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# Honors Chemistry I CEM 181H

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This 206 page Class Notes was uploaded by Ladarius Rohan on Saturday September 19, 2015. The Class Notes belongs to CEM 181H at Michigan State University taught by James Harrison in Fall. Since its upload, it has received 93 views. For similar materials see /class/207687/cem-181h-michigan-state-university in Chemistry at Michigan State University.

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Date Created: 09/19/15

Composition of air Component Nitrogen Oxygen Argon Carbon Dioxide Neon Helium Krypton Sulfur dioxide Methane Hydrogen Nitrous Oxide Xenon Ozone Nitrogen dioxide Iodine Carbon monoxide Ammonia Symb ol N2 Properties of Gases Volume 7808 2095 093 003 182 parts per million 52 parts per million 11 parts per million 10 parts per million 20 parts per million 05 parts per million 05 parts per million 009 parts per million 007 parts per million 002 parts per million 001 parts per million trace trace Gases 11 elements Pressure of gas T Press urc of mercury column Pressure of atmosphere Height proportional to atmospheric pressure Torricelli 16081647 Barometer AreaAA A Force F E dhAg dhg TABLE 4 Pressure Units SI unit pascal Pa 1 Pa 1 kgmilsi2 1 Nm 72 Pascal 16231662 Conventional units 1 bar 105 Pa 100 kPa 1 atm 1013 25 x 105 Pa 101325 kPa 1 atm 760 Torr 1 Torr 1 mmHg 133322 Pa 1 atm 1471bincir2 psi Figures in bold are exact See inside back cover for more relations N newton Note that 1 mmHg is the same as 1 Torr to Within 1 part in 107 1 atm 101324 bars 1013 millibars Consider mercury barometer height 760mm density Hg at 150C 13595 gcm3 acceleration due to gravity g 980665ms2 atmospheric pressure 101324 Nm2 11355 5 f 3992 1012119 0 g e 1 v 1 3 0 1 6 1 V quotH V 31 i uuncw39 ITquot 1 JACKSON aur i 1 V I t a i 01 I IO 1 V 3108 39 420 7 noc 7100 quot Weather map showing isobars in millibars Atmosphere Apparatus 3w pressure inside apparatus 3 open Pressure Measurements Manometers High pressure inside closed A Apparatus gt apparatus What is the pressure in kilopascals when the Hg level in the apparatus side column In an opentube manometer is 25mm lower than the level in the atmosphere side assuming that atmospheric pressure is 760 mm Hg at 150 C 1 mm hg 1 Torr 133322 Pa 785 Torr 104658 Pa What is the pressure in pascal s inside a system when a closed Hg manometer shows a height difference of 10 cm at 1500 10 cm 100 mmhg 100 Torr 13332 Pa Pressure P Gases are Characterized by 4 variables An extensive property depends on the Pressure P intenSive size extent of the system while V0Ume V extenSiVG an intensive property is independent Temperature T intensive of the size of the system quantity n extensive Boyles Law 1662 The volume of a fixed amount of gas at a constant temperature is inversely proportional to the pressure VCBP or PVCB Small volume high pressure Small volume high pressure Hg Volume V e l 95 3 1 l 39A large vo umc low pressure 938 gas Large volume low pressure IN I I l V Volume V a U urnL Volume V Charles Law For a fixed amount of gas under constant pressure the volume varies linearly with temperature VCCT Slope depends on pressure High temperature large volume gt aquot E 3 E gt 4 Low temperature small volume 273 C 0 C Temperature T Temperature T GayLussac For a fixed amount of gas in a constant volume the pressure is proportional to the temperature Avogadro S Pr39nc39ple PCGT Under the same conditions of temperature and pressure a given number of gas molecules occupy the same volume regardless High temperature of their chemical identity high pressure Molar Volume Volumemoles V n V m Pressure P Ideal gas 2241 Low temperature low pressure Hydrogen 2243 Temperature T Molar volume in litersmol at 00 C and 1 atm Ideal Gas Law an equation of state The experimental observations are all consistent with the relationship PVconstant nT And calling the constant R results in PVnRT Ideal because particles atoms or molecules a occupy no space and b do not interact with one another A limiting law TABLE 42 The Gas Constant R 8205 74 X 10 2 Latr11Ilt1rnol 1 8314 47 X 10 2 Lba1K1mol1 8314 47 LkPaK lmol l 8314 47JK1mol1 62364 LTorrIclmo O 5 D r A I b it e m 393 V Applications of ideal gas law a A 125 ml flask contains Ar at 130 atm and 77 C How many moles of Ar are in the flask P130 atm V125x1O3 L T27377350K R00820574 L atm K1mol1 n000565 moles b A 120 ml flask contains 27pg of 02 at 17 C What is the pressure in Torr P VO120 L R62364 L Torr K391 mol39l n27X1O395g32 gmol T27317290K P127x1O2 Torr or 167 x105 atm c A 200 L flask at 215 K and 20 Torr contains Nitrogen What mass of nitrogen is present in grams n V200 L T215 K P20 Torr R62364 L Torr K391 mol391 n00298 mol or 14gmol00298 mol0418 g d A 167 9 sample of Kr exerts a pressure of 100 x 102 mTorr at 440C What is the volume of the container in L V n167g838O gmol PO1 Torr T27344317K R62364 L Torr K391 mol391 V 394 X 104 L e A 26uL ampule of xenon has a pressure of 200 Torr at 150C How many atoms of Xe are present n P200 Torr V 26 X 10396 L T27315288K R62364 L Torr K391 mol1 n290 x103910 or 6023 x1023290x103910174 x1014 atoms A chemist prepares a sample of He gas at a certain temperature pressure and volume and then removes half the gas molecules How must the temperature be changed to keep the pressure and volume the same Double in K A sample of CH4 gas was slowly heated at a constant pressure of 090 bar The volume of the gas was measured at a series of different temperatures and a plot of volume vs temperature was constructed The slope of the line was 299 x104 LKl What was the mass of the sample of CH4 VnRTP and nRP slope 299x104 LKl 831447x10392 L bar K1 mol1 005 g A weather balloon is filled with He at 200C and 100 atm The volume of the balloon under these conditions is 250L When the balloon rises to where the temperature is 300C it has expanded to 800L What is the pressure of the atmosphere at that point P1V1T1P2V2T2 P20259 atm Standard ambient temperature and pressure SATP T2500 29815K and P1 bar NEW Molar Volume of an ideal gas 2479 Lmol1 Standard temperature and pressure STPTO C 27315K and P1 atm OLD Molar Volume of an ideal gas 2241 Lmol1 TABLE 43 Molar Volume of an Ideal Gas Temperature Pressure Molar volume Lmol ll 0 K 0 0 0 C 1 atm 224141 0 C 1 bar 227111 25 C 1 atm 244655 25 C 247897 1 bar 1 atm 101325 bar 1 bar 098692 atm Density of an ideal gas mass EMMPVRTMP gL1 density d volume V V V RT For a given T and P the greater the molar mass M the greater the density At constant T the density increases with pressure Raising the temperature of a gas that is free to expand at constant pressure increases the volume and therefore reduces its density Density of air at SATP is 16gL1 A gaseous fluorinated methane compound has a density of 8OgL391 at 281 atm and 300K a What is the molar mass of the compound MP dRT d 2E gL 1 s0 M T R00820574 L atm K1 mol1 701 gm0391 b What is the formula of the compound if its composed solely of C H and F Possibilities are CH3F CH2F2 CHF3 CHF3 What is the density of the gas at 100 atm and 298 K d1T1 d2T2 d2 d1ii 8gL1 287 gL1 p1 p2 P1 T2 281 298 A 115 mg sample of eugenol the compound responsible for the odor of cloves was placed in an evacuated flask with a volume of 5000 mL at 280 C The pressure that eugenol exerted under these conditions was found to be 483 Torr In a combustion experiment 188 mg of eugenol burned to give 500 mg of CO2 and 124 mg of H20 What is the molecular formula of eugenol The PVT data tells us the number of moles 483T0rr0500L n RT 553K62364LTorrK 1moz 1 70x10 4mol Given the mass and the number of moles we know the molecular weight nM 2 mass of material 3 mass 115x10 g 164gm0l moles 7Ogtlt104m0Z The combustion data combined with the mass will enable us to calculate the empirical formula and since we know the molecular weight we will be able to deduce the molecular formula In a combustion experiment 188 mg of eugenol X O CO H O burned to give 500 mg of CO2 and 124 mg of H20 2 2 2 3 moles of CO2 moles of C in X quotC1 2amp21136X10 3m015136mg 44gmol1 3 Moles of H 2 times moles of H20 formed MH 212 4X10 g 21378X10 3m015138mg 18gm01 1 mg of 0 18813614 38mg E 0238gtlt103 m0 nltHgt5 101 110 39 3 110 39 3 110 Various mole ratios Empirical formula C5H6O Molecular weight of empirical formula CSH6O 512 g mol1 61g mol1 16g mol1 82 g mol1 Molecular weight from gas measurements 164 g mol1 C10H122 Daltons Law of Partial Pressures A mixture of gases that do not react with one another behaves like a single pure gas The total pressure in a mixture of gases is the sum ofthe pressures of the individual gases so for a mixture of gases A B PPAPB P PB P PA PB 04 arm 06 arm 2 10 arm PA nBRT ampPB nBRT P nRT A a PAPBT 7 A piece of Li metal was added to a flask of water on a day when the atmospheric pressure was 7577 Torr The Li reacted completely with the water to produce 250 ml of H2g collected over the water at 23 C at which temperature the vapor pressure of water is 2107 Torr a What is the partial pressure of H2 in the flask 7577 Torr 211 Torr 7366 Torr b What mass of Li reacted Need to knOW the reaction 2 Li 2H2O 2LiOH H2 2 moles of Li reacted for every mole of H2 produced PV 2736T0rr0250L quotL022 1 1 RT 62364T0rrLK mol 296K 2 0020m01 mass 0020mol694gm0l1 0138g Mole Fractions and Partial Pressures n n ZAz Az A andso ZAnznA nAnBnc 39 n For a two component system n n JCAJKB A B 1 nAnB 77144 773 PAZnART zAnRTampPBnBRTzBI1RT V V V V SincePzanBT PAZZAPSCPBZZBP A vessel of volume 224 L contains 20 mol H2g and 10 mol N2g at 27315 K Calculate the partial pressures of H2g and N2g STP 1 1 P nRT 3mol00820574LalmK mol 27315K 3cm V 224L 2 2 1 1 ZH j 221 3 2 21 3 PH2 2atm PN2 1atm A gas mixture being used to simulate the atmosphere of another planet consists of 320 mg CH4 175 mg Ar and 225 mg N2 The partial pressure of N2 at 300 K is 152 kPa Calculate the total pressure and volume of the mixture First calculate the number of moles of each component 0320 nCH4 g 0020071101 Mol fraction of N2 16gmol1 nArLg1OOO44mol ZN quotN2 00080 0248 3995gm0f 2 nt0ta 00324 nN2 2 La 2 00080moz 2802gmoz smce PN2 ZNZP nt0tal 00200 00044 00080 2 00324moz P613 kPa 1 1 quotRT 00324831447kPa LK m0 300K 1 32L 613kPa P Diffusion The gradual dispersion of one substance 7 through another such as Krypton dispersing through a Neon atmosphere XI Time Effusion the escape of a gas through a small hole into a vacuum Grahams law of effusion At constant temperature the rate of effusion of a gas is inversely proportional to its molar mass rate of e usion 0c L average speed 0C L W W Consider two gases with molar masses MA and MB rate of e uswn OfA quot1019015195 2 MB limefor A molecules to e use rate of e usion of B molecules M A time for B molecules to e use M B It takes 30 ml of Ar 40s to effuse through a porous barrier time 2 The same volume of an unknown compound takes 120s M2 1 2 To effuse under the same conditions What is its molar mass t mel 2 M2 3995gmol1 360gmol1 What is the molecular formula for a compound of empirical formula CH that effuses 124 times more slowly than Kr at the same temperature and pressure raIeKr IMCHN 124 13N 39 l838 rate CHN l MKF N10 and C10H10 Second Empirical Observation The rate of effusion increases as the square root of the temperature rate of e usion at T 2 T 2 rate of e usion at T1 T1 This suggests that the average speed is proportional Average Speed 0C T to the square root of the temperature Because the average speed is also proportional to T the reciprocal square root of the molar mass we Average Speed 0c M may write The average speed of molecules in a gas is directly proportional to the square root of the temperature and inversely proportional to the square root of the molar mass The Kinetic Molecular Theory of Gases 1 A gas consists of a collection of molecules atoms in continuous random motion 2 Gas molecules atoms are infinitesimally sma points 3 The molecules atoms move in straight lines until they collide 4 The molecules atoms do not influence one another except during collisions This means there are no attractive or repulsive forces acting between the gaseous particles Pressure is Force per unit Area Force is the rate of change of linear momentum zmdzm M dr dt dt The number of molecules in this volume is that fraction of the total volume multiplied All of the molecules with All of these molecules in by the number Of mOIeCUIeS velocity vX and traveling the volume AvXAt will hit quot1 the SyStem towards the wall will hit it The wall in At seconds AV N in atimeinterval At number xgtlt N Before mvx After mzx Change in momentum The average number of collisions after comsion with wall with the wall during the interval At ls half the number in the volume zmv x total momentum change N NAVxAt X 2mVx 2V AVfo X number of collisions 2V total momentum change rate of change of momentum NAVxAt NmszAt 1 NmAV2 gtlt2mvx X x X 2V V At V aquot dmv d pressure force area F 2 ma 2 m dt dt dt NmAVi vai P VA V Since not all of the molecules are traveling with the same speed one needs to replace 2 further since x with an average ltv gt ltV2gt ltV2gt ltV2gt ltV2gt 3ltV2gt x y Z x V 2 12 NmltV gt 2 2 2 2 P NmVrmS V V1V2 VN rmS 3V 3V N 2 2 PV varms nMVrms 3 3 PV 2 nRT 12 T RT rms M 1 3RT 2 Ekinetic EMVrms 2 Calculate the root mean speed of N2 molecules at 200C 1930 3 R T 1 2 M28gmol28x10393kgm rms M T27320293K R83145JK 1m0391 1 1 12 3gtlt83145JK mol gtlt293K quotquotS 28x103kgm0l 511ms11533ftsec Hydrog n Calculate the molar kinetic energy in Joulesmol of Kr at 558500 and 548500 1 2 3RT Ekinelic EMVrms T T273 15 5585 329 00K and T273 15 5485 328 00K R831447 JK397mol397 E 41032 Jmol397 and E 4090 7 Jmol397 Heat capacity 4103 7 40907 12 5Jmol397 1 Fraction u Maxwell Distribution of Speeds The number of molecules with a speed between V and vdV M 32 2 2 M 2RT dN Nfvdv m 4 v e V ZERT 0020 1 0020 v 100 gmol 100 K 0015 F 50 gmol 1 i8 0390 E 300K 0 010 k g 0010 500 K 392 Li 0005 0005 20 gmol1 0 I I I I I I 0 100 200 300 400 500 600 O 100 200 300 400 500 600 Speed ms l Speed ms391 Detector gt 0020 l 39 39 1100K Slit 7quot 0015 a 55 Q a 0010 9 3 Molecular 393 390 LL beam 0005 OIIIII o 100 200 300 400 500 600 Speed ms 1 Useful integrals 2 7239 ml 9 8 Qquot N a X ml 9 X to Q3 SIH x U Q Q X M Qquot N N Q K J x Q Q X M Qquot N I 4 Do 8 DO 8 0T8 DO 8 058 J CBI 9 km Q3 II I legl 52ml Confirm normalization of MaxwellBoltzmann distribution function 32 M j 2 MV22RT fV47r2 RT V e Let a the ia32VZeavz 2RT J 00 4 3200 2 av2 4 32 J V dvz a V e dvz a 1 J 5 OI J 40532 Most probable speed Consider the MaxwellBoltzmann distribution function 32 M j 2 MV22RT fV47r2 RT V e Let a i then fV ia32VZem2 2RT J i ia32jiv26avz ia32 2aV3eaV2 2V60W2 01quot J dv J df 2 sowhen z 0V dV mp a and vm 2RT1290 mS l p M Mgmol 00 3 ax2 1 Useful integral J x 9 dx 2 2 0 20 Given the MaxwellBoltzmann distribution function fvia32v2e39m2 with a J 2RT The average speed is given by dN 00 4 3200 3 m 4 V V VfVdV a V e dV I N 6 7r 6quot 75a V a V 39M Mgm0l The average of the square of the speed 00 Useful integral e 2dng 5 0 8 0 2gt 2dNoO 2 4 3200 4 av2 V V VdV a Ve dV lt I N of f gt J of The root mean square rms speed is 3RT 3 2a Vrms quotltV2gt 2 21579 nes 1 M Mgm0l fv 0005 0004 0003 0002 0001 0000 V mp Vav lt Vrms MSOgmol T100K O l39l39l39l39l39l39l39l39l 50 100 150 200 250 300 350 400 450 500 550 600 650 vms Average kinetic energy dN w 1 w 1 E 2 IE 2 JEfVdV 2 MJ V2fVdV M ltVZgt N 0 2 0 2 ltEgtMltv2gtMZlJ Compression factor Z N O A 0 Real Gases Gases can be liquefied implies attractive forces Liquids are difficult to compress suggests repulsive forces Z Compression Factor For an ideal gas Z1 Separation 800 Pressure arm Potential energy Ep gt 1 gce Gases may be liquefied by use of the JouleThomson effect cooling brought about by expansion Heat exchanger Compressor Throttle Liquefied gas Equations of State for a Real Gas vmal Force equatlon Van der Waals equation J D van der Waals 1873 B C PVanT 1 2 quot2 VM VM Pa 2 V nbnRT V a accounts for attractions while b accounts for repulsions or molecular size TABLE 45 Van der Waals Parameters Gas 4 Lzatmq noliz b 1072 Lmol7l ammonia 4225 3707 argon 1363 3219 benzene 1824 1154 carbon dioxide 3640 4267 chlorine 6579 5622 ethane 5562 6380 hydrogen 02476 2661 hydrogen sulfide 4490 4287 oxygen 1378 3183 water 5536 3049 Volume correction in van oler Waals equation 3 Given a co4ection of molecules each with a diamezter oi two 53f them will exclude 3 N71d 7 EnNA d 71d nb a volume 3 and so N molecules will exclude 3 Vzdeat V quot5 Given b from experiment we can estimate the effective size of a molecule b NA7zd3 13 d 3 3 2x10 10b13m22b13Ang Z NA b for O2 is 319 102Lmoli so d 293 A The O2 bond length is 121 A Pressure correction in van der Waals equation well number of collisions with the wall Numzwr 0f malecules m Sysrem E Volume of system V Number of molecules in system N restraining force Volume of system V 39 W K W quotT pressure reduction 11 a V V V 2 n Pideal Pa Valenceshell electronpair repulsion model VSEPR Together with Lewis structures allows one to predict shapes for some simple molecules In particularthose with one central atom to which all other atoms are attached mawarx Linear Angular Trigonal planar Trigonal T slmped Totrahcdral pyramidal Seesaw Square planar Trigonal Square pyramidal Octahedral Pentagonal bipyramidal bipyramidal Some shapes have speci c bond angles some do not Rule 1 Regions of high electron concentration bonds and lone pairs on the central atom repel one another and to minimize these repulsions the bonds and lone pairs move as far apart as possible while maintaining the same distance from the central atom Six possible electron pair arrangements around a central atom 1095o Linear Trigonal planar Tetrahedral 90 goo 90 120 7 90 Trigonal Octahedral Pentagonal bipyramidal bipyramidal F B 5 Boron trifluoride BF3 6 Boron tr39 or39de39 BF3 Trigonal Planar EI Cl P P j j 7 Phosphorus pentachloride PCI5 3 Phosphorus PentaChIOFidea PC15 Trigonal bipyramid Be CI 000 4 Beryllium chloride BeCl2 Linear iE u E E 8 Sulfur hexafluorideSF6 2 Sum heXa uorideaSFe Octahedral Rule 2 There is no distinction between single and multiple bonds A multiple bond is treated as a single region of high electron concentration O 2 2 39IO CarbonateionCO3239 H CarbonatEiOmCogz39 Trigonal planar 0 0 2 2 O I C O O C m Carbon dioxide C02 Linear When there is no central atom consider bonding about each atom independently H H Ky C C 1205 120quot H H J 12 Ethene CZH4 39l3 Ethene CZH4 oak 39I4 Ethyne CZH2 Molecules with lone pairs on the central atom Classify molecules according to their VSEPR formula AXnEm A central Atom X attached atom ligand E lone pair 2 0 l O S O AX3E1 17 s If39t39 50239 u lelon 3 18 Sulfitei0n15032 Electrons tetrahedral Shape trigonal Pyramidal Predict the geometric structure of formic acid CHZO2 Predict structure and electron distribution of NF3 502 C102 Return to Sulfite ion 39I8 Sulfite ion 5032 21 Sulfite ionSO3239 Electron distribution tetrahedral Bond angle 0t 10950 Rule 4 The strengths of repulsions are in the order lone pairlone pairgt lone pairbondgt bondbond Bonding pairs Consider IFr AX 4E1 a b Axial lone pair Equatorial lone pair AxsEz CIFs AX4E2 XeF4 Atoms TShaPEd Atoms Square planar E39eCtr n distribution Electron distribution Octahedral trigonal bipyramidal r 739 i i i 39i i i m r A V c 77 Polar Molecule is one with a dipole moment A dipole moment is a vector 11 t 2 air units are Coulomb meter i1 qi 2 charge on I39m particle r 2 position vector to I39m particle IDebye 3336x 103930 Cm What is the dipole moment in Debyes H C16 Xpm ypm zpm 2 0 0 0 1 0 200 100 1 0 200 100 WNH 34 Dipole moment Dipole moment of HCI is 108D RHCI 128 pm What is charge on the atoms 1 2 13quot 14 15W 16IVI17N118NIII Dipole moment of HF is 191D RHF 92 pm What is charge on the atoms Non polar molecules have zero dipole moment All homonuclear diatomics H2 Cl2 N2 etc Many highly symmetric molecules are also nonpolar because bond dipoles cancel BF3 CGHG TAIL 31 Dipole Moments of Sclccged Molecules Molecule Dipole moment D I Molecule Dipole moment D HP 19 PH 058 HZ 103 ASH1 020 H 4 HB 080 SbH 012 H 042 OS 033 CO 0 12 CO 0 11 0248 151 u Naf l 900 CH CsCl no4 mEHC1CHI 190 26 Carbon leXIde C02 H10 1 85 transCHCICHCI 0 NH 47 Fnr pairs nl39 ions In 1hr gm phase nnl he hulk ionic snlicl 3 transDiChIOFOEthenerC2H2C392 29 cisDichloroetheneC2H2C2 31 TetrachloromethaneCCl4 32 TrichloromethaneCHCl3 VSEPR type Nonpolar Polar AXZE 0203 H20 AX2E2 l339 Xel2 BrlF39 VSEPR type Nonpolar Polar I31 XeF2 BrlF AX2E3 H J N none none known known 39 A AX2E4 1 53 VSEPR type Nonpolar Polar AX3E 3 AX3E2 VSEPR type Nonpolar Polar AX4E 4 VSEPR type Nonpolar Polar XeF4 AX4E2 r 13quot I PCI5 IPCI4F AX5 TK VSEPR type Nonpolar Polar IF AX5E Difluoroethenes Polarnonpolar Dipole moment F F F F H Clt ampC CC F H H H F Valence Bond Theory Walter Heitler Fritz London John Slater and Linus Pauling 039 bonds are cylindrically symmetric with respect to the internuclear line Atomic orbitals 2 2px v A JiQJ obond 6 bond o bond F2p H1 5 N2p N2p Eu N25 N25 33 HydrogenfluorideHF 34 NitrogenN2 7239 bonds are formed when electrons in two 9 orbitals overlap side by side It has a single nodal plane containing the internuclear line N2p N2p stm N25 34 Nitrogen N2 icbond N2 has both sigma and pi bonds TC2py Zpy c bond b nbonds a 62pz 2pz A single bond is a 0 bond A double bond is a a bond plus one 7r bond A triple bond is a 7 bond plus two 7r bonds Sometimes the two electrons involved in a sigma bond can come from one of the two bonded atoms This is called a dative bond HFHl H I39IIE l F s F 33 Boron trifluoride BF3 37 NH3BF3 NH3BF3 Strength ofa chemical bond is frequently correlated with the type of bond and increases single lt double lt triple TABLE 22 Bond Dissociation Energies of Diatomic Molecules ltWW CEO Bond dissociation Molecule energy H1 424 N2 932 02 484 0 0 CO 1062 F1 146 c1Z 230 Bt l 181 12 139 N N HF 543 39 39 HQ 419 H31 354 H1 287 The problem with carbon 35 Carbon He2522px 2py1 Not enough unpaired electrons on carbon to form 4 bonds in CH4 Electron Promotion requires 42 eV 36 Carbon He2512p 2py 2pz1 Can now form 3 bonds to the 2p orbitals and one to the 2s Bonds in CH4 are all equivalent Hybridization Mix the three 2p and the 2s orbitals to form 4 equivalent orbitals requires no energy SP3 38 sp3 hybridized carbon If these orbitals are to be equivalent and orthogonal they must have a tetrahedral orientation h1spxpypz hz Spxpypz h3 Spxpypz h4spxpy pz The normalized forms are h1spxpypZ2 h2spxpypz2 thdV h3Spxpypz2 h4 spxpy pz2 lz fz39j Oifz39ij Collecting the p component as pi the various hybrids become p1pxpypZJ h1sJ p12 p2pxpypZJ h2S P22 p3pxpypz h3S p32 p4pxpypZ h4S P42 1 if i1 cos 1 131095O pipjdV 13 l39fl39i j Other types of Hybridization sp2 P1Pz 5 1 s p1 p227py5pz 6 3 1 I7 6P 1P 3 y 2 h spy 1 2PzsEp2 2 2 2 6 3 1 h sgpyVEPZ s 5p3 3 E E Intermediate H20 Side View Top View Mv39Qluwm amp 35mm stxm rpmm39kv mm um my 5mm Bosh all quotglib mm a 3d2 sp3d Sp U U s s 2 s 3 v p b p c p TABLE 32 Hybridization and Molecular Shapequot Electron Number of Hybridization of Number of arrangement atomic orbitals the central atom hybrid orbitals linear 2 5p 2 trigonal planar 3 5172 3 tetrahedral 4 5173 4 trigonal bipyramidal 5 5117362 5 octahedral 6 sp3d2 6 Other combinations of 5 7 and dorbitals an give rise to the same or different shapes but these combinations are the most common How many sigma and pi bonds are there in 0C5 HCN NH3 What about hybridization AXS d Cl l CI P Cl CI 27 Phosphorus pentachloride PCI5 Characteristics of multiple bonds C N and 0 form double bonds using their p1T orbitals among themselves and with atoms of latter periods Double bonds are rarely found between elements of period 3 and higher Free rotation about CC bond Restricted rotation about CC bond sp3 hybridization on Carbon sp2 hybridization on Carbon 0Csp3Hls 0Csp3C5p3 A H C2p I H 1 c5 2 u 17 P i H 103mm F o o 439I spzhybridizedcarbon H l s35lpm H H 5Csp2H1s 6C5p2CSp2 6Csp2Csp2 44 transRetinal 43 cisRetinal Node between atoms Excited state Excite electron in 7 bond to a higher energy level x Excitation energy Break TE bond 2 Ground state Limitations of Lewis Theory Doesn t account easily for paramagnetism of O2 electron deficient compounds like BZH6 97gt H 131 pm quotquot39B B Lizo T H ii9pm H 33 mmmh g g g Molecular Orbital Theory Valence electrons move in orbitals that are often delocalized over the entire molecule and are called molecular orbitals Molecular orbitals are built from linear combinations of atomic orbitals When atomic orbitals interfere constructively they give rise to bonding orbitals When they interfere destructively they give rise to antibonding orbitals N atomic orbitals combine to give N molecular orbitals 6 Hydrogen Molecule 1 r a r r VISA 2 e A 0 A B 47mg 1 LR jlsB e rBao H B 7mg ylsA ylsB 0 WlsA VlsB 039 1 z 1s lsA lsB m lsA lsB m 01 W1A V1 B VlsA VlsB E S s s nergy A 015 VISA WISE 0 y 1s lsA WlsB m VISA WISE Ols N VISA V1sB J bonding 22AB llISA l1sB antibonding Ols If 2 ZSAB 1 e rAa0 e rBa0 1 WM 2 m l1sB m 1 r a r a SAB J WISAWISBdV 2 3J e A 06 B OdV V 7 00 V A quotA 3973 a a a r R2rA R2rB HA 4 HB R2 R2 1 rAa0 rBa0 Ra0 2 SAB lesAwlsBdV jje e 6 1Ra0Ra0 V 0 V 10 05 M A A B A B 0 R 0 R 00 Small overlap Large overlap 0 1 2 3 4 P atomic orbitals can form molecular orbitals of o and TC symmetry pzApZB pZApZB B N G GgZp W1 PmPw Z 88 75 quot J22SAB 139u2pr 7 2px Atomic p orbitals can form molecular orbitals of o and 7 symmetry 1sz Variation ofthe molecular orbital energies of rst row homonuclear diatomics L12 Be2 B2 C2 N2 02 F2 7 sz gZPXJFgZP A i l l ogzpz i l39 uzl szy l Tllr 4H 613925 4 H i 39 w agzs 3 M V T 3 H l quotri H L Liz Bez B2 c2 N2 Energy A Energy 021 F2 The atomic orbitals on main group elements on the left of the periodic table can hybridize and then form molecular orbitals consider B 152 252 2 p1 after hybridization B 1S2Sp1sp1p7lr W Spa1 51708 This is possible because of the small sp energy separation Photoelectron spectroscopy cglsmiuls G 23 Radlatlon 6 2A source nusz uzpy I 627 gl Sample chamber 40 38 4 3 L F Electrostatic analyzer Ionization energy MJmoFI 1 MJmol 1036 eVparticle hV EMO EKinen39c Detector Bond Order b is the net number of bonds allowing forthe cancellation of bonds by antibonds b12number of electrons in bonding orbitals gtilt the number of electrons in antibonding orbitals b 3Ne Ne 3 2 7 1 Bond order 0 l i l 800 400 7 307 Bond lengthpm Bond energykJmol no a B2 C2 N2 02 F2 Number of valence electrons 02 F2 Energy a Write the molecular orbital con guration for F239 F and F52 b What are the bond orders 0 Any paramagnetic d Is the highest occupied orbital a or it The ground state electron con guration of the ion C is 2 2 02sa s7z p What is the charge on the ion and what is it39s bond order Determine the bond orders and use them to predict which of the following pairs has the stronger bond a C or C2 b 02 or 0 Which of the following molecules would you expect to have the lowest ionization energy aC bC2cCi Molecular Orbitals of Heteronuclear Diatomics Atomic orbital of comparable energy will form a MO 0 antibonding 15 436 eV 499 eV 6 bonding 428 eV AH Consider the HF molecule 7176 evill F atom orbital energies H atom orbltal energies The bonding molecular orbital in a heteronuclear molecule AB W CAWA CBW B Covalent bond Ci C123 Polar covalent bond C31 gt C Pl 39b d 2 2 OOl llC on gtgtCB I For the HF molecule l CFVFp CHIlHls r 039 And because of electronegativty difference 2 2 F40 while W22 CF gt CH O Bonding in heteronuclear diatomics involves an unequal sharing ofthe bonding electr ns The more electronegative element contributes more strongly to the 39bonding orbitals whereas the less electronegative element contributes moire strongly to the antibonding orbitals t mtElkmn mgu a 5mm Alon Ammo Numbu 5mm 1 22 21 3 3 A 5p 1 a 45m 2 He 4ng 3 L 4473 4135 a a 4733 4305 5 a 77535 4455 43m 5 c 41325 4705 4533 7 N 45529 4955 45522 a o 40559 1254 532 9 F 725 33 1573 u 730 m N 42 773 7 193m 4 35a 11 N 40573 4797 45m 4132 12 Mg 49 032 7 3 753 72 232 253 13 m 535m am 3239 0393 42m 15 5 752312 5157 4255 45m 4227 15 P 4997a 77511 75401 4595 4392 15 3 42m 4an5 4533 433a 4437 A7 CA 404225 mm 4073 4a73 4307 m Ar 412511 42322 572 3273 4593 19 K 433534 45490 41520 1749 4934 4147 20 Ca 443354 45323 43529 4245 4341 4195 21 a 373m um 42m 4395 4422 4425 42a 32 a 745244 42150 45235 77191 75351 ass 4217 33 As 7432535 45330 750154 4030 421 4535 0379 35 s 453355 49552 55253 3932 4557 4337 4403 35 Hr 49mm 552m 755554 42272 77472 4353 4457 35 xx 7520154 499a 43m 40m 4331 115 4524 Only atomic orbitals of the same symmetry will mix in a given MO In HF the F p0 orbital will have a non zero overlap with the hydrogen 13 while the F p7I will have a zero overlap The bond will therefore be between the F p0 and the H13 Molecular Orbitals in Polyatomic Molecules H20 and SF6 Energy V 01 22 22 4 2 7 Energy Q Zal His1 ampH1s1 Q an 3111 r K g J 2t 7 atomic orbitals E quotg lt generate 1 fF 7 molecular orbitals Nonquot bonding 121 1511 Bonding Benzene electrons in the molecular plane are called 6 electrons while those in the p71 atomic orbitals are 7139 electrons Energy Emmy 6Csp2Hls 4W 6C5p2Csp2 0C3p2C5p2 quot39 3 222 o Ht 4 211 8 111 O Most of the chemistry and electronic spectroscopy of benzene and other aromatic molecules obtains from the TC electrons Annhundin 39 V a Bonding 3 TE Electron Theory conSIder ethene Conjugated and aromatic hydrocarbons can be treated with a TC electron approximation in which we view the TC C illil C electrons as moving in some fixed electrostatic potential due to the electrons in the o framework 391 a in if E i39 4 o framework A 2 a The effective Hamiltonian for a TC electron H I EV Ve r A The energy expression in terms W HWdV of the approximate wavefunction ll 2 I gtxlt v WV Express 11 in terms of a linear V CAVA CBVB combination of carbon p7I atomic orbitals Use the energy minimization principle select CA and CB so that the energy E is a minimum E 91 HlldV CjHAA C HBB 2CACBHAB NCACB tide C31 C 2CACBSAB DCACB HAA IWAFIVAdV ampHBB IWBFIVBdV HAB IiiAPde ampHBA IiiBILde DaNNaD DaNNaD 8E aCA aCA amp aE aCB aCB aCA D2 aCB D2 Setting a E amp 6 2 t0 0 results in 6CA 6C3 DaN N6D ampD6N 6D 6CA 6CA 6C3 6C3 aNNaD amp aNNaD N 51nceE2 aCA D aCA 8C3 DOCB 8N EaD amp aNEaD aCA aCA aCB aCB Recall NCACB C HAA C HBB 2CACBHAB amp DCACB C31 C 2CACBSAB SO 6N 6 2CAHAA2CBHAB amp a DZZCA i39ZCBSAB CA aCA So CAHAA CBHAB ECA CBSAB 0r CAHAA ECBHAB ESAB 0 Differentiating WRT CB CAHBA ESBACBHBB E 0 Dividing both equations by CA results in HAA ERHAB ESAB 0 HBA ESBARHBB E 0 Where RCACB At this point one can evaluate HA4 HA3 HBA HBB amp SBA SAB And solve for R and E and then using the normalization equation gtxlt 2 2 2 w dez CAyA CBzB dV CA CB 2CACBSAB 1 One can then solve for CA and CB separately Energy While this can be done an approximation that provides considerable insight and much simplification was suggested by Erich Huckel 1930 Simplifications and assumptions of Huckel Theory For ethene in particular SABOHAAHBBaandHAB And so the two simultaneous equations become a E R 0 it amp M Ra E 0 Eliminating R results in X a E2 2 0rEai Once one has E one nds R i1 To nd CA and CB recall Vde CAyA CBlB2 dV C31 C 2CACBSAB 1 And 5117766 R SA and S A B O Huckel approximation we have B 2 2 2 1 1 RCBCB1 SOCBZi Zal ldCAZi Z Resulting in 01S m M and 0 Z M 2 WA W3 J5 J2 EOt E a Butadiene a linear polyene 7m E a2 cos nl2N N1 0 16l83 a 0618 gt 00 5 a 7 g a 1 Lu Ha06183 aL6l8B Pm smecHrM s w u mm mm N pZ n21 W1 037 17191 06015 p 2 0601519 3 W 0601519103717p2 03717p3 4601519 4 W3 06015191 03717p 2 O3717p73 06015p4 W4 03717191 0601519 2 06015p 3 4371719 4 Electron distribution N 2 amp where an is the occupation of orbital n pm Ian O M Bonding 1 m DHZ WDJE39EJZE 5333 4 E a ZB Energy Benzene En a2 cosz n0 i1 N 2 In large molecules there are many closely lying energy levels and the HOMOLUMO gap can be quite small Such molecules are often colored Energy 46 BrCa rotene C40H5r Unoccupied LUMO Excitation HOMO 47 Lycopcnc CmN Occupied Liquids and Solids Intermolecular forces bare ion bare ion interaction E 47rgor Z e ion dipole interaction E 2 2 47i80r 47i80r hydration cation I anion Iondipole interactions are strong for small highly Charged ions Consequently cations that are small andor highly Charged are often hydrated in solid compounds Radius Na 102pm Radius Cu 87pm NaZCO31OHZO CusO45H20 Radius of NH 151pm Radius ofCl39 181 pm NH4C anhydrous With ions of similar radii hydration is more likely with higher Charge Ba2 radius 135 pm while K is 138 pm K salts are anhydrous KCI while Ba salts are often hydrated BaCl22H20 La is next to Ba and forms 3 ions and has radius 122pm LaNO33 6H20 LaSO43 9H20 Calculate the ratio of the potential energies for the interaction of a water molecule with a Ca2 ion and with an In ion Take the center of the dipole to be located at Rion 100 pm Which ion will attract water more strongly E Ion d ole P 012 g ECa2 2 801002g180j2 EIn3 1001002 31 3 2 00 2 2 M 054 Group 3 Ear 3 200 P3 52 cl radius 212 184 181 i Pm I 3 39 A53 562 B 201 250 n Wlns 222 198 196 151200 Te239 I39 101 150 221 220 51 100 1 50 DipoleDipole interactions between polar molecules Static solid l i zl Edd 3 4721901 ROtating mum and gas influences boiling points l i zl 6 6 ll 7 Edd 4723980 r Small in gas and larger in liquids Which ofthe arrangements of CHZCI2 molecules has the strongest intermolecular attractions Lama KM For which of the following molecules will dipoledipole interactions be important 02 03 C02 302 London Forces Interaction between instantaneous dipole moments Property of ALL atoms and molecules 051052 bUt ELondon 6 is most important between nonpolar molecules 4722901 Depends on the polarizability a of the substances the ease with which the electron clouds can be distorted Polarizability tends to increase with the number of electrons in the species Halogen melting point 0C boiling point 0C F2 220 188 C12 101 34 Br 7 59 2 114 184 TABLE 52 Melting and Boiling Points of Substances Substance Melting point C Boiling point C Substance Melting point C Boiling point C Noble gases Small inorganic species He Z7035 K 269 4 7 K 1 259 253 Ne i249 i246 N1 i210 i196 Ar 189 4186 01 218 i183 Kr 157 15 H20 0 100 e 112 i108 H15 86 60 Halogens NH3 78 33 F1 7220 7188 02 778s C12 101 34 soZ 76 10 Brl 7 59 Organic compounds 11 114 184 CH i182 i162 Hydrogen halides CF 150 129 HF 7 20 CCL 7 3 77 HC 114 85 CbH 6 80 HBr 789 67 CHO 794 65 H 51 35 glucose 142 d sucrose 184d Abbreviations 5 solid suhhmes d solid decomposes Under pressure Is the trend in the boiling points of HCI HBr and HI consistent with Dipoledipole or London forces Normal boiling point C 200 100 200 Hydrogen Bonding 100 H20 1 Mg H256 e42 33 NH3 785 HQ 54139 11 Hydrogen Fluoride HFI 88 PH 3 SiH47112 7164 CH4 2 3 Period OHO 10 Hydrogen bond HzTC 2 SbH3 17 H1735 SHH4 52 Elements F O N with H directly attached strongest type of interaction Requires lonepair Acetic acid dimer Which of the following molecules are likely to form hydrogen bonds PH3 HBr 02H4 HNo2 CH3OCH3 CH3COOH CH3CH20H CH3CHO In each pair indicate which substance has the stronger intermolecular forces Ne and Ar NF3 and BF3 SiH4 and GeH4 NaF and HF Liquids Short range order and long range disorder Molecules are very close but able to move past one another The ease with which they move past one another is measured by the viscosity Difficult to predict as it depends on both molecular forces and molecular shape Viscosity centipoise B Diethyl ether Surface tension the net inward attraction of the molecules on the surface for the molecules in the bulk Capillary action is the rise of liquids up narrow tubes Due to adhesion TABLE 53 Surface Tensions of Liquids at 25 C Surface tension Liquid y mNm Meniscus curved surface of benzene 2888 carbon 270 liquid in narrow tube tetrachloride ethanol 223 cohesive vs adhesive forces hexane 184 mercury 472 methanol 226 water 7275 580 at 100 C Which liquid in each of the following pairs has the greatest surface tension cis dichloroethane or transdichloroethane benzene at 20 C or benzene at 60 C Which substance has the greater viscosity in its liquid form ethanol CH3CH20H or dimethyl ether CH3OCH3 Butane C4H1O or propanone acetone CH3COCH3 The surface of glass contains many OH groups that are bonded to SiO2 If the glass is teated with SiCH33C1 Chlorotrimethylsilane HCl is given off and the SiCH33 group bonds to the O that had a H attached to it How Will the new glass surface differ in its response to H20 Types of solids Amorphous Crystalline at9m5 or mo39eF U39eS atoms or molecules are Me In a random Jumble in an orderly array class examples characteristics molecular assembles of discrete BeClzSSP 412 Relatively low melting and boiling points brittle if pure molecules ice glucose napthalene Network Atoms covalently bonded to B C Black F Hard rigid brittle very high melting point insoluble in water Solids neighbors BN SiO2 Metallic Cations held together by a sea s and d block Malleable ductile lustrous electrically and thermal conducting of electrons elements ionic Mutual attraction of cations NaCl KNO3 Hard rigid brittle very high melting points those soluble in water and anions CuSO4 SHZO give conducting solutions Molecular solids discrete molecules l2 low melting point Dono fo ago WW ggo gev oxggen l hgdrugen Network solids Strong covalent bonds hard rigid high melting and boiling points 17 0357 run diamond Metallic solids Cations in a sea of electrons Malleable ductile lustrous electrically and thermally conducting ABABAB Hexagonal close packed hcp Mg Zn Ti Co Layers A B A Hexagonal closed packed Coordination number 12 Layers A B A ABCABC Cubic closed packed Face centered cubic Layers A B A Layers C B A nit Cell The smallest unit ofa crystal that when stacked together repeatedly without any gaps and without rotations can produce the entire crystal Cubic closepack Face centered unit cell Coordination number 12 Al Cu Ag Au 39r y A a z 1 w a What is the ratio ofthe volume occupied by the atoms and the volume of the unit cell in the cubic closepacked structure 918 18 4 3 Vatoms NEW N7 chbe 83 2 r3 lz Vcella3832r3 r0270 2 437tr3 ratio 2 135 2 074 Same as hcp V atom H exagonal closepacked Mg Zn Ti 00 Atoms in unit cell N6 Body centered cubic Fe Na K Atoms in unit cell N2 AL l ofthe unit cell in the body centered cubic structure V 43Trra V a Atoms in unit cell mm be N2 diagonal of cube a4r 3 V Vubea3id ran0 0 C 5 cube a Unit Cell The smallest unit of a crystal that when stacked together repeatedly without any gaps and without rotations can produce the entire crystal Face centered cubic unit cell Body centered cubic unit cell pfO68 14 Bravais lattices All crystal structures can be expressed in terms of 14 basic patterns of unit cells Monoclinic l Monoclinic C Triclinic Hexagonal Trigonal R Polonium atomic radius 167 pm crystallizes in a primitive cubic structure a Number of atoms in a unit cell 7 r 1 b Coordination number 66 6 c Length of side of unit cell a Simple cubic b lFace centred cubic rile lBody centred cubic a 2r3 3 4 pm Three dimensional View showing the number of atoms per unit ceil Silver crystallizes in in a fee stmct39ure and has a density of 10500 gcm3 Estimate the atomic radius ong m density ce 3 cell Vcell a 2612 4r2 s0 a812r Vce a3 832r3 mce 4at0mscell10787gmol 6824x10 23 cm3 Vcell a 4086x10 80m density 105 gcm36022x1023 atomsr7101 r 1445x10 80m a 1445pm The density of Cu is 893 gem3 and its atomic radius is 128 pm Is the metal fcc Closed pack or bcc density 2 a 2 mass cell Vcezz mass N Mgmol 4 atomcellg6355gmol 24221X1023gCe NA 6022gtlt10 atomm0 3 Veg a3 2832r3 832128x10 8 cm3 V06 2 4745 gtlt10240m3 denSiZJfcc d 89Ogcm3 The density of Cu is 893 gem3 and its atomic radius id 128 pm Is the metal Closed pack or bcc density 2 a 2 mass cell Vcezz maSSZN Mgmol 2 atomcellg6355gmol 22111X1023gCe NA 6022gtlt10 atomm0 3 3 Vcellz a3 3 1232128gtlt108 cm3 6 3 6 2583x1o 24cm3 dbcc 817 gcm V C a Primitive cubic b Body centered cubic c Facecentered cubic ccp Radius ratio radius of smaller ion radius ratio radius of larger ion 167 CsCl 0923 p 181 102 NaCl 2 2056 p 181 050 88 coordination NaCI Rocksalt structure 66 coordination KBr Rbl CaO AgCI CSBI Csl TCl TBr Calculate the number of anions and cations and the formula unit 0 Ca at opposite corners of small cubes F at centers of small cubes uorite rutile CaF2 fluortite Tio2 rutile Estimate the density of CsCI mass cell density d Vcell mass cell 2 mg mCS 3545gm0113291gm0 6022 gtlt1023 atomsm0 2796gtlt1023 g xga 2rCl rCS 2181pm 170pm1012mpm 702x103910 m V8 a3 6658x10 30m3 C d 2796x1023 6658x10 30 414x106gm3 a 414gcm3 deXp 399gcm3 TiCaO3 Bragg dz raction equation co 6 25 Q Q0 0 Q 2 3 0 f 30 f 0 J59 O 9 Lattice layer 1 d Lattice layer 2 Bragg Equation 2d sin 9 2 m1 1 600 If 0 3 2 60 0N 00 P e Lattice layer 1 g Lattice layer 2 LiF crystallizes in a rock salt Na Cl motif a4018 pm What is the smallest angle at which a Xray beam with A7107pm must strike the planes making up the faces of the unit cell for the beam to be diffracted 2dsin6l 2 n1 sin6 2d 7101 sin 6 24018 008836 6 5070 QM hm 1 5mmme PART 056 PM 55 9995939121 2321 1 C53Thnmbf39s upammxS Ex m hum can 0 mi km WWW 0 mm m3 Swen kW cwz Mata Wham vwadu bl km anew 0 Thuj pass 39 n Walk SQDLBJHL mum madzk 3mm m bkm 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Ck 03mm 13 mama b but M Lam onQ gt9 Ur s vanme Lumen bL moi1d hm mgnu mw6 um rumx M og Wand mama R w owm possum mamas sum r em mum CL Mm wm M w 53thch 5101331 3 48 QEsz m momme m rm 0 m ng LABWM S w wwwmc m muxg m ixw dc magma 39p 39 mow 0xc EKLLWVm S E m Lbn LhO8x I53 WE LZAQX K Mgquot X 51me m 38109 043 E 3 Mun 3 x b 39LUm LSMDK 2 5be0 Sir E 17lb x V gt P rf f mlI39IWlb K J39 203 8 23 Pg Chemical Kinetics Concentration of reactant Measure the decrease in a reactant or the incerase in a product as a function of time Tangcn I Tim C Average reaction rate deterioration of penicillin a moiL i i7 wrrkv i 10 20 30 U 10 2U 50 Time wccksi Tum weeks Instantaneous reaction rate vt d Penicillin dt Relative rates are determined by the stoichiometry of the reaction 2H1g gtH2g12g dHI or dH2 or d12 d dt a t ldHI dH2 d12 2 dt a t dt N2g3H2g gt2NH3g 2 rate of reaction 2 Vt i lldp ldhwn a t 3 dt 2 dt aAbB gtcCa D 164A 31 l zl 2 rate of reaction 2 Vt rate of reaction 2 Vt a dt 9 dt c dt d dt J molrL l Molar concentration of N30 Rate Laws and Reaction Order Put different amounts of N205 into 5 asks all of same volume and place in a water bath at 65 C N205 vaporizes and begins to decompose 2N205g gt 4N02g 02C 7 j 4 In 4 m 008 3 Measure InItIal 7quot reaction rate I V0 and plot 3 104 006 4 versus molar Z concentration E a 2 a 3 Z 4 6 04 004 gt a 39 k 0 g mulL i squot E A 0002 2 E 39 10 l l 0 004 008 Time Molar conccnn39ution N100 mart k 00052 S391 and is called the rate constant If one measures the rate slope at any given concentration one finds that it is proportional to the concentration and is given by a N O Vt Mkjv205 a rate law a t Where k is the same constant found in the initial rate measurements Except for special cases rate laws must be determined experimentally 2N205g gt 4N02g02g 2N02g gt 2N0g 02g d N02 a t Vt k39 N022 2N02g gt 2N0g02g 0006 7 0006 TA 0005 0005 2 0004 0004 0003 0003 0002 7 0002 Rate 0F consumprinn of N01 l nolL l Rate of consumption of N02 mall Ls 0001 000 0 0 44 002 004 006 000 1 i 0002 0004 0006 0008 1 a N01 molr b N012 mull11 Vt M k NO 2 k 054M1LSI dt 2 Reaction Rate lawquot Temperature K Rate constant Gas phase HZ 13 2 HI leHIgjllgl 500 43 X 0 7 Lmol 39squot 600 44 X 10 4 700 65 x 10 2 800 26 2 HI gt H 11 1412112 500 54 x 10 9 L molil sil 600 97 X 10 700 18 gtlt 10 3 300 97 x 1039 2 2 N10 gt4 N01 01 IaLNlO5 29x 313 328 338 52 x 10 3 2 N30 8 2 N2 01 H0120 1000 076 s 1 1050 34 2 No2 gt 2 NO 01 MNOIF 573 054 Lmol is 1 c3146 8 2 CH3 wsz 973 55 gtlt 10 4 s cyclopropunc gt propane Mcyclupmpanc 773 67 X 10 4 sfL Aqueous solution 1430 OIr 2 1110 k1 I0H 298 15 gtlt 10 Lmol CHBr H gt CHXOH BK kHBruH 298 28 X 10 4 L mul l L CIZHMOH 1410 8 2 CJIUOI MCUHHOI lH 298 18 x 10 4 LmaWsquot t39For the unique instantaneous mtc TTln39cc 5igni canr gures Common Rate Laws Order in A Rate law 0 Rate I 1 Rate HA 2 Rate IAjl 2NH3g gt N2g 3H2g On a hotPt wire the rate is constant until all NH 3 is gone a zero order reaction a b Vt kA B The total order of the reaction is the sum of the exponents ab Exponents need not be integers E k 02 MM 202ltggt 2SO3ggt 12 203 gt 302g Rate 2 Rate 143022 so3 Determining a rate law 3103 aq SBF aq 6H30 aq gt 3B 2 aq 9H20l Initial concentration mol L7 1 Initial rate Experiment 150 Br 1130 mmol BrOffL lsix 1 010 010 010 12 2 020 010 010 24 3 010 030 010 35 4 020 010 015 55 dBr0 a b c a1 b1 c2 V0kBr0 Bf 1130 dt 0 total order 6 V 0 12 10 3 1L 1S 1 k x mo 12mol 3L3S 1 3037 Bf H3O T 1x10 4 moz4L 4 2AZBC gt 3G4F Initial concentrations in mmoIL1 a Order of each reactant experiment A0 B0 C0 V0mmolGL391 s391 1 10 100 700 20 a b c 2 20 100 300 40 Z 3 20 200 200 16 0 0 0 4 10 100 400 2 5 462 0177 124 a1 b22 00 b rate constant VO 2 A13 313 C 101002 c Experiment 5 WC VO kA0 BE 2x10546201772 29x106mm01Lls1 2x105mm012Lzs1 First order rate laws A gt products dA dA AM I t 7kA m kdz Ago ArklOId 1n A kt t0 AAOektt0 A10 1 Cycloprupanc C3H6 What is rate constant P Heat to 773K cltmmgt CzH motL 1mm 7111 kt7t0 A 0 150x103 76 50 0 in 4675 5 124x103 669 1 2 10 100x103 7691 h1AlnA07kZ 5 15 083x103 V7 09 1 700 0040m1n 7 770977650 5 The rate law for the reaction N202 gt 2N0 is first order in the concentration of N202 Derive an expression for the time dependent behavior of NO the product concentration zkjf202 N202N2020 6 Ct N0 2N2020 NZOZD N0 2N202O 1 e kz dN202 l dN0 dt 2 dt kN202 kN202O gm N0 2N2020 1 e k Fquot q a 395 1 4 c OJ L F C For a first order reaction the half life is independent of the initial concentration kzr0 05A0 1n kt t0 ln2 A10 f t nt A ii II III MI 0 t 5E I E Time

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