"ORGANIC COMPOUNDS SUMMARY"
SCH 4U REVIEW 1 of 39 SCH 4U REVIEW CHAPTER 1: ORGANIC COMPOUNDS organic compound – a compound that contains carbon and usually hydrogen catenation – the property of carbon to form a covalent bond with another carbon atom, forming long chains or rings functional group – a group of atoms in an organic molecule that impart particular physical and chemical characteristics to that molecule – there are three main components: multiple bonds between C atoms (e.g. – C ≡ C – ) C bonded to a more electronegative atom (i.e., N, O, OH, or a halogen) C double bonded to O BASIC NAMING NUMBER PREFIX FOR ALKYL GROUP FUNCTIONAL PREFIX FOR NAMING NAME GROUP NAMING 1 meth- methyl- -F fluoro 2 eth- ethyl- - Cl chloro 3 prop- propyl- - Br bromo 4 but- butyl- -I iodo 5 pent- pentyl- - OH hydroxy 6 hex- hexyl- - NO2 nitro 7 hept- heptyl- - NH2 amino 8 oct- octyl- 9 non- nonyl- 10 dec- decyl- ISOMERS OF ALKYL GROUPS butyl (or n-butyl) CH3 – CH2 – CH2 – CH2 CH3 – CH – CH3 isobutyl (or i-butyl) CH2 CH3 – CH – CH2 – CH3 s-butyl (secondary butyl) CH3 t-butyl (tertiary butyl) CH3 – C – CH3 SYMBOLS R, R΄, R˝ alkyl group X halogen atom (O) oxidizing agent (like KMnO4 or Cr2O7 2- in H2SO4) SCH 4U REVIEW 2 of 39 PRIORITY FOR NAMING (FROM HIGHEST TO LOWEST) OH hydroxyl NH2 amino F, Cl, Br, I fluoro, chloro, bromo, iodo CH2CH2CH3 propyl CH2CH3 ethyl CH3 methyl ALKANES a hydrocarbon with only single bonds between carbon atoms; general formula, DEFINITION CnH2n+2 prefix referring to number of carbons in the longest continuous chain + “-ane” alkyl groups are named in alphabetical order, numbered from the end that will give the lowest combination of numbers NAMING the presence of two or more of the same alkyl groups requires a “di” or “tri” prefix before the alkyl group name cyclic hydrocarbons have the carbon ring become the parent chain, and the prefix “cyclo” is used before the parent name GENERAL –C–C– FORMULA propane CH3 – CH2 – CH3 cyclohexane CH2 EXAMPLE(S) CH2 CH2 | | CH2 CH2 CH2 POLARITY non-polar FORCES dispersion forces relatively low BOILING increases with chain length POINT straight chains have higher b.p.s than branched chains SOLUBILITY immiscible in water and other polar solvents combustion C3H8 + 5 O2 3 CO2 + 4 H2O REACTIONS substitution (halogenation) 1) CH4 + Cl2 CH3Cl + HCl 2) CH3Cl + Cl2 CH2Cl2 + HCl SCH 4U REVIEW 3 of 39 ALKENES a hydrocarbon that contains at least one carbon-carbon double bond; general DEFINITION formula, CnH2n prefix referring to number of carbons in the longest continuous chain that contains the double bond + “-ene” NAMING alkyl groups are named in alphabetical order, numbered from the end that is closest to the double bond GENERAL | | FORMULA –C=C– EXAMPLE(S) propene CH2 = CH – CH3 POLARITY non-polar FORCES dispersion forces BOILING POINT relatively low SOLUBILITY immiscible in water and other polar solvents halogenation (with Br2, Cl2, etc.) room temperature ethene + bromine 1,2 – dibromoethane Br Br CH2 = CH2 + Br2 CH2 – CH2 hydrogenation (with H2) catalyst, heat, pressure ethyne + hydrogen ethane H H CH2 = CH2 + 2 H2 CH2 – CH2 H H REACTIONS hydrohalogenation (with hydrogen halides) room temperature propene + hydrogen bromine 2-bromopropane H Br | | H – C = CH – CH3 + HBr H – C – CH – CH3 | | H H PREPARATION hydration (with H2O) H2SO4 catalyst REACTION FOR 1-butene + water 2-butanol ALCOHOLS CH3 – CH2 – CH = CH2 + HOH CH3 – CH2 – CH – CH2 OH H “The rich get richer” … when a hydrogen halide or water is added to an MARKOVNIKOV’S alkene or alkyne, the hydrogen atom bonds to the carbon atom within the RULE double bond that already has more hydrogen bonds. SCH 4U REVIEW 4 of 39 a cis isomer has both alkyl groups on the same side of the molecular strucutre CH3 CH3 C=C cis-2-butene H H GEOMETRIC a trans isomer has alkyl groups on the opposite side of the molecular ISOMERS strucutre CH3 H C=C trans-2-butene H CH3 ALKYNES a hydrocarbon that contains at least one carbon-carbon triple bond; general DEFINITION formula, CnH2n-2 prefix referring to number of carbons in the longest continuous chain that contains the triple bond + “-yne” NAMING alkyl groups are named in alphabetical order, numbered from the end that is closest to the triple bond GENERAL FORMULA –C≡C– EXAMPLE(S) propyne CH ≡ C – CH3 POLARITY non-polar FORCES dispersion forces BOILING relatively low POINT SOLUBILITY immiscible in water and other polar solvents REACTIONS Same as Alkenes (resulting in double bonds rather than single) AROMATIC HYDROCARBONS DEFINITION a compound with a structure based on benzene (a ring of six carbons) consider the benzene ring to be the parent molecule alkyl groups are named to give the lowest combination of numbers, with no NAMING particular starting carbon (as it is a ring) when it is easier to consider the benzene ring as an alkyl group, we use the name “phenyl” to refer to it GENERAL FORMULA SCH 4U REVIEW 5 of 39 methylbenzene CH3 EXAMPLE(S) POLARITY non-polar FORCES dispersion forces BOILING relatively low POINT SOLUBILITY immiscible in water and other polar solvents REACTIONS Same as Alkanes (performs substitution reactions) ORGANIC HALIDES a compound of carbon and hydrogen in which one or more hydrogen atoms have DEFINITION been replaced by halogen atoms halogen atoms are considered to be attachments to the parent chain and are NAMING numbered and named with a prefix as such GENERAL R–X FORMULA H H | | 1,2-dichloroethane H–C–C–H | | Cl Cl EXAMPLE(S) Cl 1,2-dichlorobenzene Cl POLARITY polar (due to halogen atom) FORCES dispersion forces (increased strength b/c of carbon-halogen bonds) BOILING higher than the corresponding hydrocarbons POINT SOLUBILITY more soluble in polar solvents addition ethylene + hydrogen idodie 1-iodoethene H I | | H–C≡C–H + H–I H–C=C–H REACTIONS substitution ethane + chlorine 1-chloroethane + hydrogen chloride H Cl CH3 – CH3 + Cl – Cl H–C–C–H + H – Cl H H SCH 4U REVIEW 6 of 39 elimination 2-bromopropane + hydroxyl group propene + bromine ion + water H Br H H H H | | | H – C – C – C – H + OH- H–C–C=C–H + Br- + H2O H H H H ALCOHOLS an organic compound characterized by the presence of a hydroxyl functional group (OH-) - S DEFINITION O O < 109.5° H S+ the “e” ending of the parent hydrocarbon is changed to “ol” to indicate the presence of the OH- group NAMING the chain is numbered to give the OH- group the smallest possible number when there is more than one OH- group, the endings “diol” and “triol” are used, and each is indicated with a numerical prefix, however the “e” ending remain then (e.g., 1,3-propanediol) GENERAL R – OH FORMULA propanol CH3 – CH2 – CH2 – OH OH EXAMPLE(S) phenol POLARITY polar (due to hydroxyl group) FORCES hydrogen bonding and dispersion forces BOILING high (due to capacity for hydrogen bonding) POINT SOLUBILITY very soluble in polar solvents and nonpolar solvents (due to OH- group) hydration (preparation) H2SO4 catalyst 1-butene + water 2-butanol CH3 – CH2 – CH = CH2 + HOH CH3 – CH2 – CH – CH2 OH H SCH 4U REVIEW 7 of 39 dehydration (elimination) H2SO4 catalyst propanol propene + water CH3 – CH – CH2 CH3 – CH = CH2 + HOH OH H Primary Alcohol (1°) CH3 – CH2 – OH 1 other carbon group attached to carbon with OH OH 1°, 2°, AND 3° Secondary Alcohol (2°) CH3 – CH – CH3 ALCOHOLS 2 other carbon groups attached to carbon with OH OH Tertiary Alcohol (3°) CH3 – C – CH3 3 other carbon groups attached to carbon with OH CH3 ETHERS an organic compound with two alkyl groups (the same or different) attached to an oxygen atom - S DEFINITION C O C O 116° C C S+ the longer of the two alkyl groups is considered the parent chain the other alkyl group with the oxygen is considered to be the substituent group (with prefix of carbons and “oxy”) numbering of C atoms starts at the O NAMING propane ethoxy CH3CH2CH2 – O – CH2CH3 3 2 1 1 2 ethoxypropane GENERAL R – O – R΄ FORMULA EXAMPLE(S) methoxyethane CH3 – O – CH2 – CH3 POLARITY polar (due to the V-shape and C - O bonds) FORCES dispersion forces BOILING medium (higher than hydrocarbons, lower than alcohols of similar length) POINT SOLUBILITY soluble in polar solvents and nonpolar solvents SCH 4U REVIEW 8 of 39 condensation (preparation) * addition of two alcohols (same or different) * H2SO4 catalyst methanol + methanol methoxymethane + water CH3 – OH + CH3 – OH CH3 – O – CH3 + HOH ALDEHYDES an organic compound that contains the carbonyl group (–C=O) on the end DEFINITION carbon of a chain the “e” ending of the parent hydrocarbon is changed to “al” to indicate the NAMING presence of the R-C=O O GENERAL || FORMULA R[H] – C – H O || EXAMPLE(S) propanal CH3 – CH2 – C – H POLARITY polar (due to carbonyl group) FORCES dispersion forces BOILING medium (higher than ethers, lower than alcohols of similar length) POINT SOLUBILITY similar solubility to alcohols oxidation (preparation) 1° alcohol + mild oxidizing agent aldehyde + water ethanol + (O) ethanal + water OH O | || CH3 – C – H + (O) CH3 – C – H + HOH | H REACTIONS reduction (hydrogenation) aldehyde + hydrogen 1° alcohol proponal + hydrogen 1-propanol O OH CH3 – CH2 – C – H + H2 CH3 – CH2 – C – H KETONES an organic compound that contains the carbonyl group (–C=O) on a carbon DEFINITION other than those on the end of a carbon chain (in the middle) SCH 4U REVIEW 9 of 39 NAMING the “e” ending of the parent hydrocarbon is changed to “one” O GENERAL || FORMULA R – C – R΄ O || EXAMPLE(S) propanone (acetone) CH3 – C – CH3 POLARITY polar (due to carbonyl group) FORCES dispersion forces BOILING medium (higher than ethers, lower than alcohols of similar length) POINT SOLUBILITY similar solubility to alcohols oxidation (preparation) 2° alcohol + oxidizing agent ketone + water 2-propanol + (O) propanone + water OH O || CH3 – C – CH3 + (O) CH3 – C – CH3 + HOH H REACTIONS reduction (hydrogenation) ketone + hydrogen 2° alcohol butanone + hydrogen 2-butanal O OH CH3 – CH2 – C – CH3 + H2 CH3 – CH2 – C – CH3 NOTE: Tertiary alcohols do not undergo oxidation reactions; no H atom is available on the central C atom. CARBOXYLIC ACIDS one of a family of organic compounds that is characterized by the presence of a carboxyl group: DEFINITION O –C O–H the “e” ending of the parent alkane is changed to “oic acid” NAMING numbering starts with the C of the carboxyl group O GENERAL || FORMULA R[H] – C – OH SCH 4U REVIEW 10 of 39 CH2CH3 EXAMPLE(S) 4-ethyl-3-methylhexanoic acid CH3 – CH2 – CH – CH – CH2 – COOH CH3 POLARITY polar (due to carboxyl group) FORCES dispersion forces and hydrogen bonding BOILING very high (higher than hydrocarbons of similar length due to –CO and –OH POINT groups) SOLUBILITY similar solubility to alcohols dissolution in water (proof of acidity) ethanoic acid + water carboxylate ion + hydronium ion CH3COOH + H2O CH3COO- + H3O+ O– CH3 – C * shares the double bond O oxidation (with weak oxidizer) 1) 1° alcohol + weak oxidizing agent aldehyde + water 2) aldehyde + weak oxidizing agent carboxylic acid + water O REACTIONS 1) CH3 – CH2 – OH + Cr2O72- + H+ CH3 – C – H + H2O + Cr3+ O 2) CH3 – C – H + Cr2O72- + 2 H+ 3 CH3 – COOH + 4 H2O oxidation (with strong oxidizer) 1° alcohol + strong oxidizing agent carboxylic acid + water ethanol + oxidizing agent + hydrogen ethanoic acid + water + manganese dioxide O 3 CH3 – CH2 – OH + 4 MnO4- + 4 H+ 3 CH3 – C – OH + 5 H2O + 4 MnO2 NOTE: Ketones are not readily oxidizing, as in the oxidation of aldehydes. ESTERS an organic compound characterized by the presence of a carbonyl group bonded DEFINITION to an oxygen atom the group that is attached to the double-bonded O becomes the parent chain NAMING with the “e” ending changed to “oate” the other group is named as a substituent group O GENERAL || FORMULA R[H] – C – O – R΄ SCH 4U REVIEW 11 of 39 2-methylbutyl propanoate O CH3 EXAMPLE(S) CH3 – CH2 – C – O – CH2 – CH – CH2 – CH3 POLARITY less polar than carboxylic acids (loss of OH- group) FORCES dispersion forces BOILING medium (lower than carboxylic acids, higher than aldehydes / ketones of similar POINT length due to extra O) SOLUBILITY less soluble than acids condensation (formation) carboxylic acid + alcohol ester + water ethanoic acid + methanol methyl ethanoate + water O O || || CH3 – C – OH + CH3 – OH CH3 – C – O – CH3 + H2O REACTIONS hydrolysis (saponification) ester + NaOH sodium salt of acid + alcohol O O R – C – O – R΄ + Na+ + OH- R – C – O- + Na+ + R΄ – OH NOTE: Esters generally have nice odours and are used to create artificial flavours. AMINES an ammonia molecule in which one or more H atoms are substituted by alkyl or DEFINITION aromatic groups 1) nitrogen group is named as a substituent group using “amino-” NAMING 2) alkyl group is named as a substituent group from “-amine” R΄ [H] GENERAL | FORMULA R – N – R΄΄ [H] 1) 1-aminopropane NH2 2) propylamine CH2 – CH2 – CH3 N-ethyl-N-methyl-1-aminobutane EXAMPLE(S) CH2CH3 | CH3 – N – CH2 – CH2 – CH2 – CH3 POLARITY polar (not as polar as alcohols, because N is less polar than O) SCH 4U REVIEW 12 of 39 FORCES dispersion forces and N-H bonds BOILING medium (lower than alcohols of similar length, higher than hydrocarbons) POINT SOLUBILITY soluble in water formation alkyl halide + ammonia amine + ammonium halide REACTIONS iodoethane + ammonia aminoethane + ammonium iodide CH3 – CH2 – I + 2 NH3 CH3 – CH2 – NH2 + NH4I Primary Amine (1°) CH3 – N – H 1 alkyl group attached to N | H methylamine 1°, 2°, AND 3° Secondary Amine (2°) CH3 – N – CH3 AMINES 2 alkyl groups attached to N | H dimethylamine Tertiary Amine (3°) CH3 – N – CH3 3 alkyl groups attached to N | CH3 trimethylamine AMIDES an organic compound characterized by the presence of a carbonyl functional DEFINITION group (C=O) bonded to a nitrogen atom alkyl group attached to double-bonded O is considered to be the substituent NAMING group, attached to the parent “-amide” O R΄΄ [H] GENERAL || | FORMULA R[H] – C – N – R΄ [H] O || EXAMPLE(S) propanamide CH3 – CH2 – C – NH2 POLARITY slightly more polar than amine of similar length (extra O) FORCES dispersion forces and N-H bonds BOILING same as amines POINT SOLUBILITY soluble in water formation H2SO4 catalyst REACTIONS carboxylic acid + amine amide + water ethanoic acid + ammonia ethanamide + water SCH 4U REVIEW 13 of 39 O H O || | || CH3 – C – OH + H – N – H CH3 – C – NH2 + HOH EXAMPLE OF A STEPPED SYNTHESIS REACTION Write a series of equations for a method of synthesis for N-ethylethanamide from an alkane and ammonia. 1. ethane + chlorine chloroethane + hydrogen chloride CH3 – CH3 + Cl – Cl CH3 – CH2 + H – Cl | Cl 2. chloroethane + water ethanol + hydrogen chloride CH3 – CH2 HOH CH3 – CH2 + H – Cl | | Cl OH 3. ethanol + strong oxidizer ethanoic acid + water + manganese dioxide CH3 – CH2 + MnO4- CH3 – C – OH + HOH + MnO2 | || OH O 4. chloroethane + ammonia aminoethane + ammonium chlorine CH3 – CH2 + 2 H – N – H CH3 – CH2 – NH2 + NH4Cl | | Cl H 5. ethanoic acid + aminoethane N-ethylethanamide + water O H || | CH3 – C – OH + CH3 – CH2 – NH2 CH3 – C – N – CH2 – CH3 + HOH || O COMMON NAMES COMMON NAME IUPAC NAME COMMON NAME IUPAC NAME ethylene ethene formic acid carboxyl group (COOH) propylene propene acetic acid ethanoic acid acetylene ethyne toluene / phenyl methane methyl benzene formaldehyde methanal acetate ethanoate acetaldehyde ethanal acetamide ethanamide acetone propanone FLOW CHART OF ORGANIC REACTIONS SCH 4U REVIEW 14 of 39 CHAPTER 2: POLYMERS polymer – a molecule of large molar mass that consists of many repeating subunits called monomers; two types: addition and condensation monomer – a molecule or compound usually containing carbon and of relatively low molecular weight and simple structure which is capable of conversion to polymers by combination with itself or other similar molecules or compounds dimer – a molecule made up of two monomers ADDITION POLYMERS addition polymer – a polymer formed when monomer units are linked through addition reactions; all atoms present in the monomer are retained in the polymer alkene + alkene polymer | | | | | | | | | | monomer C=C + C=C –C–C–C–C– or – C– C – | | | | | | | | | | n less reactive than their monomers, because the unsaturated alkene monomers have been transformed into saturated carbon skeletons of alkanes forces of attraction are largely van der Waals attractions, which are individually weak, allowing the polymer chains to slide along each other, rendering them flexible and stretchable CONDENSATION POLYMERS condensation polymer – a polymer formed when monomer units are linked through condensation reactions; a small molecule is formed as a byproduct SCH 4U REVIEW 15 of 39 polyester – a polymer formed by condensation reactions resulting in ester linkages between monomers dialcohol + dicarboxylic acid polyester (dimer) + water O O O O || || || || HO – CH2 – CH2 – OH + C–C – O – CH2 CH2 – O – C – C – + 2 H2O | | n HO OH polyamide – a polymer formed by condensation reactions resulting in amide linkages between monomers; also known as a nylon diamine + dicarboxylic acid polyamide (dimer) + water H H O O H H O O | | || || | | || || N – CH2 – N + C–C – N – CH2 – N – C – C – + 2 H2O | | | | n H H HO OH COMPOUNDS OF LIFE protein – a large complex molecule made up of one or more chains of amino acids (an amino group and carboxyl group attached to the same carbon atom) – perform a wide variety of activities in the cell, including muscular growth, cellular repair, and serve as building blocks for all body tissue O || – NH – CH – C – | n R carbohydrate – a compound of carbon, hydrogen, and oxygen, with a general formula Cx(H2O)y – a major source of food energy, including sugars, starches, and cellulose – produced through photosynthesis in plants – provides an equivalent amount of energy as an equal mass of fatty acids fat – known chemically as a triglyceride (an ester of three fatty acids which are long-chain carboxylic acids and one glycerol molecule) – serves as a storage system, reserve supply of energy, and insulation glycerol + fatty acids fat (triglyceride) H O O | 3 || || H – C – OH HO – C – R H–C–O–C–R | | H – C – OH H O | || H – C – OH H – C – O – C – R’ | SCH 4U REVIEW 16 of 39 H O || H – C – O – C – R” | H nucleic acid – hereditary information stored in all living cells from which the information can be transferred; the chief types being DNA and RNA – DNA is created from four different nucleotides (monomer consisting of a ribose sugar, a phosphate group, and one of four possible nitrogenous bases) – carries the genetic information that encodes proteins and enables cells to reproduce and perform their functions nitrogenous phosphate base sugar CHAPTER 3: ATOMIC THEORY THE BOHR ATOMIC THEORY electrons travel in the atom in circular orbits with quantized energy – energy is restricted to only certain discrete quantities there is a maximum number of electrons allowed in each orbit electrons “jump” to a higher level when a photon (a quantum of light energy) is absorbed, resulting in absorption spectrum (series of dark lines) electrons “drop” to a lower level when a photon is emitted, resulting in bright-line spectrum (series of bright lines) ORBITS VS. ORBITALS orbit – 2-D path; fixed distance from nucleus; circular or elliptical path; 2n2 electrons per orbit orbital – 3-D region in space; variable distance from nucleus; no path and varied shape of region; 2 electrons per orbital; predicted by Schrodinger’s equation QUANTUM NUMBERS TO DESCRIBE ORBITALS n – principal quantum number or energy level l – secondary quantum number or subshell (s, p, d, or f) m1 – magnetic quantum number (direction of the electron orbit) m2 – spin quantum number (can only be +½ or –½ to describe spin of electron) VALUE OF l SUBLEVEL NUMBER OF MAX. NUMBER PRESENT AT SYMBOL ORBITALS OF ELECTRONS n= 0 s 1 2 1-7 1 p 3 6 2-7 2 d 5 10 3-7 3 f 7 14 4-7 6p 5d SCH 4U REVIEW 17 of 39 - 32 e 6s 4f 5p - 4d 18 e 5s 4p - 3d 18 e 4s 3p - 8e 3s 2p - 8e 2s - 2e 1s CREATING ENERGY-LEVEL DIAGRAMS energy-level diagram – interpretation of which orbital energy levels are occupied by electrons for a particular atom or ion; also called an orbital diagram m2 (spin) m1 (one orbital is distinguished from another at the same sublevel) e.g., 9F 2p 2s l (secondary quantum number or sublevel) 1s n (primary quantum number or energy level) RULES FOR ENERGY-LEVEL DIAGRAMS Start adding electrons to the lowest energy level (1s) and build up from the bottom until the limit on the number of electrons for the particle is reached – the aufbau principle To obtain the correct order of orbitals for any atom, start at the hydrogen and move from left to right across the periodic table, filling the orbitals; see below: SCH 4U REVIEW 18 of 39 For anions, add extra electrons to the number for the atom. For cations, do the neutral atom first, and then subtract the required number of electrons from the orbitals with the highest principle quantum number, n (i.e., you would remove the electrons from 4s, rather than 3d.) Each orbital will hold a maximum of two electrons that spin in opposite directions – the Pauli exclusion principle Electrons must be distributed among orbitals of equal energy so that as many electrons remain unpaired as possible – Hund’s rule Half-filled and filled subshells are more stable than unfilled subshells as the overall energy state of the atom is lower after the electron is promoted to a lower energy level: Predicted Actual Cr: [Ar] Cr: [Ar] Cu: [Ar] Cu: [Ar] 4s 3d 4s 3d ELECTRON CONFIGURATION electron configuration – a method for communicating the location and number of electrons in electron energy levels principal quantum number e.g., O: 1s2 2s2 2p4 3p5 number of electrons in orbital(s) S2-: 1s2 2s2 2p6 3s2 3p6 orbital shorthand electron configuration – when the electron configuration is written with the preceding noble gas placed before the subshell information (e.g., Cl: 1s2 2s2 2p6 3s2 3p5 becomes Cl: [Ne] 3s2 3p5) isoelectronic – when two atoms/ions have the same electron configuration (e.g., Ne, F-, Na+) ferromagnetism – exhibited by the metals iron, cobalt, nickel and a number of alloys that become magnetized in a magnetic field and retain their magnetism when the field is removed paramagnetism – exhibited by materials like aluminum or platinum that become magnetized in a magnetic field but it disappears when the field is removed (caused by unpaired electrons) QUANTUM MECHANIC THEORY quantum mechanics – the current theory of atomic structure based on wave properties of electrons; also known as wave mechanics Heisenberg uncertainty princple – it is impossible to simultaneously know exact position and speed of a particle SCH 4U REVIEW 19 of 39 electron probablity density – a mathematical or graphical representation of the chance of finding an electron in a given space; see below for example of a 2s orbital: CHAPTER 4: CHEMICAL BONDING chemical bond – formed as the result of the simultaneous attraction of two or more nuclei TYPES OF BONDS 1. Ionic Bonding – when one atom has low ionization energy and low En, while the other has high ionization energy and high En (i.e., metal and nonmetal) – Δ En > 1.7 – the electrostatic attraction between positive and negative ions 2. Covalent Bonding – when both atoms have high ionization energy and high En (i.e., two nonmetals) – Δ En ≤ 1.7 – the sharing of valence electrons between atomic nuclei 3. Metallic Bonding – when both atoms have low ionization energy and low En (i.e., two metals) LEWIS THEORY OF BONDING Atoms and ions are stable if they have a noble gas-like electron structure (i.e., a stable octet of electrons). Electrons are most stable when they are paired. Atoms form chemical bonds to achieve a stable octet of electrons. A stable octet may be achieved by an exchange of electrons between metal and nonmetal atoms. A stable octet may be achieved by the sharing of electrons between nonmetal atoms, resulting in a covalent bond. LEWIS STRUCTURES Lewis structure – a symbolic depiction of the distribution of valence electrons in a molecule 1. Arrange atoms symmetrically around the central atom (usually listed first in the formula, not usually oxygen and never hydrogen). 2. Count the number of valence electrons of all atoms. For polyatomic ions, add electrons corresponding to the negative charge, and subtract electrons corresponding to the positive charge on the ion. SCH 4U REVIEW 20 of 39 3. Determine the number of valence electrons all the atoms “want”, and subtract the number of valence electrons it has (result from step 2). Divide this number by 2 to determine the number of bonds the molecule will have. 4. Place a bonding pair of electrons between the central atom and each of the surrounding atoms. 5. Complete the octets of the surrounding atoms using lone pairs of electrons. Remember hydrogen (2 e-), beryllium (4 e-), and boron (6 e-) are the exceptions. Any remaining electrons go on the central atom. 6. If the central atom does not have an octet, move lone pairs from the surrounding atoms to form double or triple bonds until the central atom has a complete octet. Confirm the number of bonds is correct by comparing it to the result in step 3. 7. Draw the Lewis structure and enclose polyatomic ions within square brackets showing the ion chare. e.g., SO3 O S O3 has 6 3 (6) 6 + 18 = 24 e- wants 8 3 (8) 8 + 24 = 32 e- S difference = 8 e- = 4 bonds O O VALENCE BOND THEORY Covalent bonds form when atomic or hybrid orbitals with one electron overlap to share e-. Bonding occurs with the highest energy (valence shell) electrons. Normally, the s orbitals (sphere shape) and p orbitals (dumb-bell shape) overlap with each other to form bonds between atoms. sp3, sp2, and sp hybrid orbitals are formed from one s orbital and three, two, and one p orbital, respectively, with orientations of tetrahedral (109.5°), trigonal planar (120°), and linear (180°), respectively. End-to-end overlap of orbitals or single covalent bonds is called a sigma (σ) bond. Side-by-side overlap of unhybridized p orbitals is called a pi (Π) bond (weaker than σ bond). Double bonds have one pi bond, while triple bonds have two pi bonds. VSEPR THEORY Valence-Shell Electron-Pair Repulsion Theory – pairs of electrons in the valence shell of an atom stay as far apart as possible to minimize repulsion of their negative charges 1. Draw the Lewis structure for the molecule, including the e- pairs around the central atom. 2. Count the total number of bonding pairs and lone pairs of electrons around the central atom. 3. Use the table below to predict the shape of the molecule. # OF e- PAIRS ORIENTATION NUMBER OF AROUND OF BONDING BOND MOLECULAR SHAPE EXAMPLE CENTRAL ELECTRON AND LONE ANGLES GEOMETRY ATOM PAIRS PAIRS 2 linear 2 BP, 0 LP 180° linear BeF2 X–A–X triangular trigonal X 3 3 BP, 0 LP 120° BF3 | planar planar SCH 4U REVIEW 21 of 39 A X X X | tetrahedral 4 BP, 0 LP 109.5° tetrahedral CH4 A X X X trigonal NH3 A 4 tetrahedral 3 BP, 1 LP < 109.5° X X pyramidal PCl3 X v-shaped H2O A tetrahedral 2 BP, 2 LP < 109.5° or bent OF2 X X tetrahedral 1 BP, 3 LP 180° linear HCl A–X X X | X trigonal 120° & trigonal 5 5 BP, 0 LP PCl5 A bipyramidal 90° bipyramidal | X X X X | X 6 octahedral 6 BP, 0 LP 90° octahedral SF6 A X | X X e.g., HOF(l) .. .. O H : . . : .F. : 2 BP and 2 LP, therefore HOF(l) is v-shaped: O H F Note: – ions are treated in the same way, but square brackets are placed around the diagram with the charge placed in the upper right hand corner – double and triple bonds are treated as one group of electrons when using VSEPR theory, and most of those molecules take the same shape as their Lewis structure POLAR MOLECULES bond dipole – the electronegativity difference of two bonded atoms represented by an arrow pointing from the lower (∂+) to the higher (∂-) electronegativity nonpolar molecule – a molecule that has either nonpolar bonds (≤0.5 Δ En) or polar bonds whose bond dipoles cancel to zero (i.e., VSEPR diagram is symmetrical) polar molecule – a molecule that has polar bonds with dipoles that do not cancel to zero e.g., NH3 . . ∂- N 3.0 H H 2.1 therefore, it is a polar molecule ∂+ 2.1 H 2.1 ∂+ INTERMOLECULAR FORCES ∂+ intermolecular force – the force of attraction and repulsion between molecules; much weaker than covalent bonds SCH 4U REVIEW 22 of 39 dipole-dipole force – a force of attraction due to the simultaneous attraction of one dipole by its surrounding dipoles – the more polar the molecules, the stronger the force – the shape of the molecule also affect the dipole-dipole strength (i.e., closer together = stronger force) London dispersion force – the simultaneous attraction of an electron by the positive nuclei in the surrounding molecules – the greater the number of electrons per molecule, the stronger the force As dipole-dipole force or London dispersion force increases, the boiling point increases, allowing you to predict relative boiling point, if one of the forces is the same between the two substances, or if both forces are moving in the same direction. hydrogen bonding – the attraction of a hydrogen atom to a lone pair of electrons in N, O, or F atoms in adjacent molecules (possible because H has no valence e- to “shield” nucleus) .. .. H – .F : ------------ H – .F : . . STRUCTURE AND PROPERTIES OF CRYSTALLINE SOLIDS PROPERTIES FORCE/ SOLUBILITY CRYSTAL PARTICLES MELTING CONDUC- EXAMPLES BOND IN LIQUID/ POINT TIVITY SOLUTION Ionic medium (higher as NaCl, ions ionic Na3PO4, ionic charge yes yes (+ and -) CaF2, MgO, increases, CuSO4•5H2O size of ion decreases) metallic (fixed high nuclei with mobile (higher as Pb, Fe, Metallic cations delocalized number of yes no Cu, Al, Ag, electrons, valence Au allowing for electrons conduction increases) of heat and electricity) SCH 4U REVIEW 23 of 39 Polar Molecular low (higher as PF3, ICl, molecules number of yes CHCl3, H2O, electrons, London, polarity of NH3, SO3 dipole- molecules, no dipole, and hydrogen presence of inert gases, molecules hydrogen diatomic Nonpolar Molecular or single bonds no elements, atoms increases) CO2, CCl4, SF6, BF3 Covalent Network covalent Crystal (a 3-D very high arrangement (melting of strong point diamond, SiC, covalent increases AlN, BeO, atoms bonds no no as strength SiO2, CuCl2, between of Mg2S atoms; covalent resulting in bonds hardness and increase ) high melting point) NETWORK SOLIDS (AKA “SUPERMOLECULES”) ALLOTROPES OF CARBON: a) diamond – a 3-D network solid – each C atom is in the centre of a tetrahedron whose vertices are occupied by other C atoms – each C atom shares its valence e- with 4 other C atoms – bonding e- are tightly bound and highly localized a) graphite – a 2-D network solid – each C atom is surrounded by a 3 others in a plane – the double bond consists of delocalized electrons, therefore a good conductor – separate layers are held together by dispersion forces and are easily separated SILICA AND THE SILICATES: silicon combines with oxygen to form silicon dioxide, SiO2 (silica) it further reacts with metal compounds to produce metal silicates SCH 4U REVIEW 24 of 39 a) quartz – a 3-D network solid b) mica – a 2-D network solid c) asbestos – a 1-D network solid CHAPTER 5: THERMOCHEMISTRY thermochemistry – the study of the energy changes that accompany physical or chemical changes of matter thermal energy – energy available from 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 the system that is capable of absorbing or releasing thermal energy heat – amount of energy transferred between substances exothermic – releasing thermal energy as heat flows out of the system; negative molar enthalpy endothermic – absorbing thermal energy as heat flows into the system; positive molar enthalpy temperature – average kinetic energy of the particles in a sample of matter open system – one in which both matter and energy can move in or out (e.g., burning marshmallow) isolated system – an ideal system in which neither matter nor energy can move in or out (e.g., ideal calorimeter) closed system – one in which energy can move in or out, but not matter (e.g., realistic calorimeter) calorimetry – the process of measuring energy changes in a chemical system enthalpy change (ΔH) – the difference in enthalpies of reactants and products during a change q = quantity of heat (J) = m c ΔT m n M MOLAR ENTHALPY molar enthalpy (ΔHx) – the enthalpy change associated with a physical, chemical, or nuclear change involving one mole of a substance; examples are below: TYPE OF MOLAR ENTHALPY EXAMPLE OF CHANGE solution (ΔHsol) + NaBr(s) Na (aq) + Br-(aq) combustion (ΔHcomb) CH4(g) + 2 O2(g) CO2(g) + H2O(l) vaporization (ΔHvap) CH3OH(l) CH3OH(g) freezing (ΔHfr) H2O(l) H2O(s) neutralization (ΔHneut) 2 NaOH(aq) + H2SO4(aq) 2 Na2SO4(aq) + 2 H2O(l) formation (ΔHf) C(s) + 2 H2(g) + ½ O2(g) CH3OH(l) ΔH = n ΔHvap or sol = mcΔT ΔH = q n SCH 4U REVIEW 25 of 39 ASSUMPTIONS USED IN CALORIMETRY: 1. No heat is transferred between the calorimeter and the outside environment. 2. Any heat absorbed or released by the calorimeter materials is usually negligible. 3. A dilute aqueous solution is assumed to have a density and specific heat capacity equal to that of pure water (1.00 g/mL and 4.18 J/g•°C). LAB: COMBUSTION OF ALCOHOLS Mass of Fuel Burned: 1.41 g Mass of Water Heated: 97.00 g Mass of Water Vapourized: 0.18 g Mass of Pop Can: 16.07 g Temperature Change of Water: 22.2 °C Heat Absorbed by Water: q = m c Δt = (97.00 g) (4.18 J / g · °C) (22.2 °C) = 9.00 x 103 J Heat Absorbed by Can: q = m c Δt = (16.07 g) (0.900 J / g · °C) (22.2 °C) = 321 J Heat Used to Vapourize Water: q = m · LV = (0.18 g) (2268 J / g) = 4.0 x 102 J Total Heat Evolved by Fuel: qtotal = (9.00 x 103 J) + (321 J) + (4.0 x 102 J) = 9.72 x 103 J = 9.72 kJ Number of Moles of Fuel: n=m M = 1.41 g 60.11 g / mol = 0.0235 mol Molar Heat of Combustion of Fuel: ΔH = qtotal n = 9.72 kJ 0.0235 mol = 414 kJ/mol % error: 100% – experimental enthalpy x 100% actual enthalpy = 100% – 414 kJ x 100% 490 kJ = 15.5%, therefore unacceptable (>10%) REPRESENTING ENTHALPY CHANGES METHOD 1: THERMOCHEMICAL EQUATIONS WITH ENERGY TERMS e.g., C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l) + 2802.7 kJ SCH 4U REVIEW 26 of 39 METHOD 2: THERMOCHEMICAL EQUATIONS WITH ΔH VALUES e.g., C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l) ΔH = -2802.7 kJ METHOD 3: MOLAR ENTHALPIES OF REACTION e.g., ΔHrespiration = -2802.7 kJ/mol glucose METHOD 4: POTENTIAL ENERGY DIAGRAM e.g., C6H12O6(s) + 6 O2(g) Ep (kJ) ΔH = -2802.7 kJ 6 CO2(g) + 6 H2O(l) Reaction Progress STANDARD ENTHALPY OF FORMATION standard enthalpy of formation (ΔH°f) – the quantity of energy associated with the formation of one mole of a substance from its elements in standard state; zero for elements in standard state HESS’S LAW (ADDITIVITY OF REACTION ENTHALPIES) Hess’s law – the value of the ΔH for any reaction that can be written in steps equals the sum of the values of ΔH for each of the individual steps (i.e., ΔHtarget = Σ ΔHknown) e.g. Determine the enthalpy change involved in the formation of two moles nitrogen monoxide from its elements. N2(g) + O2(g) 2 NO(g) (1) ½ N2(g) + O2(g) NO2(g) ΔH°1 = +34 kJ (2) NO(g) + ½ O2(g) NO2(g) ΔH°2 = -56 kJ 2 x (1): N2(g) + 2 O2(g) 2 NO2(g) ΔH°1 = 2(+34) kJ -2 x (2): 2 NO2(g) 2 NO(g) + O2(g) ΔH°2 = -2(-56) kJ N2(g) + O2(g) 2 NO(g) ΔH° = +68 kJ + 112 kJ = +180 kJ USING STANDARD ENTHALPIES OF FORMATION TO DETERMINE ΔH ΔH = Σ nΔH°f (products) – Σ nΔH°f (reactants) e.g., MULTI-STEP CALCULATION If 3.20 g of propane burns, what temperature change will be observed if all of the heat from combustion transfers into 4.0 kg of water? C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) ΔH°f (CO2) = -393.5 kJ/mol ΔH°f (H20) = -285.8 kJ/mol ΔH°f (C3H8) = -104.7 kJ/mol ΔH°f (O2) = 0.0 kJ/mol mpropane = 3.20 g SCH 4U REVIEW 27 of 39 mH2O = 4.0 kg cH2O = 4.18 J/(g•°C) ΔH = Σ nΔH°f (products) – Σ nΔH°f (reactants) = 3 mol x -393.5 kJ + 4 mol x -285.8 kJ – 1 mol x -104.7 kJ + 5 mol x 0.0 kJ 1 mol 1 mol 1 mol 1 mol = -2219 kJ ΔHc (propane) = qwater n ΔHc = mcΔT ΔT = n ΔHc mc = mpropane ΔHc Mpropane mwater c = (3.20 g) (2219 kJ) (44.11 g) (4.0 kg) (4.18 J/g•°C) = 9.6°C CHAPTER 6: CHEMICAL KINETICS chemical kinetics – the area of chemistry that deals with rates of reaction rate of reaction – the speed at which a chemical change occurs, generally expressed in concentration per unit time, such as mol/(L•s) rate = Δc Δt average rate of reaction – the speed at which a reaction proceeds over a period of time; determined using slope of a secant (line between two points) instantaneous rate of reaction – the speed at which a reaction is proceeding at a particular point in time; determined using a tangent METHODS TO MEASURE RATE: pH change; conductivity; volume of gas produced; change in mass of products; change in colour RATE LAW EQUATION r = k [X]m [Y]n k = rate constant; valid only for a specific temperature [X] and [Y] = concentrations of reactants m and n = order of reaction (describes the initial concentration dependence of a particular reactant) overall order of reaction – the sum of the exponents in the rate law equation e.g., r = k[BrO3(aq)-]1 [HSO3(aq)-]2, therefore overall order is 3 (1 + 2) zeroth-order reaction – the rate does not depend on [A] e.g., if the initial concentration of A is doubled, the rate will multiply by 1 (20), and so will be unchanged SCH 4U REVIEW 28 of 39 first-order reaction – the rate is directly proportional to [A] e.g., if the initial concentration of A is doubled, the rate will multiply by 2 (21) second-order reaction – the rate is proportional to the square of [A] e.g., if the initial concentration of A is doubled, the rate will multiply by 4 (22) e.g., NO(g) + H2(g) HNO2(g) EXPERIMENT NO (MOL/L) H2 (MOL/L) INITIAL RATE OF REACTION (MOL/(L•S) 1 0.001 0.004 0.002 2 0.002 0.004 0.008 3 0.003 0.004 0.018 4 0.004 0.001 0.008 5 0.004 0.002 0.016 6 0.004 0.003 0.024 a) Write the rate law for the reaction. r = k [NO(g)]2 [H2(g)]1 The units for rate constant, k, are b) Write the overall order of the reaction. related to overall order of reaction: 3rd order first order overall, the units are 1/s c) Calculate k for the reaction (use for experiment’s values). or s-1 r = 0.02 mol/L•s r = k [NO(g)]2 [H2(g)] second order overall, the units are [NO] = 0.001 mol/L (0.02) = k (0.001)2 (0.004) L/(mol•s) or L/(mol-1•s-1) [H2] = 0.004 mol/L k = 5 x 106 L2/(mol2•s) third order overall, the units are L2/(mol2•s) or L/(mol-2•s-1) COLLISION THEORY CONCEPTS OF THE COLLISION THEORY: A chemical system consists of particles (atoms, ions, or molecules) that are in constant random motion at various speeds. The average kinetic energy of the particles is proportional to the temperature of the sample. A chemical reaction must involve collisions of particles with each other or the walls of the container. 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 formed. Ineffective collisions involve particles that rebound from the collision unchanged. SCH 4U REVIEW 29 of 39 The rate of a given reaction depends on the frequency of collision and the fraction of those collisions that are effective. activated complex activation energy – the minimum increase in potential energy activation of a system energy Ep required for products molecules to react net potential energy activated complex – an unstable reactants change, ΔH chemical species containing partially broken Reaction Progress and partially formed bonds representing the maximum potential energy point in the change; also called the transition state reaction mechanism – a series of elementary steps that makes up an overall reaction elementary step – a step in a reaction mechanism that only involves one-, two-, or three-particle collisions rate-determining step – the slowest step in a reaction mechanism reaction intermediates – molecules formed as short-lived products in reaction mechanisms rate-determining step e.g., HBr(g) + O2(g) HOOBr(g) (slow) elementary step HOOBr(g) + HBr(g) 2 HOBr(g) (fast) reaction 2 HOBr(g) + HBr(g) H2O(g) + Br2(g) (fast) mechanism 4 HBr(g) + O2(g) 2 H2O(g) + 2 Br2(g) reaction intermediate FACTORS AFFECTING RATE OF REACTION 1. NATURE OF THE REACTANTS: each reactant contains a different number of bonds, each with differing bond strengths, that must be broken for the reaction to proceed each reactant has a different threshold energy (minimum kinetic energy required to convert kinetic energy to activation energy) each reactant requires a different collision geometry that can be simple or complex 2. TEMPERATURE : an increase in temperature increases the rate of reaction as temperature rises, the reactant particles gain kinetic energy, moving faster, colliding more frequently, and thus reacting more quickly with a higher temperature, a larger fraction of the molecules will have the required kinetic energy to have effective collisions 3. CONCENTRATION: an increase in reactant concentration increases the rate of reaction SCH 4U REVIEW 30 of 39 the greater the concentration, the greater the number of particles per unit volume, which are more likely to collide as they move randomly within a fixed space 4. SURFACE AREA: an increase in reactant surface area increases the rate of reaction reactants can collide only at the surface where the substances are in contact, and by increasing the surface area, you are increasing the number of particles in an area, thereby increasing the probability of an effective collision 5. CATALYST: a catalyst is a substance that increases the rate of a chemical reaction without itself being permanently changed a catalyst provides an alternate “pathway”, with lower activation energy, to the same product formation, meaning a much larger fraction of collisions are effective the catalyst can help break the bonds in the reactant particles, provide a surface for the necessary collisions, and allow the reactants’ atoms to recombine in new ways catalysts are involved in the reaction mechanism at some point, but are regenerated before the reaction is complete CHAPTER 7: CHEMICAL SYSTEMS IN EQUILIBRIUM equilibrium – the balanced state of a reversible reaction or process where there is no net observable change; the rate of the forward reaction equals that of the reverse reaction (A ↔ B) – can be approached from either side of the reaction equation – the concentration of the reactants and products do not change (are constant) solubility equilibrium – an equilibrium between a solute and a solvent in a saturated solution phase equilibrium – an equilibrium between different physical states of a pure substance (e.g., ice over a lake) chemical reaction equilibrium – an equilibrium between reactants and products of a chemical reaction EQUILIBRIUM LAW EQUATION For the general chemical reaction: aA + bB ↔ cC + dD K = [C]c [D]d where: • A, B, C, D are chemical entities in gas or aqueous phases (liquids and [A]a [B]b solids are omitted from the equation) • a, b, c, and d are the coefficients in the balanced chemical equation • K is the equilibrium constant (temperature and pressure specific) - if K is large, reaction’s concentration of products greater than reactants - if K is small, reaction’s concentration of reactants greater than products - K is inversely proportional to the K value of the reverse reaction LE CHÂTELIER’S PRINCIPLE Le Châtelier’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 concentrations of reactants and products VARIABLES AFFECTING CHEMICAL EQUILIBRIA: VARIABLE TYPE OF CHANGE RESPONSE OF SYSTEM SCH 4U REVIEW 31 of 39 concentration increase shifts to consume added reactant decrease shifts to replace removed reactant temperature increase shifts to consume added thermal energy (away from heat term) decrease shifts to replace removed thermal energy (towards heat term) volume increase (decrease in shifts towards side with larger total number pressure) of gaseous entities decrease (increase in shifts towards side with smaller total pressure) number of gaseous entities common ion effect dissolving a compound into shifts away from common ion to consume solution that adds a the added reactant common ion VARIABLES THAT DO NOT AFFECT CHEMICAL EQUILIBRIA catalysts - no effect adding inert gases - no effect SOLVING EQUILIBRIUM PROBLEMS 1. Write a balanced equation for the reaction and list the known values. 2. If the direction the system must go to attain equilibrium is not obvious (i.e., one entity is not present initially), calculate Q with the initial concentrations and compare it to the value of K to determine which direction the system will proceed to attain equilibrium. 3. Construct an ICE (Initial concentration, Change in concentration, Equilibrium concentration) table and input the initial concentrations. 4. Let x represent the changes in concentration, multiplying it by the coefficient in the balanced equation. The reactants should all change in the same way and all the products should proceed in the opposite way. 5. Rewrite the E row using the x values. 6. Substitute equilibrium concentrations into the equilibrium constant equation. 7. Apply appropriate simplifying assumptions, if possible (e.g., 4x3 ÷ (0.4 – 2x)2 can be simplified to 4x3 ÷ (0.4)2 because x value is so small in comparison). 8. Solve for x. 9. Justify any assumptions you have made (i.e., the x value you get should be plugged into the original equation and the difference between the two must be less than 5%). 10. Calculate the equilibrium concentrations by substituting x into equilibrium concentration expressions from the E row. e.g., 4.00 mol of hydrogen and 2.00 mol of iodine are placed in a 2.00-L reaction vessel at 440°C and react to form hydrogen iodide. At this temperature, the K is 49.7. Determine the concentrations of all entities. H2(g) + I2(g) ↔ 2 HI(g) K = 49.7 [H2(g)] = 4.00 mol [I2(g)] = 2.00 mol [HI(g)] = 0.00 mol/L 2.00 L 2.00 L = 2.00 mol/L = 1.00 mol/L H2(g) + I2(g) ↔ 2 HI(g) Initial concentration (mol/L) 2.00 1.00 0.00 Change in concentration (mol/L) –x –x +2 x Equilibrium concentration (mol/L) 2.00 – x 1.00 – x 2x SCH 4U REVIEW 32 of 39 2 K = [HI] [H2(g)] = 2.00 mol/L - x [H2] [I2] = 2.00 mol/L - (0.93 mol/L) 49.7 = (2x)2 = 1.07 mol/L (2.00 – x) (1.00 – x) [I2(g)] = 1.00 mol/L - x 4x = 49.7 (2.00 – x) (1.00 – x) 2 = 1.00 mol/L - (0.93 mol/L) 0.92x2 – 3.00x + 2.00 = 0 = 0.07 mol/L [HI(g)] = 2x x = –b ± √ b2 – 4ac = 2(0.93 mol/L) 2a = 1.87 mol/L = 3.00 ± √ 9.00 – 7.36 1.84 = 2.33 or 0.93 2.33 is rejected, because concentrations cannot have a negative value (i.e., 2.00 – 2.33 = - 0.33) SOLUBILITY PRODUCT CONSTANT solubility – the concentration of a saturated solution of a solute in a particular solvent at a particular temperature; specific maximum concentration solubility product constant (Ksp) – the value obtained from the equilibrium law applied to saturated solution (remember solids are not included in the equation); omit units as with all K values – can only be determined for ionic compounds that are classified as insoluble or slightly soluble 1. Write a balanced equation and list the known values. 2. Use the solid product to write an equilibrium equation for it dissolving into ions. 3. Find the Ksp value for the solid product and write it next to the equilibrium equation. 4. Determine the number of moles of both ions, by using the mole ratios and initial concentrations of reactants. 5. Determine the concentration upon mixing, by dividing the number of moles by the new volume. 6. Plug these new concentrations into the Ksp equation to determine the Q (experimental value). 7. Compare the Q value to the Ksp to predict whether a precipitate will form (see below). USING Q TO PREDICT SOLUBILITY Ion product, Q > Ksp precipitate will form (supersaturated solution) Ion product, Q = Ksp precipitate will not form (saturated solution) Ion product, Q < Ksp precipitate will not form (unsaturated solution) e.g., 20.0 mL of 0.20 mol/L ammonium sulfate solution is added to 130 mL of 0.50 mol/L barium nitrate solution. What are the concentrations of the ions and will a precipitate form? (NH4)2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2 NH4NO3(aq) Choose the solid to write the equilibrium equation (remember any - NO3 molecule is aqueous) BaSO4(s) ↔ Ba2+(aq) + SO42-(aq) Ksp = 1.08 x 10-10 nBa2NO3 = c v nNH4NO3 = c v = (0.50 mol/L) (0.130 L) = (0.20 mol/L) (0.020 L) = 0.065 mol = 0.0040 mol = nBa = nSO4 [Ba2+(aq)] = n [SO42-(aq)] = n v v Divide by the “new” volume (i.e., 20 mL and 130 mL equals 150 mL) SCH 4U REVIEW 33 of 39 = 0.065 mol = 0.004 mol 0.150 L 0.150 L = 0.43 mol/L = 0.027 mol/L Ksp = [Ba2+] [SO42-] = (0.43) (0.027) = 0.011 0.011 > 1.08 x 10-10, therefore it will precipitate ENTROPY spontaneous reaction – one that, given the necessary activation energy, proceeds without continuous outside assistance entropy, S – a measure of the randomness or disorder of a system or the surroundings – equals 0 when the temperature is at absolute zero (0 K) FACTORS THAT INCREASE ENTROPY (S) the volume of a gaseous system increases (i.e., pressure decreases) the temperature of a system increases the physical state of a system changes from solid to liquid to gas, or liquid to gas (i.e., Sgas > Sliquid > Ssolid) fewer moles of reactant molecules form a greater number of moles of product molecules complex molecules are broken down into simpler subunits (e.g., combustion of organic fuels into carbon dioxide and water) CLASSIFICATION OF SPONTANEOUS AND NONSPONTANEOUS REACTIONS Endothermic (ΔH > 0) Exothermic (ΔH < 0) spontaneous at high temps. spontaneous Entropy increases (ΔS > 0) nonspontaneous at low temps. C(s) + O2(g) CO2(g) H2O(s) H2O(l) spontaneous at low temps. nonspontaneous Entropy decreases (ΔS < 0) nonspontaneous at high temps. 3 O2(g) 2 O3(g) 2 SO2(g) + O2(g) 2 SO3(g) Gibb’s free energy, G – energy that is available to do useful work; ΔG° = ΔH° – (T ΔS°) CHAPTER 8: ACID-BASE EQUILIBRIUM BRØNSTED-LOWRY THEORY Brønsted-Lowry acid – proton donor Brønsted-Lowry base – proton acceptor amphoteric (amphiprotic) – a substance capable of acting as an acid or a base in different chemical reactions; a substance that may accept or donate a proton conjugate pair e.g., acid HC2H3O2(aq) + H2O(l) ↔ C2H3O2-(aq) + H3O+(aq) base strong acid – an acid with a very weak attraction for protons and easily donates it to a base strong base – a base with a very strong attraction for protons The stronger an acid, the weaker its conjugate base, and vice versa. AUTOIONIZATION OF WATER SCH 4U REVIEW 34 of 39 autoionization of water – the reaction between two water molecules producing a hydronium ion and a hydroxide ion (H2O(l) ↔ H+(aq) + OH-(aq) Kw = [H+(aq)] [OH-(aq)] [H+(aq)] = Kw [OH-(aq)] = Kw = (1.0 x 10-7 mol/L) (1.0 x 10-7 mol/L) [OH-(aq)] [H+(aq)] = 1.0 x 10-14 In neutral solutions [H+(aq)] = [OH-(aq)] In acidic solutions [H+(aq)] > [OH-(aq)] In basic solutions [H+(aq)] < [OH-(aq)] PH pH – the negative of the logarithm to the base ten of the concentration of hydrogen (hydronium) ions in a solution pOH – the negative of the logarithm to the base ten of the concentration of hydroxide ions in a solution pH = –log [H+(aq)] pOH = –log [OH-(aq)] pH + pOH = 14.00 [H+(aq)] = 10–pH [OH-(aq)] = 10–pOH e.g., Calculate the pH of a solution prepared by dissolving 4.3 g of Ba(OH)2(s) in water to form 1.5 L of solution. 100% Ba(OH)2(aq) Ba2+(aq) + 2 OH–(aq) nBa(OH)2 = 4.3 g x 1 mol 171.3 g = 2.5 x 10-2 mol [Ba(OH)2(aq)] = 2.5 x 10-2 mol 1.5 L = 1.7 x 10-2 mol/L [OH-(aq)] = 2 (1.7 x 10-2 mol/L) = 3.3 x 10-2 mol/L pOH = –log [OH-(aq)] = –log (3.3 x 10-2) = 1.47 pH = 14.00 – pOH = 14.00 – 1.47 = 12.53 STRONG VS. WEAK ACIDS AND BASES strong acid – an acid that is assumed to ionize quantitatively (completely) in aqueous solution (i.e., percent ionization is > 99%); HCl(aq), HNO3(aq), and H2SO4(aq) are the only strong acids we will work with strong base – an ionic substance that dissociates completely in water to release hydroxide ions; all of the metal hydroxides are strong bases weak acid – an acid that partially ionizes in solution but exists primarily in the form of molecules weak base – a base that has a weak attraction for protons percent ionization (p) = concentration of acid ionized x 100% [H+(aq)] = p x [HA(aq)] concentration of acid solute 100 SCH 4U REVIEW 35 of 39 acid ionization constant (Ka) – equilibrium constant for the ionization of an acid e.g., Calculate the Ka and pH of hydrofluoric acid if a 0.100 mol/L solution at equilibrium at SATP has a percent ionization of 7.8%. HF(aq) 7.8% H+ - (aq) + F (aq) Ka = [H+(aq)] [F-(aq)] [HF(aq)] [H+(aq)] = (p/100) [HA(aq)] = (7.8 / 100) (0.100 mol/L) = 0.0078 mol/L HF(aq) ↔ H+(aq) + F-(aq) Initial concentration (mol/L) 0.100 0.000 0.000 Change in concentration (mol/L) –x +x +x Equilibrium concentration (mol/L) 0.100 – 0.0078 0.0078 0.0078 = 0.0922 Ka = [H+(aq)] [F-(aq)] pH = –log [H+(aq)] [HF(aq)] = –log (0.0078) = 0.00782 = 2.1 0.0922 = 6.6 x 10-4 CHAPTER 9: ELECTROCHEMISTRY (ELECTRIC CELLS) OXIDATION NUMBER oxidation number – an integer that is assigned to each atom in a compound when considering redox reactions – a positive or negative number corresponding to the apparent charge that an atom in a molecule or ion would have if the electron pairs in covalent bonds belonged entirely to the more electronegative atom RULES FOR ASSIGNING OXIDATION NUMBERS 1. The oxidation number of an atom in an uncombined element is always 0 (e.g., H2 is 0). 2. The oxidation number of a simple ion is the charge of ion (e.g., Ca2+ is +2). 3. The oxidation number of hydrogen is +1, except in metal hydrides when it is -1 (e.g., the H in NaH is -1). 4. The oxidation number of oxygen is -2, except in peroxides when it is -1 (e.g., the O in H2O2 is -1). 5. The oxidation number of Group 1 element ions is +1. The oxidation number of Group 2 element ions is +2. 6. The sum of oxidation numbers in a compound must equal 0. 7. The sum of oxidation numbers in a polyatomic ion must equal the charge on the ion (e.g., OH- is -1). OXIDATION-REDUCTION Oxidation can be defined as a reaction in which: 1) an element is chemically united with oxygen (e.g., C + O2 CO2; carbon is oxidized) 2) a metal is changed from an uncombined to a combined state (e.g., Zn + Cl2 ZnCl2) 3) an element loses electrons, and therefore has an increase in oxidation number (e.g., Ca Ca2+ + 2e-) SCH 4U REVIEW 36 of 39 Loss of Electrons is Oxidation “LEO” Reduction can be defined as a reaction in which: 1) an element loses oxygen (e.g., Fe2O3 2 FeO + ½ O2) 2) a metal is changed from a combined to a uncombined state (e.g., FeO Fe + ½ O2) 3) an element gains electrons, and therefore has a decrease in oxidation number (e.g., Cl2 + 2e- 2 Cl-) Gain of Electrons is Reduction “GER” redox reaction – a chemical reaction in which electrons are transferred between particles; two or more atoms undergo a change in oxidation number; also known as oxidation- reduction reactions – all single displacement reactions are redox, while some combination and decomposition reactions are; double displacement reactions are never redox e.g., oxidation +1 -2 0 +4 -2 +1 -2 H2S(g) + O2(g) SO2(g) + H2O(g) reduction BALANCING REDOX EQUATIONS USING OXIDATION NUMBERS this method is most appropriate when dealing with covalent compounds 1. Assign oxidation numbers to all the atoms in the equation. 2. Identify which atoms undergo a change in oxidation number. 3. Determine the ratio in which these atoms must react so that the total increase in oxidation number equals the decrease (i.e., the total number of electrons lost and gained is equal). 4. Balance the redox participants in the equation using this ratio. 5. Balance the other atoms by the inspection method. 6. Add H+(aq) or OH-(aq) to balance the charge, depending on if it is an acidic or a basic solution (the total charge on each side must be the same). 7. Add H2O(l) to balance the O atoms. e.g., Ag(s) + Cr2O72-(aq) Ag+(aq) + Cr3+(aq) oxidation: lost 1 e- (x 6) 0 +6 -2 +1 +3 6 Ag(s) + Cr2O72-(aq) + 14 H+(aq) 6 Ag+(aq) + 2 Cr3+(aq) + H2O(l) reduction: gained 2(3 e-) = 6 e- (x 1) BALANCING REDOX EQUATIONS USING HALF-REACTIONS this method is most appropriate for ionic reactions in solution and for relating to electrical processes 1. Separate the skeleton equation into the start of two half-reaction equations (one for the oxidation reaction and one for the reduction reaction). 2. Balance all species, other than O and H. 3. Balance the oxygen, by adding H2O(l) for acidic solutions or OH-(aq) for basic solutions. 4. Balance the hydrogen, by adding H+(aq) for acidic solutions or H2O(l) for basic solutions. SCH 4U REVIEW 37 of 39 5. Balance the charge on each side by adding electrons and canceling anything that is in equal amounts on both sides. 6. Multiply each half-reaction equation by simple whole numbers to balance the electrons lost and gained. 7. Add the two half-reaction equations, canceling the electrons and anything else that appears in equal amounts on both sides. 8. Check to ensure all entities and the overall charge on both sides balance. e.g., MnO4– + N2H4 MnO2 + N2 4 [MnO4– + 2 H2O + 3 e- MnO2 + 4 OH–] 3 [N2H4 + 4 OH– N2 + 4 H2O + 4 e-] 4 MnO4– + 8 H2O + 12 e- 4 MnO2 + 16 OH–] 3 N2H4 + 12 OH– 3 N2 + 12 H2O + 12 e-] 4 MnO4– + 3 N2H4 4 MnO2 + 3 N2 + 4 H2O + 4 OH– TECHNOLOGY OF CELLS AND BATTERIES electric cell – a device that continuously converts chemical energy into electrical energy; contains two electrodes (solid conductor) and one electrolyte (aqueous conductor); each electrode has a cathode (+) and anode (–); electrons flow from the anode to the cathode battery – a group of two or more electric cells connected in series voltage – the potential energy difference per unit charge; measured in volts (V), 1 J/C electric current – the rate of flow of charge past a point; measured in amperes (A), 1 C/s NAME OF CHARACTERISTICS TYPE HALF-REACTIONS CELL AND USES dry cell 2 MnO2 + 2 NH4+ + 2 e- Mn2O3 + 2 NH3 + H2O inexpensive, portable (1.5 V) Zn Zn2+ + 2 e- flashlights, radios longer shelf life, primary cell alkaline dry 2 MnO2 + H2O + 2 e- Mn2O3 + 2 OH- higher currents for – electric cell cell (1.5 V) Zn + 2 OH- ZnO + H2O + 2 e- longer periods that cannot same uses as dry cell be recharged small cell, constant mercury cell HgO + H2O + 2 e- Hg + 2 OH- voltage during life (1.35 V) Zn + 2 OH- ZnO + H2O + 2 e- hearing aids, watches completely sealed, Ni-Cad cell 2 NiO(OH) + 2 H2O + 2 e- 2 Ni(OH)2 + 2 OH- lightweight secondary (1.25 V) Cd + 2 OH- Cd(OH)2 + 2 e- power tools, shavers, cell – electric portable computers cell that can large currents, be recharged lead-acid cell PbO2 + 4 H+ + SO42- + 2 e- PbSO4 + 2 H2O reliable for recharges (2.0 V) Pb + SO42- PbSO4 + 2 e- all vehicles fuel cell – high energy density, aluminum-air 3 O2 + 6 H2O + 12 e- 12 OH- readily available electric cell cell (2 V) 4 Al 4 Al3+ + 12 e- aluminum alloys that produces electric cars SCH 4U REVIEW 38 of 39 electricity by lightweight, high a continually efficiency, can be hydrogen- supplied fuel O2 + 2 H2O + 4 e- 4 OH- adapted to use oxygen cell 2 H2 + 4 OH- 4 H2O + 4 e- hydrogen-rich fuels (1.2 V) vehicles and space shuttle GALVANIC CELLS galvanic cell – consists of two half-cells separated by a porous boundary with solid electrodes connected by an external circuit; standard cells occur at SATP and with concentrations of 1.0 mol/L cathode – positive electrode; reduction (gaining electrons because electronegativity is higher) of the strongest oxidizing agent occurs here; the half-cell that is higher on the “Relative Strengths of Oxidizing and Reducing Agents” table; “red cat on the roof” reduction at the cathode, higher half-cell anode – negative electrode; oxidation (losing electrons because electronegativity is lower) of the strongest reducing agent occurs here; the half-cell that is lower on the table inert electrode – a solid conductor that will not react with any substance present in a cell (usually carbon or platinum) Electrons travel in the external circuit from the anode to the cathode Internally, anions from the salt bridge move toward the anode and cations from the salt bridge move toward the cathode as the cell operates, keeping the solution electrically neutral CELL NOTATION electrons cathode (+) | electrolyte || electrolyte | anode (-) (reduction) (oxidation) CELL POTENTIALS standard cell potential (ΔE°) – the maximum electric potential difference (voltage) of a cell operating under standard conditions reference half-cell – a half-cell assigned an electrode potential of exactly 0.00 volts; the standard hydrogen half-cell, Pt(s) | H2(g), H+(aq) standard reduction potential (ΔEr°) – represents the tendency of a standard half-cell to attract electrons in a reduction half-reaction, compared to the reference half-cell SCH 4U REVIEW 39 of 39 standard oxidation potential (ΔEo°) – represents the tendency of a standard half-cell to lose electrons in an oxidation half-reaction; the value of the reverse reaction with an opposite sign ΔE° = ΔEr° – ΔEr° cell cathode anode A positive standard cell potential (ΔE° > 0) indicates that the overall cell reaction is spontaneous. e.g., SOA OA + Ag(s) | Ag (aq) || Cu2+(aq) | Cu(s) RA SRA 2 [Ag+(aq) + e- Ag(s) ] Cu(s) Cu2+(aq) + 2 e- Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) ΔE° = ΔEr° – ΔEr° cell cathode anode ° ΔE = ° ΔE Ag+ – ΔE°Cu2+ = 0.80 – 0.34 = 0.46 V