General College Chemistry
General College Chemistry CHE 201
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Experiment4 Preparation of alum KA1SO4 212HO y and the limited reactant concept The mechanism of formation of alum 2 Als 2 KOHaq 6 H201 gt 2 KA1OH4aq 3 Hzg 1 2 KA1OH4aq HZSO4aq gt 2 AlOH3s 2 H20 1 KZSO4aq 2 2 AlOH3s 3 st04 gt AlzSO43aq 6 H20 3 A12so43 aq KzSO4aq 24H20 gt 2 KA1SO4z12H20s 4 I 2 Als 2 KOHaq 4 H2804 22 H20 gt 2 KAlSO4z 12HzOs 3 Hzg overall reaction Stoichiometr o the overall reaction 2 moles Als gt 2 moles alum 2 moles KOHaq gt 2 moles alum 4 moles HzSO4aq gt 2 moles alum Calculations m Calculate number of moles of Als nAl m Al amp MMA1 2698 gmol Calculate number of moles of KOH nKOH M X VKOH 9 Calculate number ofmoles of H2804 nHZSO4 M x VHZSO4 4 Use stoichiometry of the balanced overall reaction to determine the limiting reactant and the theoretical yield in moles and grams Calculate the yield V39 Chagter 6 Quantum Theory and the Electronic Structure of Atoms Our world until the end of the nineteenth century The Macroscopic World Studied and characterized by Classical Physics Matter Radiation Newtonian Mechanics classical theory of Newton s laws electromagnetic Radiation Thermodynamics by Maxwell The end of the nineteenth century witnessed the birth of the quantum theory and the discovery of the microscopic world The Microscopic World For chemists it is the world of atoms and molecules Studied and characterized by quantum physics In order to understand the quantum theory and the microscopic world we need to Understand classical properties of matter and waves Waves Wavelength A Amphtude The distance between corresponding points on adjacent waves is the wavelength A E i The number of waves passing a given point per unit of time is the frequency v For waves traveling at the same velocity the longer the wavelength the smaller the frequency Electromaqnetic Radiation I i i i W I I 1020 1013 1016 1014 to12 1010 400 500 Wavelength m 10 11 10 9 10 7 10 5 10 3 10 1 101 103 I I I I l l I I I I l I q I I I I quot I I Gamma X rays IUI1m Infrared IMicrowavesI Radio frequency rays I Violet I I I I I I I I I I I I 106 104 lt Frequencys1 I 108 750 nm 600 700 All electromagnetic radiation travels at the same velocity the speed of light 0 30 X 108 ms Therefore 0 1v The Nature of Energy of Radiation The wave nature of light does not explain how an object can glow when its temperature increases Max Planck explained it by assuming that energy comes in packets called quanta The Nature of Energy of Radiation Einstein used this assumption to explain the photoelectric effect He concluded that energy is proportional to frequency E hv a where h IS Planck s constant 663 x 10 34 Js The Nature of Energy of Radiation Therefore if one knows the Wavelength or the frequency of light one can calculate the energy in one photon or packet of that light Ephoton hVphoton or C E h 1 where c v it Pay attention to units The Nature of Enemy of Radiation Another mystery involved the emission spectra observed from energy emitted by atoms and molecules Demo H He and Ne atoms Solar radiation intensity 7 n m Atomic spectra httpastroustr bqfrkoppendischargeindexhtml Emission Spectrum of the Hvdroqen Atom Quantization of Energy of Electrons 7 Niels Bohr adopted Planck s j assumption and explained these phenomena in this way 1 Electrons in an atom can only occupy certain orbits corresponding to certain energies Only a line spectrum of 8 discrete wavelengths is 200 ado 1000 2000 MWquot 0 H 2 Copyright 2000 USB Bohr Model for the energy Levels of the hydrogen atom n gt00 0 1 Electrons in an atom can only E occupy certain orbits 5 RH corresponding to certain 4 energies emission R 3 w H 2 Electrons in permitted orbits 9 have specific allowed RH energies these energies will 2 T not be radiated from the atom absorption 1 RH R n H EH 2 2 n 123 n RH218gtlt10 18J 109737 cm391 Frequencies of Absorption and Emission Transitions In The hydroqen atom Absorption Emission Ef 1 AEZEf EiZhVW AEhu AEhu w AEZEfEihV Ei 7 Ef RH n123 AER H R H En 2 2 2 n 1 f ni RH 2109737cm39 gt 1 1 AEEf EihV RH 1 de Broglie s Postulate WaveParticle Duality of Matter A V m P 2 mV Wave property Particle property De Broglie argued that both light and matter obey his equation Why the wave property of matter is important for microscopic objects while it is not important for macroscopic objects Let us carry out the following exercises a What is the value of 7 of an electron traveling at 100 of the speed of light b What is the value of 7 of a person m 626 kg moving at 1ms A nswer x szv p 6626gtlt1034Js 31 6 243A 91gtlt10 kggtlt3gtlt10 a A This wavelength is about five times greater than the radius of the H atom b k m 1O6gtlt1025 A This wavelength is too small to be detected r s I de Broglie Waves Are Observed Experimentally u w 8 htt IANwwmatteror ukdiffractionlintroductionlwhat is diffractionhtm a b a Electron diffraction Al b Xray diffraction experiment Al The Heisenberg Uncertainty Principle 2 httpIlichemedchemwisceduIJCEWWWIArticlesNVavePacketNVavePackethtml AXAP 2 g h or AXAmV 2 In many cases our uncertainty of the whereabouts of an electron is greater than the size of the atom itself Quantum Mechanics Erwin Schrddinger developed a mathematical treatment into which both the wave and particle nature of matter could be incorporated It is known as quantum mechanics Quantum Mechanics The wave equation is designated with a lower case Greek psi 11 The square of the wave equation 22 gives a probability density map of where an electron has a certain statistical likelihood x of being at any given instant in time Quantum Numbers Solving the wave equation gives a set of wave functions or orbitals and their corresponding energies Each orbital describes a spatial distribution of electron density An orbital is described by a set of three quantum numbers Principal Quantum Number n The principal quantum number 17 describes the energy level on which the orbital resides The values of n are integers 2 O Azimuthal Quantum Number I This quantum number defines the shape of the orbital Allowed values of I are integers ranging from O to n 1 We use letter designations to communicate the different values of I and therefore the shapes and types of orbitals Azimuthal Quantum Number I Valueofl O 1 2 3 Type oforbital s p d Maqnetic Quantum Number m Describes the threedimensional orientation of the orbital Values are integers ranging from I to I s m s l Therefore on any given energy level there can be up to 1 s orbital 3 p orbitals 5 d orbitals 7 forbitals etc Maqnetic Quantum Number m Orbitals with the same value of 17 form a shell Different orbital types within a shell are subsheHs Number of Total Number Possible Subshell Possible Orbitals in of Orbitals in 7 Values of l Designation Values of m l Subshell Shell 1 0 ls 0 1 1 2 0 25 0 1 1 2p 1 D 1 3 4 3 0 35 0 1 1 3p 1 0 1 3 2 3d 2 1 0 1 2 5 9 4 0 4s 0 1 1 4p 1 0 1 3 2 4d 21 0 1 2 5 3 4 f 321 0 1 2 3 7 16 s Orbitals O ml0 Spherical in shape Radius of sphere increases with increasing value of n s Orbitals Observing a graph of probabilities of finding an electron versus distance from the nucleus we see that s orbitals possess n1 nodes or regions where there is O probability of finding an electron Q Orbitals 1 m 1 0 1 Have two lobes with a node between them Z Z Z Z xy xy x y x i 3 p2 Px py d Orbitals 2 m21012 Four of the five orbitals have 4 lobes the other resembles a p orbital with a doughnut around the center Energies of Orbitals O N For a oneelectron hydrogen atom orbitals on the same gm energy level have the same energy That is they are degenerate Energy gt Energy gt Energies of Orbitals 4pm As the number of FEE electrons increases 3pm though so does the 135 repulsnon between them 2amp3 Therefore in many 2H electron atoms orbitals on the same energy level are no longer degenerate Spin Quantum Number mS In the 1920s it was discovered that two electrons in the same orbital do not have exactly the same energy The spin of an electron describes its magnetic field which affects its energy Spin Quantum Number 1778 This led to a fourth N quantum number the spin quantum number T ms The spin quantum number has only 2 allowed values 12 l N and 12 S Pauli Exclusion Principle No two electrons in the N same atom can have exactly the same energy T For example no two electrons in the same atom can have identical sets of quantum numbers Electron Confiqurations 4p5 Distribution of all electrons in an atom consist of 4 Number denoting the energy level p Letter denoting the type of orbital 5 Superscript denoting the number of electrons in those orbitals Orbital Diaqrams Each box represents one orbital 1 L 1 Halfarrows represent the electrons 15 25 The direction of the arrow represents the spin of the electron Hund s Rule For degenerate orbitals the lowest energy is attained 1 when the number of 13 25 229 electrons with the same spin is maximized Periodic Table We fill orbitals in increasing order of energy Different blocks on the periodic table II I I I I I I I then correspond to is l l l l l l rent types of Ieegetativesblock Elizggiigta ve p block orbitals El Transition metals l fBlock metals Some Anomalies Some irregularities occur when there are enough electrons to halffill s and d orbitals on a given row 1 For instance the electron Core 5 2A 3A 4A 5A 6A 7A 2 configuration for copper is 4 8 Ar 481 3d5 2 16 rather than the expected we S Ar 432 3d4 w This occurs because the 4s and 3d orbitals are very close in energy quot39 57 58 59 60 61 62 63 64 65 66 67 68 69 70 Lanthanide La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Xe series 51 653 41524 4quot652 4 651 3652 652 7652 4454139 mes 11quot 652 4f 653 if zes2 4fl 6s 4f 6lt2 vs 7 These anomalles Occur 89 9o 91 92 93 94 95 96 97 98 99 100 101 102 Ac nide series Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Rnl MI 2752 5ng 51736311 Scam 41752 5775 5976741 5975 mm 3 SJquot w 3 1 071111 Wu 7 5 5 S S In ms I I Metals Metalloids E Nonmetals Detailed and Abbreviated Electron Confiqurations Of Metals and Nonmetals OLULM Aia ymvw 391 80 5 25 th 2mm 257 2 F9 1 25 74gt 168 we 3523 n 39quot N23 35 4 11Na WC 3D539 19K Ll S E 39 l8 Ll39S Detailed and Abbreviated Electron Confiqurations Of Transition Metals T V CY 218C E rj LlSZ SC I J 3 317 30l3 57quot gorle 240 CM 45 34 ELTth 3M j 4539 S Mn Cm 457 3amp5 Znl 25 9 457 quotso539 I as 3A nvmmumm 290 EM gquot 34 z D Ar 1 45 3A 302m 1 3 Diamaqnetism and Paramaqnetism Paramaqnetic substances Any substance that possesses net nonzero electron spin quantum number and therefore attracted by a magnet 40 weal I 26Fe lgrAr14s23d6 07 W4 I U 49 3d M3 1314113101T 3S23p1 3 3quot 3739 Diamaqnetic substances Any substance that possesses net zero electron spin and therefore not attracted by a magnet rS L 3am 48Cdz36Kr5524d10 Wilmiwhrhvhc O 531 m 099 21 own board A Very Useful Tool For Understandinq Electron Confiquration of Atoms Just click on the following link and you love what you find Electron Con quration From Prentice HaII SEPARATION OF CATIONS BY PAPER CHROMATOGRAPHY INTRODUCTION Chemists are often faced with the need to separate and to identify chemical substances in a mixture You may have heard or read about the various kinds of chromatography such as paper column liquid or gas In this experiment paper chromatography will be used to separate and to identify the cations present in a mixture The separation and identification procedures take advantage of the differences in chemical properties of the various cations You will be expected to successfully carry out the separation and identifications to understand how these are achieved and to understand the basic chemistry involved In simple terms chromatography involves a moving phase such as a gas or liquid which carries the substances to be separated across a stationary phase such as paper cellulose silica SiOz or alumina Al203 Both the moving phase and the stationary phase must have a tendency to attract the substances to be separated We say that the moving phase dissolves or quotcarriesquot the substances while the stationary phase adsorbs or quotholdsquot the substances If for a particular substance the moving phase strongly attracts a substance and the stationary phase very weakly holds the substance we would have this substance suffering little or no retardation as it is moved along by the moving phase Now imagine a second substance present along with the first substance This second substance is also strongly attracted to the moving phase but unlike the first substance it is also fairly strongly attracted to the stationary phase As the moving phase travels along we find that the first substance runs ahead of the second substance and a separation has occurred Some difference in the chemical properties of the two substances has allowed this separation How do we know experimentally that the separation has occurred If substance one is a red dye and substance two a blue dye no problem If both are colorless initially but turn different colors on being sprayed with a special reagent no problem If the two substances glow when ultraviolet light is focused on the stationary phase we would have a way of detecting the separation We must take advantage of a difference in some chemical or physical property to detect the separation and to identify the substances The difference in mobility of two colored substances leads to separation and identification Suppose both substances were dyes with the same blue color and blue dye 1 ran well ahead of blue dye 2 If you studied an quotunknownquot mixture of the dyes and found two widely separated blue spots you could easily identify which was 1 and which was 2 Suppose you ran another unknown and found only one blue spot that moved almost as fast as the moving phase You would be justified in saying only 1 was present But suppose you have two blue dyes which more closely obey a corollary of Murphy s Law and do not separate so well If the unknown contained only one of the two dyes you would be hard pressed to say which one We need to be more exact about how far the substance moves relative to the moving phase In paper chromatography use is made of the Rf value defined as R distance substance moved 1 f distance solvent front moved or 2 d R 7 f D For a given moving phasestationary phase system each substance will have a unique and fixed Rf value The term quotfixedquot as used here means that the Rfvalue will be the same from one experiment to the next and will be the same regardless of how far the solvent travels The ambiguity in identification in the above experiment could be removed by calculating an R value and comparingto the known R values forthe suspected substances In this experiment you will calculate R values and use them as a means of identification In the procedure to be used in this experiment the stationaw phase is paper and the moving phase is a solvent consisting of acetone water and HCI or hydrochloric acid Acetone has the formula CHSCOCHS is an organic solvent and is a component offingernail polish remover Since HCI is a strong acid it will be present as H and Cl39 ions The three substances to be separated are the cations Fe Cub and Nib The source of the cations will be the nitrate salts which are soluble in water as for example H10 CuN02s a Cu2aq 2N0339 mi 31 As the various cations move across the paperthe nitrate anions also move alongto maintain electroneutrality The cations do have unique but faint colors and in orderto produce more intense colors the following reactions will be carried out at the end ofthe chromatographic experiment Fe 30H gt Fe0H 4 pale yellow colorless redbrown precipitate 2 4NH3 gt CuNH342 5 pale blue colorless deep blue complex ion HL CH1 Ni 2NH3 2 HONkNOH gt 6 CH3 vew pale green colorless dimethylglyoxime colorless red complex The student is expected to fully understand these reactions In the red complex solid lines between atoms are covalent bonds dashed lines are hydrogen bonds and arrows are a special type of bond coordinate covalent wherein each nitrogen atom donates an electron pairto nickel The source of NHS and OH39 in reactions 4 6 is an aqueous solution of ammonia gas in which the following reactions occur NHgas S NHaqueous H20 3 NHAOH 3 NH OH39 7 SAFETY PRECAUTIONS In addition to the general safety precautions please note the following 0 Acetone A volatile very flammable substance Do not have any flames or ignition sources near y Ammonia Atoxic irritating gas Avoid breathingthis gas Hydrochloric Acid A corrosive substance avoid exposure Metal Salts These may be corrosive andortoxic avoid excessive exposure WASTE DISPOSAL 0 Chromatography solvent and all liquid waste should be disposed of in a designated container 0 Dw chromatography strips may be disposed of in the trash can PROCEDURE 0 tain two strips of a er two 39 capillawLuue and two 250 mL flasks with cork stoppers from the stockroom Using a pencil label the top of one strip with a K for known and the top of a second strip with a U for unknown Nearthe other end bottom of each strip make a small pencil mark at the side and 3 cm from the end Thoroughly mixtogether 10 drops of each ofthe three cation solutions in a small testtube and using a micropipette or a capillaw tube spot a small amount ofthe mixture on the K paper strip atthe bottom end nearthe center ofthe strip and in line with the pencil mark This is your known chromatography strip Now obtain 10 drops of an unknown solution in a small test tube and prepare an unknown chromatography strip in a similar manner Be sureto use a clean pipet or capillary tube Immediately record the unknown number or letter on the report sheet on the space indicated Carefully study the chromatographic setup which is shown in Figure 1 below Place the K strip in a clean dw 250 mLflask with the bottom end down and close to but not touchingthe bottom of the flask Bend the labeled end overthe top rim ofthe flask so that the strip hangs inside the flask Remove the strip from the flask Repeatthis process usingthe U strip in the other flask Obtain about 100 mL of solvent in a clean dw beakerand using a funnel pour about 50 mL into each flask so that msolvent splatters onto the upperwalls ofthe flask Now carefully hang one strip in each flask and place the stopper in the flask making sure that the end ofthe paper is immersed in the solvent but that the sample spot isnot immersed Figure 1 STOPPER quotKquot or quotuquot LABEL PAPER STRIP FLASK SAMPLE SPOT 0 SOLVENT Figure 1 Equipment for paper chromatography When the solvent front has moved to within about one inch of the stopper remove the strip from the flask be sure to draw a pencil line along the solvent front before it evaporates so that the D value can be measured restopper the flask and wave the strips around in the air until dry about one minute At this point try to identify the various cations one ach strip and calculate Rfvalues using d and D values in mm and using as the starting point the point where the initial spot was placed at the bottom of the strip If the sample appears as a fairly large spot measure d to the middle of the spot and if the solvent front is not straight measure D to some average point Record the data as directed on the report sheets Now carry out detection reactions 4 and 5 by holding each strip in a beaker of ammonia NH vapor in the hood Do not allow the strip to dip into the ammonia solution If the expected colored spots do not appear spray the paper with a small amount of water and expose to NH3 again Reaction 5 occurs almost directly as NH3 is absorbed by the paper whereas reaction 4 occurs after the sequence of reactions 7 which generates OH needed for reaction 4 If you do not understand these reactions and the manner in which each is made to occur consult the laboratory assistant or the instructor Detection reaction 6 is carried out by spraying each strip with a solution of dimethylglyoxime in the hood Calculate Rf values and record all data as directed on the report sheets By observation and comparison of the known and unknown strips and by Rf values you should be able to identify the cations present is your unknown Record your conclusion in the space indicated on the report sheet Remove the completed report sheets from the manual attach your chromatographic sheets and turn in to the laboratory assistant KEYWORDS o Cation A positively charged ion such as H Na or Fe3 o Anion A negatively charged ion such as NO and Cl 0 Chromatography A means of separating and identifying the components of a mixture The method is based on a mobile phase such as a gas or liquid and a stationary phase such as paper 0 Phase A distinct and separate entity in a mixture For example ice water consisting of ice cubes in water has two phases a solid phase the ice cubes and a liquid phase the water 0 Known A mixture of known identifiable substances used as a standard in a procedure for identifying which of the substances might be present in any given combination of the substances 0 Unknown A mixture containing some combination or only one of the substances present in the known 0 Ammonia A molecular substance consisting of uncharged molecules having the formula NH3 Ammonia is a gas at ordinary conditions 0 Ammonium on A cation having the formula NH4 0 Complex on As opposed to simple monatomic ions like Cl or land simple polyatomic ions like N03 or 5042 a complex ion often contains a central metal atom such as Nib surrounded by fairly complex molecules or ions such as that depicted in reaction 5 The complex ion exists as an entity and any net charge is considered as belonging to the whole combination and not to any particular atom or atoms REPORT SHEETl NAME CHE 201 SECTION DATE LAB ASSISTANT Paper Chromatography Unknown Instructions Under each section of A and B below draw a rough sketch of the paper chromatographic strip horizontal and include the color and cation identity of each spot Show the calculation of the Rf value for each spot A Known 1 Preliminary Results before detection reactions 2 Final Results after detection reactions B Unknown 1 Preliminary Results as above REPORT SHEET 2 2 Final Results after detection reactions C List the cations present in your unknown EXERCISE SHEET 1 CHE 201 SECTION NAME 1 In a paper chromatographic experiment X moved 30 mm while the solvent front moved 50 mm What is Rf for X in this system 2 Assume that two substances A and B can be separated in a particular paper chromatographic system a Write Equation 2 for A and for B Use these two equations to derive a third equation for dA dal b Define dAi dB c Using your derived equation in a explain what is meant by the statement llthe longer the run the better the separationquot EXERCISE SHEET 2 d Using your derived equation in a explain what is meant by the statement quotA and B are difficult to separate because they have similar Rf values RfA and RfBquot 3 Write a balanced equation for the dissolving of solid iron III nitrate ferric nitrate in water 4 What is the net charge on the red complex in reaction 6 Explain how you arrived at your answer Chapter Six Thermochemistrv Thermodynamics Einstein said It is the only physical theory of universal content concerning which I am convinced that within the Framework of the applicability of its basic concepts it will never be overthrown Thermochemistrv Thermodynamics is the science of the relationships between heat and other forms of energy Thermochemistiy is the study of the quantity of heat absorbed or evolved by chemical reactions Energy There are three broad concepts of energy Kinetic energy is the energy associated with an object by virtue of its motion Potential energy is the energy an object has by virtue of its position in a field of f0 rce G0 to slide 61 Internal energy is the sum of all energies of the particles making up a substance 80393quot Energy Goto slide 59 Gom W Hydroelectric Energy Radiation Energy Chemical Energy G0t0slide64 We will look at the first three forms of energy in details Energy Kinetic Energy An object of mass m and speed or velocity v has kinetic energy Ek equal to 1 2 Ek va This shows that the kinetic energy of an object depends on both its mass and its speed A Problem to Consider Consider the kinetic energy of a person whose mass is 130 lb 590 kg traveling in a car at 60 mph 268 ms Ek 590 kggtlt 268 ms2 Ek 212x104kgm2 s2 Ek 212x104J The SI unit of energy kgm2s2 is given the name joule J Energy Potential Energy This energy depends on the position such as height in a field of force such as gravity For example water of a given mass m at the top of a dam is at a relatively high position h in the gravitational field 9 of the earth E1 mgh A Problem to Consider Consider the potential energy of 1000 lbs of water 4536 kg at the top of a 300 foot dam 9144 m 2 E1 4536 kggtlt980 ms x9144 m Ep 406x105kgm2 s2 E1 406x 105J Energy Internal energy is the energy of the particles making up a substance it involves energy on the microscopic scale The total energy of a system is the sum of its kinetic energy potential energy and internal energy U Some times is denoted as E Etot EkEpU Internal Energy Microscopic Energy Does a glass of water sitting on a table have an energy Micmecnpie 393 WilmaHt r kinetic energy magi mm 39 quot 39 5 39 is part ef internal mgr an 3 energy macmscwic Molecular attractive 39 u quot 393 fences are maimed with mutantial energy Internal Energy Microscopic Energy Systems with the eeme temperature El Translational kinetic meneaterriie 39 39 39 39 energy gas mommlar Internal Vibratienal and retetienal 935 339 Energy kinetic energyr I F39otanlial energyfmm intermolecular forces liquid or solid Ideal gases intermolecular forces are negligible Internal Energy and Phase Transition heating steam A vaperizing Mylar 100 h h H 100 calfgm P 35E 3 3 93 DDIIJI39IQ point E 4156 sng i i a heat waiar 0 C tun watt E in E mEi u Ug Ice Heat of vapurizatiun 0 C 39 cell39ng f 2260 mm quot 39HEd il i V heating ice i 39 1399 calfgrn Heatul P Energy added at constant rate 334 M39ng fusion Time gt WORK Work is a form of energy and therefore it has the units of energy Mechanical work W Fgtlt d Newtonm joule Electrical work PV WORK W F X 1 P X m2 X m Work is done by the system against 3 a xed external pressure and Pm PAV therefore a fixed external force P2AV Since P2 Pext and V is a state function then W PextAV ext V2 V1 Pay V2T G0 to slide 67 a h Expansion of an isothermal gas against a piston of mass m and an area Assume that all external pressure is exerted by the piston mass Heat Heat is defined as the energy that flows into or out of a system because of a difference in temperature between the system and its surroundings Heat flows from a region of higher temperature to one of lower temperature once the temperatures become equal heat flow stops Heat lHeat transfers as a result of temperature difference it transfers from a hotter body to a colder object lThe amount of heat q that is transferred between the system and its surroundings is expressed as Siqn Convention for Work and Heat I 5q 5W The First Law of Thermodynamics The Law of Conservation of Energy energy may be converted from one form to another but the total quantity of energy remains constant The First Law of Thermodynamics is The Law of Conservation of Energy AEuniV AEsys AEsurr 0 where AE AEsurr q W sys The internal energy is a state function therefore AESys 2E2 E1qW 0139 AU U2 U1qW sys For convenience the subscript sys will be dropped out Heat of Reaction In chemical reactions heat is often transferred from the system to its surroundings or vice versa The substance or mixture of substances under study in which a change occurs is called the thermodynamic system or simply system The surroundings are everything in the vicinity of the thermodynamic system Heat of Reaction Heat is denoted by the symbol q The sign of q is positive if heat is absorbed by the system The sign of q is negative if heat is evolved by the system Heat of Reaction is the value of q required to return a system to the given temperature at the completion of the reaction G0 to slide 62 Heat of Reaction Exothermicity Endothermicity out of a system into a system Surroundings Surroundings Aqlt0 Heat of Reaction The heat absorbed or evolved by a reaction depends on the conditions under which it occurs Usually a reaction takes place in an open vessel and therefore at the constant pressure of the atmosphere The heat of this type of reaction is denoted qp the heat at constant pressure Enthalpy and Enthalpy Chanqe Enthalpy denoted H is an extensive property of a substance that can be used to obtain the heat absorbed or evolved in a chemical reaction An extensive property is one that depends on the quantity of substance Enthalpy is a state function a property of a system that depends only on its present state and is independent of any previous history of the system G0 to Slide 66 Enthalpv and Enthalpv Chanqe The change in enthalpy for a reaction at a given temperature and pressure called the enthalpy of reaction is obtained by subtracting the enthalpy of the reactants from the enthalpy of the products AHH H products reactants Enthalpv and Enthalpv Chanqe The enthalpy of reaction is equal to the heat of reaction at constant pressure AHqp AE5q W AEzAH PAV For an ideal gas at constant pressure and temperature AB 2 AH AnRT If AV lt 0 gt 5W gt 0 gt work is done on the system If AV gt 0 gt 5W lt 0 gt work is doneby thesystem Thermochemical Equations A thermochemical equation is the chemical equation for a reaction including phase labels in which the equation is given a molar interpretation and the enthalpy of reaction for these molar amounts is written directly after the equation N2g 3H2g gt 2NH3g AH 318 kJ Thermochemical Equations In a thermochemical equation it is important to note phase labels because the enthalpy change AH depends on the phase of the substances 2H2g 02g gt 2H2 o AH 4837 kJ 2H2g 02g gt 2H2 AHquot 5717 kJ Thermochemical Equations The following are two important rules for manipulating thermochemical equations When a thermochemical equation is multiplied by any factor the value of AH for the new equation is obtained by multiplyinq the AH in the original equation by that same factor When a chemical equation is reversed the value of AH is reversed in siqn Examples For Manipulating Thermochemical Equations 2H2g 02g 2H20g AH 4837 kJ If we divide the equation by 2 then H2 g Ozg gt H20g Ago 2 24185 kJInol If we reverse the equation then H20g gt H2 g Ozg AHO 24185 kJmol Applyinq Stoichiometrv and Heats of Reactions Consider the reaction of methane CH4 burning in the presence of oxygen at constant pressure Given the following equation how much heat could be obtained by the combustion of 100 grams CH4 CH4g 202g gt C02g 2H20l AH 8903 kJ 1111016 8903 Id 11 moles Id 11 of CH4 100 g CH4 gtlt1rr101CH4 0625 mole 160g AH 0625 mol x 8903 kJmol 556 kJ qp 2 AH 556kJ Measurinq Heats of Reaction qsgtltmgtltATCAT Joule Mass 9 Heat capacity JOC1 Specific Heat Jg391 C1 Tf Ti C See table 62 A Problem to Consider Suppose a piece of iron requires 670 J of heat to raise its temperature by one degree Celsius The quantity of heat required to raise the temperature of the piece of iron from 250 C to 350 C is q CAT 670 J0 CX 350 0C 250 0C q610J A Problem to Consider Calculate the heat absorbed when the temperature of 150 grams of water is raised from 200 C to 50000 The specific heat of water is 4184 Jg 00 q s x m x AT q 4184gcn 150gx 500 2000C q 188x103J Heats of Reaction Calorimetry A calorimeter is a device used to measure the heat absorbed or evolved during a physical or chemical change Calorimetry at constant pressure A coffee cup calorimeter See Figure Calorimetry at constant volume A Bomb calorimeter MW A Problem to Consider When 236 grams of calcium chloride CaClZ was dissolved in water in a calorimeter the temperature rose from 250 C to 387 C lfthe heat cagacity of the solution and the calorimeter is 1258 J C what is the enthalpy change Qer mole of calcium chloride Heats of Reaction Calorimetry First let us calculate the heat released by the reaction using the heat absorbed by the calorimeter qrxn Ccalx AT 12580igtlt 387 0C 250 0C C qm 172gtlt104J Now we must calculate the heat per mole of calcium chloride Heats of Reaction Calorimetry Calcium chloride has a molar mass of 1111 gmol so 1 mol CaCl 2 236 CaCl gtlt g 2 111g 2 0212 mol CaC12 Now we can calculate the heat per mole of calcium chloride gm qr 472k 811kJmol molCaC12 0212 mol Standard Enthalpies of change The term standard state refers to the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data 1 atm pressure and the specified temperature usually 25 C The enthalpy change for a reaction in which reactants are in their standard states is denoted AH delta H zero or delta H naughF Standard Molar Enthalpies of Formation The standard molar enthalpy of formation of a substance denoted AH is the enthalpy change for the formation of one mole of a substance in its standard state from its component elements in their standard state Note that the standard enthalpy of formation for a pure element in its standard state is zero See table 64 Calculation of Standard Enthalpies of Reactions Direct Method From Standard Molar Enthalpies of Formation AHO Z MAE products Z MAE reactants Where the summations are over all products and reactants For a general reaction aAbE 3c ch AHO cA g C mm D aA g A bA 13 Example Large quantities of ammonia are used to prepare nitric acid according to the following equation 4NH3g 502g gt 4N0g 6H20g What is the standard enthalpy change for this reaction Use Answer 4NH3 g 502 g gt 4N0g 6H20g 44590 50 49029 6 2418 AHO Z ViA fQ product Z ViA fQ reactants Using the summation law AH0 4 m019029 6 mol 2418 m0 m0 4 mol 4590 kJ 5 m010 kJ m0 m0 AH0 906 kJ Be careful of arithmetic signs as they are a likely source of mistakes II Indirect Method Hess s Law Hess s laW of heat summation states that for a chemical equation that can be written as the sum of two or more steps the enthalpy change for the overall equation is the sum of the enthalpy changes for the individual steps Mm Hess s Law For example suppose you are given the following data Ss 02g gt 802g quot 2961 kJmol 2803g gt 2802g 02g AH 198 kJ Could you use these data to obtain the enthalpy change for the following reaction 25s 302g gt 2803g AH Hess s Law If we multiply the first equation by 2 and reverse the second equation they will sum together to become the third 25s 202g gt Mg AHO 2961 kJmol gtlt 211101 Z SCLdg 02g gt 2803g AHO 198 kJ gtlt1 ZSS3Ozg gt2SO3g AH0 2 7902 k The enthalpy of solution AH soln AHsoln soln 39 Hcomponents Cannot be measured Can be measured The Solution Process for NaCI e e b O o a e b Na and Cl inns in the gaseous stat 03 m a 95 3 93 as a lt9 5e B 6 5 a 9 AHsoln Step 1 Step 2 788 784 4 kJmol Hydmwdmmdc ons 6 7 ames5 Heats of Solution of Some Ionic Compounds Which substances could be AlIlsoln Compound kJmol used for melting ice LiCl 371 C302 828 Wthh substances could be Nam 40 used for a cold pack KC1 172 NH4C1 152 NH4NO3 262 Fuels A fuel is any substance that is burned to provide heat or other forms of energy In this section we will look at Foods as fuels Fossil fuels Coal gasification and liquefaction Foods as Fuels Food fills three needs of the body It supplies substances for the growth and repair of tissue It supplies substances for the synthesis of compounds used in the regulation of body processes It supplies energy About 80 of the energy we need is for heat The rest is used for muscular action and other body processes Carbohydrates A typical carbohydrate food qlucose C6H1206 undergoes combustion according to the following equation C6H1206S 602g gt 6C02g6H201 AHO 2803 kJmol One gram of glucose yields 156 kJ 373 kcal when burned Note in context of nutrition 1 Cal 1000 cal 4184 J 4184 kJ E A representative is glyceryl trimyristate C45H86O6 The equation for its combustion is C45H8606lts 1 3702ltggt a 45 C02g 43H201 AHO 27820 kJmol One gram of fat yields 385 kJ 920 kcal when burned Note that fat contains more than twice the fuel per gram than carbohydrates contain Go to slide 78 Fossil Fuels Fossil fuels account for nearly 90 of the energy usage in the United States hard coal the oldest variety of coal contains about 80 carbon C graphite 02 gt C02g AHO 393 kJ mol Natural Gas Fuels Purified natural gas is primarily methane CH4 but also contains small quantities of ethane C2H6 propane CgHs and butane C4H10 We would expect the fuel value of natural gas to be close to that for the combustion of methane CH4g 202g C02g 2H20g AHO 802 kJmol Petroleum Fuels Petroleum is a very complicated mixture of compounds Gasoline obtained from petroleum contains many different hydrocarbons one of which is octane 08H18 CgH18lt102ltggt a 8CO2g 9H2Og AHO 5074 kJmol This value of AH0 is equivalent to 444 kJgram Coal Gasification and Liquefaction With supplies of petroleum estimated to be 80 depleted by the year 2030 the gasification of coal has become a possible alternative First coal is converted to carbon monoxide using steam Cs H20g gt C0g H2 2 The carbon monoxide can then be used to produce a variety of other fuels such as methane C0g3H2g gt CH4g H20g Rocket Fuels In second and third stages of Saturn Vl vehicle that sent a three man Apollo crew to the moon used Hydrogenoxygen system H2 g o2 g gt H20g AHO 242 kJmol The landing module for the Apollo mission used a fuel made of hydrazine and a derivative of hydrazine with dinitrogen tetraoxide as an oxidizer 2N2H4ltggt N204ltggt 3N2ltggt 4H20ltggt AHO 1049 1d G0 to slide 80 Operational Skills Calculating kinetic energy Pot E and Work First Law of Thermodynamics Writing thermochemical equations Manipulating thermochemical equations Calculating the heat of reaction from the stoichiometry Relating heat specific heat and heat capacity Calculating heat of reaction from calorimetric data Calculating AH from calorimetric data Calculating the enthalpy of reaction from standard molar enthalpies of formation Applying Hess s law and its applications for calculation of enthalpy of reactions Animation Kinetic Molecular TheoryHeat Transfer Return to Slide 4 Animation Hess s Law Return to Slide 44 Figure Conversion of Light Energy to Kinetic Energy in a SolarPowered Vehicle Bd StaiBS Mmgm i Enatg y L V M Source University of MissouriRolla Solar CarTeam Figure Conversion of Light Energy to Kinetic Energy in a Solar Powered Lawn Mower Cont d Figure Potential Energy and Kinetic Energy Back to slide 4 Source Tennessee Valley Authority Figure Illustration of a Thermodynamic System next Surroundings Endothermic Heat System Heat y 7 Endothermic Surroundings Exothermic Heat System Heat V 1 Exothermic Ammonium Dichromate Decomposes with the Release of Heat in a Fiery Reaction Source Joseph P SinnotJFundamental Photographs New York An Analoqv to Illustrate a State Function 1 Campsite B altitude 4800 ft n Distance not state fun Altitglilde state function Campsite A altitude 1200 ft L ction Figure PressureVolume Work Before Heat 3686 k NaOHaq After Return to slide 13 Figure StrikeAnywhere Matches Figure Single Lit Match Cont d Table 62 Speci c Heats and Molar Heat Capacities of Some Substances Molar Heat Speci c Heat Capacity Substance Jg C Jmol C Aluminum A1 0901 243 Copper Cu 0384 244 Ethanol CZHSOH 243 1122 Iron Fe 0449 251 Water H20 418 753 gtkValues are for 25 C Go back to slide 32 ConstantPressure Calorimetrv AHqp A Simple Coffeecup Calorimeter Clrxm QCal Qwater Clcal ltlt Clwater Clrxn QWater Cllrxn m X S X ATgtwater Aern qrxn Return to slide 35 Thermometer Stirrcr Styroloam cups l Reaction f a mixture No heat enters or leaves BLE 3 Heels of So I Reactions Measured at C lanl Pressu AH Reaction Example 39kJmol Heat of neutralizaiion HClaq Na0Haq gt NaClwq H200 562 Heat of ionization H100 gt Haqj OHaq 562 Heat of fusion HZOU HZOU 6 0 Heal 0139 vaporizalion HQOU gt H20g 441quot Heal of reaction MgCIEO 2Na gt ZNaClx Mg 802 Measured ax 25 C AA 100T the value is 4079 k1 ConstantVolume Calorimetry A Bomb Calorimeter Determination of heat of combustions qrxn q cal Thermometer Current for Ccal X ignition coil Ccall is a constant that includes Stirring motor All parts of calorimeter and water AB q 5w 232122 At constant volume 0 AE qv qrxn Cutaway of steel bomb Aern 72 qrxn Ignition Graphite coil sample Figure Enthalpy Diagram Illustrating Hess s Law Enthalpy H kJ Reactants 2 mol C graphite 2 mol 02g AH3 2210 k Products 2 mol COg 1 m0102g D AHl 7870kJ AH2 5660 kJ 2 mol C02g Return to 44 Table 64 Standard Enthalpies of Formation at 25 C Substance or Ion AH kJmol Substance or Ion AH kJmol Substance or Ion AH kJmol equotg 0 Carbon HCO3 aq 6920 Bromine Cg 7167 Hydrocarbons BIG 1 1 19 Cs diamond 1897 CH4g 7437 Braq 1215 Cs graphite O C2H4g 5247 Brg 2190 CC14g 9598 C2H6g 8468 Brzg 3091 CCI4 1354 C6H6l 490 Br2l 0 COg 1 105 Alcohols HBrg 3644 C02g 3935 CH3OHI 2387 Calcium C032quotaq 6771 CZHSOHU 2777 Cas 0 CSgg 1 169 Aldehydes Ca2 aq 5428 C820 8970 HCHOg 117 CaCO3s calcite 12069 HCNCE 1351 CH3CHO quot 1661 CaOs 6351 HCNI 1089 CH3CHOI 1918 See Appendix B for uddilional values Table 64 Cont d Standard Enthalpies of Formation at 25 C Substance or Ion AH kJmol Substance or Ion AH kJmol Substance or on AH kJmol Chlorine Hydrogen Pbs 0 C1g 1213 Hg 2180 Pb2a 17 C an 1672 H aq 0 PbOs 2194 Clquotg 2340 Hg 15362 PbSs 9832 C123 0 H2 Z 0 Nitrogen HC1g 9231 Iodine Ng 4727 Fluorine 1g 1068 N28 0 Fg 7939 naq 5519 4590 Fg 255l Ig 1946 NH4a 1325 Faq 3326 125 0 9029 F2g 0 Hlg 2636 N03g 3310 HFg 2725 Lead HN03aq 2074 Sce Appendix B for additional values Table 64 Cont d Standard Enthalpies of Formation at 25 C Substance or Ion AH kJmol Substance or Ion AH kJmol Substance or Ion AH kJmol Oxygen Silver Na g 6093 0g 2492 AgS 0 Na2C03s 1 1308 I 0212 I 0 Agaq 1056 NaCls 4111 032 1427 AgBrS 1004 NaHCO3s 9508 OH a 2300 AgCs 1271 sulfur 2418 AgFs 2046 Sg 2770 2 2858 ASKS 6l84 Ss monoclinic 0360 Silicon Sodium Ss rhombic 0 Sis 0 Nag 1073 525 1286 SiCl41 6870 Nas 0 502g 2968 SiF4g 16149 Nz1aq 240 1 H2519 2050 81025 quartz 9109 Sce Appendix B for additional values Figure Sources of Energy Consumed in the United States 1996 Return to slide 51 Hydroelectric Nuclear power and other electric 44 power The Launching of the Columbia Space Shuttle Return to slide 55 Source NASA Conceptual Problems with Line ArtIllustrations Conceptual Problem When white phosphorus burns in air it produces phos phorusV oxide P4s 502g gt P4010s AH 3010 U What is AH for the following equation P4OI05 P4S 502C Conceptual Problem Carbon disul de burns in air producing carbon dioxide and sulfur dioxide CSzl 3025 gt C02g 2302g AH 1077 k What is AH for the following equation lcsza 02g a lCOxg 802g cs2 Chapter 3 Stoichiometry Calculations with Chemical Formulas and Equation Or Ratios and Combination What Do We Need to Know Chemical Equations ethanol oxygen pacetic acid Water Or on two steps as ZCZHSOH 02 gt 2CH3COH 2H20 2CH3COH 02 2CH3COOH 1C2H50H 202 2CH3COOH szo 1 Knowing the balanced equation grams 10 grams or 10 grams grams How In the grocery store Chemical reactions oranges and lemon atoms and molecules count the numbers count the numbers Eggs atoms and molecules package of eggs package of atoms or molecules a dozen of eggs moles Peanuts and candy atoms and molecules weight by grams weight in grams Information from Balanced Chemical Eguations ethanol oxygen acetic acid Water V 1 Molecular Formula 2 Relative number of atoms and molecules in the reaction Or Formula Unit 1 Molecular Weight and Formula Weight Molecular weight MW MW substance W Ni 9 775 a re p a r a a a In A39le m L sL We glamorous weight or all 111 a 10363 ecu or lEl JCti substance MW H20 2gtlt10 amu 1gtlt160 amu 180 amu Formula weight FW FW substance li i n Aft 3 a n 3 quot L P g L m weight of our FW NaCl lgtlt2299 amu 1gtlt3545 amu 5844 amu Examples Calculate the formula weight of each of the following substances to three signi cant gures CH3Cl and Fe2SO43 FW CH3C1 1gtlt120 amu 3gtlt10 amu 1gtlt 3545 amu 5045 amu 504 amu FW Fe2SO43 2gtlt558 amu 3gtlt321 amu 12gtlt160 amu 3999 amu 400 x 102 amu Problems In the laboratory We cannot calculate absolute number of atoms or molecules in any sample in the lab Why 100 g of ethanol contains 131gtlt1023 molecules It would take 415 billion years to count that number With a device that counts million molecules second We cannot weigh atoms and molecules in amu unit or in grams Why Mass of an atom E 103923 grams E 103926 Kg How can we proceed The Mole Concept A mole is de ned as the number of atoms in exactly 12g sample 0fC 12 The number of atoms in exactly 12g sample of C12 equal 60221367x1023 Avogadro s number N A 60221367gtlt1023 602 gtlt1023 3 sig Fig 1 mole of any substance contains N A units Mole and Molar Mass 1 mole of any substance contains N A units 1 mole of atoms contains N A atoms 1 mole of molecules contains N A molecules 1 mole of ions contains N A ions 1 mole of contains NA Molar Mass gmol The mass of one mole of a substance in grams Formula Weight and Molar Mass F or all substances The molar mass in gmol is numerically equal to the Formula weight in atomic units Formula Unit Formula Weight Molar Mass amu g mol391 0 160 160 H 10 10 H20 180 180 NaCl 5844 5 844 CH3Cl 1194 1194 Importance of the Mole Unit Units atoms molecumE gtltNA N nmoles unit NA number of moles of atoms molecules and ions massmolar mass 11 molesx molar mass masses of atoms molecules and ions in grams Mole Calculations Formula unit Molar mass number of molec number of moles Mass g mol nmolm n mole m g H20 180 1 mole g 0166 x1023 299 x 103923 nmole nmolec molec N A molec mol39l 1 molec 602 x1023 molec mol39l 0166 x1023 mol m n mol x M g mol39l 0166 gtlt103923molgtlt 180 g mol391 299 x 103923 g Mole Calculations Formula unit Molar mass number of molec number of moles Mass g mol nmolm n mole m g H20 180 1 mole g 602 gtlt1023 180 Nmolec n mol x N A molec mol39l 1 mol x 602 x1023 molec mol39l 602 x1023 molecules m n x M 1 mol x 180 g mol39l m n molgtlt180 g mol39l 180 g mole Determininq Chemical Formulas Kno in MW MF iAranCk W g Mole ratio n m k If we know mass percent of each element we know mass ratio Formula with simplest mole ratio we know mole ratio we know empirical formula 1 Mass Percentaqe and Mass Ratio MF AB Ck m x Similarly we calculate mB and mC mass rat10 1s mA mB mC 2 Mole Ratio and Empirical Formula MF AanCk Knowing n1A mB Inc and Knowing M A MB M can calculate nA nB no where 3th 11 an L Ll 2 H MA gfnimn z 3 The Empirical Formula Get the simplest ratio of n A nB nC How Determine the smallest number among n A nB nC Assuming that 11A is the smallest number then D1V1de each of n A nB nC by n A I1A11B11C The empirical formula HA HA HA An Example for Determining the Empirical Formula From Masses of Elements A substance contain the following 175 grams of Na 397 grams of Cr 428 grams of 0 What is the empirical formula Empirical formula is N212Cr207 Element Mass M gmol No of moles Slmplest mole ratio Na 175 230 Cr 397 520 x2 2 pl 0 428 160 3572 7 An Example for Determining Empirical Formula from Composition An Example A substance give the following mass 175 Na 397 Cr 428 0 What is the empirical formula A Key Point If we have mass assume that the total mass is 100 g Then mNa 175 g mCr 397 g mo428g Empirical Formula Empirical formula is N212C1 207 Element Mass Mgm01 N0 of moles SlmpleSt mole rat10 39 1 Na 175 230 Cr 397 520 1 quot 1 xz 0 428 160 o 39 3502 Mass Percentage Percentage Composition from the Molecular Formula Calculate the percentage composition in formaldehyde CHZO M of formula 2 1x 120 2 x10 1gtlt 160 gmol 300 gmol Elements n of element M of element Mass mole gmol C 1 120 1x120 x100400 300 H 2 10 2x13901x100 673 30 0 1 160 1x160 x100 533 300 always check that sum ofpercentage 100 Calculating the Mass of 21n Element in g Given Mass of a Compound From Formula Weights From masses in grams Elemental Analeis Calculating the Percentage Masses of Elements Example combustion of acetic acid C H O CO2 H20 424 mg 621 mg 254 We want to calculate the following C H 0 Applying the law of conservation of mass then mC in reactants mC in CO2 mH in reactants mH in H20 mO in reactants mO in H20 mo in CO2 m9 from air 1 Mass Ratio in one Mole C H 0 CO2 H20 424 mg 621 mg 254 o sample mass MF mass ratlo 1n one mole mass MF Element No Moles of Mass ratio element in MP in one mole CO C 1 120 gmol391 C 2 440 gmol391 C02 1 H20 H 2 2X 101 gmol H 18O gmol391H20 2 Mass 00 sample mass y mass rat10 1n one mole mass 0 mass of C in C02 mass C in sample mass of H in H20 mass H in sample mass O in sample 100 399672 534 Empirical Formula From Mass Percentage Assume that the total mass is 100g Empirical formula is CHZO Element Mass Mgm01 N0 of moles Simplest mole rat10 C 399 120 1 H 67 101 2 O 534 160 Molecular Formula from Empirical Formula In the previous example If we know that the MW is 600 amu calculate the molecular formula molecular weight 6O amu empirical formual weight 3O amu Molecular formula 2 n x Empirical formual Molecular Formula 2 2 x Empirical formual 2 X The sample is acetic acid Stoichiometrv Quantitative Relations in Chemical Reactions Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction Stoichiometry is based on the following 1 The balanced chemical equation 2 The relationship between mass and moles Such calculations are fundamental to most quantitative work in chemistry Molar Interpretation of a Balanced Chemical Equation and the Law of Conservation of Mass N2g 3H2g gt 2NH3g 1 molecule 3 molecules gt 2 molecules 1 mole 3 moles gt 2 moles An Example for Balancinq Chemical Equations Combustion of Methanol CH3OH1 02 g gt C02 g H20g Balance the equation 2 moles 3 moles 2 moles 4 moles 3xeejeeag 2x28g 4 2g 1609 1609 Calculation of Number of Moles of Products from Number of Moles of Reactants Method I Suppose we wished to determine the number of moles of NH3 we could obtain from 48 mol H2 N2g 3H2g gt 2 NH3g Because the coefficients in the balanced equation represent moletomole ratios the calculation is simple 48 mol H2 x 211101NH3 32 mol NH3 3 mol H2 Calculation of Number of Moles 0f Reactants from Number of Moles of Products Method 11 N2g 3H2g gt 2 NH3g 1 mole 3 moles 2 moles moles 64 mole 3 H101 H2 moles 64 mol NH3 x 2 mol NH3 96 mol H2 Calculation of Masses of Reactants from Masses of Products N2g 3H2g gt 2NH3g 1 mole 3 moles 2 moles I 140 lat Calculation of Masses of Products from Masses of Reactants N2g 3H2g gt 2NH3g 1 mole 3 moles 2 moles Limitinq Reactant LR n m 3 a con uence con uence onouuuuuw ouoauooum nonooooo4 onoonooo 5 E d 0 0 fl A A A a C A B 16 tires remain in excess limits the number of assembled cars Limiting Reactant or Limiting Reagent and the Theoretical Yield The limiting reactant or limiting reagent is the reactant that is entirely consumed when the reaction goes to completion The limiting reactant ultimately determines how much product can be obtained The theoretical yield of product is the maximum amount of product that can be obtained from The limiting reactant An Example on the Limiting Reactant and the Theoretical Yield Zns 2 HClaq ZnC12aq H2g 1 mol 2 moles 1 quot 039 What is the LR 030 m0 03952 m0 What is the theoretical yield of H2g 030 mol 030 mol 271 is the QXCQSS reactant 052 mol 026 mol 1 HCl The limiting reactant The theoretical yield Example 2 Combustion of Methanol MCI ICE 302 3303 5E39Ig3911D Consider combustion of 100 ml of methanol in 200L of air What it is the limiting reactant LR Which reactant remains in excess What is the theoretical yield of water How much reactant remains in excess The density of methanol is 0791 gml the density of O2 is 131 gL Answer work it out on the board Download answer PDF The Experimental Actual Yield ls Never Greater Than The Theoretical Yield Why Experiments are never perfect Yield Experlment a1 y161d x100 Theoret1ca 1y1eld Consider the previous experiment on combustion of methanol If the experiment yield of water is 195 grams what is the percent yield of water Yield 0 water 229x100 50 g Operational Skills Calculating the formula weight from a formula or model Calculating the mass of an atom or molecule Converting moles of substance to grams and vice versa Calculating the number of molecules in a given mass Calculating the percentage composition from the formula Calculating the mass of an element in a given mass of compound Calculating the percentages C and H by combustion Determining the empirical formula from percentage composition Determining the molecular formula from percentage composition and molecular weight Relating quantities in a chemical equation Calculating with a limiting reactant Stoichiometry and the Ideal Gas Law Preparation of Nitrogen Gas and Identification of the Cation Present in an Alkali Metal Nitrite Salt OBJECTIVE To carry out a procedure capable of determining the identity of an unknown nitrite salt NaNOz or KNOZ The procedure calls upon the reaction of sulfamic acid H2N503H and the unknown nitrite salt 1 and use of the pertinent ideal gas law PV RT MNOZ aq H2N503H aq 9 NM H20 MHSOMaq 1 INTRODUCTION I I I u Numerous experiments done in the and centuries led to an empirical description of the physical properties of gases in terms of the thermodynamic variables pressure P volume V number of moles and absolute temperature T Pressure is usually expressed in atmospheres atm volume in liters L and temperature in Kelvin K R is a constant of proportionality termed the universal or ideal gas constant 00821 atm LmoIK As the name quotidealquot suggests strict adherence to this law occurs only in selected cases at conditions of moderate to high temperatures and moderate to low pressures PURPOSE amp CALCULATIONS Your purpose in this experiment is to identify the alkali metal cation M a quotspectator ionquot Li Rb and Cs are possible cost and safety of handling limit actual unknowns to Na and W accompanying the reactive nitrite ion and which produces quothalfquot of the nitrogen gas product The other nitrogen arises from the sulfamic acid coreagent which also produces hydrogen sulfate anion coproduct in 1 above Sulfamic acid is used in excess so that the number of moles of nitrogen formed and calculated from the ideal gas law above equals the number of moles of unknown nitrite salt used 2 mol MNOus mol Nug PNZ atmVN2 L00821 atm Lmol KTN2 K 2 Once that mole number is obtained it can be used with the mass of unknown measured to 0001 g precision to calculate the cation molar mass as in 3 below g Mimol M g unknown 46 gmomo MNOZ SJmol MNOZN 3 APPARATUS The apparatus Fig 1 shown on the following page will be used in this experiment The reaction between aqueous MNOZ and sulfamic acid will occur in the 250 mL Erlenmeyer flask The volume of nitrogen gas produced in the 250 mL Erlenmeyer flask will equal the volume of water 100200 mL displaced from the 500 mL Florence flask and then collected in the 250 mL beaker The volume of that displaced water 100200 mL can be directly measured although the precision is not high using a graduated cylinder Medicine dropper lip Figure 1 Apparatus forthe nitritesulfamic acid reaction with displacement and collection of water A second way to get the volume of displaced water is to weigh the dry 250 mL beaker then weigh the beaker plus displaced waterfollowingthe reaction noting its temperature Subsequent division of the mass difference bythe density of water atthat temperature and conversion to liters gives a more precise displaced watervolume 4 VNZ L displaced water L g beaker Sq displaced water g beaker flii lOOO n1L L density ml water at T gmL 4 The nitrogen gas temperature TNZ C K is assumed to be that ofthe water displaced Obtainingthe nitrogen gas pressure PM mm Hg is a bit more complex But following Dalton s Law of Partial Pressures it may be easily obtained by subtractingthe watervapor pressure PHZOS mm Hg at that collection temperaturequot from the ambient air pressure P2 mm Hg that day 5 which will be provided foryou at the beginning ofthe lab period Of course any calculation involving 2 must include a 760 mm Hgatm unit factor conversion Pm mm Hg Pam mm Hg Pms mm Hg 5 The watervapor pressure PHZOS mm Hg at that collection temperature TNZ C K may be obtained from the listing TNZC K 17290 18291 19292 20293 21294 22295 23296 24297 PW mm Hg 145 155 165 176 187 199 211 Safety Preca utions 0 Protective eyewear approved by your institution must be worn at all times while you are in the laboratory 0 Use the proper technique and great care when inserting glass tubing through a rubber stopper Serious cuts to the hand can occur quite easily if this task is performed incorrectly Be sure that the stopper hole is lubricated and wrap the glass tube in a towel or lab apron before attempting to insert Grasp the glass rod near the point of insertion to minimize torque on the glass EXPERIMENTAL PROCEDURE Assemble the apparatus shown in Fig 1 Make sure each flask is securely clamped CAUTION Use the proper technique and great care when inserting glass tubing through a rubber stopper Serious cuts to the hand can occur quite easily if this task is performed incorrectly Be sure that the longer piece of glass tubing entering the 500 mL Florence flask extends nearly to the bottom Clean the 250 mL Erlenmeyer flask thoroughly and rinse it with distilled water It need not be dry The Florence flask will serve as a water reservoir Nitrogen will be produced in the Erlenmeyer flask and then pass through the rubber tubing into the Florence flask forcing water into the 250 mL beaker The volume of displaced water collected in the 250 mL beaker equals the volume of nitrogen gas formed A medicine dropper tip should be inserted into the exit end of the rubber tubing this will help to prevent water from flowing out of the rubber tube into the 250 mL beaker Clean dry and weigh the 250 mL beaker Fill the 500 mL Florence flask almost completely with water Remove the stopper from the Erlenmeyer flask and attach a pipette bulb to the glass tubing extending through the stopper Halffill another 250 mL beaker with water With the pinch clamp off the rubber tubing pipette bulb forcing water from the flask through the rubber tubing and out into the beaker Pinch the rubber tubing leading from the Florence flask to the 250 mL beaker to dislodge bubbles of air Remove the pipette bulb Verify that there are no air bubbles in the rubber tubing glass tubing and stopper components The presence of any air bubbles in these components will adversely affect the accuracy of the experiment Ifany air bubbles are stillpresent dislodge them by pinching or forcing water through the tubing by squeezing the pipette bulb Obtain a 10 x 75 mm glass culturetest tube Test the fit of this tube inside your 250 mL Erlenmeyer flask while both are dry and empty This tube must be too long to lay flat across the bottom of the Erlenmeyer flask If it isn t replace either appropriately Once this proper fit has been established remove this tube from the Erlenmeyer flask Now add 250 mL of 00825 M sulfamic acid solution to the Erlenmeyer flask followed by 65 mL of distilled water Obtain a vial of unknown nitrite salt and weigh it to 0001 g accuracy Place all of the unknown in the small glass test tube Weigh the empty vial The difference 0306 g is the mass of the unknown Add 34 mL distilled water to this small test tube and being careful not to splash tap the side of the tube repeatedly to dissolve the unknown tilt the Erlenmeyer flask now containing 100 mL of sulfamic acid solution to an angle and carefully slide the small tube now containing 4 mL of unknown solution down into it so that no contents ofthe tube prematurely mixes with the solution in the flask Fig 2 Figure 2 An quotangledquot sulfamic acid solutioncontaining 250 mL Erlenmeyerflaskwhich has just received the small test tube holding unknown nitrite solution Carefully set the Erlenmeyerflask down Raise the watercontaining 250 mL beaker in orderto siphon enough water back into the Florence flask so that the water level in it is up to its neckbottom as shown Then quickly and securely tightly stopperthe Erlenmeyerflask reactorwith the rubber stopperquot An accurate determination ofthe amount of N2 gas formed requires that all connections in the quotwater and gas line system be airtight Test for leaks by lifting and loweringthe medicine droppertip out of the water remaining in the beaker When there are no leaks in the system no more than 23 water drops will be observed Now close the pinch clamp on the rubbertubing and carefully remove the quotmedicinedroppertip from the watercontaining beaker carefully placing it in the dw previously weighed 250 mL beaker last change to weigh it if you have not already done sol Open the pinch clamp on the rubbertubing Unclamp the Erlenmeyerflask from its ring stand tilting the flask enough to get just a little bit ofthe solution in the small test tube to mix with the sulfamic acid solution Do not mix tooast orthe rapidlyproduced N2 gas may form so quickly that it will blow the stopper out of the tube spraying acid on you and your neighbors Water will start flowing into the previouslydiv 250 mL beaker as soon as N2 gas bubbles begin to form After about a minute repeat the process swirlingthe reagents once more Continue this sequence systematically until both solutions have been thoroughly mixed and no more N2 gas bubbles form When all the unknown solution has been reacted adjust the pressure inside the system until it equals that in the lab To do this raise or lowerthe 500 mL Florence flask andor beaker until all upperwater surface levels are at a uniform height Now close the pinch clamp Finally carefully remove the quotmedicinedroppertip from the beaker measure the water temperature and finally weigh the beaker and water Aftenlvard with a graduated cylinder measure the volume of water in the beaker From the water mass and its density at the recorded temperature you may confirm or correct the apparent volume of water displaced Perform all need calculations Identifythe cation Na or K in your unknown nitrite reagent and complete your report sheet Make sure your TA initials your data sheet before you leave laboratow that day Stoichiometry and the Ideal Gas Law Preparation of Nitrogen Gas and Identification of the Cation Present in an Alkali Metal Nitrite Salt Student Name Partner Name RESULTS 1 MNOZ unknown number 2 Mass of vial and Unknown 3 Mass ofempty vial 4 Mass of unknown 5 Volume of displaced water 6 Volume of displaced N2 gas 7 Atmospheric pressure that day 8 Displaced water r 9 Displaced water vapor pressure 10 Displaced N2 gas pressure 11 Calculated number of mo Nzgas 12 Calculated number of mo MNOZW 13 Calculated molar mass M 14 Error in calculated molar mass M Error 15 Identity of alkali metal cation M CHE 201 Section T A Initial mL mm Hg mm Hg K also N2 gas temp mm Hg mol N2 moMNOZ S g Mmo M include the sign 100 lg Mmo M observed value g Mmo M true value The calculation used to obtain 11 above g Mmo M true value Common Cations and Anions Monoatomic Cations H Hydrogen Lil Lithium Na Sodium Kl Potassium Cs Cesium Agl Silver Cul Copperl cuprous Sr2 Strontium C02 CobaltH cobaltous Mn2 ManganeseH manganous Ni2 NickleH nickelous Sn2 TinH stannous Monoatomic Anions H Hydride F Fluoride Cl Chloride l Iodide Polyatomic Cations Hg22 Mercuryl mercurous H3O Hydronium Polyatomic Anions N027 Nitrite 3032 Sulfite HSO4 Hydrogen sulfate CN Cyanide HPO42 Hydrogen phosphate C0327 Carbonate ClO Hypochlorite C1037 Chlorate C2H3027 Acetate MnO4 Permanganate Cr2072 Dichromate NHJ Calcium Aluminum Barium Zinc Ironal ferrous Ironall ferric Magnesium Cadmium CopperH cupric MercuryH mercuric LeadH plumbous ChromiumHI chromic Oxide Sulfide Bromide Nitride Ammonium Nitrate Sulfate Hydroxide Phosphate Dihydro gen phosphate Hydrogen carbonate Chlorite Perchlorate Peroxide Chromate Gases Gases Important Parameters V volume T temperature N number of moles P pressure Pressure Pressure is the force per unit area of surface P Forceunit area F A Newton s law I ma Kgms2 Kg constant acceleration of gravity 981 msz P KgmS392m2 Kgms2 Pascal Pa Atmospheric Pressure Patmz h mmHg Vacuum Mercury I39Ig h mull Atmospheric p ssure 1 Units of Pressure 1 mmHg torr is the pressure exerted by a Column of Hg 1mm high at 000 C 1 atm 760 mmHg 760 torr 147 psi Units of Pressure Important Units of Pressure Unit Pascal Pa Atmosphere atm mmHg or torr Relationship or De nition kgm 2 1 atm 101325 x 105 Pa 100 kPa 760 11111ng 2 1 atm latm 101325 bar proportional Vacuum quot Gas Pressure to gas pressure Gas pressure 1 m7 Pgas Ah mmHg on mercury surface A Flask Equipped with a ClosedTube Manometer The Empirical Gas Laws Boyle s Law See Fiqure and Animgtion Boyle s Law2 V OL lP constant moles and T Pfof PigtltVi Animation Microscopic illustration of Boyle s Ialv Go to 13 End053d Encl oged gas gas E i I f q a 22 E l I Hg Hg HSEU mm Enclosed gas A E S I P Pgas760 mmHg Pgas1520 mmHg Pgas2280 mmHg An Example A sample of chlorine gas has a volume of 18 L at 10 atm lfthe pressure increases to 40 atm at constant temperature what would be the new volume using Pf x Vf Pi x Vi Pi x Vi 10 atmx 18 L Pf 40 atm Vf 045 L Vf The Empirical Gas Laws Chares s Law See Animation Charles s Law See Video Liquid Nitroqen and Balloons V on Tabs constant moles and P or constant moles and P L Ti Go to 17 An Example A sample of methane gas that has a volume of 38 L at 50 C is heated to 860 C at constant pressure Calculate its new volume Vf Vi use T f Ti V Vifo 38L359K f Ti 278K Vf 49 L See Video Collapsinq Can2 The Empirical Gas Laws Combined Gas Law In the event that all three parameters P V and T are changing their combined relationship is defined as follows VociVocT gt VocI gt P P P PVOCTgt Vconstant PiVi Pf Vf T Tf l An Example A sample of carbon dioxide occupies 45 L at 30 C and 650 mm Hq What volume would it occupy at 800 mm Ho and 200 C PV Pfo usmg JE39 T 1 f V P V Tf 650 mm Hg45 L473 K f PfTi 800 mm Hg303 K Vf 57L The Law of Combininq Volumes GavLussac At a qive Tand P the volumes of gases that react with one another are in the ratio of a whole numbers An example N2g 3H2g gt 2NH3g I I 1 3 2 The Empirical Gas Laws Avoqadro s Law Equal volumes of any two gases at the same temperature and pressure contain the same number of molecules The volume of one mole of gas is called the molar gas volume Vm Volumes of gases are often compared at standard temperature and pressure STP chosen to be 0 C and 1 atm pressure Avoqadro s Law At STP Vm 224 L For all gases The Ideal Gas Law constant T The constant is independent of temperature pressure but does depend on amount of the gas C t 39lt The constant is independent of temperature pressure and amount of the gas wn mta1tR nT Ideal gas law PV 2 nRT The Ideal Gas Law and the Molar Gas Constant PV 2 nRT For one mole of a gas at STP P 1atm T273K R PV 100 atm x 224 L HT 1 m01gtlt 273 K R 0082 Latm molK Using the Ideal Gas Law Haw many grams of oxygen are there in a 500 L gas cylinder at 21 C when the oxygen pressure is 157 atm PV 2 nRT 1 atm L mol 0082 K O C 273 n a 157 atm x 500 L RT 0082 atm L mol391K391 x 294 K m n x M gtlt m104gtlt103 g O2 325mol Animations for ideal gas law Gas Density from Ideal Gas LaW What is the density of O2 in gL at 25 CC and 0850 atm V 1L P 0850 atm T 25273K 298 K Using the ideal gas law we can calculate n then we can calculate the density PV 2 nRT n 0850 atm gtltl L RT 0082 Latm mo1391K 1 x 298 K 00347 mol m n X M 00347 m01gtlt 320 gm01391 111 g 02 d OfO2 at 25 CC and 085 atm 111 gL Gas Density and Molar Mass Determination PVnRT m n M PV3gtltRT gt PMERT M V PM dRT Molar mass of gas Density of gas An Example What is the density of O2 in gL at 25 CC and 0850 atm PM dRT Molar mass of gas Density of gas PM 0850 atm x 320 gmol391 d RT 0082 Latmmol391K391 x 298 K 111 gL Similarly we can calculate M if we know d Example At STP 0280 L of a diatomic gas weighs 0400 g Calculate the molar mass and the density of the gas using the ideal gas law Use the calculated molar mass and the periodic table to identify the gas Answer Using the ideal gas law PVnRT2PV RT MmRT M PV RT AtSTP T 273 K P 1 atm knowing that V 0280 L m 0400 g l l M mRT 0400g gtlt 00821Latmmol K gtlt 273K 320 gmorl PV l00atm gtlt 0280L M 100atm gtlt 320gmol1 1 1 143 gL 1 RT 00821Latmmol K gtlt273K The gas is an 02 gas Stoichiometrv Problems Involvinq Gas Volumes Use the ideal gas law V G 11 Example Air bags are in ated upon collision with N2 gas using The following fast reaction 6NaN3 s Fe203s gt 3Na20s 2Fes 9N2 g g of NaN3 are required to provide 750 L of N2 at 25 C and 748 mmHg Answer 6NaN3s FezO3 s gt 3Na20s 2Fes 9N2g 6 1110168 9 moles mol 750 L 302 mol Use the ideal gas law PV 2 nRT 748 760 atmx 750 L 0082 Latmmol 411 x 25 273K 11N2 302molofN2 mol OfNaN3 302gtlt69 201 mol NaN3 g OfNaN3 201 mol NaN3 x6501 gmol NaN3 131 g NaN3 Partial Pressures of Gas Mixtures Dalton 3 Law of Partial Pressures the sum of all the pressures of all the different gases in a mixture equals the total pressure of the mixture Where each gas in the mixture obeys the ideal gas law PAV nART A qas Mixture and Dalton s Law of Partial Pressures N HZPnmng i1 The gas mixture as well as each individual gas in the mixture occupies the same volume and obeys the ideal gas law where PtV ntRT RV niRT i123N n Xi 1 mole fraction Example A gas mixture at constant pressure 1 atm and temperature as air consists of 79 N2 20 02 and 1 C02 by volume a Calculate the average molar mass of the mixture b Calculate the mole fraction of each gas Answer At constant P and T I CC V gt by volume mole Average molar mass ZNZ gtlt MN2 102 gtlt MO2 mol C02 gtlt MCO2 79 20 1 4m gtlt280 gmol 1mgtlt320 gmol 1mgtlt440 gmol 1 290 gmol1 mol 2 E X 100 nt xi gtltlOO Check the answer 2 xi 1 i AXNZ x02 XCO2 0790200101 l Examgle 100 L of dry air at 25 C and 786 mmHg contains 0925 9 N2 other gases b What is the mole fraction of N2 in air P 1mmH m N2 6 3 g O780 2 P 786mmHg a1r c What is the mole percent of N2 in air N2 0780 x 100 780 of N2 Collectinq Gases Over Water A useful application of partial pressures arises when you collect gases over water See Figure As gas bubbles through the water the gas becomes saturated with water vapor The partial pressure of the water in this mixture depends only on the temperature See Table Animations for collection of gas over water A Problem to Consider Suppose a 156 mL sample of H2 gas was collected over water at 19 C and 768 mm Hg What is the mass of H2 collected First we must find the partial pressure of the dry H2 PHZ Ptot PH20 From tables the vapor pressure of water at 19 00 as 165 mm Hg PH2 768 mm Hg 165 mm Hg PH2 752 mm Hg PHz 752 mm Hg x la tm 0989 atm 760mmHg V 156 mL 0156L T19273292K n n g 0989 atm0156 L RT 00821 Latm292 K mol K n 000644 m0 Nextconvert moles of H2 to grams of H2 m 2000644 molH gtlt H2 2 H2 00130 g H2 The Van der Waals Equation of State n2 m a zjw nb nRT V 21112 V2 take account of the attractive forces nb corrects for the finite volume of individual molecules Effects of Attractive Forces on the Pressure of A Real Gas Effects of attractive forces and speed on the pressure of a gas As the attractive forces and the speed increase the pressure of the gas increases a z z D m I w EU a u CD HG m Z Liquot Er mam 32 m Him 33 a we 33 32 33 3 23 E 2 SEE N NE A A q A ES Em uz A 3 3 m 39 in IF 5 an 41 lquot 1 1 5 H quotI r 5 H Fl fquot Lf l 1 H m 11 m quot5 39339 F 39139quot f r M 139quot w rg b r n rI a FE 139 394 Egg WEZDE HLEE 1H3 TEE m4 F35 gucn jm Ee mnam Enm E maEni mezzo E EE m tn 555 emu Em ribs EEG a 65 m uo 05me mowgtn ms The Kinetic Molecular Theory of Gases The ideal gas and real gas equations of states provide equations that allow us calculating gaseous parameters but they do not provide physical explanation for gaseous behavior on the molecular level For example the kinetic molecular theory of gases provides answers for following questions How the pressure of a gas is related to the motion of individual molecules What happens to the gas when it is heated The Model The model concerns a gas that consists of a very large number of particles atoms or molecules with the following assumptions iaiThe particles possess mass but their volume is negligibly small The particles are in constant random motion collision between particles and walls of the container are elastic collisions See Figure iiDistance between individual atoms and molecules are much larger than sizes of the particles Electrostatic forces attractive and repulsive are neglected but elastic collisions between individual gaseous molecules and walls of the container are considered iiThe average kinetic energy of a collection of particles is proportional to the absolute temperature K See animation Kinetic molecular theory of qases The Maxwell distribution law Cig 70 an g mam NMQEHQ g mquot Tzsun K Na 28 02 9 mm Numberm mniemie Numberm mubcuiss Ha m an g mm D 5am mm mun ltan Maiacuiarspead m squot Maia2mm weed m 1 mi Go to slide 43 How the pressure of a gas is related to the motion of individual molecules let us consider a gas consisting of N particles moving in a three dimensional space in this case the average pressure of the gas can be expressed as N 2 P 2 m0 gt mu 2 Average of squares VNgtOf velocities Volume The average kinetic energy or average translational energy of an N particle system is related to the average of the squares of velocities as 1 2 KE 2 Etrans Emu 1 2 3 PV KE Etrans Emu gt Etrans EkT PVNkT k RNA 2N P W Etrans What happens to the gas when it is heated The average kinetic energy of a gas depends on temperature and mass 1 3 Etrans Emu EkT Energy per molecule of a gas Energy per mole of a gas U V See Simulation slide 502 The various speeds of Maxwell Distribution cm 2RT Most Probable Speed p M E SRT Average Speed 72M 01mg 2 152 3113 Root Mean Square Speed GO to slide 40 Example Calculate the values of 0 mp c and elmS for 02 at 300K Answer 2 8314 K39l 1 1 Cmp x J m0 300K2395m51 003200 Kgmol39 Similarly E lgR T 446 ms391 72M l3RT 1 Crms Y HIS J SI unit of energy R 8314 JK391mol1 J Kgmzs2 Molecular Speeds Diffusion and Effusion Diffusion is the transfer of a gas through space or another gas over time See Animation Diffusion of a Gas also see animation slide 52 Effusion is the transfer of a gas through a membrane or orifice See Animation Effusion of a Gas see slide 53 The equation for the rms velocity of gases shows the following relationship between rate of effusion and molecular mass Rate of c hsion oC L N Molecular Speeds Diffusion and Effusion According to Graham s law the rate of effusion or diffusion is inversely proportional to the square root of its molecular mass Rate of effusion of gasquotAquot M of gas B Rate of effusion of gas quotBquot M of gas A A Problem to Consider How much faster would H2 gas effuse through an opening than methane CH4 Rate of H2 MCH4 Rate Of 4 Rate of H2 160 gmol 2 8 Rate of CH 4 20 gmol o 80 hydrogen effuses 28 times faster than CH4 Summary for Kinetic theory of gases 3 1 E mU2 NkT Energy per N molecule of a gas trans 2 2 3 Energy per n moles of a gas Etrans EMU EDRT Rate of effusion of gasquotAquot M of gas B Rate of effusion of gas quotBquot M of gas A J SI unit of energy R 8314 JK391mol1 J Kgm2s2 Collection of Gas over Water P Hydro en g partial pressure 752 mmHgl HXglOcmorw wixh waler vapor ac partial pressure 7 mmHg Return to Slide 32 Figure 516 Elastic collision of steel balls The ball is released and transmits energy to the ball on the right Photo courtesy of American Color Return to Slide 39 Animation Kinetic Molecular Theory Return to Slide 42 Molecules in Motion II httpmc2CChemBerkeleyEDUJavamol eculesindexhtm Animation Diffusion of a Gas Return to Slide 45 Animation Effusion of a Gas Return to Slide 45 ChaQter 4 Reactions In Aqueous Solutions What is a solution A homogeneous mixture of two or more substances When we dissolve a substance in a liquid The substance is the solute and the liquid is the solvent When the solvent is water The solution is an aqueous solution Examples of Aqueous Solutions A Cup of Coffee Coffee Sugar Water Sea Water Salt other solutes Water Concentration The general term concentration refers to the quantity of solute in a standard quantity of solution Molar Concentration Molar concentration or molarity M is defined as the moles of solute dissolved in one liter cubic decimeter of solution moles of solute Molarity M o llters of solutlon n moles M IndL Z V liters An Example Calculation of molaritv from number of moles of Solute and volume of the solution Asample of 00341 mol iron chloride FeCI3 was dissolved in water to give 250 mL of solution What is the molarity of the solution number of moles of FeCl3 Volume of solution in liters Molarity M 00341 mole of FeCl3 00250 liter of solution 136M Feel3 An Example Calculation of mass of the solute from molaritv and volume of the solution How many grams of NaCI are required to prepare 500 ml of 02M aqueous solution M 3 molL V nMgtltV moles n O2molL1 gtlt Lml 01 mole 1000m1L391 mass g nmoles gtlt MMgmol391 mass O1molegtlt 585 gmol39l 585 g Dilutinq Solutions Upon addition of water to a solution dilution the number of moles of a given substance remains constant An Example Preparation of dilute solutions from concentrated stock solutions For 21 12M stock solution of HCl how many ml of this solution gives 500 ml of 100 M solution Mixliszfo Mf fo 100 molL391gtlt500 ml Mi 12 molL 3911000mlL391 42 mL 2 0042 L Vi Solution Stoichiometerv An Exalee How many ml of 300 M HCI are required to react with 3000 g of Zns Answer Zns 2 HC1300M aq gt ZnC12aq H2g 1 m0 2 mole 1 mol 1 mol 3000 g mole gt V 04588 moles 09176 moles 3000 g Zn 1 6539 gmol 04588 mole Zn x 2 mole HCl 1 mole Zn 2 09176 mole 09176 mole HCl 300 molL391 HCl 2 04588 mole nHC1 nMgtltVgtV 0306L306 m1 Types of Solutions What happens on the microscopic level When a compound is dissolved in water Nature of the compound Ionic Compound Molecular Compound H2O Type of Solution Ionic Compounds 39 Cl Na A model of a portion of sodium chloride crystal What Happens when sodium chloride is dissolved in water NaCl gt Na Cl Dissociation of ionic compounds in water make their solutions good conductors to electricity electrolytes Molecular Compounds CH3OH Methanol p O D c 39 3 L C o Methanol molecular compound dissolves in water but does not dissociate to ions nonelectrolytes Weak Electrolytes Produce small number of ions in aqueous solutions Acetic Acid completely dissolves but partially dissociates in water CHsCOOH H20 H30 CHscoo Stronq Electrolytes Strong acids Strong bases Soluble ionic salts Precipitation Reactions Examgle Reaction of AgN03aqwith NaCI aq AQNOB aq KCI aq gt AgCI S KNO3 3 I Insoluble ionic compound AgCI Gravimetric Analysis Includes all analytical methods in which the final stage involve weighing Gravitational methods that employ precipitation reactions are commonly used in laboratories l Precipitation followed by isolation and weighing the dry precipitate An Example Gravimetric Analysis Consider the problem of determining the amount of lead in a sample of drinking water Adding sodium sulfate NaZSO4 to the sample will precipitate leadll sulfate NaZSO4aq Pb2aq gt 2Naaq PbSO4s The PbSO4 can then be filtered dried and weighed Soluble and insoluble ionic compounds Solubilitv quidelines for common ionic compounds in water Soluble Ionic Compounds Important Exceptions Compounds containing N037 321 1302 Cl None None Compounds of Ag Hg22 and Pl2 Compounds of Ag Hg22 and Pb2 Compounds of Ag Hg22 and Pb2 Compounds of 812 Ba2 Hg22 and Pb2 Insoluble Ionic Compounds Important Exceptions Compounds containing 82 C032 PO43 OH Compounds of NH4 the alkali metal cations and Ca2 512 and Ba2 Compounds of NH4 L and the alkali metal cations Compounds of NHr and the alkali metal cations Compounds of the alkali metal cations and NH4 Ca2 Sr2 and Ba2 Metathesis Exchanqe Reactions Metathesis comes from a Greek word that means to transpose AQNOB aq KCI aq gt AgCI S KNO3 6 l Insoluble ionic compound Molecular Equation The molecular equation lists the reactants and products in their molecular form AgNO3 aq KCI aq gt AgCI s KNO3 aq