the gas? V 2 = P 1 V 1 / P 2 = (1.00 atm)(2.00 L) / (4.00 atm) = L
Gas Law Problem 2 To what temperature must a L sample of O 2 gas at K be heated to raise the volume to L? T 2 = V 2 T 1 /V 1 = (10.00 L)(300.0 K) / (3.000 L) = K
Gas Law Problem 3 A 3.00 L sample of NH 3 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 = P 1 V 1 T 2 / P 2 T 1 = (100.0 kPa)(3.00 L)(300. K) / (80.0 kPa)(500. K) = 2.25 L
Gay Lussacs Law of Combining Volumes N2N2 1 volume H 2 3 volumes 2 NH 3 2 volumes When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.
Gay Lussacs Law of Combining Volumes When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.
Avogadros Law Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.
Volume of one mole of any gas at STP = 22.4 L.STP 22.4 L at STP is known as the molar volume of any gas. Mole-Mass-Volume Relationships
V P nT atmospheres
Determination of Density Using the Ideal Gas Equation Density = mass/volume D = MP/ RT
Mole fraction (X): Ratio of moles of one component to the total moles in the mixture (dimensionless, similar to a %) 184 Ex: What are the mole fractions of H 2 and He in the previous example?
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: P T = P gas + P H2O P gas = P T – P H2O 185 See Appendix B for vapor pressures of water at different temperatures.
Grahams 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 = M B Rate of effusion of B M A
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 H 2 SO 4 are needed to completely neutralize 30.0 mL of M KOH?
Concentration Molarity Parts per Million Percent by Mass Percent by Volume
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
Parts Per Million 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 (its OK!)
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%
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%
Colligative Properties Vapor Pressure Lowering B.P. Elevation DT f = m. k f (or D T b = m. k b ) 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 C 6 H 12 O 6 or 1M Na 3 PO 4 ?
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 234 cm cm2 _ cm cm cm5 Measurement# of Sig Figs cm cm cm cm cm5
What Precision? A numbers precision is determined by the furthest (smallest) place the number is recorded to mL : thousands place mL : ones place mL : tenths place 5.30 mL : hundredths place 8.7 mL : tenths place mL : thousandths place
Rounding with addition and subtraction Answers are rounded to the least precise place.
Rounding with multiplication and division Answers are rounded to the fewest number of significant figures.
Metric Conversions Determine how many powers of ten difference there are between the two units (no prefix = 10 0 ) 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
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 T Solving for T
Solving for q How many joules are absorbed by grams of water in a calorimeter if the temperature of the water increases from 20.0 o C to 50.0 o C? q = mC T = (100.0 g)(4.18 J/g o C)(30.0 o C) = J
Solving for m A sample of water in a calorimeter cup increases from 25 o C to 50. o C by the addition of joules of energy. What is the mass of water in the calorimeter cup? q = mC T, so m = q / C T = (500.0 J) / (4.18 J/g o C)(25 o C) = 4.8 g
Solving for T If a 50.0 gram sample of water in a calorimeter cup absorbs joules of energy, how much will the temperature rise by? q = mC T, so T = q / mC = ( J)/(50.0 g)(4.18 J/g o C) = 4.8 o C If the water started at 20.0 o C, what will the final temperature be? – Since the water ABSORBS the energy, its temperature will INCREASE by the T: 20.0 o C o C = 24.8 o C
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.
Heat of Reaction Reactions either absorb PE (endothermic, + H) or release PE (exothermic, - H) Exothermic, PE KE, Temp Endothermic, KE PE, Temp Rewriting the equation with heat included: 4 Al(s) + 3 O 2 (g) 2 Al 2 O 3 (s) kJ N 2 (g) + O 2 (g) kJ 2 NO(g)
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 (q p ) of a reaction at constant pressure H = E + PV H = E + P V(at constant P) H = (q p + w) + (-w) H = q p 205
Sign conventions H > 0 Heat is gained from surroundings + H in endothermic reaction H < 0 Heat is released to surroundings - H in exothermic reaction 206
5.4: Enthalpy of Reaction ( H rxn ) Also called heat of reaction: 1.Enthalpy is an extensive property (depends on amounts of reactants involved). Ex: CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) H rxn = kJ – Combustion of 1 mol CH 4 produces 890. kJ … of 2 mol CH 4 (2)(-890. kJ) = kJ What is the H of the combustion of 100. g CH 4 ? 207
2. H reaction = - H reverse reaction CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) H = kJ CO 2 (g) + 2 H 2 O (l) CH 4 (g) + 2 O 2 (g) H = kJ 208
5.6: Hess Law If a rxn is carried out in a series of steps, H rxn = ( H steps ) = H 1 + H 2 + H 3 + … 209 Ex. What is H rxn of the combustion of propane? C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) 3 C (s) + 4 H 2 (g) C 3 H 8 (g) H 1 = kJ C (s) + O 2 (g) CO 2 (g) H 2 = kJ H 2 (g) + ½ O 2 (g) H 2 O (l) H 3 = kJ H rxn = ( ) + 4( ) = kJ 3[ ] 3( ) 4[ ] 4( ) Germain Hess ( ) C 3 H 8 (g) 3 C (s) + 4 H 2 (g) H 1 = kJ
5.7: Enthalpy of Formation ( H f ) Formation: a reaction that describes a substance formed from its elements NH 4 NO 3 (s) Standard enthalpy of formation ( H f ): forms 1 mole of compound from its elements in their standard state (at 298 K) C 2 H 5 OH (l) H f = kJ – H f of the most stable form of any element equals zero. H 2, N 2, O 2, F 2, Cl 2 (g) Br 2 (l), Hg (l) C (graphite), P 4 (s, white), S 8 (s), I 2 (s) 210 Ex: 2 N 2 (g) + 4 H 2 (g) + 3 O 2 (g) 2 2 C (graphite) + 3 H 2 (g) + ½ O 2 (g)
Hess Law (again) 211 Ex. Combustion of propane: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) Given:Compound H rxn (kJ/mol) C 3 H 8 (g) CO 2 (g) H 2 O (l) H 2 O (g) H rxn = [3( ) + 4( )] – [1( ) + 5(0)] = kJ
5.5: Calorimetry Measurement of heat flow Heat capacity, C: amount of heat required to raise T of an object by 1 K q = C T Specific heat (or specific heat capacity, c): heat capacity of 1 g of a substance q = m c T 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 T = (40.0 g)(0.45 J/(g K))(100.0 – 0.0 ºC) = 1800 J 212
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) – H products - H reactants = H EXAMPLESEXAMPLES
Making a PE Diagram X axis: Reaction Coordinate (time, no units) Y axis: PE (kJ) Three lines representing energy (H reactants, H activated complex, H products ) Two arrows representing energy changes: – From H reactants to H activated complex : Activation Energy – From H reactants to H products : H ENDOTHERMIC PE DIAGRAM EXOTHERMIC PE DIAGRAM
Endothermic PE Diagram If a catalyst is added?
Endothermic with Catalyst The red line represents the catalyzed reaction.
Exothermic PE Diagram What does it look like with a catalyst?
Exothermic with a Catalyst The red line represents the catalyzed reaction. Lower A.E. and faster reaction time!
19.1: Spontaneous Processes Reversible reaction: can proceed forward and backward along same path (equilibrium is possible) Ex: H 2 O 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 219
Entropy Entropy (S): a measure of molecular randomness or disorder – S is a state function: S = S final - S initial + S = more randomness - S = less randomness – For a reversible process that occurs at constant T: – Units: J/K 220
Examples of spontaneous reactions: Gases expand to fill a container: Ice melts at room temperature: Salts dissolve in water: 221 Particles are more evenly distributed Particles are no longer in an ordered crystal lattice Ions are not locked in crystal lattice
19.3: 3 rd Law of Thermodynamics The entropy of a crystalline solid at 0 K is 0. How to predict S: S gas > S liquid > S solid S more gas molecules > S fewer gas molecules S high T > S low T Ex: Predict the sign of S for the following: 1.CaCO 3 (s) CaO (s) + CO 2 (g) 2.N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 3.N 2 (g) + O 2 (g) 2 NO (g) 222 +, solid to gas -, fewer moles produced ?
19.5: Gibbs free energy, G Represents combination of two forces that drive a reaction: H (enthalpy) and S (disorder) Units: kJ/mol G = H - T S G° = H° - T S° (absolute T) 223 Called free energy because G represents maximum useful work that can be done by the system on its surroundings in a spontaneous reaction. (See p. 708 for more details.) Josiah Willard Gibbs ( )
Determining Spontaneity of a Reaction If G is: reaction is spontaneous (proceeds in the forward direction Positive Forward reaction is non-spontaneous; the reverse reaction is spontaneous ZeroThe system is at equilibrium 224
19.6: Free Energy & Temperature G depends on enthalpy, entropy, and temperature: G = H - T S H S G and reaction outcome -+Always (-); spontaneous at all T 2 O 3 (g) 3 O 2 (g) +-Always +; non-spontaneous at all T 3 O 2 (g) 2 O 3 (g) --Spontaneous at low T; non-spontaneous at high T H 2 O (l) H 2 O (s) ++Spontaneous only at high T ; non-spontaneous at low T H 2 O (s) H 2 O (l) 225
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. Solubility 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. Supersaturated Unsaturated: There are still solvent molecules available to dissolve more solute, so more can dissolve. Unsaturated How ionic solutes dissolve in water: polar water molecules attach to the ions and tear them off the crystal. How ionic solutes dissolve in water:
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 H 2 O at that particular temperature. At 40 o C, the solubility of KNO 3 in 100g of water is 64 g. In 200g of water, double that amount. In 50g of water, cut it in half.
Supersaturated If 120 g of NaNO 3 are added to 100g of water at 30 o C: 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 24 o C (to 54 o C), the excess solute would dissolve.
Unsaturated If 80 g of KNO 3 are added to 100g of water at 60 o C: 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 12 o C (to 48 o C), the solution would become saturated
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
Formulas, Naming and Properties of Acids Arrhenius Definition of Acids: molecules that dissolve in water to produce H 3 O + (hydronium) as the only positively charged ion in solution. HCl (g) + H 2 O (l) H 3 O + (aq) + Cl - Properties of Acids Naming of Acids Formula Writing of Acids
Properties of Acids Acids react with metals above H 2 on Table J to form H 2 (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 (NO 2, SO 2, CO 2 ) react with water for form acids. This is how acid rain forms from auto and industrial emissions.
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. Dont forget the (aq) after it…its only an acid if its in water! Hydrosulfuric acid: H +1 and S -2 = H 2 S (aq) Carbonic acid: H +1 and CO 3 -2 = H 2 CO 3 (aq) Chlorous acid: H +1 and ClO 2 -1 = HClO 2 (aq) Hydrobromic acid: H +1 and Br -1 = HBr (aq) Hydronitric acid: Hypochlorous acid: Perchloric acid:
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
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.
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 NH 3 (aq) is the same thing as NH 4 OH: NH 3 + H 2 O NH 4 OH Also called ammonium hydroxide.
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 =
Neutralization H +1 + OH -1 HOH Acid + Base Water + Salt (double replacement) HCl (aq) + NaOH (aq) HOH (l) + NaCl (aq) H 2 SO 4 (aq) + KOH (aq) 2 HOH (l) + K 2 SO 4 (aq) HBr (aq) + LiOH (aq) H 2 CrO 4 (aq) + NaOH (aq) HNO 3 (aq) + Ca(OH) 2 (aq) H 3 PO 4 (aq) + Mg(OH) 2 (aq)
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.
16.2: Dissociation of Water Autoionization of water: H 2 O (l) H + (aq) + OH - (aq) 241 K W = ion-product constant for water H 3 O + (aq) or H + (aq) = hydronium
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.
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. HNO 3 + H 2 O H 3 O +1 + NO 3 -1 – Since HNO 3 lost an H +1 during the reaction, it is an acid. – Since H 2 O gained the H +1 that HNO 3 lost, it is a base.
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: NH 3 + BF 3NH 3 BF 3 Lewis baseLewis acidLewis salt 6 CN - + Fe 3+ Fe(CN) 6 3- Lewis baseLewis acidCoordination compound 244 Gilbert N. Lewis (1875 – 1946)
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, P X – Solutes in liquids: molarity, [X] 245
Reversible Reactions and Rate 246 Reaction Rate Time Backward rate Forward rate Equilibrium is established: Forward rate = Backward rate When equilibrium is achieved: [A] [B] and k f /k r = K eq
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 (K eq ): 247 K eq is strictly based on stoichiometry of the reaction (is independent of the mechanism). Units: K eq is considered dimensionless (no units) Cato Guldberg Peter Waage ( ) ( ) or
Relating K c and K p Convert [A] into P A : 248 where n = = change in coefficents of products – reactants (gases only!) = (c+d) - (a+b)
Magnitude of K eq Since K eq [products]/[reactants], the magnitude of K eq predicts which reaction direction is favored: – If K eq > 1 then [products] > [reactants] and equilibrium lies to the right – If K eq < 1then [products] < [reactants] and equilibrium lies to the left 249
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 QK, reaction proceeds to left, until Q=K 250
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)= K 1 X K 2 251
15.3: Types of Equilibria Homogeneous: all components in same phase (usually g or aq) N 2 (g) + H 2 (g) NH 3 (g) Fritz Haber (1868 – 1934)
Heterogeneous: different phases CaCO 3 (s) CaO (s) + CO 2 (g) Definition:What we use: 253 Even though the concentrations of the solids or liquids do not appear in the equilibrium expression, the substances must be present to achieve equilibrium. Concentrations of pure solids and pure liquids are not included in K eq expression because their concentrations do not vary, and are already included in K eq (see p. 548).
15.4: Calculating Equilibrium Constants Steps to use ICE table: 1.I = Tabulate known initial and equilibrium concentrations of all species in equilibrium expression 2.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 3.E = Calculate the equilibrium concentrations 254
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 [OH 1- ] is 4.64 x M. Calculate K eq at 25ºC for the reaction: NH 3 (aq) + H 2 O (l) NH 4 1+ (aq) + OH 1- (aq) 255
NH 3 (aq) + H 2 O (l) NH 4 1+ (aq) + OH 1- (aq) Initial Change Equilibrium M - x M 0 M + x 4.64 x M NH 3 (aq) H 2 O (l) NH 4 1+ (aq)OH 1- (aq) X X X x = 4.64 x M
Equilibrium (c) 2006, Mark Rosengarten When the rate of the forward reaction equals the rate of the reverse reaction.
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. – H 2 O (l) H 2 O (g) Phase equilibrium: At the melting point, the rate of solid turning to liquid equals the rate of liquid turning back to solid. – H 2 O (s) H 2 O (l)
Le Châteliers 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 NEXT
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 NEXT
Examples For the reaction N 2 (g) + 3H 2 (g) 2 NH 3 (g) + heat – Adding N 2 will cause the equilibrium to shift RIGHT, resulting in an increase in the concentration of NH 3 and a decrease in the concentration of N 2 and H 2. – Removing H 2 will cause a shift to the LEFT, resulting in a decrease in the concentration of NH 3 and an increase in the concentration of N 2 and H 2. – 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.
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.
16.6: Weak Acids Weak acids partially ionize in water (equilibrium is somewhere between ions and molecules).HA (aq) A - (aq) + H + (aq) 263 K a = acid-dissociation constant in water Weak acids generally have K a < See Appendix D for full listing of K a values
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) + H 2 O (l) HA (aq) + OH- (aq) – Cation (B + ) = a conjugate acid B + (aq) + H 2 O (l) BOH (aq) + H + (aq) 290
17.1: Common Ion Effect Addition of a common ion: solubility of solids decrease because of Le Châteliers principle. Ex: AgCl (s) Ag + (aq) + Cl - (aq) – Addition of Cl - shifts equilibrium toward solid 291
17.4: Solubility Equilibria Dissolving & precipitating of salts – Solubility rules discussed earlier are generalized qualitative observations of quantitative experiments. Ex:PbCl 2 (s) Pb 2+ (aq) + 2 Cl - (aq) K sp = [Pb 2+ ][Cl - ] 2 = 1.6 x K sp = 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. 292
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: HC 2 H 3 O 2 + NaC 2 H 3 O 2 – Or consist of a weak base and its conjugate acid (usually in salt form) Ex: ammonia and ammonium chloride: NH 3 + NH 4 Cl 301
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? M KOH
Titration of a Weak Base and Strong Acid -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 pH Volume of HCl added (mL)
Redox: Reduction occurs when an atom gains one or more electrons. Ex: Oxidation occurs when an atom or ion loses one or more electrons. Ex: LEO goes GER Copper metal reacts with silver nitrate to form silver metal and copper nitrate: Cu + 2 Ag(NO 3 ) 2 Ag + Cu(NO 3 ) 2.
Identifying OX, RD, SI Species Ca H +1 Cl -1 Ca +2 Cl H 2 0 Oxidation = loss of electrons. The species becomes more positive in charge. For example, Ca 0 Ca +2, so Ca 0 is the species that is oxidized. Reduction = gain of electrons. The species becomes more negative in charge. For example, H +1 H 0, 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.
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?
Agents Ca H +1 Cl -1 Ca +2 Cl H 2 0 Since Ca 0 is being oxidized and H +1 is being reduced, the electrons must be going from the Ca 0 to the H +1. Since Ca 0 would not lose electrons (be oxidized) if H +1 werent there to gain them, H +1 is the cause, or agent, of Ca 0 s oxidation. H +1 is the oxidizing agent. Since H +1 would not gain electrons (be reduced) if Ca 0 werent there to lose them, Ca 0 is the cause, or agent, of H +1 s reduction. Ca 0 is the reducing agent.
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.
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 H 2 O 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.
Writing Half-Reactions Ca H +1 Cl -1 Ca +2 Cl H 2 0 Oxidation: Ca 0 Ca e - Reduction: 2H e - H 2 0 The two electrons lost by Ca 0 are gained by the two H +1 (each H +1 picks up an electron). PRACTICE SOME!
Practice Half-Reactions Dont forget to determine the charge of each species first! 4 Li + O 2 2 Li 2 O Oxidation Half-Reaction: Reduction Half-Reaction: Zn + Na 2 SO 4 ZnSO Na Oxidation Half-Reaction: Reduction Half-Reaction:
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.
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 E cell. The standard for measuring the cell potentials is called a SHE (Standard Hydrogen Electrode). Description of SHE (Standard Hydrogen Electrode) Reaction2H + (aq, 1M) + 2e - H 2(g, 101kPa) E 0 = 0.00 V
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?
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.metals 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.
Metal Activity 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 + FeCl 3 3 KCl + Fe WILL happen, because K is being oxidized, and that is what Table J says should happen. The reaction Fe + 3 KCl FeCl K will NOT happen. 3 K 0 + Fe +3 Cl -1 3 REACTION Fe K +1 Cl -1 NO REACTION
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
Voltaic Cells Produce electrical current using a spontaneous redox reaction Used to make batteries!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)
Making Voltaic Cells
Electrolytic Cells Use electricity to force a nonspontaneous redox reaction to take place. Uses for Electrolytic Cells: – Decomposition of Alkali Metal Compounds Decomposition of Alkali Metal Compounds – Decomposition of Water into Hydrogen and Oxygen Decomposition of Water into Hydrogen and Oxygen – Electroplating 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
Decomposing Alkali Metal Compounds 2 NaCl 2 Na + Cl 2 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)
Decomposing Water 2 H 2 O 2 H 2 + O 2 The H + is reduced at the (-) cathode, yielding H 2 (g), which is trapped in the tube. The O -2 is oxidized at the (+) anode, yielding O 2 (g), which is trapped in the tube.
Electroplating The Ag 0 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 Ag 0, which coats on to the ring.
Spontaneity of Redox Reactions: E 0 = E 0 red( reduction process-cathode) – E 0 red (oxidation process-anode) A positive value of E 0 indicates a spontaneous process and a negative value of E 0 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 E 0 Ex:
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 Cs, saturated) Alkanes – Alkenes (1 double bond between 2 Cs, unsaturated) Alkenes – Alkynes (1 triple bond between 2 Cs, unsaturated) Alkynes These are called aliphatic, or open-chain, hydrocarbons. Count the number of carbons and add the appropriate suffix!
Alkanes CH 4 = methanemethane C 2 H 6 = ethaneethane C 3 H 8 = propanepropane C 4 H 10 = butanebutane C 5 H 12 = pentanepentane To find the number of hydrogens, double the number of carbons and add 2.
Methane Meth-: one carbon -ane: alkane The simplest organic molecule, also known as natural gas!
Ethane Eth-: two carbons -ane: alkane
Propane Prop-: three carbons -ane: alkane Also known as cylinder gas, usually stored under pressure and used for gas grills and stoves. Its also very handy as a fuel for Bunsen burners!
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!
Pentane Pent-: five carbons -ane: alkane Your Turn!!! Draw Hexane: Draw Heptane:
Alkenes C 2 H 4 = EtheneEthene C 3 H 6 = PropenePropene C 4 H 8 = ButeneButene C 5 H 10 = PentenePentene To find the number of hydrogens, double the number of carbons.
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!
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!
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.
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.
Alkynes 4 C 2 H 2 = EthyneEthyne 4 C 3 H 4 = PropynePropyne 4 C 4 H 6 = ButyneButyne 4 C 5 H 8 = PentynePentyne 4 To find the number of hydrogens, double the number of carbons and subtract 2.
Ethyne Now, try to draw propyne! Any isomers? Lets see!Lets see Also known as acetylene, used by miners by dripping water on CaC 2 to light up mining helmets. The carbide lamps were attached to miners 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 3000 o C!
Propyne This is propyne! Nope! No isomers. OK, now draw butyne. If there are any isomers, draw them too.butyne
Butyne Well, heres 1-butyne! And heres 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!pentyne
Pentyne 1-pentyne 2-pentyne Now, draw all of the possible isomers for hexyne! Naming: Check this link out es/351/orgnom/ja vanom/nomenclat ure2StartPage.ht m
Isomers Isomers are compounds that have same molecular formula (same number of atoms) but a different structure. There are three types of isomers: 1.Structural Isomers: Same number of atoms, arranged differently. 2.Geometric Isomers (Cis- trans-): Happens in = or triple bonded compounds since these are inflexible bonds. Ex. 3.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.
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 Alkyl groups – Halide groups Halide groups – Other functional 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.
Organic Families Each family has a functional group to identify it. – Alcohol (R-OH, hydroxyl group) Alcohol – Organic Acid (R-COOH, primary carboxyl group) Organic Acidprimary – Aldehyde (R-CHO, primary carbonyl group) Aldehydeprimary – Ketone (R 1 -CO-R 2, secondary carbonyl group) Ketonesecondary – Ether (R 1 -O-R 2 ) Ether – Ester (R 1 -COO-R 2, carboxyl group in the middle) Ester – Amine (R-NH 2, amine group) Amine – Amide (R-CONH 2, amide group) Amide These molecules are alkanes with functional groups attached. The name is based on the alkane name.
Alcohol On to DI and TRIHYDROXY ALCOHOLS
Di and Tri- hydroxy Alcohols
Positioning of Functional Group PRIMARY (1 o ): the functional group is bonded to a carbon that is on the end of the chain. SECONDARY (2 o ): The functional group is bonded to a carbon in the middle of the chain. TERTIARY (3 o ): The functional group is bonded to a carbon that is itself directly bonded to three other carbons.
Organic Acid These are weak acids. The H on the right side is the one that ionized in water to form H 3 O +. The -COOH (carboxyl) functional group is always on a PRIMARY carbon. Can be formed from the oxidation of primary alcohols using a KMnO 4 catalyst.
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.*
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 KMnO 4 or PCC catalyst.*
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.etherfication
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.esterification
Amine - Component of amino acids, and therefore proteins, RNA and DNA…life itself! - Essentially ammonia (NH 3 ) with the hydrogens replaced by one or more hydrocarbon chains, hence the name amine!
Amide Synthetic Polyamides: nylon, kevlar Natural Polyamide: silk! For more information on polymers, go here.go here.
Combustion Happens when an organic molecule reacts with oxygen gas to form carbon dioxide and water vapor. Also known as burning.
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. C 2 H 6 + Cl 2 C 2 H 5 Cl + HCl The second Cl can then substitute for another H: C 2 H 5 Cl + HCl C 2 H 4 Cl 2 + H 2
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
Etherification* Alcohol + Alcohol Ether + Water A dehydrating agent (H 2 SO 4 ) removes H from one alcohols 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.
Esterification Organic Acid + Alcohol Ester + Water A dehydrating agent (H 2 SO 4 ) removes H from the organic acid and removes the OH from the alcohol. The two molecules join where there H and OH were removed.
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.
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
Addition Polymers Joining monomers together by breaking double bonds Polyvinyl chloride (PVC): vinyl siding, PVC pipes, etc. Vinyl chloride polyvinyl chloride n C 2 H 3 Cl -(-C 2 H 3 Cl-)- n Polytetrafluoroethene (PTFE, teflon): TFE PTFE n C 2 F 4 -(-C 2 F 4 -)- n
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.
Rubber The process of toughing rubber by cross-linking the polymer strands with sulfur is called...called...