Chem 121 Notes PDF

Millikan Experiment Oildropexperiment Mainfindings chargeoftheelectron Massofthe electron Oildropletsinjected 2 chargedplates bottom is negative topispositive Inbetween theseplatesthere is an adjustable electricfield Overtime oil dropswill fallintothech...

Millikan Experiment Oildropexperiment Mainfindings chargeoftheelectron Massofthe electron Oildropletsinjected 2 chargedplates bottom is negative topispositive Inbetween theseplatesthere is an adjustable electricfield Overtime oil dropswill fallintothechamberthroughthehole in thefirstplate In the chamber the droplets willpickupa charge hencewill berepelled bynegativeplate Tofind charge 2 methods Electronshover inspace when electric field repelling the electronmatcherstrength Stationary of gravitational field pulling droplet down a.at 5 iiiIiieon electrical fieldstrength dictated D M pxu F calculated willvarybetweendropletsastheypickupdifferentmagnitudes in the chamber but themagnitud willalways be a multipleof the fundamentalunit ofcharge of an election 159 1019 c tangan Chargeofelectron n.ee qq.IEaiii ii mass arteritis i Therechargethenusedto JJ Thompson gelm ratio find mass of the electron using iii Solve forme kg siitis 9 109 10 3 kg  JJ Thompson keyfindings chargemass ratio of an electron Electron not through Anode tobecollimated rayproduced by cathode Electron beam deflected byelectricfieldcreated by chargeddeflectionplates electron beamdeflected awayfrom negativeplate Introduction of magneticfield which too maydeflect beam in oppositewaytoelectric Electronbeamlights upphosphorous on thescreen when ithits ionizes measure thedisplacement of thebeam Usethedisplacement S and thefield strengths tofind que fy Horce EE iIigtioaigjantacamanonelectron beam onelectrons ay Tt usedisplacement rightafterdeflectionplates timerequired totravel detectionplatedistance front 1 fit Tra S displacement of beam measured onglass unsth centreof deflectionplatestoscreen g 12 length of deflectionplates 1 it trial But wedonthave lusal velocityof electrons Wedothisbynull experiment equate electricfield ielaieno ifeng.tt  the I petition eEtelv o overall force is 0 hence 1024 Subbackinto S 5 a E E E 1.7588 10 C Te es acceptedvalue kg iii FFF. deflected electionbeam cathode magtie anode field deflectorplates ecentric Vacuum electricity passed at lowpressure Ionizes gas causing lightrays tobeemitted communenningggammermanaggg Disproving theplum Rutherford Scattering puddingmodel pouffu Use a decaying source to produce alphaparticle t use accelerator lived at atom thick goldfoil  Findings alphaparticles mostly forward scatter passed throughemptyspace withminor deflection some a were backstattered must be area inside theatomswhere mare is concentrate ie a nucleusincenter some were repelled something positively charged ie nucleus atomic = protons Repulsive coulombic force zzejffiist.fm fpermitivityofvaicum atdmic.at maTiie aeiigaiig min distance ofclosest approach closestpointto thenucleus where repulsiveforce still active toget 1loserwewouldDrea thefore min isthusthenotest approximation to thenuclearradius actual value will besmaller repulsionoccurswhen the ke of particle is completely transformed intopotential energy zZe 4thEormin 22 2x omar min inmate massalphaparticle Mr inkg NA hatamoto Neutron uncharged particle  Visualization of atoms use a Scanning Tunnelling Microscope Probe electron chargedensity measures ofatoms atsurface Movesharp tip charged oversolidsurface of atoms seecharge of each atom Ionization Energy Tendency to give up an electron to a neighbour 1st IE minimum energyrequired to detach an electron from a nettral gaseousatom g Xtig e Endothermic need input ofenergy to remove an electronfrom its attraction to a nucleus Higher1st IE across a period as increase in mass number nuclearcharge wo a significant increase in shielding up a group Ces shielding fullelectronshells Lower 1st E down a group move shells shielding valence electron from positive pull of the nucleus p orbital electrons higher than S easier to remove ergy p3 as repulsion e e multiple Ionization of Atoms g suaaveionizatilonenegi.es 2g to endothermic ie increase as gets harderto get more remove from positiveions more attraction between e and Xt Largest IE jumps between major shells  Electron Affinity g to X'g Minimum energy required to add an election to a gaseous atom Exothermic a bond releases making energy No group 18 EA fullvalence shell group 1 high EA as filling s orbital is beneficia X and e EA may be successive endothermic as repulsion between eeawnegamy.tdf.IT eeanssaeaonieac on large electronegativity high E and EA halogens Low elertronegativity low E and EA alkali metals Mulliken EN 0.5 EAT IET Pauling EN ΔEaa AAbond dissociation energy geometric mean ofbond energy ΔEbb BBbond dissociation AB Bond dissociation originatesfromthe geometricmeanof Eaa Ebb asatoms identical covaliptomentofar dead Tenet tonic component of AB d deals Fadets plugin Δ to solvefor XA XB 5 electronegativity difference XA XB 010240 Difference in electronegativity solved byfinding 4 in KTmol and thensowing for X  s y g verylonic if the ENdifference is 2 largely lovalent 0.4 inteumediat.Y.fi genxageloniclhaF n F.EE iiiiiiiiiiiiiiiiiiiiianel Dipolemoment in Debye D ID 3 336 10 30 c m Columbmeters Percentage lonic Character D R A 0.2082A 8 µ solve forioniccharacter fraction ofcharge Q ef Dipole moments arevectorquantities magnitude direction Arrowfaceto more ENpositive Dipole moments canal in classi USFPR whenatomssamearoundcentre Dipole Moments Finding dipolemoment of an individual bond in a molecule solving Dipolemoment OH bond p gR It forthewholemolecule thedipolemoment maybedifferenttuetogeometry Moleculardipolemoment is thesum of all individual dipolemoments  c M I coulombs 1 Determinedipolemoment eachbond of To 212 2 3D pionaans Px µ cos 0 2 N2cos 0 21 H Ny Uisin 0 2 42cos 0 2 H molecule hotfy pythagoras Moleculardipole water is 1185Δ of similar dipoleof on is 1.529 µ acoso dsino  tinttesth Asbonds intheace inlength theydecrease instrength In carbon thisrule breaksdown ie not proportional In resonance structures bondorderbetween integers formalthaus Lewisstructure lookgroupnumber Circleatom If numberovergroupthen change if it containslonepairthats2 if it cuts a bond that I double bonds cut are 2 Resonance Hybrids morethan I way todrawhewisstructurethatall have sameformal charge dispersion bondorder 15 03 2 same lewis structures to is 0 Breakdown OctetRule group 3 and belowexpand octet due to empty d and f orbitals promote to empty bonds subshell you createmore unpaired electrons tocreatemore Strange owes  NO radical Bf electrondeficient nocharge in I I Origins ofBonding Antibonding Bonding attraction fromeathnucleito e Tint Antibonding f electronnot in internuclearregion butstillattracted tonuclei o Covalent Bonding in Hydrogen electrons shared between protons e electronsrepeleachother protons repel eachother eachnullers attracted to bothelectrons pt U potential energy will be thesumof attraction repulsion antigen t.E.la tiisolat attraction: 1/r where r is the distance in An hitconversion hepulsion as two make positive Ionic Bonding CoulombStabilization  FERAL l Ed isthesum of theionizationenergy and desoul If KFstayed neutral wesee no attraction or repulsion at far R distancesmuch betwe As R shrinks in nucleigetcloser together there isminorattraction to on an other trough issuper shallow henceweak bond covalent bondingfor Rfnot prettered as shallownotstable asymptote At R shrinks we see mast repulsion Hence ionic bonding we see deep trough as R decreases Thisdepthconfersstabilityof a molecule t hence ionicbonding ispretrewed overcovalent Hence large dissociation energy when ionsgettooclose they will repel crossover pointbetween ionicand covalent electronjumpsfrom Kto f lausing keepenergy drop increased stability AE Q Q2 4HER distance between charge depth oftrough iiiiiiiii.ir electronpairsin valenceshellsrepel each other hence molecules takeshape to reduce it  t.ir 1k planar tetrahedral fftrigygjam.ua I so octahedral SteneNumber numberof environments bonds and lonepairs are each Lonepairs lonepairs bring electrons Koserto centralatom distorting overall shape of themolecule due to repulsion onotheratoms willdistortbond angle by 2.5 eachlonepair 1 no trigyfam.am infdpin seesaw if Tshape Linear removefrom equilateralplane Tonditometert itorm air Hynffy Typhon inactionArea g mercury a 9.8m s acceleration duetogravity g p mass gravity MI Trea  P mgn v my 1 D An Volume crosslectionArea height ofmercury meters P mgh V m u mass volume p density P pgh Pressure density gravity height of mercury meters Wgf 103 g cm Pg hm s atm 101325Pa bar 105Pa latm totor.com gistemmne pav pressure C and volume constant ie inverselyproportional foundwhen increased amount of mercury increasing pressure on trappedair ledto the volume of trapped air moved proportionally toincreasedpressure Small deviations fromideal gaslaw occur for REALgatesat higherpressures imamervolumes finitevolume of thegas moleculesmust be considered and the interactionsbetweenthem At higher temperatures Vmal Videal  CharlesLaw U T volume directlyproportional totemperature At lowKelvinTemp all gases expand bythesame amount if theystartand endwith same temperature as eachother volume of gas at 00C T c Vp Yolume gas lonstantfor allgases 1k 27315C V V0 1 7 27315C Idealgaslaw BOYLE V α volumeinverselyproportional to pressure when t and h are constant IMARLES U α T volume directlyproportional to temperature inKelvin When Pand n constant Vαn volumedirectly proportional to number ofmoleswhen Pand Tconstant Hence V α R PV constantan gases proportionality ionstant 0.08206 Latmmot or 831 Jmork 212,5  it Patton's Law Outtome Total pressure is the sum of Partial Pressures of the Individual Gases Ptot PA PB Ptot Ntotf tomorefraction off f pressureproportional KA PA APtot Partialpressure moretraction total pressure Kinetic Theory Gases 1 Pure gases consistof large number ofidentical molecules separated bydistances that aregreat compared to theirsize 2 Gas molecules constantlymoving inrandomdirections with a distribution of speeds 3 Molecules exertforces on one anotherbetween collisions moving in straightlines still elastic with constantvelocity 4 Collision of molecules w walls is elastic no energy lost in lollision Elasticity of louision p MU willhave direction vector momentum mass x velocity when a molecule hits the wall it will travel backwards w thesame momentum as when it hitthewall no energylost v2 Ox t Uy t iz Ye willreverse Ap final momentum initial momentum 2mVx allotherdirections VyandV2 Stayconstanthence cancelled Timebetween collisions 21 55in iaveneninxairection Magnitude of momentum transferred to wall persecond t.IT Eid met Thisisalsotheforceexertedonthewall f ma met I mF Totalforce an by moneygles Nmett F P NmyI f meansquarespeedofgasmolecules P NayI naff PV NMTX setn'dofdirectionpretterence pu 4 aug net sumi artica E RT EiggBoltz Na ehergyofparticle  by meanSquarespeed 3ft meansquarespeed isproportional to the temperature and inverselyproportional to mass hence allmoleculesmove faster at higher temperature lighter moleculesmove fasterthanheavierones at thesame temperature Maxwell Boltzmann KB R Na tofindparticleproportion fifty Integration f u s flu 4 ÉÉ re 2ft velocity Yin e q Jan 1 sour collingiting synthronized rotatingsectors key knowhowmanyparticles travelling at a certain speed particles fromboxescape through a hole collimated into a beam beam throughrotatingsectors speed of beammust match openingangle speed ofsector known and controlled beam hitsdetector at end  I Increase temperatureshifts graphdown andright increasing proportion of particles with ke Ea Areaunder is the same same ofparticles Speeds probable up Mostprobablespeed Lhi Mean speed Mt man Vp FI derive f u tofind Max J FI integrate v v between O and inftofindVav Rms FI integrate V2 f ul between O and infto find Urms Effusion Reliesof collision w walls Zw collision frequency w walls 2W Area proportional toarea 2W Vare are speed Zw NV numberdensity movemolecules 2W α I JA FIA 2W phroportionality constant  rareoretion F Enrichment Enrichment factor MB MA MIA If enrichment factorgreaterthan 1 numerator willincrease If enrichment factor less than I denominator will increase IsotopicRatio Isotopic Ration enrichmentfactor weep Heavy water vs light waterreactions Heavy are better absorbfew neutrons higher moderatingpower Slowdownneutrons verywell usesnatural uraniumdioxide asfuel noneed toenrich to 7 Light absorbmoreneutrons slow moderatingratio need 7 enricheduranium dioxide as fuel Moderators Heavy use Heavy wateras moderator Dao light normal water Coolant Heavy Uses Meavywater as a coolant light normal water or liquid sodium or CO2  Diffusion Gaseous diffusion through a porous membrane Lightercomponent enriched with factor Mfa lighter move collisions w porous membrane at same temperatures Collisions Frequency ofMolecule Molecule Atoms as it movessweepsout a cylinder of volume byhittingotherparticles w the same diameterout of theway length of cylinder Vaue I second volume ofcylinder Tyave Id crosss ection area ofcylinder using d ofmolecule Collision Rate 21 dependent on mass 21 Viet Nf Vcyl Nf Id 2 8 Nfd tf mime mi t ma 21 forpuregas noted't a 4yÉf kg aim event I Ntf from as n ya 0 08206 outinto if weuse 8.31 we willget m L m 100  Uvel E Jave MeanFreePath Diffusion 21 rate of collisions 1 2 I rate time betweencollisions mean freepath distance 2 distance speed time E NIv azy E Ñ Lm mustbe bigger than thediameterofthemolevies in idealgas Displacement Ar i meansquare displacement shortestdistance betweencollisions Ar displacement 4r 4712 t 122 sum of all changes in direction ay are GDE D diffusionconstant m 5 Fart JOE D A GeneralTrends D X Vave Proportionality constant 3 1 16 in liquid solids increase indensity decreasediffusionconstantdue to tight D 3 I rave hi molecules to movearound how Δ 3 fqzy.fm downmobility overalldiffusion gases also decreased w increaseddensity D molecules collidemoreoften travel 1m az shorterdistances between solutions gg g g  CompressibilityFactor Idealgas law breaksdown at highpressureand low volume and temperature 2 PVMRT tompressibility factor PV ART ideal gases have a ratio of l a increases 2 decreases below 1 attraction makes compression easierduetotransient austering b increases 2 increases above 1 volumes ofparticleslargerthenharder tocompress duetorepulsion At hightemperature boyle'slaw breaksdown PV not constant At low temperatures charle'slaw breaks down V notproportional to T Hence adjust PV ART for real gases a f v no net finite volume ofparticle tractiveforce ya toqueering fewer ofparticle man rateof collisionswith wall pairwiseattraction 2 it b a Rt a r t w second virial coefficient BCT attractive repulsive forces between 2 molecules pairwise on gas T beingpositive means repulsiveforcesdominate negative means attractive forces dominate CIT tripletofmolecules DCT 4molecules 3rdvirialcoefficient 4thvirialcoefficient  b a RT n v drops to 0 solve for T which is BCTI where the realgas iiiiii ii.i.mn distance unitoflength Ucs R 42 f EG attractiveforce depthofb ondingtrough J repulsiveforce potentiale nergy when R is largetherewillbenoattraction when R decreases therewill be an attraction when R is too small there will berepulsion small shallowtrough meanslow attraction between molecules Tofindforce exertedupon oneanotherfrom this equation we derive it iiiiii.it E c.mn f Éif Atlowtemp attractiveforcesstrengthe pushingdeviation below1 Athightemp repulsiveforcesoccur Boyle Temperature region where slope pushing deviationabove ofrealgases is similar to slope of idealgas properties similar  Solids Liquids PhaseTransitions Solids liquids havesimilardensities as solid changevolume 2 10 whenmeltinginto a liquid Hence intermoleculardistancesare about thesame Here intermolecular attractions repulsions are balanced Compressibility Appliedpressure fractionally decreases volume ofsolidsandliquidsie they are fairlyincompressible littlesparespace betweenatoms molecules Gaseshoweverare compressible Thermalexpansion fractional increase in volume per degreeof temperature increase Isobaric Coefficient of thermal expansion f TF willbesimilar for all gasesas 0366 Gaseswillexpandthesame amountas one another if theystartand end atsame temperature In liquids soudsthere is much smaller change per temperaturedegreeincrease Fluidity Rigidity liquids gases experience fluidity solids are rigid maintaining shape understress Shear Viscosity Resistance of a material to macroscopic flow on a micronopiclevel its resistance when I lay is draggedacrossanother layer Hardness resistance to indentation Elasticity capacity to recovershape diffusion migration ofatoms in amedium mixing atomiclevel connectionwhere whole regions movetogether  surface tension boundaries between phases Notension inside a liquid particles are surrounded by likeparticles forces are similar around homogenous distribution Surface Tension on thesurface forces are not equally distributed as no layerofparticles on the top spatial distribution not homogenous tension Spherical shape maximizesthe water mercury head up due to highsurface energy used as a minimum SA thus maximum stability Intermolecular forces R distance between particles potential energy will decrease as R increases Ion Ion Nat CI Potential energy ie PE decreases slowly as distanceincreases Deepestwell depth dipole dipole polarpartially charged molecules PE ie PE decreases fasterwith increased distance Ion dipole polar molecule and ions interact PE Shell formation of Oppositepoles on induceddipole Presence of charged ion distorts electronchargedistribution ofatom causinginduceddipole PE α weak Induced dipole induceddipole London dispersionforces Electrons fluctuate in an atom driving correlated function in otheratoms Net attractive Larger heavier atoms more polarizable ie stronger LDF PE fastest decrease in PEwithincreased R manifffit Stronger theattractiveforces higher the boiling point more energy to overcome bonds to causephasetransition with LDF mr will increase BP Hydrogen bonding dramatically inineases BP waterhigh BP 2 sets of hydrogen bondsforming dimers Hydrogen Bonding H2O HF NH Needed H bonded to O N or F O NF have high enough electronegativities to deshield hydrogen which can then be A bonded to lone pairs Directional between Hydrogen and lone pair Hydrogen Bonding in Water Hydrogenbonding causes high BP without Mbonds waterwould boilat150k Oxygen bonded chemically to 2 hydrogens and hydrogen to 2 more formingtetrated bonded Thisforms a crystaninestructurewhen frozen icelessdensethanwater L  repeated hexagonal structure with space between opennetwork freezing liquid density increases when looted whenfrozendensitydrops tofreeze Asice looted density will increaseslightly PhaseEquilibria Dynamic Equilibrium rate of forwardreaction rate of backwardreaction evap condensation hate properties at equilibrium independent of howthesystemgotthere Vapor pressure dependent on temperature highertemperature would increase theamount of molecules in thevaporphase Phasesin 1120 Supercooling allowingwatertocool below 0 withoutfreezing duetonucleation baby icecube grows homogenously within a liquid or the side of the bottle Metastable ie not happy Superweating waterpersists as liquid above BP Metastable iiii maistisuimasi o I.EE It's.IE atm latm 0.006 below normal condition ftp.o TB negativecope increasepressuredecreases me f mostthingssolidifyunderpressure butwaterliquifiesdueto openstructure oficecrystals  CO2 Triplepointabove atm Therefore innormal atmosphericcondition coz sublimestogas Carbon Turns fromgraphite to diamond at highpressure and persists whenreducedbackto latm Threshold for reversal is high Tih Whitetinturns to gray tin powdery at temperaturebelow 13.2 C and latmpressure Stainedglass inimtan sowumticssnowoverannii.gg System part of theuniverse we are considering Closed system boundarythatprevents theflow of matter between system surroundings Open System no boundary to flow of matter between system and surroundings surrounding region that can exchange matter energy w the system Extensive Property sum of properties ofsub systems Energy Volume Mass Depend on mailpresent Intensive Property sub systems T and P Density BP Dont depend on matterpresent Thermodynamic state macroscopic condition of asystemwhereproperties are determinedby surroundings constraints Equilibrium after a system relaxes tomeet its constraint  startingpoints ie can meet this condition fromdiffering is independent ofpath taken togetthere Processes Thermodynamic process process which leads to a change in thermodynamic state of a system physical or chemical Reversibleprocess path on surface of equilibriumStates there are infinite of reversible paths which connect thermodynamic States A and B Irreversible Intermediatestages not described bypoints on equation of state Statefunction statefunction is independent of path Change in a Startandendimportant d volume work not statefunction Pressure q not state function Temperature InternalEnergy 1stLaw of Tunnodynamics au Ftw workdone Changeinternal energy work physical myth Δ Pr pressurevolumework mass x gravity 4h 4m kg mtg Pressure VolumeWork we PextAV workdone on thesystem Putting I'm cam Jones  Unification ofMeat Work I calorie amountof heatneeded toraise the temperature of 1gram of H2Ofrom 145to its Heat is proportional toworkdon mm Asobjectfalls thepaddlewheelspins doingwork on thesystem raising its fpm app gang temperature g I showed that temperature of a substance could be raised byadding heat to it insulation Thermometer work penticmeatcap Kal or 4185 m q anyunit as change 4201g q and w not statefunctions buttheycombine together to form one AU q tw qsys qsurr ways Wsurr Ausys Dusurr douniv Ausys AUsurr 0 CONSERVATIONOFENERGY Energy lostfromsystem gained bysurroundings Total change in universe is 0 energy conserved  Heat19 Internal Energy U Totalenergy of a system includingpotentialenergy betweenmolecule kineticenergy of molecules and chemical energystoredin chemical bonds Heat Amount of energy transferred between 2objects initially at different temperatures Ice Calorimeter measured FEE specific heat capacity Amount of heat energy required to raisethe Tof 1grammass by 1degreek transferred fromone body toanother q q lost gained Cp specificheatc apacity under constant P cu V c molarheatcapacity up n or culn qp ncpdT qr ncu4t qpositive ifheatadded tosystem q negative if heat transferred awayfromthesystem It one thingloresheat anotherhas togain it Use this to find final temperatures M Cs 4T Me Cs etc Heatcapacity at constant volume BOMB CALORIMETER Container tightlysealed volume cannot change with no volumechange no work isdone w PextAV thermometer if f inflation burningsample steel bomb at constant volume will be quency 4T q AU 9th hence do gu nerd T no workhere Enthalpyastatefuntiont Heat transfer at constantpressure au Epi Ip Pextav internalenergy Itpressureconstantthen 40 qp PAV Ep Aut PAU Δ PV Pav as pressureconstant Enthalpy AH du APV An ncpdt Ep  Kinetics Ekin 32 ART Monoatomic gases AV 322ndAt Constant volume do qu neuat 3 2ARAT noudt 3 28 Cv for monoatomicgases or If calculating do in constant pressure use noodtstill as doonlyimpactedbytemperature find CVfrom cp AU at constant pressure do qp tw acudt ncpat p V2 Vi ᵗ 99t_m cu p Cp SLR gu nav at Ep nap at AU neutt both Pand Vconstantuse this Constantpressure q 40 as work isdone AU isless than q as qinputused todowork onsurrounding toexpandgas Constant volume noworkdone so I U q  Heat Work foridealGas A C AC B isobaric isochovic P ADB vochone isobaric m mntwandqaaa msn.fi no work PV hence do q acudt non constantvolume 10 9 Wu PAV aiqfdb Ea.upneaxcap.ci Monoatomic Translational RT 2 R L É É s qyyyyas.pe g  I calculate cv based of thesedegrees of freedom add tofind cp Cpdeviates at lower temperatures with increase in rotational Dofdue toquantumeffects vibrational impact measured value of CP vibrational contribution measured value op valuewithoutvibrationaladdition 100 percentage of op areto vibration Heat capacity solids of heavy oskilators highm populate vibration movefully at low temperatures causingsharprisein MIDTERM 2 Thermochemistry Enthalpy heattransferred atconstantpressure In du PV an qp ncpdt man Hess's Law Two or more chemical processesadded togetherto giveanother wemustadd enthalpyasstate function Enthalpybased on molarquantity  40 ingases AK du 4 PV du 4H d DV PV NRT APV dart RTAng AU AK RTang calculation of internal energy at constant temp and pressure IDEAL Phase Change Molar Enthalpy of fusion melting endothermic to breakbonds AHfusion Alefreeze Molar enthalpyofvaporization probsuse du AK dart endothermic to break bonds asgasmoves 4Hrap Akcondensation Standard Enthalpy of formation Ahf enthalpyofformation permol formation fromelementsinstandardstates Use formation to findoverallenthalpy change Hfproduct EMFreactants molarquantities Bond Enthalpy Average bonds overmany moleculeslessaccurate AH Booken made  Pext p Reversible no IdealGas Isobaric P constant Sochovic Vconstant Isothermal T constant Adiabatic qnottransferred Isothermal Constanttemperature System in largerelevoir at a fixed temperature Heat transferred between system relevan to ensureconstant T Temperature of reservoir unchanged dueto itssize do her isothermal In nepat ffm.tw just q as volumeconstant hence Etw q w tomake duo of reverse as we art w net in quo averave q net In V2 du willnotbe 0 If a phasechangeoccurs Isothermal expansion ifthegasexpands it implies work done is negative and q ispositive ie work done againstsurroundings heat transferred intosystem to keep Tconstant Isothermal compression surroundings dowork on thesystem ie w 70 thesystem mustthenloseheatto the reservoirto maintain a constant T  Summary for Isothermal qtW AU O vie w net af and change sign for q an 0410 0 Adiabatic 9 0 no transfer of heat in or out of the system place adiabaticwarearound system to prevent heat flow Hena in order for a gas to expand adiabatically work theenergymustlonne from internal energy of thegas decreasing its temperature overtime o AV w nerdt q TV Tzu y cp or RV PzUzY summary A hence du w novdt Volume Pressure in Isothermal and Adiabatic Isothermal involves heat q exchange with surroundings to maintain a constant T Adiabat processes do not exchange q w surroundings hence temperature ofsystem will dropwhen work isdoneingasexpansion Adiabatic PV constant ExternalPressure isothermal PVconstant 1am  Comparing Isothermal to Adiabatic expansion P t V initialsame final volume of isothermal is 2x larger than adiabati workoutput foradiabatic is 40 that of isothermal Adiabaticline four moverapidly w increasing volume than isothermal Thisis because no heat is transferred in adiabatic expansion so work must comefrom internal energy of thegas hence temperature ofgas declines Spontaneity Natureof spontaneouschange Achangethatoccurs withoutoutside intervention ithasdirection spontaneityis connected to entropy which increases in the direction of spontaneity Increased entropy stability more waysenergycan bedistributed spontaneity connected to Gibbs Free Energy It do applicable under constant P and T Probability molecules Probofarrangement Microstates R Microscopic mechanicalstates A number ofpossible microstates possible tombination of position and momenta available to Nmoleculesthatcomprise a given system with energy U and volume V I α f U VN  Entropy 5 KB In a possiblearrangments Rina 51k entropywill increase Increase in microstates Entropyreliant on Volume of microstateswilldouble when volume doubles Numberofparticles N Doubling N doubles S particles S KD n r Stebin arrange if we double 5 Kb In arrange 5 2 kbIn arrange www.iif Entropy Meat SteamEngines 1 Hotsteaminjected at highpressure into a piston cylinder 2 Gas Immediately expands pushingpiston outwards againstexternal load doingwork 3 Reciprocatingmechanismsreturn arrangement to original positions Irreversible thesteamcools when expanded and is exhausted from cylinder at endof stroke  URCE engine offline seek to minimizeenergyloss toenvironment ie want work done to be largerproportionof heat input heatflowout consÑ why enginegainsenergy in formof heat from high temperaturereseroi engine uses part of thatenergy to perform work on thesurroundings expand remainder of the thermal energy isdischarged lostbyengine tolowertemperature reseroir Idealization of the Engine CarnotCycle setsfundamental limit onefficiency reversiblemodel tooslow tobe useful Generalization of Clausius nodevicethatcan transferheat fromloldertowarmerreseroir without netexpenditureofwork I heat always flowsspontaneously fromhotto cold 2 Workalwaysrequired to refrigerate a body Generalization of LordKelvin nodevicethatcantransform heatwithdrawfrom a reservoir completelyinto workw no othereffect fitdgy As Spinal Sinitial Entropy is a statefunction In a reversibleprocess theentropy of theuniverse is constant In an irreversibleprocess the entropy of theunivers CarnotAnalysis 2Isotherms 2Adiabatic eachstepis reversible  III v complete cycle v expansion compression AB isothermal expansion CD Isothermalcompression BC Adiabatic expansion pA Adiabatic compression Isotherm urging q w Adiabatic w du ncudt o q Wnet all 4works or whet nR Th T tuff deriving wnet ARTIn art 14 tf foradiabats 1 BC DA 11 11 and Adiabati Isotherms  HeatEngine Efficiency a ear wa a to in e This II Toget100 efficiency ofthe engine Thwould be infinity difference greatestpossibletemperature I wouldbe 0 NB e.EE iasa qCD a o isothermal compression isthe energydischarged StandardstateEntropies S Kbln 2 Entropy at equilibriumstate is 0 AS n f at entropychange when raisingfromlow to highertemperature St Sf AT entropychangefrom 0 T notachangejustentropy readet Sofproducts moves ESofreactants moves  asystem Asurroundings ΔStot 0 spontaneous Astot 0 reversible Astot 40 nonspontaneous Derivation of Astot 49 III assurroundings PHASE CHANGE g astot Assyst Assure 11 Assys 4nys I.gs astot dfmyiITSsys G M TS Constant T do an This Astot 4K Atconstant T or P an q Attsy q Insurr q IS ff fifaquer grey Entropy at constant. T agrees KImoI ff  GibbsFreeEnergy 1H Tds GibbsfreeEnergy is 0 when 4h Tds solvet At the normal freezingpointofwater In Tds 40 0 do willbebelow 0 ie spontaneous when Tds da da will beabove 0 ie not spontaneous when an Tds Chemical Equilibrium As a chemical reactionproceeds therateofchange decreasesuntilthere is essentiallynochange dynamicequilibrium Rateconstant 44,5 1 k Equilibriumstates displays nomacroscopicevidence ofchange reached viaspontaneous process dynamic balance offorward backward process same state reached regardless of approach Steadystates productsformedby process removed byanother 1 2 3 in jiteady Equilibrium constant ei  K Largerthan 1 equilibriumshifted forwards making moreproducts k smaller than I equilibriumdrifted backwards maning more reactants e 1 perfect equilibrium Q larger than Kc Qhasmoreproductshence need toshift backwardstoreactants toreach Kc Q concentration anytime KKN Temperaturedependent only ism.HN Q 7k 40 70 sportchange R bothsportchangestofix L Qand directionofChange an iii PureReactants PuveProducts neChatelier A system at equilibrium thatissubject to stress will react in a way thattendsto countera the stress ΔVolume look at ofgasmoles changed reduced volume increasedpressure favoursidewill lessgasmoles A temp increasetemperaturefavors endothermic process VantMoff Equation derivation Temperature dependence of Equilibrium constants RTink do Ak Tds ink 2 1ft In ki 2 subtract InR2 If y in Weassume 4h constant filetodoover ATsmall butheatcapacitychangesoverATlarge so notgood Kdependence vs IT forexo t endo exothermic Increase temperature favours III them rose endothermic inseaet  4 some enthalpy slope intercept 4 some entropy VanHoffAplication onVapor Pressure Temperature dependence Pressure of Vapor 1120 1 1120 g of T then Assume Invap and I swap are independent in in my Vaporpressure Jones Inp annual ethuilibrium 831 appercure atm Pwilldiffer from literaturevalue 4Huapdoeschangew temperature an effect neglected in calculation  Entopy AS fit dyev all reversibleprocesses Is 9 As at constant T dstrans as 112 neverablephase transitions 9T constant P and T Is ARIn Idealgas Change of V at constant T AS n culn T2 T Constant V no phase changes AS n cpIn T2 T constant P no phase changes Assuro surroundinglargeheat bath constant d dstot AssystAssure largerthan 0 its a spontaneous If process so IS f  do 46 0 at equilibrium Ideal gases 4G ARTIn Pressure of idealgaschanged at constant temp at Δ n TS OH TAS Tds as nrinky new arm on nein t do ARTInP2p 4G ART InIep given or latm NRting AGand K AG RTIn R do RTIn 4 liquids M 10 16 t RTInQ RTlnktRTINQ.AT nIQIK  Referencecycles INO g Naog NO2191 3 NO Pro Neo g Puzo t Noz PNE I da tout do for da do paying nondrying INO Pref Nao Prof Noz Pref 40T ChargereferenceP AG ARTIN P2 p 3 2 7 In Pref no 402 6 Changeatreferences do netindys Rtln 1 Ran Ran I 81 AGoverall AG tdoztdb aa aotrtinfff.IE Noteof k forward reaction has k backwards reactionhas Rz K Kz I If the coefficients I 8 1 20 4 02 are change eg 9  Thenraise thechangeie R ki Addreactions multiply K Minusreactions divide Kout hewill belarger if 180 ispositive and large and dh negative and large Increase of 1 and decrease In favorlarge k as 160 RTINK ink 4 da an Tas 2 as K exp Conaptopactivity Non Ideal Mixtures Solutions as well as multiple phases do ARTIn Ihnen Pretaatm activity effective concentration orpressure Δ G ARTIn a activitycoefficient yi of non idealspecies as ai altivity coefficientyi forsolute insolution of concentration i Cret 1M ai 9k  K loe treat mixedphases thisway CO NM 2 g t 2Wh g t 1120 I CO2 aq E m K ff.es n as solids useconcentration gases we ret I latm wfffÉ BOILING POINT ELEVATION are s 111 ATD Kb m ng ITS µm BPwill increasewithadditionof NB molality particles 1g reduce the mole fraction of the solvent which is meant to be evaporating, reducing the vapor pressure.  l If Kfm vaporpressure guy mm mm mm ay ff otygy.fm identified where upsolidsolventintersects Temperature upsolvent in solution determiningions fromdissolution solvefor molality m 4ft find molarity it nodissociation m g Iwu divide the two tosee howmanyions needed OsmoticPressure I minimum pressure whichneeds to beapplied to a solution topreventtheinward flowof itspure lomentacross a semipermeable membrane soution density P 94 p.mn gpsntin.ki.na.im I CRT TV ART Y.fi iiiiater seminars I waterina sugarcannot pass 0908206 thoughnonsuan sugar Reverse Osmosis Applypressure tosetwater via desalinization probably solve for number of moles of salt or somehting - concentration is ions i think, not particles. Equate nct and pgh  RaoultsLaw dilutesolutions lowsolute moletractionofsolentclosetoone soluteparticlescannot add to vapor pressure non volatile Prap solventPsolvent positivedeviation weaksalient solute interaction Negative deviation stronger solvent solute interaction henry's law proportionality constant partialvapor P morefraction pressure P2 K2 1 Xi Vaporpressureof volatiledissolved substance isproportional tomolefractionofthe substance in solution Liquid liquid Gassolute Liquidsolvent NB When Raoultslaw is valid for solvent reenry law validforsolute can so owning no NONIDEAL manging en yy  Y Liquify Vaporizeremove f mm mm ummm cnn.ua.mg 61svolatile in liquid Find more traction frompartial pressures p between theBootcomponent temp After Original 01312 016 2 Azeotropes Non idealsolutions Cannotreach 100 purity as both components behave as I TbC Tboc 100 100 1 Iron y y If no 0.096 Helson HCl Heo me Cuzcoon Maximum boiling Minimum boilingAzeotrope azeotrope Solute solventforces solute solute or solvent solvent attractive and stronger than the weaker strongerthan solute solute solutesolvent or solvent solute Colloidal Suspensions Milk Fatdroplets in water 10 9 106m sizetypical colloidal particles are charged balanced byionsin solution Flocculation Addsaltstodiminish electrostatic repulsion ofsuspended colloidal particles leading to aggregation sedimentation Particles always moving dueto Brownian motion  desalinationtwaravash Desalination Intakes extract sourcewater t convey Pretreatment removesolids t biologicals Desalination Posttreatment Concentrate Managment handling t disposal reuseofresiduals DesalinationProcess Membrane desalination requires intrinsic minimum available energy caved exergy needed ReverseOsmosis Thermal Heat saline solution I generatewatervapor collect Ion exchange vapor oncoldsurfaces Electrodialysis 112AtmP waterboils650C ie lowerp ressure boilslower  Energyneededfordesalination Bestcase reversibleidealizedprovers dosaltremoval dGaddingsalttosolution At equilibrium Hydrolic Pressure Needed OsmoticPressure dw TosdV differential amountofthemwater dittee t bosmotic pressure W Tosdu workincreases withocean temperature increase Energy Requirements vs Extent of 1120recovery Water recovery r Vi V2 Vi w Wo in w in I minimumwork forzerorecovery Zerorecovery whereare of freshwaterismade from brackish wesee that itgets harder themorewater we wishto use as itssalinityincreases w water extraction D CRT TV ART mgh disbu concentration if Nad Nat a  Solving for then find semipermeablearea w f distance thetenth Pressure arm is kgcm km plosiveosmotic 278kg m Li 27805 Ii 1kcal 42005hence 0.66kcal ti minimumtomake 1 Iiofsalt 244 freewater Reverse Osmosis Implementation Seawater Brackishwater lowerwork membrane Materials Microporous polysulfone on thinpolyamide layer layer supported Issues WaterFlux permeability membrane lifetime Saltrejection themembrane foulingof Disposing concentrate Energyusage Atmospheric waterHarvesting Atmospherecontainssomuch water Waterdesal requires aicestoSalimaubrackish H2O Atmospheric can bedone in inland areas alless waterexistsin atmosphere in 2 forms droplets vapour  fogcollectingcatchesliquid Mesh suspended perpendicularto wind to catch droplets in fog poresite optimized to make arons tall notclog vertical wines or hyamphob.iesurfaces wind geographical limitations Requirespersistant fog and Acid Base Equilibria Bronstead CowryAcids Bases A Donate Mt Bronstead Lowry B Accepts Mt Cuzcoon t no Ch Coo t Mzot A B Conjugate Base ConjugateAcid Amphoteric Act as a base and acid NCOs KCO t H2O H2O OH BASE NCOs 1120 032 On ACID CrwitAcid Base M if I I IN H I 1H If Filament c e  initiate Diprotic Acids Ea's A a ma anita Kazsmaller httightlyboundt negativeion Strongthen weak ie 42504 if strong 112804 HSOy t Mt 2H t 5042 Autoionization of water 420 I MIO I I 430 t on ACID BASE ACID BASE 21120 not On NOT Mt On t fi E n Kw not on Kw Kt on Kw at250C 1 0 1014 pre 7.0 StrongAaas Bases Strong fulldissociation 44 Mr HI Mason AND or Nhi M Naoh mom na Lon 1 10 m Info o on Naoh 1H30 1 10 M fwy ph log 430T WeakAcid Baser ka ionizationconstant Notpulldissociation ionization ra a.IE 4 KIs COnY4h Kw KbKa PhatPKD PKW Indicators MO 1 Motag 499 faq ka 4 44 Indicators changes at differentph  Buffer solution Ht Ka weakacid conjugatebase bothpresent Eff Want tohave ratioconstant in large concentrations of each Hf it Then smallamount baseis added of MA changesslightly and A thussmallchange Moon 420 Not MCOO Initial M 1.00 0 0.500 de y ty ty Equi m t osty y y ka 177 104 Y dissociation is small so we write y 4 177 10 yC 3 54 10 4 M not y ph 3.45 Designing BufferSolutions pH pka logoftp Plugindefined ph and pka of weak acid solve Acid ConjugateBase Ratio for  Titration Curves Strong Acid Strong Base It how Whenadd base neutralise equal more acid fifth overall I volume Initial pHjustacid HA Solvent Mt A Solvent PKa Pled log ka K'a 160 1601 RT Inka Ink'a 2.3RT pka pka 46 16 440 4H01 T 15 1501 Changingstructure oforganicacid hassmallimpact on entropy ofionization butlarger influence on enthalpy of ionization How large a change in enthalpy is needed for an acid byoneunit ionization to lowerpka at 25C Assume entropy ofionizationunchanged Entropy unchanged 15 45 so an Ah 160 doo a a si a i mn  Lowering pka means increasing Ka AcidRain Alreadyacidic naturally 1120 1 021g McCO lag 502 NOX from fossilfuel combustion Nagt02 ZNOg NOg t 1202cg NO21g killfish 3NOz 1420 MN t NOg Leach Al into to 03g g deforestation sowedin rain Damagetobuildings and 502 9 503cg Miso 42504 14505 Ht Hson 5042 Mt Damage to buildings Caco g t Mason qq Castagt 50m 99 t H2o CO2 Abatement Clemistry Metal Smelters SulfideOnes roasted toseparate 2Nis s t 3021g 2N O s t 25021g 4652 s 110219 2620 s 85021g b Recover 502as a liquid 502 forresale Mason production rubbing 50219 t 1202 5031g Fo 2k250m 503g 112504 503 liq cig 0205catalyst Dissolve in concentrated 42504 PowerPlantFireGasDesultorination Ca Oh folia 5021g Caso 4201 2Naong 5021g Na 503 t Mao I Caso t Casoy 2h20m 120,2g 2h29 Naz503 s t 12021g Nazson s  WetScrubbing Caco 5021g Caso t 021g CaCon p t 5021g Caso s t Mao i stray haratodisposeof so t 2k20g t 12021g Cason 2470151 q Policy OECD project Andraintransboundaryproblem OECDReport1977 2014 PM2.5 Annex to USCA agreement 502 4T Foodsupplyimpacts Oceans Climate Change Cozig t H2O l 4260,4g CARBONIC ACID der of 42603 datCo2 for 237.1kt 100 Cozlag 100 Hz 03Ga  hecomes usedtoreadylevels Increasing throughthe years Impacts Global TemperatureRise 1.14 C since 19thcentury Hottest yearonrecordoverand over warmingOcean 90 of energy top100m Warming 0.6 F Shrinting IceSheets Antartica 148 billiontons yearlost Uksnowtakingworldwide freshwatercreation slowrelease water Sea levelrise accelerating everyyear 8inch last century South FaitAsia OceanAcidification Acidity up 30 CarbonicAcid Permafrost than release Chy and CO2 Clathrates Trapmolecules in itscrystalline structure ClathratehydratesrepresentEarths Largest Hydrocarbon Source Cut andCO2 clathrate toumation Thermodynamics Myendothermic CO2exothermic  Mechanism Chy occupiescases first Chy clathrate decomposes Ozoccupies M cages anyneoccupies s cages CarbonCapture CO2 capture using sorbents post combustion CO2 adsorbs to amine containing polymers on silica substrate Capture at source Amine Appended MOF Adsorbents Metal Organic Framework verylargeSA MOF clinically heatedand cooled CO2stickstoMof at lowtemps CO2 desorbs from Mof athigh temp CO2 capturevia wetScrubbing 400 ppm CO2 in ambient air CO2 in air reacts with ROH toform Kzco K2603reactsto form Caco Call decomposes at high temperatures toform Cao and Co2 Selective membranes for CO2 capture 02 selectivelyabletopass through a membrane CO2 isfacilitated through membrane by amine lauriers Selectivity AGAINST Nzas these don'treact w amines Pressure 2 Uatm partial vacuum 0.2 0.3atm usedtopush 102throughfiller  balancing redox: Acid: Balance all atoms bar O,H Balance Oxygen by adding water to other side Then balance H as a result Use e- to balance charge after that Base: Split reaction into its two parts Balance everything bar O and H Balance O with water Balance H+ as a result Then add same amount of OH- as H+ to BOTH sides Combine OH and H where they exist on the same side into H2O Use e- to balance