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. First, system reaches equilibrium Then, system continually exchanges products and reactants, while maintaining equilibrium distribution. Reactants Product At equilibrium the amount of reactants and products are constant, but not necessarily equal
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: [ C] K [ A] c a [ D] [ B] d b - small superscript letters are the stoichiometry coefficients - [A] concentration chemical species A relative to standard state
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 P gas, emphasize pressure instead of concentration 1 bar 10 5 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: K 1 + [ H ][ A [ HA] Reversing the reaction results in a reciprocal equilibrium reaction: ' 1 1 / K + 1 [ HA] K [ H ][ A ] ]
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 K 1 + [ H ][ A [ HA] ] [ CH K 2 + [ H + ] ][ C] + K 3 [ A ][ CH ] [ HA][ C] K3 K1K2 + [ H ][ A [ HA] + ] [CH ] + [ H ][ C] + [ A ][ CH ] [ HA][ C]
Equilibrium Constant 3.) Manipulating Equilibrium Constants Example: Given the reactions and equilibrium constants: K w 1.0 x 10-14 K NH3 1.8 x 10-5 Find the equilibrium constant for the reaction: Solution: K 1 K w K 2 1/K NH3 K 3 K w *1/K NH3 5.6x10-10
Equilibrium and Thermodynamics 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 H products H reactants Standard enthalpy change (DH o ) all reactants and products are in their standard state. DH o negative heat released - Exothermic - Solution gets hot DH o positive heat absorbed - Endothermic - Solution gets cold
Equilibrium and Thermodynamics 3.) Entropy Measure of a substances disorder Greater disorder Greater Entropy - Relative disorder: Gas > Liquid > solid DS S products S reactants DS o change in entropy when all species are in standard state. - positive product more disorder - negative product less disorder DS o +76.4 J/(K. mol) at 25 o C More disorder for aqueous ions than solid
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
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 Free energy: DG DH -TDS where T is temperature (Kelvin)
Equilibrium and Thermodynamics 4.) Free Energy Example: Is the following reaction favored at 25 o C? DH o -74.85 x 10 3 J/mol DS o -130.4 J/K. mol Free energy: DG DH TDS (-74.85x10 3 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
Equilibrium and Thermodynamics 5.) Free Energy and Equilibrium Relate Equilibrium constant to the energetics (DH & DS) of a reaction Equilibrium constant depends on DG: K e D G o RT where R (gas constant) 8.314472 J/(K. mol) T temperature in kelvins The more negative DG larger equilibrium constant Example: DG -35.97 K e D G o RT e ( 35.97 x10 3 J / mol )( 8.314472J /( K. mol )( 298.15K ) 2.00 x10 6 Because K is very large, HCl is very soluble in water and nearly completely ionized
Equilibrium and Thermodynamics 5.) Free Energy and Equilibrium If DG o is negative or K >1 the reaction is spontaneous - Reaction occurs by just combining the reactants If DG o 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
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
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 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
Le Châtelier s Principal 2.) Example: Consider this reaction: K [ Br - ][ Cr [ BrO 2-3 O 2-7 ][ Cr ][ H 3+ ] 2 + ] 8 1 10 11 at 25 o C At one equilibrium state: [H [Br + ] 5.0 M ] 1.0 M [Cr 2 O [BrO 2-7 - 3 ] 0.10 M ] 0.043 M [Cr 3+ ] 0.0030 M
Le Châtelier s Principal 2.) Example: What happens when: 2- [Cr2O7 ] 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: - 3 2-7 [ Br ][ Cr H Q 2O ][ - 3+ [ BrO ][ Cr ] 2 + ] 8 ( 1.0)( 0.20)( 5.0) ( 0.043)( 0.0030) 8 2 2 10 11 K
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
Le Châtelier s Principal 3.) Affect of Temperature on Equilibrium Combine Gibbs free energy and Equilibrium Equations: K e e DG o RT ( DH DS ) + o RT e o ( DH TDS ) R RT e DH o RT e DS o R Only Enthalpy term is temperature dependent: K(T ) e D H o RT
Le Châtelier s Principal 3.) Affect of Temperature on Equilibrium 1. Equilibrium constant of an endothermic reaction (DH o +) increases if the temperature is raised. DH + 2. Equilibrium constant of an exothermic reaction (DH o -)decreases if the temperature is raised. D D DH -
Le Châtelier s Principal 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? DG - spontaneous Diamonds Graphite - While reaction is spontaneous, takes millions of years to occur
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 2+ - 2 18 sp [ Hg2 ][ Cl ] 1.2 10
Solubility Product 1.) Saturated Solution contains excess, undissolved solid 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 1.0x10-6 M. Note that Cu 4 (OH) 6 (SO 4 ) gives 1 mol of SO 2-4 for 4 mol of Cu 2+? K sp 2. 3 10 69
Solubility Product 2.) If an aqueous solution is left in contact with excess solid, the solid will dissolve until the condition of K sp is satisfied Amount of undissolved solid remains constant Excess solid is required to guarantee ion concentration is consistent with K sp 3.) If ions are mixed together such that the concentrations exceed K sp, 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
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 MgF 2 by the addition of NaF PbCl 2 precipitate because the ion product is greater than K sp.
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 [Cu 2+ ] in a solution saturated with Cu 4 (OH) 6 (SO 4 ) if [OH - ] is fixed at 1.0x10-6 M and 0.10M Na 2 SO 4 is added to the solution.
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
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
Complex Formation 3.) Affect on Solubility Formation of adducts increase solubility Implies low Pb 2+ solubility: K sp K 2+ - 2 9 sp [ Pb ][ I ] 7.9 10 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
Complex Formation 3.) Affect on Solubility Total concentration is dependent on each individual complex species 2+ + 2 Pb Pb + PbI + PbI ( aq ) + PbI + PbI total 2 3 4 Total solubility of lead depends on [I - ] and the solubility of each individual complex formation.
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 3.2x10-7 M. Zn(OH) 2 (s) K sp 3.0x10-16 Zn(OH) + b 1 2.5 x10 4 Zn(OH) - 3 b 3 7.2x10 15 Zn(OH) 2-4 b 4 2.8x10 15
Acids and Bases 1.) Protic Acids and Bases transfer of H + (proton) from one molecule to another Hydronium ion (H 3 O + ) combination of H + with water (H 2 O) Acid is a substance that increases the concentration of H 3 O + 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 + Extended to non-aqueous solutions or gas phase Acid proton donor Base proton acceptor acid base salt
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 +
Acids and Bases 5.) Autoprotolysis acts as both an acid and base Extent of these reactions are very small water + Kw [ H ][ OH ] 1.0 10-14 - H 3 O + is the conjugate acid of water - OH - is the conjugate base of water - K w is the equilibrium constant for the dissociation of water Acetic acid K 3.5 10 15
Acids and Bases 6.) ph negative logarithm of H + concentration Ignores distinction between concentration and activities (discussed later) ph log[ H + ] ph + poh log[ Kw ] 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 - ] 10-7 M or ph 7
Acids and Bases 6.) ph ph values for some common samples
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 +?
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
Acids and Bases 7.) Strengths of Acids and Bases Equivalent weak acids react with water by donating a proton - only partially dissociated in water - equilibrium constants are called K a acid dissociation constant - K a is small K a K a K a + [ H ][ A [ HA] ] Equivalent weak bases react with water by removing a proton - only partially dissociated in water - equilibrium constants are called K b base dissociation constant - K b is small + [ BH ][ OH K b K b [ B] K b ]
Some Common Weak Acids (carboxylic acids) ACID FORMULA K a pk a ACID FORMULA K a pk a 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 2 0 4 ) 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 CrO 4 1.82 E-1 0.74 phosphoric acid (2) (H 2 PO 4 ) - 6.17 E-8 7.21 chromic acid (2) (HCrO 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
Some Common Weak Acids (Metals cations)
Some Common Weak Bases (amines) BASE FORMULA K b pk b 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 The K a or K b of an acid or base may also be written in terms of pk a or pk b pk a log( K ) pk log( K ) a b b As K a or K b increase pk a or pk b decrease - a strong acid/base has a high K a or K b and a low pk a or pk b
Acids and Bases 8.) Polyprotic Acids and Bases can donate or accept more than one proton K a or K b are sequentially numbered - K a1,k a2,k a3 K b1,k b2,k b3
Acids and Bases 8.) Relationship Between K a and K b K a + [ H ][ A [ HA] ] K b [ HA][ OH [ A ] ] K w + K a K b [ H ][ A ] [ HA][ OH ] + [ H ][ OH ] [ HA] [ A ] K w K a K b
Acids and Bases 8.) Relationship Between K a and K b Example: Write the K b reaction of CN -. Given that the K a value for HCN is 6.2x10-10, calculate K b for CN -.