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This 71 page Class Notes was uploaded by Dr. Drew Flatley on Friday October 30, 2015. The Class Notes belongs to CHEM 112 at University of Massachusetts taught by Jeanne Hardy in Fall. Since its upload, it has received 13 views. For similar materials see /class/232352/chem-112-university-of-massachusetts in Chemistry at University of Massachusetts.
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Date Created: 10/30/15
Turn on Camtasia Recording The exams were dropped off to be graded Wednesday morning The data are not yet back We are starting chapter 19 the cocktail party chapter What does your room look like mpowgt Perfectly orderly A few things out of place Pretty messy A complete disaster A tornado hit 2nd Law of Thermodynamics The universe tends to greater and greater disorder Molecules are like undergraduates me me to leave their options open A reaction is spontaneous if disorder entropy for the universe is positive Spontaneous change results in the dispersal of energy Why does EnTropy exisT There are so many ways to be messy and only one way to be tidy Place for everything and everythin Wh does Entro39 exist There are so many ways to be messy and onl one wa to be tid Place for everything and everything in it Statistics Poker cards Directionoli ry of Reactions Probability suggests that a spontaneous reaction will result in the dispersal of or of or Matter Dispersal Due last night in OWL lst Law of Thermodynamics What is lost by me is gained by you Energy is neither created or destroyed The law of conservation of energy Change in the internal energy of the system is the sum of the heat transferred to or from the system and the work done on or by the system What is the de nition of system nrl39IPOFDP a System Surroundings Exothermic Endothermic Enthalpy State Fu nction H w Gives off heat Consumes heat Heat of formation at constant pressure Part of the universe we are studying The rest of the universe 6 Function that is only determined by the initial and final states but is independent of path What is the de nition of Exothermic nrl39IPOFDP a System Surroundings Exothermic Endothermic Enthalpy State Fu nction H w Gives off heat Consumes heat Heat of formation at constant pressure Part of the universe we are studying The rest of the universe 6 Function that is only determined by the initial and final states but is independent of path What is the de nition of State Function A SYStem 1 Gives off heat B Surroundings 2 Consumes heat C Exothermic 3 Heat of formation at constant pressure D39 EndOthermlc 4 Part of the universe we are E Enthalpy studying F State Function 5 The rest of the universe 6 Function that is only determined by the initial and final states but is independent of path a a AG AH TAS Spontaneous Tells us about the drive to equilibrium Tells us nothing about the rate A chemical system at equilibrium will never spontaneously change in a way that results in the system no longer being in equilibrium 12 Directionolity of Reactions How probable is it that reactant molecules will react PROBABILITY suggests that a spontaneous reaction will result in the dispersal of energy or of matter or ofenergy amp matter 13 Directionoli ry of Reactions Probability suggests that a spontaneous reaction will result in the dispersal of energy or of matter or both Energy Dispersal 14 39 f Directionolity of Reactions Energy Dispersal Exothermic reactions involve a release of stored chemical potential energy to the surroundings The stored potential energy starts out in a few mOIecuIes out IS tInaIIy cuspersed over a great many molecules The final state with energy dispersed is more probable and makes a reaction spontaneous 15 v 6 2 2 a 132523 6 24 23 44 34 33 6543210 A 5230 gtwmmzm 6 quanta of energy Number of different ways to achieve this arrangement A 390 I39M 0 IE 9 I39M 0 I39M 9 II LL 9 km 0 lib 0 I39M 0 II L KDLnQ39mNH 0 IO vmvnn AMEN 19 E gt u E E vacuummun Gas expands into a new container with twice the volume Energy levels for a gas in a container Shading indicates the total energy available 2 a gt o w z w Energy levels for a gas in a new container with twice the volume More energy states are now available with the same total energy The states are closer together ozmswamcummm 21 En rropy S S JlKmol H20 li 6995 mamas S gases gt S liquids gt S solids 22 En rr opy and S ra res of Ma r rer S Br2 liq lt S Br2 gas S H20 sol lt S H20 liq Entropy S Entropy of a substance increases with temperature Molecular motions Molecular motions of of heptane C7H16 heptane at different temps Heptane C H15 g at 1501 K 5 3 921 ms moi 24 En rropy S Increase in molecular complexity generally leads to increase in 39 in J b J Methane CH4 AL J methane ethane propane 5 Jill mol 1863 2292 2703 25 Entropy S Entropies of ionic coulombic attractions S JIKmol 35quot H X39s 390 M90 269 Jyr M 39 39 NaF 515 M92 amp 0239 Na amp F39 26 En rropy S Entropy usually increases liquid or solid dissolves in a solvent 27 Turn on Camtasia Recording gm GredhL n4 OWL What kind of chemical reaction is found in batteries AOxidation BAcidBase xidationReduction DThermodynamic EKinetic as Balancing OxidationReduction Reactions 1 Assign oxidation numbers 2 Separate into oxidation and reduction half reactions ASK is Morrow basic 3 Balance each half reaction using the following steps 906 FD Balance all elements excep Balance oxygen by addin Balance hydrogen by addi Balance charge by adding e ectrons Electrons go on me muHT product side for UAILJAI ION reactions Electrons go on the LEFT reactant side for REDUCTION reactions In BASIC solution his additional step For every H ad 0 BOTH sides ofthe reaction Combine H OH into H20 Cancel out any waters that appear on both sides gen or hydrogen You should now have a balanced half reaction 3 4 Multiply balanced half reactions so an equal number of electrons are consumed aw produced 5 Add together half reactions 6 Clean up Combine identical substances and reduce coefficients to the lowest terms 7 CHECK Atom and charge must balance v0 zn gt eduction 0fV02 e vogidn isyeudw in acid midi with Zn Zn added with i39me the with time the ue v0 inn iinauy dim v ion is m ydudw veg ion is idddced is midi ieduced m gieen vednned tn vium v2 inn tn this v0 inn v inn Add zii Balancing Equations Balance the following in acid solution v02 Zr gt v02 Zn2 EStjquot E tZJquot Step 1 Write the halfreactions Ox 2quot a 2 Red V024 Vow Step 2 Balance each halfreaction for mass OX 2 WI Red 2quot V02 gtVo 0 Add sz w vdeficient side and add H on other side for Hbalance Balancing Equations Step 3 Balance halfrea l ig for charge OX Zn 3 ZWquot739 Red 744 VDr a ID H10 5quot 4 0 3amp6quot Step 43 an appropriate factor 3am W Ox 2m 4 2w 2 Wales Red 4H 2V02 azvo FZH o 4 2V0 gt 2 szon 42H Step752 J it d balancefnaITreact fens b Jim I 7 9 homemade VH 0 Ba Iancing Equations Never add 02 O atoms or 0239 to balance oxygen MA HzO Never add H2 or H atoms to balance hydrogen H Be sure to write the correct charges on all the ions Check your work at the end to make sure mass and charge are balanced PRACTICE In the reaction we just balanced which species is being reduced 2v02 Zn 4H gt 2vo2 2H20 Zn2 v02 W 7 V B Zn 0 v02 D Zn2 as 8 under onditions Write the balanced REDUCTION half reaction 4 47 b me N Cur s N at WWquot 0 0 HNo3 Cr3 a NO CrO4239 39 59 422 m 43 391 1 Identify the species being reduced HNO3 or Cr3 3o HNOS 5 NO Questio he following skeletal oxidationreduction reaction occurs Conclusion N is being raiqu Cris being 0144mm Rahohm Bal ce each half reaction using the following steps a Balance all elements excer t ox fen or h droien fBalance oxygen by adding H20 c Balance hydrogen by adding H A Balance charge by adding electrons Electrons go on the RIGHT product side for OXIDATION reactions Electrons go on the LEFT reactant side for REDUCTION reactionslt IC solution do this ad 39 39 step For every OTH sides of the reaction out any waters that appear ozn both sides 5 4 3 Hquot Ar HN03 36 quot NO ZHZO 10 Skeleton Reaction HNO3 Cr3 NO CrO4239 Oxidation 1 Reaction Cr3 9 042 g 7 4 OX CV 4 3 J4 Hzo H 4 amp 5 Combine Half Reactions 6quot g g HNO3 5quot 5939 9 NO ZHZO N 7 CHECK Atom and Charge must balance HNO3 Cr3 2H20 a NO CrO4239 5H Reactants Products H moles H moles N mole N mole O moles O moles Cr mole Cr moe Net Charge Net Charge The Masses amp Charges Balance 12 Electrochemistry Alessandro Volta 1745 1827 Italian scientist and inventor Luigi Galvani 17371798 Italian scientist and inventor 13 CHEMICAL CHANGE gt ELECTRIC CURRENT 2quot metal With time Cu plates out onto Zn metal strip and Zn strip disappears Electrons are transferred from Zn to Cu but there is no useful electric current 7 25quot Oxidation 505 3 2 4 Reduction Cu ch 7 146 can Zn Lv 39 CJ JS 14 CHEMICAL CHANGE gt ELECTRIC CURRENT To obtain a useful current we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire This is accomplished in a GALVANIC or VOLTAIC cell A group of such cells is called a battery 15 Electrochemical CG Zn gt Zn2 2e wire Cu2 2e gt Cu O Lidazh39ov39 Oxidation Reduction Anode Zn salt Cathode Negative Positive Zn ions Electrons travel through thale WW ISalt bridge allows LVMDMS 96mm to move between electrode compartments LU Terms Used for Voltaic Cells 2v 00 gtzwn at k Direction of etectmn flow 9 f Electmde Electmde 0 3 3 rd 7 Salt bridge 0 Cad Electrolyte ions in SDLU OM 7quotquot lt Am39nns Uxidized r Reduced 391 39 species Eilth species r u 7 V 777quot r ANOD E compartment EMHDDE compartment DEIDATIDN occurs EDUETIDN mccurs 39 Q n 17 The Cu Cu2 and Ag I Ag Cell i H D111 anode Eu bndge comm MEMOS fla N Lquot Net reat on Cu s 2 Ag aq gt Cu iaq 2 Agtsl 4 A 3 ltugt 18 What OCCUFS at the anode A Oxidation B Reduction C Electrons are produced D Electrons plate onto the anode E Electrons reaction e we V reduce the metal that is the anode Eli Turn on Camtasia Recording What is the pH of an aqueous solution of 0532 M dimethylamine a weak basbg with the Kformula 39CH32NH39ud 395qu aCuaNH 4amp0 3 LOH H3NH P x 0552 M o O C 7 vlt ht 0532 x KF 4r I L Kb 2 57 539439 Iquot 045139 N H 05517 MWhat is themqueous solution of 0532 M dimethylamine a weak base with the formula CH3ZNH hm M Q Mm a Lu1uu AM LOH gqumu kQ l 0551 a 0 7 E o 5324 t x k Cwlcsmw b EWLNH 5quot 0 X Txloquot39 W mm 9 log W a Uquotquot J quotquot5 14 f Pu mm 0532 51mquot 4 0557 J ou Iq lns lzlS L Vhahs memqueaus somuon 010532 M dimel1ylaminea weak v 4551 uj 077M quot75 PH 7 I123 4an 1H 175 quotH Kw LOHO H Cotk3 CHM may 4011 556940 1 J L111 H ogt ao39 DJ CREEK 5 IZZS P 39 What is the pH of a QQQJJVI formic acid solution What is the concentration of formic acid at equilibrium R imp4 maul we mco 4430 90M quot o o c E oooI 3 X 4 93131 j g igxmo Lulaquot ER 0x f 41110quot M f 5 lB I 1 Tl I aim digto V W39Wh 5W nu Whatis the pH Ufa 009 VI formic acid solution What is the concentration of formic acid at equilibrium R 13 up lira H We 1 Hen vHao l D 0 rcs w N w lm wimvu mm W MK 5 M f E LODV V vi X a W31quot i AmmoW n w quot at Juww zi 411104 39 gnl u l slap apart 1 lmyquot y 32m 6 a 32m 6W3 ll quotquot m izrw 9 X 1 Mao N 239 LVN 1 L to Midf Him m at will um rawam WW wmm um cman Wt NW mu um i 7 7 A 4 gnw tlapw t 7 L on l writ 1311 X quot 32 x 39w a gnu Iquot 391 ej g i l6331YNquot J 39FH 000 1 auto 4 winquotM The hydronium ion concentration ofan aqueous solution of 0313 M ammonia is A 727x103913 B 421x103912 C 856x 103910 D 501x 10399 E 473X 10395 The hydronium ion concentration of an aqueous solution of 0313 M ammonia is Polyprotic Acids 3 Mum steps 19 mspmc mi ia 7 H3704 quot 10 7 30 HLFD 15110 3 HzPoq H50 139 H30 4 Hm i quot39 39 3 Lily0398 1Lwr393 39 P EM succew wc Hquot mm sbp 66v quotF A 5 w l0 Fwnes umquot dif culi 7 gym Run39iu fwn 39 item LL lb W M W 92 W W subsewir s bps YVWTI w v 39 w tow 17me RUM aw beams Hg lb t gubwwt39 steps Brons tmok39 W Q39 Add H GUM Em W MW Lewis Acids amp Bases Lewis acid a substance that Ed an electron pair EFL the boron stem 1 surrounded by my three electrun pairs Lewis base a substance that J gtU OM41 an electron pair NH3 me N alum has three band pairs and nne lone pair at etectruns we pair acceptor Mowes w inn iMWMpHc acid Budd emu Bahia charged Mm 032 39 Few Makeuch wiiM fol iosz J an in rmquot 5 rrrrr ma hy mm mm mm mm Lewis bases 6 pair donor Wcs w pupil39s uFo J KWI39MS or OH l 04 2323333335731115 13 Lewis Acids amp Bases Formation of hydronium ion is also an excellent example a a H 4 039 H quota H 39 I H l5 Eectron pair of the new OH bond originates on the Lewis base H Lewis AcidBose Reoc rion H l Hquot 4 NH3 L335 n A L H CC WY N H l H H Lawisdg b equot on Y Which of the following compounds are lewis acids A BF3 B CH4 C CI D Fe3 E H2O 5a Which of the following compounds are lewis bases W 6quot A BF3 Add imampld o oc l wl39 B CH4 WW c Cl39 Em g D FefaW a R AM A mE H2O im 15 H A B C D E Which of the following compounds can act as a lewis acid or base BF3 CH 4 Cl Fe3 H2O Turn on Camtasia Recording What is the pH of an aqueous solution of 0532 M dimethylamine a weak base with the formula CH32NH The hydronium ion concentration of an aqueous solution of 0313 M ammonia is A 727x103913 B 421x103912 C 856x103910 D 501X10399 E 473X10395 The hydronium ion concentration of an aqueous solution of 0313 M ammonia is Polyprotic Acids Lewis Acids amp Basas Lewis acid a substance that an electron pair Lewis base a substance that pair an electron s w M13 me u Mum m mm ma um and an inn m nuiecirnns Lewis acids e39 pair acceptor an in m by an r Lewis bases e39 pair donor 7 q 4 gt1 M13 in u and an Lewis Acids amp Bases Formation of hydronium ion is also an excellent example Eectron pair of the new OH bond originates on the Lewis base Lewis AcidBose Reoc rion 43 Which of the following compounds are lewis acids A BF3 B CH4 C D E CI39 Fe3 H2O Which of the following compounds are lewis bases A BF3 CH4 CI Fe3 H2O mpow Which of the following compounds can act as a lewis acid or base A BF3 B CH4 C Cl D Fe3 512E H2O Turn on Camtasia Recording We have completed chapter 17 and are moving on to chapter 18 If we add the same amount of HCI to l L OT water or to l L OT blood which will have the lower pH A Water B Blood C They will both be the same 51 The Common Ion Effect QUESTION What is the effect on the pH of adding NH4C to 025 M NH3aq NH3aQ H209 NH4aQ OH39aQ The pH will go up 1 down 2 no change 3 QUESTION What is the effect on the pH of adding 01M NH4C to 025 M NH3aq NH3aQH20ltgt NH4aQ OH39aQ Buffer Solutions HCI is added to pure water HCI is added to a solution of a weak acid H2PO439 and its conjugate base H PO 2 Buffer Solutions A buffer solution is a special case of the common ion effect The function of a buffer is to resist changes in the pH of a solution Buffer Composition Weak Acid Con39 Base HOAc H2P0439 NH Buffer Solutions Consider HOAcOAc to see how buffers work ACID REACTS WITH ADDED OH OAC39 H20lt gtHOAC OH39 Kb56X103910 Therefore the reverse reaction of the WEAK ACID with added OH has 1Kb 18 x109 Kreverse is VERY LARGE so HOAc completely consumes l Buffer Solutions Consider HOAcOAc to see how buffers work CONJ BASE ADDED H HOAc H20 lt gt OAc H3O Ka 18 x105 Therefore the reverse reaction of the WEAK BASE with added H has Kreverse 39l l a 00 x 39iu4 Kreverse is VERY LARGE so OAC39 completely consumes l Problem What is the pH of a buffer that has HOAC 0700 M and OAC39 0600 M HOAC H20lt gt OAc H30 Ka 18x 105 Problem What is the pH of a buffer that has HOAC 0700 M and OAC39 0600 M HOAC H20lt gtOAC39 H30 Ka18x10395 Buffer Solutions Notice that the expression for calculating the H conc ofthe buffer is Notice that the H or 0H concs depend on 1 K and 2 the ratio of acid and use concs HendersonHasselbalch Equation Take the negative log of both sides of this equation The pH is determined largely by the pKa ofthe acid and then adjusted by the ratio of acid and conjugate base 12 Adding an Acid to a Buffer Problem What is the pH when 100 mL of 100 M HCI is added to a 100 L of pure water before HCI pH 700 b 100 L of buffer that has HOAc 0700 M and 0A0 0600 M pH 468 Preparing a Buffer You want to buffer a solution at pH 430 This means H3O 10PH Choose an acid such that H3O is about equal to or pH Get the exact H3O by adjusting the ratio of acid to conru ate base Acid H 0 K g I 3 1 Con base a Preparing a Buffer You want to buffer a solution at pH 430 or H301 50 x10395 M POSSIBLE ACIDS Ka HSO439I 5042 12 x 102 HOAo OAc 18 x105 HCN CN39 w A 1010 Preparing a Buffer You want to buffer a solution at pH 430 or H301 50 x10395 M Oxidation States of Atoms in Elements and Compounds Oxidation Number Examples 1 Atoms in their elemental state 0 Fe Hz 02 2 Monatomic ions charge F39 Na Fe 3 Group 1A 1 NaCl KN03 4 Group 2A 2 MgO 5 Fluorine 1 HF ClF 6 Hydrogen 1 H20 7 Oxygen 2 02 HClO4 8 Group 7A 1 HCl 9 Group 6A 2 Psz Balancing OxidationReduction Reactions 1 Assign oxidation numbers 2 Separate into oxidation and reduction half reactions 3 Balance each half reaction using the following steps a Balance all elements except oxygen or hydrogen b Balance oxygen by adding H20 c Balance hydrogen by adding H d Balance charge by adding electrons Electrons go on the RIGHT product side for OXIDATION reactions Electrons go on the LEFT reactant side for REDUCTION reactions e In BASIC solution do this additional step For every H add OH39 to BOTH sides of the reaction Combine H and OH39 into H20 Cancel out any waters that appear on both sides You should now have a balanced half reaction 4 Multiply balanced half reactions so an equal number of electrons are consumed and produced 5 Add together half reactions 6 Clean up Combine identical substances and reduce coefficients to the lowest terms 7 CHECK Atom and charge must balance Hints Never add 02 O atoms or 0239 to balance oxygen Never add Hz or H atoms to balance hydrogen Be sure to write the correct charges on all the ions 1am 201 39 39 mm HIVReam s m Fgy 2 g 2 my 257 NDau2H a My 4mm 177 m s 4 5a m 4 H39aq 2 4550414 an 1635 Mum aq 3 H pq 39 a mm a mum 131 away we Aum wzg quotMum mu nUy AqkH ltW4e 392EI aq7110m 133 munm M 72mm H229 wane 28 may we nu ma 4 Mm a g 7 mug 2 mom um u1aqHumze vu aw an N may f Nu ltw2 935 3 my may nux wnze z 1739 Wm E mm 393 lg a a 0335 g mmvzmum we 2 mm a uquotliq2e 3 mm a u va 1 l2e g 015 a w 39 F g w W m E Niquotaq22 vlWu FbSOAM Z A thY 50 W1 Ld aq 2 7 v mm itquotmn 2 Fem 2mm 21 vlnm mum 2 wz an 1 may 3 7 mm mm 29 7 W a qu quot33 K HHHE M31 h m mm 4n m m mm m Mm ml mmquot mm 2ans Emukslcab nnmm
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