lOMoARcPSD|12129262 Gr.12 chemistry - Exam Review Chemistry (High School - Canada) StuDocu is not sponsored or endorsed by any college or university Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 1 of 67 Chemistry - Reference page Partial Charges: • 𝒮+ (partial postive) • if something is a partial positive it’s electronegativity will be less than the other elements electronegativity • if the molecule has more than two elements you look at the individual bonds • 𝒮- (partial negative) • if something is a partial negative it’s electronegativity will be higher than the other elements Partial Charges: • Bonds • to know if a molecule has polar or non-polar bonds you look at the difference in electronegativity • if the difference is above 0.3 its polar bonds • if the difference is below 0.3 its non-polar bonds • Molecules • to know if a molecule is polar or non-polar, you can look at its symmetry • if its symmetrical the molecule is non-polar • if its asymmetrical the molecule is polar Forces: • London • if the molecule is non-polar you will only have london forces • Dipole Dipole • if the molecule is polar you will have dipole dipole forces (only if you don’t have hydrogen bonds) Hydrogen Bonds • • if you only have a H with a 𝒮+and a N, O, or F with a 𝒮- you will have a hydrogen bond Density: Example; d = Mass ——— Volume dwater = 1.00g/mL ( = 1.00g/mL³) ( = 1mL = 1 g/cm³) Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 2 of 67 Chemistry History of Atomic Theories: (https://www.youtube.com/watch?v=-4Us5PTb4j8) Scientist Name Dates Theory of Model Diagram of Model - proposed that all matter was Dalton 1803 composed of atoms, indivisible and indestructible building blocks - he was the first to discover a sub atomic particle, the electron Thompson 1897 - proposed the plum pudding model of the atom in 1904 - in Thompson’s model the atom is composed of electrons - gold foil experiment in which he Rutherford 1911 demonstrated that the atom ha a tiny and heavy nucleus - depicts the atom as a small, positively Bohr 1913 charged nucleus surrounded by electrons that travel in circular orbits around the nucleus Ions: ion - an atom that loses or gains electrons (to form a full valence shell) and results in a charge. It is a charged particle. • Cations • lose electrons • become more positive • Anions • gains electrons • becomes more negative orbitals - 3D space around a nucleus where there is a probability of finding electrons. Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 3 of 67 Bohr Rutherford Diagram: Line Spectra - Bohr’s Evidence for Electron Energy Levels Energy Shells & Sub-Shells: Principal Energy Level (shell) Bohr’s First Postulate: • Main energy levels for electrons • electrons do not radiate energy as they orbit the • Corresponds to the period on the periodic table nucleus. each orbital corresponds to a state of constant energy (stationary state) • example; period 1 is principal energy level 1 Bohr’s Second Postulate: Electron Energy Sub-Levels (sub-shells) • electrons can change their energy only by • orbital - 3D space mourned a nucleus undergoing a transition from one stationary state where there is a probability of finding to another. electrons Summary: Orbital Principal (SubEnergy Shell) Level • electrons absorb and emit very specific amounts Shape # of orbitals Max # of electrons S 1+ spherical 1 2 P 2+ dumb-bell like 3 5 D 3+ shape can vary 5 10 F 4+ shape can vary 7 14 of energy. photon - a quantum of light energy quantum - a small, discrete (specific), indivisible quantity. Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 4 of 67 Energy Level Diagrams & Electron Configurations: - electrons in atoms have different energies. Each (neutral) type of atom has a unique number of electrons. However, many elements have similar properties. scientists has studied atomic structure (electron configuration or arrangement) in order to understand properties. 3 Rules for Energy Level Diagrams: • Aufbau Principal (means “building up”) • each electron is added to the lowest energy level available. Hund’s Rule • • for orbitals at the same energy level, one electron occupies each sub orbital before the electrons pair up. • Pauli Exclusion Principal • only two electrons (with opposite spins) can occupy each orbital. Electron Configuration For Ions: Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 5 of 67 Lewis Bonding Theory: • atoms/ions are stable if they have an octet of electrons • electrons are most stable in pairs • atoms form chemical bonds (ionic or covalent) to become stable Linking Lewis Theory to Quantum Mechanics: • the “octet” of electrons in Lewis Theory comes from the maximum of 2 electrons in the S orbital and 6 in the P orbital. Explaining Ionic Charges: • Zinc forms a 2+ ion. The electron configuration for this is: [Ar]4s², 3d¹⁰ this shows that there are 12 valence electrons. If another atom were to take the 4s electrons this would leave zinc with a filled 3d orbital - a relatively stable state. Electron Configuration for Ions: Rules for Drawing Lewis Structures: STEP 1: arrange the atoms symmetrically around the central atoms - usually first atom - usually not oxygen - never hydrogen STEP 2: count the numbers of total valence electrons (for polyatomic ions, add or substance as necessary) STEP 3: place one bonded pair of electrons (straight line) between the central atom and each surrounding atom STEP 4: complete the octet of each surrounding atom using lone pairs of electrons (any remaining electron pairs go on central atom) STEP 5: if central atom doesn’t have an octet, move lone pairs of electrons from a surrounding atom to form double/triple bonds STEP 6: draw the final lewis structure. (polyatomic ion: enclose in square brackets and write ion charges outside of bracket at top right). Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 6 of 67 Exceptions to the Octet Rule: BeCl2 (*NOT ionic*) BCl3 (*some elements have too few electrons to form an octet (H, Be, B)*) SF6 (*some elements have more than an octet*) PCl5 (*◣ direction is back into the page*) (*\\\ direction is out of the page (toward you)*) (add 3-D concept to 2-D drawing) XeF6 (*some noble gases can form compounds*) Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 7 of 67 The VSEPR Theory: • VSEPR (Valence Shell Electron Pair Repulsion) • valence electrons stay as far away as possible to minimize repulsion • when looking at a molecule we look specifically at the central atom (the one that has the most bonding electrons) to determine the 3-D geometry Rules of VSEPR: • • • • • only the valence shell electrons of the central atom(s) are important to molecular shape valence shell electrons are paired or will be paired in a molecule or polyatomic ion bonded pairs of electrons and lone pairs of electrons are treated approximately equally valence shell electron pairs repel each other electrostatically the molecular shape is determined by the positions of the electron pairs when they are a maximum distance apart (with the lowest repulsion possible) Representing 3-D Structures: • the solid wedge represents atom/pairs of electrons coming TOWARDS the viewer • the dashed wedge represents atom/pairs of electrons going AWAY from the viewer How to Use VSEPR: 1. draw lewis structures of the molecule, including the electron pairs around the central atom 2. count the total number of bonding pairs and lone pairs around the central atom 3. use the chart provided to determine the geometry (shape name) and draw a shape diagram Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 8 of 67 Polarity: Polar Bonds: • Polar bonds are created if there is an equal attraction of an atom for electrons (electronegativity). • Use periodic table to look up electronegativity values for each element in a bond, then determine the difference. • < 0.45 → non-polar bond (covalent) • 0.45-1.7 → polar bond (covalent) • 1.7+ → ionic bond (metal and nonmetal - involve electron transfer, not sharing) Examples; HBr BH3 ▲ENEG = 3.0 - 2.2 = 0.8 ∴ there is polar bonds the molecule is polar as well ▲ENEG = 2.0 - 2.2 = 0.2 ∴ there is non- polar bonds the molecule is non-polar as well Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 9 of 67 Polarity: Polar Molecules: • Bond Dipole - the electronegativity difference of two bonded atoms represented by an arrow pointing from the lower (𝒮-) to the higher electronegativity (𝒮+) • Polar Molecule - a molecule that has polar bonds who dipoles that do not cancel to zero • overall a polar molecule has a pastiche end and a negative end • if the molecule is asymmetrical (for example has a lone pair of electrons) it will be polar Non-Polar Molecules: - a molecule that has either non-pear bonds, or polar bonds whose bond dipoles cancel to zero - non-polar molecules that have polar bonds are completely symmetrical molecules Theoretical Prediction of Molecular Polarity: STEP 1: draw a lewis structure for the molecule STEP 2: use the number of electron pairs and VSEPR rules to determine the shape around each central atom STEP 3: use electronegativity differences to determine the polarity of each bond STEP 4: determine if there is an overall dipole for the molecule (example; is it symmetrical or asymmetrical?) STEP 5: classify as polar or non-polar Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 10 of 67 Intermolecular Forces: • intermolecular forces are forces of attraction (and repulsion) between molecules • they involve an electrostatic force (positive and negative charges) • overall they are weak compared to intramolecular forces (ionic and covalent bonds). if we say that a covalent bond has a strength of 100 then intermolecular forces would range from 0.001 to 15. London Dispersion Forces: • Simultaneous attraction of electrons in one molecule for the nuclei in a different molecule • all types of molecules exhibit london forces • they are the weakest of all the intermolecular forces • they are very weak because electrons are always moving so the attraction comes and goes as they move around. Dipole Dipole Forces: • exist between polar molecules. (a polar molecule is created when there is enough of a difference in electronegativity between atoms in a molecule). • occurs when the dipole of one molecule is attracted to the dipole in a different molecule • are stronger than london forces • an ion-dipole exists between a dipole in a molecule and an ion Hydrogen Bonds: • exists when a strong enough dipole is created in molecules (hydrogen atoms are bonded to highly electronegative atoms like N, O, or F) • the hydrogen (+ charge) on one molecule is attracted to a pair of lone electrons (- charge) on an adjacent molecule • are the strongest of all the intermolecular forces Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 11 of 67 Types of Solids: • Ionic Solid - a 3-D lattice of cations and anions, held together by ionic bonds (strong electrostatic forces), with an overall neutral charge. - particles involved: - ions (cations & anions) - forces involved: - ionic (both within & between) - properties: - hard - strong force (crystal lattice) - brittle - once initially broken ions easily repel each other - high melting point - strong force, lots of energy needs to break the bonds - soluble in water - ions form forces of attraction with water (polar) - conduct electricity - can easily transfer electrons Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 12 of 67 • Metallic Solid - a lot us of nuclei surrounded by a “sea of electrons”. Held together by metallic bonds. - particles involved: - cations (with a sea of electrons) - forces involved: - metallic (both within & between) - properties: - soft - very hard - flexible, but strong bonds (sea of electrons) - conduct electricity - solids conduct {valence electrons can freely move around} - liquids conducts - ductile/malleable - non-directional bonds mean planes of atoms can slide over each other and remain bonded - lustrous - valence electrons absorb and emit energy from light Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 13 of 67 • Molecular Solid - separate molecules attracted by week into molecular forces called van der Waals forces. If molecules are polar, they are attracted by stronger intermolecular forces, called dipoledipole forces. - particles involved: - molecule - forces involved: - intramolecular forces (covalent bonds - intermolecular forces (london, dipole-dipole, & hydrogen bonds) - properties: - soft - weak intermolecular forces - low melting point - not much energy required to break these forces - some soluble in water - can form adequate electrostatic forces (which is whats needed to dissolve in water - non conducting electricity - individual particles are neutral (no ions) Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 14 of 67 Covalent Network Solid - large molecules held together by continuous covalent bonds. - particles involved: - atoms (neautral) - forces involved: - covalent (both within & between) - properties: - very hard - strong force between all atoms (crystal structure) - very high melting point - strong force between all atoms (crystal structure) - not soluble in water - there bonds are too strong to break - non conduct electricity - neutral particles Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 15 of 67 Organic Chemistry Organic Chemistry: organic chemistry - is the study of structure, properties, and reactions of organic molecules (contains carbon and hydrogens). • it is important because: • they make up living things • medical applications • huge variety of compounds • different structures • different properties • different uses • carbon features • can form 4 covalent bonds • can form single, double and triple bonds IUPAC Naming: • international union of pure and applied chemistry • specific naming system Chemical Formula: • C5H10O2 • CH3(CH2)3COOH Structural Formula: or Organic Family: organic family - a group of organic compounds with common structural features that import certain physical and chemical properties Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 16 of 67 Functionals Groups: functional group - a structural arrangement of atoms hydrocarbons - only hydrogens and carbons example; methane, ethene, ethyne Alcohols Aldehydes Ketones Structure DICTATES Function Carboxylic Acid Esters Aromatic Ethers Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 17 of 67 Naming Hydrocarbons: hydrocarbon - compound that only contains hydrogen and carbon Alkane - hydrocarbon with only single carbon bonds Alkene - hydrocarbon with at least one double carbon bond Alkyne - hydrocarbon with at least one triple carbon bond Cyclic Hydrocarbon - closed ring structure Aromatic Compound - structure based on benzene Properties of Benzene Uses of Benzene Colorless manufacturing of: Flammable • plastics toxic • dyes melting point: 5.5℃ • rubber boiling point: 80.1℃ • medications CH4 C2H6 C4H10 C7H16 Downloaded by robert joel (jordanbradford2004@icloud.com) >> >> >> >> methane (natural gas) ethane butane (lighters) heptane lOMoARcPSD|12129262 18 of 67 Hydrocarbons can be straight chained or branched: General prefixes used in IUPAC Naming: meth - 1 hex - 6 eth - 2 hept - 7 prop - 3 oct - 8 but - 4 non - 9 pent - 5 dec - 10 4. use the prefix that corresponds to the number of carbon atoms in the chain 5. use the suffix “ane” example; Example - Structural Diagrams of Hexane: Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 19 of 67 Rules for Branched Alkanes: • branches alkanes have alkyl groups • these groups are named using the organic prefixes, with the suffix “yl” • • • • • CH3 CH2 - CH3 CH2 - CH2 - CH3 C7H15 C9H19 >> methyl >> ethyl >> propyl >> heptyl >> nonyl Rules Examples 1. identify the longest carbon chain. (this may travel through one or more branches - need to check all possibilities). Heptane (chain) 2. number the carbons starting with the end closest to the branches 3. name each branch (alkyl group) and identify its location on the parent chain (always wants the lowest numbering) • ethyl off carbon 5 • methyl off carbon 3 4. write the name according to the general format: (number of location) - (branch name)(parent chain) * if more than one branch is present, list them in alphabetical order according to ally group * is the same alkyl group is present more than once use di, tri, tetra etc. before the alkyl name to indicate how many there are. * list each number location on the parent chain at the start of naming. 5 - ethyl - 3 - methylheptane Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 20 of 67 Naming Alkenes & Alkynes: * the rules are similar to alkanes, but use the endings “ene” for double bonds and “yne” for triple bonds Rules Examples 1. find the longest parent chain that includes the double or triple bond 2. number the carbon atoms starting at the end closest to the double or triple bond 3. the location of the multiple bond is indicated by the number of the C atom that starts the bond • C = C off C1 4. identify any branches and name accordingly • methyl off carbon 3 5. write the full name following proper conversions * hyphens are used between numbers and words 3 - methyl - 1 - butene Aromatic Hydrocarbons: * benzene ring is the parent molecule * when a branch is not easily named, benzene can become the branch. it is then called phenyl Name of Molecule Structure methyl benzene 1,2 - diethyl benzene 1,3 - diethyl benzene 1,4 - diethyl benzene 2 - phenylbutane Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 21 of 67 Reactions of Hydrocarbons: • Combustion fuel hydrocarbon + oxygen gas → carbon dioxide + water C6H14(l) + 19O2 → 12CO2 + 14H2O • Substitution • reaction with a halogen: → + + Hydrogen Bromide ethane Bromoethane • reaction of an aromatic hydrocarbon with a halogen: → + cyclohexane chlorine Downloaded by robert joel (jordanbradford2004@icloud.com) + hydrogen acid chlorocyclohexane lOMoARcPSD|12129262 22 of 67 • Addition • halogenation (with Br2 or Cl2) → + bromine ethene 1,2 - dibromoethane • hydrogenation (with H2) → + ethyne hydrogen gas ethane • hydrohalogenation (HBr) → + propene hydrogen bromide 2 - bromopropane • hydration (with H2O) + propene → water 2 - hydroxypropane How do you know which Carbon to put the C and OH on? Markovnikov’s Rule “The rich get richer” (with reference to hydrogens atoms) * when a hydrogen halide or water is added to an alkene or akin the hydrogen atoms bonds to the carbon that already had the most hydrogen atoms Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 23 of 67 Naming Alcohols: alcohol - OH group Rules For Naming: • “ol” is added to the end of the parent chain example; Common Alcohols: • ethanol (wine) • glycerol • isopropanol • 2 - propanol (rubbing alcohol) • retinol (vitamin A) • cholesterol methanol - meth = 1 carbon - an = single carbon bonds - ol = alcohol (OH) • if more than 2 carbon atoms or more than two —OH groups are present use the numbering system to indicate location example; example; 1 - propanol - prop = 3 carbons - an = single carbon bonds -ol = alcohol (OH) - brake fluid 2 - propanol (isopropanol) - 2 (iso) = OH coming off the second carbon - an = single carbon bonds - ol = alcohol - cleaning cuts Sub-Categories of Alcohols: Primary - one other carbon Secondary - 2 other carbons Downloaded by robert joel (jordanbradford2004@icloud.com) Tertiary - 3 other carbons lOMoARcPSD|12129262 24 of 67 Poly-Alcohols: - more than 1 OH group - use di, tri, tetra, etc - use numbering Cyclic Alcohols: 2 - isopropyl - 5 - methyl - 1 - cyclohexanol Reactions of Alcohols: preparing alcohol: alkene → alcohol Hydration Reaction: Follows Markovnikov’s Rule * the simplest (methanol) is toxic to humans. even small amounts or swallowed can lead to blindness and death. Combustion Reaction: C3H8O propanol + 5O2 → 4H2O oxygen water + 3CO2 + carbon dioxide Elimination Reaction: (dehydration (opposite of hydration reaction) Downloaded by robert joel (jordanbradford2004@icloud.com) energy lOMoARcPSD|12129262 25 of 67 Naming Ethers: ether - R - O - R’ (R & R’ could be different or the same) Rules for Naming: Properties of Ethers: 1. add foxy to the end of the smaller group, then name the larger group 2. name each group alphabetically and add ether to the end * if two alkyl groups are the same, the prefix • volatile (easily evaporates) • flammable • used in he olden day as an anesthetics • used as solvents for fats and oils “di” is used example; example; 1. methoxyethane 1. ethoxyethane 2. ethyl methyl ether 2. diethyl ether Reactions of Ethers: formed when two alcohols react Condensation Reaction: Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 26 of 67 Naming Ketones: Functional Group - Ketone - carbonyl group bonded to two other carbon atoms * insects use ketones to communicate with each other Rules for Naming: Example; 1. replace “e” ending of the name of an alkane with “one” propanone Naming Aldehydes: functional groups - aldehyde - carbonyl group on a terminal carbon * aldehydes are detectable over one distances by our sense of smell. Rules for Naming: Example: 1. take parent alkane name, drop the final methanal “e” and add “al” ethanal Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 27 of 67 Reactions of Aldehydes & Ketones: oxidation - historically, any reactions involving oxygen - broadened to all chemical processes that involves a loss of electrons - partner molecule that undergoes reduction (gain of electrons) - oxidation mean gain of oxygen or loss of hydrogen (alcohols gain oxygen) Making an Aldehyde: + alcohol → aldehyde (need a primary alcohol) (O) → H H3C - C = O + H2O Making a Ketone: alcohol → ketone (need a secondary alcohol) + (O) → * tertiary alcohols do not undergo this oxidation reaction. WHY NOT? * there are no other H to form water * there aren’t enough bonds to have a double bonded O (O=) Making an Alcohol From Aldehyde: H H3C - C = O + H2 → Making an Alcohol From Ketone: + H2 → Downloaded by robert joel (jordanbradford2004@icloud.com) + H 2O lOMoARcPSD|12129262 28 of 67 Naming Carboxylic Acids: Functional Group - carboxyl group (this is made up of conbonyl and hydroxyl groups When wine is left exposed to air, it turns “sour”. In this case the alcohol in the wine has turns into a carboxylic acid. - carboxyl acids are weak acids (only partially ionize) found in citrus fruits, rhubarb, crab apples found in sour milk and yogurt (lactic acid) lactic acid also form in muscles when you workout, and you need to give your body time to get rid of it to recover Rules for Naming: 1. take the longest alkane/ alkaline parent chain and replace “e” ending with “oic acid” Example; methanoic acid Downloaded by robert joel (jordanbradford2004@icloud.com) 2 - methylbutanoic acid lOMoARcPSD|12129262 29 of 67 Reactions of Carboxylic Acids: Example 1: + aldehyde (O) → oxidizing agent - ethanal carboxylic acid - ethanoic acid Example 2: + butanal (O) → oxidizing agent butanoic acid Series of Equations: + methanol (O) oxidizing agent + methanal → (O) + methanal → oxidizing agent methanoic acid Downloaded by robert joel (jordanbradford2004@icloud.com) water lOMoARcPSD|12129262 30 of 67 Naming Esters: Functional Group - formed from an alcohol and a carboxylic acid • occurs naturally in plants • responsible for odours of fruits and flowers • synthetic esters are added to flavouring and cosmetics/perfumes Examples: Odour Name Apple Methyl butanoate Cherry Methyl benzoate Rum Ethyl methanoate Pineapple Ethyl butanoate Formula Rules for Naming: Rules for Drawing: 1. first part of the name is alkyl group 1. when drawing, the acid is drawn first from alcohol 2. second part of the name is from the acid 3. ending is “oate” Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 31 of 67 Reactions of Esters: Example 1: Esterification (condensation) → + butanoic acid + H2O ethyl butanoate ethanol Example 2: + H2O → + benzoic acid propyl benzoate propanol Hydrolysis (reversal of esterification): + Ester NaOH Base → + alcohol Downloaded by robert joel (jordanbradford2004@icloud.com) R - C - O- Na+ salt of an acid lOMoARcPSD|12129262 32 of 67 Polymerization: Polymers - are large molecules composed of a repeating sequence of monomers. A monomer is usually a small molecule or compound. A common monomer is ethene (ethylene), C2H4. Polymers are typically produced using two types of reactions; addition polymerization and condensation polymerization reactions. Polymerization - is a process of reacting monomer molecules together in a chemical reaction to form polymer chains or three-dimensional networks. Polyethylene: polymer of ethene PolyPropene: polymer of propene ** You need to have an alkene because you need to have the electron potential from the double bond. Polyvinyl Chloride: polymer of vinyl chloride Polystyrene: polymer of styrene Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 33 of 67 Organic Solvents: A solvent is a substance that dissolves a solute (a chemically distinct liquid, solid, or gas) resulting in a solution. A solvent is usually a liquid it but can also be a solid or gas. The quantity of solute that can result in a specific volume of solutes varies with temperature (the higher the temperature, the more solute can dissolve because the electrons have more energy and are moving around more which give it the ability to form attractions with water). Use Solvent Used Dry cleaning Terechloroethelene Paint cleaners Toluene, Turpentine ** they have low boiling points because they’re nonpolar Nail polis remover, glue solvents Acetone, Methyl acetate, Ethyl acetate Spot removers Hexene Detergents Citrus terpenes Perfumes Ethanol Properties: • Many solvents are volatile (many solvents are liquid with low boiling points, which means that they will evaporate form vapour in the air when containers are left open. There vapour can be breathed in and lead to health problems). • Flammable (flashpoint below 38°C) flashpoint - lowest temperature • Combustible (flashpoint about 38°C) at which vapours will ignite (given a source) Health Concerns: • Exposure to different solvents can cause a host of health issues: toxicity to the nervous system, reproductive damage, kidney and liver damage, respiratory impairment, cancer, and dermatitis. Safety Considerations: • eye protection - prevents splashes from getting in eyes • have proper ventilation - prevents vapours from being breathed in • store properly - prevents volatile solvents from being vaporized, and also prevents spills • read the safety information - be aware of all safety concerns for each solvent • wear gloves and proper footwear- prevents from skin contact and getting in cuts • wash hands after handling solvents - keep from prolonged skin exposure Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 34 of 67 Organic Family Forces (intramolecular/ intermolecular) Hydrocarbons • intramolecular • ethane • covalent (strong) • -89℃ • Intermolecular H3C - H3 • london force (weak) Alcohol Boiling Point (based on 2 carbon molecule) Solubility in Polar Solubility in NonSolvents Polar Solvents H2O Very low solubility High solubility • Intermolecular • hydrogen bonds (strong) • dipole-dipole • ethanol • 78.5℃ Ethers • Intermolecular • dipole-dipole • london force • dimethylether • small chain • small chain methoxymethane high • • • low solubility -23℃ solubility long chain • • long chain high • • H3C - O - CH3 solubility • low solubility Aldehydes • small chain • polar • dipole-dipole • long chain • non-polar • ethanal • -21℃ • small chain • small chain • high • low solubility solubility • long chain (1-3) • high solubility • long chain • low solubility Ketones • small chain • polar • dipole-dipole • ethanal • -21℃ • small chain • small chain • high • low solubility solubility • long chain • long chain • high solubility • low solubility Carboxylic Acid • Intermolecular • hydrogen bonds (strong) • dipole-dipole • ethanoic acid • vinegar • 118℃ • small chain • small chain • high • low solubility solubility • long chain • long chain • high solubility • low solubility Esters • Intermolecular • dipole-dipole • methyl ethanoate • small chain • 31.8℃ • some solubility H3C - CH3 • small chain • small chain • high • low solubility solubility • long chain • long chain • high solubility • low solubility Downloaded by robert joel (jordanbradford2004@icloud.com) • long chain • high solubility lOMoARcPSD|12129262 35 of 67 Isomers: Isomer - each of two or more compounds with the same formula but a different arrangement of atoms in the module and different properties Isomers of C6H14 • hexane • 2 - methylpentane • 3 - methylpentane • 2- dimethylbutane • 2,3 - methybutane Isomers of C6H12 • cyclo hexane • 1 - hexene • 2 - hexene • 3- hexene Molecules Molecule Name Structural Diagram Common Name IUPAC Name Natural Gas Methane Vinegar (CH3COOH) Acetic Acid or Ethanoic Acid Acetone Propanone Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 36 of 67 Energy Changes & Rates of Reaction: Energy Changes in Matter: Thermochemistry - the study of energy changes that company physical or chemical changes in matter Thermal Energy - energy available form a substance as a result of the motion of its molecules Chemical System - a set of reactants and products under study, usually represented by a chemical equation Surroundings - all matter around a system that is capable of absorbing/ releasing thermal energy Temperature - the average kinetic energy of the particles in a sample of matter Heat - the amount of energy transferred between substances Exothermic - releasing thermal energy as heat flows out of a system Endothermic - Absorbing thermal energy as heat flows into a system Open System - matter and energy can move in and out of a system Closed System - energy can move in and out of a system, but not matter Isolated System - an ideal system; neither energy nor matter can move in or out of a system Enthalpy - an amount of heat energy (usually measured in kilojoules) Physical Change - a change in the form of a substance in which no chemical bonds are broken Chemical Change - a change in the chemical bonds between atoms, resulting in the rearrangement in an atom, resulting in the formation of new atoms Nuclear Change - a change in the protons or neutrons in an atom, resulting in the formation of new atoms Activated Complex - intermediate sate that is formed during the conversions of reactants to products (high energy) Activation Energy - the minimum energy required for a reaction to occur Catalyst - substance that alters the rate of a chemical reaction without being consumed or permanently changed Hess’s Law - the value of the ∆H for any reaction that can be written in steps, equals the sum of the values of ∆H for each individual step Specific Heat Capacity - the heat required to raise the temperature of the unit mass of a given substance by a given amount Reaction Mechanism - a series of elementary steps that makes up an overall reaction Reaction Intermediates - molecules formed as short-lived products in section mechanisms Rate-Determining Step - the slowest step in a reaction mechanism Kinetic Energy - (motion) moving electrons, vibrating atoms and molecules, rotation of molecules Potential Energy - nuclear potential (protons/neutrons), electric potential of atoms in bonds, stored chemical energy, energy is stored and released as the position of particles change Effective Collisions - one that has sufficient energy and correct orientation of the colliding particles so that bonds can be broken and new bonds can be formed Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 37 of 67 Types of Enthalpy Changes: Physical Chemical Nuclear How is Energy Used? • energy is used to overcome or allow intermolecular forces to act fundamental particles • remain unchanged at the molecular level • energy changes overcome the electronic structure and chemical bonds within the particles (atoms or ions) • energy changes overcome the forces between protons and neutrons in the nuclei Range of Typical Enthalpy Changes (kJ/ mol) ∆H = 100 - 102 kJ/mol ∆H = 102 - 104 kJ/mol ∆H = 1010 - 1012 kJ/mol Measurement Number of Sig Figs 32.07 m 4 0.0041 g 2 5 x 105 kg 1 6400 s 2 204.0 cm 4 10.0 kJ 3 100 people infinite Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 38 of 67 Calorimetry: - the technological process of measuring energy changes in a chemical system ideal system what we will see Measuring Energy Changes: 3 Factors: 1. temperature change (how much energy (heat) was put in/out) 2. mass (surroundings) (given the same amount of heat a small mass will undergo a larger temperature change than a bigger mass ) 3. types of substance (different substances can absorb different amounts of energy) relationships between amount of heat transferred. mass, type of substance, temperature change q = heat transferred (to/from surroundings) → Joules m = mass → grams q = mc∆T c = specific heat capacity → J/g·℃ ∆T = temperature change (T2 - T1) → ℃ Density of water = 1g/mL (1mL = 1cm2) Specific Heat Capacities of Substances (table 1 on page 301) Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 39 of 67 Enthalpy Change (∆H): ∆H - the difference in enthalpies of reactants and products during a chemical change → broad category of types of energies: • Kinetic energy: - moving electrons - vibrating atoms and bonds - rotation of molecules • Potential energy - nuclear potential (protons & neutrons) - electric potential of atoms in bonds Exothermic Reactions: the release of energy the the surroundings ∆H ⊖ (system) ∆H ⊕ (surroundings) Endothermic Reactions: the absorption of energy from the surroundings ∆H ⊕ (system) ∆H ⊖ (surroundings) Law of Conservation of Energy - the total energy of the system and surroundings remains the same Reverse processes have an equal but opposite change in enthalpy example; H2O(l) → H2(g) ½O2(g) ∆Hdecomposition H2(g) + ½O2(g) → H2O(g) ∆Hcombustion = -285.8kJ = +285.8kJ * to controls for variables, energy changes in chemical systems are measured at STP or SATP Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 40 of 67 Molar Enthalpies (∆Hx): enthalpy change is associated with a physical, chemical, or nuclear change involving one mole of a substance example; combustion of hydrogen H2(g) + ½O2(g) → H2O(g) + 285.8kJ 1 mol ½ mol 1 mol release of energy Exothermic Reaction (table 1 - page 306: molar enthalpies reaction) example; vaporization & condensation of water H2O(l) + 40.kJ → H2O(g) ∆Hvapoization = +40.8kJ H2O(g) → H2O(l) + 40.kJ ∆Hcondensation = -40.8kJ * your breaking up the intermolecular forces NOT the covalent bonds *molar enthalpies are obtained EMPIRICALLY and listed in reference tables Generally: To BREAK bonds, energy has to be put into a system which is ENDOTHERMIC To FORM bonds, energy is released out of a system which is EXOTHERMIC Using Molar Enthalpies in Heat Calculations: a common refrigerant (Freon-12, molar mass 120.91g/mol) is alternately vaporized in tubes inside a refrigerator, absorbing heat. and condensed in tubes outside the refrigerator, releasing heat. this results in energy being trans formed from the inside to the outside of the refrigerator. the molar enthalpy for the refrigerant is 34.99kJ/mol. if 500.0g of the refrigerant is vaporized what is the expected enthalpy change ∆H? ∆Hvaporization = 34.99kJ/mol m = 500.0g Molar mass = 120.91g/mol ∆H = ? nFreon12 = 500.0g ÷ 120.91g/mol = 4.135mols ∆H = 4.135mols x 24.99kJ = 144.7kJ ∆H = n∆Hx Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 41 of 67 Representing Enthalpy Changes: Method 1 • include an energy value as a term in the thermochemical equation CH4O(l) + 3/2O2(g) → CO2 + 2H2O(g) + 726kJ Method 2 • write a chemical equation and state the enthalpy change CH4O(l) + 3/2O2(g) → CO2 + 2H2O(g) ∆H = -726kJ Method 3 • state the solar enthalpy of a specific reaction ∆Hcombustion = -726kJ/mol CH4O(l) Method 4 • potential energy diagram Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 42 of 67 Calorimetry of Physical and Chemical Changes: - The values for molar enthalpies are obtained by carefully designed experiments and precise measurements/ - when we investigate energy changes we base out analysis on the law of conservation of energy: the total energy change of the chemical system is equal to the total energy change of the surroundings. ∆Hsystem = qsurroudings H → enthalpy of system q → surroundings * one quantity is ⊖, gives off heat * one quantity is ⊕, absorbs heat Assumptions used in calorimetry: 1. no heat is transferred between calorimeter and the outside environment 2. any heat absorbed or released by the calorimeter materials, such as the container, is negligible 3. a dilute aqueous solution is assumed to have a density and specific heat capacity equal to that of water (1.00g/mL and 4.18J/g℃) Example; in a calorimetry experiment, 7.46g of potassium chloride is dissolved in 100.0mL (100.0g) of water at an initial temperature of 24.1℃. The final temperature of the solution is 20.0℃. What is the molar enthalpy of the solution of potassium chloride? system: KCl(s) → K+(aq) + Cl-(aq) c = 4.18J/g℃ mass = 100.0g ∆T = -4.1℃ ∆Hsolution = ? 1. q=mc∆T q = (100.0g)(4.18J/g℃)(-4.1℃) q = -1713.8 J q = -1.7kJ 2. n = m÷M = 7.46÷74.55 = 0.1000670 3. ∆Hsolution = 1.7kJ÷0.100067mols = 17kJ Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 43 of 67 Potential Energy Diagrams: - a graphical representation of the energy transferred during a physical or chemical change Potential Energy - stored chemical energy, energy is stored or released as the portion of particles change. - bonds breaking requires energy - bonds forming released energy A - exothermic reaction combustion reaction B - exothermic reaction because the products have less EP than the reactants (∆H = ⊖) C - endothermic because the products have more EP than the reactants (∆H = ⊕) Activation energy (Ea): the minimum energy required for a reaction to occur Activated Complex: intermediate state that is formed during the conversion of reactants to products (high energy) ∆H: enthalpy change of the system Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 44 of 67 Hess’s Law: - Not all chemical reactions can be analyzed directly from calorimetry experiments. Some reactions are very slow or other produce too many products to clearly distinguish between in a calorimeter. Hess’s Law: The value of the ∆H for any reaction that can be written in steps, equals the sum of the values of ∆H for each individual step. For example; consider the following Potential Energy Diagrams: If two or more equations with known enthalpy changes can be added together to from a new “target” equation, their enthalpy chnages may be added together to yield the overall enthalpy change to the “target” equation. ∆HTARGET = ∑∆HKNOWN Rules for Manipulating Equations: 1. if a chemical equation is reversed, the sign of ∆H changes. 2. if coefficients are altered by multiplying or dividing, ∆H is altered in the same way. example; What is the enthalpy change (∆H) for the formula of two moles of nitrogen monoxide from its elements? Referenece Equations: (1) ½N2(g) + O2(g) → NO2(g) (2) NO(g) + ½O2(g) → NO2(g)) ∆H = +34kJ ∆H = -56kJ Target Equation: N2(g) + O2(g) → 2NO(g) ∆H = ? Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 45 of 67 Multistep Energy Calculations: - in practise, energy calculations rarely involve only a single step calculation of heat or enthalpy change. Several energy calculations may be required, involving a combination of energy change definitions such as: - Heat flows: q = mC∆T - Enthalpy Change: ∆H = n∆Hx - Hess’s Law: ∆Htarget = ∑∆Hknwon - In these types of problems, ∆H is often found by using standard molar enthalpies or Hess’s Law and then equated to the transfer of heat, q. As well, conversions between moles and mass may be required to find the china in enthalpy of a specific amount of a substance. example; in the productions of sodium carbonate one step in the endothermic decomposition of sodium hydrogen carbonate: 2 NaHCO3 + 129.2kJ → Na2CO3(g) + CO2(g) + H2O(g) what quantity of chemical energy ∆H, is required to decompose 100.0kg of sodium hydrogen carbonate? 100kg = 100,000gNaHCO3 MNaHCO3 = 84.01g/mol n = m÷n = 100,000 ÷ 84.01 = 1190.33 mols 129.2kJ ÷ 2 = 64.6kJ ∆H = n • ∆Hx = (1190.33mols) (64.6kJ) = 76,895.3 kJ Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 46 of 67 Enthalpies of Formation: - Standard Enthalpy of Formation: the quantity of energy associated with the formation of one mole of a substance from its element in their standard states. Writing Formation Equations: 1. write one mole of product in the state that has been specified. 2. write the reactant elements in their standard states (solid, liquid, gas) 3. choose equations coefficients for the reactants to give a balanced equations yielding one mole of product (fractions may be used) example; write the formation equation of liquid ethanol 2C(s) + 3H2(g) + ½O2(g) → C2H6O(l) *∆Hf of an element already in its standard state is 0kJ Use pg. 799-800 for enthalpies ∆H = ∑n∆H°f(products) - ∑n∆H°f(reactants) example; the main component in natural gas used in home heating or laboratory burners is methane. what is the molar enthalpy of combustion of methane fuel? C2H4 + 2O2 → 2CO2 + 2H2O ∆H = [∑n∆H°f(products)] - [∑n∆H°f(reactants)] ∆H = [1mol(-393.5kJ/mol) + 2mol(-285.8kJ/mol)] - [1mol(-74.4kJ/mol) + 2 (0kJ/mol)] ∆H = -890.7kJ Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 47 of 67 Collision Theory & Factors Affecting Reaction Rate: Collision Theory: • a chemical system consists of particles (atoms, ions, or molecules) that are in constant random motion at various speeds • a chemical reaction must involve collisions of particles with each other • an effective collision is one that has sufficient energy and correct orientation of the colliding particles so that bonds can be broken and new bonds can be formed • ineffective collisions involve particles that rebound from the collision, essentially unchanged in nature The rate of a reaction depends on: 1. frequency of collisions 2. fraction (proportion) of those collisions that are effective Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 48 of 67 Factors that Affect the Rate of a Reaction: • Temperature - if you increase the temperature, you’re also increasing the potential energy, which causes more collisions, then that increases the probability of a higher frequency of effective collisions • Surface Area of Reactants - by increasing the surface area your getting more particles that can participate in collisions (at a faster rate) - many small pieces have a greater surface area than just one large piece • Catalyst - a catalyst is a substance that alters the rate of a chemical reaction without being consumed or permanently changed - catalysts work by orienting the reactions molecules properly, by orienting the reactants in the proper way, less energy is needed - enzymes are proteins that speed up chemical reactions in the body • Concentration of Reactants - if you increase the concentration, then you have more reactants, which means more collisions, which increases the chances of getting more effective collisions • Nature of Reactants - different elements naturally react at different rates - some react very slow (gold/silver) - some are very reactive (sodium) Analogy of Video Factor Affecting Reaction Rate Hire a match maker catalyst Increase population of school concentration Shrink size of hallways concentration Half the time between classes temperature Break up students in packs surface area Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 49 of 67 Reaction Mechanism: • If you have a collision involving one particle this means you are decomposing or breaking it down • The single particle reaction mechanisms involve a step in which the single molecule hits container walls or other particles • This is converting kinetic energy to potential energy • The collision of these particles simultaneously is much less frequent then he two particle collision • Most chemical reactions occur as a sequence of elementary steps Elementary Steps - a step in a reaction mechanism that only involves one-, two-, or three-particle collisions • This overall sequence is called the reaction mechanism Reaction Mechanism - a series of elementary steps that makes up an overall reaction • Increasing the concentration of the fast steps will have no effect on the chemical reaction • Increasing concentration concentration at the slow steps should increase the rate of chemical reaction Rate Determining Step - the slowest step in the reaction mechanism Reaction Intermediates - molecules formed as a short-lived products in reaction mechanisms example; HBr(g) + O2(g) → HOOBr(g) HOOBr(g) +HBr(g) → 2HOBr(g) 2{HOBr(g) + HBr(g) → H2O(g) + Br2(g)} 4HBr(g) + O2(g) → 2H2O(g) + 2Br2(g) slow fast fast Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 50 of 67 Chemical Systems & Equilibrium: Dynamic Equilibrium - a balance between forward and reverse processes occurring at the same time Chemical Equilibrium - a chemical equilibrium between reactants and products of the chemical reaction in a closed system Phase Equilibrium - a dynamic equilibrium between different physical states of a pure substance in a closed system Homogenous Equilibrium - one in which all of the reactants and products are present in a single solution Heterogenous Equilibrium - a system whose reactants, products, or both are in more than one phase Ionization - is the process by which an atom or a molecule acquires a negative or positive charge by gaining or losing electrons to form ions Dissociation - the splitting of a molecule into smaller molecules, atoms, or ions, especially by a reversible process Concentrated - high molarity Dilute - low molarity Strong - 100% ionization/dissociation (NO equilibrium) Weak - less than 50% ionization/disassociation (equilibrium form) Equilibrium Constant - a number that expresses the relationship between the amounts of products and reactants present at equilibrium Acid - proton donor Base - proton acceptor Conjugate Acid - contains one more + charge than the base that formed it Conjugate Base - contains one more - charge than the acid that formed it Auto-ionization - the reaction between two water molecules producing a hydronium ion and a hydroxide ion Hydronium Ion - the ion H3O+ Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 51 of 67 Chemical Equilibrium: Reference Video: Crash Course Chemistry #28 (up to 4:25) Throughout this unit when we speak of equilibrium reactions, we are usually referring to a closed system: the system can exchange energy, but not matter (no atoms can escape). Equilibrium Overview: Up to this point, you have generally been told that reactions occur in one direction only: reactants turn into product or products turn into reactants. This is actually not the case. In many natural chemical reactions, forward and reverse processes are both occurring simultaneously. example; 3H2(g) + N2(g) ⇌ 2NH3(g) Forward reaction: left-to-right reaction → Reverse reaction: right-to-left reaction ← The rates of these forward and reverses reactions are dependant on several factors, but mostly depend on the concentration of the substances. Typically when a reaction begins, the forward and reverse reaction rates are not equal. (If you start with only reactants, then the rate of forming the products will be fast due to the high amount of reactants particles available to collide and react. As the concentration of these reactant particles decreases, and the concentration of the product particles increases, the rate of the forward reactions flows, and threat of the reverse reaction speeds up.) Eventually a balance is formed between the forward and reverse reactions and the rate of the forward reaction matches the rate of the reverse reaction. When this happens, we say that the system is at equilibrium. Qualitatively, at equilibrium the system will look stable. We will se no observable changes (ie. colour change, gas bubbles forming etc). This is because the number of particles changing into products will equal the number o particles changing into reactants. The systems as a whole will look constant. It is important to note that equilibrium does not mean there are equal concentrations of reactants and products (we very rarely see this). Dynamic Equilibrium: Solubility Equilibrium: example; NaCl(s) ⇌ Na+(aq) + Cl-(aq) dissolving and crystallizing taking place at the same rate A dynamic equilibrium between different physical states of a pure substance in a closed system Phase Equilibrium: example; A balance between forward and reverse processes occurring at the same rate A dynamic equilibrium between a solute and a solvent in a saturated solution in a closed system H2O(l) ⇌ H2O(g) evaporation and condensation at the same rates Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 52 of 67 Equilibrium Stoichiometry: ICE tables (math strategy) I → initial concentrations of reactants and products [2.0mol/L] (square brackets imply concentration) C → change in concentrations E → equilibrium concentrations of reactants and products example 1; H2(g) + F2(g) ⇌ 2 HF(g) **balance** the reaction begins with 1.00mol/L of H2(g) and F2(g), calculate the concentrations of the H2(g) and HF(g) at equilibrium if the concentration of F2(g) at equilibrium is 0.24mol/L mol/L H2(g) F2(g) 2HI(g) I 1.00 1.00 0 C -x -x +2x E 1.00-x 1.00-x F2(g) 0.24 = 1.00 - x x = 0.76 mol/L 2x H2(g) 1.00 - x = [H2(g)] 1.00 - 0.76 = [H2(g)] 0.24 mol/L = [H2(g)] example 2; 2NH3(g) ⇌ 2 3H2(g) + N2(g) 2HF(g) 2 (0.76) = [HF(g))] 1.52 mol/L = [HF(g))] **balance** when 4.0mol of NH3(g) is introduced into a 2.0L container and heated to a particular temperature, the amount of ammonia changes to 2.0mol. Determine the equilibrium concentrations of the two entities mol/L 2NH3(g) 3H2(g) N2(g) I 2.0 0 0 C -2x +3x +x E 2.0-2x 3x x [NH3(g) initial] = 4.0mol ÷ 2.0L = 2.0mol/L [NH3(g) equil] = 2.0mol ÷ 2.0L = 1.0mol/L NH3(g) 1.0 = 2.0 - 2x x = 2.0 - 1.0 x = 0.5mol/L H2(g) [H2(g)] = 3x [H2(g)] = 3 (0.5) [H2(g)] = 1.5mol/L Downloaded by robert joel (jordanbradford2004@icloud.com) N2(g) [N2(g)] = x [N2(g)] = 0.5mol/L lOMoARcPSD|12129262 53 of 67 Equilibrium Constant (K): When chemical reactions take place in a closed system, forward and reverse reactions occur continuously and the system always contain products and reactants. In previous studies, you used stoichiometry to calculate amounts of reactants and products. In equilibrium systems, equilibrium law is used: For the general chemical reaction aA + bB ⇌ cC + dD K = [C]c[D]d ———— [A]a[B]b A,B,C,D are: chemical entities in gas/aqueous phases (concentration) a,b,c,d are: coefficients in the balanced equations (products) (reactants) K is constant called the equilibrium constant. Each chemical system has a unique K value that is dependant on the temperature. H2(g) + I2(g) ⇌ 2HI(g) Exp. # Initial concentrations Equilibrium Concentrations [H2(g)] [I2(g)] [HI(g)] [H2(g)] [I2(g)] [HI(g)] 1 2.000 2.000 0 0.442 0.442 3.116 2 0 0 2.000 0.221 0.221 1.560 3 0 0.010 0.350 0.035 0.045 0.280 Experiment Ratio of Equilibrium Concentrations K = 1 K = 2 K = 3 Value of K [HI(g)]2 ————— [H2(g)][HI(g)] (3.116)2 ————— = 49.7 (0.442)(0.442) 49.7 [HI(g)]2 ————— [H2(g)][HI(g)] (1.560)2 ————— = 49.8 (0.221)(0.221) 49.8 [HI(g)]2 ————— [H2(g)][HI(g)] (0.280)2 ————— = 49.7 (0.035)(0.045) 49.8 Note the following characteristics of equilibrium law: - the molar concentrations of the products are always multiplied by one another and written in the - numerator, and the molar concentrations of the reactants are always multiplied by each other and written in the denominator the coefficients in the balanced chemical equation are equal to the exponents of the equilibrium expression the concentrations in the equilibrium law expression are the molar concentrations of the entities at equilibrium (mol/L - for gases, this means the moles of gas per litre of space occupied) Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 54 of 67 Example 1: write the equilibrium law expression for the reaction of nitrogen gas with hydrogen gas to produce ammonia gas in a closed system. N2(g) + 3H2(g) ⇌ 2NH2(g) K = [NH3(g)]2 ————— [N2(g)][H2(g)]3 Example 2: calculate the value (K) of the equilibrium constant for the reaction of nitrogen gas with hydrogen gas to produce ammonia gas if the equilibrium concentrations are 1.50 x 10-5 mol/L, 3.45 x 10-1 mol/L, and 2.00 x 10-4 mol/L respectively. N2(g) + 3H2(g) ⇌ 2NH2(g) K = [NH3(g)]2 ————— [N2(g)][H2(g)]3 ** favours (2.00 x 10-4)2 ——————————— (1.50 x 10-5)(3.45 x 10-1)3 K = 0.065 or 6.49 x 10-2 reactants Forward & Reverse Reactions: - K represents the equilibrium constant for the forward reaction - Kʹ represents the equilibrium constant for the reverse reaction Forward Reaction Example; N2(g) + 3H2(g) ⇌ 2NH3(g) [NH3(g)]2 K = ————— [N2(g)][H2(g)]3 = 6.49 x ⇌ 2NH3(g) ⇌N2(g) + 3H2(g) 10-2 [N2(g)][H2(g)]3 Kʹ = ————— [NH3(g)]2 = 15.4 K=1 1 — = — = 6.49 x 10-2 Kʹ 15.4 The equilibrium constant of a forward reaction and the equilibrium constant of the reverse reaction are reciprocal quantities Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 55 of 67 Equilibrium Law in Chemical Reactions: Limitations of Equilibrium Constants and Precent Reaction Values: - the position of the equilibrium is a measure of the extent to which reactants become products in a closed system - the value of the equilibrium constant, K, depends on the temperature example: percent reaction values are dependant on temperature and concentration N2(g) + 3H2(g) ⇌ 2NH3(g) K = 4.26 x 108 at 25℃ K = 1.02 x 10-5 at 300℃ K = 8.00 x 10-7 at 400℃ Homogenous Equilibria: - the equilibria in which all entities are in the same phase (all gases or all aqueous solution) - however in some systems reactants and produce are in different phases, this is called heterogenous equilibria - Heterogeneous Equilibria: - equilibrium systems can involve (s), (l), (g), (aq), most will be homogenous solutions - equilibria in which reactants and products are in more than one phase example; 2H2O(l) + 2H2(g) ⇌ O2(g) [H2(g)]2[O2(g)] K = ————— [H2O(l)]2 = 6.49 x 10-2 - liquid water written in the denominator is a problem because the concentration of - liquid cannot change; it is fixed and equal to the substances density 1L of liquid water has a mass of 1.00kg = 55.5mols ∴ the [H2O(l)] = 55.5mol/L adding.removing water doesn’t change the concentration adding/ removing H2(g) or O2(g) does change the concentration the [H2O] is a constant , its value is incorporated into the K value shown example; K[H2O(g)] = [H2(g)]2[O2(g)] K 55.52 = [H2(g)]2[O2(g)] K = [H2(g)]2[O2(g)] Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 56 of 67 - the concentrations of entities in a condenses state (s) or (l) are not included as variables in the equilibrium law expression, but rather incorporated into the value of the equilibrium constant - water vapour is a gas just like H2 or O2, its concentration varies - equilibrium law expressions are always written from the net ionic form, balanced with simplest numbers Magnitude of K: - the magnitude of the equilibrium constant provides a measure of the extent to which the reaction has gone to completion when equilibrium is reached Equilibrium Constant: Limitations of Equilibrium Constant - the position of equilibrium is a measure of the extent to which reactants become products in a closed system - it is important to note that the value of the equilibrium constant, K, depends on temperature N2(g) + 3H2(g) ⇌ 2NH3(g) K = 4.26 x 108 at 25℃ Generally, any calculation involving an equilibrium law equation must specify a temperature. K = 1.02 x 10-5 at 300℃ K = 8.00 x 108 at 400℃ Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 57 of 67 Heterogeneous Equilibria equilibrium systems can involve solids, liquids, gases, and aqueous solutions Homogenous Equilibria - all entities are in the same phase Heterogeneous Equilibria - reactants and/or products are in more than one phase most of the systems we will study will be homogenous equilibria. however, in some systems, the reactants and products are in different phases. the concentration of solids, and pure liquids (ie. water) cannot change, their concentrations are fixed and equal to their densities. because these substances have fixed concentrations, their values become part of the value K, and not written in the equilibrium law expression. only substances in the gas or aqueous phases are included, as their concentrations can change. example; Zn(s) + CuCl2(aq) ⇌ Cu(s) + ZnCl2(g) Zn(s) + Cu2+(aq) + 2Cl-(aq) ⇌ Cu(s) + Zn2+(aq) + 2Cl-(g) Zn(s) + Cu2+(aq) ⇌ Cu(s) + Zn2+(aq) K = [Zn2+(aq)] ———— [Cu2+(aq)] Magnitude Explanation ∴ product favoured K >> 1 [product] much greater than [reactants] K=1 [product] and [reactants] are equal K << 1 [reactants] much greater than [products] ∴ reactant favoured Calculations involving equilibrium systems in the equilibrium law expression, the variables are K, and the concentrations of reactants and products in the gas or aqueous phases. as long as you know all values except one, you can mathematically calculate the one unknown example; sulfur trioxide gas is produced when sulfur dioxide gas and oxygen gas react. calculate the equilibrium concentration of oxygen gas if 1.50mol/L sulfur dioxide and 3.50 mol/L sulfur trioxide are found in an equilibrium mixture at 600℃. the equilibrium constant for the reaction at 600℃ is 4.30 2SO2(g) + O2(g)⇌ 2SO3(g) [SO2(g)]equil = 1.50 mol/L [SO3(g)]equil = 3.50 mol/L K = 4.30 @ 600℃ K= [SO3(g)]2 [O2(g)] = 3.50 ÷ 1.50 —————— = 1.26 mol/L [SO2(g)]2[O2(g)] Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 58 of 67 Le Chatelier’s Principle: when a chemical system at equilibrium is disturbed by a change in a property, the system adjusts in a way that opposes the change. Equilibrium Shift: movement of a system at equilibrium, resulting in a change in the concentration of reactants and products. generally, Le Chatelier’s Principle predicts that if a product (or reactant) is added (or removed), the system will undergo an equilibrium shift to restore the equilibrium. (either more reactants will be produced, or more products will be produced). example; a pop bottle has an equilibrium between CO2(g) and CO2(aq), when the cap is removed CO2(g) escapes, but when the cap is put back on the system reestablishes equilibrium and the concentrations of CO2(g) and CO2(aq) are lower. Variables Affecting Chemical Equilibrium: Concentration if more reactant is added: the system will use some of that added reactant and change it into product to re-establish the equilibrium if more product is added: the system will produce more reactant to re-establish equilibrium if reactant is taken away: system shifts left to produce more reactants if product is take away: systems shifts right to produce more products Temperature the energy in a chemical equilibrium can be treated as a reactant or product endothermic reaction: reactants + energy ⇌ products exothermic reaction: reactants ⇌ products + energy energy can be added or removed from a system by heating or cooling the container. in either situation, the equilibrium shifts to minimize the change. if the system is cooled, the system tries to “warm” itself and the equilibrium shifts the direction that produces heat. if heat is added, the equilibrium shifts in the direction that absorbs heat. Volume increasing volume (decreasing pressure): shift happens toward the side with the larger total amount of gaseous entities decreasing volume (increasing pressure): shift happens toward the side with the smaller total amount of gaseous entities Variables That Do Not Affect Equilibria: Catalysts - alter the activation energy required, do not affect concentrations of entities. Inert Gases - gases that cannot react, do not change the concentrations of other entities. Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 59 of 67 The Reaction Quotient: this tells us that the larger the value of K, the more the reaction (as written) favours the products. a very small K means the reactants are favoured. if we know the concentrations of the substances in a closed chemical system, we may want to determine wether the system is at equilibrium — and if not, in which direction it will shift to reach equilibrium. if a chemical system begins with only reactants, it is obvious that the system will shift right to produce products. if both reactants and products are present, the direction is less obvious. in such case, we can substitute concentrations into the equilibrium law expression to produce a trial value called the reaction quotient (Q). Reaction Quotient (Q): a test calculation using measured concentration values of a system on the equilibrium law expression Interpreting Q Q=K system is at equilibrium Q>K too much product so the system shifts left to produce more reactant Q<K too much reactant so the systems shifts right to produce more product example 1; the following reaction occurs in a closed container at 445℃. the equilibrium constant, K, is 0.020. 2HI(g) ⇌ H2(g) + I2(g) if the concentration of hydrogen iodide is 0.20 mol/L, hydrogen is 0.15 mol/L and iodine is 0.09 mol/L, determine if the system is at equilibrium and if not, the direction the reaction will proceed K = 0.020 ∴ reactant favoured Q = [H2(g)][I2(g)] Q = (0.15)(0.09) ————— ————— [HI(aq)]2 (0.20)2 Q = 0.3375 K = 0.020 Q > K too much product ∴the system will shift left to produce more reactants Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 60 of 67 Equilibrium Calculations: example 2; carbon monoxide reacts with water vapour to produce carbon dioxide and hydrogen. at 900℃, K is 4.200. calculate the concentration of all entities at equilibrium if 4.000 mol of each entity is initially placed in a 1.000-L closed container. Step 1: write a balanced chemical equation and list known values CO(g) + H2O(g) ⇌ CO2(g) + H2(g) V = 1.000L K = 4.200 nall entities = 4.000mol Step 2: calculate concentrations if necessary [all] = 4.000mol ÷ 1.000L [all] = 4.000mol/L Step 3: calculate Q and determine the direction of shift Q = [CO2(g)][H2(g)] —————— [CO(g)][H2O(g)] Q = (4.000)(4.000) —————— (4.000)(4.000) Q = 1.000 K = 4.000 K > Q too much reactant ∴ the system will shift right to produce more product Step 4: ICE table CO(g) H2O(g) CO2(g) H2(g) I 4.000 4.000 4.000 4.000 C -x -x +x +x E 4.000 - x 4.000 - x 4.000 + x 4.000 + x Step 5: sub into equilibrium law expression and find x 4.200 = (4.000 + x)(4.000 + x) ————————— (4.000 - x)(4.000 - x) → 8.198 - √4.200 - x = 4.000 + x → 4.198 = 3.049x →→ x ≐ 1.38 √4.200 = √(4.000 + x)2 √4.200 = 4.000 + x → ————— → ——— ———— √(4.000 - x)2 1 4.000 - x Step 6: use x value to solve for the equilibrium concentrations [CO(g)] = [H2O(g)] = 4.000 - x [CO2(g)] = [H2(g)] = 4.000 + x = 4.000 - 1.38 = 4.000 + 1.38 = 2.623 mol/L Downloaded by robert joel (jordanbradford2004@icloud.com) = 5.377 mol/L lOMoARcPSD|12129262 61 of 67 The Solubility Product: this is a special case of equilibrium where excess solute is in equilibrium with its aqueous solution. Solubility: the concentration of a saturated solution of a solute in a particular solvent at a particular temperature; a specifi c maximum concentration Ksp: the solubility product constant the value obtained from the equilibrium law applied to a saturated solution copper (I) chloride is dissolved in water. (has a low solubility, so even a small amount results in a saturated solution) Balanced Equation: CuCl(s) ⇌ Cu(aq)+ + Cl(aq)Solubility Equilibrium Law Expression: Ksp = [Cu(aq)+] [Cl(aq)-] Calculating Solubility Using Ksp Values: example; calculate the Ksp for magnesium fluoride at 20℃, given a solubility of 0.00172 g/100mL MgF2(s) ⇌ Mg(aq)2+ + 2F(aq)1. n = m÷M = 0.00172g ÷ 62.31g/mol = 2.76 x 10-5 mol 2. 100mL x 1L —— 1000mL = 0.1L 3. C = 2.76 x 10-5 mol ÷ 0.100mL = 2.76 x 10-4 mol/L 5. Ksp = (2.76 x 10-4)(5.52 x 10-4)2 = 8.3 x 10 -17 4. MgF2(g) Mg2+(aq) 2F-(aq) I — 0 0 C — +x + 2x E — x x Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 62 of 67 Q>K • supersaturated • ∴ precipitate formed Predicting Precipitation: find Q (trial in product) and compare to K example 1; CuCl(s) ⇌ Cu+(aq) + Cl-(aq) Q=K • saturated solution • ∴ at equilibrium Ksp = [Cu+ (aq)][Cl-(aq)] Q = (____)(____) = _____ Q<K • unsaturated • ∴ no precipitate formed example 2; The Ksp of AgCl(s) is 1.8 x 10-10 at 25℃ in each case determine the type of solution and whether a predicate forms. [Ag+(aq)] = 1.6 x 10-7 a) [Cl-(aq)] = 1.3 x 10-5 Q = (1.3 x 10-5) (1.6 x 10-7) Q = 2.08 x 10-12 Q < Ksp ∴ unsaturated b) [Cl-(aq) ] = 1.5 x 10-5 Q = 3 x 10-9 [Ag+(aq)] = 2.0 x 10-4 Q > Ksp ∴ supersaturated Common Ion Effect: - when equilibrium exists in a solution, the equilibrium can be shifted by dissolving into the solution a compound that adds a common ion example; - saturated solution of sodium chloride - in equilibrium with undissolved sodium NaCl(aq) ⇌ Na+(aq) + Cl- (aq) - if a few drops of HCL is added, additional crystals of sodium chloride will form - use Le Chatelier’s principal to explain the HCL releases large numbers of chloride ions into the solution HCl(aq) ⇌ H+ + Cl-(aq) - those additional ions increase the concentration of chalice ions, shifting the NaCl equilibrium to the left - causing NaCl to precipitate out - a similar result would happen if a solution containing Na+(aq) ions were added instead - in this case, the common ion would be the sodium ions Common Ion Effect - a reduction in the solubility of a salt caused by the presence of another salt having a common ion Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 63 of 67 Acids & Bases: - Acids and bases are electrolytes that for aqueous solutions with unique properties: Properties of Acids & Bases Acids Bases Property sour bitter taste <7 7 - 14 pH forms H+ in solution forms OH-(aq) turns litmus red turns litmus blue litmus used for cleaners Recall the Arrhenius definition of acids and bases: Acids - solutes that produce H+(aq) Bases - solutes that produce OH-(aq) Problem - does not explain how NH3 has basic properties In 1923 a chemist in Denmark and a chemist in England (independantly) recognized that in most acid-base interactions, a proton (H+ ion) is transferred from one reactant to another A Bronsted-Lowry ACID is a proton-donor A Bronsted-Lowry BASE is a proton-acceptor Bronsted-Lowry Theory: Example; H2O(l) + base HCl(aq) acid ⇌ H3O+(aq) + Cl-(aq) hydronium ions Amphoteric (amphiprotic) - a substance capable of acting as an acid or base in different chemical scenarios In proton transfer at equilibrium, both forward and reverse reactions involve BronstedLowry acids and bases. Conjugate acid-base pair - two substance whose formulas only differ by a proton example; identify the conjugate acid-base pair in the following reactions: CH3COOH(aq) acid + H2O(l) base ⇌ CH3COO-(aq) conjugate base Downloaded by robert joel (jordanbradford2004@icloud.com) + H3O+(aq) conjugate acid lOMoARcPSD|12129262 64 of 67 Properties of Strong & Weak Acids & Bases: Properties Acids Bases Strong Weak Strong Weak Attraction for protons very weak strong attraction strong attraction weak attraction % ionization 100% ionization low (less than 50%) 100% ionization low (less than 50%) Nature of conjugate partner weak conjugate base strong conjugate weak conjugate base acid strong conjugate acid pH << 7 (lesser) <7 > 7 (greater) >> 7 Auto-ionization of Water: H2O(l) + HCl(aq) base acid Kw = 1.0 x 10-14 Kw = [H3O+(aq)][OH-(aq)] 1.0 x 10-14 = [H+(aq)][OH-(aq)] ⇌ H3O+(aq) + conjugate acid [product ÷ reactant] Cl-(aq) conjugate base example; if we had pure water, what would be the concentration of H+ and OH-? pH = -log [H+(aq)] [H+(aq)] = 5.3 x 10 -6 [H+(aq)] = 6.2 x 10 -3 [H+(aq)] = 7.8 x 10 -6 1.0 x 10-14 = [ x ][ x ] √1.0 x 10-14 = √x2 1.0 x 10-7 =x pH = 5.28 pH = 2.21 pH = 5.11 Dissolving an Acid - dissolving an acid increases the concentrations Dissolving a Base - dissolving a base increase the concentration of OH-(aq) - in neutral solutions [H+(aq)] = [OH-(aq)] - in acidic solutions [H+(aq)] > [OH-(aq)] - in basic solutions [H+(aq)] < [OH-(aq)] - since Kw = [H+(aq)][OH-(aq)], if you know one of the concentrations, you can find the other example; a 0.15 mol/L solution of HCl is at SATP, calculate the concentration of hydroxide ions. H2O(l) + HCl(aq) ⇌ H3O+(aq) + Cl-(aq) C = 0.15 mol/L [H+(aq)] = 1.0 x 10-7 mol/L Kw = [H+(aq)][OH-(aq)] 1.0 x 10-7 (0.15)x ———— = ——— 6.67 x 10-7 mol/L = x 0.15 0.15 ** strong acid ∴ 100% ionization Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 65 of 67 pH: pH = -log [H+(aq)] [H+(aq)] = 10-pH 0 - 2 → strong acid ; 3 - 6 → weak acid ; 8 - 11 → weak base ; 12 - 14 → strong base example; a sample of sodium hydrozide of 1.0 mol/L, determine the concentration of H+ ions + OH-(aq) NaOH → Na+(aq) c = 1.0 mol/L c = 1.0 mol/L + 1.0 x 10-14 = [H+]1.0 Kw = [H (aq)][OH (aq)] ———— ———— [H+] = 1.0 x 10-14 Terms: • strong - 100% ionization/dissociation (NO equilibrium) • weak - <50% ionization/dissociation (equilibrium formed) • concentrated - high molarity • diluted - low molarity pOH: pOH = -log [OH-(aq)] [OH-(aq)] = 10- 0 - 2 → strong base ;3 - 6 → weak base ; 8 - 11 → weak acid ; 12 - 14 → strong acid ** this is true if all solutions have the same concentrations example; a) calculate the pOH of a solution that has a hydroxide ion concentration of 2.5 x -13 10 mol/L pOH = -log -2.5 x 10(-13) pOH = 12.60 b) calculate the pH of this solution 12.60 + 1.40 = 14.00 Kw = [H+(aq)][OH-(aq)] -13 -14 + **pH and pOH always add to 14 1.0 x 10 = [H (aq)]2.5 x 10 -2 + 4.0 x 10 mol/L =[H (aq)] Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 66 of 67 Percent Ionization of Weak Acids: weak acids - only a little bit ionizes (∴ equilibrium forms) % ionization = [H+(aq)] ———— x 100% [OH-(aq)] example; calculate the % ionization of methanol acid if a 0.10M solution has a pH of 2.38 HCO2H(aq) ⇌ + H+(aq) CO2H-(aq) 0.10 mol/L pH = 2.38 [H+] = 10-2.38 = 4.17 x 10-3 mol/L % ionization = (4.17 x 10-3 ÷ 0.10) x 100% = 4.2% ** ∴ reactant favoured Acid Ionization Constant, Ka: Ka - equilibrium constant for a weak acid example; calculate the Ka of acetic acid if a 0.1000M solution at equilibrium at SATP, has a percent ionization of 1.3% CH3COOH(aq) ⇌ Ka = [H+][CH3COOH-] ——————— [CH3COOH] 0.013 = [H+] —— 0.1000 + H+(aq) CH3COOH-(aq) % ionization = [H+] ————— [CH3COOH] ** ∴ reactant favoured = 1.3 x 10-3 mol/L Ionization Constant for Weak Bases: B - weak base (a base will have at least one lone pair of electrons) B(aq) base + H2O(l) acid ⇌ BH+(aq) conjugate acid + OH-(aq) conjugate base Kb = [BH(aq)][OH-(aq)] ——————— [B(aq)] Downloaded by robert joel (jordanbradford2004@icloud.com) lOMoARcPSD|12129262 67 of 67 Relationship Between Ka & Kb: Kw = 1.0 x 10-14 Kw = Ka · Kb ← for an acid base conjugate pair (each reacting with water separately) NH3(aq) + base ⇌ H2O(l) acid NH4+(aq) conjugate acid OH-(aq) conjugate base Kb = [NH4(aq)][H2O(l)] ——————— [NH3+(aq)] NH4+(aq) acid + ⇌ H2O(l) base NH3(aq) conjugate base H3O(aq) conjugate acid Ka = [NH3(aq)][H3O(aq)+] ——————— [NH4+(aq)] Kw = Ka · Kb LS = Kw = [H3O+][OH-] RS = [H3O+][NH3] [NH4+][OH-] —————— · ————— [NH4] [NH3] = [H3O+][OH-] example; find the value of Kb for the acetate ion C2C3O2, if the Ka of acetic acid is 1.8 x 10-5 Kw = Ka · Kb 1.0 x 10-14 = 1.8 x 10-5 · Kb 5.6 x 10-10 = Kb Downloaded by robert joel (jordanbradford2004@icloud.com)