Principles Chemistry II
Principles Chemistry II CHM 212
Popular in Course
Popular in Chemistry
verified elite notetaker
This 0 page Class Notes was uploaded by Houston Kovacek on Sunday November 1, 2015. The Class Notes belongs to CHM 212 at Marshall University taught by Staff in Fall. Since its upload, it has received 64 views. For similar materials see /class/233289/chm-212-marshall-university in Chemistry at Marshall University.
Reviews for Principles Chemistry II
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 11/01/15
Chapter 17 Additional Aspects of Aqueous Equilibria In the last chapter we talked about acid and base solutions in particular weak acid and base solutions That discussion was ne for laboratory work but it falls short of describing the more general nature of solutions Many real solutions contain several species in addition to the solute and solvent We39ll talk about three types of equilibria in this chapter buffers slightly soluble salts and complex ions 171 The Common Ion Effect Consider a solution of a weak acid say acetic acid CH3C02H HOAc HOAc Hf OAc39 What happens to the equilibrium if we add some NaOAc Le Chatelier39s principle tells us the equilibrium will shift left to remove some of the added ion What is the effect on pH when this happens There are fewer hydrogen ions present so the pH goes up This should come as no surprise since we just added a base to the solution This simple observation can help avoid mistakes when working problems Try to write out a Le Chatelier39s principle argument for the change in pH EX What is the pH of a solution that is initially 0500M in both HOAc and NaOAc HOAci 0500M HOAce 0500 x M Hm 0M Hue xM OAci 0500M OAce 0500 XM Ka W 13 X10s 0500 x Since 0500 gt 100Ka we can assume X is negligible in the calculation 18 X105 x0500 39 0500 X 18 X 10395 thus He 18 X 10395M gt pH 474 This is the same pH obtained for a solution that is 0300 M in HOAc and NaOAc EX 171 p 662 We will see later that in general for a weak acid if HA A39 then H Ka MiXing an acid and its conjugate base eg acetic acid and sodium acetate as described above can generate these kinds of solutions or they may be generated by either adding some strong base eg sodium hydr0Xide to a weak acid solution or by adding a strong acid to a weak base solution We will eXplore this further when we reach titrations in Section 173 HOAcWD 12 NaOHaq gt 12 HOAcWD 12 NaOAcGuD EXample 172 p 663 demonstrates the following rule If a solution contains both a strong acid and a weak acid the pH is determined m by the concentration of the strong acid The strong acid dissociates completely and the hydrogen ions released retard the dissociation of the weak acid This makes the weak acid a negligible contributor to the hydrogen ion concentration It is widely and incorrectly believed that orange juice can cause an upset stomach because of its acidity While many people have trouble tolerating orange juice a calculation like that in Problem 172 shows that a change in pH is not the cause of the distress All of the discussion above applies to basic solutions as well All you have to do is switch the word base for acid and Kb for Ka 172 Buffered Solutions Solutions that resist a change in pH are called buffers These solutions contain both a weak acid to consume OH39 and a weak base to consume H Most commonly they are a conjugate acidbase pair eg acetic acidsodium acetate They undergo small pH changes when small amounts of strong acid or base are added It turns out the math is a little simpler for buffered solutions than for weak acids or bases dissolved in water Because signi cant amounts of both an acid and its conjugate base are present ionization is suppressed Again let s use acetic acidsodium acetate HOAc Ht OAc39 In the case of pure acetic acid HOAce HOAci 7 X He OAc39e x In contrast in a buffer the concentrations look like this HOAce HOAci X OAc39e OAc39i 7 X He X But according to Le Chatelier39s principle the addition of OAc39 retards the ionization of HOAc This results in an X that can be neglected in both its subtraction from HOAci and addition to OAc39i This may be clearer while looking at the problem on p 10 of the Chapter 16 notes The result is HOAce HOAci and OAc39e OAc39i for a buffer solution Consider the hypothetical acid HA HA H A39 Ka and assume there is an appreciable amount of both HA and A39 present in solution so we can neglect ionization Ka Hum HA H Ka Hl A now take the negative logarithm of each side HA lo H lo K g g a A HA H lo lo p gKa g A A H K lo P P a g HA base In general pH pKa log 1W This is called the HendersonHasselbach equation It is very convenient if one works a lot of buffer problems although it isn t necessary One may still work the problem as for a weak acid and I ll work it both ways in the next example For a pH change of 1 unit 90 of the original weak acid or base must be consumed Prove this to yourself One common mistake using the equation is forgetting which goes in the numerator the acid or the base Here s a simple way to remember Adding a base increases pH and pH will only increase if the base goes in the numerator Addition of Strong Acids or Bases to Buffers It is probably best to address this subject through an example so we ll go over one here Ex A buffer solution is made containing 050 moles of acetic acid and 050 moles of sodium acetate The volume totals 10 L What is it s pH After adding 010 moles of hydroxide ion 040 moles of hydroxide ion total a b Short method HOAci 050molHOAC 050M L 04cm 050 n olom 050M pKa log 18 x10395 474 pH 474 WM 474 050 HOAC molHOAc 050 molHOAc 7 010 molOH 040 mol HOAc 040molHOAC 040M L molOAc 050 molHOAc 010 molOH 060 mol HO 060M HOAC 060 1101 060M pH 474 IogA 492 040M HOAC Long method See Chap 16 notes p 10 for a similar calculation The calculation of the initial concentrations is same as for the rst column HOAce 040 4 XM He XM 0Ac39e 060 XM X060 x 040 x 18 X 105 Since HOACL gt 100Ka assume 040 X m 040 and 060 x 060 18 X 105 040 X H 12 X 10395M Check if assumption ok pH logl2 X 10395 492 If no buffer had been present ie pure water the pH would have been 1300 c This is worked the same as b The nal pH is 569 80 ofthe original acid is consumed here The capacity of a buffer is the amount of strong acid or base that can be added without a signi cant change in pH This de nition may sound nebulous but what constitutes signi cant change depends on the situation For example if blood pH changes only i005 pH units see p 651 from its expected value a medical condition will develop A change of 05 units usually results in death In more typical chemical reactions maintaining a pH range of 01 pH units is typically unnecessary so significant carries a different meaning What is always true is that the more buffering agent present in solution the higher the buffer capacity Thus 1 L of a buffer solution that is 10 M in both weak acid and weak base will have more capacity than 1 L of a buffer solution that is 01 M in both acid and base 173 AcidBase Titrations You should review the section on titrations at the end of Chapter 4 p 140 before going on You will be doing several titrations in CHM 218 and you need to be familiar with the theory and calculations before going into the laboratory Recall that a titration is a procedure in which a solution of known concentration is reacted with a known volume of a solution of unknown concentration in order to measure the latter s concentration Historically the end of reaction was signi ed with an organic dye that changed color when the reaction was complete Today pH meters allow the endpoints to be determined with greater precision Strong Acid Strong Base Titrations Consider the titration of a known concentration of the strong base NaOH e g 010 AI into a solution of an unknown concentration of the strong acid HCl The most convenient method of following a titration is to make a plot of the volume of added titrant vs pH Initially the pH of the initial acid solution is quite low for a 010 M HCl solution pH 100 The pH will increase slowly at the beginning When the number of added moles of base begins to approach the number of moles of acid in the unknown solution the rate of pH increase will accelerate Typically within a few tenths of a milliliter of the equivalence point pH will change several units The equivalence point occurs when the moles of titrant equal the moles of unknown The pH will continue to increase rapidly for a short distance before beginning to level off below the pH of the base solution remember it is diluted by the water in the acid solution so the final solution will never achieve a pH of 1300 This experiment is shown graphically below pH OmeOOONb o 40 Vol OH mL Before going further here s a question to think about How do we know when the equivalence point occurs In real life we don t calculate the curve an experiment is performed in which we measure pH as a function of added acid or base Today we use a pH meter but that wasn t available a hundred years ago Before pH meters and even today when only the endpoint is needed organic dyes indicators were used to signal when the titration was complete These dyes dramatically changed color on going from a protonated to deprotonated form For example phenolphthalein changes from colorless C20H14O3 to bright pink C20H13O339 at a pH of 82 7 10 It is interesting to note that phenolphthalein is used to determine the endpoints of strong acid and base titrations even though their endpoints occur at pH 7 Here are two questions for you 1 Why is it all right to use phenolphthalein here and 2 One hundred years ago how did they know to use phenolphthalein instead of methyl orange pH change from 32 7 44 red acid form yellow base form for strong acidbase titrations Find the answers below We should now calculate some of the points on the displayed titration curve Figure 176 p 672 The initial point is simply the pH of a 0100 M hydrochloric acid solution Let s look at the addition of 10 mL of 0100 M sodium hydroxide solution and 499 mL Try calculating the pH after adding 4999 and 4999 mL of the sodium hydroxide solution on your own We begin by determining the number of moles of acid in the original solution molHinitial 00500 Lsoln M 000500 molH L 1 molHcl soln Now write out the balanced equation and determine the number of moles of base added HCIW NaOHW gt H20 NaCIW 101 00100 L 31110100mam 11110101 000100 mol K1 molN OH J OH soln The amount of acid remaining in solution equals the difference between these numbers then calculate concentration and finally pH molHend 000500 molH 000100 molOH 000400 molH Hq 000400 molH 00667M 00600 L501quot pH log00667 118 In an exactly analogous manner the pH of the solution after adding 499 mL of sodium hydroxide solution is 400 It might strike you as really odd that the pH increases so dramatically on addition of such small amounts After all the rst 10 mL of base increases pH by 018 units but the 09 mL increase from 490 mL to 499 mL results in a 100 unit increase see book for the 49 mL value Remember the pH scale is logarithmic As the titration approaches equivalence H approaches zero Thus very small amounts of base cause a large percentage change in remaining acid concentration Before answering the two questions raised on the previous page we need to go over a definition The endpoint of a titration is the volume of titrant which causes the indicator to change color In principle this can lead to a difference between the volume read at the endpoint and that required for the equivalence point If you repeat the calculations above for 50001 mL and 5001 mL of added base you ll see that it takes a negligibly small amount of base to trigger the indicator and so we can use the endpoint volume as the equivalence point volume Phenolphthalein works because even though it doesn t change color until pH 5 85 the titration curve is almost vertical here Only a small fraction of a drop moves pH from 7 to 85 Methyl orange on the other hand will require several mL of titrant to effect the color change One requirement for an indicator is that the color change occurs rapidly They knew that the dyes worked because they could use authentic samples of an acid and its conjugate base to track color changes Weak AcidBaseStrong BaseAcid Titrations There are two significant differences between these titrations and those where both components are strong The first is the appearance of the curve Because the acid or base being titrated is weak the vertical portion of the curve is compressed somewhat see below The second difference is that the final pH is lower than 700 for a strong acidweak base titration and higher than 700 for a strong baseweak acid titration This is because the products of a strong baseweak acid titration are water and the conjugate base of the weak acid Remember that this 10 conjugate base will also be a weak base Your book uses the word neutralization here In this context neutralizing the weak acid means completely removing its proton not that the solution has a neutral pH The calculations are more involved because one must determine the equilibrium value each time Fortunately the HendersonHasselbach equation can be used for these titrations HCI curve 4o 60 Vol OH mL Finally the weaker the acid or base the less pronounced is the break in the titration curve until at a Ka or Kb of much smaller than 10397 it becomes dif cult to see where it occurs see Figure 1711 p 658 Everything said so far about weak acids applies in reverse to weak bases Read Titrations of Polyprotic Acids on your own 174 Solubility Eguilibria So far all of the equilibria we have discussed in the past two chapters have been homogeneous equilibria We will now explore heterogeneous equilibria Recall that all ionic compounds are strong electrolytes but some have low solubilities These slightly soluble salts establish an equilibrium with the solution For the most of the rest of this section we will use magnesium hydroxide as an example Mg0H2 s Mg2aq 2 OHaq 11 The equilibrium constant for this type of equilibrium bears a special name the solubility product constant Ksp In this case Ksp Mg2l0H39l2 Remember the parent magnesium hydroxide does not appear in the equilibrium because it is in a different phase see Chapter 153 Unlike acid and base solutions where everything is soluble Ksp is used to determine the solubility of the salt under a variety of conditions We ll look at several over the next few pages At this point it is important for you to recall that solubility refers to the mass of solute that will dissolve in a given volume of solvent frequently gL Molar solubility is molarity M Solubility and the solubility product constant are related but not necessarily linearly Thus it is incorrect to say that because the Ksp of some salt is smaller than that of another it is necessarily less soluble We ll come back to this shortly Ex What is the Ksp of magnesium hydroxide if a saturated solution contains 165 x 10394 mol of Mg2 ions per liter of solution Ksp Mg l OH392 165 x10394 molMg2 if Mg 1 00 L 165 x10394M 1mol 2 1 then OH39 165 x10394molMg2 amp j 330 x10394M 1molMg2 1molMgOH2 100 L Ksp 165 x 10394330 x 103942 180 x 103911 Now let s look at the same problem in reverse Ex What is the molar solubility of a saturated magnesium hydroxide solution Ksp Mg2f OH12 180 x 1011 Let Mg2f X then OH39 2X from stoichiometry X2X2 180 X 103911 4X3 180 X 103911 X 165 X 10394 MgOH2 165 X10394 molMg21molMgOH2 165 X 10394M L K 1molMg2 soln It is a very common mistake to forget to multiply the hydr0Xide X by 2 or to square the 2X but it there is twice as much hydr0Xide as magnesium ion and it must be taken account of in both places EX What is the solubility of magnesium hydr0Xide 4 165X10 mol 5832 solubility L MgOH2 J gMgOH2 J 000962 gL soln K 1InOIMgOH2 175 Factors That Affect Solubility There are a number of factors that affect the solubility of slightly soluble salts They can range from a simple application of Le Chatelier s principle to properties inherent to the salt in question We begin with the first case The Common Ion Effect It has long been known indeed it predates Le Chatelier s principle that adding a salt that contains an ion present in the slightly soluble salt depresses the salt s solubility This is just what you d eXpect based on Le Chatelier s principle Skip A Closer Look on p 682 EX What is the molar solubility of magnesium hydr0Xide in a solution which is 10 M in magneSiU m Chloride This is the amount of added Mg2 ion from the MgClz plus the amount that dissolves from the MgOH2 Mg210XM OH39 2XM 18 X 10 11 10 X2X2 We can now make an assumption similar to that done for weak acidbase equilibria Since 10 gtgt Ksp we will assume X is negligible in the calculation 18 X 10 11102X2 X 21 X 10396 the assumption is correct 21X10396 mol 1mol J 21X10396M soln 1 In01Mg2 This is much less than the 165 X 10394M found when magnesium hydr0Xide is dissolved in pure water Solubility and pH Now in addition to being a salt magnesium hydr0Xide is also a base Therefore you d eXpect that if we raised the pH of the water we were dissolVing the MgOH2 into the latter would be less soluble another Le Chatelier s principle effect As a reference point saturated MgOH2 has a pH of 1052 EX What is the molar solubility ofmagnesium hydr0Xide at pH 1100 at pH 1000 pH 1100 3 pOH 1400 71100 300 3 OH39 antilog300 100 X 10393M 18 X 10 11Mg2100 X 103932 Mg218 X 10395M MgOH2 18 X 10395 molMgz JflmolMng2 18 X 10395M Lsoln K 1m01Mg2 pH 1000 2 pOH 1400 71000 400 2 OH39 antilog400 100 x 10394M 18 x 10 11Mg2fl00 x1042 Mg218 X10393M 3 MgOH2 18 X 10 molMg2 l molMgOH2 J 18 X 10393M L soln 1InOIMg2 This calculation shows us that the solubility of magnesium hydroxide drops changes rapidly changing pH Not surprisingly lowering the pH raises the solubility of MgOH2 pH can still have an effect even when the common ion effect plays no role Consider the solubility of silverI cyanide in acidic solution AgCNS x Agaq CN39aq Ksp 12 x103916 Hf quotQ CN a F HCqunmx K 20 X 109 this is UK AgCNS Hfaq Agaq HCNW K24X10397 24 x10397 Ag HHCN H We see from the above equilibrium eXpression that silverI cyanide becomes more soluble as pH decreases This is reasonable because there are two different species competing to bind with the cyanide ion Ag and Hf SilverI ions bind more strongly than do hydrogen ions but as the concentration of hydrogen ions increases they tie up more and more of the cyanide ions by weight of numbers In general the solubilities of slightly soluble salts that contain the conjugate base of a weak acid increases as pH decreases the hydrogen ions convert some of the anion to the weak acid Formation of CompleX Ions We learned in Chapter 16 that metal ions are Lewis acids Water halide ions cyanide ions and ammonia are just a few of many Lewis bases Different metal ions show different levels of 15 affinity for different Lewis bases In the presence of halide ions silverI forms a very stable lattice and precipitates from solution as AgX X F39 Cl39 Br39 I39 we ll discuss this more in the next section In contrast in the presence of ammonia silverI quickly binds to two ammonia molecules to form a complex ion The term complex refers to the contrast with the simple elemental ion Ag A complex ion consists of a metal ion to which more than one Lewis base is covalently bound Agaq f 2 NH3 AgNH32aq AgNH3 l 17x 107 AgNH32 f where Kf is the formation constant Formation constants can be quite large as can be seen in Table 171 p 687 meaning that on mixing a soluble salt with the appropriate Lewis base the complex ion forms readily Such equilibria can also be used to dissolve slightly soluble salts Ex How much silverI chloride will dissolve in 10 L of a solution that is 010 M in NH3 at equilibrium What is the free silverI ion concentration AgC1S Agfaq Cl39aq Ksp 18 x 103910 A211 f 2 NH e A2NH1O quot9 Kg 17 x 107 AgC1s 2NH3 AgNH32faq Cl39aq K 31 x 103 a molAgC1dissolved molC1in solution 7 We use Clquot because while most of the W Ag r is bound in the complex some K 31 x 10393 2 NH3 exists as free ion let AgNH32e Cl39le X X2 31 x 103 2 010 X 56 X 10393 Lsoln 111101Cl 1 mOIAgCl 56X10393 mol lmolA c1 1433 gA c1 Ina39SSAgCl 1390 Lsoln C1 g g J 080 gAgC1 b Ksp 18 X 103910 Ag56 X 10393 Ag 32 X 10398M Saturated silverI chloride solution has Ag 13 X 10395 M so 998 of the silverI ion released into solution is tied up as a compleX ion Amphoterism Many metal hydr0Xides and 0Xides are slightly soluble in neutral water but are surpisingly soluble in both acidic w basic solutions These substances are called amphoteric Since they are metal hydr0Xides and 0Xides their solubility in acidic solutions probably comes as no surprise Hydrogen ions protonate the hydr0Xide and 0Xide ions to water leaving the metal ions free in solution MOH Ht gt M H20 M20 2H gt M H20 In basic solution one or more hydr0Xide ions acts as a Lewis base and coordinates to the metal ion forming a soluble compleX ion In general MOHHS OH39aq MOHn139aq So why does this happen and why to only certain metal 0Xides and hydr0Xides There are two parts to this answer one electronic the other structural First for a substance to be amphoteric the metal ion must bind to 0Xidehydr0Xide much more strongly than to water Generally the interaction of 0Xides and hydr0Xides will be stronger because of electrostatic 17 attraction If the cation interacts with water at a comparable level as oxidehydroxide the compound will dissolve e g NaOH Metal ions with large charges or small sizes are more apt to yield amphoteric oxides and hydroxides The other feature is structural If metal ions develop Lewis acidbase interactions with neighboring ions they can crosslink making a large structure that is difficult to dissolve Oxide and hydroxide have multiple lone pairs of electrons It can now bind to several metal ions ie crosslinking When a solution hydroxide replaces the bridging interactions monomeric complexes form and they are generally soluble 176 Precipitation and Separation of Ions Since some salts are only slightly soluble in water it serves to reason that if one mixes solutions containing ions that are components of a slightly soluble salt there is a good chance that the salt will precipitate from solution Going back to Chapter 15 we know that if Q gt K Sp a salt will precipitate more concentrated solutions will yield a larger value and higher probability of precipitation This feature has important practical considerations Many expensive metals are isolated by precipitation of their salts We ll go over a couple of examples here Ex If 250 mL ofa 20 x 10393Mmagnesium chloride solution mixes with 150 mL ofa 10 x 10393 M barium hydroxide solution will a precipitate form We first need the concentrations of the magnesium and hydroxide ions 7 7 It is important to rem ember that you must add VT 7 25390 mL 15390 mL 7 40390 mL or 0390400 L the volumes ofthe solutions and recalculate Hm mnoenrrqrinnq 20 x10393 molMgCl2 1molMg2 L soln InOlMg2 Lsoln J 50 X 105 IIIOlMgZJr 1 molMgC12 50 x10quot5 molMg2 M 2 g 00400L 125 x10393M soln E 1X1 10x10393 mol 2 1 molOH 00150 Lsohp 30 X 105 m010H soln BaOH2 30 x105 mol OH39 Mg2 75 X10394M 00400 Lsoln Q Mg20H2 125 x10375 x1042 70 x1010 Since Q gt K S magnesium hydroxide does precipitate It will continue to do so until the ion concentrations drop to a concentration where Q Ksp Suppose you had a solution that was 010 M each in copperI goldI and silverI ions Could they be efficiently separated and how The solubility product constants of their chlorides are KspCuCl 12 X 10396 KspAgCl 18 X 1039 KspAuCl 20 X 1039 These are each separated by 34 orders of magnitude and are worth a try The general equilibrium is MClS Maq Cl39aq Ksp MCl39 If MCl39 lt Ksp for a compound then no salt will precipitate We want to add chloride ion to the solution until just before the second ion begins to precipitate GoldI chloride will precipitate rst all solubility product formulae are qualitatively the same and AuCl has the smallest Ksp Thus we need to nd out how much chloride ion will just begin to precipitate silverI chloride AgCl 010Cl3918 X 103910 Cl39 18 X 10399M As long as Cl39 gt 18 X 10399 M no silverI chloride will precipitate But how much of the gold will have been removed from solution at this point AuCl Au18 X 10399 20 X 103913 Aut 14 X 10394M Recall the initial concentration of goldI was 010M so 999 of the gold is removed from solution prior to the commencement of silverI chloride precipitation Likewise for copperI chloride Cl39 12 X 10395M and Ag 15 X 10395 M so 9998 of the silver precipitates before copperI chloride begins to come out of solution Thus this is a pretty good method of separating coinage metal ions from solution and from each other 177 quot quot Analvsis for Metallic Elements This topic was first introduced in Section 42 p 118 which provides a short review Frequently we are more interested in whether a particular ion is present in some minimum amount than eXactly how much is there For this purpose a series of tests has been developed The process of determining if a particular ion is present in a solution is called gualitative m Figure 1722 gives a full set of rules if you want to learn them You must learn the rules below To use them you would add one of the anions in 2 If a precipitate formed you would know Ag Pb2 or Hg22 was present If no precipitate formed then add one of the anions from 3 and so forth These rules include a few oversimplifications but they will work in most situations and are easier to remember than the more comprehensive list in the book 1 All ionic compounds containing alkali metal cations NH4 NO339 and acetate are soluble 2 Cl39 Br39 and I39 salts are soluble eXcept with Ag Pb2 Hg22 3 EXcept for 1 carbonate and phosphate salts are insoluble 4 Except for l and Group HA sul des and hydroxides are insoluble March 3 2005 20 Chapter 15 7 Chemical Eguilibrium Chemical equilibrium can be de ned two different ways i It is the condition in which the concentrations of all reaction species do not change with time or ii It is the condition of opposing reactions proceeding at the same rate You saw this concept introduced brie y in Chapter 4 last semester but it wasn t discussed in any detail In this chapter we will explore how and why chemical equilibria occur 151 The Concept of Equilibrium Before getting to the mathematical description of equilibrium we should talk about its molecular basis In a sense that a reaction should go to completion or not go at all is fairly intuitive If the products are more stable than the reactants the reaction goes if they re not the reaction doesn t That a reaction should go part way and apparently stop is a little surprising Consider an equilibrium between A and B in solution where each reaction is unimolecular In each case interconversion occurs when an A or B molecule undergoes a collision of sufficient force to provide the activation energy necessary for reaction What if A and B are of nearly equal stability with B just slightly more stable than A Even though formation of B is favored there will always be collisions that result in regeneration of A The equilibrium results when the forward and reverse rates equalize F rateforward kfA ratereverse krB Another way of expressing this is looking at a plot of the activation energy of the reaction Chapter 14 In Figure l we see that the activation energy of the forward reaction A gt B is nearly equal to the activation energy for the reverse reaction B gt A In this situation there is enough background energy to allow both reactions proceed at an appreciable rate so both materials are present in the reaction vessel at all times Eaforward Ea revers e reactants wmeonm products Reaction progress gt Figure 1 If we start with pure A in a container initially the forward rate will be large and the reverse rate zero The forward rate will then slow A is used up and the reverse rate will accelerate B is formed until they become equal If pure B is used then the reverse rate will be large and the forward rate will be zero Likewise the large rate will slow down and the small rate will increase until they equalize At equilibrium kfA krB The following graphs show representative plots for the change in concentration for two species with time Figure 2 and the change in the rates of A disappearance and B appearance as the system approaches equilibrium Figure 3 concentration rate 0 5 10 15 20 25 30 0 5 10 15 20 25 30 time time Figure 2 Figure 3 The previous equality can be rearranged to produce a useful equation Ic Af 9 r kf K where Keq is the equilibrium constant Always remember that equilibria are dynamic processes both the forward and reverse reactions are always occurring 152 The Eguilibrium Constant The previous equation can be generalized for any combination of materials in solution at equilibrium aA bB cC dD C Dd eq Aa Bb If the same reactants were in the gas phase the partial pressures of the gases are used in place of concentrations and the equilibrium constant becomes It is important to use the eq as a subscript to help differentiate between the equilibrium and rate constants handwriting sometimes makes distinguishing a capital K from a little k difficult Example N2 g 3 H2 g 2 NH3 g 2 7 PN39H3 K eq PNZ PH23 Thus the equilibrium constant expression depends only on the reaction stoichiometry For this reason it is convention to write out the stoichiometry using only whole numbers Using fractions can change the equilibrium constant For example C A 2BC Keq VABAVC Ke 2 2 Sq Ar1131 Now let A 2M B 3M and C 4M then V2 Keq ZO222 Keq39 0471 23 9 2 23 3 This is one of the reasons that you were told in CHM 211 to balance equations with only whole number coefficients Section 31 Equilibrium constants have no units They are temperature dependent since rate constants are temperature dependent so the temperature must be speci ed when an equilibrium constant is given Since equilibrium constants depend only on stoichiometry and temperature mechanism is irrelevant This is important because catalysts change mechanisms Chapter 14 As we will see at the end of this chapter adding a catalyst causes the system to reach equilibrium more quickly but the amounts of the various species aren t affected The Magnitude of Equilibrium Constants There are 3 possible scenarios for the size of equilibrium constants large small or about one When equilibrium constants are very large gt 1000 reactions are essentially complete if the product can be removed quickly If they are very small lt 0001 then essentially no product is formed It is only if the equilibrium constant is near one that an appreciable amount of all reactants and products eXists Using the book s example of phosgene cog Cl2 g x COC12g Keq SOLIS 149 X108 100 0C C0 C12 If PCOC12 100 atm and FCC PC12 then what are the pressures of CO and C12 149 X 108 X2 x PCO PCB 819 X10395M As you can see the pressures of CO and C12 are about 0001 that of COC12 For all practical purposes there is no reactant material left in the reaction vessel Likewise for reactions with extremely small equilibrium constants there is little product formed In many situations we can think of these reactions as either going to completion or never starting Nevertheless these small amounts of material can be important we ll discuss why in Section 156 Historical note In World War I phosgene was the major component of mustard gas If the equilibrium constant is closer to l as in the next reaction there may be a signi cant amount of each substance present P P cog H20g F co2 g H2 9 Keq PCOZP H2 CO H20 510 830 0C If at equilibrium PCO PHZO PCO2 100 atm then PH2 510 atm If Keq lt 1 the equilibrium favors the reactants lies to the left If Keq gt 1 the equilibrium favors the products lies to the right The Direction of the Chemical Equation and K q Since equilibria can be approached from either side the way the reaction is written is somewhat arbitrary When an equilibrium is reversed Keq is inverted For the previous example if the equation is written in the opposite direction the following results are obtained 002 g H2 g f 00g H20g Keq L 0196 830 0c PCOZPHZ 510 Other Wavs to 39 39 39 Chemical metinn andK q Values An interesting situation occurs when two equilibria occur in sequence we ll see examples of this later We ll go through Sample Exercise 154 as a way of producing some generalities about combining equilibrium expressions Consider the two weak acids HFaq and H2C204aq at 25 0C and a mixture of them HlllF HFltaq g Hltaq F ltan Keq HF H2C 042 H20204 2 Haq C2042 aq Keql 2 HZCZO4 H C O F 2 2HF C02 HCO2F K aq 2 4 aq f 2 2 4 aq eq2 HF2C2042 The question is What is the relationship between the rst two equations and the third We generate the third equation by reversing the second equilibrium and adding it to the rst which must be doubled to give the correct stoichiometry 2 f 39 K llF7l 2 HF Ht F rH aq aq 9 eq1 K HF J O 39 H C O 2 H aq C242aq H2C201 Keqz l 2 2 4 H2C20427 Z HlllF H C O 2HF 2H C02 2H 2F HCO K 7 aq W 2 4 W r aq W 2 2 4 Sq HF J qumzofJ which simpli es the reaction and equilibrium expression we sought HCZOF 12 sq HF2C042 2 HFWD C2042an H2C204 2 F39aq This example and the one in the book illustrate three facts about equilibria and their associated equations 1 When an equilibrium is reversed Keq is inverted 2 When an equilibrium occurs more than once the equilibrium constant is raised to that power 3 When two equilibria are added their equilibrium constants are multiplied
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'