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Final Exam Study Guide

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by: Kayla Kerch

Final Exam Study Guide Cem 141

Marketplace > Michigan State University > Economcs > Cem 141 > Final Exam Study Guide
Kayla Kerch
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Everything covered in CEM 141 this year and everything you need to know to ace the final! Stoichiometry problems along with explanations of all chapters.
General Chemistry
A. Pollock
Study Guide
cem 141
50 ?




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"I had to miss class because of a doctors appointment and these notes were a LIFESAVER"
Joelle Wisozk

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This 21 page Study Guide was uploaded by Kayla Kerch on Friday December 11, 2015. The Study Guide belongs to Cem 141 at Michigan State University taught by A. Pollock in Fall 2015. Since its upload, it has received 39 views. For similar materials see General Chemistry in Economcs at Michigan State University.


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Date Created: 12/11/15
CEM 141 Final Study Guide Ch. 1 Claim, Evidence & Explanation/Reasoning  Why coffee cools down when milk is added o Claim: coffee cools when milk is added o Evidence: temperature of milky coffee is between milk & coffee o Reasoning: Higher temp. molecules collide w/lower temp. molecules, as they collide E is transferred, hot molecules slow down & cold molecules speed up Atoms  Matter is made of atoms  Smallest building blocks, can’t be split  Positively charged nucleus surrounded by e- in orbitals o Most of the mass is in the nucleus  Contain protons (p+), neutrons (n0) & electrons (e-) o They are neutral as a whole Dalton’s Atomic Theory  Atoms are in constant motion-Brownian motion  Elements are composed of small, indivisible indestructible sphere particles called atoms  Atoms of elements are different from atoms of another element  Compounds are combinations of atoms of two or more elements  Atoms are neither created/destroyed during a reaction (chemical reactions are rearrangements of atoms) JJ Thomson Experiment (cathode rays) st  1 subatomic particle discovered was the e-  Ray of particles emerged from cathode toward the other end (anode)  Particles were deflected by the electromagnetic fields in a direction that indicated they were negatively charged  The metal that the cathode was made of did not affect the behavior of the ray so, the nature of the e- is independent of the metal  Findings: the electron in atoms Thomson’s Plum Pudding Model  Atoms contain e- embedded in the atom  Problem: model didn’t justify Rutherford’s experiment (most alpha particles went through golf foil, some deflected back) Existence of electrons  Claim: atoms contain e- embedded in atom  Evidence: Thomson’s experiment showed us that particles could be deflected/bent by magnetic fields  Reasoning: the metal the cathode was made out of had no effect on the rays being bent, the composition of the ray is independent of the element it came from Ernst Rutherford Experiment (gold foil)  Radioactive source emitting alpha particles (He atom/nucleus) with lots of E at gold foil  Most alpha particles went straight through (some deflected back)  The positive charges that came close to the nucleus were deflected o Evidence for: positive charge must be in the nucleus  Findings: there is a small dense positively charged nucleus in the center of atoms Rutherford’s Planetary Model  Orbits contain e-  Problem: e- aren’t stable & would emit E & orbits would collapse Existence of Small, Dense, Positively Charged Nucleus  Claim: atoms contain a positive nucleus  Evidence: Rutherford’s experiment showed some atoms were deflected off gold foil and therefore had a positive nucleus  Reasoning: those positive charges that came close to the nucleus were deflected, most of the volume is where e- exist Neutrons  Last subatomic particle discovered > they are harder to detect because they have no charge/neutral  Located in nucleus, slightly heaver then p+ Model of Atom we use now  Electrically neutral  Cloud of e- take up most of the space o Cloud can be shifted, e- move Coulomb’s Law  Positive & negative (opposite) charges attract & like charges repel  Force of attraction= (q1 x q2)/ r2 o Charges on particles (q) o Distance between particles (r) Gravitational Forces vs. Electromagnetic Forces *Similarity for both forces Gravitational Forces (gravity) Electromagnetic/Electrostatic Force Attraction between objects that have Attraction/repulsion between objects mass w/electric charge Mediated by fields* Mediated by fields* Requires 2 or more objects* Requires 2 or more objects* Weaker Stronger Always attractive Can be attractive/repulsive Decreases as distance increases* Decreases as distance increases* Holds us on the earth Increases as charge increases Stops us from falling through the earth Throwing/Holding a Ball:  When the ball is in the air, the only force acting on it is gravitational force  Force of attraction: o Increases as mass of interacting objects increases o Decreases as distance between objects increases  When holding the ball, forces acting on it are gravity & electrostatic Kinetic Energy (KE)  KE= 1 m v 2 E associated with motion 2  As ball moves toward the ground, KE increases (velocity is increasing) Potential Energy (PE)  E associated w/ the position of system of objects in field  Can’t have PE without a field, or without 2 or more objects  As ball moves toward the ground, PE of system (ball & earth) decreases  Gravitational interaction between earth & ball: o Cause PE to decrease o Cause ball to fall down  Total E of the system (ball & earth) stays the same *E can be transferred from one object to another & E can be transformed (from PE to KE or KE to PE)* What Makes Atoms Stick Together Solid Liquid Gas -Touching in very organized -touch, but there is more -Completely separated & ordered state disorder & they take the shape of the container London Dispersion Forces (LDF)  Forces are not very strong, they are a type of intermolecular force (IMF)  Caused by fluctuation of e- density in molecule/atom  Between 2 atoms/molecules  Partially positive side of atom being attracted to partially negative side on the other atom  As atoms approach KE increases (PE decreases) until they get too close  If they get too close their e- clouds repel each other  Larger e- cloud= stronger LDF o Larger/floppier e- cloud is stronger because their protons can get closer together He atom Interactions (noble gas)  Held together by LDFs  When they hit the wall of a container total E decreases o It transfers to the wall  E transferred through collisions  Increased temp = move faster = more collisions = more E = break apart  Why they move toward each other: o Attractive electrostatic force o LDF  Why they oscillate: o Cloud of e- repel each other and LDF attracts them to each other  To keep them close together: o Lose E by bringing in 3 atom o E transferred by collisions to 3 atom  To form stable interaction: o Remove E  Increasing temperature: o Increase E (move faster) o Knew it was warmer by colliding, E is transferred from container to atoms o If enough E is transferred to the atoms they break apart o If thermal E increases, temperature increases He Atoms- PE Graph Atoms Approaching/coming closer together:  Electromagnetic attraction & LDF bringing them together  KE increases, PE decreases At the well:  Electromagnetic attraction = repulsion  Most stable  Depth of well tells you: o How strong the interaction is & how much E is needed to get the atoms out of well o Deeper well requires more E o Deeper wells have higher boiling points  Position of well (left/right) tells you: o Internuclear distance- distance between atoms at the most stable point o Farther right the well is means they have a larger atomic radius Atoms Very/Too Close Together:  The e- clouds start to overlap and the negative charges repel each other  PE increases, KE decreases He vs. Xe Atoms  2 Xe atoms can’t get as close together as 2 He atoms because their e- clouds will repel o The LDF are larger between the Xe atoms then the He atoms because they have more e- so they are further away from the nucleus making the e- cloud larger o Xe has larger LDFs o Xe has larger boiling point (stronger LDFs require more E to break the forces) Breaking bonds  E is absorbed (when atoms absorb enough energy the forces between them are overcome) Forming bonds  E is released (when atoms lose E they come closer together and do not have enough E to overcome to attractive force between them so they stick together) H Atom Interactions  2 H atoms approach w/stronger attraction then He atoms it forms a covalent bond  When something collides into H atoms, E is transferred from H atoms to that atom & that forms a covalent bond between them  2 H atoms= H2 molecule H2 Molecules (H-H)  When stable, H2 stays in PE well  When they collide and absorb enough E, H2 separate into H atoms  Takes 6000 K to break 2 H atoms and 4K to break 2 He atoms o 2 He are held together by weaker interactions (LDF)  Under normal circumstances H exists as H2 diatomic molecules o Other examples: O2, N2, F2, Cl2, Br2, I2 H @ 5K H @ 15K H @ 30K *H @ more than 6000K looks like gas monoatomic picture Stoichiometry Problems Stoichiometry  Allows us to calculate how much can be produced in a reaction  Coefficients tell you mole ratio Mole 23  6.022 x 10  Use for mass to mole conversions Stoichiometry Calculations: 1. Write a balanced equation 2. Draw map of where you start & how you get to the end 3. Write out calculation w/units 4. Calculate Balance the equation: C6H 14 + O2 > CO 2 + H 2 O 2 C6H 14 + 19 O2 > 12 CO 2 + 14 H 2 O Calculate the molar mass of: H 2 O 2H = 2 x 1 g/mol = 2 g/mol 1O= 1 x 16 g/mol = 16 g/mol 2+16= 18 g/mol ^atomic mass Mass > Mole conversions Ex) How many moles of (Ca( NO 3 2 ) are in 325 g? 1 Ca= 1x40 g/mol 1 mol 2N= 2x14 g/mol 325 g Ca( NO 3 2 x = 1.98 mol Ca( NO 3 2 164 g 6O= 6x16 g/mol ^= 164 g/mol How many moles of O atoms are in the Ca( NO ¿ ? 3 2 NO 3 2 1.98 mol Ca( NO 3 2 x 1molCa¿ = 11.88 mol O 6molO ¿ Ex) Which is biggest? 10g C H 4 , or 10g C2H 6 ? 10 g C H x 1mol = .625 mol C H 4 16g 4 10 g C H x 1 mol = .333 mol C H 2 6 30 g 2 6 Ex) How many moles of NH 3 would be produced if 6 moles of H2 reacted w/excess N 2 ? 2mol NH 3 H + N > 2 NH 6 mol H x 3 = 4 mol NH 2 2 3 2 3mol H 2 3 How many grams of N2 would be required to produce 2.75 moles of NH 3 ? 1mol N 28gN 2.75 mol NH 3 x 2 x 2 = 38.5 g N2 2mol NH 3 1molN 2 Limiting Reagents Ex) Find the limiting reagent starting with 3 moles C 2 4 O and 5 moles H 2 to make C2H 6 2 C2H 4 O + H 2 > C 2 O6 2 1molC H2O 6 2 3 mol C2H 4 O x 1molC H O = 3 mol C 2 O6 2 * C 2 4 O is the 2 4 limiting reagent because it can C H O only make 3 mol 2 6 2 1molC H2O6 2 5 mol H 2 x = 5 mol C 2 6 2 1mol H 2 Ex) How many moles of NH 3 would be produced if 6 moles H 2 reacted with 3 moles N2 ? N 2 + 3 H 2 > 2 NH 3 2mol NH 3 3 mol N 2 x = 6 mol NH 3 1mol N 2 H 2mol NH 3 NH 6 mol 2 x 3mol H = 4 mol 3 * would be produced because it is the 2 limiting reagent Using the equation: CH + 2 O > CO + 2 H O 4 2 2 2 Ex) How many moles of CO would be produced from 1 mole of O ? 2 2 1molCO 1 mol O x 2 = .5 mol CO 2 1molO 2 2 What mass in grams of CO 2 would be produced from 16g O 2 ? 1molO 1molCO 44gCO 16g O2 x 2 x 2 x 2 = 11g CO 2 32gO 2 2molO 2 1molCO 2 If you had 16g O and 16g CH , how much water would be produced in grams? 2 4 1 molO 2mol H O 18g H O 16g O 2 x 2 x 2 x 2 = 9g H 2 *because it is the 32 gO2 2 molO2 1 mol 2 O limiting reagent 1molCH 4 2mol H2O 18gH O2 16g CH 4 x x x = 36g H 2 16gCH 4 1molCH 4 1molH 2 actualamount produced(given) Percent Yield = theoretical amount(calculated) 100 If the reaction produced 8g CO what would the percent yield be? 2 8gCO 2 x 100 = 73% 11gCO 2 If you had 10g CH 4 and 10g O 2 , what is the maximum amount of CO 2 that could be produced? 1molCH 4 1molCO 2 44gCO 2 10g CH 4 x x x = 27.5g CO 2 16gCH 4 1molCH 4 1molCO 2 1 molO 1 molCO 44gCO 10g O 2 x 2 x 2 x 2 = 6.9g CO 2 *because it is 32 gO2 2mol O 2 1 molCO 2 the limiting reagent Ch. 2 Electromagnetic Spectrum 2  Log scale (2 objects 10 m apart are 100m apart)  Highest Frequency (Lowest λ) o Gamma o X-rays o UV o Visible Spectrum  Blue  Red  *Red has longer wavelength then blue o IR o Radio  ^Lowest frequency (highest λ) Light is a Wave  Wavelength λ- distance from peak to peak  Frequency v (Hz)- # of wave fronts per second  Amplitude- height of peaks (intensity)  C = λv, c=3x 108 m/s  Energy increases as frequency increases (wavelength decreases)  Large wavelength has small frequency & small amplitude 18 Ex) Determine λ (in nano meters) of an x-ray with a frequency of 3 x 10 Hz 1nm (3 x 1 08 m/s) = (λ)( 3 x 10 18) 1 x 1 0−10 m x −9 = 1 x 10 nm 1x10 m λ= 1 x 1 0−10 m Diffraction  When waves hit a barrier with a slit, the wave that goes through the slit is diffracted Interference  Constructive o Two waves the same reinforce & create higher crests & troughs  Destructive o Two opposite waves cancel out (become straight line) Electromagnetic Radiation as a Wave  Claim: E/m radiation is a wave  Evidence: E/m radiation is diffracted & shows patterns of interference  Reasoning: When e/m goes through a barrier w/ a slit the resulting waves are diffracted, they can be combined either constructively or destructively Photoelectric Effect  Metals emit e- when light shines on their surface. The certain threshold frequency must be met otherwise no e- will be ejected no matter how intense the light is. There must be enough E (a photon) to emit an e-  E is transferred as a photon, each photon has definable E  One photon ejects one e-  If the photon does not ha−34enough E, then no e- is ejected  E= hv, h= 6.626 x 10 J/s 18 Ex) What is the E of a photon of frequency 4 x 10 4 x1018 = (6.626 x 10−34J/s )(E) 51 E= 6 x 1 0 J What is the λ of a photon of E 6.2 x −8 10 J 6.2 x10 J 25 −1 E=hv −34 = 9.36 x 10 s 6.626x10 J.s 8 Λ=c/v 3x10 m/s = 3.2 x 1 0−18 m 9.36x10 s −1 E/m Radiation is a Particle  Claim: E/m radiation is a particle.  Evidence: The photoelectric effect says that metals emit e- when light shines on their surface  Reasoning: The light is transferring energy to the e- which is transformed into KE which gives e- enough E to be emitted. The photons of E absorbed have to be more than the threshold or no e- will be ejected. If a certain amount of E is absorbed then the e- will be ejected. Electrons in Atoms Have Quantized E Levels  Different atoms contain different e- that each have specific energy levels that are quantized. It requires the exact amount of energy the e- needs to move to the next energy level.  Emission spectra has lines where the e- are moved from higher to lower energy levels and emit a photon  Absorption spectra have lines where the e- are moved from a lower to higher energy level from absorbing a photon. o Spectra are created by absorbing and emitting photons. It takes a specific amount of energy to move and e- to the next energy level and specific amount of energy must be emitted to move it down an energy level. o Spectra show light only of specific energies Electrons (Wave Properties)  Energy diagrams, e- transition between E levels by absorbing or emitting photons  Atoms absorb/release specific amounts of E to transition, these specific amounts emit that specific wavelength  Core are right elements which are harder to remove than valence e- bc they are attracted to the positive nucleus, valence are outermost e- which are the easiest to remove Bohr  e- are most likely in their orbital, only works for Hydrogen  Orbits have definite energies so e- in atoms are quantized  E of photons corresponds to the difference in E levels of e-  If e- loses E it is de-excited and moves to a lower E orbit closer to the nucleus  Atoms of elements can’t absorb/emit any λ of light (only certain wavelengths that correspond to that element) E Diagrams  Each E level has a quantum #  Higher # = more E, E levels are not orbits  E- transition between E levels by absorbing/emitting photons Effective nuclear charge  Core e- cancel out the positive charge from the same number of p+ (Zeff: p+ minus core e-) Periodic Trends  Atomic radius increases down a row, and decreases across a row left to right  Ionization E: decreases down a group, and increases across the row left to right o When there are more e-, the IE is smaller bc they are easier to remove the valence e- when there are many of them so it takes less E to remove them  Zeff: increases across a row, high Zeff hold onto their e- more tightly bc their e- are more strongly attracted to the nucleus (this is why atomic radius decreases across a row bc they hold onto e- more tightly so they are smaller) Ch. 3 Big Bang to Atoms  Started from one and it burst into more at very high temperatures, as it cooled quarks and leptons formed, as it cooled further p+ and n0 formed, a few minutes later when it cooled more H+, D+, He2+ and Li3+ formed through fusion  Atoms from your body come from one star before the big bang  The number of atoms in the universe is constant  Everything in the universe is moving away from us (Doppler effect)  Evidence: CMBR Cosmic Background, red shift (red on spectra has longer wavelengths showing its moving away from us) Nuclear fusion, fission and radioactive decay  Fusion (adding two nuclei together)  Fission (breaking apart nuclei, forms lighter more stable nuclei +Energy) *U in equations, chain reactions  Decay (emitting particles, alpha, beta or gamma) Balance nuclear equations:  Mass # on top  Atomic # on bottom (number of p+ and e-)  Isotopes have diff # of n0  E=mc^2  m=(mass defect)(amu) Atomic Atomic Atomic Mass of 1 Massof 2+ Mass of 1−Massof 2 = ¿lement1of 1 ¿ ¿Element2of 2 ¿ ¿ of 1−Atomic¿of 2¿ determine element Nano-particles, and larger macro scale materials  Atoms can be joined together and have emergent properties  Surface area to Size ratio affects properties  Boiling/melting points happen when more than one atom interact together (interactions between particles have to be overcome)  Nano, don’t have a state Melting/Boiling Points of Diatomic Molecules vs. Diamond  Hydrogen bonds to form H2, 2 e- in the bonding MO make it bond to itself  He, has 2 e- in the anti bonding MO so it cancels out the 2 e- that are in the bonding MO  To break the covalent bond between 2 H atoms, enough energy has to be added to raise e- to the anti bonding MO  Covalent bonds (Diamon) have high boil/melt points, molecules (H2) have low melt/boil points bc it is easier to break the LDFs then the covalent bonds so they are still in the form of molecules when they are boiled it takes a lot more E to break their covalent bonds  Diamond/graphite have such high melting points because they would just disintegrate if they were melted, it also takes a lot more E to break the 3D structure of all of the covalent bonds Forming Bonds  Electrostatic forces are there (PE) when two atoms are approaching, when their nuclei get a nuclei in distance apart the strong nuclear force takes over and PE is no longer acting on the atoms (during fusion)  Forming a bond, E is released to the surroundings Molecular Orbital Bond Model (metals)  Combine atomic orbitals (n atomic orbitals=n molecular orbitals)  Each orbital can contain up to 2 e-  Lower 2 in model are bonding MO, higher 2 are anti bonding  If anti bonding are filled up this cancels out the bonding between the atoms (that is why He doesn’t form a covalent bond just LDF) Valence Bond Model (everything else)  Atomic orbitals overlap to from bonds  The greater the overlap the stronger to bond  Each bond is made of 2 e-  e- are localized in the bond Metals  Conduct electricity: e- can move freely around  Malleable: atoms can move w/respect to one another  Shiny: absorbs photons & moves to higher E level then immediately re-emits it & moves to lower E level  Interacts with many wavelengths so the metal is white/colorless, silvery Ch. 4 Diamond  Carbon forms 4 identical bonds  Sp3 hybrid orbital, tetrahedral geometry  Localized e-, sigma bonds  Hard o 3D network of strong covalent bonds  High melting point o Takes lot of E to overcome strong covalent bonds  Doesn’t conduct o Localized e- in bonds aren’t free to roam & there’s a large “band gap” Graphite  Each C attached to 3 C, forms sheets of graphite  Sp2 orbitals form sigma bonds and leftover p orbital is a pi bond  Slippery o Sheets held together by weak LDFs so they slide over each other  Conducts electricity o Delocalized pi bonds over structure allow e- to move freely  Shiny o Absorbs and re-emits photons (like metals) Carbon  Bonds to C, H, O, N, S, P  Emergent properties  Methane, CH4 o 4 identical C-H bonds equidistant from one another o C has sp3 hybridization Lewis Structures  Calculate # of valence e- o H=1, B=3, C=4, N=5, O=6, F=7  How many bonds it usually forms o H=1, B=3, C=4, N=3, O=2, F=1  Write skeleton using 2 e- for each bond  Make sure each atom (except H) has 8 e- by adding lone pairs  Add multiple bonds (double, triple) if there aren’t enough  Don’t give 3D info Isomers  # of carbons > # of isomers it has o 0, 1, 2, 3 > 1 isomer o 4 > 2 isomers o 5 > 3 isomers o 6 > 5 isomers o 7 > 9 isomers Sigma bonds  Allow free rotation of bonded atoms Pi bonds  Not freely rotatable, pi bond would break Formal Charge 1  F.C.= (# of valence e-)-(dots or lone pair e-)-(sticks 2r bonding e-)  NH4 (draw out Lewis structure first) o FC (N)= (5)-(0)-(4)= +1 VESPR (Valence Shell e- Pair Repulsion) Groups of e- Geometry Bond Hybridizatio Examples around Angle n center atom 2 e- Linear 180 Sp CO2 3 e- Trigonal Pyramid 120 Sp2 BF3 4 e- Tetrahedral 109 Sp3 CH4, H2O 5 e- Trigonal 90/120 Sp3d PCl5 Bipyramid 6 e- Octahedral 90 Sp3d2 SF6 *For geometry count all groups of e- surrounding that atom (double/triple=1 e-) *For shape count groups of e- surrounding atoms but ignore lone pairs Geometry/Shape  Ignore lone pairs  To figure out shape: o “Take off” lone pairs from geometry and the remaining structure is the shape  H2O (draw Lewis Structure) o Geometry: tetrahedral (3 bonds, 1 lone pair= 4 e- groups) o Shape: bent (3 bonds, ignore lone pairs = 3 e- groups)  CO2 o Geometry: linear (2 bonds (double/triple=1), no lone pairs= 2 e- groups) o Shape: linear (2 bonds, ignore lone pairs= 2 e- groups) Electronegativity  Ability of an element to attract e- to itself when its in a bond  Increases across table left > right  Decreases across table top > bottom  Noble gases don’t form bonds so they can’t be polar  H & C are so similar they we look at them as the same electronegativity Polar Bonds  2 atoms with different electronegativities bond o Unequal sharing of e- o Results in a dipole  Dipole points towards the more electronegative element in the bond o Ex) H-F it would point towards F since F is more electronegative o Ex) H-C is not polar since they have same electronegativities  Shape is important! o Dipole arrows can cancel each other out if they are facing the same way  Ex) CO2, linear: arrows pointing left and right cancel each other out so CO2 is nonpolar  Ex) H2O, bent: arrows pointing upward toward O do not cancel out, they combine to form one larger dipole so water is polar London Dispersion Forces  Weakest IMF  Temporary fluctuating dipoles  Only force present in nonpolar molecules Dipole-dipole Forces  In polar substances (along with LDFs)  Stronger then LDFs  The more electronegative element is partially negative and the less electronegative element is partially positive  Dipole-dipole forces happen when the partially negative element in one molecule is attracted to the partially positive element in another molecule o Ex) H-Cl, Cl of one molecule is attracted to the H of another molecule Hydrogen Bonding (IMF)  Strongest IMF  Compounds with H-bonds also had dipole-dipole and LDFs  Between molecules with H covalently bonded to either O, N, or F and a lone pair of e- on either O, N or F o H hydrogen bonds to the O, N or F with the lone pairs o (O, N or F)-H {hydrogen bond} :(O, N or F) o Ex) CH3OH Strengths of Bonds & IMFs Strongest >Covalent Bond Ionic Bond Hydrogen Bond (IMF) Dipole-dipole Forces (IMF) Weakest > London Dispersion Forces (IMF) Water  Anomalous properties  High mp/bp & specific heat  Water freezes at 0 degrees C, water boils at 100 degrees C  Low vapor pressure  Density of ice < density of liquid water o Ice takes hexagonal crystal from which creates “holes” Ionic Bonding  Transfer of valence e- creates ions (each ion achieves noble gas configuration)  Between nonmetals (high electronegativity) and metals (low EN)  Metals lose e- to nonmetals  Metals form cations +  Nonmetals form anions –  Ionic compounds are neutral  Electrostatic forces holds ions together NaCl  Na+ is bigger then Cl- o Na has more e- so it has more repulsion which = larger size  Forms colorless crystals  Conducts electricity in liquid state but not solid state Charges on Ions  K > +1 S > -2  Mg > +2 Al > +3 Formulas for Ionic Compounds  Na+ & Cl- > NaCl  Mg 2+ & O 2- > MgO  Ca 2+ & Br - > CaBr2 Lattice Energy  E released when ionic lattice forms from ions in gas phase  Force of attraction, Coulomb’s law  Strong attraction= more charge, smaller ions  Weaker attraction= less charge, larger ions Ch. 5 Temperature  Measure of hotness  E always moves from the hotter object to the cooler object  1 drop boiling water vs. 1 bucket boiling water = same temps.  Depends on average KE of molecules  T=KE Kinetic Energy  KE= 1/2mv^2 KE= 3/2kT  An individual gas particle can not have a temperature  If average velocity increases, temperature increases  Heavier & lighter molecule = same KE  Lighter molecule has higher average velocity Thermal Energy  1 drop boiling water vs. 1 bucket boiling water = not same thermal E  Depends on how much material you have  Added through collisions  Moves faster, translate, vibrate/rotate Maxwell-Boltzmann Distribution  2 curves, left one is higher and short & right one is shorter but wider  Higher curve: lowest temperature, lowest velocity & highest mass  Lower/longer curve: higher temperature, high velocity & small mass Gas Molecules  Speed of gas o Distributed, random amounts of E are transferred through collisions creating a distribution of speeds  When gas molecules collide they don’t stick together: o Have a high KE, enough E to overcome the attractive interactions between them  Why wouldn’t you smell it immediately across a room if you opened container: o Takes time for molecules to move o Collide with each other and other molecules in air o Don’t travel in straight lines to your nose, when they collide their direction changes o Random movement (Brownian motion) Heat  Transfer of thermal E from one place to another by collisions Thermal E  E associated w/motion of particles Specific Heat  E it takes to raise temperature of 1 gram by 1 degree C  Lower specific heat raises temperature the fastest o Lower sp. Heat means its easier to raise the temperature  Water specific heat is high o Lots of molecules per gram, low molecular weight o Lots of IMFs Phase Changes  Solid > liquid > gas o Absorbs E, breaks interactions  Gas > liquid > solid o Releases E, makes interactions  Temperature doesn’t change until change of phase is complete  Extra E being added is used to overcome the IMFs State Functions  Depend on initial & final stakes  Don’t depend on path taken o Ex) Elevation of campsite  T, E, H, S, G, P, V or altitude Path Functions  Depend on the path taken  Depend on how the change takes place o Ex) Distance travelled  Lower q, w First Law of Thermodynamics  E cannot be created/destroyed  E can only be transferred/transformed Exothermic  E goes out of the system  q- sign  Heat goes out of the reaction and the surroundings get hot Endothermic  E goes into the system  q+ sign  Heat goes into the reaction and the surroundings get cold Enthalpy ΔH  Change in H or q  Equation: q=m x c x change in T (Tf-Ti)  Mass x specific heat x change in temp.  Ex) 50g metal heated to 100 degrees C added to 50g water at 20 degrees C, highest temp. water reaches is 27.5 degrees C, what is specific heat? o Water q = (50g)(4.18)(7.5)=1567.5 o Metal q= (50g)(c)(72.5)=3625 o Compute: 1567.5/3625 = .43 J/g C Entropy  Disorder  Measure of # of possible arrangements for a given state  More possible arrangements = higher entropy 2nd Law of Thermodynamics  For any change the total entropy of universe must increase  You can’t get as much E back as you put in (some is lost-spread out at thermal E)  Δ Stotal Δ Ssystem+ Δ Ssurroundings Mixing  Dye mixes because it spreads out (possible arrangements increases) & entropy increases  It will never un-mix because it is more probable that they will stay mixed Entropy & Phases  Solid o Low entropy o Molecules are fixed in place & can’t move relative to one another (one possible arrangement)  Liquid o Medium entropy o H-bonds, molecules can move relative to one another not completely free moving though o Some possible arrangements (more then solid, less than gas)  Gas o High entropy o No forces holding molecules together so they are free to move independently (most possible arrangements) Ex) Which has more entropy? Deck of cards in order or Shuffled deck CaO (s) + C O2 (g) or CaC O3 (s) *making a gas, entropy is always higher H 2 (l) @ 25°C or H2O (l) @ 50°C *molecules can move faster & some LDFs are broken allowing more arrangements Hot block or Cold block *entropy increases with temperature ΔH (change in enthalpy) of Phase Change Which has higher Δ H vap? C H3OH or C H3C H3 -H bonds -LDF -Dipole *harder to break all the bonds -LDF Δ S = ΔH surroundings surroundings T Freezing Vs. Melting Δ Ssystem Equal to Δ S system Δ Hsystem- Equal to Δ Hsystem Δ Hsurroundings Equal to Δ Hsurroundings Δ Ssurroundings Less than Δ S surroundings *same magnitude just opposite directions Freezing Water  Exothermic  Δ Ssurroundingsends on temperature o Higher temp = smaller the effect on entropy Δ S total  Tells us whether change will occur o If total entropy increases the change will happen o If it decreases it will not happen  Use Gibbs free energy to measure ΔG Gibbs Free E Solid > Gas ΔG = ΔH – T ΔS *T is always positive For this process to happen ΔG must be (-) = (+) –(+)(+) negative so you want a high temperature ^depends on how high the temp is. Gas > Liquid ΔG = ΔH – T ΔS *Want T to be small so ΔG can be negative (-) = (-) - (+)(-) ^system loses E when you make those interactions


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