Chemical Equilibrium Introduction 1.) Equilibria govern diverse phenomena Protein folding, acid rain action on minerals to aqueous reactions 2.) 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, while maintaining equilibrium distribution. First, system reaches equilibrium 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: 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: Reversing the reaction results in a reciprocal equilibrium reaction:

Chemical Equilibrium Equilibrium Constant K1 K2 K3 3.) Manipulating Equilibrium Constants If two reactions are added, the new K is the product of the two individual K values: K1 K2 K3

Chemical Equilibrium Equilibrium Constant Example: 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 Equilibrium and Thermodynamics reaction. 1.) Equilibrium constant derived from the thermodynamics of a chemical reaction. deals with the relationships and conversions between heat and other forms of energy 2.) Enthalpy DH – is the heat absorbed or released when the reaction takes place under constant applied pressure DH = Hproducts – Hreactants Standard enthalpy change (DHo) – all reactants and products are in their standard state. DHo – negative  heat released - Exothermic - Solution gets hot DHo – positive  heat absorbed - Endothermic - Solution gets cold

Chemical Equilibrium Equilibrium and Thermodynamics 3.) Entropy Measure of a substances “disorder” Greater disorder  Greater Entropy - Relative disorder: Gas > Liquid > solid DS = Sproducts – Sreactants DSo – change in entropy when all species are in standard state. - positive product more disorder - negative  product less disorder DSo = +76.4 J/(K.mol) at 25oC More disorder for aqueous ions than solid

Chemical Equilibrium Equilibrium and Thermodynamics 3.) Entropy Increase in temperature results in an increase in Entropy (S) Increase occurs for all products and reactants Primarily concerned with DS, which is only weakly temperature dependent - generally treat DS and DH as temperature independent

Chemical Equilibrium Equilibrium and Thermodynamics 4.) Free Energy Systems at constant temperature and pressure have a tendency toward lower enthalpy and higher entropy Chemical reaction is favored if: - DH is negative  heat given off and - DS is positive  more disorder Chemical reaction is not favored if: - DH is positive and DS is negative Gibbs Free Energy (DG): determines if a reaction is favored or not when both DH and DS are positive or negative - A reaction is favored if DG is negative where T is temperature (Kelvin) Free energy: DG = DH -TDS

Chemical Equilibrium Equilibrium and Thermodynamics 4.) Free Energy Example: Is the following reaction favored at 25oC? DHo = -74.85 x 103 J/mol DSo = -130.4 J/K.mol Free energy: DG = DH –TDS = (-74.85x103 J/mol) – (298.15K)(-130.4 J/K.mol) DG = -35.97 kJ/mol  DG negative  reaction favored Favorable influence of enthalpy is greater than unfavorable influence of entropy

Chemical Equilibrium Equilibrium and Thermodynamics 5.) Free Energy and Equilibrium Relate Equilibrium constant to the energetics (DH & DS) of a reaction Equilibrium constant depends on DG: where R (gas constant) = 8.314472 J/(K.mol) T = temperature in kelvins The more negative DG  larger equilibrium constant Example: DG = -35.97 Because K is very large, HCl is very soluble in water and nearly completely ionized

Gas flows towards a vacuum. A vacuum does not naturally form. Chemical Equilibrium Equilibrium and Thermodynamics 5.) Free Energy and Equilibrium If DGo is negative or K >1 the reaction is spontaneous Reaction occurs by just combining the reactants If DGo is positive or K < 1, the reaction is not spontaneous - Reaction requires external energy or process to proceed Gas flows towards a vacuum. spontaneous A vacuum does not naturally form. nonspontaneous

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: To return to equilibrium (balance), some (not all) CO and H2 are converted to CH3OH Add excess CO(g): 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: At one equilibrium state:

Chemical Equilibrium Le Châtelier’s Principal 2.) Example: What happens when: 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:

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 Combine Gibbs free energy and Equilibrium Equations: Only Enthalpy term is temperature dependent:

Chemical Equilibrium D D Le Châtelier’s Principal DH = + DH = - 3.) Affect of Temperature on Equilibrium 1. Equilibrium constant of an endothermic reaction (DHo = +) increases if the temperature is raised. 2. Equilibrium constant of an exothermic reaction (DHo = -)decreases if the temperature is raised. D DH = + D DH = -

Chemical Equilibrium Le Châtelier’s Principal DG = - 4.) Thermodynamics vs. Kinetics Thermodynamics predicts if a reaction will occur - determines the state at equilibrium Thermodynamics does not determine the rate of a reaction - Will the reaction occur instantly, in minutes, hours, days or years? - While reaction is spontaneous, takes millions of years to occur DG = - spontaneous Diamonds Graphite

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:

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 SO42- for 4 mol of Cu2+?

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 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 adduct Lewis acid Lewis base

Chemical Equilibrium Complex Formation 3.) Affect on Solubility Formation of adducts increase solubility Solubility equation becomes a complex mixture of reactions - don’t need to use all equations to determine the concentration of any species Ksp Implies low Pb2+ solubility: 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 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 zinc-containing 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) Ksp = 3.0x10-16 Zn(OH)+ b1 = 2.5 x104 Zn(OH)3- b3 = 7.2x1015 Zn(OH)42- 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 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 acid base salt

Chemical Equilibrium Acids and Bases 3.) Salts – product of an acid-base reaction 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 4.) 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 - 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

Chemical Equilibrium Acids and Bases 6.) pH – negative logarithm of H+ concentration Ignores distinction between concentration and activities (discussed later) 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 pH values for some common samples

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 weak bases react with water by removing a proton - equilibrium constants are called Kb – base dissociation constant - Kb is small Ka Equivalent Ka Kb Equivalent Kb

Chemical Equilibrium Some Common Weak Acids (carboxylic acids) 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 uric acid H(C5H3N4O3) 3.89 hexanoic acid H(C6H11O2) 1.41 E-5

Chemical Equilibrium Some Common Weak Acids (Metals cations)

Chemical Equilibrium Some Common Weak Bases (amines) FORMULA Kb 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” 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

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-.