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Gupta 2014 Credit: Google Images for the pictures Chapter 1 Summary Notes Main Concepts Elements: substances that cannot be decomposed into simpler substances Compounds: substances composed of two or more elements Law of Constant Composition or law of definite proportions: the relative masses of elements are fixed in a given chemical substance. Law of Multiple Proportions: Applies ONLY when two elements combine to form two or more compounds. The masses of one element which combine with a fixed mass of the second element are in a ratio of whole numbers Mixtures: combinations of two or more substances Techniques for separating mixtures: filtration, distillation, chromatography Properties: Physical vs. chemical: Did the sample (really) change? Intensive vs. extensive: Does the measurement depend on quantity of sample? Explanations -Law of Constant Composition: Ex. In pure H2O, H and O combine in a 1:8 mass ratio. Does law of constant composition hold good for CuSO4.5H2O? Why or why not? -Law of Multiple Proportions: Explain the following in terms of multiple proportions: -Separation Techniques: Hand Separation- for mixtures that can be visually differentiated based on mass, color, shape etc. Filtration: Fitrate, precipitate, heterogeneous mixtures Separating Funnel: For immiscible liquids, layers separate with lesser density layer on top. Centrifugation: Separates particles of different masses based on centrifugal force. Heavier particles at the bottom and the lighter particles on top. Distillation- uses differences in the boiling points to separate a homogeneous mixture. Chromatography-separates homogenous mixtures (mostly inks) based on the differences in solubility of the mixture in a solvent. There is a stationary and a mobile phase Summary of the Chapter and Important things to remember: Gupta 2014 Credit: Google Images for the pictures Chapter 2 Summary Notes Main Concepts Naming Compounds Review: Before naming a compound, it is important to know its type because naming depends upon the type. For naming purposes, we classify compounds as ionic compounds, molecular compounds, and acids. Ionic Compounds can be identified by the presence of a metal in it. (generally solids) Ex. NaCl, K2SO4, PbSO4 Molecular compounds are made up of all non metals. (generally liquids and gases) Ex. H2O, N2O5 Acids begin with H (generally present as aq solutions or gases) Ex. HCl, H2SO4, HClO3 Coordination compound: compound that contains a complex ion or ions. Ex. [Cu(NH3)4]Cl2 1. Name cation before anion; one or both may be a complex. (Follow standard nomenclature for noncomplexes.) 2. Within each complex (neutral or ion), name all ligands before the metal. -Name ligands in alphabetical order -If more than one of the same ligand is present, use a numerical prefix: di, tri, tetra, penta, hexa, … -Ignore numerical prefixes when alphabetizing. In any (uncharged) atom: The #f protons= atomic number (Z) # of e= # of p Mass # (A)= atomic #- # of neutrons Atomic Symbol: 6C12 Isotopes are atoms of the same element containing different numbers of neutrons and therefore having different masses. Explanations Naming Compounds Flow Chart Does the formula start with H? NO YES Does it begin with a metal that has more than one oxidation number? (e.g. Fe, Ni, Cu, Sn, Hg) NO YES Name the first element followed by its oxidation number (Roman Numeral) or “old school” –ic or –ous endings. Does the formula contain a polyatomic ion? NO YES Are both elements nonmetals? NO YES Name the first element, Then the second element with an –ide ending. Name the first element, then the polyatomic ion. If two elements are present, name both, then the polyatomic ion (e.g. NaHCO3 is sodium hydrogen carbonate). Name the first element using the proper prefix (never mono–). Name the second element with the proper prefix (including mono–) and –ide ending. 1 = mono– 4 = tetra– 7 = hepta– 10 = deca– 2 = di– 5 = penta– 8 = octa– 3 = tri– 6 = hexa– 9 = nona– (not nano–) It is an acid (must be aqueous). Does the acid contain a polyatomic ion? NO YES Does the acid end with a polyatomic ion? –ite –ate Name the polyatomic ion, replacing the –ate ending with –ic. Add the word acid. Name the polyatomic ion, replacing the –ite ending with –ous. Add the word acid. Write the prefix hydro–, then the name of the second element with –ic ending. Add the word acid. Practice: 1. Sodium Sulfide (Na2S), Potassium Nitrate (KNO3), Ferrous Sulfate Fe (SO4), Ammonium Chloride (NH4Cl), Phosphoric Acid (H3PO4) 2. What is the O.N. of P in PO43- ion? 3. What is the O.N. of Fe in Fe(NO3)3? Summary of the Chapter and Important things to remember: Gupta 2014 Credit: Google Images for the pictures Chapter 3 Summary Notes Main Concepts Atomic mass units: 1 amu = 1.66054 x 10-24 g or 1 g = 6.02214 x 1023 amu 1 atom of 12C isotope defined as weighing exactly 12 amu Explanations Ex. 12C: 98.892% x 12 amu =11.867 13 C: 1.108% x 13.00335 amu =+ 0.1441 AW = 12.011 amu Therefore 12.011 g C = 1 mol C Percent Composition Moles: Avogadro’s Number = 6.022 x 1023 = atoms in exactly 12.000 g of 12C = 1 mol -Moles important because they can be used for ratios. Mass cannot be used for ratios. Converting: grams → moles → molecules Determining Empirical Formula Start with the number of grams of each element, given in the problem. -If percentages are given, assume that the total mass is 100 grams so that the mass of each element = the percent given. -Convert the mass of each element to moles using the molar mass -Divide each mole value by the smallest number of moles calculated. -Round to the nearest whole number. This is the mole ratio of the elements and is represented by subscripts in the empirical formula. -If the number is too far to round (x.1 ~ x.9), then multiply each solution by the same factor to get the lowest whole number multiple. If one solution is 1.5, then multiply each solution in the problem by 2 to get 3.If one solution is 1.25, then multiply each solution in the problem by 4 to get 5. Summary of the page and Important things to remember: Ex. What is the percent of O in CuSO4.5H2O? 57.7% Ex. How many molecules of H2O in 100.0 g H2O? 3.343 x 1024 molecules H2O Ex. What is the empirical formula of a compound that is composed of 80.% Carbon and 20.% Hydrogen? If the molar mass is found to be 30 g/mol, what is the molecular formula? CH3, C2H6 Gupta 2014 Credit: Google Images for the pictures Chapter 3 Summary Notes Contd. Main Concepts EF from Combustion Data: To find out the EF of a compound, the compound is burned in the air. The equation for combustion of these compounds is all follows. For Hydrocarbons: CxHy + O2 CO2 + H2O For Compounds Containing CHN: CxHyNz+ O2 CO2 + H2O+ NO2 For Compounds containing C, H and O CxHyOz + O2 CO2 + H2O Explanations Ex. A 0.6349 g sample of the unknown produced 1.603 g of CO2 and 0.2810 g of H2O. Determine the empirical formula of the compound. Ans. C7H6O2 Ex. When carbon-containing compounds are burned in a limited amount of air, some CO(g) is produced as well as CO2 (g). A gaseous product mixture is 35.0 mass % CO and 65.0 mass % CO2. What is the mass % C in the mixture? 32.7% C Ex. If 6.0 g hydrogen gas reacts with 40.0 g oxygen gas, what mass of water will be produced? OR 2H2 (g) + O2 (g) → 2 H2O (g) Ex. A mixture of morphine (C17H19NO3) and an inert solid is analyzed by combustion with O2. The unbalanced equation for the reaction of morphine with O2 is C17H19NO3 + O2 → CO2 + H2O + NO2 The inert solid does not react with O2. If 4.000 g of the mixture yields 8.72 g of CO2, calculate the percent morphine by mass in the mixture. 83.1% Limiting Reactant: The reactant that runs out first. LR can be determined by comparing the mole ratios of the reactants. LR determines the amount of the products formed. % Yield Ex. If in the previous example, only 40.0 g water were formed, what is the percent yield and percent error? Actual yield 100 Theoretica l yield Accepted value - Measured result % Error Determining the Formula of a Hydrate: To determine the formula of a hydrate, a certain mass of hydrate is heated to drive off the water. Then the mass of water driven-off is calculated. Now the mole ratio of water to the compound is calculated. Accepted value 100 Summary of the page and Important things to remember: Ex. When 2.000 gram of Na2CO3.xH2O was heated, 0.914 gram of anhydrous residue remained. What is the formula of this compound? Ans: Na2CO3.7H2O Sodium carbonate heptahydrate Gupta 2014 Credit: Google Images for the pictures Chapter 4 Summary Notes Main Concepts Explanations Reactions: To be able to successfully write reactions, you will need to know the Synthesis Reactions: following: Solubility Rules , Nomenclature Types of Reactions (explained later in this worksheet), How to write net ionic equations MUST KNOW For AP Chemistry reaction prediction: 1. Always write balanced net ionic equations (meaning dissociate soluble compounds (based on solubility rules), 2. Metal are insoluble and are atomic, written as (s) in these equations. Ex. Mg(s) 3. Molecular compounds such as gases (CO2, H2S Etc.) are written as (g) and will not dissociate into ions. 4. Water is written as (l) and does not dissociate. 5. Ionic compounds may or may not dissociate depending on solubility rules. Ex. PbSO4 insoluble and NaNO3 soluble. 6. Even a soluble ionic compound may NOT dissociate if it is in solid form (meaning no water present to actually dissociate the ions.) 7. Weak acids and bases partly dissociate or ionize and are written with a reversible arrow. 8. Remember PSHOFBrINCl. Phosphorus occurs as P4, Sulfer as S8 and rest as diatomic. 9. While we are reviewing, remember the difference between Zn and Zn2+ and Cl2 and 2 Cl10. Strong acids (HCl, HBr, HI, HNO3, H2 SO4 (first dissociation only!), HClO4 and HClO3) and strong bases (Group 1 alkali metal hydroxide and Ca, Ba, Sr hydroxides from group 2) dissociate in aq. Solutions. Weak acids and bases are not dissociated in net ionic equations. Solubility Rules Always soluble: alkalies, NH +, NO -, C H O 4 3 2 3 Decomposition Single Replacement 2 Types of Reactions: Double displacement. Precipitation, neutralization, gas forming. H2CO3 in water = H2O & CO2 Single displacement or redox replacement: (metals displace metals and nonmetals displace nonmetals) Combination or synthesis = two reactants result in a single product • Metal oxide + water metallic hydroxide (base) • Nonmetal oxide + water nonbinary acid • Metal oxide + nonmetal oxide salt Decomposition = one reactant becomes several products • Metallic hydroxide metal oxide + water • Acid nonmetal oxide + water • Salt metal oxide + nonmetal oxide • Metallic chlorates metallic chlorides + oxygen • Electrolysis decompose compound into elements (water in dilute acids or solutions of dilute acids) • Hydrogen peroxide water + oxygen • Metallic carbonates --> metal oxides + carbon dioxide • Ammonium carbonate ammonia, water and carbon dioxide. Hydrolysis = compound reacting with water. • Watch for soluble salts that contain anions of weak acid the anion is a conjugate base and cations of weak bases that are conjugate acids. Reactions of coordinate compounds and complex Combustion Reactions: Gupta 2014 Credit: Google Images for the pictures Chapter 5 Summary Notes • Complex formation by adding excess source of ligand to transitional metal of highly charged metal ion such as Al3+ Al =4 ligands others 2X ox # • Breakup of complex by adding an acid metal ion and the species formed when hydrogen from the acid reacts with the ligand Redox = change in oxidation state= a reaction between an oxidizer and a reducer. 1. Familiarization with important oxidizers and reducers 2. “added acid” or “acidified” 3. an oxidizer reacts with a reducer of the same element to produce the element at intermediate oxidation state Molarity (M) = moles solute volume of solution = Precipitation mol L Titration is a method to determine the molarity of unknown acid or base. In titration, an acid or base of unknown molarity is titrated against a standard solution (whose M is known) of acid or base.The end point in a titration is indicated by a color change by the indicator. Indicators are weak acids or bases and are added in small quantity (1-3 drops) to indicate the end point. At equivalence point (which should be close to end point), moles of H+= moles of OHM1V1= M2 V2 (sometimes used to get moles , M= moles/L , so moles= M XV) -What other ways can you get the moles- for a solid acid or base? For a gas? Electrolyte: substance which, in aqueous solution, ionizes and thus conducts electricity. Ex: salt in water. Non-electrolyte: substance which, in aqueous solution, does not dissociate and thus does not conduct electricity Strong & weak electrolytes: conductivity depends on degree of dissociation and equilibrium position: HA (aq) ↔ H+ (aq) + A- (aq) Strong = nearly completely dissociated Weak = partially dissociated Molecular equation: shows complete chemical equation with states of matter, undissociated BaCl2 (aq) + Na2SO4 (aq) → 2 NaCl (aq) + BaSO4 (s) Complete ionic equation: shows complete chemical equation with states of matter, dissociated if appropriate Ba2+(aq) + 2 Cl-(aq) + 2 Na+(aq) + SO42-(aq) → 2 Cl-(aq) + 2 Na+(aq) + BaSO4 (s)Spectator ions: present in reaction but do not “participate”; depend on solubility rules Cl- (aq) and Na+ (aq) 3. Net ionic equation: shows chemical equation without spectator ions Ba2+ (aq) + SO42- (aq) → BaSO4 (s) Summary of the page and Important things to remember: Ex. How many mL of a 3M NaOH solution are required to completely neutralize 20.0 mL of 1.5M H2SO4? (Start by writing a balanced equation!) Ans. 20.0 mL Ex. How many g of NaOH is required to completely react with 100. mL of 1M HCl? Gupta 2014 Credit: Google Images for the pictures Main Concepts 0th and 1st Laws of Thermodynamics: 0. 2 systems are in thermal equilibrium when they are at the same T 1. Energy can be neither created nor destroyed, or, energy is conserved Internal Energy Includes translational, rotational, vibrational energy Change (ΔE = Efinal - Einitial) is often measured o ΔE > 0: Energy of system increases (gained from surr.) o ΔE < 0: Energy of system decreases (lost to surr.) o ΔE = q + w q = heat added/liberated from system q > 0 : heat added to system q < 0: heat removed from system w = work done on or by the system w > 0 : work done to system w < 0 : system does work on surr. Calorimetry: Measurement of heat flow, experimental technique used to measure the heat transferred in a physical or chemical process o Calorimeter: the apparatus used in this procedure; two types: constant pressure (coffee cup) and constant volume (bomb calorimeter) o Coffee Cup Calorimeter: system in this case is the “contents” of the calorimeter and the surroundings are cup and the immediate surroundings o qrxn + q solution = 0 qrxn: heat gained/ lost in the chemical reaction qsolution: the heat gained/lost by solution Heat Capacity, C: Amount of heat required to raise T of an object by 1 K o q = CΔT Specific Heat (or Specific Heat Capacity), c: heat capacity of 1 g of substance o q = mcΔT Explanations Ex. Octane and Oxygen gases combust within a closed cylinder in an engine. The cylinder gives off 1150J of heat and a piston is pushed down by 480J during the reaction. What is the change in internal energy of the system? (Ans: ΔE = -1630J) Ex. How much energy is required to heat 40.0 g of iron (c = 0.45J/gK) from 0.0oC to 100.0oC? (Ans: q = 1800J) Ex. 0.500g of Mg chips are placed in a coffee-cup calorimeter and 100.0mL of 1.00M HCl is added to it. The reaction is: Mg(s) + 2HCl(aq) H2(g) + MgCl2(aq) The temp. of the solution increases from 22.2C to 44.8C. What’s the enthalpy change for the reaction, per mole of Mg? Assume specific heat capacity of solution is 4.20J/gK and density of the HCl solution is 1.00 g/mL. (Ans: ΔH= -4.64*105 J/molMg) Gupta 2014 Credit: Google Images for the pictures Chapter 5 Summary Notes Contd. Main Concepts Enthalpy, H: change in heat content of a reaction at constant P o H = E + PV ΔH = ΔE + PΔV ΔH = (qp+w) + (-w) ΔH = qp qp = heat content ΔH > 0: heat gained from surr. + ΔH in endothermic reaction ΔH < 0: heat released to surr. + ΔH in exothermic reaction Enthalpy of Reaction, ΔHrxn: heat of reaction, extensive property, depends on states of reactions and products o ΔHrxn = -ΔHreverse rxn Hess’s Law: If a rxn is carried out in a series of steps: ΔHrxn = Σ(ΔHsteps) = ΔH1 + ΔH2 + ΔH3 + · · · Enthalpy of Formation, ΔHf: heat needed to form substance from its elements o Standard Enthalpy of Formation, ΔHfo: forms 1 mole of compound from its elements in their standard state (at 298K) o ΔHf of pure element (C, O2, H2, etc) is 0. ΔHrxno = Σ [n*ΔHfo(products)] - Σ[n*ΔHfo(reactants)] Bond Enthalpy: Amount of energy required to break a particular bond between two elements in gaseous state o Indicates the “strength” of a bond ΔHrxn ≈ Σ [ΔHbonds broken] - Σ[ΔHbonds formed] o NOTE: this is the “opposite” of Hess’s Law Summary of the page and Important things to remember: Explanations Ex. What is the ΔH of combustion of 100g CH4 if ΔHrxno = -890kJ? (Ans: -5550kJ) Ex. What is ΔHrxno of the combustion of propane? C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Givens: 3C(s) + 4H2(g) C3H8(g) ΔH1 = -103.85kJ C(s) + O2(g) CO2(g) ΔH2 = -393.5kJ H2(g) + ½O2(g) H2O(l) ΔH3 = -285.8kJ o (Ans: ΔHrxn = -2219.8kJ) Ex. What is ΔHrxno for the combustion of liquid benzene? C6H6(l) + 15/2O2(g) 6CO2(g) + 3H2O(l) Givens: ΔHfo(C6H6(l)) = +49 kJ/mol ΔHfo(CO2(g)) = -394 kJ/mol ΔHfo(H2O(l)) = -286 kJ/mol (Ans: ΔHrxno = -3268 kJ/mol) Ex. What is ΔHrxno for the following reaction? CH4(g) + Cl2(g) CH3Cl(g) + HCl(g) Given: ΔH/mol of C-H bond: +413 kJ/mol ΔH/mol of H-Cl bond: +431 kJ/mol ΔH/mol of C-C bond: +348 kJ/mol ΔH/mol of Cl-Cl bond: +242 kJ/mol ΔH/mol of C-Cl bond: +328 kJ/mol ΔH/mol of C=C bond: +614 kJ/mol (Ans: ΔHrxno ≈ -104 kJ/mol) Gupta 2014 Credit: Google Images for the pictures Chapter 6 Summary Notes Main Concepts Electromagnetic Spectrum: radiant energy can travel without matter λν c = speed of light = 3.0 x 108 m/s λ = wavelength (m) ν = frequency (Hz) Planck’s Theory: Blackbody radiation can be explained if energy can be released or absorbed in packets of a standard size called quanta h c h = Planck’s constant = 6.63 x 10-34 J-s E h Photoelectric Effect: As first explained by Einstein in 1905, the photoelectric effect is the spontaneous emission of an electron from metal struck by light if the energy is sufficient Atomic Emission Spectra: spectrum for specific wavelengths of light emitted from pure substances Bohr’s Model of the H Atom: Bohr applied idea of quantization of energy transfer to atomic model, theorizing that electrons travel in certain “orbits” around the nucleus Allowed orbital energies are defined by: R 2.178 10 18 n = principal quantum number = 1, 2, 3 ... En 2 H n n2 Explanations Atomic Emission Spectra for H Line Series Transition down to (emitted) or up from (absorbed)… Type of EMR Lyman 1 UV Balmer 2 Visible Paschen 3 IR Brackett 4 Far IR RH = Rydberg’s constant = 2.178 x 10-18 J Line Series: transitions from one level to another Heisenberg’s uncertainty principle: The position and momentum of a particle cannot be simultaneously measured with accuracy. Schrödinger’s wave function: Relates probability ( 2) of predicting position of e- to its energy. h2 d 2 d E U ih 2 2m dx dt Matter as a Wave: m = h / c Particles (with mass) have an associated wavelength h / mc Waves (with a wavelength) have an associated mass and velocity Probability Plots for 1s, 2s, and 3s Orbitals Gupta 2014 Credit: Google Images for the pictures Pauli Exclusion Principle: no two charges in an atom can have the same set of four quantum numbers n, l, m1, ms. Effective Nuclear Charge: the net positive charge acting on the outermost electron. Shielding Effect: inner electrons shielding the outer electron from the full charge of the nucleus. Electron Configuration: the way the electrons are distributed among the various orbitals of an atom. The most stable, or ground, electron configuration is one in which the electrons are in the lowest possible energy states. Hund’s Rule: for degenerate orbitals (orbitals with the same energy), the lowest energy is attained when the number of electrons with the same spin is maximized. The periodic table is your best guide to the order in which orbitals are filled. s-block and p-block contain the representative (main group) elements. The ten columns in the middle that contain transition metals, elements in which d-orbitals are being filled. f-block metals are the ones in which the f-orbitals are being filled. Diamagnetic: paired electrons Paramagnetic: unpaired electrons Mass Spectroscopy: Helps identify # and abundance of isotopes and structures of different compounds. Chlorine has two isotopes, 35Cl and 37Cl, in the approximate ratio of 3 atoms of 35Cl to 1 atom of 37Cl. You might suppose that the mass spectrum would look like this but that is not the case because Chlorine consists molecules that Ex: Give ground state electron configurations for the following: Ni2+ and Ni3+ Ans: Ni2+ = [Ar]3d8, Ni3+= [Ar]3d7 E x: Is Cupric ion dia or paramagnetic? Why? Paramagnetic, due to an unpaired d e. Mass Spectrograph for Chlorine (Cl2) fragment (chemguide.co.uk) You could have the following mass fragments--35 + 35 = 70, 35 + 37 = 72, 37 + 37 = 74. So the actual mass spectrograph will look like the one on the right. (chemguide.co.uk) Summary of the page and Important things to remember: Gupta 2014 Credit: Google Images for the pictures Photoemission Spectroscopy (PES) In a photoelectron spectroscopy experiment any electron can be ionized when the atom is excited. Unlike the first ionization, in this experiment any electron can be removed, not just the electron that requires the least amount of energy. PES gives insight into the structure of atom. Each peak in PES indicates the number of electrons and the position of the peak indicates the amount of energy required to remove those electrons. Note that s electrons will require more energy than p electrons due to higher ENC hence s electrons will be farther out on the energy axis. http://www.chem.arizona.edu/chemt/Flash/photoelectron. html Summary of the page and Important things to remember: Gupta 2014 Credit: Google Images for the pictures Chapter 7 Summary Notes Main Concepts Explanations Periodic Trends Key Words Principal Energy Level: more energy levels means bigger atoms Nuclear Charge (# of p in an atom): attraction of electrons to nucleus; increased nuclear charge causes atomic radius to decrease Shielding Effect: inner electrons increase atomic size by reducing the attractive force on outermost electrons Atomic Radius Effective Nuclear Charge: force of attraction felt valence e from nucleus; a high ENC means smaller ionic radius (greater attraction to outermost electrons) Atomic Radius: half distance between two covalently-bonded atoms Periodic Trends Atomic Radius: increase going down, decrease going right (more energy levels and shielding going down, higher ENC going right) Size of Ions: cations smaller than anions (fewer e-, less e- repulsion), bigger going down,smaller going right Isoelectronic series have same # of electrons, different # of protons Ionization Energy: the amount of energy required to remove an electron from the ground state of a gaseous atom or ion. A(g) A+ + e Bigger going right, smaller going down Exceptions: between Group 2&13, Group 5&6 Isoelectronic Series Ionization Energy Zeff = Z − S Zeff = ENC Z = atomic number S = screening constant (# of inner electrons) Gupta 2014 Credit: Google Images for the pictures Chapter 7 Summary Notes Main Concepts Periodic Trends (cont’d) Electron Affinity: amount of energy required to add an electron to a gaseous atom: Cl(g) + e− Cl− Larger going right, smaller going down Exceptions: between Group 1&2, Group 14&15 Electronegativity: tendency to attract electrons in a covalent bond Increases going up and right Metals and Nonmetals Metals: tend to form cations, metal oxides are basic Nonmetals: tend to form anions, nonmetal oxides are acidic, poor conductors of electricity Metallic character increases down a group, decreases across a period Explanations Electron Affinity Electronegativity Alkali metals (1A)—The most reactive metal family, these must be stored under oil because they react violently with water! They dissolve and create an alkaline, or basic, solution, hence their name. Alkaline earth metals (2A)—These also are reactive metals, but they don’t explode in water; pastes of these are used in batteries. Halogens (7A)—Known as the “salt formers,” they are used in modern lighting and always exist as diatomic molecules in their elemental form. Noble gases (8A)—Known for their extremely slow reactivity, these were once thought to never react; neon, one of the noble gases, is used to make bright signs. http://www.sparknotes.com/testprep/books/sat2/chemistry/chapt er4section6.rhtml Summary of the Chapter and Important things to remember: Gupta 2014 Credit: Google Images for the pictures Chapter 8 Summary Notes Main Concepts Intramolecular Bonding o Ionic: electrostatic attrction between oppositely charged ions. Generally solids. Ex: NaCl, K2SO4 o Covalent: sharing of e- between two atoms (typically nonmetals). Generally gases/liquids. Ex: CO2, SO2 o Metallic: “sea of e-”; bonding e- relatively free to move throughout 3D structure. Generally solid. Ex: Fe, Al o Covalent Network: large number of atoms/molecules bonded in network through covalent bonding. Ex: SiO2, Si, Ge, Diamond, Graphite Ionic Bonding o Results as atoms lose or gain electrons to achieve a nobel gas electron configuration. Typically exothermic. Bonded state lower in energy (more stable) Opposite charges create electrostatic attraction, which determines the strength of the ionic bond o Occurs when difference in electronegativy is > 1.7 o Use brackets when writing Lewis symbols of ions. Lattice Energy: Measurement of the strength of the ionic bond o Hlattice = energy required to completely separate 1 mole of solid ionic compound into its gaseous ions o Electrostatic attraction (and thus lattice energy) increases as ionic charges increase and as ionic radii decrease. DH lattice µ Q+Qr+ + r- Covalent Bonding: Atoms share electrons to achieve noble gas configuration that is lower in energy (therefore more stable) o Occurs when difference of electronegativity is ≤ 1.7 Polar covalent: 0.3 < diff in EN ≤ 1.7 Nonpolar covalent: 0 ≤ diff in EN ≤ 0.3 Coordinate covalent: shared pair contributed by only one of the two sharing species. Ex: Lewis acids and bases Explanations Ex. Draw the Lewis symbol for fluoride. Ex. In the following pairs, which has a greater lattice energy and why? NaCl or KCl NaCl or MgS Covalent bond strength is measured by bond energy. Bond energy is calculated by = Energy of reactant bonds- Energy of product bonds Gupta 2014 Credit: Google Images for the pictures Metallic Bonding: Forms between metal atoms because of the movement of valence electrons from atom to atom to atom in a “sea of electrons”. The metal consists of cations held together by negatively charged electron “glue”. o Results in excellent thermal and electrical conductivity, ductility, and malleability o A combination of 2 metals is called an alloy The Octet Rule: Atoms tend to gain, lose, or share electrons until they are surrounded by 8 electrons in their outermost energy level (filled s&p shells), and are thus energetically stable Lewis symbols (electron-dot symbol) o Shows a dot for valence electrons of an atom/ion o Places dots at top, bottom, right, and left sides and in pairs only when necessary (Hund’s rule). o Primarily used for representative elements only (Groups 1A – 8A) o Transition metals typically form +1, +2, & +3 ions Transition metal atoms first lose both “s” electrons, even though it is a higher energy subshell. Ex. Cr2+, Cr3+ Most lose electrons to end up with a filled or half-filled subshell. Ex. Cu+ ion Lewis structures: used to depict bonding pairs and lone Ex. Draw the Lewis structures for the following pairs of electrons in the molecule 1. Total the # of valence electrons in the system. Add using steps. Show work! a) Cl2 the total negative charge if you have an ion, b) CH2Cl2 subtract the charge if you have a cation. c) NH3 2. Number the electrons if each atom is to be “happy” d) NaCl (8 electrons for octet rule, or 2 for hydrogen) e) HCN(SO4) 3. Calculate the number of bonds in the system. f) H2O2CNS Covalent bonds are made by the sharing of electrons. # of bonds = (electrons in step 2 – electrons in step 1) / 2 4. Draw the structure. The central atom is usually the atom with the least electronegativity. 5. Double check your answer by counting total number of electrons Gupta 2014 Credit: Google Images for the pictures Exceptions to the Octet Rule o Odd-electron molecules. Ex. NO or NO2 o Incomplete octet. Ex. H2, He, BeF2, BF3 o Expanded octet (occurs in molecules when the central atom is beyond the third period. The empty 3d subshell is used in hybridization). Ex. PCl5, SF6 Formal Charge: the numerical difference between # of valence electrons in the isolated atom and # of electrons assigned to that atom in the Lewis structure. Doesn’t represent real charges, just a useful tool for selecting most stable Lewis structure 1. Assign unshared electrons (usually in pairs) to the atom on which they are found 2. Assign one electron from each bonding pair to each atom in the bond (split the electrons in a bond) 3. Subtract the electrons assigned from the original number of valence electrons o Used to select the most stable (and therefore most likely) structure when more than one structure is reasonable according to rules o Most stable: Has FC on all atoms closest to zero Has all negative FC on most EN atoms Resonance Structures: Equivalent Lewis structures that describe a molecule with more than one likely arrangement of electrons o Notation: use double-headed arrow between all resonance structures o “Real” electron structure of the molecule is an “average” of all resonance structures Bond Order: Indication of bond strength and bond length Bond Enthalpy: Amount of energy required to break a particular bond between two elements in gaseous state. Indicates “strength” of a bond. Hrxn ≈ (Hbonds broken) - (Hbonds formed) Summary of the page and Important things to remember: # Valence electrons in free atom – # Non-bonding electrons – ½ (# Bonding electrons) Formal Charge Ex. Draw at least 2 Lewis structures for each, then calculate the formal charge for each atom in each structure. a) SCNb) N2OBF3 Ex. Draw the Lewis structure and determine the bond S-O, C-C, and C-H bond orders. c) SO3 d) C6H6 Ex: CH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g) DHrxn = ? Bond Ave DH/mol Bond Ave DH/mol C-H 413 Cl-Cl 242 H-Cl 431 C-Cl 328 C-C 348 C=C 614 Ans: Hrxn ≈ -104 kJ/mol Gupta 2014 Credit: Google Images for the pictures Chapter 9 Summary Notes Main Concepts Valence Shell Electron Pair Repulsion Theory: helps construct molecular 3-D shape from 2-D Lewis Structures Electron Domains: areas of valence electron density around the central atom Includes bonding electron pairs and lone electron pairs A single, double, or triple bond counts as one domain Basis for VSEPR: each group of valence electrons (electron domains) around a central atom tend to be as far as possible from each other to minimize repulsion and this determines molecular geometry of molecule lonepair-lonepair repulsion > lonepair-bondpair repulsion > bondpair-bondpair repulsion Molecular Dipole Moment: molecules with polar covalent bonds might have a net dipole moment depending on the compound's 3-D geometry and symmetry Check if individual bonds are polar and if individual dipole moments cancel out due to symmetry Polarity: polar substances are soluble in water and non-polar substances are soluble in non-polar solvents like benzene and oil Valence Bond Theory: predicts bond strengths based upon orbital overlap for covalent bond formation Basis for VB: covalent bond forms when orbitals of two atoms overlap An orbital can have a max of two electrons with opposing signs The bond strength depends on the attraction of nuclei for the shared electron, so greater the Explanations Gupta 2014 Credit: Google Images for the pictures overlap, the stronger the bond Sigma Bond: end to end overlap of orbital which allows free rotation of parts of molecule (single bonds) Draw hybrid orbital diagram for C2H4. Pi Bond: side to side overlap of orbital which restricts rotation (2nd and 3rd bonds in double and triple bonds) Summary of the page and Important things to remember: Gupta 2014 Credit: Google Images for the pictures Hybrid Orbital Theory: an extension of VB theory, where atomic orbitals "hybridize" to form new hybrid orbitals; explains the bonding in terms of quantum mechanical model of atom (s, p, d, f orbitals) Basis for HO: valence atomic orbitals in the molecule are very different from those in isolated atoms The number of hybrid orbitals obtained equals the number of atomic orbitals mixed The type of hybrid orbitals obtained varies with types of atomic orbitals mixed ns and np give two sp hybrids ns and two np give three sp2 hybrid orbitals ns and three np give four sp3 hybrid orbitals Delocalized Bonds are present in compounds with resonance structures Molecular Orbital Theory: is based on the wave nature of the electrons and is a better model to explain paramagnetism of oxygen Basis for MO: Molecular orbitals form through the combination of atomic orbitals Bonding MO: stable orbital that forms between nuclei Antibonding MO: less stable orbital that forms behind nuclei Sigma MO: orbital forming from a combination of two 1s or 2s orbitals form different atoms or two 2pz orbitals from different atoms Pi MO: orbital forming from a combination of two 2px or 2py from different atoms (do not appear until B2) Diamagnetism: all electrons paired; no magnetic properties Paramagnetic: at least 1 unpaired electron; drawn into exterior magnetic field since spins of atoms become aligned; unlikely to retain alignment when field is removed Ferromagnetism: occurs primarily in Fe, Co, and Ni; drawn into exterior magnetic field since spins of atoms become aligned; likely to retain alignment when field is removed Summary of the page and Important things to remember: Gupta 2014 Credit: Google Images for the pictures Chapter 10 Summary Notes Main Concepts Characteristics of Gases Particles in a gas are very far apart, and have almost no interaction. Gases expand spontaneously to fill their container (have indefinite volume and shape.) Pressure Pressure = a force that acts on a given area F P A Atmospheric pressure: the result of the bombardment of air molecules upon all surfaces 1 atm = 760 mm Hg = 760 torr = 101.3 kPa = 14.7 PSI Barometer: measures atmospheric P compared to a vacuum * Invented by Torricelli in 1643 * Liquid Hg is pushed up the closed glass tube by air pressure Manometers: measure P of a gas *Closed-end: difference in Hg levels (Dh) shows P of gas in container compared to a vacuum * Open-end: Difference in Hg levels (Dh) shows P of gas in container compared to Patm Gas Laws Boyle’s Law: the volume (V) of a fixed quantity (n) of a gas is inversely proportional to the pressure at constant temperature (T). Explanations Closed V constant 1 P P1V1 P2V2 Gupta 2014 Credit: Google Images for the pictures Main Concepts Charles’ Law: V of a fixed quantity of a gas is directly proportional to its absolute T at constant P. Gay-Lussac’s law: P of a fixed quantity of a gas is directly proportional to its absolute T at constant V. Seen as derivative of C’s and B’s laws Avogadro’s hypothesis: Equal volumes of gases at the same T & P contain equal numbers of molecules Combined Gas Law Explanations V constant T V1 V2 T1 T2 P constant T P1 P2 T1 T2 PV constant T Summary of the Chapter and Important things to remember: P1V1 P2V2 T1 T2 Chapter 11 Summary Notes 2014 Credit: Google Images Explanations for the pictures MainGupta Concepts INTRAmolecular Forces: the forces holding atoms together to form moleculs INTERmolecular Forces: Forces between molecules between ions, or between molecules and ions Intermolecular Forces (IMF) -IMF < intramolecular forces (covalent, metallic, ionic bonds) -IMF strength: solids>liquids>gases -Types of IMFs Ion-Ion Forces, Ion-Dipole Forces, Dipole-Dipole Forces, H-bonds extreme dipole-dipole, LDFs Types of IMF Electrostatic Forces: act over larger distances in accordance with Coulomb’s Law Ion-Dipole: between an ion and a dipole (a neutral, polar molecule has separate partial Boiling points and melting points are good indicators of relative charges) IMF strength -Increasing with increasing polarity of molecule and increasing ion charge Ion-Permanent Dipole -Water is highly polar and can interact with positive ions to give hydrated ions in water QQ -Attraction between ions and dipole depends F 2 d on ion charge and ion-dipole distance Dipole-Dipole -Weakest electrostatic force (not all IMFs, LDFs weaker than dipole-dipole); exist between neutral polar molecules -Increase with increasing polarity (dipole moment) of molecule Hydrogen Bonds (H-bonds) -H is unique among elements because it has a single e- that is also a valence e-When e- is “hogged” by a highly electronegative atom (very polar covalent bond), the H nucleus is partially exposed and becomes attracted to e- rich atom nearby -Explains why ice floats on water, has latticelike structure, explains why molecules with Hbonds have higher boiling points, H-bonding in water, O -H bond is very polar H-bonding in biology -DNA bases bind to each other due to specific hydrogen bonding between Lewis Bases Summary of the page: Gupta 2014 Credit: Google Images for the pictures Inductive Forces arise from induced distortion of e- cloud London Dispersion: between polar or nonpolar molecules or atoms, but is generally mentioned for non polar molecules when other forces are absent. Very weak, motion of ecreates an instantaneous dipole moment which induces a dipole in an adjacent atom Nonpolar molecules can dissolve in water due to LDFs. Water induces a dipole in electron cloud. Solubility increases with mass of gas due to greater distortion. When induced forces between molecules are very weak, the solid will sublime (solid to gas) Liquids Molecules are in constant motion, molecules close together Liquids are almost incompressible Evaporation -To evaporate, molecules must have sufficient energy to break IMFs -Condensation is reverse (remove energy and make IM bonds) Vapor Pressure Heat of Vaporization heat required (at constant P) to vaporize the liquid Equilibrium vapor pressure & the Clausius-Clapeyron Equation o Used to find ∆vapH˚ o Logarithm of vapor pressure P is proportional to ∆vapH˚ and to 1/T o lnP = -(∆vapH˚/RT) + C o Surface Tension leads to spherical liquid droplets Properties resulting from IMFs Viscosity: resistance of a liquid to flow Surface Tension: energy required to increase the surface area of a liquid Intermolecular forces lead to capillary action and concave meniscus for a water column -Capillary Action: movement of water up a piece of paper depends on the H-bonds between H2O and the OH groups of the cellulose in the paper Summary of the page: Gupta 2014 Credit: Google Images for the pictures Cohesion: attraction of molecules for other molecules of the same compound Adhesion: attraction of molecules for a surface Meniscus: curved upper surface of a liquid in a container; a relative measure of adhesive and cohesive forces London Dispersion Forces -Increase with increasing molar weight, increasing # of e-, increasing # of atoms -“Longer” shapes (more likely to interact with other molecules) Phase Changes -Endothermic: melting, vaporization, sublimation -Exothermic: condensation, freezing, deposition Structures of solids Amorphous: without orderly structure (ex. Rubber, glass) Crystalline: repeating structure; have many different stacking patterns based on chemical Molecular formula, atomic or ionic sizes, and bonding Types of Crystalline Solids: -Atomic: Properties: poor conductors, low melting point -Molecular: Properties: poor conductors, low to moderate melting point -Ionic: Properties:hard and brittle, high melting point, poor conductors, some solubility in H2O -Covalent (a.k.a. covalent network): Properties: very hard, very high melting point, Ionic Covalent Network Metallic generally insoluble, variable conductivity Credits: Google Images -Metallic: Properties: excellent conductors, malleable, ductile, high but wide range of melting points Summary of the page Gupta 2014 Credit: Google Images for the pictures Chapter 13 Summary Notes Main Concepts Vocabulary Molarity: measure of concentration in solutions, mol/L M mol solute Liters solution Solvation: dissolving; the interactions between solute and solvent Crystallization: process by which solute particles leave solvent Explanations Describe the steps with the lab equipment needed to make 0.1 M 100 ml NaOH? (Hint: volumetric flask) How are supersaturated solutions created? Dissolve solute with heat, then cool solution slowly. Solution is “tricked” into appearing unsaturated Saturation Saturation: solution that is in equilibrium with undissolved solute Unsaturated solution contains less solute than saturated solution Supersaturated solution contains more solute than saturated solution but appears unsaturated Factors Affecting Solubility Miscible liquids mix; both are polar or both are nonpolar Covalent network solids do not dissolve in polar or nonpolar solvents Increasing temperature increases solubility for most solids, but decreases solubility for gases Summary of the Chapter and Important things to remember: At 40oC, the solubility of KNO3 in 100g of water is 64 g. What is the solubility of KCl at 10oC? 30 g At which points would an unsaturated solution appear? Supersaturated solution? How much KClO3 needs to be added to 10g of KClO3 at 60oC to make a saturated solution? 20 g What is normal boiling point of ethanoic acid? 117 oC Gupta 2014 Credit: Google Images for the pictures Chapter 14 Summary Notes Main Concepts Reaction rate: A measure of the (average) speed of a reaction Reaction rate is affected by: 1. Concentration of reactants 2. Temperature of the reaction 3. Presence/absence of a catalyst 4. Surface area of solid or liquid reactants and/or catalysts -Average Rate: Rate of a reaction over a given period of time -Instantaneous Rate: Rate of reaction at ONE given point of time. -Initial Rate: Rate of reaction at t=0. (its instantaneous rate at t=0) Reaction order: the exponents in a rate law (can be fractions) Rate law: shows how the rate of reaction depends on the concentration of reactant(s); determined experimentally. Cannot be determined by the coefficients of a balanced reaction (unless in an elementary step) Rate = k[A]m[B]n *Units of k change w/order of the rxn To Find Rate Laws: 1. Using Initial Rates 2. Integration Method: Determining Rate Law by Determining the Change in Concentration of reactants over time gives rate law either graphically or by calculations. Explanations Rate with regards to other 1 [C ] 1 [ A] 1 [ B ] Rate reactants and products c t a t b t 2N2O5 (g) → 4 NO2 (g) + O2(g) If D[O2]/Dt = 5.0 M/s, what is D[N2O5]/Dt?-10.0 M/s Orders of Reactions & Related Equations: Using Initial Rates Method: data given Expe [A] [B] Initial Rate of Formation of rimen C in M t 1 0.60 0.15 6.3´10-3 2 0.20 0.60 2.8´10-3 3 0.20 0.15 7.0´10-4 Integration Method (Using graphs or equations) Zero Order Rate k[ A] 0 Rate [A]t = -kt + [A]0 (k) = M/s First Order ln[A]t = -kt + ln[A]0 or log[A]t = -kt / 2.303 + log[A]0 (k) = s-1 Summary of the Chapter and Important things to remember: A t Gupta 2014 Credit: Google Images for the pictures The Collision Model -Reactants must collide, and with the right Second Order orientation and energy for an effective collision -Elementary steps: a single event or step (reaction) in a multi-step reaction -Molecularity: the # of molecules participating as reactants in an elementary step -Catalyst: Substance that changes the rate of a reaction without undergoing a permanent chemical change itself 1 = kt + 1 [A] [A]0 Check for Permissible Rxn. Mechanism 1. Balanced eq. 2. Rate Determining Step (RDS) is the slow one. -Radioactive decay follows a first order kinetics. Half life (t1/2) can be calculated, if rate constant is known or vice versa. -Activation energy ( Ea ): minimum energy required to initiate a chemical reaction Half Life ln 2 0.693 k t1/2 t1/2 Activation Energy For a graph between ln k and 1/T, the slope can be used to calculate Ea— activation energy. Remember to use 8.314 j/mol. K for R. Slope Summary of the Chapter and Important things to remember: Ea R Gupta 2014 Credit: Google Images for the pictures Chapter 15 Summary Notes Explanations Main Concepts Chemical Equilibrium Occurs when rate of forward reaction = rate of reverse reaction. Ex. Vapor pressure: rate of vaporization = rate of condensation, Saturated solution: rate of dissociation = rate of crystallization. Conc. of reactants and products does not have to be equal at the equilibrium, only the rates of forward and reverse rxn become equal. Express concentration in Partial Pressure for gases and molarity for solutes in liquids Rate = kforward [A] Rate = kreverse [B] at equilibrium [ A] PA RT [B] k f [A] k r [ B] or PB RT PA RT P Rate k r B RT Rate k f PB k f constant K eq PA k r -If Kc > 1, then more products at equilibrium -If Kc < 1, then more reactants at equilibrium -If Kc = 1, then almost equal concentrations of products and reactants There is a spontaneous tendency towards equilibrium. (spontaneous ≠ quickly, spontaneous = always moving towards equilibrium) It is possible to force equilibrium one way or the other temporarily by altering the reaction conditions, but once this “stress” is removed, the system will return to its original equilibrium. The equilibrium expression is: Law of Mass Action : aA+bB↔cC+dD Concentrations of pure solids and pure liquids are not included in Keq c d (P ) (P ) [C]c [D]d Kp C a D b Kc or (PA ) (PB ) [A]a [B] b For a heterogeneous equilibrium: CaCO3 (s) ↔ CaO (s) + CO2 (g) K eq Summary of the page and Important things to remember: [CaO] (PCO 2 ) [CaCO3 ] K p PCO 2 Gupta 2014 Credit: Google Images for the pictures