Chemical Equilibrium Introduction 1 Equilibria govern diverse phenomena

Chemical Equilibrium Introduction 1. ) Equilibria govern diverse phenomena Ø 2. ) Protein folding, acid rain action on minerals to aqueous reactions Chemical equilibrium applies to reactions that can occur in both directions: Ø Ø Ø 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 distribution. Reactants Product At equilibrium the amount of reactants and products are constant, but not necessarily equal

Chemical Equilibrium Constant 1. ) The relative concentration of products and reactants at equilibrium is a constant. 2. ) Equilibrium constant (K): Ø 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 Constant 2. ) Equilibrium constant (K): Ø Ø 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 Constant 3. ) Manipulating Equilibrium Constants If two reactions are added, the new K is the product of the two individual K values: K 1 K 2 K 3

Chemical Equilibrium Constant 3. ) Manipulating Equilibrium Constants Ø Example: Given the reactions and equilibrium constants: Kw= 1. 0 x 10 -14 KNH 3= 1. 8 x 10 -5 Find the equilibrium constant for the reaction: Solution: K 1= Kw K 2=1/KNH 3 K 3=Kw*1/KNH 3=5. 6 x 10 -10

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

Chemical Equilibrium and Thermodynamics 3. ) Entropy Ø Ø Measure of a substances “disorder” Greater disorder Greater Entropy - Relative disorder: Gas > Liquid > solid DS = Sproducts – Sreactants Ø 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 25 o. C More disorder for aqueous ions than solid

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

Chemical Equilibrium and Thermodynamics 4. ) Free Energy Ø Ø 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 Ø Chemical reaction is not favored if: - DH is positive and DS is negative Ø 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 Free energy: where T is temperature (Kelvin) DG = DH -TDS

Chemical Equilibrium and Thermodynamics 4. ) Free Energy Ø Example: Is the following reaction favored at 25 o. C? DHo = -74. 85 x 103 J/mol DSo = -130. 4 J/K. mol Free energy: DG = DH –TDS = (-74. 85 x 103 J/mol) – (298. 15 K)(-130. 4 J/K. mol) DG = -35. 97 k. J/mol DG negative reaction favored Favorable influence of enthalpy is greater than unfavorable influence of entropy

Chemical Equilibrium and Thermodynamics 5. ) Free Energy and Equilibrium Ø Ø 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 Ø Ø 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

Chemical Equilibrium and Thermodynamics 5. ) Free Energy and Equilibrium Ø If DGo is negative or K >1 the reaction is spontaneous - Ø 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? Ø Change concentration, temperature, pressure or add other chemicals Ø 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? Ø 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 H 2 are converted to CH 3 OH If all added CO was converted to CH 3 OH, 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 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 Le Châtelier’s Principal 4. ) Thermodynamics vs. Kinetics Ø Thermodynamics predicts if a reaction will occur - Ø 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? DG = spontaneous Diamonds - Graphite While reaction is spontaneous, takes millions of years to occur

Chemical Equilibrium Solubility Product 1. ) Equilibrium constant for the reaction which a solid salt dissolves to give its constituent ions in solution Ø Solid omitted from equilibrium constant because it is in a standard state Ø Example:

Chemical Equilibrium Solubility Product 1. ) Saturated Solution – contains excess, undissolved solid Ø Ø Solution contains all the solid capable of dissolving under the current conditions Example: Find [Cu 2+] in a solution saturated with Cu 4(OH)6(SO 4) if [OH-] is fixed at 1. 0 x 10 -6 M. Note that Cu 4(OH)6(SO 4) gives 1 mol of SO 42 - for 4 mol of Cu 2+?

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 Ø Ø 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 Ø Ø 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 Mg. F 2 by the addition of Na. F Pb. Cl 2 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 Ø Ø Equilibrium re-obtained following Le Châtelier’s Principal Reaction moves away from the added ion Find [Cu 2+] in a solution saturated with Cu 4(OH)6(SO 4) if [OH-] is fixed at 1. 0 x 10 -6 M and 0. 10 M Na 2 SO 4 is added to the solution.

Chemical Equilibrium Complex Formation 1. ) High concentration of an ion may redissolve a solid Ø Ø 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 Ø Ø Ø 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 Ø Formation of adducts increase solubility Implies low Pb 2+ solubility: Ø Ksp 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 Pb 2+ in solution Concentration of Pb 2+ 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 Ø 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 Ø 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. 2 x 10 -7 M. Zn(OH)2 (s) Zn(OH)+ Zn(OH)3 Zn(OH)42 - Ksp = 3. 0 x 10 -16 b 1 = 2. 5 x 104 b 3 = 7. 2 x 1015 b 4 = 2. 8 x 1015

Chemical Equilibrium Acids and Bases 1. ) Protic Acids and Bases – transfer of H+ (proton) from one molecule to another Ø Hydronium ion (H 3 O+) – combination of H+ with water (H 2 O) Ø Acid – is a substance that increases the concentration of H 3 O+ Ø Base – is a substance that decreases the concentration of H 3 O+ - base also causes an increase in the concentration of OH- in aqueous solutions acid 2. ) Brønsted-Lowry – definition does not require the formation of H 3 O+ Ø Extended to non-aqueous solutions or gas phase Ø Acid – proton donor Ø Base – proton acceptor acid base salt

Chemical Equilibrium Acids and Bases 3. ) Salts – product of an acid-base reaction Ø Ø Ø 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 Ø Extent of these reactions are very small water - H 3 O+ 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. ) p. H – negative logarithm of H+ concentration Ø Ø 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 p. H if [H+]=[OH-] - This occurs when [H+] = [OH-] = 10 -7 M or p. H = 7

Chemical Equilibrium Acids and Bases 6. ) p. H Ø p. H values for some common samples

Chemical Equilibrium Acids and Bases 6. ) p. H Ø Example: What is the p. H of a solution containing 1 x 10 -6 M H+? What is [OH-] of a solution containing 1 x 10 -6 M H+?

Chemical Equilibrium Acids and Bases 7. ) Strengths of Acids and Bases Ø Ø 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 Ø 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 Ø 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 Kb

Chemical Equilibrium Some Common Weak Acids (carboxylic acids) ACID FORMULA Ka p. Ka acetic acid H(C 2 H 3 O 2) 1. 74 E-5 4. 76 hydrocyanic acid HCN 6. 17 E-10 9. 21 ascorbic acid (1) H 2(C 6 H 6 O 6) 7. 94 E-5 4. 10 hydrofluoric acid HF 6. 31 E-4 3. 20 ascorbic acid (2) (HC 6 H 6 O 6)- 1. 62 E-12 11. 79 lactic acid H(C 3 H 5 O 3) 8. 32 E-4 3. 08 boric acid (1) H 3 BO 3 5. 37 E-10 9. 27 nitrous acid HNO 2 5. 62 E-4 3. 25 boric acid (2) (H 2 BO 3)- 1. 8 E-13 12. 7 octanoic acid H(C 8 H 15 O 2) 1. 29 E-4 4. 89 boric acid (3) (HBO 3)= 1. 6 E-14 13. 8 oxalic acid (1) H 2(C 204) 5. 89 E-2 1. 23 butanoic acid H(C 4 H 7 O 2) 1. 48 E-5 4. 83 oxalic acid (2) (HC 2 O 4)- 6. 46 E-5 4. 19 carbonic acid (1) H 2 CO 3 4. 47 E-7 6. 35 pentanoic acid H(C 5 H 9 O 2) 3. 31 E-5 4. 84 carbonic acid (2) (HCO 3)- 4. 68 E-11 10. 33 phosphoric acid (1) H 3 PO 4 6. 92 E-3 2. 16 chromic acid (1) H 2 Cr. O 4 1. 82 E-1 0. 74 phosphoric acid (2) (H 2 PO 4)- 6. 17 E-8 7. 21 chromic acid (2) (HCr. O 4)- 3. 24 E-7 6. 49 phosphoric acid (3) (HPO 4)= 2. 09 E-12 12. 32 citric acid (1) H 3(C 6 H 5 O 7) 7. 24 E-4 3. 14 propanoic acid H(C 3 H 5 O 2) 1. 38 E-5 4. 86 citric acid (2) (H 2 C 6 H 5 O 7)- 1. 70 E-5 4. 77 sulfuric acid (2) (HSO 4)- 1. 05 E-2 1. 98 citric acid (3) (HC 6 H 5 O 7)= 4. 07 E-7 6. 39 sulfurous acid (1) H 2 SO 3 1. 41 E-2 1. 85 formic acid H(CHO 2) 1. 78 E-4 3. 75 sulfurous acid (2) (HSO 3)- 6. 31 E-8 7. 20 heptanoic acid H(C 7 H 13 O 2) 1. 29 E-5 4. 89 uric acid H(C 5 H 3 N 4 O 3) 1. 29 E-4 3. 89 hexanoic acid H(C 6 H 11 O 2) 1. 41 E-5 4. 84

Chemical Equilibrium Some Common Weak Acids (Metals cations)

Chemical Equilibrium Some Common Weak Bases (amines) BASE FORMULA Kb p. Kb alanine C 3 H 5 O 2 NH 2 7. 41 E-5 4. 13 Ammonia NH 3 (NH 4 OH) 1. 78 E-5 4. 75 dimethylamine (CH 3)2 NH 4. 79 E-4 3. 32 ethylamine C 2 H 5 NH 2 5. 01 E-4 3. 30 glycine C 2 H 3 O 2 NH 2 6. 03 E-5 4. 22 hydrazine N 2 H 4 1. 26 E-6 5. 90 methylamine CH 3 NH 2 4. 27 E-4 3. 37 trimethylamine (CH 3)3 N 6. 31 E-5 4. 20 Ø The Ka or Kb of an acid or base may also be written in terms of “p. Ka” or “p. Kb” Ø As Ka or Kb increase p. Ka or p. Kb decrease - a strong acid/base has a high Ka or Kb and a low p. Ka or pkb

Chemical Equilibrium Acids and Bases 8. ) Polyprotic Acids and Bases – can donate or accept more than one proton Ø Ka or Kb are sequentially numbered - Ka 1, Ka 2, Ka 3 Kb 1, Kb 2, Kb 3

Chemical Equilibrium Acids and Bases 8. ) Relationship Between Ka and Kb

Chemical Equilibrium Acids and Bases 8. ) Relationship Between Ka and Kb Ø Example: Write the Kb reaction of CN-. Given that the Ka value for HCN is 6. 2 x 10 -10, calculate Kb for CN-.