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See course web page at

by: Esteban Reynolds

See course web page at CHGN 124

Marketplace > Colorado School of Mines > Chemistry > CHGN 124 > See course web page at www bradherrick com
Esteban Reynolds
Colorado School of Mines
GPA 3.77


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This 29 page Class Notes was uploaded by Esteban Reynolds on Monday October 5, 2015. The Class Notes belongs to CHGN 124 at Colorado School of Mines taught by Staff in Fall. Since its upload, it has received 16 views. For similar materials see /class/219629/chgn-124-colorado-school-of-mines in Chemistry at Colorado School of Mines.


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Date Created: 10/05/15
A Chapter IQ Chemical Kinetics 1 De nitions v r Kinetics Study of speed of chemical reactions and the mechanisms by which they occur i Sgd Rate change in the concentration of product or reactant per unit of time It is always apositive mrmber3939hereactanrsovertimedisappearsandtheprodnctappears Hypothetical Reaction A ge 1 1 00 PM of me 39 A 084 numbers of moles of A and B 5 0 8 o 74 as a function of reaction time M 39 0 60 0quot0 078 for the othetical reaction Average rate At 2 0396 054 i 7 some zero there are 39 a 045 100 mol A and 0 mol 3 The A l A 39239 04 x 0 40 number of moles of A decreases mo 85 0 I and the number of moles of B A Z 02 03926 39 03930 03922 increases as the reaction 016 proceeds G O o 10 20 3o 40 so 50 Time min The rate depends on the nature of the reactants ex some metals are more reactive than others 39 quotquotquot39 Ex 2 asubstancethatis nelygroundordissolved ratctsrapidlyduetoagreatsnr oearea exposed for reaction concentration ofrea tant in general as the concentration of reactants 39 increases theprobabilityofcollisionsincreasesand theprobabiligofreactionsoccurringincreasesand the rate increases temperature as the temperature increases more and more reactant quotquotquot 39 rnolecules have the necessary energy Ea activation en to react and formproduct molecules upon collisions catalyst speed up reaction rates without being consumed v 2Rate and concentration Consider the general reaction aAbB we dD eE rr Its rate law expression is X Y Rate M01545 1 A B C Time p Z Where kisthe speci c rate constant ABC are reactants is concentration in mail liter M x is the order of the reaction with respect to A y is the order of the reaction with respect to B z is the order of the reaction with respect to C u l in general Icy32 have values of 0 12 xyz is the overall order of the reaction Note x yz have nothing to do with the stoichiometric coef cients in the equation for the reaction they must be determined by experiments Hi Note 2 aleggg 5132031231 igtegy 9f 39 g change of concentmu39on Interval oftime Instantangougr3tegt Wye slope of the tangent that touches the concentration Curve at the given time qmcuaq H300 gt cH0Haq HClaq 01 Instantaneous 009 39 rate at t 0 initial rate 008 007 C4H9Cll M p 0 9 o o o o S 8 8 52 8 8 Speci c Rate Constant k m 4 C4H9Cl l 100200300400500600700800900 Time 5 Concentration of butyl chloride CH9Cl as a function of time Lines are drawn that are tangent to the curve at t O and t 600 s The slope of each of these tangents is de ned as the vertical change divided by the horizontal change that is ACH9Cl t The reaction rate at any time is related to the slope of the tangent to the curve at that time Because CHCl is disappearing the rate is equal to the negative of the slope Thereactionratehasunitsof concentlation time molesec Theunitsofktherefore dependon theoverallorderofthereaction falsity 0f Units ofk Exam Fla 0 39 Ms 391 RainL k 1 51 Roll K Eel 2 M IS l kart K 9 3 M434 2 45 k 933 To detemu39ne value for k in an example use the data from 39 riment in that exam 1e 111 into the rate law expression and solve for 1c any ewe p r p g I39 Halflife andtheintegratedrateequation r The halflife of a reactant is the time required for halfof the reactant to be consumed in the reaction r L1 11 two We can 3593333993 A firstorder reaction is one whose rate depends on the conCentration of a si subStltut A 5 gle reactant raised to the first power For a reaction of the sort A produc the rate law may be first order AIA Rate k A two kt gt At l quot 12 A10 Using calculus this equation can be transformed into an equation that relaz 1n ktm the concentration ofA at the start of the reaction Ao to its concentration at an E lnl t1239 ic392quotquot Notice that t1 2 for a firstorder rate 0 693 k law is ind dent of the initial concentration of reactant ssa Pressure CH3NC tort N o O O other time t A lnIA 1nAo kt or The function quotInquot is the natural logarithm In terms of base10 or common logarithms kt A Id logrAl loglAlo m or fA L 2363 The factor 2303 arises from the conversion of natural to base10 logarithms This equation has the form of the general equation for a straight line y mx b m which m is the slope and b is the yintercept of the line lnA 1m 1mm 1 mt b y Thus for a first order reaction a graph of lnA versus time gives a straight lin Witha slope of k and a yintercept of lnAo 52 U 50 in 48 U 46 a 44 I 3 4392 3 Variation in the partial 9quot 40 39 pressure of methyl isonltrile 9 38 CHgNC with time at E 35 1989 C during the reaction 39 34 CH3NC CH3CN b A 10000 20000 30000 0 10000 20000 30000 plot of the natural logarithm Time s Time s of the CH3NC pressure as a function of time 3 lb lshows how the partial pressure of methyl isonitrile varies witl time as it rearranges in the gas phase at 1989 C We can use pressure as a unit 0 concentration for a gas because from the idealgas law the pressure is directl proportional to the number of moles per unit volume 39 the halflife of a secondorder 139 3912 two t the halflifeof a second order reaction is dependent on the initial concentration of reactant cg Plots of the 394398 r the reaction a 50 Mom N09 0267 I a m E 52 connecting the data points in a 5 5 4 plot of lnNOz versus time a 39 I is not linear consequently we 56 39 39can conclude that the reaction is not rst order In N02 The 58 plot of 1 NO versus time b o 100 200 300 Is linear the reaction is second T113196 order in N02 tion dbeys a secondo SEE LOE39EL39EEEEE39E Relying on calculus this rate law can be used to derive the following equatiox Rate 35 my 1 L A k A10 I Q This equation has the form of a straight line 1 in b If the reaction is second order a plot of 1A versus twill yield a straight line with a slope equal to k and a y intercept equal to 1 Ao One way to distinguish between first and secondorder rate laws is to graph both 1nA and 1 A against 1 If the InIA plot is linear the reaction is first order if the 1A plot is linear the reaction is second order rder rate law Rate ItNOjz e 250 1N92 500 100 200 I Time 5 line graph we have that it 0543 M squot for the disappearance of N02 Summary of kinetics for reactions of form aA gt productsl 3915 0365 sustkewi trespgcue ng crews ex we can use the rst order equation if A9C and the Rx is rst order in A A B C and the Rx is rst order in A and zero order in B 39 Second otdet Zero First RateLaw Ratek RatekA RatekA2 Integrated Rate Law A kt A mm kt mmo i E kt 1 39 A Ala 39Plot for Straight line A vs time anA vsrttme 7 vs time Relationship of Rate to Slope k Slope 1 Slope k Slope Halflife tquot 111 t 133 1 2 I U2 k 912 A10 l I 39 quot 300 only the plot of 1 N02 versus time is linear Thus the mac From the slope of this straight J fr The half fe Tmis constant only in the reactions of order one Tm 122 k For tg rst order reaction AbC Ex If A o 2M and k 35 1s how long would it take for 75 ofA to v React Answer in Fl 4 kt 1 1 C30 9 75 A o hammocks 3520Ml9 in 025 A0 0351 4 In no a ln25xa 903550 Una lh05 O3 tlha t X L Swan twogdzs 39 a 7570 met arm amt We TsJ quot9 T11 n39q 148 06 a 2Tb RXLCB 3556024 035 ELI ifthe T a ofthe rslorder reaction A9C is l 2 sec and k 35 13 anti 4 O 2 M calculate the initial concentration of A if it took 4 seconds for 75 to 39 react answer in ll kk lhf o3 9 475 A itth MQLGJDQSGemw in 02590 036 x43 in Fo was 41 44 1 Qon39i wnk bxquot lad CL Wai 61d ULQLL39IWanci WATV Miami Ma 1670 any Gimmi U W Q Mad CLLAU l Lf a TV oco EGO J 0 M The halflife Tm is not constant in a reaction of order two Tm 1 K A0 39 When have to calculate consecutive T1z as A0 is changing 39 For the second order reaction AsyC Ex If A o 2M and k 35 Lm s how long would it take for 75 of A toreact L 035 xt JP 0 answer Ho l A 3 055624 4 s bags Dec 05 9 Dom U mtg 8150 T 1 O W I how I202 39Ew De 8 a 1 035 6 Frach39mg a 509 v adirp W g00 Ll l smc c km W i new W Mg T 1 2260 035xi 711W PnJSf b3 hwwor 9575 WUZ I43 26 236 02 LIL518 Dec EXZ Calculate the initial concentration of A A o for the second order reaction A gt C o n 5 kx4 knowmg that it takes 4 seconds for 75 to react and Its F 075 Lm s 39 W ma Wm m gamet 39 3ij 75 lidVT Matted aTya 9 Qa Tyl gQQc 391 I 39 l qc kTv We LiaC another OPUM Alix 397 L 3 A 035 4 o75x4 RC Hg 5OIQ L A0 0amp6 tngWFGF 4 one 3 9 095 A0 a o 9 d R 0amp0 we we mg wmab We 8W4 WM tag CmCQntuahM o Lina 075 gt40 L 390 H 90 R 39 M LLg a A 015 o Q Lyn W O 95 S CDC9 0156 W be 160 m lsouaedxq O Two meoriesareusedto affect reaction rates if Collision theory ofReaction tatesusedforlstepreactions Premise inorderforareactiontooccm ratomsmoleculesorions A must collideoomeimooontact B musthaveane ectiveoollision it c Transition State Theory Ergngsgz in a lstep reaction the reactants are rst converted into a hi reac ngspeciesmnsthaveaminimumenergywhichwillbetmnsferred into breaking old bonds and making new ones Reacting species must have proper orientation called a transition state before forming products Energy pro e for the rearrangement of methyl isonitrile The molecule must surmount the activation energy barrier before it can form the product acetonitrile Wernight imagine passing through an intermediate state in which the NEC portion of the molecule is sitting sideways n c HSC XEC gt H3Culll H3C CEN The change in the energy of the molecule during the reaction is shown The diagram shows that energy must be supplied to stretch the bond between the H3C group and the NEC group so as to allow the NSC group to rotate After the NEC group has twisted sufficiently the C C bond begins to form and the energy of the molecule drops Thus the barrier represents the energy necessary to force the molecule through the relatively unstable interme diate state to the final product The energy difference between that of the start ing molecule and the highest energy along the reaction pathway is the activation energy E The particular arrangement of atoms at the top of the barrier is called the activated complex or transition state The conversion of H3CNEC to H3C CEN is exothermic therefore shows the product as having a lower energy than the reactant The en ergy change for the reaction AE has no effect on the rate of the reaction The rate depends on the magnitude of En generally the lower l5I is the faster the re action C Activated H3O I ll complex Energy 6 inquot H3C CEN Reaction pathway gh energy intermediate state eaction Mechanisms and the Ratelaw expression Most reactions occur in a series of steps called the mechanism of a reaction These steps are thought to be primarily unimolecular decomposition ex A 93 C bimolecular collisions ex A B 9C note trimolecular collisions are rare since it is very improbable that 3 or more species will be at the same place at exactly the same time to collide and react Ex N2 2 02 2 N02 is not likely acne step reaction the slggvgst step determinesthe rate of the overall reaction It is the step with the highest activation energy POM Ema Ductqu a 3skprtad139m lh Q skp ii fk slap Ow kw E The experimentallyderived ratelaw expression Rat k A quot B y is used to postulate a mechanism for the reaction The value of x and y are related to the coef cients of the reactants in the slowest step in uenced in some cases by earlier steps When the slow step in a mechanism is the rst step the order of a reaction wr to a particular reactant the exponent indicates the number of molecules of the reactant that appear as reactants in that rst slow rate determining step 1 9 themsiia pramszim haeisehthe 90119612116 for a reaction atleasLLc tszie must be met the sum ofall the steps must be the same as the overall reaction when like terms are cancelled the rate law expression of the mechanism must match the true experimentallyderived rate law expression If the slow step is the rst step then the can eg Jhareactant is the exponent in the rate law expressmn A Rate k each reactant e If the criteria are met the mechanism MAY be correct q 3 Effect of temgratnre changes on the reaction rates 39 Rate kA B quotConsiderthelstepreaction A B yC 39 possible potential energy diagram 39139ransition state as s 2 an AF 1 Progress of reaction As the temperature increases more and more reactant molecules have the necessary energy Ea activation energy to react and form product molecules upon collisions B does not change with the temperature it only changes if mechanism changes Arrhenius developed the mathematical relationship between temperature the rate constant k which does depend on 39I and the activation energy Ea which does NOT depend on T Where k speci c rate constant A equency factor related to nature of reactants and of effective collisions r Ea activation energy R8314 JmolK T absolute temperatmeinKelvin 39 i E Taking the log and considering 2 di erent temperatures T1 and T2 ll 4 Effect of Catalysts v Catalysts are substance speed up reaction rates without being consumed They appear in an equation h A B catalyst 39 C catalyst or may only appear over the arrow Most catalysts appear to function by somehow lowering the activation energy of the origmal39 reaction by providing an alternate reaction pathway Ea quot Ell at earl kl we a 39 Energy rofilesfo th ta d p r em I725 Uncatalyze d reaction decomposition of hydrogen peroxide and for the reaction as catalyzed by Br39 The catalyzed reaction involves two successive steps each of which has a lower activation energy than the uncatalyzed reaction Notice that the energies of reactants and products are unchanged by the catalyst Energy gt H202 21120 Br Iznzoo There are two categories cf catalysts Reaction pathWay a hgqoggngqys thequot catalyst exists in same as reactants ex strong acids H b gagggegggugajh catalyst exists in a dam phase than reactants It is usually a solid Which loyr s the activation energy layproviding a surface upon which the reactions can occur lt4 w EggLTHERMODYNALIICSJ 39 D v K DEFINITIONS a Spontaneous Processes A process that occurs without outside help Spontaneous Nonspontaneous Pants fall to oor Pants hang themselves in closet Raw egg becomes hard boiled in water Hardboiled egg becomes raw egg b ReVersible Processes A reversible process is a process that is always in equilibrium Example Ice melting at 0 C H20 5 5 H20 1 Equilibrium is adjusted by adding or subtracting heat Water can go from solid to liquid to solid to liquid etc C Irreversible Processes A process not in equilibrium Exmnple Ice melting at 25 oC Process only happens in one direction d Jill 51 calorie kg m2 1 cal 4184 J exact new de nition of cal de ned as U 1 52 old de nition of calorie energy needed to raise the temperature 1 m2 of1gofH201 C quot K E2mv2 gt kgs 2 aside 1 dietary calorie 1 Cal 1 kcal SI unit a Heat 714 choatic molecular motionj L7 r relate dto temperaturequot heating increases or decreases molecular motion in all directions q heat gained by system q heat lost by system Work 5 39 quot4 iconCerted motion Work Force X distance gt w F X d movement against force is work I work increases or decreases molecular motionin a speci c direction in gases work is related to changes in volume xquot PAW if quot A I 7 a i g v v M Tipsy I 9 a w work done on system compress1on w work done by system expansion 3 39 Internal Energy is the sum of kinetic and potential energy in a thermodynamic system ISipeQ39nggrgy energy Ofmovement example thermal energy molecules moving KE 12le EateritialEIem energy stored that is released through motion example ball in air ll FIRST LAW OF TIIERMODYNAMCS The change in the energy of a system is due to heat andor work I q represents heat and w represents work In other words the sum of the kinetic energy change and the potential energy change of a system is equal to the heat and the work a the gas is con ned to the right haif of the cylinder by a partition When the partition is Movable Partition removed b the gas spontaneously Grreversibly expands to ll the whole cylinder No work is done by the system during this expansion In c we can use the piston to compress the gas back to its original state Doing so requires that the surroundings do work on the b C system an 7 N Piston Note that pressure is a force divided b an area and that the change in volume is a distance These can be used to de ne wart w as w 1 AV For an expanding gas AV is a positive quantity because the volume is increasing Thus AV and w must have opposite signs so we have a negative sign in front of our equation because it is doing work on our surroundings enthalpy at constant pressure is the measurement ofh at a 3 F qp AH meXAT 5 AHM D ZnPAlI products anAH reactants J 7 Elements are not included in the calculation because elements require no change in form V HESS LAW The total heat of reaction can be written as a sum of minireactions V AH1 AHz AH3 lt7 E SECOND LAW OF THERMODYNAMCS In any spontaneous processes the total entropy of the universe increases for irreversible processes and is zero for reversible processes K AS 0 2 l1an For a thermodynamic process the universe can be broken into two parts 1 System contains what we re interested in 2 Surroundings everything else ASsys ASsurr 2 0 ENTROPY Entropy is disorder When hat exchange is ngt involved1 spontaneous processes always increase entropy I Consider diffusion of gas in two connected gas bulbs open stopcock gt A A 4 A 139 atm 0 atm lntm Vatm Gas spontaneous expands increasing entropy Thermodmg ic de nition of entropy Entropy is de ned via a change in heat i e a change in choatic motion 39 Lev AS T A qrev heat in a reversible process AS E n8 productsl 2 mS reactants 1 Miscellaneous notes on entropy and molecular motion 9 Entropy increases during phase changes 39 S gt 1 1 gt g 39 S gt g o Entropy increases when number of particles increases Entropy increases when temperature or de eeipf freedom increase Number of modes of motion increases Types of molecular motion 0 Eaasnaoaal Particle moves in a straight line i e particle is translated Accounts for most entropy in m Accounts for substantial amount of entropy in liquids Has no contribution to the entropy of solids b Yibrational 661 s Tn a bond vibrate as if on a spring Accounts for substantial part of entropy of ligg39d vibrations occur within molecules vibrations occur between molecules librations Accounts for all entropy in solid Vibrations of water c i BOEtiBIeL Molecule rotates on axis Entropy measures with a ratio the tendency of energy to spread out to become less concentrated in one physical localization or one energetic state Molecules are incessantly speeding through space colliding rotating gas liquid and vibrating solid They tend to spread out their energy by moving rotating vibrating in as many ways as possible Forexample consider the reaction between nitric oxide gas and oxygen gas to form nitrogen dioxide gas 2N0g 028 39 2N02g 39 39 Tire entropy change for this reaction is negative AS lt 0 because 3 molecules of gas react to form 2 molecules of gas The formation of thin w bonds decreases the number of degtegof rgedom or forms of motion available to the atoms 39 quot If we decrease the hergal energy of a system by lowering the temperature the energy stored in translational vibrational and rotational forms of modem decreases 39 Entropy changes that occur as the temperature of a crystalline solid rises from absolute zero Absolute entropy S Temperature K gt 12 l THIRD LAW OF TIIERMODYNAMICS quotSubstance 5 JmolK Gases Hzg 1306 N2g 1915 023 2050 Unlike enthalpies of formation the standard molar entropies of elements H20g 1838 are not zero NH5g 1925 The standard molar entro i tan all 39 39th39 39 1a CHsOHQ 2376 0 pesg er ymcreasewr increasm mo mass of the substance g I C5H5g g 2692 The standard molar entropies generally increase with the number of atoms 3 39Liql ds 1n the formula 9f the substance 9 the number and importance of the vibrational degrees of freedom of molecules increase with increasing mass and increasing C 3 1 17239s number of atoms r 61 161 Solids Fes 272 sedge 1423 NaCl5 723 As less energy is stored the entropy 39of the system decreases If we keep lowering the temperature do we reach a state in which these motions are essentially shut down a point of perfect order I The entropy of a perfect crystal is zero at absolute zero 0 K Consequences of the Third Law Absolute zero is unattainable Entropy of all substances at absolute zero is zero At temperature above zero crystal will not be perfect Vibrational motion introduces imperfections 39 To remove imperfection takes some sort of motion Reconsider 2nd Law of Thermo Ast ASsurr 2 0 Using the thermodynamic de nition of entropy the entropy change of the surroundings can be related to the heat of the system ll qsgt i ASsurr T T At constant pressure C1sys AHW5 Therefore the second law can be rewritten as AH AS Tmzo gt TAS AHZO sys At constanttemperature AKGrAHTAS l gt AH TAS SO gt AGE 0 AG is use il to decide if a reaction Occurs AG lt 0 nm is spontaneous AG 0 system is at ethbrhnn Recall ASuniv O for reversible process AG gt O rxn is nonspontaneous ie rm does not happen 5 AG AH TASquot l z r NotewAH is the change of energy that occurs in a chemical reaction difference between the 39 energy required to break the chemical bonds in the reactants and the energy produced by the formation of the chemical bonds of the products a AG it is the portion of the energy change of a spontaneous reaction that is ee to do useful work The remainder of the energy enters the environment as heat T A S sys q sys T of free can be summarized in table Standard FreeEnergy Changes 39 t State Standard J 39 I E of Matter State 39 39 I 39 Solid Pure solid AG 2 nAGS products E mAGf reactants 1 Liquid Pure liquid 39 Gas 1 atm pressure I Solution 1 M concentration v C A We de ne the reaction quotient Q D3 EATquot LB Forageneral reaction aAbB Sc C dD a direction of RX Reaction AG Q K In QK Spontaneous make more product lt 1 Non spontaneous make more reactants gt 1 Equilibrium no change 1 w There is a direct relationship between AG and In QK AG RT 111ng a I D 17 4 1513 j t O We generally refer to reactions where reactants and products are in their standard states gt cc 1m P1ann A 557 A k QTz 1V 6 AG RT an RT an AG RT1n1 RTanO RTan AGO RT In K makes it possible to calculate K knowing G0 Makes it possible to calculate Go knowing K K gt1 means AGO lt0 the reaction starting with reactants and products in their standard states is Spontaneous to make more products K lt1 means AGO gt0 the reaction starting with reactants and products in their standard states is Non spontaneous to make more products V J 1 r L at R fx39 quotW 39 Ki Spontaneous l AG lt 0 K if a mixture has too much N2 and H2 relative to NH3 the equilibriumlies too far to the left Q lt K and the mixture will react to form NH spontaneously If there is to much NH3 in the mixture the equilibrium lies to the right Qgt K and the NH3 wlll decompose spontaneously into N and H2 Free energy At equilibrium Q K and the free energy is at a minimum A6 0 Pure Equilibrium Pure N2 H2 mixture Q K AG 0 L Chm ACIDS BASES AND SALTS 9 Naming acids Binagy or hydrohalic acids HF HCl HBr H H28 etc hydro ic acid HF is the only weak hydrohalic acid Although the H F bond is very polar the bond is so strong due to the small F atom that the acid does not completely dissociate K Oxyacids contain a polyatomic ion A Most common form CF ic endin strong acids 1 HNO3 nitric from nitrate N03 2 HzPos phosphoric from phospate 3 HZSO4 su1furic from sulfate H HClogchloric from chlorate Acids with 1 less oygen than the MCF ous ending weal eis acids B l HNOZ nitrous from nitrite 2 HgPOgphosphorous from phosphite 3 stogsulfurous from sulfite 4 HCngchlorous from chlorite C Acids with 2 less ox ens than the MCF hypo ous 39Very weak acids 1 HNOhyponitrous 39 2 H3P02hypophosphorus 5 3 HClO hypochlorous D Acids with I more ox en than the MCF per ic ve39ry strongjacids Kc 1 HC104 perchloric acid 0 E Organic acids haveCarboxvmroup ECOH 1 usually weak acids 2 acetic acid HC2H302and benzoic acid C7H5COOH Acid Base Theories A Arrhenius Fahd 39 Ciddonates IT proton in water BC HOH gt H30 Cl39 2 Arrhenius basedonates OH ion in water NaOH gt Na OH fr N lt B BronstedLowg Theogy lAltid ates a proton H in water 3 Baseaccepts a proton in water HCl HOH 9 1130 Cl39 Acid base 39 Conjugates products Conjugate basewhat is left of acid after is loses a proton Conjugate acidwhat the base becomes after it gains a proton v NH3 HOH gt NH OH Base acid conjugate acid conjugate base C Lewis AcidBase Theory acids are electron pair acceptors Mgrquot BC13 B only has 6 electrons quot 39 Bases are electron pair donors NH rj 7gt39 33 C M 7 31 Strength of Acids de ned by eguilibrium positions W f 7 39 f V 3 J39 J of L A Strong acidsequilibrium lies far to the right A strong acid completely dissociates in water 39 HCl H20 9 H3O Cl39 HNO3 H20 9 1130 N0339 A strong acid yields a wk conjugate base much weaker than H20 memorize list HCl HBr HI HNO3 H2804 HClO3 HClO4 B Weak acidsequilibrium lies far to the left Very little of a weak acid dissociates Most of it stays in its molecular form HC2H302 H20 H30 Cngoz I A Maud has a strong conjugate base stronger than water C Acid dissociation constant CK 39 Forthe reaction HA H206 H30 A39 then39 H20 is gt part of the equilibrium equation the concentration of water is so much larger than any other component of these solutions that we can build it into the equilibrium constant Because the concentration of water is so large the reaction between an acid and water is a pgeydojritgrder reaction that only depends on the concentration of the base or the acid xi Weak acid strength is compared by the K values of the acids The smaller the K the weaker the acid Strong acids do not have K1 values because the HA is so small that it cannot be measured 1 o Amphoterism Amphoteric substancesubstance that can act as an acid or as a base Ex H20 Autoionization of water H20 H20 H30 OH39 w Base acid conjugate conjugate q I f I 4 p Ilr Acrd base 39 zzy p a Z 53 j M 56 4 MmEmf 7 Ign product constant for water Kw Kw H300H1 a quot 39 39 Kwlffll0Hl 7 7 117351 0 At 25 c Kw 10X103914 molzLZ because H OH39 10X10 7M 74 44 2 we matter what an aqueous solution contains at 25P C H MOH 10 x 1039 I V w H Neutral solution H OIIj 1 44 7 7 51 Acrdic solution H gt OH H 1 nilL we W 4 7g Basxc solution H lt OH39J g6 R 3 tgt 0 1 39 Kw varies with temperature 95 4quot 37 ill 39 K Bases Strong Base one which dissociates completely in water and has large Kb values Alkali metal hydroxides BaOH2 SrOH2 CaOH2 MgOH2 limited solubility longer 30 gt lower laws emeraj gt dissed ahm General reaction for a base B and water is given by Baq H200 BHaq OH39aq Base Acid Conjugate Conjugate Acid Base The equilibrium constant for this equation BHOH39 K b B Weak Base small Kb values ex NH H20 6 NH OH Base acid conjugate acid conjugate base The lone pair on N forms a bond with a H Most weak bases involve N Relationship between Ka and Kb 5 Consider Kb of ammonia KbNlI3 18 x 10 NlI3 aq H20 1 SNH4 aq OH aq NHJ lOH l NHgl Now consider K of conjugate acid NH4 NH4 MD H20 1 SNH all H30 21G H30iiNH3i quot NHfl Kb Now multiply Ka and Kb lNHalleo xlNHXlOH l NH m3 H30OH39 KaXsz TE The pH Scale pH Scale provides a convenient way to represent solution acidi pH 409 H f Scales is from O to 14 actually slightly greater and less than this l0 1415 acidic pH 7 neutral basic l39pH pOH 14 3 To nd H 39om pH 11 antilogpH and to nd OH from pOH OHI antilog pOH pH and pOH are logarithmic functions The pH changes by 1 for every power of 10 change in H pH decreases as H increases To calculate the pH for a strong acids pH log V A if the acid concentration is less than 10 x 107 the water becomes the important source of H and the pH is 700 The pH of an acidic solution cannot be greater than 7 at zs cmm Another exception is calculating the pH of H2804 solution that is more dilute than 10 M At this concentration the If of the HSO439 must also be calculated b strong bases pOH log OH39 539 Weak Acid Solution Step 1 List the major species in the solution I I Step 2 Choose the species that can produce H and write a balanced equation for the reaction producmg H Step 3 Using the values of the equilibrium constant for the reactions you have written decide which equilibrium will dominate in producing H R Step 4 Write the equilibrium expression for the dominant equilibrium L Step 5 List the initial concentrations of the species participating in the dominant equilibrium 6 Step 6 De ne the change needed to achieve equilibrium that is de ne x E Step 7 Write the equilibrium concentrations in terms of x Step 8 Substitute the equilibrium concentrations into the equilibrium expression Step 9 Solve for x the quoteasyquot way that is by assuming HA0 X HA0 Step 10 Using the 5 rule verify whether the approximation is valid Step 11 Calculate Hquot and pH 500 RUIB x HA X 100 lt 5 then the approximation is acceptable 0 39 Adxjrl f W W A We g 39 V concentration of ions ionization 39 mo 1 concentration of acid K 39 percent of ionization li Wquot quot7 When the K of an acid is aboutg ythis usually means they are ionizing less taxi09 unless it is a very dilute solutiog 39 a 39 The percent of ionization of a weak acid 6390 Increases with decreased concentration 390 EX CH3COOH H20 23 CH3CUO39 y H30 S 0 According to Lechatelier more water will make the reaction shift to the product side more ions Bercent ionize O 005 010 015 Acid concentration M mas61 WM Maia 39 d Polyprotic Acids Polygrotic Acid acids which can furnish more than one proton Characteristics of Week Polvbrotic Acids 1 Typically successive Ka values are so much smaller than the rst value that only the rst dissoeiation step makes a significant contribution to the equilibrium concentration of H This means that the calculation of the pH for a solution of a typical weak polyprotic acid is identical to that for a solution of a weak monoprotic acid 2 Sulfuric acid is unique in being a strong acid in its rst dissociation step and a weak acid in its second step For rTe latively concentrated solutions of sulfuric acid 10 M or higher the large concentration of H from the 5 rst dissocration step represses the second step Which can e neglected as a contributor of H ions For dilute solutions of sulfuric acid the second step does make a signi cant contribution and the quadratic equation must be used to obtain the total H concentration AcidDissociation Constants of Some Common Polyprotic Acids Z Name 4 Formula K KHZ 39 Kn3 39 v a r 391 39 Va 3 mg 4 Ha I a i AscogE Hzc H o6 80 x 107 16 x 10 I 9 Carbo 1c H2C03 43 x 10 7 56 x lo39j 39 s Citric H3c61 1507 74 x 10 4 17 x 10 40 x 10quot a 4 Oxalic HJCZO 459 x 10392 64 x 10quot I 3 8 x 10l3 7 z7 45 Phosp J H3PO4 75 x 10 1 62x 10 I 4 41 gt74 KN H 00 6 7 7 Sulfur H2803 17 x 10quot 64 x 10b I 1 3 7 Sulfuric H3804 Large 12 X 10 J P A Z Piaf l iJgafrartan Hzc mo6 10 x 10 3 46 x10quot fess Z L JLvavW rt 1 3 t Note as long as the successch Ka va differ b a factor of 10 we 1 15 V possible to obtain a satis actory estimate of the pH 39 a polyprotlc aCId solution b considering 9m Kai 6 Determination of the pH of a mixture of weak acids Only the acid with the largest Ka value will contribute an appreciable H0 Determine the pH based on this acid and ignore any others Y Predicting the direction a reaction will proceed at equilibrium the reaction proceeds in thedii btion of the weaker acid and baseRemer b r the stronger the acid the weaker the conjugate base and the stronger the base the weaker its conjugate acid CH3COOH CN39 7 CH3COO39 HCN Acid base conjugate base conjugate acid Looking at Table HCN is a weaker acid than CH3COOH and CH3COO39 is a weaker base than CN so at equilibrium the reaction favors the right side notice the arrows K HQ mem K name s CH3on u39 ACID BASE PROPERTIES OF SALT SOLUTIONS 1 Hydrolysis Reaction of a substance with water which creates acidic or basiS conditions Substance causes water to split apart lyse 9 3531le ideas salt l Salts don39Ved 39om mixing strong acid and strong base do not affect pH Neither ion hydrolyzes water Examples pH of KBr solution is neutral 2 Salts derived from strong base and a weak acid raise pH ie cause basic conditions Anion hydrolyzes water Example pH ofNaC2H302 solution is basic I Strong base attracts H from HOH C2H302 H OH 22 C2H30H OH 3 Salts derived from strong acid and a weak base lower pH ie cause acidic conditions Cation hydrolyzes water Example pH of NH41 solution is acidic 131 aq H20 1 SNH3 aq H3O aq 139 aq 7 Strong acid gives away HTL NH4 1 H NH3 6 Note When cation is pan of an insoluble hydroxide the cation will hydrolyze the water to create acidic condition pH of AlCl3 solution is acidic AlCl aq 3 H20 1AlOH3 s 3 HF aq 3 Cl aq Remember the production of a precipitate shifts the equilibrium to the right v o Acidic salts are also produced when the cation is a hydrated metal ion usually with a Charge of 3 or more These behave as weak acids The higher the charge on the metal the stronger the acidity of the hydrated ion The electrons are pulled away om the 0H bond and toward the positivelycharged metal ion Ex AlCl3 AlNO33 F eC13 and Ala120 K values will be given to you in this type of problem Ex A1c13 M6H20 gt AlI2053 3C1 H20 Mutant 9 window H These are usually complex ions Set1pm 4 Salts derived from Weak acid and weak base Both cation and anion hydrolyze water pH depends on which ion has stronger acidbase properties if base is stronger ie Kb gt Ka then pH gt 7 if acid is stronger ie K gt Kb then pH lt 7 k Example NH4N02 K 100 gtlt10quot14 KbltN02 KAI 1102 45gtlt1O 4 22 gtlt10 11 Ka gt Kb thus pH lt 7 NH4N02 solution is acidic 5 CalculatinoI the pH of SaltsWhen determining the exact pH of salt solutions we can use the Ka of the weak acid formed to nd the K of the salt 0R we can use the K of the weak base formed to nd the KI of the salt IKXKbKw Steps for nding the pH of a salt 39 l Ionize the salt Look at both the cation and the anion Determine which ion comes from either a weak acid or a weak base Ignore the neutral ion 2 React that ion with water 39 quotx 3 Calculate the K3 or Kb for that ion by dividing the Kw by the Ka of the weak acid or Kb of the weak base from which the ion was derived 4 Set up the eguilibrium expression using the reaction from 2 above 5 Solve for x 6 Calculate 911 or pOH VII CHEMICAL STRUCTURE or ACIDS AND BASES 9 Easieslhatsftect cid SJTEILgthi The strength of an acid is often the combination of three factors 5 8 The polarity ofthe HX bond it 4quot 4 7 4 I r 839 42 k 4 x Note ionic hydrides such as NaH are basic A 7 quot H is aproton acceptor NaH HOH Na OH H2 The strength of the HX bond very strong bonds ex H F are less easily dissociated than are weaker ones HF is weak acid whereas the other hydrogen halides HCl are strong acids in water The stability of the conjugate base X In general the greater the stability of the conjugate base the stronger is the acid b E39923 The strength of an H X bond tends to decrease as the element X increases in size the acidity of the binary acid increases H are more easily donated HC gt HF Moving from left to right in a row the bend strength is not the determinant factor of acidity The bond polarity electronegativity of X prevails the acidity increases with the electronegativity of X from left to right H F gt H ac 5A 7A 5quot CH 13 Acid base 4 NH3 H20 HF properties of the binary i No acid or I it compounds of hydrogen and base properties weak base weak and 99 4 nonmetals of periods 2 andJ3 3H FE d 3 4 PH3 H28 HCl g 4enO No acid or n it base properties Weak base Weak acrd Strong acrd g Inueasirig gquot 39 vi v 93 139 1 cOxyacids i base Strength acids in whichOH groups and possibly additional oxygen atoms are bound to a central atom Y o H v i Non metal For oxyacids that have the same number ofOH groups and same number of O atoms Acid strength increases with increasing electronegativity of the central atom gtthe 0H bond becomes more polar thereby favoring the loss of H H Cl C gt H Bro a For oxyacids that have the same central atom Y acid strength increases as the number Of oxygen atoms attached to Y increases gtT he additional 0 atoms pull electron density from the 0 H bond further 369 drift 1ncreasmg its polarity H Hypochlorous Chlorous Chloric Perchloric Cl39 2 W0 H CZ Cit HICIgt CII IQ H Q q c H OI Cll Ici O K 30 X IlO E K 11 X 10 Strong acid Strong acid lnn easing acid strength d Ersatz 263 The acid strength of carboxylic acids increases as the number of electronegative atoms in the acid increases g EX tri uoroacetic CF3COOH more acidic greater Ka than acetic acid CH3COOH I if x A A i 139quot 7 E 39 2 R LLU LEWIS ACIDS AND BASES I 3 De nitions 539 k 39i i 5 Lewis acid electron pair acceptor W 39 Lewis base electron pair donor ex Consider H30 aq OH aq gtHZO 1 H is electron pair acceptor Lewis acid 0 396 g H O 0 H o O h H OH39 is electron pair donor Lewis base 39 3 VV 7t q H F H F L 39 M 3 t 5 if 7 lt3 1 Fm 1544 y 1 l I V 2 I est H N 4 39 F gt H N B F r 39 l l I l 7 A 5 quot H F F Base Acid Vil R Lewis acids includemolecules that like BF3 have an incomplete octet of electrons Compounds with multiple bonds Ex reaction between C02 and H20 H 0 Hquot O H O an I 1 Q C H Q C gt H Q C H II N H o O 0 Water behaves as an electronpair donor and C02 as an electronpair acceptor n 0h 1 Lewis acidbase theory is useful to explain structure of metal ions in aqueous solution V giggination complexes y 63 Consider Cuz ion CuSO4 s H20 1 a CuSO4 aq very pale blue bright blue What has changed In the solid Cu2 is surrounded by 804239 a In the solution Cu2 is surrounded by H20 OJH30 3 2 H Cu acts as Lew15 acrd H20 acts as Lewis base 2 r H H 1 Fir6 539 quotgvti f 39 quotl 0 1 Kmquot Z W 39 x If H H i w 00 g r I H H L f o o la H H in V y 6 r 95 l l 39 Hints Coordination Numbers are usually 4 or 6 Acids break complex ions apart Look for transition metals and Aluminum especially Zn Fe Ag These will be hints that it is a complex ion reaction 9 Ag 2NH3 AgNH32r LCWIS Cid Lewis base g H H H N Ore H The stability of a complex ion in aqueous solution can be judged by the size of the equilibrium constant for its formation from the hydrated ion g Aga lH Agaq ZNHs AgNH3gt2 ltan Kr AgNH3Z 17 x107 The complex ion has a lower energy more stable than the separated metal ion and ligands molecules 39 DOLE a 39 Crystal eld theory I repulsive force between the outermost electrons on the metal and the negative charges on the ligands 3 b c d c m a An octahedral array of negative charges approaching a metal ion bl The orientations of the d orbitals relative to the negatively charged ligands Notice that the lobes of the d1 and d orbitals b and c point toward the charges The lobes of the dxy d and dx orbitals d f point between the charges Due to their different orientation and shape the d orbitals of the metal ion do not all behave in the same way under the in uence of the crystal eld this will cause the splitting of the energies of metal ion 1 orbitals In the isolated metal ion the ve d orbitals are equal in energy A transition metal complex will therefore be able to absorb visible Light which will excite an electron from a lower energy d orbital into a higher energy one f The magnitude of the gap of energy between the d orbital and consequently the gglgr of a complex depend on both the metal and the surrounding ligands nob 5 Acidity of hydrated metal ion f FeHzo63ltaq FeCHzO5OH2 aq 39 H39ltaq 39 The ow of electron density from the water molecule to the metal ion causes the OH bond to become more polarized as a result those water molecules donate If more easily The release of a HP acidity of hydrated ion will increase 39 and smaller size OH bond more polar Interaction of a water molecule with a cation of 1 charge or 3 charge The interaction is much stronger with the smaller ion of higher charge causing that hydrated ion to be more acidic Ex a soluti of Al N03 is more solution Weak electrostatic interaction eak electron 10 M solutions of a Salt NaNO3 calN03 I series of nitrate salts 39 Estimated pH 1 70 69 x r q H r 0 l x 0399 V 3 6 V A lt k0 I a e a l With a39greater metal charge or acidic a same concentration NaNOg Strong electron interaction o gt Strong electron shift Amen Aime 55 V 35 12 5 6 2 HRDgtG 3 44110 PilHQOSOH1 4H 3039 1


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