Chemical Equilibrium Introduction 1.) Equilibria govern diverse phenomena 2.) Protein folding, acid rain action on minerals to aqueous reactions Chemical equilibrium applies to reactions that can occur in both directions: reactants are constantly forming products and vice-versa At the beginning of the reaction, the rate that the reactants are changing into the products is higher than the rate that the products are changing into the reactants. When the net change of the products and reactants is zero the reaction has reached equilibrium. Then, system continually exchanges products and reactants, maintaining First, system reacheswhile equilibrium equilibrium distribution. Reactants Product At equilibrium the amount of reactants and products are constant, but not necessarily equal Chemical Equilibrium Equilibrium Constant 1.) The relative concentration of products and reactants at equilibrium is a constant. 2.) Equilibrium constant (K): For a general chemical reaction Equilibrium constant: K c d [C ] [ D] [ A]a [ B]b Where: - small superscript letters are the stoichiometry coefficients - [A] concentration chemical species A relative to standard state Chemical Equilibrium Equilibrium Constant 2.) Equilibrium constant (K): A reaction is favored when K > 1 K has no units, dimensionless - Concentration of solutes should be expressed as moles per liter (M). - Concentrations of gases should be expressed in bars. ► ► express gas as Pgas, emphasize pressure instead of concentration 1 bar = 105 Pa; 1 atm = 1.01325 bar - Concentrations of pure solids, pure liquids and solvents are omitted ► are unity ► standard state is the pure liquid or solid 3.) Manipulating Equilibrium Constants Consider the following reaction: [ H ][ A ] K1 [ HA] Reversing the reaction results in a reciprocal equilibrium reaction: K '1 [ HA] [ H ][ A ] 1 / K1 Chemical Equilibrium Equilibrium Constant 3.) Manipulating Equilibrium Constants If two reactions are added, the new K is the product of the two individual K values: K1 K2 K3 [ H ][ A ] K1 [ HA] K2 [CH ] [ H ][C ] [ A ][CH ] K3 [ HA][C ] [ H ][ A ] [CH ] [ A ][CH ] K 3 K1K 2 [ HA] [ HA][C ] [ H ][C ] Chemical Equilibrium Equilibrium Constant 3.) Manipulating Equilibrium Constants Example: Given the reactions and equilibrium constants: Kw= 1.0 x 10-14 KNH3= 1.8 x 10-5 Find the equilibrium constant for the reaction: Solution: K1= Kw K2=1/KNH3 K3=Kw*1/KNH3=5.6x10-10 Chemical Equilibrium Le Châtelier’s Principal 1.) What Happens When a System at Equilibrium is Perturbed? Change concentration, temperature, pressure or add other chemicals Equilibrium is re-established - - Reaction accommodates the change in products, reactants, temperature, pressure, etc. Rates of forward and reverse reactions re-equilibrate Chemical Equilibrium Le Châtelier’s Principal 1.) What Happens When a System at Equilibrium is Perturbed? Le Châtelier’s Principal: - the direction in which the system proceeds back to equilibrium is such that the change is partially offset. Consider this reaction: At equilibrium: Add excess CO(g): To return to equilibrium (balance), some (not all) CO and H2 are converted to CH3OH If all added CO was converted to CH3OH, then reaction would be unbalanced by the amount of product Chemical Equilibrium Le Châtelier’s Principal 2.) Example: Consider this reaction: K [ Br - ][Cr2O72- ][ H ]8 [ BrO3- ][Cr 3 ]2 1 10 11 at 25 o C At one equilibrium state: [H ] 5.0 M [Cr2O72- ] 0.10 M [Cr 3 ] 0.0030 M [Br ] 1.0 M [BrO 3- ] 0.043 M Chemical Equilibrium Le Châtelier’s Principal 2.) Example: What happens when: [Cr2O72- ] increased from 0.10 M to 0.20 M According to Le Châtelier’s Principal, reaction should go back to left to off-set dichormate on right: Use reaction quotient (Q), Same form of equilibrium equation, but not at equilibrium: Q [ Br - ][Cr2O72- ][ H ]8 [ BrO3- ][Cr 3 ]2 1.0 0.20 5.0 8 0.0430.0030 2 2 10 11 K Chemical Equilibrium Le Châtelier’s Principal 2.) Example: Because Q > K, the reaction must go to the left to decrease numerator and increase denominator. Continues until Q = K: 1. If the reaction is at equilibrium and products are added (or reactants removed), the reaction goes to the left 2. If the reaction is at equilibrium and reactants are added ( or products removed), the reaction goes to the right Chemical Equilibrium Le Châtelier’s Principal 3.) Affect of Temperature on Equilibrium 1. Equilibrium constant of an endothermic reaction (DHo = +) increases if the temperature is raised. D DH = + 2. Equilibrium constant of an exothermic reaction (DHo = -)decreases if the temperature is raised. D DH = - Chemical Equilibrium Solubility Product 1.) Equilibrium constant for the reaction which a solid salt dissolves to give its constituent ions in solution Solid omitted from equilibrium constant because it is in a standard state Example: K sp [ Hg 22 ][Cl- ]2 1.2 10 18 Chemical Equilibrium Solubility Product 1.) Saturated Solution – contains excess, undissolved solid Solution contains all the solid capable of dissolving under the current conditions EXAMPLE: Find [Cu2+] in a solution saturated with Cu4(OH)6(SO4) if [OH-] is fixed at 1.0x10-6M. Note that Cu4(OH)6(SO4) gives 1 mol of SO42for 4 mol of Cu2+? K sp 2.3 10 69 Chemical Equilibrium Solubility Product 2.) If an aqueous solution is left in contact with excess solid, the solid will dissolve until the condition of Ksp is satisfied Amount of undissolved solid remains constant Excess solid is required to guarantee ion concentration is consistent with Ksp 3.) If ions are mixed together such that the concentrations exceed Ksp, the solid will precipitate. 4.) Solubility product only describes part of the solubility of a salt Only includes dissociated ions Ignores solubility of solid salt Chemical Equilibrium Common ion effect – a salt will be less soluble if one of its constituent ions is already present in the solution. Decrease in the solubility of MgF2 by the addition of NaF PbCl2 precipitate because the ion product is greater than Ksp. Chemical Equilibrium Common Ion Effect 1.) Affect of Adding a Second Source of an Ion on Salt Solubility Equilibrium re-obtained following Le Châtelier’s Principal Reaction moves away from the added ion EXAMPLE: Find [Cu2+] in a solution saturated with Cu4(OH)6(SO4) if [OH-] is fixed at 1.0x10-6M and 0.10M Na2SO4 is added to the solution. Chemical Equilibrium Complex Formation 1.) High concentration of an ion may redissolve a solid Ion first causes precipitation Forms complex ions, consists of two or more simple ions bonded to each other ppt. formation Complex forms and redissolves solid Chemical Equilibrium Complex Formation 2.) Lewis Acids and Bases M+ acts as a Lewis acid accepts a pair of electrons X- acts as a Lewis base donates a pair of electrons Bond is a coordinate covalent bond ligand Lewis acid adduct Lewis base Chemical Equilibrium Complex Formation 3.) Affect on Solubility Formation of adducts increase solubility Implies low Pb2+ solubility: Ksp K sp [ Pb2 ][ I - ]2 7.9 10 9 Solubility equation becomes a complex mixture of reactions - don’t need to use all equations to determine the concentration of any species Only one concentration of Pb2+ in solution Concentration of Pb2+ that satisfies any one of the equilibria must satisfy all of the equilibria All equilibrium conditions are satisfied simultaneously Chemical Equilibrium Complex Formation 3.) Affect on Solubility Total concentration is dependent on each individual complex species Pbtotal Pb 2 PbI PbI 2 ( aq ) PbI 3 PbI 42 Total solubility of lead depends on [I-] and the solubility of each individual complex formation. Chemical Equilibrium Complex Formation 3.) Affect on Solubility EXAMPLE: Given the following equilibria, calculate the concentration of each zinccontaining species in a solution saturated with Zn(OH)2(s) and containing [OH-] at a fixed concentration of 3.2x10-7M. Zn(OH)2 (s) Zn(OH)+ Zn(OH)3Zn(OH)42- Ksp = 3.0x10-16 b1 = 2.5 x104 b3 = 7.2x1015 b4 = 2.8x1015 Chemical Equilibrium Acids and Bases 1.) Protic Acids and Bases – transfer of H+ (proton) from one molecule to another Hydronium ion (H3O+) – combination of H+ with water (H2O) Acid – is a substance that increases the concentration of H3O+ Base – is a substance that decreases the concentration of H3O+ - base also causes an increase in the concentration of OH- in aqueous solutions acid 2.) Brønsted-Lowry – definition does not require the formation of H3O+ Extended to non-aqueous solutions or gas phase Acid – proton donor Base – proton acceptor acid base salt Chemical Equilibrium Acids and Bases 3.) Salts – product of an acid-base reaction 4.) Any ionic solid Acid and base neutralize each other and form a salt Most salts with a single positive and negative charge dissociate completely into ions in water Conjugate Acids and Bases Products of acid-base reaction are also acids and bases A conjugate acid and its base or a conjugate base and its acid in an aqueous system are related to each other by the gain or loss of H+ Chemical Equilibrium Acids and Bases 5.) Autoprotolysis – acts as both an acid and base Extent of these reactions are very small water K w [ H ][OH - ] 1.0 10 14 - H3O+ is the conjugate acid of water - OH- is the conjugate base of water - Kw is the equilibrium constant for the dissociation of water Acetic acid K 3.5 10 15 Chemical Equilibrium Acids and Bases 6.) pH – negative logarithm of H+ concentration Ignores distinction between concentration and activities (discussed later) pH log[ H ] pH pOH log[ K w ] 14.00 at 25 o C A solution is acidic if [H+] > [OH-] A solution is basic if [H+] < [OH-] An aqueous solution has a neutral pH if [H+]=[OH-] - This occurs when [H+] = [OH-] = 10-7M or pH = 7 Chemical Equilibrium Acids and Bases 6.) pH Example: What is the pH of a solution containing 1x10-6 M H+? What is [OH-] of a solution containing 1x10-6 M H+? Chemical Equilibrium Acids and Bases 7.) Strengths of Acids and Bases Depends on whether the compound react nearly completely or partially to produce H+ or OH- strong acid or base completely dissociate in aqueous solution - equilibrium constants are large - everything else termed weak Strong no undissociated HCl or KOH Chemical Equilibrium Acids and Bases 7.) Strengths of Acids and Bases weak acids react with water by donating a proton - only partially dissociated in water - equilibrium constants are called Ka – acid dissociation constant - Ka is small Ka Equivalent Ka weak bases react with water by removing a proton - only partially dissociated in water - equilibrium constants are called Kb – base dissociation constant - Kb is small Kb Equivalent [ H ][ A ] Ka [ HA] Kb [ BH ][OH ] Kb [ B] Chemical Equilibrium Some Common Weak Acids (carboxylic acids) ACID FORMULA Ka pKa ACID FORMULA Ka pKa acetic acid H(C2H3O2) 1.74 E-5 4.76 hydrocyanic acid HCN 6.17 E-10 9.21 ascorbic acid (1) H2(C6H6O6) 7.94 E-5 4.10 hydrofluoric acid HF 6.31 E-4 3.20 ascorbic acid (2) (HC6H6O6)- 1.62 E-12 11.79 lactic acid H(C3H5O3) 8.32 E-4 3.08 boric acid (1) H3BO3 5.37 E-10 9.27 nitrous acid HNO2 5.62 E-4 3.25 boric acid (2) (H2BO3)- 1.8 E-13 12.7 octanoic acid H(C8H15O2) 1.29 E-4 4.89 boric acid (3) (HBO3)= 1.6 E-14 13.8 oxalic acid (1) H2(C204) 5.89 E-2 1.23 butanoic acid H(C4H7O2) 1.48 E-5 4.83 oxalic acid (2) (HC2O4)- 6.46 E-5 4.19 carbonic acid (1) H2CO3 4.47 E-7 6.35 pentanoic acid H(C5H9O2) 3.31 E-5 4.84 carbonic acid (2) (HCO3)- 4.68 E-11 10.33 phosphoric acid (1) H3PO4 6.92 E-3 2.16 chromic acid (1) H2CrO4 1.82 E-1 0.74 phosphoric acid (2) (H2PO4)- 6.17 E-8 7.21 chromic acid (2) (HCrO4)- 3.24 E-7 6.49 phosphoric acid (3) (HPO4)= 2.09 E-12 12.32 citric acid (1) H3(C6H5O7) 7.24 E-4 3.14 propanoic acid H(C3H5O2) 1.38 E-5 4.86 citric acid (2) (H2C6H5O7)- 1.70 E-5 4.77 sulfuric acid (2) (HSO4)- 1.05 E-2 1.98 citric acid (3) (HC6H5O7)= 4.07 E-7 6.39 sulfurous acid (1) H2SO3 1.41 E-2 1.85 formic acid H(CHO2) 1.78 E-4 3.75 sulfurous acid (2) (HSO3)- 6.31 E-8 7.20 heptanoic acid H(C7H13O2) 1.29 E-5 4.89 uric acid H(C5H3N4O3) 1.29 E-4 3.89 hexanoic acid H(C6H11O2) 1.41 E-5 4.84 Chemical Equilibrium Some Common Weak Acids (Metals cations) Chemical Equilibrium Some Common Weak Bases (amines) BASE FORMULA pKb alanine C3H5O2NH2 7.41 E-5 4.13 Ammonia NH3 (NH4OH) 1.78 E-5 4.75 dimethylamine (CH3)2NH 4.79 E-4 3.32 ethylamine C2H5NH2 5.01 E-4 3.30 glycine C2H3O2NH2 6.03 E-5 4.22 hydrazine N2H4 1.26 E-6 5.90 methylamine CH3NH2 4.27 E-4 3.37 trimethylamine (CH3)3N 6.31 E-5 4.20 The Ka or Kb of an acid or base may also be written in terms of “pKa” or “pKb” pK a log( K a ) Kb pK b log( K b ) As Ka or Kb increase pKa or pKb decrease - a strong acid/base has a high Ka or Kb and a low pKa or pkb Chemical Equilibrium Acids and Bases 8.) Polyprotic Acids and Bases – can donate or accept more than one proton Ka or Kb are sequentially numbered - Ka1,Ka2,Ka3 Kb1,Kb2,Kb3 Chemical Equilibrium Acids and Bases 8.) Relationship Between Ka and Kb [ H ][ A ] Ka [ HA] Kb [ HA][OH ] [ A ] K w Ka Kb [ H ][ A ] [ HA][OH ] [ H ][OH ] [ HA] [ A ] K w Ka Kb Chemical Equilibrium Acids and Bases 8.) Relationship Between Ka and Kb EXAMPLE: Write the Kb reaction of CN-. Given that the Ka value for HCN is 6.2x10-10, calculate Kb for CN-.