2 EquilibriumThe state where the concentrations of all reactants and products remain constant with time.
3 Reactions are reversible A + B C + D ( forward)C + D A + B (reverse)Forward and Reverse Rxns can be shown by double arrowA + B C + D
4 A + B C + DInitially there is only A and B so only the forward reaction is possibleAs C and D build up, the reverse reaction speeds up while the forward reaction slows down.Eventually the rates are equalSo concentrations of the reactants and products no longer change with time
6 Static or Dynamic?At equilibrium, forward and reverse reaction rates are equalMay seem like no changes are occurring but there are changesNo NET changesOn the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.chemical reactions take place ,but concentrations of reactants and products remain unchanged
7 Analogies and Metaphors to think about In a football game, the number of players on the field is constant although exchange of players (substitution) changes actual persons.Connected fish bowl analogy . Two fish tanks are connected by a tube large enough to allow passage of fish. A number of fish are placed in one of the tanks. At equilibrium, the number of fish in each tank will eventually become unchanged.Two jugglers analogy.Drinking fountain line:Ten students waiting in line to get a drink of water on a hot day. As each gets a drink, the same student reenters the line (equilibrium in a closed system).(b) Same situation as "a," except as each student gets a drink and leaves, a new student enters the line (steady-state in an open system).Picture a number of horses and wranglers in a corral. As each wrangler mounts a horse, the wrangler is bucked off. The equilibrium is:Horse + Wrangler Mounted wrangler
8 Molecular SimulationIn this simulation two gaseous reactants collide to produce a more dense solid.A B Cgaseous R dense P
9 Homo vs HeteroHomogeneous Equilibria all reacting species are in the same phasegas phaseequilibrium constant can be expressed in terms of pressure or concentration, Kp or KcSolution (aqueous) phaseconcentration term for the pure liquid does not appear in the expression for the equilibrium constant but aqueous substance concentrations do appearHeterogeneous Equilibria all reacting species are not in the same phaseconcentration term for solid or liquid does not appear in the expression for the equilibrium constant
10 Equilibrium Summarized Forward and Reverse rates are equalConcentrations are not.Rates are determined by concentrations and activation energy.Molecular Motion is frantic and constantly changingMacroscopically no net change is occurring (We can’t observe any changes)
12 Can you identify when the system reached equilbrium?
13 Distinguishing between Physical and Chemical Equilibrium As with physical and chemical changes, physical and chemical equilibrium follow the same rules:Physical no changes to the chemical properties of the substances involvedEx. equilibrium of water vapor with liquid water in a partly filled sealed bottleChemical involve changes in the chemical composition of substances. Bond breaking and bond formation is involved.Ex. dissociation of acetic acid water into acetate and hydronium ion
14 Activity Model Dynamic Equilibrium with Coins Now lets plot the data using excel
15 Law of Mass Action For a reaction: aA + bB ⇄ cC + dD equilibrium constant: KPure liquids and pure solids have concentrations of 1.c
16 Playing with K If we write the reaction in reverse. cC + dD ⇄ aA + bB Then the new equilibrium constant isc
17 They are simply the inverse of one another. Forward ReactionaA + bB ⇄ cC + dDSo we call this K1And K1= 1 = K2-1K2Reverse ReactioncC + dD ⇄ aA + bBSo we call this K2And K2= 1 = K1-1K1
18 The units for KAre determined by the various powers and units of concentrations.They depend on the reaction.will always have the same value at a certaintemperature (Why will the T effect it?)no matter what amounts are initially addedratio at equilibrium will always be same
19 K Has no units Is constant at any given temperature. Is affected by temperature.Equilibrium constants are reaction, phase, temperature and pressure dependentThere is a K for each temperature.Equilibrium constant values are thus established for a specific reaction in a specific system and will be unchanging (constant) in that system, providing the temperatures does not change.
21 What does the size of my K mean? Large K > 1products are "favored“Ex. 1 x 1034K = 1neither reactants nor products are favoredSmall K < 1reactants are "favored“Ex. 4 x 10-41
22 Now Let’s Calculate Your K Using your data from the simulation calculate K
23 Different Equilibrium Constants (All of these are known as Keq) Kc is the most used general form with molar concentrations.Kp can be used with partial pressures when working with a gas phase reaction.Ka is used for the dissociation of weak acids in water.Kb is used for the dissociation of weak bases in water.Kw is the equilibrium expression for the dissociation of water into its ions.Ksp is used for the dissociation into ions of sparingly soluble solids in water.
24 Practice Writing the Equilibrium Expression 4NH3(g) + 7O2(g) NO2(g) + 6H2O(g)First write the equilibrium expression using no concentration values.What is the value for K if the concentrations are as followsNH3 1.0 MO MNO MH2O 1.8 M
25 Equilibrium with Gases Equilibria involving only gases can be described using pressures or concentrationsIf using pressures,use pA not [A]KP not KCbe sure all pressure are in the same unitsN2(g) + 3H2(g) NH3(g)
26 Calculating Kc from Kp where Δn is the difference in moles of gas on either side of the equation (np – nr)R is the gas law constant:T is Kelvin temperatureFor: N2(g) + 3H2(g) NH3(g)Δ n = (2) – (1+3) = -2h
27 Practice Setup the expression for KP in terms of KC, R and T 2NO(g) + Cl2(g) NOCl(g)
28 What the equilibrium constant tells us… if we know the value of K, we can predict:tendency of a reaction to occurif a set of concentrations could be at equilibriumequilibrium position, given initial concentrationsIf you start a reaction with only reactants:concentration of reactants will decrease by a certain amountconcentration of products will increase by a same amount
29 Using this we can make ICE charts The following reaction has a K of 16. You are starting reaction with 9 O3 molecules and 12 CO molecules.Find the amount of each species at equilibrium.
33 Practice 1 ICE Charts Consider the following reaction at 600ºC 2SO2(g) + O2(g) SO3(g)In a certain experiment 2.00 mol of SO2, 1.50 mol of O2 and 3.00 mol of SO3 were placed in a 1.00 L flask. At equilibrium 3.50 mol were found to be present. CalculateThe equilibrium concentrations of O2 and SO2, K and KP
34 Practice Problem 2 ICE Charts Consider the same reaction at 600ºCIn a different experiment .500 mol SO2 and .350 mol SO3 were placed in a L container. When the system reaches equilibrium mol of O2 are present.Calculate the final concentrations of SO2 and SO3 and K
35 The Reaction Quotient (Q) Tells you the directing the reaction will go to reach equilibriumCalculated the same as the equilibrium constant, but for a system not at equilibriumQ = [Products]coefficient [Reactants] coefficientCompare value to equilibrium constant
36 What Q tells us IF THEN Q = K reaction is at equilibrium Q > K too much products,left shiftQ < K too much reactants,right shift
37 Example 1 Reaction Quotient For the synthesis of ammonia at 500°C, the equilibrium constant is 6.0 x Predict the direction the system will shift to reach equilibrium in the following case:
39 Example 2 In the gas phase, dinitrogen tetroxide decomposes to gaseous nitrogen dioxide:Consider an experiment in which gaseous N2O4 was placed in a flask and allowed to reach equilibrium at a T where KP = At equilibrium, the pressure of N2O4 was found tobe 2.71 atm.Calculate theequilibriumpressure of NO2.
40 Example 3At a certain temperature a 1.00 L flask initially contained mol PCl3(g) and 8.70x10-3 mol PCl5(g). After the system had reached equilibrium, 2.00x10-3 mol Cl2(g) was found in the flask.PCl5(g) PCl3(g) + Cl2(g)Calculate the equilibrium concentrations of all the species and the value of K.
42 ApproximationsIf K is very small, we can assume that the change (x) is going to be negligible compared to the initial concentration of the substancescan be used to cancel out when adding or subtracting from a “normal” sized number to simplify algebra
43 Example 4At 35°C, K=1.6x10-5 for the reaction 2NOCl(g) ⇄ 2NO(g) + Cl2(g)Calculate the concentration of all species at equilibrium for the following mixtures2.0 mol NOCl in 2.0 L flask1.0 mol NOCl and 1.0 mol NO in 1.0 L flask2.0 mol NO and 1.0 mol Cl2 in 1.0 L flask
47 Le Chatelier’s Principle can predict how certain changes or stresses put on a reaction will affect the position of equilibriumhelps us determine which direction the reaction will progress in to achieve equilibrium againsystem will shift away from the added component or towards a removed component
48 Change Concentration equilibrium position can change but not K system will shift away from the added component or towards a removed componentEx: N2 + 3H NH3if more N2 is added, then equilibrium position shifts to right (creates more products)if some NH3 is removed, then equilibrium position shifts to right (creates more products)
49 Adding Gasadding or removing gaseous reactant or product is same as changing concentrationadding inert or uninvolved gasincrease the total pressuredoesn’t effect the equilibrium position
50 Change the Pressure by changing the Volume only important in gaseous reactionsdecrease Vrequires a decrease in # gas moleculesshifts towards the side of the reaction with less gas moleculesincrease Vrequires an increase in # of gas moleculesshifts towards the side of the reaction with more gas molecules
52 Change in Temperatureall other changes alter the concentrations at equilibrium but don’t actually change value of Kvalue of K does change with temperatureif energy is added, the reaction will shift in direction that consumes energytreat energy as areactant: for endothermic reactionsproduct: for exothermic reactions
58 Another Dynamic Equilibrium Equilibrium occurs when the solution is saturated
59 Ksp The solubility product constant (Ksp) is similar to the Keq. When a solid is added to water, some dissolves (and splits into ions) while some remains a solid.Ex: NaCl(s) Na+(aq) + Cl-(aq)The mass action expression for this is:Ksp = [Na+][Cl-](remember… solids are 1)
60 Common Ion EffectThe solubility of a solid is lowered if the solution already contains ions common to the solidDissolving silver chloride in a solution containing silver ionsDissolving silver chloride in a solution containing chloride ions
61 Precipitation Opposite of dissolution Can predict whether precipitation or dissolution will occurUse Q: ion productEquals Ksp expression but doesn’t have to be at equilibriumQ > K: more reactant will form, precipitation until equilibrium reachedQ < K: more product will form, dissolution until equilibrium reached
68 Qualitative AnalysisProcess used to separate a solution containing different ions using solubilitiesExample ProblemA solution of 1.0x10-4 M Cu+ and 2.0x10-3 M Pb2+. If I- is gradually added, which will precipitate out first, CuI or PbI2?
73 Solutions of Acids or Bases Containing a Common Ion Ion provided in solution by an aqueous acid (or base) as well as a salta. HF(aq) and NaF (F- in common)b. HF(aq) H+(aq) + F-(aq) Excess F- added by NaFEquilibrium shifts away from added component. Fewer H+ ions present.pH is higher than expected.a. NH4OH and NH4Cl (NH4+ in common)b. NH3(aq) + H2O(l) NH4 +(aq) + OH-(aq)Equilibrium shifts to the left. pH of the solutiondecreases due to a decrease in OH- concentration
74 Equilibrium Calculations Consider initial concentration of ion from salt when calculating values for H+ and OH-
75 Acid + Base Salt + Water Acid + Base Conjugate Base + Conjugate AcidSome solvents are amphiproticWater can act as an acid and a base!Methanol can act as an acid and a base!AutoprotolysisSome solvents can react with themselves to produce an acid and a baseWater is a classical exampleWeak acids dissociate partially, weak bases undergo partial hydrolysis. Strong acids and bases are strong electrolytes.
77 Kw (Dissociation of Water) Water is amphiprotic it also undergoes autoprotolysisKw = 1.0E-14 at about 25 ˚CThis is where the pH scale we commonly use originates from!What is the concentration of hydronium and hydroxide ions in neutral solution? What is the pH? What is the pOH?
78 Weak Acid & Weak Base Equilibria Weak acids produce weak conjugate bases, and weak bases produce weak conjugate acidsKa is a “special” equilibrium constant for the dissociation of a weak acid (found in standard tables)Kb is a “special” equilibrium constant for the hydrolysis (or dissociation” of a weak base.
82 Calculations……..What is the pH of a 1.0 M solution of acetic acid (HAc)?What assumption can you make?If [acid] is about 1000 times the Ka value, it’s concentration in solution won’t change much!Use an “I-C-E” table to look at this.The text goes into a more elaborate discussion of approximations. I will allow approximations if the concentrations or pH values do not change in the hundred’s decimal place.
83 What is the pH of a 4.0 M solution of phosphate ion? Write reactionCalculate KbSetup “I-C-E” tableMake assumptionsSolve algebraically.
85 BuffersBuffers resist the change in pH because they have acid to neutralize bases and bases to neutralize acids.Made from a weak acid (HA) and the salt of its conjugate base (A-, where the counter ion is gone for example), or a weak base and the salt of its conjugate acid.
86 Features of BuffersBuffers work best at maintaining pH near the Ka of the acid component, usually about +/- 1 pH unit. This is their buffer capacity .Buffers resist pH changes due to dilution.All seen when we use the “Buffer Equation”
87 Henderson-Hasselbalch (Buffer) Equation A modification of the equation for the dissociation of a weak acid.The pH is the pH of the buffered solution, pKa is the pKa of the weak acid.What is the pH of a buffer solution made from 1.0 M acetic acid and 0.9 M sodium acetate?You add .10 moles of sodium hydroxide to the above solution? What is the new pH?
88 H-H Equation & Buffers…. If [A-] = [HA] pH = pKa!This is what we see at half-way to the equivalence point in the titration of a weak acid with a strong base!Dilution does not change the ratio of A- to HA, and thus the pH does not change significantly in most cases
89 How do you prepare buffers? Select a buffer ‘system’ based on the pH you want to maintain.Use the H-H equation to calculate how much acid and conjugate base you needTake one of three approaches:Mix acid and base forms, measure pH and adjust with strong acid or strong baseStart with a solution of weak acid, and add strong base until you reach the desired pHStart with a solution weak base and add strong acid until you reach the desired pHREMEMBER, STRONG Acids or Bases react completely!Dilute as necessary, adjust pH further if needed with strong acid or base.
90 You want 1L of a buffer system that has a pH of 3.90? What acid/conjugate base pair would you use?How would you go about figuring out how much of each reagent you might need?How would you prepare and adjust the pH of this solution?