General Chemistry I (GT
General Chemistry I (GT CHEM 111
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Popular in Chemistry
This 66 page Class Notes was uploaded by Marilou Hyatt Jr. on Tuesday September 22, 2015. The Class Notes belongs to CHEM 111 at Colorado State University taught by Dana Johnson in Fall. Since its upload, it has received 20 views. For similar materials see /class/210345/chem-111-colorado-state-university in Chemistry at Colorado State University.
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Date Created: 09/22/15
3 September Announcements Last day for survey easiest homework points Morecolostateeduclass Recitation still meets next week no lab though ALEKS due date for 1st Objective Tuesday 97 at 5 pm Makeup Toledo for people who registered late Wednesday 98 at 7 pm room TBA Electromagnetic EM Spectrum Visible light a part of a continuum of radiation Wave nature of light Speed all EM radiation moves at the same speed in a vacuum namely the speed of light C 300 x 108 m s Wavelength A distance from one point on a wave to the corresponding point on the next units of nm pm or angstroms A 103910 m Frequency v number of waves per second units of 1 sec sec l or hertz All 3 related by 9 cxv CAxv A and v are inversely related Clicker question IR radiation is used by chemists to determine what type of bonds are in a molecule A C 0 bond will absorb IR radiation of wavelength 96 pm What is the frequency of that radiation in Hz c 300 x 108 m s 3125Hz 31 x 1013 Hz 32 x 1039 Hz 320x 1039 Hz 3125 x 1013 Hz mpow Clicker solution CAxv9 solveforv9vc C 300x 108m s A 96 pm x 10 6 m um 96x 10 6 m v 300x 108 m s 96x 10396 m v 3125 x 1013 s1 or Hz v 31 x 1013 Hz 2 sig figs Other wave properties Amplitude height of a crest of a wave connected to intensity of color not energy of the wave Refraction when a wave passes into different media its speed changes which changes its angle of motion Diffraction seen when a wave bends around the edge of an object can result in interference Constructive interference the waves are in phase add up to larger amplitude Destructive interference the waves are out of phase cancel each other out to little or no amplitude Particle properties Blackbody radiation Blackbody idealized object in physics that absorbs any radiation that falls on it hot coals or ovens are real examples Radiation referring to the energy a blackbody can absorb or emit was not found to be continuous as expected by the wave picture of light Max Planck 1858 1947 Summarized the data of blackbody radiation into the following formula E nhv E energy in J n positive integer 1 23 ampc h Planck s proportionally constant 6626x 1034 s v frequency in s l Implications of E nhv n quantum number implied the energy could only be absorbed or emitted in distinct packets of energy These packets came to be called quotquantaquot If the radiation is due to the atoms that make up the blackbody 9 the atoms themselves were quantized were separated into distinct energy levels Rather than seeing the atom as a simple sphere of electrons around a nucleus there were now quantized energy levels Clicker question An xray has a wavelength A of 13 What is the energy in J associated with this wavelength A 103910 m h 6626x 1034 s c 300x 108 m s A 153 x 10451 B 15 x 10441 C 153 x 10441 D 15 x 10451 Clicker solution We know E hv but there s no A in that equation ButcAv9vc If we substitute in for v we get E hc h 6626 x 10341 s c300x108ms A 13 xx 10 10 m NS 13x 103910 m E 6626x 1039 3 x 108 13 x 103910 E to 2 sig figs 15 x 10451 5 November Announcement No class Friday before Thanksgiving 11 19 Mark your calendar Friday after Exam V Clicker question You have two oneliter containers each filled with a different gas These containers are connected via a closed port There are 025 moles of gas A at a pressure of 030 atm in one container and there are 050 moles of gas B at a pressure of 060 atm in the other When the port is opened so the gases freely mix what is a correct statement A The total pressure is now 090 atm B The total mass of the gases is 075 g C XA 05 D The volume has effectively doubled E None of these are correct Solutions intro Solutions homogeneous mixtures Mixed on molecular level Can separate by physical means Solvent larger component Solute smaller component Solute dissolves in the solvent Water as a solvent Strong dipoles Bent shape gives polarity Ubiquitous everywhere Ionic compounds dissolving To dissolve an ionic compound means the strong ionic forces must be broken Water molecules will rearrange themselves for many iondipole interactions which break apart the ions Each ion then has a quotshellquot of water molecules in solution Solubility Measure of a compound s ability to dissolve Different ionic compounds have different solubilities Solvent and temperature must be specified Iondipole interactions not as strong as ionic bonds Units vary amount of solute volume of solution Indicates maximum amount of material that will dissolve NaCl in H20 at 20 C 365 g L NaCC03 in H20 at 20 C 100 g L Maximum amount that will dissolve in solution is saturated Dissolving reactions Can write an equation to depict solvation NaCl s Na aq Cl aq S 9 solid Aq aqueous H20 over arrow to indicate solvent not a part of reaction Note one mole of NaCl 2 moles of ions Coefficients give moles of the reaction Clicker question How many total moles of ions are in solution assuming that 19042 g of MgClz dissolves completely in water 1 moles 2 moles 3 moles 4 moles 6 moles mpowP Ions in solution Electrolytes dissolved ions that can conduct electricity Nonelectrolytes will dissolve but not ionize do not conduct electricity 8 December Final exam room Chemistry A103 Announcements 2 lectures left Final model assignment Due beginning of class on Friday Can certainly turn in early ALEKS extended This week s due date is tomorrow at 5 pm Office Hours Friday in Yates 103 Check RamCT for Finals week office hours Final Exam rooms Calculating AHrxn Remember H is a state function AHrxn is only based on intial and final states Therefore we can picture any process for a reaction The simplest is to think of all bonds breaking then reforming This is not what actually happens but it works The AHrxn formula Aern EAl39limndsbroken ZAl39liwndsformed Recall 2 sum of Signs are essential Clicker question If a reaction is endothermic what does that say about the relative stability of the products vs the reactants A You cannot make a statement regarding their stability B Products are more stable C Products are less stable Low energy 9 more stable Endothermic 9 takes energy to get to products Aern example CH4 g 202 g 9 C02 g 39l39 ZHZO g ExothermicAHrxn is negative Clicker question What is the Aern of the following reaction and is the reaction endo or exothermic PClg I Clz g 9 PC5 s 419 kJ exothermic 243 kJ endothermic 243 kJ exothermic 419 kJ exothermic porn P Cl 331 kJ mol Cl Cl 243 kJ mol AH reactants 1236 k AH products 1655 k Thermochemical equation A balanced equation with its AHrxn As in ZHZO I 9 2H2 g Oz g Aern 572 kJ Sign of AHrxn is connected to direction of the reaction So 2H2 g 02 g 9 ZHZO I AHrxn 572 kJ Additional stoichiometry Thermochemical equations connect energy to amount AHrxn is determined by number of bonds present Therefore if ZHZO I 9 2H2 g Oz g Aern 572 kJ Then H20 II 9 H2 g y 02 E Aern 286 kJ everything divided by two Sample caIcuIation Given MgO COZ9MgC03AHrxn 1173 k What is AH when 355 g of COD reacts with excess MgO 355 g x 1 mol COz44 g x 1173 kJl moI C02 946 k 1 November Announcements Model assignment Intermolecular forces Due beginning of class 113 Refer to form to help shape your discussion Gas observations Low viscosity Does not resist flow Moves more easily than liquids and solids Generally low densities Measured in grams per liter Liquids and solid 9 grams per milliliter Miscible Mix together easily in any proportion Liquids depend on polarity Kineticmolecular theory Postulate 1 particle volume Volume of the particles is much less than the volume of the container Gas is a collection of point masses with empty space between Postulate 2 particle motion Particles are in constant motion They move in straight lines until they collide with something Postulate 3 particle collisions Collisions are elastic Energy exchanges and direction changes Particles do not attract or repel one another 2 factors from the above Speed of motion Collisions give a variety of speeds at any one time Higher temperatures 9 faster average particle motion Origin of pressure Gas molecules move until they hit something Those collisions exert energy Pressure Z the energy of gas colliding with the container Gas variables State of a gas can be described as Volume V Pressure P Temperature T Amount n P T and n were shown to have specific linear relationships with V I As a result these are interdependent variables that combine to form the quotideal gas law Units of pressure Pressure force area Newton N SI unit of force kgm s2 Pascal Pa S unit of pressure M m2 Atmosphere atm atmospheric pressure at sea level at 0 C 1 atm 101325 kPa 1 mmHg 1 torr 1760 atm 1 bar 100 kPa Volume vs pressure Boyle s Law V is inversely proportional to P As one gets larger the other gets smaller eg if V triples then P drops to 13 A linear relationship between V and 1 P V is proportional to 1 P T and n held constant Charles s Law V directly proportional to T As one increases so does the other eg ifT doubles in Kelvin V doubles as well Linear relationship between V and T V is proportional to T P and n held constant Clicker question What would the final volume of a gas in terms of an initial volume V be after 1 the pressure on the gas is doubled 2 the temperature of the gas is lowered from 300 to 100 K A 23 V B 16 V C 6 V D 32 V E No way to know Absolute zero Different gases still show the same linear relationship between V and T If those lines are extended they converge This temperature became known as llabsolute zeroquot Lord Kelvin discovered the kelvin scale based on gas expansion by temperature Volume vs amount Avogadro s Law V is directly related to n As one gets larger so does the other eg if n doubles so does V A linear relationship between V and n V is proportional to n T and p held constant Clicker question If you know the starting n P V and T of a sample of gas and you note that the pressure of your sample fell by half what could NOT explain what happened The size of the container doubled The temperature fell from 150 to 75 C The amount of gas doubled A amp B B amp C mpow 11 October Clicker question from Friday What is the molecular formula for vitamin C from the information that was provided C3H403 CsHsos C5H705 C1H1301 didn t do it mpow Molar mass of vit C 17612 g mol Mass of vit C is 1000 g Mass of C02 15 g Mass of H20 041 g Mass ofC 1201 4401 x 15 g 0409 g Mass of H 1008 18016x 041 g 0046 g Mass of O 1000 g 0409 g 0046 g 0545 g Moles ofC 0409 gx 1 mol 1201 g 00341 moles Moles of H 0046 gx 1 moi1008 g 00456 moles Moles of O 0545 gx 1 mol 16 g 00341 moles C00341 00341 H 0 0455 00341000341 00341 C1 H 133 01 X 3 C3H403 Compare to molar mass 3 x 1201 4 x 1008 3 x 16 88 1761288 2 CngOs EN trends Pauling s scale puts F at the top with a value of 40 Inverser related to atomic size Electrons pulled by larger Zeff smaller size Atom behaves the same in a bond Partial charges result 6 and 6 Generally increases as you move up and to the right onic and covalent bonds Form a continuum Cutoff between the two labels is arbitrary Usually made using AEN Lewis structures Way to visualize bonding relationships in a molecule Built up from Lewis dot symbols Based on the octet rule Basic process single bonds Place least electronegative atom in the center Lower group number higher period number Typically this atom will need to make the most bonds Generally listed first in molecular formulae Count number of valence electrons Account for any charges Draw single bond between central atom and any atoms surrounding it Subtract 2 e for each bond drawn Use remaining electrons to fill each atom s octet CBr4 4 74 32 valence electrons 8 used in bonds 24 remain fill octets All used PCI3 Valence electrons 9 5 73 26 6 used in bonds 20 remain fill octets 2 remain fill phosphorus Things to remember H and F are never central atoms In general H F and halogens as surrounding atoms form one bond 0 forms 2 bonds N forms three bonds C forms four bonds Lewis structure does not indicate shape Higher bond orders If the central atom still has an incomplete octet use lone pairs to increase the number of bonds NOS Valence e 5 63 1 negative charge 24 6 in bonds 18 remain N does not have a full octet Clicker question In the correct Lewis structure for C2F4 how many lines have you drawn for bonds F1909quot GnuIwa 15 September Announcements Exam tonight 7 9 pm Double check Clark room Only pencil calculator TI30Xa only and ID No hats and no bathroom breaks Cell phones must be off Reflection exercise Recitation grade about your study skills Accessible through RamCT Can access 9 pm tonight through 6 pm Friday You have 2 hours to finish Grades posted after de Broglie wavelength All matter shows wave properties de Broglie wavelength is calculated using the following A h mv A wavelength h Planck s constant m mass of the object in kg u speed of the object in ms Planck s constant in different units J kg m2 s2 SohJskgm2s Clicker question Can you see the de Broglie wavelength of a football when you throw it No Heisenberg uncertainty principle Macroscopic particle wave distinction easy Moving particle has definite location Predictable motion Microscopic difficult to distinguish properties Moving particle spread out like a wave Becomes impossible to know both speed and location Uncertainty principle results AxmAu2 h 4n Ax uncertainty in position Au uncertainty in speed m mass in kg h Planck s constant in kgm2 s Example with uncertainty How accurately can an umpire know the position of a baseball m 0142 kg moving at 1000 mih i 100 447 ms i 100 AxmAu 2 h 4n 9Ax 2 h mAu4n Au 44700100 0447 ms Ax 2 6626x 1039 kgm2s 0142 kg 0447 ms 4n 2 831 x1034 m Small number 9 low uncertainty Quantum mechanics Relatively new field developed out of the particle wave duality and Heisenberg s uncertainty principle Makes statements about wave like objects at the atomic level whose positions are impossible to know exactly Shro39dinger s equation describes the position of the electron in 3 dimensions HllJ EllJ H Hamiltonian operator E Energy of the atom W Wave function description of the electron Each solution is a different quotorbitalquot each is a different energy state completely different than the Bohr quotorbitquot llJ vs W2 Lu Wave function Mathematical description of the electron in 3 dimensions No physical correlation Probability density Gives us an idea of where the electron would most likely be Correlated with our quotelectron cloud concept Quantum numbers Remember that each solution of Schro39dinger s equation gives a different orbital Each orbital is specified by 3 quantum numbers in a hierarchical relationship Principal quantum number n Indicates size of the orbital 9 distance from the nucleus Positive integers 1 2 3 ampc The higher the n the higher the energy In Indicates the atom s energy level or quotshel Angular momentum quantum number I Indicates the shape of the orbital Only have integer values from 0 to nl If n 3 then you can only have I 0 1 2 Notice of l values n Indicate energy sublevels or subshellsquot Correlated with letters I 0 9 s sublevel s for sharpquot I 1 9 p sublevel p for principalquot I 2 9 d sublevel for dif fusequot I 3 9 fsublevel for fundamentalsquot Sublevels are named by joining the energy level number n with the letter abbreviation forl 9 1s 2p 4d ampc Magnetic quantum number m Indicates the orientation of the orbital in space Integer values from 2 through 0 to 2 IfI 2 then m 2 1 0 1 2 For each energy shell n the number of orbitals in that shell n2 Clicker question Can you have a 2d sublevel Yes No it has the wrong magnetic quantum number No it has the wrong angular momentum number No it has the wrong principal quantum number porn Clicker solution For a 2d suborbital n2 2 But I can only have integer values from 0 to nl In this case I 0 It cannot equal 2 so wrong angular momentum number The s orbital Spherical in shape Larger n 9 larger sphere further out from the nucleus The p orbital Dumbbell shaped Nodal plane at the nucleus Oriented on the 3 axes The d orbital 4 of the 5 possibilities have 4 lobes each The 5 h has a donut llRidiculously shaped 12 November Announcements Exam sections Bond amp molecular polarity intermolecular forces gas laws solutions chemical equations precipitation reactions 95 103 121 123 51 53 54 56 41 42 43 33 Problem session in class Monday 11 15 no clicker points optional day Review session Monday 11 15 6 pm in A101 Refined model assignment Form on RamCT Due by Wednesday 11 17 Clicker question When you mix K3P04 aq and CaN032 aq to form Ca 3PO42 s and KNOg aq what is the stoichiometric ratio of Ca 3PO42 s to KNOg aq in the molecular equation A 12 16 C 23 D 21 E 31 2K3PO4 aCIl l39 3C3lNO3l2 MI 9 C33PO42 5 l39 6KN03 aCIl Precipitation reactions Two soluble ionic reactants forming a precipitate The precipitate is the insoluble solid that falls out of solution Precipitation rxns cont d General formula AB aq CD aq 9 AD s CB aq Ions switch places with each other quotDoubledisplacementquot reaction If one of the combinations is insoluble you have a precipitation reaction Insolubility chart Once you know the possible products refer below to determine if any of them are insoluble Chapter 43 Page 148 Acidbase reactions Use simplest definitions of and Acid produces H ions when dissolved in water HA 9 H A Base produces OH39 ions when dissolved in water BOH 3 OH39 Acid base strength The terms quotstrongquot and quotweakquot are applied to acids and bases depending on the completeness they ionize dissolve in solution Strong acids bases completely ionize HCl HBr H2504 sulfuric acid HN03 nitric acid Weak acids bases ionize very little HC2H302 acetic acid HF These terms do not apply to reactivity HF is a weak acid that can etch glass The H Acid dissociation can be written 2 ways HCI g H aq Cl39 aq But the H is a proton Higth charged in a small volume Reacts with the solution Haq H20 I 9 H30 aq Therefore HCI g H20 I 9 H30 aq Cl aq 25 October Announcements Review session tonight A103 8 10 pm Recitations this week Exam preparation All Tuesday slots are open Orbital overlap review Bonds form when partiallyfilled orbitals overlap Atomic orbitals can overlap or form hybrids first Hybridization is given by the number of electron groups H C EC H C has 2 groups 2 hybrid orbitals 9 2 atomic orbitals Sp hybridized Clicker question For the molecule TeF539 which statement below is false The Te atom is sp3d2 hybridized The bond angles are all lt 90 The Te F bond is formed from 2 different orbitals overlapping There are no lone electron pairs around Te The orbital diagram for Te would have unhybridized d orbitals 5399575 Polar bond definitions Nonpolar covalent bonds result when the atoms of a bond have the same electronegativity electrons shared equally Polar covalent bonds result with a moderate difference of electronegativity unequal sharing partial charges The polarity of the bonds of a molecule affects macroscopic characteristics Molecular polarity Bond polarity affects the polarity of the molecule Molecules are polar if there is a net inbalance of charge Polar bonds do not guarantee polar molecules Relating shape to polarity is essential Dipole moment u measurement of polarity charge times distance in units of debye D Nonpolar molecules u 0 D Polar molecules ugt 0 D Examples Nonpolar molecules C02 SiCl4 SF5 Note symmetries Polarity cancels out Polar molecules H20 NH3 CHCI3 Note asymmetries Net charge on the molecule Results in dipole moment 1 September Isotopic abundance Isotopes atoms of an element with a different of neutrons can vary the mass of the elements most are chemically equivalent exist in different amounts For Carbon 12C carbon 12 most abundant 6 no 9889 13C carbon 13 7 n 111 C carbon 14 s n lt001 Affects weight on periodic table Reported weight is an average of all isotopes taking into account their abundance llAtomic massquot or llatomic weight Used for calculations Calculating atomic mass Atomic mass unit amu standard of measure for all atomic masses 112 the mass of the carbon 12 atom also called the Dalton Da Total weight 2 portions of each isotope Portion of each isotope mass of isotope x abundance Clicker question Copper has 2 naturally occurring isotopes 63C with a mass of 629396 amu 65C with a mass of 642978 amu If copper has an atomic weight of 63546 amu what is the percent abundance of each isotope Clicker solution The abundances of the 2 isotopes must equal 100 So set one equal to x and the other to 1 629396x 6429781 x 63546 629396x 642978 642978x 63546 629396x 642978x 63546 642978 19882x 13818 only4sig figs though X 13818 19882 X 0695001 9 06950 4 sig figs 9 6950 X 1 03050 9 3050 Modern atomic theory vs Dalton s 1 All matter is composed of atoms the smallest unit to retain the characteristics of the element divisible into smaller units vs Dalton s indivisible billiard ball theory 2 Atoms of one element cannot be converted to those of another via chemical means This is possible via nuclear reactions Dalton did not know about nuclear chemistry 3 All the atoms of an element have the same of p0 and e which determines reactivity The number of n0 changes which affects atomic mass Dalton assumed unchanging mass Intro to the periodic table As more and more elements were discovered a need arose for an organizing scheme Scientists noticed recurring or periodic characteristics appearing in the elements Dmitri Mendeleev 1836 1907 published the most successful organizing scheme at the time elements were arranged by increasing atomic mass By accounting for the periodic elements he was able to predict elements yet undiscovered Current table arranged by atomic number The table Periods horizontal rows different properties Groups vertical columns similar properties Group A main group elements Group B transition elements or metals nner transition elements or metals Common groups to know lA alkali metals 2A alkaline earth metals 7A halogens 8A noble gases General classification Metals left side generally shiny and solid conduct heat and electricity well malleable and ductile focus of inorganic chemistry Nonmetals right side gases and dull solids conduct poorly focus of organic chemistry Metalloids on the stair step line properties in between semi conductors 27 September Ch 8 problems 6 9 11 20 21 25 29 33 37 41 53 54 64 74 7s 86 Announcements Model assignment due Wednesday 929 Beginning of class 2 parts final refined model of the atom and initial bonding Early Check Scores are incorrect Corrected scores are up now All 3 exam versions are now posted Main group ions Remember there is stability in sets of full orbitals The noble gases are unreactive for this very reason Main group ions form to be isoelectric with have the same electron configuration as the noble gases Mainly involves groups 1A 2A 6A and 7A 1A amp 2A lose electrons 6A amp 7A gain electrons Ne 1s22s22p6 Na 152252263519 lSZZSZZps e F ls22s22p5 e ls22s22p6 Light atoms in 3A 4A and 5A Ionization to noble gas configuration involves a lot of energy C group ions 4 or 4 almost never form B group ions 3 and N group 3 Not common but possible These atoms tend to share electrons to gain a full set of orbitals Heavy atoms in 3A 4A and 5A The d block alters how these atoms ionize Too many electrons to lose to get to noble gas config They will lose electrons to a full d10 Called a llpseudonoble gas configurationquot In Kr 5s24d105p1 In Kr 5524d10 In3 Kr 4d10 Transition metal ions Main group ions form by adding or removing electrons from the orbitals in the highest energy shell Same thing is true here even though we have a partially filled d block First remove electrons in the n s orbital Then remove electrons in the nl d orbital Most common ions will give sets of partially filled d orbitals Example Fe Ar 456 Fe Ar 3d6 Fe Ar 3d5 Pa ramagnetism and Diamagnetism Clue to electron configuration Paramagnetism has unpaired electrons is attracted to a magnetic field Cu Ar 451w Diamagnetic has fully paired electrons is not attracted to a magnetic field Cu Ar 3d10 Ionic size All related to 29 and electron repulsion Cations Loss ofelectrons 9 less repulsion 9 higher 29 Nucleus pulls the fewer electrons closer Therefore Cations are smaller than parent atoms In the same period higher charge 9 smaller ion 1 gt 2 Anions More electrons 9 more repulsion 9 lower Zeff Nucleus cannot hold more electrons as tightly Therefore Anions are larger than parent atoms In the same period lower charge 9 larger ion 2 gt 1 Clicker question Which species has the largest radius A Ar B Cl C CI D K E 5239 A B D and E are all isoelectronic so they all have the same number of electrons but E has the fewest protons 9 holds the electrons the least 9 largest radius Also C is larger than A but smaller than B 3 December Announcements 4 lectures left MSLQ up on RamCT CLASS instructions on RamCT Outside website Please read instructions fully Final Exam stuff 50 50 old new material 2 hour exam 79 am 12 17 NOT in normal rooms Will be announced when known Clicker question SiCl4 reacts with H20 to produce SiOz and HCl If 156 moles of SiCl4 react with 220 moles of H20 what is the maximum amount of HCl that can be produced A 156 moles B 220 moles C 376 moles D 440 moles E 624 moles 156 moles SiCl4 x 4 moles HCl 1 mole SiCl4 624 moles HCl 220 moles H20 x 4 moles HCl 2 moles H20 44 moles HCl Acid Base reactions What volume of 120 M NaOH will be needed to neutralize 225 mL of 30 M H2504 2NaOH HZSO42HZO NaZSO4 225 mL H2504 X 1 L HZSO41000 mL H2504 X 3 mol HZSO4l L H2504 X 2 molNaOH 1 mol H2504 X 1 L NaOH 12 molNaOH 11 L NaOH p 451 list of strong acids and bases Gas phase reactions Given the following reaction at 273 K and 1 L 4NH3 g 502 g 9 4N0 g 6HZO I 4 moles of NH3 and 5 moles of 029 4 moles of NO What is the initial pressure P nRT V P9x00821x2731 P 201 atm What is the final pressure P nRT V P4x 00821x 2731 P 897 atm This is the pressure for gases liquids and solids are not taken into account Gas phase reactions 4NH3 g 502 g 9 4N0 g 6HZO I Pressure is based solely on moles of gas vs liquids or solids Must account for all gas present Recall Dalton s Law when dealing with limiting reagents Clicker question 2C4H10 g 1302 g 9 8C02 g 10H20 g CHM and Oz are reacted in a closed container at a constant temperature If no reactant remains at the end of the reaction what can be said about the final pressure of gas in the container A P nal Pinitial 3 PfinalgtPinitial C P nalltPinitial D No statement can be made about the final gas pressure if the exact initial gas pressure isn t known Excess reagents A mixture of 00375 g of H2 and 00185 moles of Oz in a closed container at constant temperature is sparked to ignite the formation of water 2l39lz g l39 02 g 9 ZHZO g How much 02 is left after the reaction Must determine starting amount And how much reacted 00375 g H2 x 1 moi2016 g 00186 mol 00186 mole x 1 mol 02 2 mol H2 000930 mol Oz used 00185 000930 000920 mol in excess Energy definitions System defined area for study part of the universe Le a reaction flask Surroundings broadly everything else the rest of the universe more narrow everything else relevant to the system ie the lab with the reaction flask E sum of all the energies within the system AE change in E E nal Einilial 0quot Eproducls 39 Ereaclants Energy transfer Total energy must be conserved An energy change on the system will have an opposite change on the surroundings Clicker question You have a flask of acetonitrile at 290 K that you allow to cool to room temperature How would you describe the AE of the system A POEquot AE 0 AE lt 0 AE gt 0 Not enough information 20 October Announcements Practice Exam here tonight 7 9 pm Clicker based review session Sat 1023 at 3 pm in this room Exam will cover the following sections 27 amp 28 covalent bonding 3032 93 covalent bonding 95 electronegativity 100102 110112 Refined bonding model assignment Form on RamCT Due at the beginning of class Monday 1025 M conception of bonding Discuss differences from first bonding model Review Valence Bond theory helps explain molecular shapes Bonding orbitals hold 2 electrons They are stronger with more overlap They hybridize Hybridization when atomic orbitals mix to form new hybrid orbitals Spg from one s and 3 p orbitals tetrahedral class sz from one s and 2 p orbitals trigonal planar class Sp from one s and one p orbital linear class Spgd hybridization Trigonalbipyramidal class 1 s 3 p and 1 d 9 sp3d orbitals First of the expanded octet classes Must have d orbitals for this 9 only period 3 and larger Sp3d2 hybridization Octagonal class 1 s 3 p and 2 d 9 6 sp3d2 orbitals Again must have d orbitals for this Is it just the valence electrons that hybridize m Procedure for molecular info Formula SF4 Lewis structure of electron groups 5 groups determines hybridization Molecular class trigonalbipyramidal Hybridization sp3d hybridization of lone pairs or of bonded atoms 1 lone pair Shape of molecule see saw Lone pairs always take an equatorial position Clicker question For N02 What is the hybridization of N What is the shape of the molecule A sz bent B Sp3 tetrahedral C SpZ trigonal planar D Sp3 bent Can t be B or C not enough groups How many groups 3 Hybridization is sp2 Clicker question For the best resonance form of COCl2 C is the central atom what is the hybridization on and what is the shape of the molecule A sz tetrahedral B SpS trigonal planar C sz trigonal pyramidal D Sp3 Tshaped E sz trigonal planar Clicker question For the I3 anion what is the shape of the molecule and how many d orbitals were hybridized A Bent 1 B Linear 2 C Linear 1 D Linear 0 E Bent 2 4 Octo be r Announcements Exam Wednesday Clark building 7 9 pm Same rooms as last time Must know First 4 periods of the periodic table S block and yellow p block ions Fig 218 pg 65 Bold polyatomic ions Table 25 p 68 Recitations tomorrow All are open for exam preparation optional Check class schedule for options Lattice energy calculated Dependent on Coulomb s Law again Energy q1 qz distance Or energy cation charge anion charge sum of radii Notice gradual difference in energies when charges remain the same Charge difference makes a greater change HF 1050 k moles MgO 3923 k moles Clicker question Which compound has the highest lattice energy A LiBr B NaBr C KBr D RbBr E CsBr All charges are equal smallest radius Covalent bonds As in ionic bonds also entail the atom obtaining a full outer shell octet rule However these bonds form by sharing electrons vs transferring electrons Typically between nonmetals Because the octet rule is satisfied only by sharing electrons covalent compounds form molecules vs formula units Depicting covalent bonds Lewis dot symbols Each atom counts each shared electron toward the octet Shared electrons are quotbonding pairsquot typically shown as a line Unshared electrons are quotlone pairs or quotnonbonding electrons Bond order number of electrons shared between atoms Single bonds bond order of 1 Oz CZHZ double bonds bond order 2 N2 HCN triple bonds bond order 3 Covalent bond formation 2 H atoms incomplete octets Attraction for each other s electrons Optimal distance for sharing electrons ie bond length Nuclei repel each other if too close PPN Note Bond Energy Bond Energy BE Energy applied to break or released to form a bond Breaking requires energy BE gt 0 Forming releases energy BE lt 0 BE is larger for stronger bonds related to bond length and bond order In single bonds longer bond lengths 9 weaker bonds C F 453 kJmol C CI 339 kJ mo C Br 276 kJmol C l 216 kJmol Bonding electrons held more weakly easier to break Bond order as it increases bond length decreases BE increases C2 bonds Naming binary covalent compounds Common names H20 NH4 Otherwise Lower group number comes first with no charge Except when 0 is bonded to Cl Br or If both are in same group higher period names first Second element gets idequot ending Greek prefixes are used to indicate number First word omits mono l Second rarely omits Clicker question What is the correct name for 03 Monosulfur trioxide Sulfur trioxygen Trioxygenmonosulfide Sulfur trioxide Sulfite mpow Formula molecular mass Mass of compounds given in terms of formula units or molecules Basic formula Formula molecular mass Z ofall atomic masses in the formula unit or molecule NaCl 5844 amu PbN032 Pb 2N 60 22 September Announcements Practice exam tonight A103 this room 7 9 pm Early Check In the gradebook on RamCT S vs U After every exam Configuration of groups Take note of similar configurations within a group Group SA Noble Gases He Ne Ar All have full shells Group 1A alkali metals Li Na K All have one electron in their outermost shell Group 4A carbon group C Si All have 4 electrons in their outermost shell Their common configurations 9 similar reactivity give the reason for the periodicity seen among the elements Connection to the periodic table The table is constructed around electron configurations Periods energy shells Atoms generally grouped by sublevel S block vs p block vs d block vs fblock Sublevels are ordered by their energy level Top down as you read Careful of 3 d elements on down Helpful mnemonic to remember order What can be filled for each n Clicker question What is the correct electron configuration of Ge Ar 3d w4s24p2 wrong orbital order Ar 4s24p2 missing 3d Kr 3d w4s24p2 wrong noble gas amp wrong order Ar 4s23d104p2 Kr 4s23d lu4p2 wrong noble gas mpow Transition metals Follows aufbau principle and Hund s rule generally 3d orbitals fill between 4s and 4p Vanadium V Ar 4523d3 Arsenic As Ar 4s23d104p3 2 major exceptions Chromium Cr Ar4s3ei 455 Copper Cu Ar 452 349 451w Result sets of half filled or filled orbitals are particularly stable Inner transition metals These are the f block elements I value 3 m 3 2 1 0 1 2 3 7 total orbitals Lathanides 4f and artinides 5f Filling order 6s 9 4f 9 5d 9 6p 7s 9 5f 9 6d 9 7p Categories of electrons nner core electrons 9 can be condensed using noble gas terms fill the orbitals lowest in energy not used in bonding Outer electrons 9 those in the outer most shell highest n value Valence electrons 9 for main group elements same as outer electrons for transition elements these may include d electrons as well used in bonding Clicker question How many unpaired electrons are in the orbital diagram for sulfur W999 wal O Electrons in a shell Despite various subshells orbitals share the same space within the same shell These orbitals combine to give us a spherical picture of the atom Affects atomic size radius 22 October Announcements Clicker review tomorrow Sat 1023 at 3 pm In this room Revised bonding model assignment Form on RamCT in Recitation folder Due beginning of class Monday 10 25 Exam will cover the following sections 27 amp 28 covalent bonding 30 32 93 covalent bonding 95 electronegativity 100 102 110 112 2 types of bonds So far Valence bond theory says orbitals overlap to form bonds There are many types of orbitals s vs sp3 vs spgdz These overlap only in 2 ways 0 sigma bond Formed from endtoend overlap Every bond we ve drawn contains a 0 bond n pi bonds Formed from sidetoside overlap Only part of double and triple bonds 0 bonds are stronger Endtoend overlap Cylindrically symmetric High density along the bond 9 strong bond Free rotation possible in single bond n bonds Sidetoside overlap Electron density above and below the bond Weaker than 0 bonds Unhybridized p orbitals Seen with sp2 and sp atoms Cannot exist without a 0 bond Sp molecules Double bonds 9 one o and one n bond Triple bonds 9 one 0 bond and two n bonds Two areas of electron density Bond comparisons Rotation Single bonds freely rotate due to endtoend overlap Double and triple bonds have restricted rotation due to position of n bond Strength n is weaker than 0 but Triple bond gt double bond gt single bond Double contains a o and a n bond similar for triple Clicker question What would be an incorrect understanding of n bonds in relation to 0 bonds a They are weaker since they form sidetoside overlap b They are weaker since they are made of p orbitals c They are only found with bond orders gt 1 d They restrict the rotation of the molecule e They have gt 1 area of electron density Polar covalent bonds Differientiated from ionic bonds by A EN More electronegative atom attracts electrons to itself Gives that end of the bond a 6 charge Polarity marked with I 9 29 November Clicker question In the following reaction what is being reduced 30 S C0 g 9Pb S C02 g A PbO Pb goes from 2 to 0 gain of electrons B CO C Pb D co2 PbO CO Pb C02 Oxidation numbers are 2 2 2 2 9 0 4 394 Clicker question In the following reaction what is the oxidizing agent PbO s CO g Pb s C02 g A PbO Oxidizing agent is reduced B CO C Pb D C02 Review of reaction coefficients The only numbers changed when balancing equations Indicate amounts of material in a complete reaction Thought of as numbers of moles or molecules Can be used as conversion factors Simple examples Must use coefficients to convert among compounds in an equation remembering these are moles 2N2H4 l N204 ll 9 3N2 g 4H20 g 2 moles N2H49 3 moles N2 4 moles H20 9 1 mole of N204 0317 moles of N2H49 0476 moles of N2 More complicated Many problems will be in terms of mass Must convert to moles before solving Mass example Given the following equation what mass grams of water is produced when 245 g of CrZO3 reacts completely Cr203 s 3HZS g 9 Cr253 s 3HZO I 245 g Cr203 x 1 mol Cr203 152 gCr203 x 3 mol HZOl mol Cr203 x 18016g H20 1 mol H20 871 g H20 Solution example 400 mL of a 2M solution of Hg2N032 is added to a solution of Kl What is the mass of the resulting precipitate assuming complete precipitation Hg2N032 ZKI ngl2 2KN03 400 mL Hg2N032 x 1 L Hg2N0321000 mL Hg2N032 x 2 mol Hg2N032 1 L Hg2N032 x 1 mol nglz 1 mol Hg2N032 x 655 g nglzl mol nglz 524gHg22 Basic process 1 Complete and balance the equation 2 Convert given information to moles i Solutions use concentration ii Gases use gas law iii Masses use molar mass 3 Use equation coefficients to convert 4 Convert to desired units Clicker question Calculate the mass in grams of H25 that forms if 368 g of aluminum sulfide reacts completely with water A253 HZO 9AOH3 gst A 110 Grams B 251 grams C 835 grams D 278 grams 368 g Al253 x 1 mol Al253 15017 g Al253 x 3 mol HZS 1 mol Al253 x 34086 g HZS 1 mol HZS 2505896g Net reactions When more than one reaction occurs in sequence the equations can be added to show overall reaction A product of the first is a reactant of the second ZCUZS 3029 2Cu20 2502 Cu20 C 9 2Cu CO Coefficient of common compound must match ZCUZS 3029 2Cu20 2502 2Cu20 2C 9 4Cu 2CO Add reactants products ZCUZS 302 2Cu20 2C 9 2Cu20 2502 4Cu 2C0 Eliminate common compound on either side of the arrow 2Cu2 302 2C 9 2502 4Cu 2C0 6 December Announcements 3 lectures left 3 surveys left all due 12 15 MSLQ in assessments tab CLASS the instructions on RamCT Final reflection coming soon Final model assignment Final refined model of molecular interactions Due beginning of class this Friday 12 10 Form on RamCT Conservation of energy Total energy of the universe must be conserved AEuniverse AEsyslem AEsurroudings 0 1st Law of Thermodynamics State function State function a property dependent only on its initial and final states independent of the path taken Akin to elevation change no matter the path taken you still summit the mountain E as a state function Chemically E is only dependent on the initial and final states of the compounds No matter the path the AE is the same AP and AV are also state functions Heat and work Energy is transferred in two ways heat q and work w Therefore AE q w Both in units of Joules J or calories cal Joule SI unit of energy 1Jlkgm2s2 1 J 4184 cal Heat Both terms are in relation to the syst For q Heat flowing out of a system becoming cooler q Heat flowing into a system becoming hotter q Work For w Work done by a system losing energy w Work done on a system gaining energy w W PAV Most important work considered in this class is the expanding of a gas Pressureexlema x AV The gas pushes back the atmosphere 9 work done by the system 9 negative sign Therefore w PAV Clicker question If you ran a gas expansion reaction at sea level with an atmospheric pressure of 147 psi and then in Denver with a pressure of 121 psi how would the calculated change A AV would increase B AV would decrease C AV is independent of pressure D Need to know the reaction W PAV W will be the same P1AV1 42sz Enthalpy H Enthalpy H allows one to calculate energies without calculating q and w separately AE q w If w PAV AE q PAV Q A PAV AH AH change in heat at constant pressure Enthalpy cont d AH AE PAV E P and V are all state functions 9 H is also a state function AH H nal Hintial AH Hpmducls Hreactants Referred to as lenthalpy of reaction Endothermic vs exothermic reactions Endothermic AH gt 0 absorbs heat Exothermic AH lt 0 releases heat AH of common reactions Heat of formation AHf 1 mole of a substance is produced from its elements Na s 72 Clz g 9NaCI s Heat of fusion AHms 1 mole of a substance melts HCZHSO2 s 9 HCZHSO2 I Heat of vaporization AHvap 1 mole of a substance vaporizes H20 9 H20g 17 November Kinetic energy is related to temperature Speed depends on molar mass amp T Acids and bases in electrolytes Compounds that ionize in solution will conduct electricity Amount of current related to amount of ionization Strong acids 9 strong electrolyte HCI g 9 Hi aq Cl aq Weak acids 9 weak electrolyte HCszoz I M Haq CzHgoz aq Acid base reactions In every acid base reaction we will consider the products are water and a salt HX aq MOH aq 9 H20 l MX aq Example with equations HCI aq NaOH aq 9 Molecular equation HCI aq NaOH aq 9NaCl aq H20 l Total ionic Hi 31 Cl 31 Nal 31 0H W 9 Nal 31 Cl aq H20 Net ionic Haq OH39aq 9 H20 l Clicker question Is there ever a time when the total ionic and net ionic equations are identical A Never B Sure If the salt that is formed is insoluble there are no spectator ions Redox reactions Redox oxidationreduction reactions where there is an exchange of electrons occurs in both ionic and covalent reactions Redox terminology Oxidation loss of electrons Reduction gain of electrons Oxidizing agent that which does the oxidizing Reducing agent that which does the reducing The oxidizing agent is reduced The reducing agent is oxidized Loss of Electrons Oxidation Gain Electrons Reduction Leo the lion says grr Oxidation number ON Assigned to each atom in a formula based on electronegativities The charge an atom would have if all electrons were transferred completely Related closely to ion charge MgOvsHCl Oxidation number rules General Atoms in elemental form 0 Monatomic ions charge on the ion Sum ofall ON s in a neutral molecule or formula unit must equal zero Sum of all ON s in a polyatomic ion must equal the charge ON rules cont d Group 1A 1 Group 2A 2 Hydrogen 1 when bonded to nonmetals 1 when bonded to metals and boron Fluorine 1 Oxygen 1 in peroxides OO single bond 2 in everything else except with F Group 7A 1 with metals and nonmetals except 0 or halogens in higher periods Examples Remember electrons are negatively charged Loss of electrons 9 ON goes up becomes more positive Gain ofelectron 9 ON goes down becomes more negative reduces 4Fe s 302 g 9 2Fe203 s 0 0 6 3x2 6 2x3 Fe is oxidized reducing agent 0 is reduced oxidizing agent 25 August Basic Definitions Chemistry the study of matter its changes and the associated energy of those changes Matter anything with mass and volume 2 types of change Physical change occurs within the substance itself without changing into or interacting with another substance Chemical change occurs when a substance interacts with or changes into another substance Energy ability to do work total of potential energy and kinetic energy Potential energy energy of position of an object or the bonds and atoms of a molecule Kinetic energy energy of motion Measurements Units and Conversions All measurements have a quantity and a unit This class has 200 students Conversion factors are used to move from one unit to another All conversion factors 1 Format desired unitoriginal unit Conversion practice Mt Sherman is 14036 feet high What is its height in miles if 5200 feet 1 mile A marathon is measured to be 262 miles What is its length in meters SI units French llSysteme Internationalquot meter gram vs English sytem foot ounce Main units we will use Length meter Time second Volume liter Not an SI unit m3 though used with decimal prefixes Mass kilogram Mass quantity of matter which does not change Weight depends on gravitational pull earth vs moon Density mass volume Derived unit ratio of base units Example of an intensive property irrespective of the amount of material Extensive property dependent on the amount of material Temperature kelvin note absence of Measure of hotness in relation to a standard intensive Heat amount of total energy extensive nterconversion formula K C 27315 OR C K 27315 F95C32 OR CF 32 59 First model assignment Form on RamCT in Recitation Folder Use both words and pictures Describe your understanding of what an atom is Due beginning of class this Friday 827 Don t look things up work from memory Graded for completeness Will make a refined model before first exam 10 September Announcements Chapter 7 problems 2 7 9 11 13 16 20 22 23 29 31 ALEKS quizzes Exam next Wednesday 815 7 9 pm Rooms TBA but all in Clark Through 72 Recitations next week Tuesday exam review open to everyone Thursday go over the exam you will take on Wednesday Revised model of the atom Use the form on RamCT Explain your concept of the atom With words and pictures do not cut and paste Talk about how it has changed from your initial model Give reasons why things have changed Where did new ideas come from If it hasn t talk about what is the same Make a copy to study for the exam Due the beginning of class Monday 913 or before The photon Einstein s light particle Bundles of electromagnetic energy Connected to Planck s quantum number
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