AP Chemistry Study Guide (Part 1 of 3) Chapter 1 - Basics Metric System Prefixes Conversions Definitions Methods of Separating Mixtures Table of Contents Chapter 2 - Nomenclature/History Nomenclature Binary Greek (Molecular) Roman Latin Acidic Basic Hydrates Dead Guys John Dalton Guy-Lussac Avagadro J.J. Thompson Rutherford Robert Millikan Origins of Radioactivity Definitions Chapter 3 - Stoichiometry Stoichiometry Empirical Formula Molecular Formula Gravimetric Analysis Determining the Limiting Reactant Percent Yield/Percent Error Chapter 4 - Types of Reactions/Solutions Parts of Solutions Hydration Three Types of Solutions Acidic and Basic Solutions Measuring Solutions Making Solutions Dilution Types of Reactions Titration Redox Reactions Definitions Chapter 5 - Gas Laws Table of Gas Laws Boyles, Charles, Guy Lussac, and Avagadro's Law Combined Gas Law Ideal Gas Law Dalton's Law AP Chemistry Study Guide (Part 1 of 3) Corrected Values - Van der Waal's Forces Graham's Law of Effusion Mole Fraction Root Mean Square Velocity Kinetic Molecular Theory Stoichiometry of Gases Chapter 6 - Thermochemistry Basics of Thermodynamics General Picture Direction of Energy Heat and Work (Basics) The Laws of Thermodynamics The Kinds of Energies Entropy Changes Spontaneity Systems at Equilibrium Putting =H>, =S>, and =G> Together Heat and Work (Detailed) Hess' Law In-Depth Look at Hess' Law Common Exothermic/Endothermic Reactions Enthalpy of Hydration ΔHhyd Definitions Chapter 7 - Atomic Structure and Periodic Trends Basics - Parts of a Wave Equations/Constants Electromagnetic Waves Hydrogen Spectrum The Bohr Model Electron Configuration Exceptions to Electron Configuration Quantum Numbers Polyelectronic Atoms and Effective Nuclear Charge Heisenberg Uncertainty Principle Quantum Energy Einstein Periodic Trends Cations vs Anions Chapter 8 - Bonding Ions Calculating Lattice Energy Partial Ionic Character Ionic Bonds Energy of Ionic Attraction or Repulsion Covalent Bonding Dipole Moments How It's Drawn Covalent Bond Energies Exceptions to Octet Rule Resonance/Formal Charge VSEPR AP Chemistry Study Guide (Part 1 of 3) Chapter 9 - Hybridization and Molecular Orbitals Sigma and Pi Bonds Bond Order Pi Bonds Hybridization Magnetism The Molecular Orbital Model Homonuclear Diatomic Molecules Let's Screw With Our Brains... Chapter 10 - Liquids and Solids The 2 Forces Types of Intermolecular Forces Dipole-Dipole Forces Hydrogen Bonding Forces London Dispersion Forces Van der Waal's Forces Liquids Surface Tension Beading Capillary Action Viscosity Solids Crystals and Types of Unit Cells Special Atomic Solid (Carbon) Molecular Solids Molecules Without Dipole Ionic Solids Vapor Pressure Condensation Dynamic Equilibrium What is This? Wait...What? What We Can Conclude Phase Diagrams Chapter 11 - Solutions and Solubility Review Energy of Making Solutions Types of Solvent and Solutes Structure and Solubility Pressure Effects (Gases Dissolving in Liquids) Everyday Example Henry's Law Temperature Effects Vapor Pressure of Solutions Raoult's Law Deviation Colligative Properties Boiling Point Elevation/Freezing Point Depression Electrolytes in Solution/Van Hoff Factor Osmotic Pressure Equations Involving the Van Hoff Factor Other Definitions AP Chemistry Study Guide (Part 1 of 3) Chapter 12 - Chemical Kinetics Factors That Affect the Rate of Chemical Reactions Concentration Temperature Catalysts Homogeneous Catalysts Heterogeneous Catalysts Reaction Mechanisms Important Terms Requirements Integrated Rate Law Rate Law Graphs Arrhenius Equation/Derivation of Arrhenius Equation Final Notes on Kinetics Chapter 13 H Chemical Equilibrium Beginning Notes Special K Table Relationship Between Keq and Kp Reaction Quotient (Q) Le Châtlier's Principle Concentration Pressure/Volume Temperature Catalysts Reverse Reactions What if it isn't at Equilibrium?!?!?!? In-depth look at ICE Charts Chapter 14/15 - Acids, Bases, Titration, and Equilibrium Acids Bases Strong Acids Strong Bases Definitions Molecularity of Acids/Bases Electronegativity and Acids Oxidation Numbers Generalizations of Acids/Bases Acid-Base Reactions and Equilibrium Neutral Solutions Acidic or Basic Favored Common-Ion Effect Acid-Base Reactions and Calculations Major Species Acid/Bases: Return of the Sig Figs Titration and Equilibrium Equivalence Point In-depth look at Titration Reaction 3 Sets of Important Steps for Titration Reactions Titration Curves Buffer Solutions and the Hendersen-Hasselbach Equation Example Hendersen-Hasselbach Equation Percent Ionization AP Chemistry Study Guide (Part 1 of 3) Solubility Product Constant Q and Ksp Relation to pH Complex Ions (Ligands) Complex Ion Nomenclature Chapter 16 - Spontaneity, Entropy, and Free Energy Equations (New and Old) Relationship Between ΔG> and ΔG Chapter 17 - Electrochemistry Basics Galvanic Cells (Cell Diagram) Standard Reduction Potential (SRP) Chart Calculating Voltage From a Picture Spontaneity Table Nernst Equation Application Batteries Dry Cell or Leclanché Cell Mercury Batteries Lead Storage Batteries Fuel Cell Solid State Lithium Electrolysis Application Chapter 18 - Nuclear Chemistry Basics Why Nuclei are Unstable Nuclear Trends Magic Numbers Isotopes Half-Life Calculations Types of Radioactive Decay Band of Stability Synthesis of Nuclei Binding Energy Example Chapter 1919-21 Aren't Important Chapter 22 - Organic Chemistry Basics Nomenclature Alkanes Cycloalkanes Alkanes Continued Carbon Numbering Oh Snap, 2 Numbers! Relation Between Carbon Number and Substituents But Wait! There's More! (Isomers) Alkanes and Alkynes Structural Formulas Acetylene AP Chemistry Study Guide (Part 1 of 3) Additional Alkane Information Addition Reactions of Alkenes Markovnikov's Rule Anti-Markovnikov's Rule Hydro-boration Oxidation Reaction Radical Addition Reaction Exceptions to Radical Additions Advanced Nomenclature Cis- vs. TransExceptions to the Longest Chain Rule Cycloalkenes Special Alkene, Benzene Benzene Nomenclature Ortho-, Meta-, ParaReactions Involving Benzene Basic Function Groups Ester Nomenclature Organic Chemistry Overview Miscellaneous AP Chemistry Study Guide (Part 1 of 3) Metric System Mass - kilograms (kg) Length - centimeters (cm) Time - seconds (s) Volume (L) Prefixes Tera (T) Giga (G) Mega (M) Kilo (k) Deci (d) Centi (c) Milli (m) Micro (μ) Nano (n) Conversions Volume 1mL = 1cm3 1L = 1dm3 22.4L = 1 mole (at STP) 1L = 1.06qt 28.32L = 1ft3 Energy 1J = 1Nm (Newton-meter) 4.184 J= 1cal 4.184 kJ = 1Kcal Chapter 1 - Basics Temperature - Kelvin (K) or Celsius (>C) Electric Current - ampere (amp, A) Amount of a Substance - (mole) 1,000,000,000,000 1,000,000,000 1,000,000 1,000 .1 .01 .001 .000001 .000000001 1012 109 106 103 10-1 10-2 10-3 10-6 10-9 Mass 454g = 1lb 1kg = 2.205lbs Other Important Conversions 1Latm = 101.3J 1J = kg P m2/s2 Pressure 1atm = 760 torr 1atm = 760mmHg 1atm = 101.325 kPa 1 atm = 14.7psi Length (English-Metric) 1 in = 2.54cm 1m= 1.094yd 1m = 3.28ft 1.6km = 1mi 5280ft = 1mi 1 Angstrom (A0) = 1 P 10-10m Temperature K = >C + 273 > C = K - 273 R >C = (>F T 32) Speed S S >F = >C + 32 Warp Factor 1.71= 15.00 x 108 m/s R STP: Standard Temperature and Pressure - 1.00atm 0°C Room temperature: 25>C Definitions Precise - How repeatable something is Accurate - How close/correct to the actual value Mass - Resistance to change in motion Weight - Force of gravity Random Error - Results fluctuate each time Systematic Error - Same thing wrong every time Theory - Changeable. Explains how things behave the same in different systems Law - Unchangeable. Summary of observations Law of Conservation of Mass - The mass on the left is equal to the mass on the right Homogeneous Mixture - A mixture that can be separated through chemical means only (filtration, etc.) Heterogeneous Mixture - A mixture that can be separated by physical means Substance - Something with constant composition (uncontaminated) Physical Change - A change in the state of matter (solid - gas...Still contains same compounds) Chemical Change - A change in the composition of the molecules (New compounds formed) AP Chemistry Study Guide (Part 1 of 3) Methods of Separating Mixtures Distillation - Vaporizes a liquid (based on volatility*), condenses the most volatile vapors, that then condenses goes through a condenser to change it back into a liquid. *Volatility - How readily substances become gases (The lower the temperature when the substance becomes gaseous, the more volatile that substance is) Filtration - Used to separate a mixture containing a solid and a liquid. Pour the mixture on a mesh (usually filter paper) and the liquid will filter through, while the solid stays on the mesh. Chromatography - Separates a system with 2 phases (mobile and stationary). Phases are based on the affinities of the components (They move throughout the system at different rates). - High affinity for mobile phase moves quickly through chromatography system, high affinity for solid phase moves slowly though chromatography system. Paper Chromatography - Uses a strip of porous paper (such as filter paper) for the stationary phase. Then a drop of the mixture that needs to be separated is placed on the paper, and dipped in liquid of the mobile phase which will travel up the paper. - The components with the weakest attraction to the paper will travel up faster than those that cling to the paper. Nomenclature Chapter 2 - Nomenclature/History Binary Greek (Molecular) Roman Latin Acid Base Hydrates IA, IIA, IIIA attached to VA, VIA, VIIA IVA, VA, VIA, VIIA attached to IVA, VA, VIA, VIIA Transition Metals Transition Metals Aqueous Aqueous Any compound with water attached Binary Metal name stay same and goes first Nonmetal ending --> -ide If Nonmetal is polyatomic ion, keep name the same Greek (Molecular) Prefixes: Mono, Di, Tri, Tetra, Penta, Hexa, Hepta, Octa, Nona, Deca (1-10) Metal is placed first. If there is only 1 of the metal, prefix is not placed. Nonmetal placed after, and always contains a prefix. Roman First find oxidation charge of the transition metal. Name the metal Place the charge (in Roman Numerals) in parenthesis after the metal Place the nonmetal after, using binary rules. Latin Cuprous Co+2 Cobaltous Cu+1 +2 Cu Cupric Co+3 Cobaltic Plumbous Sn+2 Stannous Pb+2 Plumbic Sn+4 Stannic Pb+4 Ferrous Fe+3 Ferric Fe+2 Metal is named first, followed by nonmetal using Binary rules AP Chemistry Study Guide (Part 1 of 3) Acidic MUST be aqueous Acidic if the H+1 ion is at the beginning of the compound If the acid does not contain oxygen add the prefix hydroName the nonmetal by using Binary rules and changing -ide --> -ic If the nonmetal is polyatomic and ends in -ate, change -ate --> -ic If the nonmetal is polyatomic and ends in -ite, change -ite --> -ous Basic MUST be aqueous Basic if an (OH)-1 ion is attached to the compound Name the metal using Binary rules, or if transition metal Roman rules Add hydroxide at the end Hydrates Used to account for water attached to some compounds Name the compound and then select the Greek prefix that corresponds to the number of moles of water attached and add -hydrate. Dead Guys John Dalton Atomic theory stated: All elements are made of atoms Atoms of each element are identical Chemical compounds are formed when atoms combine Atoms are not changed in chemical reactions Guy-Lussac Under constant temperature and pressure, compounds react in whole number ratios by volume Avagadro Under constant temperature and pressure, volumes of a gas contain the same number of particles J.J. Thomson Cathode-Ray experiment, discovered electron Rutherford Gold foil experiment, discovered atom is mostly empty and contains a small, dense nucleus Robert Millikan Oil drop experiment, found exact mass of electron Origins of Radioactivity Discovered by Bequerel (accidently) Alpha - Helium Nucleus, Beta - High-speed electron, Gamma - High energy light Definitions Law of Definite Proportions - Compounds have a constant composition by mass (Mass is proportional on both sides) Law of Multiple Proportions- 2 elements that form more than 1 compound can be reduced to a small whole number ratio by combining the masses of the second element with 1 gram of the first element. -Water has 8g of oxygen per 1g of hydrogen -Hydrogen peroxide has 16g of oxygen per 8g of hydrogen -16:8----2:1 H2O: 1g H2 W X YZ[\ ]^ _.`_a ]^ H2O2: 1g H2 W bW X YZ[\ ]^ _.`_a ]^ X YZ[\ c X YZ[\ ]^ bW bW X YZ[\ c^ X YZ[\ ]^ Xd.``a c X YZ[\ c bW b=8g O e_.``a c X YZ[\ c^ b= 16g O AP Chemistry Study Guide (Part 1 of 3) Chapter 3 - Stoichiometry Stoichiometry Mole - 6.022 x 1023 of anything - Number of atoms in exactly 12g of C-12 Molar Mass - Add up mass of all elements on periodic table Percent Composition - Percent of each element in a compound fghh Zi j[\Y\kl q 100 fghh Zi mZYnZokp Empirical Formula Based on mole ratios Given: Percents of elements Assume: 100%=100g 1.) Find the moles of each element 2.) Divide by lowest moles - If given a .5 after dividing, multiply all moles by 2 - If given a .333 or .666 after dividing, multiply all moles by 3 - If given a .25 after dividing, multiply all moles by 4 Finds the lowest ratio Molecular Formula Finds the formula for actual molecule Given: Molar mass of the molecule 1.) Find Empirical Formula 2.) Find molar mass of Empirical Formula 3.) Molar Mass/Empirical Formula molar mass=ratio 4.) Multiply all elements in Empirical Formula by ratio found Gravimetric Analysis Basically, just a fancy phrase for using stoichiometry to find the mass of a ppt Must write Net Ionic Equations to do these problems Example: What mass of Fe(OH)3 would be produced by reacting 75.0 mL of .105M Fe(NO3)3 with 125 mL of .150M NaOH? Fe(NO3)3(aq) + 3NaOH(aq) --> 3NaNO3(aq) + Fe(OH)3(ppt) NaOH is limiting reactant because it has higher mole ratio Find moles of NaOH by multiplying molarity and volume (in liters) Use mole to mole stoich to find out how many moles of Fe(OH)2 there are Use mole to mass stoich to find out how many grams of Fe(OH)2 there are Determining the Limiting Reactant of a Reaction Quick way: Look at which reactant has higher mole ratio - C3H8 + 5O2 --> 3CO2 + 4H2O [50.0g C3H8 75.0g O2 How many grams of CO2 ] -O2 is the limiting reactant because O2 has 5 moles, while C3H8 has 1 Long way (safe): Do the stoichiometry for both reactants. Whichever gives the smallest number is the L.R. X uvwx yz {| e uvwx y }}.`X~ y b WX uvwx y {^ b WX uvwx y^ b = 150.g CO2 }}.XX~ yz {| z | ^ 50.0g C3H8 W X uvwx ^ e uvwx y }}.`X~ y b W S uvwx ^ b WX uvwx y^ b = 61.9g O2 e_.``~ ^ ^ ^ 75.0g O2 W Percent Yield/Percent Error Given: Theoretical Yield log[ \[p q 100 % Yield = \Z\lg[ \[p % Error = 100% - % Yield Chapter 4 - Types of Reactions/Solutions Parts of Solutions Solution - homogenous mixture Solute - What gets dissolved Solvent - The stuff doing the dissolving Soluble - Can be dissolved Miscible - Liquids that dissolve each other Hydration The process of breaking ions of salts apart with water molecules -Water is good solvent because it has polar molecules AP Chemistry Study Guide (Part 1 of 3) Three Types of Solutions Strong electrolytes - Completely breaks apart into ions. Conducts electricity well. Weak electrolytes - Partially breaks into ions. Conducts slight electricity. Non-electrolytes - Does not fall apart. Does not conduct. Acidic and Basic Solutions Acids form H+ ions when they fall apart Bases form (OH)- ions when they fall apart Strong acids: H2SO4, HNO3, HCl, HBr, HI, HClO4 Strong bases: Group IA, IIA metal attached to an (OH)- ion Measuring Solutions Concentration - How much is dissolved fZ[\h Zi hZ[ol\ Molarity - = - = l\h Zi hZ[olZk fZ[\h Zi hZ[ol\ fZ[gl - = q Making Solutions Describe how to make 250. mL of a 2.0M copper (II) sulfate dihydrate solution 2.0M = uvwx ._S` XRS.d}_~ y P _{^ b= X uvwx y P_{^ .50mole CuSO4 P 2H2OW 98g CuSO4 P H2O Weigh out 98g of CuSO4 dissolved in 250. mL of H2O to obtain a 2.0M CuSO4 solution Dilution The process of adding more solvent to a known solution M1V1 = M2V2 (Moles are the same) Stock solution is a solution of known concentration used to make more dilute solutions 18.5mL of 2.3M HCl is added to 250 mL of water. What is the concentration? (2.3M)(.0185L) = (.250L)(X) .04255 mole = .250X M=.17M Take 18.5 mL of 2.3M solution of HCl to 250. mL of H2O to obtain a 0.17M solution Types of Reactions Precipitation Reactions - 2 aqueous solutions of ionic compounds poured together to form a solid (ppt) -Molecular: 3NaOH(aq) + FeCl3(aq) ---> 3NaCl(aq) + Fe(OH)3(s) -Ionic: 3Na+1(aq) + 3(OH)-(aq) + Fe+3(aq) + 3Cl-1(aq) ---> 3Na+1(aq) + 3Cl-1(aq) + Fe(OH)3(s) -Net Ionic: Fe+3(aq) + 3(OH)-(aq) ---> Fe(OH)3(s) - All other ions are known as spectator ions Acid-Base Reactions: Acid - proton donor; Base - proton acceptor (Transfer of H+ ions) -Strong acid + Strong Base ----> Salt + water Titration - A reaction where the indicator changes color = endpoint # of H+ P MAVA = # of OH- P MBVB Redox Reactions - The transfer of electrons - Oxidation states Element 0 VA, VIA, VIIA ion -3, -2, -1 respectively Oxygen -2 (Peroxide -1) Hydrogen +1, -1 only when attached to alkali metal VIIA -1 up to +7 (Fluorine ALWAYS -1) Total charge of compound 0 Total charge of ion Charge of ion AP Chemistry Study Guide (Part 1 of 3) Titration Volumetric analysis - Technique used for determining the amount of a certain substance through titration. Titrant - The solution of a known concentration Analyte - The solution containing the substance being analyzed Equivalence point - The point where enough titrant has been added to react exactly with the analyte. - Also known as the stoichiometric point Indicator - A substance added before titration, which changes the color of the analyte at (or very near) the equivalence point Endpoint - The point where the indicator changes color Equivalence point Endpoint Redox Reactions Acidic Medium: 1.) Write skeletal equation with oxidation numbers 2.) Write the Oxidation/Reduction half reaction 3.) Balance atoms that are NOT oxygen in each half reaction 4.) Balance the O2 atoms by adding H2O to the side that needs it 5.) Balance the Hydrogen's by adding H+ ions to the side that needs it 6.) Balance the charges by adding electrons 7.) Put the half reactions together and cancel Basic Medium: 1.) Write skeletal equation with oxidation numbers 2.) Write the Oxidation/Reduction half reaction 3.) Balance atoms that are NOT oxygen in each half reaction 4.) Balance the O2 atoms by adding H2O to the side that needs it 5.) Balance the Hydrogen's by adding H+ ions to the side that needs it 6.) For every H+ ion added, add an OH- ion to BOTH sides* 7.) Balance the charges by adding electrons 8.) Put the half reactions together and cancel Oxidation loses electrons/gains charge Reduction gains electrons/loses charge Oxidizing Agent - Electron acceptor Reducing Agent - Electron donor Oxidation 1/2 ALWAYS has electrons on products Reduction 1/2 ALWAYS has electrons on reactants *Wherever there is an H+ and OH- ion, it will condense into H O 2 Definitions Aqueous solution - A solution where water is the solvent Hydration - The process of using water to dissolve a salt into its ions ]^ c(¡) NH4NO3(s) ¢££¤ NH4+(aq) + NO3-(aq) Solubility - The capability of a substance to be dissolved Electrical conductivity - A solutions ability to conduct an electric current AP Chemistry Study Guide (Part 1 of 3) Chapter 5 - Gas Laws Remember that temperature is ALWAYS in Kelvin when dealing with gases Boyle's Law: Guy Lussac's Law: Avogadro's Law: Combined Gas Law: §¨ ¨ ¥X ¦X = ¥_ ¦_ ©¨ k¨ = §¨ ©¨ ©^ k^ ¨ Dalton's Law: Mole Fraction: ´X = k¨ kµ¶·¸¡ = §¨ §µ¶·¸¡ = ¬f­ agh _ ¬f­ agh X = ©^ ^ Ideal Gas Law: §^ ©^ PV = nRT ^ Corrected Pressure: k _ V' = V - nb Graham's Law of Effusion «gl\ Zi \iiohZk Zi agh _ = ©¨ ¨ Corrected Volume: PTotal = P1 + P2... + Pn «gl\ Zi \iiohZk Zi agh X = Charle's Law: §^ ^ P' = Pobs + aW b © "Real" Gas Law: k _ ®¥Z¯h + W b ° q (¦ T ±) = ²³ © Root Means Square Velocity: ¹Yh = º e« f Gases at STP: Elements: H2, N2, O2, O3 (ozone), F2, Cl2, He, Ne, Ar, Kr, Xe, and Rn Compounds: HF, HCl, HBr, HI, CO, CO2, NH3, NO, NO2, N2O, SO2, H2S, HCN Note that the compounds are not in solution, so they are gases Diffusion vs. Effusion Diffusion - The process of gas molecules spreading throughout a room (cooking scents) Effusion - The process of gas molecules escaping a narrow exit into a room (aerosol) Boyles, Charles, Guy Lussac, and Avagadro's Law P = Pressure (atm) V = Volume (L) T = Temperature (K) n = Moles (mole) When multiplied, there is an inverse relation -Boyle's Law: Pressure increases, Volume decreases (vice versa) When divided, there is a direct relation -Guy Lussac: Pressure increase, Temperature increases (vice versa) -Charles: Volume increase, Temperature increases (vice versa) - Avagadros: Volume increase, moles increases (vice versa) Ideal Gas Law P: Pressure (atm) V: Volume (L) n: Moles (mole) R: Universal Gas Law ddConstant .0821 Latm/Kmole T: Temp. (K) Corrected Values Apostrophe is just notation for corrected value, not derivative Values for a and b will be given and are dependent upon the gas Pobs = Pobserved = Pressure if it were ideal Graham's Law MW = Molecular Weight Mole Fraction Unitless Root Mean Square Velocity R = 8.3145 J/Kmole M = mass in kilograms Units: J/kg What's the Difference? How can "R" be 2 different values? If we are working with gases, we will use the .0821 Latm/Kmole If we are working with Energy, we will use the 8.31 J/moleK If you go back to Chapter 1, one of the important conversions was that 1Latm = 101.3 J So: .0821 x 101.3 = 8.31 AP Chemistry Study Guide (Part 1 of 3) Combined Gas Law Combination of Boyle's, Guy Lussac's, and Charle's Laws Ideal Gas Law In most cases, gases are not ideal and are real gases Gases only approach ideal behavior at low pressure (< 1 atm) and high temperature -Use law anyways, unless told otherwise Dalton's Law Either have to find PTotal when given the partial pressures, or find a partial pressure when given PTotal and some of the partial pressures PTotal can be found by altering ideal gas law equation: ( Zlg[ )²³ ¥Zlg[ = ¦ Corrected Values - Van der Waals Forces Real molecules interact with each other and take up space The actual volume is less because of the particle sizes More molecules will have a greater effect Molecules will be attracted to each other which will make the observed pressure less than ideal Depends on the number of molecules/liter Graham's Law of Effusion Gives us best estimate of the rate in which a gas will spread throughout a room -As gases spread throughout a room, collisions with other molecules will slow down the rate Mole Fraction Gives us the ration of moles of the substance to the total number of moles Units will cancel out Root Mean Square Velocity Gives us the average amount of kinetic energy exerted by the gas particles Kinetic Molecular Theory Assumptions (Only apply to ideal gases): - Particles are so small, we can ignore volume - Particles are in constant motion and their collisions cause pressure - Particles do not affect each other, neither attracting or repelling - Average kinetic energy is proportional to the Kelvin temperature -Two gases at same temperature have same kinetic energy Tells us the average kinetic energy: 3 ¼½¾\ = ²³ 2 R = Gas Law Constant 8.3145 J/Kmole T= Temperature in Kelvin (K) Stoichiometry of Gases 1 mole of any gas at STP = 22.4L Just like regular stoichiometry, but may need to use gas laws in order to find a missing value AP Chemistry Study Guide (Part 1 of 3) Chapter 6 - Thermochemistry Basics of Thermodynamics Energy - The ability to do work - Made of heat and work Heat - Energy transferred between objects because of temperature differences Work - A force acting over a distance Exothermic Process - Releases heat/energy Endothermic Process - Absorbs heat/energy General Picture The universe is divided into 2 halves: The system and the surroundings System - Part of the world we want to study Surroundings - Everything outside the system Open system - Exchanges both matter and energy with surroundings - Open flask - Chemicals (matter) capable of escaping, Temperature (heat - A kind of energy) of reaction can affect be affected by stuff outside the flask Closed system - Exchanges only energy with surroundings - Sealed flask - Chemicals (matter) cannot escape, Temperature (heat - A kind of energy) of reaction can affect or be affected by stuff outside flask Isolated system - Exchanges neither matter or energy with surroundings - Thermos flask - Chemicals (matter) cannot escape, Temperature (heat - A kind of energy) of reaction cannot affect or be affected by stuff outside the thermos flask Direction of Energy Measured in Joules or calories Number tells how many Joules or calories there are Sign tell the direction - Negative sign means the reaction is exothermic - Positive sign mean the reaction is endothermic Heat and Work (Basics) Heat - Energy transferred from one object to another object (both at different at temperatures) through some kind of contact Work - Energy transferred that takes place when an object is moved against an opposing force Thermal motion - Random molecular motion that is stimulated by heat energy Uniform motion - Constant molecular motion that is stimulated by work energy The Laws of Thermodynamics First Law of Thermodynamics - Energy of the universe is constant (cannot be created or destroyed) - Also known as the Law of Conservation of Energy States that: =½ = ¿ + À q is heat and w is work Second Law of Thermodynamics - The disorder or entropy of a system tends to increase - Low entropy = low disorder, high entropy = greater disorder Third Law of Thermodynamics - Entropy of a pure crystalline solid at absolute zero is zero The Kinds of Energies Energy Internal Energy (U/E) Enthalpy (H) Entropy (S) Gibbs Free Energy Bond Disassociation Energy* Equations =¹ = ¿ + À =H>rxn = Á = H> (Products) T Á = H> (Reactants) =S>rxn = Á S>§Zpolh T Á Ã>«\glgklh =G>rxn = Á = G> (Products) T Á = G> (Reactants) =H = Á ÄÅ ½Æ«\glgklh T Á ÄÅ ½Æ§Zpolh *Remember that Bond Disassociation Energy is: REACTANTS - PRODUCTS AP Chemistry Study Guide (Part 1 of 3) Internal Energy (U/E) - Energy that enters/leaves a system - Some energy can return to surroundings as expansion work - Ç =U < ¿ Enthalpy (H) - A state function: H = E + PV - E = Energy, P = Pressure, V = Volume - =H>i = Heat of formation - =H = q - Measured in kJ/mole Entropy (S) - Measured in J/K mole - If =S > 0 then disorder is increasing, becoming less organized, and more random - Thermodynamically favorable - If =S < 0 then disorder is decreasing, becoming more organized, and less random =ÃhoZokpkah = È=] (Remember to convert to Joules) -Depends directly on heat flow (=H) and inversely on temperature (T) -If heat flows to the surroundings, Kinetic Energy is increased in the surroundings and the entropy will increase. Entropy will decrease in the system. - A constant heat flow at different temperatures will have a different impact dependent upon a low or high temperature. - High temperature will cause a lower entropy change on surroundings -Low temperature will cause a higher entropy change on surroundings Entropy Changes of Physical Changes and Chemical Reactions S(S) < S(l) < S(g) - Can quickly determine whether or not the entropy of a reaction is positive or negative by looking at the phases - If phases are the same, then look at what side has more moles 2O3(g) ---> 3O2(g) More moles are produced so the entropy is increased (+) -Aqueous to solids will have a negative entropy (become more ordered) - Entropy will increase if: Volume increases Temperature increases, Kinetic Energy increases, particles dissolve, or solid --->liquid--->gas Gibbs Free Energy - Energy free to do useful work =G = =H - T=S - If =G> = negative number, reaction is spontaneous - If =G> = positive number, reaction is non-spontaneous - If =G> = 0, reaction is at equilibrium Bond Disassociation Energy - The enthalpy change required to break a particular bond - 2 useful things to remember: Draw the molecular structure for the compounds REACTANTS - PRODUCTS Spontaneity A chemical reaction is spontaneous if there is an increase in the total entropy of the system and surroundings Spontaneous change - A change that occurs without being driven by an external influence Non-spontaneous change - A change that occurs only when driven by an external influence Spontaneous exothermic reactions are common because they release heat, increasing entropy of surroundings Spontaneous endothermic reactions occur only when entropy of the system increases enough to overcome the decrease in entropy of the surroundings Systems at Equilibrium A+BHC+D No overall change in disorder =S É 0 (No entropy change) AP Chemistry Study Guide (Part 1 of 3) Putting =Ê>, =Ë>, ÌÍÎ =Ï> ÐÑÒÓÔÕÓÖ =×> - =Ã> + - + + + - =Ø> - at all temperatures - at low temperatures + at high temperatures + at low temperatures - at high temperatures + at all temperatures Spontaneity Spontaneous Dependent upon * =Ø> Dependent upon ** =Ø> Non-spontaneous Using the equation =G = =H - T=S and the table we can assume: =G = large negative - (T x small negative) We know =H is a large negative because it is measured in kJ and needs to be converted to Joules. We know =S is a small negative because it is already in Joules and no conversion is required. So , if the temperature is low, meaning the temperature (in Kelvin) has a point in which when multiplied by the small negative (=S) will result in a number that, when added to the large negative (=H), the =G will be negative. If the temperature is above that point then we know that =G will be positive. -Hypothetically: =H = -1000J/mole and =S = -100J/Kmole T = 9K =G = -1000J /mole - (9K x -100 J/Kmole) =G = -1000J/mole - (-900J/mole) =G = -1000J/mole + 900 J/mole = -100J/mole -In this case we can see that a temperature above 9.99K will result in either equilibrium or a positive =G ** Using the equation =G = =H - T=S and the table we can assume: =G = large positive - (T x small positive) We know =H is a large positive because it is measured in kJ and needs to be converted to Joules. We know =S is a small positive because it is already in Joules and no conversion is required. So , if the temperature is low, meaning the temperature (in Kelvin) has a point in which when multiplied by the small positive (=S) will result in a number that, when subtracted to the large positive (=H), the =G will be positive . If the temperature is above that point then we know that =G will be negative. -Hypothetically: =H = 1000J/mole and =S = 100J/Kmole T = 9K =G = 1000J /mole - (9K x 100 J/Kmole) =G = 1000J/mole - (900J/mole) =G = 1000J/mole - 900 J/mole = 100J/mole -In this case we can see that a temperature above 9.99K will result in either equilibrium or a negative =G * Heat and Work (Detailed) Work - A force acting over a distance w = P=V - If the volume increases (expands) then work is negative - If the volume decreases (contracts) then work is positive Heat - In a system that can't expand w = 0 =U = q (at constant volume) w q + + + + + - = =E E + Depends on size of q and w Depends on size of q and w - AP Chemistry Study Guide (Part 1 of 3) Hess' Law The process of using several reactions' =H> in order to find the =H> of one desired reaction 2 rules: If the reaction is reversed, flip the sign on that reactions =H> If the reaction is multiplied or divided, so is the =H> Hidden Rule: Don't be afraid of fractions =H> = Standard Enthalpy (Enthalpy change for a reaction under standard conditions) 1.) Start by writing the "goal" equation 2.) Then write the other reaction and label (1, 2, 3, ...n) 3.) Look for unique compounds in the "goal" equation and find it in any of the reactions. 4.) Make those unique compounds match the "goal" equation by reversing the reaction or multiplying or dividing. 5.) Cross out equations that have already been manipulated. 6.) Repeat steps 3-5 until all equations have been crossed out 7.) Rewrite the new equations and add up the =H> values In-Depth Look at Example Goal: C6H4(OH)2(aq) + H2O2(aq) ---> C6H4O2(aq) + 2H2O(l) Eq 1.) C6H4(OH)2(aq) ---> C6H4O2(aq) + H2(g) =H> = 177.4 kJ/mole Eq 2.) H2(g) + O2(g) ---> H2O2(aq) =H> = -191.2 kJ/mole X Steps 1 and 2 =H> = -241.8 kJ/mole Eq 3.) H2(g) + O2(g) ---> H2O(g) _ =H> = -43.8 kJ/mole Eq 4.) H2O(g) ---> H2O(l) Eq 1.) C6H4(OH)2(aq) ---> C6H4O2(aq) + H2(g) =H> = 177.4 kJ/mole Eq 2.) H2O2(aq) ---> H2(g) + O2(g) =H> = 191.2 kJ/mole Eq 4.) 2H2O(g) ---> 2H2O(l) =H> = -87.6 kJ/mole Unique compounds: C6H4(OH)2(aq) and C6H4O2(aq) Both compounds are in Eq1 and nowhere else, making them unique compounds Because they are already on the correct side, and have the correct number of coefficients, nothing is required to manipulate equation 1 Unique compounds: H2O2(aq) Located only in Eq2 which makes it a unique compound. However, looking at the "goal" equation, we see that H2O2(aq) is on the reactants side, not the products, so we need to reverse Eq2 AND flip the sign for the =H> of Eq2 (Rule 1) Unique compounds: H2O(l) Located only in Eq4 which makes it a unique compound. However, there are 2 moles of H2O(l) in the "goal" equation and only 1 mole in Eq4, which means we need to multiply Eq4 by 2 AND multiply the =H> of Eq4 by 2 (Rule 2) Eq 3.) 2H2(g) + O2(g) ---> 2H2O(g) =H> = -483.6 kJ/mole Unique compounds: none When looking at this equation we see that there are no unique compounds (phases matter). So, when there are no unique compound we must look at the other equations to see what we will have to do to manipulate this equation. We can approach this case 1 of 2 ways. 1.) We see that there is an excess O2(g) on equation 2 that is not in the "goal" equation so we need to make it so the O2(g) will cancel, which means multiply Eq3 by 2. 2.) We see that there is 2H2O(g) as a reactant in Eq4, but it is not located in the "goal" equation. So we need to make it so the H2O(g) will cancel, which means we must multiply Eq3 by 2. Either way, we need to multiply Eq3 by 2 AND the =H> of Eq3 by 2 (Rule 2/Hidden) Steps 3-6 AP Chemistry Study Guide (Part 1 of 3) Eq 1.) C6H4(OH)2(aq) ---> C6H4O2(aq) + H2(g) Eq 2.) H2O2(aq) ---> H2(g) + O2(g) Eq 3.) 2H2(g) + O2(g) ---> 2H2O(g) Eq 4.) 2H2O(g) ---> 2H2O(l) =H> = 177.4 kJ/mole =H> = 191.2 kJ/mole =H> = -483.6 kJ/mole =H> = -87.6 kJ/mole Step 7 177.4kJ/mole + 191.2 kJ/mole - 483.6 kJ/mole - 87.6 kJ/mole = -202.6 kJ/mole Common Exothermic/Endothermic Reactions Vaporization - Endothermic Melting - Endothermic Freezing - Exothermic Combustion - Exothermic Photosynthesis - Endothermic Breaking bonds - Endothermic Forming bonds - Exothermic Enthalpy of Hydration =Êéêë The enthalpy change involving the hydration of gas-phase ions Gas--->Aqueous If ions dissolve and solution heats - Exothermic If ions dissolve and solution cools - Endothermic Definitions Definitions Potential Energy - Energy due to position and composition (water behind a dam) Kinetic Energy - Energy due to motion of the object (dependent upon mass and velocity) Pathway - The specific conditions in which energy is divided between work and heat State function - Property of a system that depends only on the present state (Also known as state property) Internal Energy - The sum of the Potential and Kinetic energies (represented as U or E) Calorimetry - The science of measuring heat (Based on temperature change observations) Calorimeter - Device used to measure the heat associated with a chemical reaction (Fancy word for 2 Styrofoam cups nestled within each other) Heat capacity (C) - The energy required to raise a substance by 1ì (measured in J/ì) Specific heat capacity (c, s) - The energy required to raise 1g of a substance by 1ì (measured in J/gì) Molar heat capacity - The energy required to raise 1 mole of a substance by 1ì (measured in J/moleì) Constant-pressure calorimetry - The process of measuring heat using a calorimeter under constant pressure Constant-volume calorimetry - The process of measuring heat using a calorimeter under constant volume Chapter 7 - Atomic Structure and Periodic Trends Basics - Parts of a Wave Quantity Wavelength Frequency Symbol (NOT unit) Unit of Measurement Ú 1 cycle/sec (Hz, or 1/s) Û Meters (m) Equations ä = ÛÚ Related. Plug 1st equation into 2nd ½ = Ùä/Û equation to get the 3rd equation ½ = ÙÚ Δ½ = ÙÚ Ü = Ù/(Ûä) Û = Ù/ÜÝ* 2.178 q 10ÈXá àâ _ å½ = i _ T _ =æ P =(Üç) > Ù/4è _ ½ = T(6.02 Þ 10_e ßà/Ü)(2.18 Þ 10ÈXá à)(â ã_ )** Definition Distance between 2 crests of a wave How many waves produced in 1 second Constants c = 3.0 x 108 m/s h = 6.626 x 10-34 J P s Ý in this case is velocity, not wavelength **Energy required to remove 1 mole of electrons * AP Chemistry Study Guide (Part 1 of 3) Electromagnetic Waves Gamma raysàX raysàUV raysàvisibleàInfrared raysàmicrowavesàradio waves (Strongest - weakest) -Visible Light Spectrum: Red 650-700nm Orange 600-650nm Yellow 580-600nm Green 490-580nm Blue 490-450nm Indigo 450-410nm Violet 410-400nm Hydrogen Spectrum Can only give off certain energies Electrons attracted to the nucleus because of opposite charges, but didn't fall in because they were moving Putting energy into the atom moved the electron away, when it returned it emitted light The Bohr Model n = Energy level n=1 is called the ground state Z = nuclear charge 2.178 q 10ÈXá àâ _ å½ = i _ T _ å½ = ½ikg[ T ½klg[ Electron Configuration 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p…. s-2, p-6, d-10, f-14 Aufbäu Principle - As the protons are added by 1, the electrons fill up hydrogen-like orbitals in order of energy level. Hund's Rule - The lowest energy configuration for an atom is the one that has the max number of unpaired electrons in the orbital. Valence electrons - The outermost energy levels, not including the d orbital Noble gases have filled energy levels Left side of Periodic Table fills the s orbital Right side of Periodic Table fills the p orbital Transition metals fill the d orbital Lanthanide or actinide series fills the f orbital Elements in the same column have the same order shell electron configuration Exceptions to Electron Configuration Cr = [Ar] 4s1, 3d5 Cu = [Ar] 4s1, 3d10 La = [Xe] 6s2, 5d1 Ce = [Xe] 6s2, 4f1, 5d1 Pm = [Xe] 6s2, 4f3 Gd = [Xe] 6s2, 5d1 4f7 Lu = [Xe] 6s2, 5d1, 4f14 Quantum Numbers Principal quantum number (n) - Size and energy of an orbital - Integer values > 0 Angular momentum quantum number (ð) - Shape of orbital - Integer values 0 to n-1 Magnetic quantum number (Üð ) - Tells direction in each shape - Integer values -ð to ð ñ Value 0 1 2 3 4 Orbital s p d f g AP Chemistry Study Guide (Part 1 of 3) Electron spin quantum number (ms) - Tells direction electron is spinning X - Integer values ò _ Pauli exclusion principle - No two electrons in a given atom can have the same set of quantum numbers Only two electrons with opposite spins can occupy a given orbital 65.) Give the maximum number of electrons in an atom that can have these quantum numbers a.) n = 4 b.) n = 5, mð = +1 X c.) n = 5, ms = + _ d.) n = 3, ð = 2 e.) n = 2, ð = 1 a.) n = 4. Because it doesn't give a specific ð value, we use the n-1 = ð to find that ð = 3. With this we can see that there is a total of 16 orbitals when n = 4 (1 from 4s, 3 from 4p, 5 from 4d, and 7 from 4f). And we also know through the Pauli exclusion principle that one orbital can only carry a maximum of 2 electrons with opposite spins. So 16 x 2 = 32 electrons. b.) n = 5, mð = +1. This problem goes into the imaginary orbital g. We also know that the mð value ranges from + ð to - ð. And when n = 5, it will have all the previous ð and mð values along with an additional row which will be: ð mð n Subshell Orbitals in Orbitals in Subshell Shell 5 0 4s 0 1 1 4p 1, 0, -1 3 2 4f 2, 1, 0 , -1, -2 5 3 4d 3, 2, 1, 0, -1, -2, -3 7 25 4 4g 4, 3, 2, 1, 0, -1, -2, -3, -4 9 We were also given that mð = +1. From this we can say that there are only 4 subshells capable of containing that specific mð value (4p, 4d, 4f, and 4g). 4 x 2= 8 electrons. X c.) n = 5, ms = + . Again, we go back to the imaginary g orbital. This time, we are not given any _ specific ð value, so we again use n -1 = ð to find that ð = 4, With this we see that there is a total of 25 orbitals when n = 5. So 25 x 2 = 50 electrons. However, it X told us ms = + which means to account for only the electrons spinning in a _ positive direction. So we divide 50 by 2 to get a total of 25 electrons. AP Chemistry Study Guide (Part 1 of 3) d.) n = 3, ð = 2. Following the chart, we see that this takes us to the 3d row. And there is a total of 5 orbitals in that subshell. 5 x 2 = 10 electrons. e.) n = 2, ð = 1. Following the chart, we see that this takes us to the 2p row. And there is a total of 3 orbitals in that subshell. 3 x 2 = 6 electrons. Polyelectronic Atoms and Effective Nuclear Charge Contains more than one electron Has 2 energy contributions -Kinetic Energy of moving electrons -Potential Energy of attraction between nucleus and electrons -Potential Energy from repulsion of electrons Find energy by using the effective nuclear charge equation: å½ = _.Xóá qX`ô¨| õö ^ k÷ ^ È kø ^ Find energy required to remove 1 mole of electrons from a substance _ ½ = T(6.02 Þ 10_e ßà/Ü)(2.18 Þ 10ÈXá à)(â ã_ ) Heisenberg Uncertainty Principle Basically, just accounts for what percent/how often the electrons will actually be in the position predicted. Uses the equation: =æ P =(Üç) > Ù/4è Quantum Energy Discovered by Max Planck who said ΔE came in chunks with a size hù Use equation: Δ½ = ÙÚ -n is an integer -h is Planck's Constant Einstein Stated Electromagnetic radiation is quantized in particles called photons Equation to find mass of photon: Ü = Ù/(Ûä) Equation to find Energy of photon: ½ = Ùä/Û ½ = ÙÚ Periodic Trends Ionization Energy: Across a row - Increases; Down a column - Decreases -Need to take into account the effective nuclear charge and when you are removing electrons form half-filled or full orbitals (will be harder to remove) Atomic Size: Across a row - Decreases; Down a group - Increases -Measured in angstroms (1 x 10-10 meters). Influenced by shielding and charge on nucleus. Electron Affinity: Across a row - More negative; Down a group - More positive Ionic Size: Across a row - Decreases; Down a group - Increases Electronegativity: Across a row - Increases; Down a group - Decreases -The tendency for an atom to attract electrons to itself when chemically combined with another element. Corresponds directly with electron affinity. Metals: Low ionization energy, low effective nuclear charge, low electronegativity. Nonmetals: More negative electron affinity, high electronegativity (except noble gases) AP Chemistry Study Guide (Part 1 of 3) Periodic Trends Continued Metals lose electrons and have lowest ionization energy Nonmetals gain electrons and have the most negative electron affinity Alkali Metals: Decrease in ionization energy Increase in atomic radius Decrease in density Decrease in melting point Behave has reducing agents React explosively with water (except Li) Reducing Ability: The lower the ionization energy, the better the reducing agent -Cs > Rb > K > Na > Li -Only works for solids, not as solutions -In solutions: Li > Na > K - Reason is the water, when the solid metal is dissolved, there is an energy change known as hydration energy Oxidation Number Trends: Group IA IIA IIIB - IB IIIA IVA VA VIA VIIA VIIIA Oxidation Multiple +2 or Number +1 +2 oxidation +3 -3 -2 -1 0 +4 numbers Cations Vs. Anions Metals lost electrons, become smaller and are positively charged (Cation) Nonmetals gain electrons, become large and are negatively charged (Anion)