Popular in Chemistry 2
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Date Created: 10/15/15
Chapter 15 Chemical Equilibrium 915924 Dynamic Equilibrium Nzo e 2N02g Colorless brown The Reaction Quotient and Equilibrium Constant In general aAdbB cCdD Reaction quotientQc C C D g lProducts AT TBb lReactants Equilibrium constantKc C C D Q when system at equilibrium Aquot Bb Concentration molL KC and Qcare unitless Qc Problem 1a worked example 151 Write the QC for the following reactions a Nzezlg 2N02g QcN022N204 b Cd2aq 4Braq CdBr4239aq QcCdBr4239Cd2Br4 C H2C204aCI 2HaCI C204239aCI QcH2C204239H2C204 1b Write the equation for the reaction with the following QC Heterogeneous equilibria o NH3aq H20 OHaq NH4aq o OH39NH4 NH3 Problem 2 The following reactions have the indicated equilibrium constants at 700K kc1540 for H2g I25 2Hg kc2104x10394 for N2g 3mg 2NH3g What is the kC for a 2Hgtg H2g 2g kc185x10392 b NH g 12 N2 9 32 H2 9 kc981 c 2NH3g 3amp9 6HgN2g kc151X109 Gaseous Equilibria rate of forward rate of reverse aAbB cCdD kCCCDg QCCCDg Aquot Bb Aquot Bb All gaseous Kp l Q BQP PAaPBb Ppartia pressure Relationship KC and Kp KpKcRTAn Anmoes of gaseous productmoes of gaseous reactant R008206LatmK mol o TTemperature Kelvin o KpKC only when An0 Problem 3 For the equilibrium 2503g 2502g 02g KC is 408 x 10393 at 1000 K R00821 Latmmolk A does equilibrium favor the formation of 03 or 02 and 02 at this temperature B write the equilibrium expression Kc C calculate the value for Kp Answer A reactants because the kC is a very small B Kc X 5032 C Kp KCRTA 408 x 1039300821 x 100010335 Chemical Equilibrium and Free Energy In general aAbB cCdD kc ccog Aquot Bb a reaction w K very large gtgt1 has negative AG a reaction w K very small ltlt1 has positive AG Consider H2gl2g 2Hg Kc543 at 430 C 703 K if we start an experiment w a mixture of 0243 mole of H2 0146 mole of l2 and 198 mole of HI in a 100 L container at 430 C 703 K would more HI be formed or consumed QC 1982 392040035478111 H22 02430146 we want 111 to be 54 so the top number needs to decrease and the bottom needs to increase therefore HI needs to be consumed reduced In General aAbB cCdD lf QgtK the reaction proceeds from right to left until equilibrium is reached lf QltK the reaction proceeds from left to right until equilibrium is reached lf QK the reaction is at equilibrium Relationship AG and AG Reaction not at equilibrium AGAG RTan Rgas constant 8314 JK mol Tabsolute temperature K Qreaction quotient Problem 4 Consider H2g2s 2Hlg Given AG at 25 C260 kJmol PH240 atm PHI1O atm write Qp calculate AG and determine if HI will be produced or consumed in order to reach an equilibrium oQp 10 3 025 40 0 AG 260kJmo8314 X 298 X n025 083kJmo Relationship AG and K Reaction at equilibrium AGO and QK OAG R7inKor AG R7inK klt1 nKnegative AG positive reactants are favored not spont kgt1 nKpositive AG negative products are favored spont Problem 5 Practice prob 1598 The Kf for AgNH32aq at 25 C is 15 x 107 Agaq 2NH3aq AgNH32aq Using this and data from appendix 2 calculate the value of A61 for AgNH32aq AGf Ag AGf NH3 265 15 X 107 7712652 277042 17kJmol 409kJmol Problem 6 worked example 1512 KC for the reaction of hydrogen and iodine to produce hydrogen iodide H2g 2g 2Hg is 543 at 430 C What will the concentrations be at equilibrium if we start w I2000414M HI00424M H2000623M 004242000623000414 543 004242X2000623X0004l4X 543000623X 000414X004242X2 543258X10395 000623X 000414X X2004242X004242X 503X2 0735X 400X103940 l Xbib24ac 2a X525X10394 or 0151 H2000623525X10394000676 M l20004l4525X10394000467 M Hl004242525X10394004l4 M Le Chatelier s Principle Principle when a stress is applied to a system at equilibrium the system will respond by shifting in the direction that minimizes the effect of the stress Stress o additionremoval of a reactant or product 0 volume or pressure change 0 temperature change AdditionRemoval Effect N2g 3H2g 2NH39 Equilibrium l KNH32N2H23 Removal of NH3 Shifts right l in order to minimize the effect more reactants Addition of NH3 Shifts left Addition of N2 Shifts right l Removal of H2 Shifts left Volume Change Effect When volume is decreased the equilibrium is driven toward the side w the smallest number of moles of gas 0 Mg 3H2g 2NH3g o Pljw V shifts right l o Pljw V shifts left addition inert gas that is neither reactant nor product H2g l2g 2Hlg no effect Temperature Change CoC4239aq 6H20 CoH2062aq 4C39aq heat Blue Pink 0 Endothermic reactants heat products 0 Exothermic reactants products heat 0 Tl shifts left reactants o Tl shifts right products Problem 8 ConQden 4NH3g 5029 4NOg 6H20g AH904 kj How does each of the following changes affect the yield of NO at equilibrium Answer w increase decrease or no change a addition of NH3 increase yield NO b increase H20 decrease yield NO c removal of 02 decrease yield of NO AcidsBases Arrhenius Acid substance when dissolved in water products H Base OH39 Bronsted de nition 0 Acid proton donor Must have a removable acidic proton 0 HCg H20 C39aq H3Oaq acid base conjugate base conjugate acid 0 Base proton acceptor Must have a pair of nonbonding elections 0 NH3g H20 NH4aq OH39aq base acid conjugate acid conjugate base Conjugate acidbase pair acid amp base that differ only in one proton Conjugate AcidsBases Species Conjugate acid NH3 NH4 H20 H30 OH39 H20 H2NCONH2 H2NCONH3 Problem 1 a what is the conjugate base of H20 HCO339 PH4 OH CO3 PH3 b what is the conjugate acid of CN39 HCO339 0239 HCN H2CO3 HOquot Factors Affecting Acid Strength 1 bond strength stronger bond stronger it is to release proton 2 bond polarity Hydrohalic Acids HX Bond Enthalpies for Hydrogen Halides amp Acid Strengths for Hydrohalic Acids Bond Bond enthalpy klmoi Acid strength HF 5628 Weak HCl 4319 Strong HBr 3661 Strong HI 2983 Strong HltHBrltHCltHF increasing bond strength HgtHBrgtHCgtHF decreasing acid strength Oxvacids Increased electronegativity of the central atom causes increased acidity 0 HCO3 vs HBrO3 HCO3 stronger Increased number of bonded oxygen atoms causes increased acidity 0 HC03 vs HCO4 HCO4 stronger Which is the stronger acid HzSeO3 vs H2503 stronger acid H2503 HzSeO4 vs HzSeO3 stronger acid HzSeO4 Carboxylic Acids IonProduct Constant of water KW Amphoteric species that can behave either as a Bronsted acid or a Bronsted base H20 HCO339 Autoionization of water H20 H20L H3Oaq OH39aq 0 or H20 Haq OH39aq 0 KC H3OOH39 0 KW HOH39 The equilibrium expression KcKwH30OH39HOH39 Kw ionproduct constant for water At 25 C KW1O x 10 3914 n pure water 10x 10 397M Problem 2 in notebook pH and pOH scale pHogH3OogH and HJ 1O39pH since HOH3910 x 103914 OgHOH39Iog10 x 1039 pH pOH1400 p stands for og In pure water OH39H3O10 x 10397M pHogH3Oog10 x 10397700 Other pscales pOHogOH39 Conjugate acids and bases 0 A strong acid weak conjugate base 0 A weak acid Strong conjugate base 0 A strong base weak conjugate acid 0 A weak base strong conjugate base HCliaq I Haq C39aq HFaq Htiaq F39aq Acid conj base acid conj base C39aq H20 7 F39aq H20 HFaq OH39aq NaCI l Na Cl39 quotprotonA Relationship Ka and Kbacidbase conjugate pair NH3 H20 NH4 OH39 kb18 x 10394 NH4 H20 NH3 H3O ka56 x 103910 2H20 H30 kw10 X 10 14 ka X kb 0 the product of the aciddissociation constant Ka for an acid and the basedissociation constant for its conjugate base Kb equals the ion product constant of water 0 ka x kb kw kb CO3 X ka HC0339 Ka kw 10 x 10 56 x 103911 kb X ka HCO339 56 x 103911 and kb HCO339 23 x 10398 will HCO339 act as an acid or base in water base Polyprotic Acids If K is 1000 times greater than Kaz etc the acid can be treated as monoprotic when determining the pH AcidBase properties of salt solutions NaClNa Cl39 1 Anion conjugate base of a strong acid 0 Neutral eg Cl39 spectator ion 2 Anion conjugate base of a weak acid 0 Basic eg F39 H20 F39 H20 HF OH39 NaF l Na F39 l basic 3 Cation conjugate acid of a weak base 0 Acidic eg NH4 4 Metal cations of group 1A and 2A Ca2 Ba and Sr 0 Neutral eg Ca2spectator ion 5 Other metal cations undergo hydrolysis 0 Acidic eg Fe3 Cr3 Al3 Be2 In general if the salt contains an anion and cation that both hydrolyze the pH depends on the relative strengths of the weak acid Ka and base Kb 0 If kbgtka the solution is basic 0 If kbltka the solution is acidic o If kbka the solution is neutral or nearly neutral Oxides of Metals Metal oxides 0 Metal oxide water metal hydroxide basic solution Na20s H20 l 2NaOHaq 0 Metal oxide acid l salt water Na20s 2HNO3aq l 2NaNO3aq H20 o All the alkali and alkaline earth metal hydroxides except BeO are basic 0 Many other metals oxides including BeO are amphoteric Oxides of Nonmetals o Nonmetal oxides o Nonmetal oxide water acid acidic solution I H2C03aq 503H2504 N205HNO35 P4010H3PO4 C207HCO4 Nonmetal oxide base salt water C02g 2K0HaCI I K2CO3aCI H20 Lewis acids and bases 0 Lewis acid electronpair acceptor atoms w an empty valence orbital can be Lewis acids 0 Lewis base electronpair donor BrQnsted base talking about protons Common lon Effect 0 Addition of KN02 to an aqueous solution of HNOZ 0 Initially HN02aq l N0239aq Haq 0 Addition KN02aq l N0239aq Kaq 0 eq shifts l pH increases ionization decreases o CommonlonEffect the extent of ionization of a weak acid decreases in the presence of a strong electrolyte that shares a common ion the same holds for a weak base Buffers Buffer a solution of a weak acid and its conjugate base or a weak base and its conjugate acid 0 Buffers resist changes in pH when a limited amount of acid or base is added Eg CH3COOHCH3COONa or NH3NH4CI or HN02KN02 in notebook HendersonHasselbalch equation 0 Consider the buffer HAMA HAA39 HAlaqI1 Haq A39laq Ka mm kaHA HKaM HA HA A39 ogHlogKa og HA pHpKa ogA A39 HA pH pKa ogconjugate base HendersonHasselbach eq weak acid Effective Buffers 1 10 2 conjugate base 2 01 weak acid 2 pH pKa 1 buffer range Problem 4 a how would you prepare a buffer with pH 30 answer mix 10 mol nitrous acid WI 045 mol sodium nitrite in 10 L water b if you add 0010 mol HCl to buffer in a 4a calculate its pH answer pH299 c if you add 0010 mol HCl to 10 L water what is the pH Strong Acid Strong Base Titration HClaq NaOHaq l NaClaq H20 Weak Acid Strong Base Titration
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