Gen Chemistry COMPLETE StudyGuide
Gen Chemistry COMPLETE StudyGuide General Chemistry MCAT
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Popular in Chemistry
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o The emission spectrum shifts to a slightly longer wavelength Quantum numbers I m s and number of quantum states electrons per orbital o I is the angular momentum quantum number I are integers that range from O to n 1 spdf IO123 for spdf respectively spdf designates subshells s subshells hold 1 orbital p holds 3 d holds 5 f holds 7 each orbital holds a maximum of 2 electrons s subshells hold a maximum of 1x22 electrons p 3x26 d 5X210 f 7x214 A generalized formula for the above pattern for any subshell 4I2 electrons can be held for a given shell higher subshells have higher energy a low shell with a high subshell may be higher in energy than a higher shell with a low subshell o m is the magnetic quantum number m are integers that range from to I including zero 0 s is the spin quantum number s is either 12 or 12 Common names and geometric shapes for orbitals s p d 3d then 4p Fifth row 5s 4d then 5p Sixth row 6s 4f 5d then 6p Last row 7s 5f then 6d The pattern we get from looking at the periodic table is exactly in the order of increasing energy For a given subshell the columns represent how many electrons are in that subshell For example the fifth column of the d subshells contain elements that have 5 electrons in that subshell The number of columns for each subshell indicate the maximum number of electrons that subshell can hold For example the d subshells have 10 columns showing that d orbitals can hold 10 electrons total Conventional notation for electronic structure of equot If E Ep lf 0 Conventional notation l39Jlr ital diagram for rIsE2s2E 5 El IEEK L9 L1 25 3 L4 15 1 L2 a with spill we L gt 2 inquotlith stpin 1aquotE E is filllled orbital o Orbital diagrams o Metals are to the left of metalloids o Nonmetals are to the right of metalloids o Metalloids diagonal line from Boron to Polonium B Si As Te Ge Sb Po quotChemical properties quotMetals Non metals ikes to lose electrons to gain a xidation state good reducing agent ikes to gain electrons to form a xidation state good oxidizing agent ower electronegativity partially ositive in a covalent bond with non etal igher electronegativity partially egative in a covalent bond with metal quotForms basic oxides Forms acidic oxides quotPhysical properties lkiood conductor of heat and electricity Poor conductor of heat and electricity alleable ductile luster solid at room empexcept Hg olid liquid or gas at room temp rittle if solid and without luster o Oxygen group 0 The group column that contains oxygen 0 Oxygen and sulfur are chemically similar if a question asks you what element you can substitute for oxygen and still keep the same chemical reactivity then choose sulfur 0 Se Te Po nonmetal metalloid metal or metalloid The periodic table variations of chemical properties with group and row Electronic structure a repeat of electronic structure section above 0 representative elements Representative elements include the s block and the p block of the periodic table No free flowing loosely bound outer d electrons lonizatiott Energy Trend He Ionization energy decreases as you go down because of increasing radii Ionization energy increases as you go right because of decreasing radii Highest peaks are noble gases Lowest troughs are alkali metals Local maxima occurs for filled subshells and halffilled p subshells Second ionization energy is always higher than the first ionization energy usually a lot higher Alkali metals and hydrogen first ionization energy very low Second ionization much higher Alkaline earth metals first ionization energy low Second ionization energy also low Electron affinity 0 definition electron affinity is the amount of energy released when something gains an electron how easily it can gain an electron 0 Variation with group and row Non metals are more electronegative than metals Covalent bond is a sharing of electrons between elements The more electronegative element in a covalent bond gets a larger share of the electrons and has a partial negative charge The less electronegative more electropositive element in a covalent bond gets a smaller share of the electrons and has a partial positive charge If the electronegativity difference is too great an ionic bond occurs instead of a covalent one Ionic bonds result from a complete transfer of electrons from the electropositive element to the electronegative element Electron shells and the sizes of atoms 0 Electron shells Electron shells are defined by the principle quantum number the n value Going down the periodic table means jumping to the next shell As you fill to the next shell Ne to Na the effective nuclear charge decreases because the old shell stands in between the nucleus and the new shell Filling to the next shell causes a jump in atom size because of decreased effective nuclear charge As you go down a group Na to K the atomic size increases even though the effective nuclear charge stays the same because higher shells have a larger radius than lower shells Going across the periodic table means filling up the same shell by going through subshells As you fill up a shell the effective nuclear charge increases because the atomic number protons is increasing while the same shell electrons you add do not shield one another Frequently the negative sign is dropped and only the magnitude of the electrostatic energy is used The greater the magnitude of electrostatic potential the stronger the ionic bond Strong ionic bonds are promoted by high charge magnitudes q values that are close together small r value Ions that form strong ionic bonds have high charge density that is the charge to size ratio is high o Electrostatic energy on lattice energy 0 O Lattice energy measures the ionic bond strength Lattice energy is the energy required to break the ionic bond The larger magnitude of the lattice energy the stronger the ionic bond and the harder it is to break The lattice energy is proportional to the electrostatic attraction between the ions o Electrostatic force or q1q2r2 O Coulomb39s law F kq1q2r2 Larger charge magnitudes charges being closer together gt greater electrostatic force The Coulomb39s constant k is 9E9 Opposite charges attract negative F same charges repel positive F If q1 doubles the electrostatic force doubles If r halves the electrostatic force increase by a factor of 22 4 Coulomb39s law is analogous to the universal law of gravitation F Gm1m2r2 G is analogous to k and m is analogous to q The big difference is that G is tiny compared to k because gravitational force is weaker compared to the much stronger electrostatic force The covalent bond valence shell electron pair repulsion and the prediction of shapes of molecules eg NH3 H20 CO2 Molecularquot GEnmetriesa lrJ LiI1crlr 39 t lfltj 39ilIII E 1 dL3ij E lcctrolt1 P airs iME TRavcwlorg 4 P x V WM 39 Tn ECL Wa1 L clttt j1lI 1 tlEll1f1liI tlEll 2 1lTJLFu39ili1 r J E 39 TETl 39jll39l 1 Eccsriw Tslrlslparl tLiI1cslr hijrral11iiIlitl1 T llif llfl tiwiitll quot Iquoti39 El394quotl4E at IIJiIIiIJII I qEIJLPliINi1I V Ut t lil i fl ul 539l5l A31T39f Equ iic Tshsipci l Litr1rir pgr139i lu n11rJl lu plla rm r ll 1 E 3 4 oi Litrtbotrtrlcrl cllcct1ir39l pairs o In short it is the VSEPR theory o The VSEPR theory is used to predict the geometry of molecules H ie E R Carbocation j E R B H Carbanion o Nitrogen Lewis structures H N HE RH AmineAmmonia T A Wig HquotffE39 H H Ammonium Pfw39 H Imine 0 Oxygen Lewis structures Molecular oxygen Water alcohol and ethers o The molecule spends most of its time in the most stable resonance structure Stable properties 0 Octet rule is satisfied in every atom except for boron group and hydrogen o No formal charges 0 If there must be formal charges like charges are apart and unlike charges are close together formal charge o Formal charge valence electron in the unbonded atom electron in the bonded atom o Electron in the bonded atom dots around the atom lines connected to the atom o The dots around the atom represent electrons that are held entirely by the atom o The lines connected to the atom represent bonding electron pairs in which the atom only gets one of the two electrons o Formal charges other than 0 must be labeled next to the atom with the formal charge Common formal charges 0 Oxygen with only a single bond 1 0 Oxygen with no bond but have an octet 2 Oxygen usually exists as the diatomic O2 and have a double bond to themselves 0 Carbon with only 3 bonds either 1 if carbocation or 1 if carbanion o Nitrogen with 4 bonds 1 o Halogen with no bonds but have an octet 1 Halogens usually exist as a diatomic and have a single bond to themselves such as Clg o Boron with 4 bonds 1 eg BH4 Lewis acids and bases o Lewis acid accept electron pairs They don39t have lone pairs on the central atom eg BF3 o Lewis bases donate electron pairs They have lone pairs on their central atom eg NH3 0 E kQ1Q2d O O O 0 Energy Electrostatic potential x charge kQ1d x Q2 kQ1Q2d E is negative because Q1 and Q2 are opposite in charge The more negative E is the stronger the ionic bond Strong ionic bonds are promoted by high charge magnitudes Q values that are close together small d value c E lattice energy 0 O O The name used for E is the lattice energy and it measures the ionic bond strength Lattice energy is the energy required to break the ionic bond The larger magnitude of the lattice energy the stronger the ionic bond and the harder it is to break o Force attraction Rnen edquot2 The above equation describes the force of attraction between the cation n and o the anion n at a distance d apart o R is Coulomb39s constant usually written as k o ne charge of cation in coulombs positive charge n times coulombs per electron e o n e charge of anion in coulombs negative charge n times coulombs per electron e o The elementary charge or coulombs per electron e is 16E 19 but you don39t have to memorize it The MCAT will give it to you o The Coulomb39s constant is 9E9 o The official Coulomb39s law states F kQ1Q2r2 Gas phase Absolute temperature K scale When performing P FA calculations make sure that F is in Newtons A is in meter squared and the resulting P will be in Pascals You can then convert the Pascals to whatever units the answer choices are in Molar volume at 0 degrees Celsius and 1 atm 224 Lmol You must memorize this ideal gases occupy 224 L per mol of molecules o Do not get this mixed up it is 224 liters per mole not the other way around The way to remember this is that the mol is a huge number 602E23 molecules These gazillions of molecules occupy a lot of space 224 L to be exact Another way you can remember this is to look at the periodic table Air is made up mostly of nitrogen which has an atomic mass of 14 In the diatomic form N2 weighs 14x2 28 grams per mol Now air is really light In order for you to grab 28 grams of air you need more than just a bottle of air you need a huge tank totaling 224 L ldealgas de nmon 0 An ideal gas consists of pointy dots moving about randomly and colliding with one another and with the container wall The ideal gas obeys the kinetic molecular theory of gases and has the following properties Random molecular motion No intermolecular forces No negligible molecular volume Perfectly elastic collisions conservation of total kinetic energy 0 You can treat gases as ideal gases at Low pressures High temperatures 0 Deviation from the ideal occurs at high pressure and low temperature At these conditions the gas molecules are quotsquishedquot together When the gas molecules are so close together they experience intermolecular interactions Also the molecular volume becomes significant when the total volume is squished down so much The intermolecular attractions will cause collisions to be sticky and inelastic At the extremely high pressures and low temperatures gases cease to be gases at all they condense into liquids 0 Ideal gases behave according to the ideal gas law MET if ufi HE pul5icn Att rauztiun F min E32 0 b for bounce The term with the constant b is the repulsion term The greater b is the more repulsion which leads to greater pressure 0 a for attraction The term with the constant a is the attraction term The greater a is the more attraction which leads to less pressure Partial pressure mole fraction Partial pressure a component of the total pressure exerted by a species in a gas mixture The total pressure of a mixture of gas The sum of all the partial pressures Mole fraction a component fraction of the total mols that belongs to a species in a gas mixture Mole fraction for species A mols of A mols of the entire gas mixture mols of A 2 mols ofA B C iT Dalton39s law relates partial pressure to mole fraction Dalton39s law relating partial pressure to composition Pi Xi39PtotaI Ptotal Zpi 2Xi39Ptotal Pm is total pressure P is partial pressure of species i X is the mole fraction of species i Liquid phase intermolecular forces o Partially negative atoms are also called hydrogen bond acceptors They are most commonly F O or N Do ethers form hydrogen bonds with other ethers Ans no because ethers do not have a partially positive H donor The more polar a bond is the stronger the hydrogen bond The H F bond is the most polar followed by the H0 bond and lastly the H N bond Dipole interactions E 3 Z EIiu1e llipJ rse MCtTRctic1t oi g All polar molecules exhibit dipoledipole interactions This is where the polar molecules align such that opposites attract Dipoledipole interactions increase the boiling point though not as significantly as hydrogen bonding Dipole interactions are stronger the more polar the molecule is o on dipole interactions are similar to dipoe dipole interactions but it39s stronger because it is no longer an interaction involving just partial charges Instead it is an interaction between a full charge ion and a partial charge dipole o on dipole interactions get stronger when you have larger charge magnitude of the ion and large polarity of the dipole molecule Van der Waals39 forces London dispersion forces o Also called dispersion forces o Dispersion forces exists for all molecules but are only significant for non polar molecules For polar molecules dipole forces are predominant o Dispersion forces result from induced and instantaneous dipoles Induced dipoles when a polar molecule interacts with a non polar molecule then polar molecule induces a dipole in the non polar molecule Critical point the temperature and pressure at which liquids and gases become indistinguishable Critical temperature the temperature above which you can no longer get a liquid no matter how much pressure you press on it Water phase diagram is different from others because the soid iquid boundary is slanted to the left This is because water liquid is more dense than ice solid and if you increase the pressure at a given temperature then you turn ice into water Mnemonic for remembering which section of the phase diagram is for gases quotgas comes out this wayquot Freezing point melting point boiling point condensation point Freezing point temperature at a given pressure that liquids begins to freeze into a solid Melting point temperature at a given pressure that a solid begins to melt into a liquid Boiling point temperature at a given pressure that a liquid begins to turn into a gas Condensation point temperature at a given pressure that a gas begins to condense into a liquid Freezing point and melting point are the same they can both be found along the solid liquid phase boundary Boiling point and condensation point are the same they can be found along the liquid gas boundary Sublimation conversion of a solid directly into a gas Conditions for sublimation can be found along the solid gas boundary Molality Molality is a measure of the concentration of solutes in a solution Molality is given the symbol m don39t confuse the small case m with the large case M that is molarity Molality mols of solute mass in kg of solvent Compare molality mol solutekg solvent to molarity mol soluteL solution Colligative properties Colligative properties properties that depend on the of solute particles but not on the type Solute particles in solution like to keep the solution in liquid phase This is why it makes it harder to boil raises its boiling point and also makes it harder to freeze lowers the freezing point Lowering the vapor pressure is just another fancy name for raising the boiling point Van39t Hoff Factor i all colligative properties take into consideration of the Van39t Hoff factor Basically it means convert concentration to reflect the total number of particles in solution For example glucose has i of 1 because it doesn39t break up in solution NaC has i of 2 because in solution it breaks up into 2 particles Nat and Cl 0 vapor pressure lowering Raoult39s law P XsoIvent39P soIvent AP XsoIute39P soIvent P is the vapor pressure AP is the decrease in vapor pressure Xsolute mol fraction of the solute mols of solute total mols of both solute and solvent Xsolvent mol fraction of the solvent mols of solvent total mols of both solute and solvent P SoVem is the vapor pressure of the pure solvent alone When you are calculating xsme make sure you take into account of van39t Hoff ie 1 mols of NaCl in solution is actually 2 mols of particles 0 boiling point elevation detaTb kbm i AIb pC ATb is the increase in boiling point kb is the molal boiling point constant like almost every other constants the MCAT will give it to you m is the molality mol solutekg solvent i is van39t Hoff factor 0 freezing point depression deltaTf kfm i AIf 39kf39m 0 o k is a constant 0 solute is the solute concentration in solution o The partial pressure of a solute just above the soution39s surface is directly proportional to its concentration Molecular weight o Molecular weight is numerically equal to molecular mass amu o 1 amu 1 gmol o Carbon has 12 amu and weighs 12 gmol Empirical formula versus molecular formula l olecular structure olecular formula mpirical formula EH13 ll c ll ll r ll HEH lC6H12O6 ICH2O lljfii lrl rtiErflll o empirical formula is what you get after dividing everything in the molecular formula by the highest common factor Metric units commonly used in the context of chemistry o Molarity M molL o molality m molkg o mass kg molar mass gmol Description of composition by mass omass mass of species of interest total mass 100 Mole concept Avogadro39s number 1 mole 1 mol 1 Avogadro39s number 602E23 molecules Definition of density density mass volume kgm3 often in chemistry specific gravity is used specific gravity number of times the density of water density of substance density of water density of water 1 gmL 1 gcm3 specific gravity of water 1 gcm3 1 gcm3 1 density of lead 11 gcm3 specific gravity of lead 11 gcm3 1 gcm3 11 specific gravity is unitless Oxidation number common oxidizing and reducing agents quotoxidizing agents reducing agents Xygen O2 Ozone 03 Permanganates ydrogen H2 metals such as K ZIUHC1 nO439 Chromates CrO4239 Dichromates nHC1 LAH Lithium Aluminium Hydride r2072 peroxides H202 lewis acids stuff aBH4 Sodium Borohydride lewis bases ith a lot of oxygens tuff with a lot of hydrogens disproportionation reactions 0 An element in a single oxidation state reacts to form 2 different oxidation states 0 Disproportionation can occur when a species undergo both oxidation and reduc on o For example 2Cu gt Cu Cu redox titration O 0 Here the Cut acts as both oxidizing and reducing agent and simultaneously reduce and oxidize itself The oxidized Cut becomes Cu2 The reduced Cut becomes Cu Some terms and concepts A analyte stuff with the unknown concentration that you want to find out by titration Aox analyte that is an oxidizing agent analyte in its oxidized state Ared analyte that is a reducing agent analyte in its reduced state T titrant stuff that you add drip by drip to determine how much of it is needed to complete the titration Tox titrant that is an oxidizing agent titrant in its oxidized state Tred titrant that is a reducing agent titrant in its reduced state S standard something with an accurately known amount or concentration You use it in a reaction that accurately stoichiometrically produces a known amount or concentration of I2 Sex standard that is an oxidizing agent standard in its oxidized state Sred standard that is a reducing agent standard in its reduced state X reactions intermediate a species that is not present in the net equation of the overall reaction Xox intermediate that is an oxidizing agent intermediate in its oxidized state Xred intermediate that is a reducing agent intermediate in its reduced state lodimetric titration Iodine is used in redox titrations because in the presence of starch I2 is dark blue while I is colorless You can only accurately titrate something going from dark to colorless I2 gt 2 but not the otherway round A redox titration does not necessarily need the presence of Iodine As long as some type of color change can be seen at the equivalence point of the redox reaction then it will work For example 39 5H2O26H2MnO4 5O22Mn28H2O Goes from purple to colorless because of MnO4 gt Mnzt transition Redox titrations are similar to acid base titrations except instead of measuring pH you look for a color change Practice question Sex 5Xee gt 3Sree 3XeX 3XeX Aredimiting reagent gt 3Xee Aex Xoxeft over 2 gt 2Xee I2 4 I2 2Ted gt 2 Tex after a long time doing drip by drip titration you finally saw the dark color change to colorless You noted down the initial and final volume reading of your pippette to be 300 mL and 200 mL respectively The concentration of the titrant you used was 10 M You dissolved 12 mols of the standard to begin with How much analyte was there First convert everything to mols amount n MV For the titrant Ted it is 10 M x 03 L 02 L 1 mol For the standard Sex it is already given to you in mols However if it39s not you have to convert it to mols We know from the notes above that Xex XeXIeft over the amount of analyte after taking into account of stochiometric ratios Here are the stochiometric ratios From step 4 I2 I 2Ired O O O Coefficient an equation with coefficients is a balanced equation Direction A single head arrow denotes the reaction goes to completion in the direction of the arrow A double sided arrow denotes a reaction in equilibrium A double sided arrow with one side larger than the other denotes an equilibrium in favor of the side of the larger arrow Charge Denotes charge and magnitude for example 2 5 etc Neutral charges are not denoted balancing equations including oxidation reduction equations 0 0 balance the combustion of propanol C3H8O 02 gt CO2 H20 pick out the atom or group that is the easiest to balance usually represented in only 1 term on both side of the equation In this case it is carbon C3H3O 02 9 4 H20 The next easiest to balance is hydrogen C3H3O 02 9 Pj Leave the hardest to last oxygen 0 is present in every term of the equation so if we tried to balance 0 first we39d be having a hard time However now that we balanced every other term this leaves only one term left that contains 0 and that we haven39t balanced yet Do a quick count of oxygen atoms there39s 1 from C3H8O 3x2 from 3CO2 and 4x1 from 4H2O Set up this equation 1 2x 3x2 4x1 where x would be the coefficient of our last term 02 Solve for x C3H3O 9202 9 Even though we balanced out every term we39re not done yet We need to get rid of any fractions so multiply every term by 2 2C3H3O 902 9 Balancing oxidation reduction redox equations Separate into half reactions There will be 2 half equations one will be oxidation the other reduction Half equations contain only species of interest those containing the atom that undergoes a change in oxidation state Anything that is not covalently attached to the atom is not part of the species of interest Anything that does not undergo a change in oxidation state is a spectator ionspecies Balance each of the half reactions Balance both charge and atoms To balance one oxygen atom Under acidic conditions add H20 to the side that needs the oxygen atom then add Ht to the other side Under basic conditions add 2OH39 to the side that needs the oxygen atom then add H20 to the other side The lon Electron Method you balance out the atoms first then charge The Oxidation State Method treat the species of interest as a single atom those that undergo a change in oxidation number and then balance it Recombine the half reactions Multiply each half reaction by a factor such that when you add them together the electrons cancel out It39s like you39re trying to solve a simultaneous equation and you want to eliminate the electron term Finishing touches 1 2Cr6 6e gt 2Cr3 Do the same thing for the oxidation half reaction Oxidation state method 1 Cl gt CI2 2 2CI39 gt CI2 3 2CI39 gt 2CI 4 2C gt 2C 2e Recombine the half reactions 2Cr6 6e gt 2Cr3 2C gt 2C 2e Multiply everything in the second equation by 3 2Cr6 6e gt 2Cr3 6C gt 6C 6e add the two equations together 2Cr6 6e 6C gt 2Cr3 6C 6e Finishing touches Except for the electrons there are no like terms to combine or cancel at this time 2Cr6 6C gt 2Cr3 6C Convert the atoms of interest into species of interest by referring back to the original equation K2Cr2O7 6HC gt 2CrCl3 3Cl2 Now unlike the ion electron method where the equation is balanced and you only at back spectator ions at this stage of the game the oxidation state method requires you to balance the equation again This is because after you convert the atoms of interest back to their species of interest the equation is no longer balanced Next take the amount in mols of the limiting reactant 9 mols according the the above calculation and do the stoichiometry to get to how many mols of 3Xed this will yield 9 mols of A6 3 mols of Xed per 1 mol of Aed 27 mols Lastly convert mols to grams 27 mols 10 gmol 270 g The theoretical yield for the above reaction is 270 g of Xed 0 Say you did an actual experiment of the above reaction and you managed to obtain 243 g Xred then the experimental yield is 243 g 0 Percent yield experimental yield theoretical yield x 100 o For the above experiment the percent yield would be 243270 x 100 90 o Energy changes in chemical reactions thermochemistry o Thermodynamic system state function o A thermodynamic system is just a fancy name for the system that you are studying Isolated system no exchange of heat work or matter with the surroundings Closed system exchange of heat and work but not matter with the surroundings Open system exchange of heat work and matter with the surroundings o A state function is path independent and depends only on the initial and final states 0 State functions include AH enthalpy AS entropy AG free energy change AU internal energy change 0 State function is also called state quantity or function of state Conservation of energy 0 The total energy of an isolated system remains constant 0 The total energy of a closed or open system plus the total energy of its surroundings is constant 0 Total energy is neither gained nor lost it is merely transferred between the system and its surroundings o Endothermicexothermic reactions 0 Endothermic energy is taken up by the reaction in the form of heat AH is positive Exothermic energy is released by the reaction in the form of heat AH is negative enthalpy H and standard heats of reaction and formation enthalpy or H is the heat content of a reaction Mnemonic H stands for heat AH is the change in the heat content of a reaction means heat is taken up means heat is released Standard heat of reaction AHX is the change in heat content for any reaction Standard heat of formation AHf is the change in heat content a formation reaction A formation reaction is where a compound or molecule in its standard state is formed from its elemental components in their standard states The standard state is where things are in their natural lowest energy state For example oxygen is 02 diatomic gas and carbon is C solid graphite The unit for enthalpy is in energy J or it can be expressed as energy per mol Jmol Hess law of heat summation AHX AAHf sum of AH products sum of AH reactants o Bond dissociation energy as related to heats of formation 0 Bond dissociation is the energy required to break bonds AHX Bond dissociation energy of all the bonds in reactants bond dissociation energy of all the bonds in products AHX Enthalpy of formation of all the bonds in products Enthalpy of formation of all the bonds in reactants 0 Second law concept of entropy O The 2nd law states that the things like to be in a state of higher entropy and disorder An isolated system will increase in entropy over time An open system can decrease in entropy but only at the expense of a greater increase in entropy of its surroundings The universe as a whole is increasing in entropy AS 2 q T q is the heat transferred T is the temperature in Kelvin For reversible processes AS q T For irreversible processes AS gt q T Real processes that occur in the world are never reversible so entropy change is always greater than the heat transfer over temperature Because of the irreversibility nature of real processes as long as anything occurs the entropy of the universe increases o Temperature scales conversion 0 O L K c F Absolute zero 0 273 460 I f Freezing point of watermeting point o 273 O 32 ce quotRoom temperature 298 25 77 quotBody temperature 310 37 99 oiling point of watercondensation of 373 100 212 team K C 273 O F CX1832 o Heat transfer conduction convection radiation 0 Conduction heat transfer by direct contact Requires things to touch Convection heat transfer by flowing current Need the physical flow of matter Radiation heat transfer by electromagnetic radiation commonly in the infra red frequency range Does not need the physical flow of matter can occur through a vacuum o Heat of fusion heat of vaporization 0 Also called latent heat of fusion enthalpy of fusion and latent heat of vaporization enthalpy of vaporization Heat of fusion AHfu3 the energy input needed to melt something from the solid to the liquid at constant temperature Heat of vaporization AHVap the energy input needed to vaporize something from the liquid to the gas at constant temperature Latent heats can be expressed as molar values such as J mol The energy it takes to melt a solid is AHS x mos of that solid The energy it takes to vaporize a liquid is AHVap x mos of that liquid Latent heats can also be expressed as J mass where energy can be obtained by multiplying the latent heats by the mass of the substance Energy is released when either a gas condenses into a liquid or when a liquid freezes into a solid The energy released is the same as the energy of their reverse processes see formula above PV diagram work done area under or enclosed by curve For the phase transition use heat of fusion q AHfu3 x mos of icewater where AHfu3 is in energy per mol note if the heat of fusion is given in energy per mass then you should multiply it by the mass to get energy For the water phase from 0 C to 37 C use q mcWaterAT where AT is 37 Old topics The topics below are outdated They have been either modified or replaced by the most recent aamc publication o Measurement of heat changes calorimetry heat capacity specific heat specific heat of water 1 cal per degrees Celsius 0 Heat capacity the amount of heat required to raise the temperature of something by 1 C Molar heat capacity heat capacity per mol mom Specific heat capacity heat capacity per mass 900 Celsius can be replaced by Kelvin here because a change in 1 C is the same as a change in 1 K o It takes 1 cal or 42 J of heat energy to raise the temperature of 1 gram of water by1 C 0 1 calorie 42 J 1 Calorie with capital C 1000 calorie 4200 J o For water 1 gram 1 cubic centimeter 1 mL o First law AE Q W conservation of energy 0 1st law of thermodynamics is based on the principle of conservation of energy and it basically says that the change in total internal energy of a system is equal to the energy absorbed as heat minus the energy lost from doing work 0 AEQW 0 AE is the same thing as AU which is change in internal energy A M B M C M llrate Ms 1 1 1 1 2 1 1 4 1 2 1 2 1 1 2 1 0 F kAXByCZ 0 From this table a 2x increase in A corresponds to a 4x increase in the rate 2 4 so x 2 o A 2x increase in B corresponds to a 2x increase in the rate 2V 2 so y 1 o A 2x increase in C corresponds to 1x no change in rate 22 1 so z O 0 r kA2B C o r kA2B o rate constant 0 The k in the rate law is the rate constant 0 The rate constant is an empirically determined value that changes with different reactions and reaction conditions reaction order 0 Reaction order sum of all exponents of the concentration variables in the rate law 0 Reaction order in A the exponent of A O on Type 39on Order Laws nimolecular kA imolecular kA2 r kA B ermolecular kA3 r kA2B r kABC order reaction k Rate determining step o The slowest step of a multi step reaction is the rate determining step The rate of the whole reaction the rate of the rate determining step The rate law corresponds to the components of the rate determining step Dependence of reaction rate on temperature Activation energy 0 Activated complex or transition state Activated complex what39s present at the transition state In the transition state bonds that are going to form are just beginning to form and bonds that are going to break are just beginning to break The transition state is the peak of the energy profile The transition state can go either way back to the reactants or forward to form the products You can39t isolate the transition state Don39t confuse the transition state with a reaction intermediate which is one that you can isolate 0 Interpretation of energy profiles showing energies of reactants and products activation energy AH for the reaction FlEiavti39on Rate Attitra1 ioan lEnErgly FlEacti on Rate llEmp EraturE 0 When activation energy approaches zero the reaction proceeds as fast as the molecules can move and collide 0 When temperature approaches absolute zero reaction rate approaches zero because molecular motion approaches zero Kinetic control versus thermodynamic control of a reaction o A reaction can have 2 possible products kinetic vs thermodynamic product Derivation AG O at equilibrium AGAG RTlnQ O AG RT In Oat equilibrium AG RT In Oat equilibrium At equilibrium AG O 39 rforward rbackward Q Keq o Keq is a ratio of kforward over kbackward If Keq is much greater than 1 For example if Keq 103 then the position of equilibrium is to the right more products are present at equilibrium If Keq 1 then the position of equilibrium is in the center the amount of products is roughly equal to the amount of reactants at equilibrium If Keq is much smaller than 1 For example if Keq 103 then the position of equilibrium is to the left more reactants are present at equilibrium 0 The reaction quotient Q is the same as Keq except Q can be used for any point in the reaction not just at the equilibrium If Q lt Keq then the reaction is at a point where it is still moving to the right in order to reach equilibrium If Q Keq the reaction is at equilibrium If O gt Keq then the reaction is too far right and is moving back left in order to reach equilibrium 0 The reaction naturally seeks to reach its equilibrium application of LeChatelier39s principle 0 LeChatelier39s principle if you knock a system off its equilibrium it will readjust itself to reachieve equilibrium O O x oxg100gxg100mL x ppm x parts per million x mg kg x mg L Solubility product constant the equilibrium expression 0 O Solubility product constant Ksp AgC S ltgt Agt aq Cl aq Ksp for AgC AgC Ag2SO4 s lt gt 2Ag aq SO42 aq Ksp for Ag2SO4 Ag2SO42 Ksp values are found in a table Ksp for AgC 18 X 1010 Ksp for Ag2SO4 12 x 105 Ksp is simply Keq for dissolutions The higher the Ksp the more the reaction products dominate in a saturated solution at equilibrium What is the solubility of MX2 if given Ksp MX2 lt gt M2 2X Ksp lV2liXil2 Ksp M22M22 because for every M22 there39s two times as much Xquot I Ksp 4M23 Solve for M2 Solubility is the same thing as M2 because you used Q Ksp for a saturated solution If you solved for X39 instead divide your results by 2 If you were given solubility and asked to solve Ksp then know that solubility M2 X392 Common ion effect its use in laboratory separations For pure water pH og1O397 7 Acidic pH lower than 7 Neutral pH 7 Basic pH higher than 7 pOH logOH39 pH pOH 14 Conjugate acids and bases eg amino acids 0 quotAcid Base lt gt ltonjugate base Conjugate acid H20 H20 sgt0H39 Hs0 R COOH H20 lt gtR Coo H3O H20 IIRNH2 sgt0H IIRNH o More acidic lt H3NCH2COOH lt gt H3N CH2COO lt gt H2N CH2 COO39 gt more basic Strong acids and bases common examples eg nitric sulfuric 0 quotStrong acid Formula quotPerch1oric acid IIHCIO4 quotHydroiodic acid HI quotHydrobromic acid IIIIBr quotSulfuric acid IIII2SO4 quotHydrochloric acid HC1 quotNitric acid IIHNO3 quotIIydronium ion I130 or H 0 Strong acids completely dissociate in solution 0 Complete dissociation occurs because the conjugate base anion is highly stable nium hydroxide OH ater 20 0 Weak bases partially dissociate in solution 0 Partial dissociation occurs because the conjugate acid is fairly stable 0 dissociation of weak acids and bases with or without added salt CH3COOH will dissociate less in a solution containing CH3COONa salt NH4OH will dissociate less in a solution containing NH4C salt This is due to Le Chatelier39s principle the hydrolysis of salts of weak acids will produce the their conjugate bases which reduces dissociation Likewise hydrolysis of salts of weak bases will produce conjugate acids 0 hydrolysis of salts of weak acids or bases Salt of weak acid CH3COONa lt gt CH3COO39 Nat CH3COO39 H20 lt gt CH3COOH OH Salt of weak base NH4Cl lt gt NH4 Cl NH4 H20 NH3 H30F o calculation of pH of solutions of salts of weak acids or bases Salt of weak acid Let39s say a solution contains M molar of CH3COONa CH3COO39 H20 lt gt CH3COOH OH As M molar of CH3COO39 start to abstract protons from the solvent CH3COO39 M x CH3COOH x OH39 x Kb KWKa CH3COOHOH39 CH3COO X2M x o Ka Hln39 H ln o Indicators behave just like weak acidsbases o The indicator is present in such a small amount that it doesn39t affect the solution39s pH 0 When the solution has a low pH high Hl the indicator is mostly in the H In form which is of one color 0 When the solution has a high pH low Hl the indicator is mostly in the In form which is of another color Neutralization Acid Base Salt Water Interpretation of titration curves Adding lquotlauDH to HCl nquotV equivalence point pH lbasie lanaidll lddlng IlEl ti MHI3 l4plg I IIll T eqIIiIIaileI Ie I I P lrllt I I Ia IidI 2 lquotl E ba 5E acid case At the point of inflection buffer arrow the acid conjugate base or base conjugate acid pH pKa and titrant 12 weak acidbase The buffer region has pH values of pKa 1 AIliI39Ig hla H t lllgCiZ I3 s 1 39 P lance H I IIILI1 EIr iI 39hE395i l ta FREE j I JIIIinItE I I I I W emu itralenie C I cirIt39lI D E I 39 I I39II p I I lIIII1 fe I I I I I 7 I I I r 39 I I I 39 I I I I I I I I l I I I I I I E E E E E E E E E E E E 51 quot 51 quotquot E H II II in II E quotE E E E E Polyprotic acids have multiple pK Points of inflection and equivalence points Like monoprotic acids each point of inflection corresponds to the pKa for the acidic species At each pKa acidic species conjugate base of the acidic species For H2CO3 pKa1 H2CO3 HCO339 Equivalence point 1 almost everything is HCO339 pKa2 HCO339 C033 Equivalence point 2 almost everything is C032 Redox titration 0 While Bronsted acid base titrations involve proton transfers redox titrations involve electron transfers 0 Redox reduction oxidation species A gains electrons species B lose electrons 0 Reduction reduction in charge decreased oxidation number gaining electrons o Oxidation increase in charge increased oxidation number losing electrons O 5H202 2MnO4 6H 2Mn2 502 Z Normally oxygen has an oxidation state of 2 but in peroxides it is 1 The reactants here include a peroxide Oxygen and anything else in its elemental state has an oxidation number of O The product 02 is one such case Hydrogen is always 1 unless it is a hydride in which case it39s negative 1 For this reaction all hydrogens are 1 Doing some math we find that the reactant Mn has an oxidation number of 7 The half reactions reactions that depict electron transfer only are as follows Reduction Mn 5e gt Mn2 Oxidation O gt 0 e M LT REviEwm Eleclrulgr c el F I quot5 MC T QEviaw rquot Galhmniir fnlltalc argge Flaw Eta NaC aT QEwiEwPrg 2239 T r quot1 NuC aT QEwiawmrg Red Cat REDuction CAThode o Anode shoots out electrons Cathode takes in electrons electrolyte o Ions electrolyte o Electrolytes conduct electricity by the motion of ions 0 Without electrolytes there won39t be a circuit because electricity Won39t be able to travel Faraday39s law relating amount of elements deposited or gas liberated at an electrode to current 0 Current coulombs of charge per second I qt o Faraday39s constant coulombs of charge per mol of electron total charge over total mols of electrons F qn o q It and q nF thus We get 0 It nF 0 Current x time mols of e X Faraday39s constant 0 Using this equation you can solve for n mols of electrons Then using the half equation stoichiometry you can find out how many mols of element is made for every e transferred For example 1 mol of Cu is deposited for every 2 mols of electrons for the following half reaction Cu2 2e gt Cu electron ow oxidation and reduction at the electrodes 0 Electrons shoot out of the anode because oxidation occurs there to lose electrons M gtMquote39 o Electrons travel into the cathode Where it crashes into the cations on the surface of the cathode This is because reduction occurs at the cathode to receive electrons M 39 e gt M o Mnemonic Oil Rig Oxidation Is Losing e Reduction Is Gaining e o For example the cell potential for the galvanic cell shown in the diagram is ction potential table uction Potential V I 799 uII 337 pecies Reduction half reaction 2Ag 2e gt 2Ag Reduction potential 0799 Oxidation half reaction Cu gt Cu 2e Oxidation potential 0337 x 1 0337 Cell potential 0799 0337 0462 V The cell potential for all galvanicvoltaic cells is positive because the voltaic cell generates potential 0 Another example the cell potential for the electrolytic cell shown in the diagram 1S1 Reduction half reaction Cu 2e gt Cu Reduction potential 0337 Oxidation half reaction 2Ag gt 2Agquot 2e Oxidation potential 0799 x 1 Cell potential 0337 0799 0462 V The cell potential for all electrolytic cells is negative because the electrolytic cell requires potential input direction of electron ow
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