Lecture Notes CHEM 142
Popular in General Chemistry
Popular in Chemistry
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Date Created: 12/06/14
Atomic Structure 26 Protons exists in the nucleus with a positive charge equal in magnitude to the eectron39s negative charge Neutrons have virtually the same mass as a proton but not charge High density 0 Accounts for almost all of the atom39s mass 0 Relative small size The number of protons is the same for atom of a given element 0 Ex Hydrogen atoms contain one proton All carbon atoms contain 6 protons etc o Atomic number Z is equal to the number of protons de nes an element o Mass number A is equal to the number of protons and neutrons AzNa o atoms will have the same number of protons but different numbers of neutrons Demonstrate essentially identical chemical properties ex Bonding same pattern of reactivity o can be determined by adding up the masses of the sub atomic particles making up the atom Molecules and Ions 27 o formed with atoms bind together The type and number of bonds an atom forms depends on both the number of electrons and protons the atom possess o 2 types of bonds the sharing of electrons between atoms gt Ions are not produced I the transfer of valence electrons from one atom to another gt Overall charge of the compound must be zeros determines the relative number of cations to anions in the formula unit 0 Result of the electron transfer is the formation of an ion 0 Positive ion more protons than electrons 0 Negative ion more electrons than protons o Ions are not formed by the transfer of protons gt on an atom or group of atoms that has a net positive or negative charge 0 Cation positive ion 0 Anion ion with a negative charge o Chemical bond forms when the electrons on two atoms are able to interact Energy is released when formed o Oxidation state charge Use Roman numerals 0 Example Iron Fe Fe2 ironll cation Fe3 ironlll cation 0 Anions Xm root of atom name ide same as before 0 Covalent Two nonmetals I Type III non metals 0 The rst element in the formula is named rst and the full element name is used 0 The second element is named as though it were an anion Pre xes are used to indicate the number of atoms present 0 The pre x mono is never used for naming the rst element o ie CO is carbon monoxide not monocarbon monoxide 0 Diatomic molecules BrlNCHOF 0 H2 N2 02 F2 CI2 BT2 I2 0 Exceptions P4 S8 C50 0 Compounds must be charge neutral 0 Polyatomic ions 0 Referred to as a group when naming a compound Prefix Indicated mono 1 di 2 tri 3 tetra 4 penta 5 hexa 6 hepta 7 octa 8 nona9 deca 10 Lecture 6 31 33 Atomic Masses 31 Carbon standard mass unit 0 Carbon 12 de ned as exactly 12 atomic mass units amu Atomic masses of isotopes 0 Most accurate method to measure the comparative masses of atoms and isotopes is mass spectrometry o By comparing the trajectory of different atoms through the mass spectrometer their relative masses can be determined 39 ll 2 sa39rnp le oil tar39bUn is 39i39ntletted into quota mass spectr39039nne39ter the mas5 of other imtnpes can be mmgpazred to that of it Drlingl this or1393Cwe ind that 13 v masis C at V l 083 6129 mazsrs Ci 39 quotThesrefoere the amass ol 13C in amu is E imply llamu C 13 3 E ra ltl 111335 0f l B836129x111 0K 0 Atomic mass AM atomic weight AW product of mass and acceleration due to gravity 0 Intensity amount or abundance of the isotope in a sample directly related to concentration 0 Atomic mass table at beginning of textbook average given the abundance of the isotopes x their masses 0 AM a x abund AM b x abund average of element The Mole 32 and Molar Mass 33 The Mole the number of carbon atoms in exactly 12 grams of Carbon 12 ie Avogadro39s Number I E 39 l 5quot W 3 p W 391 g amu 1 P 139 1L1 lt 6i22x Lecture 7 35 37 Percent Composition 35 Two types of decomposition processes 0 To the elements H2 02 C 0 To the substances H20 CO2 We can use mass percentages to determine elemental composition 0 Mass produced x molecular mass of element molecular mass of compound mass of element in compound 0 Mass of element in compound initial mass of compound x 100 composition of element Compound Formulas 36 Molar ratio 0 Find smallest whole number ratio to determine the relative molar composition Empirical formula provides to simplest whoe number ratio of the atoms in the compound but is not necessarily the actual formula for the compound 0 Each element is multiplied by how many of it there are by its atomic mass 0 Total mass of the empirical formula 0 No molar mass no transition to molecular formula Molecular formula Compare mass of the compound given to the mass of the empirical formula o Gives you ratio to nd quotexactquot or molecular formula If the molar and empirical formula masses are identical then the formulas are also identical Empirical and molecular formulas are expressed s integer amounts of atoms No fraction subscripts Chemical Equations 37 Chemical reaction involves the reorganization of atoms from a set of initial elements andor compounds to a nal set of elements andor compounds 0 Reorganization involves the breaking and forming of chemical bonds between atoms 0 Mass is conserved all atoms present in the reactants must show up in the products o A properly written chemical equation describing a reaction will be balanced be consistent with the conservation of mass Chemical Eduation Contains two pieces of information o Composition of the species involved in the reaction 0 Physical state of the species involved in the reaction Critical Solid liquid gas aqueous solution Lecture 10 Precipitation Reactions 44 45 Precipitatio synthesis of an ionic solid 0 A solid precipitate insoluble salt forms when aqueous solutions of certain ions are mixed Precipitate must have a net charge of zero 0 To determine precipitate study combinations of reactions where you know that one of the potential precipitates is soluble o How precipitation or quotsolubilityquot rules have been determined 1quotIn llEat EI1iir J1 I39ES39iEiJ39Jcqmfl HillE are ietii IJi39ilre r1a t mills trununng 1Eh1i rltiuzsj ilIjI1rE Lil Mai quot E E p a l JquotEI l T1lil39 nr1c1I tI1 uIniu cm nj l rEI39I Ell E siriiuitlgr 39lrilet EjTE39il39FfisIiIHI 395E llr39niquotai1aii1riia e ant ina39lirllce salts 39iJI393IlJi39IIil2 Z l39Earn ljallailE euiierii1iu sn na 1i39e igfl IjI2 equotquot l39II mg illil ilJTI I ag ll PH 5 Lrrit P u l1nel iiIl39 M5 5 Lilllifti i1iri i4flil1395q 1quot3Iirlizflllltlll iEIcLIfIIiiI m E39iti EJt5 D u FTfIEl39 3 1 H E35314 iinJl 39aaEZi4 aismzat l1ii I39sImiultar 2i I1IEF L139sr4 Duly 5illighly 5maluihl The in11lI 39rtIriI E53I39utE 39ia3939lquot lI39i J39iE39LiilillE M12 Ljililiisl iiiuil liT1I1I The IquotII39iifIc1lfJi7 ir Iili2 iIi39aH Z3aIZIill13 L irEZIalZ i31 mril iZ a LET1lEriZ irE nIingii39ilr39 IalL1TIin ridlat Euliilt I15 1 aLrEHgrit1 Li E iifquotE l i5 l FIZiTEL lIi CE E EL1i nrl1amp5ltuLi P313 in 51llH Higher charge on ions results in a higher probability of insolubility more electrons involved in transfer Acid base proton transfer reactions Acid donates a proton to a base forming a molecule water or another weak acid and an aqueous alt 0 Acid proton donor 0 Base proton acceptor Oxidation reduction electron transfer reactions 0 Electron transfer from one species to another causing a change in the oxidation state of the two species OIL RIG Oxidation ls Loss of e Reduction ls Gain of e Lecture 11 0 Monatomic hydride ion H is 1 0 reduced 0 oxidized The oxidation numbers of the atomsions in a species must sum to the total charge of the species Half reaction provides reliable method of balancing redox reactions in aqueous solution Lecture 13 De nitions 51 Gas Can treat particles as individual molecules 0 Due to space between molecules Pressure 0 Force per unit area P FA SI units pressure is expressed in newtons per square meter Nmquot2 the PascaPa o 1 Pa 1Nm 2 How to measure the pressure of a gas o Barometer Pressure height acceleration due to gravitydensity P hgd Measuring height of column will tell you what the atmospheric pressure is Units mm Hg millimeters of mercuryO 0 Aka quottorr after Evangelista Torricelli 1 atm of pressure 760 mm Hg torr At sea level 0 Higher up less air pushing down therefore less torr required o Manometer quotJ tubequot connected to a gas to measure pressure Open tube Pgas htorr Patm 0 Gas pressure Pgas less than atmospheric pressure Closed tube Pgas Patm htorr 0 Gas pressure greather than atmospheric pressure The First Gas Laws 52 0 Gases are relatively easy to measure and observe in a lab o Made the physical properties of gases a popular subject 1719 centuries Boyle Charles Avogardo and Amontons determined fundamental connections between P V T and n for gases Boyles Law Boyle studied the connection between P and V of gases T and n held constant P V a constant for xed temperatures V is inversely proportional to 1P quotldeaquot behavior PV will be held constant as pressure increases PiVi PfVf 0 Go down in pressure go up in volume Charle39s Law c 1800 Lecture 14 54 56 Gas Density 54 0 We can use PV nRT to determine density of gases 0 Units Massvolume Density will depend strongly on T P and the mass of the gas molecules o Contrast with liquids and solids densities depend somewhat on T but far less on P 0 PV nRT or n PVRT o Using n and molar mass of elementcompound to convert moles to grams in order to calculate density gL PMM dRT o The density of an ideal gas depends on T P V and molar mass 0 We can use the density of a gas to determine its molar mass Dalton39s Law of Partial Pressure 55 0 Ideal gas law gas molecules are non interacting point particles Increasing the number of point particles increases the pressure by an amount that is proportional to the number of particles o For a mixture of ideal gases in a container Total pressure the sum of the individual gas pressures ie PHe 200 torr Par50O torr Ptotal 700 torr Can determine total pressure by adding partial pressures or added the moels of each gas to determine total pressure within PnRTV see example in notebook Mole Fraction and Partial pressure 0 Mole fraction x ratio of the number of moles of a component in a mixture to the total number of moles in the mixture 0 The fraction of moles of a certain gas in a mixture is equal to the ratio of its partial pressure to the total pressure of the mixture Pi XiPtota Dalton39s Law See notebook example Kinetic Molecular Theory 56 o The gas laws of Boyle Charles and Avogadro are empirical meaning they are based on observation of a macroscopic property o These laws offer a general description of behavior based on many experiments o They can tell you what happens to an ideal gas under certain conditions but not why KMT is a theoretical moecuar eve model of ideal gases which can be used to predict the macroscopic behavior of a gaseous system Postuates o Gas particles are so small that their volume is negligible 0 Gas particles are in constant random motion Lecture 14 54 56 This motion is associated with an average kinetic energy that is directly proportional to the Kelvin temperature of the gas 0 Gas molecules constantly collide with each other and with the contain walls The collisions of the particles with the container walls are the cause of the pressure exerted by the gas Collisions are elastic o The particles are assumed to exert no forces on each other they neither attract or repel their neighbors Central Points We can describe temperature and pressure from a molecular perspective 0 Pressure arises from molecules banging into the container walls 0 Temperature is directly related to the kinetic energy of the gas molecules The more KE they have the greater their temperature Total KE in 1 mol of gas 32 RT Average KE per molecule 12 muquot2 12muquot2 1NA 32RT Urms root 3RTMM The speed of a molecule that has the average KE 0 Gives us a formal connection between average gas speed T and MM 0 Uses 8314 mo 1 K1 for R 0 MM in kg Distribution of Molecular Speeds Maxwe botzmann curve Um most probable speed Uav average speed Urms the speed of a molecule with the average molecular kinetic energy Increase T increased average KE increased Urms Increased MM decreased Urms Lecture 15 0 Za 14 NNA root 8RTpieMM 0 Particle Density 0 NN PRT Real Gases 510 0 There are conditions under which a gas will behave ideally 0 Low P 0 Moderate to high T 0 At high P the volume of the individual gas molecules becomes non negigibe 0 Macroscopic gas is compressible individual gas molecules are not 0 Under high p conditions the space available for a gas molecule to move through is decreased by its neighbors so the volume of the system is reduced relative to the ideal case Veff Videal nb volume occupied by moecuesn number of moles of gasb empirical constant different for each gas 0 Ideal gas theory we assume that gas molecules do not interact o But under high P gas molecules get very close to each other and interact o At low T the molecule sped drops also increasing the importance of intermolecular interactions o Under high P andor low T the molecule don39t collide with the container as frequently so the pressure of the system is reduced relative to the ideal case o Pobs Pideal anNquot2 Pobs anVquot2V nb nRT 0 B generally increases with the size of the molecule 0 A generally increases with the strength of intermolecular forces 0 vdW equation corrects two major aws in ideal gas theory o gas molecules have nite volume which becomes important at high P 0 gas molecules have nontrivial attractions that become important at low T and high P Lecture 16 See example in lecture PowerPoint 0 If reaction is reversed equilibrium constant will be the inverse 1original constant 0 If coefficients in reaction is multiplied by a factor n Knew Koriginaquotn o Raised to the power the factor the reaction is multiplied by 0 n 2 reaction is multiplied by 2 and Kcquot2 0 Always the same value for the equilibrium constant no matter what the initial concentrations are Lecture 17 Pressures and K 63 Kc Kp RTquot de cb Kp RTquotAn A is the difference in moles products minus reactions of gaseous species only Heterogeneous Equilibrium 65 0 Reference states for solid species or liquids solid or liquid species themselves 0 Become 1 0 Can quotignorequot all solids or pure liquids when constructing K 0 Must look at phase of all species in order to determine the most reasonable reference state to employ 0 Pressure 1 atm o 1 M o Liquidssolid cancel out Elementary Applications 66 o The reaction duotient Q the value of the express for K known as the 3939law of mass actionquot at any time during the reaction o K is at equilibrium Q is at any time during the reaction Can compare Q and K to determine if system is at equilibrium Q K system is at equilibrium QltK reaction is proceeding towards products too many reactants QgtK reaction is proceeding towards reactants too many products 0 OOOO Q productsreactantstime K productsreactantstimegttime equilib Lecture 18 Solving Equilibrium Problems 67 0 Determining Equilibrium Concentrations 0 Comparing Q and K ICE table initial concentration change in concentration equilibrium concentration quotX represents the change in composition between the initial conditions and equilibrium X will be negative for reactants loss to products side Coef cient determined by balanced equation Equilibrium constant initial concentration change in concentration 0 Substitute equilibrium concentrations into the equilibrium expression and solve Solve last equation for x o Quadratic equation use quadratic formula Choose x value that makes physical sense in regards to the equilibrium concentration 0 Plug in x into the equilibrium concentrations Look for approximations if Kc is small QgtKp therefore reaction is proceeding from products to reactants
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