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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.

lillian-griffith
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Chemical Equilibrium

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  1. 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 Chemical Equilibrium

  2. 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 (Rf )  [A]a[B]b = kf [A]a[B]b a A + b 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 Similarly, in the backward reaction Rate of backward reaction (Rb )  [C]c[D]d = kb [C]c[D]d

  3. At equilibrium: rate of forward reaction = rate of backward reaction kf [A]a[B]b = kb [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 [C]c[D]d kf [A]a[B]b kb

  4. Consider the reaction: HA H+ + A- K1 = 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/K1 [H+][A-] [HA] [HA] [H+][A-] Manipulating Equilibrium Constants

  5. Equilibrium constant for sum of reactions: K3 = K1K2 = x = If two reactions are added, the new K value is the product of the two individual values: HA H+ + A-K1 H+ + C CH+ K2 HA + C A- + CH+ K3 [H+][A-] [CH+] [HA] [H+][C] [A-][CH+] [A-][CH+] [HA][C]

  6. Example : The equilibrium constant at 25oC for the reaction: H2O H+ + OH- Kw = 1.0 x 10-14 NH3(aq) + H2O NH4+ + OH- K = 1.8 x 10-5 Find the equilibrium constant for the reaction: NH4+ NH3 + H+ NH3 The 3rd reaction can be obtained by: H2O H+ + OH- Kw NH4+ + OH- NH3 + H2O 1/K NH4+ H+ + NH3 K3 = Kw /K = 5.6 x 10-10 NH3 NH3

  7. 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 (DH) : Hproduct - Hreactanct DH = + heat is absorbed endothermic DH = - heart is liberated exothermic Equilibrium and Thermodynamics • 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 Enthalpy of a reaction : heat absorbed or released by the reaction Entropy of a reaction : degree of disorder of reactants and products

  8. DS = Sproduct - Sreactanct DS = + products are more disordered than reactants DS = - 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 Entropy

  9. The change in energy of a system at a constant temperature is DG = DH -TDS The equation combines the effects of DH and DS. Hence : DG = + reaction is disfavored DG = - reaction is favored Gibbs Free Energy 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 DH or a positive value of DS, or both. The combined effects of these is given by the Gibbs free energy, G: G = H – TS where T = temperature in Kelvins G is a measure of the energy of the system, and a system spontaneously tends toward lower energy state.

  10. DGo is related to the equilibrium constant of a reaction by: K = e-DG /RT or DGo = -RTlnK where R = gas constant = 8.314 JdegK-1mol-1 From the equation: a large equilibrium constant results from a large negative free energy o Standard enthalpy, Ho, standard entropy, So, and standard free energy, Go 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 (DH negative), (ii) an increase in disorder (DS positive), (iii) DGo is negative or, equivalently, if K > 1.

  11. Example : BrO3- + Cr3+ + 4H2O Br - + Cr2O72- + 8H+ for which the equilibrium constant is given by K = =1 x 1011 at 25oC In a particular equilibrium state of this system, the following concentrations exist: [Br -][Cr2O72-][H+]8 [BrO3-][Cr3+]2 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.

  12. 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 1011 > 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 [H+] = 5.0 M; [Cr2O72-] = 0.10M; [Cr3+] = 0.003M; [Br -] = 1.0M; [BrO3-] = 0.043M Dichromate is added to the solution to increase the concentration of [Cr2O72-] to 0.20M. In what direction would the reaction proceed to reach equilibrium?

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

  14. 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: • Ksp = • = • Since AgCl(s) is the pure solid [AgCl(s)] = 1 [Ag+] [Cl-] [AgCl(aq)] [AgCl(s)] [AgCl(aq)] [Ag+] [Cl-] [AgCl(s)] Solubility Product

  15. Example: What is the solubility of Hg2Cl2, in g/l, if the solubility product is 1.2 x 10-18 ? Hg2Cl2(s) Hg22+ + 2Cl- Ksp = [Hg22+ ][Cl-]2 = 1.2 x 10-18 Let s represent the molar solubility of Hg2Cl2. Then [Hg22+ ] = 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 Hence : Ksp = [Ag+] [Cl-] This relationship measures the compound’s solubility. It holds under all equilibrium conditions at the specified temperatures

  16. Hg2Cl2(s) Hg22+ + 2Cl- • Initial conc solid 0 0.030 • Final conc solid x 2x+ 0.030 • Ksp = [Hg22+ ][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, [Hg22+ ] = 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 Common Ion Effect What will be the concentration of Hg22+if a 2nd source of Cl- was added to the solution (eg 0.030M NaCl) ?

  17. Precipitation reactions can sometimes be used to separate ions from each other. Example : Consider a solution containing lead(II)(Pb2+) and mercury(I)(Hg22+) ions, each at a concentration of 0.010M PbI2(s) Pb2+ + 2I- Ksp =7.9 x 10-9 Hg2I2(s) Hg22+ + 2I- Ksp =1.1 x 10-28 Is it possible to completely separate the Pb2+ and Hg22+ by selectively precipitating the latter with iodide? Separation by Precipitation

  18. Consider lowering the Hg22+ concentration to 0.010% (ie .010% of 0.010M = 1.0 x 10 –6 M) of its original value without precipitating Pb2+. Let x be the concentration of I- at equilibrium with 1.0 x 10 –6 M [Hg22+] Hg2I2(s) Hg22+ + 2I- Initial Conc 0 0.010 0 Final Conc solid 1.0 x 10 –6x Ksp = [Hg22+ ][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 - causePb2+ to precipitate? Q = [Pb2+][I-]2 = (0.010)(1.0 x 10-11)2 = 1.0 x 10-24 <Ksp for PbI2

  19. 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 MX2- In complex ions such as MX2-, 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 electrons Lewis base Example : Pb++ + I - [Pb I ]+ adduct Room to accept electrons Room to donate electrons dative or coordinate covalent bond Complex Formation

  20. At low I- concentrations, the solubility of lead is governed by precipitation of PbI2 : Pb2+ + 2I- PbI2(s) Ksp = 7.9 x 10-9 However at high I- concentrations, complex ion formation occurs as the reaction is driven to the right : Pb2+ + I- PbI+ K1 =1.0 x 102 Pb2+ + 2I- PbI2(aq) b2 =1.4 x 103 Pb2+ + 3I- PbI3 –b3 =8.3 x 103 Pb2+ + 4I- PbI4 –b4 =3.0 x 104 Effect of Complex Ion Formation on Solubility Example : Find the concentration of PbI+, PbI2(aq), PbI3–and PbI42- in a solution saturated with PbI2(s) and containing dissolved I- with a concentration of 1.0M.

  21. Ksp = 7.9 x 10-9 = [Pb2+][I-]2 = [Pb2+] [PbI+] = K1 [Pb2+][I-] = (1.0 x 102)(7.9 x 10-9)1 = 7.9 x 10-7 M [PbI2(aq)] = b2[Pb2 +][I-]2 = (1.4 x 103 )(7.9 x 10-9) = 1.1 x 10-5 M [PbI3–]= b3[Pb2 +][I-]3 = (8.3 x 103 )(7.9 x 10-9) = 6.6 x 10-5M [PbI42-]= b4[Pb2 +][I-]4 = (3.0 x 104)(7.9 x 10-9) =2.4x 10-4M The total concentration of dissolved lead is: [Pb]total = [Pb2 +] + [PbI+] +[PbI2(aq)] + [PbI3–]+ [PbI42-] = 3.2 x 10-4M

  22. 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 Pb2+ alone The concentration ofPb2+ that satisfies any one of the equilibria must satisfy all the equilibria. % dissolved Pb in PbI42- = (2.4x 10-4)/(3.2 x 10-4) x 100 = 75%

  23. 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 Brønsted and Lowry classification: acid proton donor eg HCl + H2O H3O+ + Cl- bases proton acceptors eg HCl + NH3 NH4+Cl- Salts an ionic compound product of an acid-base neutralization reaction typically strong electrolyte Protic Acids and Bases

  24. Example : acetic acid methylamine acetate ion methyl- ammonium ion acid base base acid Conjugate Acids and Bases In the Brønsted 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? 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. Some compounds are able to undergo self-ionization, in which they act as both an acid and a base Example : H2O H+ + OH- 2 H2O H3O+ + OH- 2 NH3 NH4+ + NH2- What compounds undergo autoprotolysis? • Compounds with a reactive H+ protic solvent • Examples of protic solvents: • CH3 CH2 OH (ethanol) • CH3 COOH (acetic acid) • Compounds without a reactive H+ aprotic solvent (eg CH3 CH2 O CH2 CH3 , CH3 CN) Autoprotolysis

  26. pH  -log[H+] • value can lie from –2 +16 • pH = -1 means -log[H+] = -1.00 or • [H+] = 10M • very concentrated strong acid 6 -2 0 2 4 10 14 16 8 12 acidic basic neutral For water : H2OH+ + OH- Kw =[H+][OH-] = 1.0x 10-14 at 25oC log Kw = log [H+] + log[OH-] - log Kw = -log [H+] - log[OH-] 14.00 = pH + pOH pH

  27. 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 H2O ie HA + H2O H3O++ A- or HA + H2O H++ A- Ka = [H+][A-] [HA] acid dissociation constant – value is small for weak acids Strengths of Acids and Bases

  28. Weak bases (B) react with water by abstracting a proton from H2O ie B + H2O BH++ OH- Kb = [BH+][OH-] [B] Base hydrolysis constant – value is small for weak bases Compounds that can donate more than one proton polyprotic acids Compounds that can accept more than one proton polyprotic bases HO Acids and Bases

  29. Kb1 = 1.4x 10-2 Kb2 = 1.59x 10-7 Kb3 = 1.42x 10-12 HO Kf = Kb1 .Kb2 .Kb3

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