1. Chemical Equilibrium In a chemical reaction, the reactions never go in only one direction In this chapter, we will study: - the equilibrium concept - equilibrium constant and its calculation - activity of ionic species – ion effect complex formation - acid-base reactions a A + b B c C + d D

2. The Rate Concept Similarly, in the backward reaction Rate of backward reaction (R b )  [C] c [D] d = k b [C] c [D] d If A is a solid or liquid molar concentration of A If A is a gas pressure of A in atmosphere Solvents are omitted from the equilibrium constant rate constant: dependent on temperature, pressure and presence of catalyst The Rate Concept Rate of a chemical reaction is proportional to the “ active masses ” of the reacting substances present at any time Rate of forward reaction (R f )  [A] a [B] b = k f [A] a [B] b a A + b B c C + d D

3. [C] c [D] d [A] a [B] b kfkf kbkb At equilibrium: rate of forward reaction = rate of backward reaction k f [A] a [B] b = k b [C] c [D] d Rearranging: = = K K can be evaluated by measuring the concentrations of A, B, C and D at equilibrium. The larger the K value the farther to the right is the reaction at equilibrium the more favorable the rate constant for the forward reaction relative to the backward reaction

4. Manipulating Equilibrium Constants Consider the reaction: HAH + + A - K 1 = If the direction of a reaction is reversed, the new value of K is simple the reciprocal of the original value of K H + + A - HA K’ 1 = = 1/ K 1 [ H + ][A - ] [ HA] [ H + ][A - ] [ HA]

5. Equilibrium constant for sum of reactions: K 3 = K 1 K 2 = x = [ H + ][A - ] [ HA] [ CH + ] [ H + ][C] [A - ][CH + ] [HA][C] [A - ][CH + ] If two reactions are added, the new K value is the product of the two individual values: HAH + + A - K 1 H + + C CH + K 2 HA + C A - + CH + K 3

6. Example : The equilibrium constant at 25 o C for the reaction: H 2 O H + + OH - K w = 1.0 x 10 -14 NH 3 (aq) + H 2 O NH 4 + + OH - K = 1.8 x 10 -5 Find the equilibrium constant for the reaction: NH 4 + NH 3 + H + NH 3 The 3 rd reaction can be obtained by: H 2 O H + + OH - K w NH 4 + + OH - NH 3 + H 2 O 1/K NH 4 + H + + NH 3 K 3 = K w /K = 5.6 x 10 -10 NH 3

7. Equilibrium and Thermodynamics Enthalpy of a reaction : heat absorbed or released by the reaction Entropy of a reaction : degree of disorder of reactants and products contribute to the degree to which the reaction is favoured or disfavoured entropy and enthalpy are related to equilibrium constant Enthalpy change for a reaction (  H) : H product - H reactanct  H = + heat is absorbed endothermic  H = -heart is liberatedexothermic K value : -does not tell us how fast a reaction will proceed toward equilibrium -tells us the tendency of a reaction to occur and in what direction

8. Entropy  S = S product - S reactanct  S = + products are more disordered than reactants  S = - products are less disordered than reactants Entropy of a substance, S : is the degree of disorder. Greater the disorder greater the entropy eg gas is more disorder (has higher entropy) than a liquid which is turn is more disordered than a solid

9. A system will always tend toward lower energy and increased randomness ie lower enthalpy and higher entropy ie a chemical system is driven towards the formation of products by a negative  H or a positive value of  S, or both. The combined effects of these is given by the Gibbs free energy, G: G = H – TSwhere T = temperature in Kelvins G is a measure of the energy of the system, and a system spontaneously tends toward lower energy state. Gibbs Free Energy The change in energy of a system at a constant temperature is  G =  H -T  S The equation combines the effects of  H and  S. Hence :  G = + reaction is disfavored  G = - reaction is favored

10.  G o is related to the equilibrium constant of a reaction by: K = e -  G /RT or  G o = -RTlnK where R = gas constant = 8.314 JdegK -1 mol -1 From the equation: a large equilibrium constant results from a large negative free energy o Standard enthalpy, H o, standard entropy, S o, and standard free energy, G o represent the thermodynamic quantities at standard state (ie 1 atm, 298K and unit concentration) Summary: A chemical reaction is favored by (i) the liberation of heat (  H negative), (ii) an increase in disorder (  S positive), (iii)  G o is negative or, equivalently, if K > 1.

11. Le Châtelier’s Principle Suppose a system at equilibrium is subjected to a change that disturbs the system (eg the equilibrium concentrations of reactants and products are altered by changing the temperature, the pressure or the concentration of one of the reactants), the effects of such changes can be predicted from Le Châtelier’s Principle which states: when a change is applied to a system at equilibrium, the equilibrium will shift in a direction that tends to relieve or counteract that change. Example : BrO 3 - + Cr 3+ + 4H 2 O Br - + Cr 2 O 7 2- + 8H + for which the equilibrium constant is given by K = =1 x 10 11 at 25 o C In a particular equilibrium state of this system, the following concentrations exist: [ Br - ][Cr 2 O 7 2- ][H + ] 8 [ BrO 3 - ][Cr 3+ ] 2

12. [H + ] = 5.0 M; [Cr 2 O 7 2- ] = 0.10M; [Cr 3+ ] = 0.003M; [Br - ] = 1.0M; [BrO 3 - ] = 0.043M Dichromate is added to the solution to increase the concentration of [Cr 2 O 7 2- ] to 0.20M. In what direction would the reaction proceed to reach equilibrium? According to Le Châtelier ’ s Principle, the reaction should move to the left to partially offset the increase in dichromate. This can be verified by setting up a reaction quotient, Q: Q = = 2 x 10 11 > K (1.0 )(0.20)(5.0) 8 ( 0.043)(0.0030) 2 Because Q > K reaction must move to the left to decrease the numerator and increase the denominator until Q = K

13. Hence to achieve equilibrium and: If Q < K reaction must proceed to the right If Q > K reaction must proceed to the left What if the temperature is changed ? temperature dependent temperature independent From: K = e -  G /RT = e -(  H - T  S )/RT = e -  H /RT.e -  S /R o ooo o If  H o is negative, e -  H /RT decreases with increasing temperature in an exothermic reaction, K decreases with increasing temperature If  H o is positive, e -  H /RT increases with increasing temperature in an endothermic reaction, K increases with increasing temperature o o

14. Solubility Product When substances have limited solubility and their solubility is exceeded, the ions of the dissolved portion exist in equilibrium with the solid material AgCl(s) AgCl(aq) Ag + + Cl - - the substance will have a definite solubility at a given temperature - a small very amount of undissociated compound usually exists in equilibrium in the aqueous phase and its concentration is constant - the overall equilibrium constant for the solubility can be written for the stepwise equilibrium: K sp = = Since AgCl(s) is the pure solid [AgCl(s)] = 1 [AgCl(aq)] [AgCl(s)][AgCl(aq)] [ Ag + ] [ Cl - ] [AgCl(s)]

15. Hence : K sp = [ Ag + ] [ Cl - ] This relationship measures the compound ’ s solubility. It holds under all equilibrium conditions at the specified temperatures Example: What is the solubility of Hg 2 Cl 2, in g/l, if the solubility product is 1.2 x 10 -18 ? Hg 2 Cl 2 (s) Hg 2 2+ + 2Cl - K sp = [Hg 2 2+ ][Cl - ] 2 = 1.2 x 10 -18 Let s represent the molar solubility of Hg 2 Cl 2. Then [Hg 2 2+ ] = s and [Cl - ] = 2s Thus: (s)(2s) 2 = 1.2 x 10 -18 s = 6.7 x 10 – 7 M Solubility in g/l = 6.7 x 10 – 7 x 472.08g/mol = 3.162 x 10 – 10 g/l

16. Common Ion Effect What will be the concentration of Hg 2 2+ if a 2 nd source of Cl - was added to the solution (eg 0.030M NaCl) ? Hg 2 Cl 2 (s) Hg 2 2+ + 2Cl - Initial conc solid 0 0.030 Final conc solid x 2x+ 0.030 K sp = [Hg 2 2+ ][Cl - ] 2 = x (2x + 0.030) 2 = 1.2 x 10 -18 (x)( 0.030) 2 = 1.2 x 10 -18 x = 1.3 x 10 -15 M Without the NaCl, [Hg 2 2+ ] = 6.7 x 10 – 7 M - addition of a product displaces the reaction toward the left This application of Le Châtelier ’ s Principle is called the common ion effect : a salt will be less soluble if one of its constituent ions is already present in the solution

17. Separation by Precipitation Example : Consider a solution containing lead(II)( Pb 2+ ) and mercury(I)( Hg 2 2+ ) ions, each at a concentration of 0.010M PbI 2 (s) Pb 2+ + 2I - K sp = 7.9 x 10 -9 Hg 2 I 2 (s) Hg 2 2+ + 2I - K sp = 1.1 x 10 -28 Is it possible to completely separate the Pb 2+ and Hg 2 2+ by selectively precipitating the latter with iodide? Precipitation reactions can sometimes be used to separate ions from each other.

18. Q = [Pb 2+ ][I - ] 2 = (0.010)(1.0 x 10 -11 ) 2 = 1.0 x 10 -24 < K sp for PbI 2 Hg 2 I 2 (s) Hg 2 2+ + 2I - Initial Conc0 0.010 0 Final Concsolid 1.0 x 10 – 6 x K sp = [Hg 2 2+ ][I - ] 2 =( 1.0 x 10 – 6 )(x) 2 = 1.1 x 10 -28 x = [ I - ] = 1.0 x 10 -11 M Will this amount of I - cause Pb 2+ to precipitate? Consider lowering the Hg 2 2+ concentration to 0.010% (ie.010% of 0.010M = 1.0 x 10 – 6 M) of its original value without precipitating Pb 2+. Let x be the concentration of I - at equilibrium with 1.0 x 10 – 6 M [Hg 2 2+ ]

19. Complex Formation If anion X - precipitates metal M +, it is sometimes observed that a high concentration of X - causes solids MX to redissolve. This phenomenon can be attributed to the formation of complex ions, such as MX 2 - In complex ions such as MX 2 -, X - is known as the ligand of M +. A ligand is defined as any atom of group of atoms attached to the species of interest. M + accepts electrons Lewis acid X - donates electronsLewis base Room to accept electrons Room to donate electrons Example : Pb ++ + I - [Pb I ] + adduct dative or coordinate covalent bond

20. Effect of Complex Ion Formation on Solubility At low I - concentrations, the solubility of lead is governed by precipitation of PbI 2 : Pb 2+ + 2I - PbI 2 (s) K sp = 7.9 x 10 -9 However at high I - concentrations, complex ion formation occurs as the reaction is driven to the right : Pb 2+ + I - PbI + K 1 = 1.0 x 10 2 Pb 2+ + 2I - PbI 2 (aq)  2 = 1.4 x 10 3 Pb 2+ + 3I - PbI 3 –  3 = 8.3 x 10 3 Pb 2+ + 4I - PbI 4 –  4 = 3.0 x 10 4 Example : Find the concentration of PbI +, PbI 2 (aq), PbI 3 – and PbI 4 2- in a solution saturated with PbI 2 (s) and containing dissolved I - with a concentration of 1.0M.

21. K sp = 7.9 x 10 -9 = [Pb 2+ ][I - ] 2 = [Pb 2+ ] [PbI + ] = K 1 [Pb 2+ ][I - ] = (1.0 x 10 2 )(7.9 x 10 -9 )1 = 7.9 x 10 -7 M [PbI 2 (aq)] =  2 [Pb 2 + ][I - ] 2 = (1.4 x 10 3 )(7.9 x 10 -9 ) = 1.1 x 10 -5 M [PbI 3 – ] =  3 [Pb 2 + ][I - ] 3 = (8.3 x 10 3 )(7.9 x 10 -9 ) = 6.6 x 10 -5 M [PbI 4 2- ] =  4 [Pb 2 + ][I - ] 4 = (3.0 x 10 4 )(7.9 x 10 -9 ) =2.4 x 10 -4 M The total concentration of dissolved lead is: [Pb] total = [Pb 2 + ] + [PbI + ] +[PbI 2 (aq)] + [PbI 3 – ] + [PbI 4 2- ] = 3.2 x 10 -4 M

22. % dissolved Pb in PbI 4 2- = (2.4 x 10 -4 )/(3.2 x 10 -4 ) x 100 = 75% Each of the reaction is governed by an equilibrium and an equilibrium constant. The total concentration of dissolved lead is considerably greater than that of Pb 2+ alone The concentration of Pb 2+ that satisfies any one of the equilibria must satisfy all the equilibria.

23. Protic Acids and Bases H + is known as proton because it is what remains when a hydrogen atom loses its electron Protic refers to the chemistry that involves the transfer of H + from one molecule to another In the Br ø nsted and Lowry classification: acid proton donor egHCl + H 2 O H 3 O + + Cl - basesproton acceptors egHCl + NH 3 NH 4 + Cl - Saltsan ionic compound product of an acid-base neutralization reaction typically strong electrolyte

24. Conjugate Acids and Bases In the Br ø nsted and Lowry classification: the products of a reaction between an acid and a base are also classified as acids and bases. What does this mean? Example : acetic acid methylamineacetate ion methyl- ammonium ion acidbase acid Acetic acid and the acetate ion are said to be a conjugate acid-base pair Conjugate acids and bases are related to each other by the gain or loss of one H +

25. Autoprotolysis Some compounds are able to undergo self- ionization, in which they act as both an acid and a base Example : H 2 O H + + OH - 2 H 2 O H 3 O + + OH - 2 NH 3 NH 4 + + NH 2 - What compounds undergo autoprotolysis? Compounds with a reactive H + protic solvent Examples of protic solvents: - CH 3 CH 2 OH (ethanol) - CH 3 COOH (acetic acid) Compounds without a reactive H + aprotic solvent (eg CH 3 CH 2 O CH 2 CH 3, CH 3 CN)

26. pH pH  -log[H + ] - value can lie from –2 +16 - pH = -1 means -log[H + ] = -1.00 or [H + ] = 10M very concentrated strong acid -2024 6 8 10 12 1416 neutral acidicbasic For water : H 2 O H + + OH - K w = [H + ][OH - ] = 1.0 x 10 -14 at 25 o C log K w = log [H + ] + log[OH - ] - log K w = -log [H + ] - log[OH - ] 14.00 = pH + pOH

27. Strengths of Acids and Bases Acids and bases are commonly classified as strong or weak, depending on whether they react nearly “ completely ” or only “ partially ” to produce H + or OH - Example of strong acid: HCl(aq) H + + Cl - Example of strong base: KOH(aq) K + + OH - Weak acids and bases react “ partially ” to produce H + or OH - Weak acids (HA) react with water by donating a proton to H 2 O ie HA + H 2 O H 3 O + + A - or HA + H 2 O H + + A - K a = [ H + ][A - ] [ HA] acid dissociation constant – value is small for weak acids

28. Weak bases (B) react with water by abstracting a proton from H 2 O ie B + H 2 O BH + + OH - K b = [ BH + ][OH - ] [B][B] Base hydrolysis constant – value is small for weak bases Acids and Bases Compounds that can donate more than one proton polyprotic acids Compounds that can accept more than one proton polyprotic bases HOHO

29. K b1 = 1.4 x 10 -2 K b2 = 1.59 x 10 -7 K b3 = 1.42 x 10 -12 K f = K b1.K b2.K b3 HOHO