Chemical Equilibrium 1/26/12. Equilibrium Constants. Equilibrium Constants. Equilibrium Constants. Thermodynamics. Thermodynamics

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1 Equilibrium Constants Chemical Equilibrium For a generic chemical reaction, the equilibrium constant is defined as: aa + bb cc + dd (1) c [C] [D] K eq1 a [A] [B] d b The equilibrium constant, K eq, for a chemical reaction indicates whether the reactants or the products will be favored in an equilibrium process Equilibrium Constants For the reverse reaction cc + dd aa + bb (2) the equilibrium constant is the inverse of the forward reaction: 1 [A] K eq2 K [C] eq1 a c [B] [D] b d Equilibrium Constants When reactions are added to produce a net reaction, the net equilibrium constant is the product of the K eq s for each reaction: HA H + + A - (1) H + + C HC + (2) HA + C HC + + A - (net) + - [H ] [A ] K1 [HA] + [HC ] K2 + [H ][C] + [HC ][A ] Knet K1K 2 [HA][C] Enthalpy, ΔH, is a measure of the change in heat content between reactants and products CH 4 + 2O 2 CO 2 + 2H 2 O(g) ΔH o rxn kj Methane releases heat to surrounding when combusted NaCl(s) Na + (aq) + Cl - (aq) ΔH o rxn 3.87 kj Sodium chloride takes heat from surroundings when dissolved Entropy, ΔS, is a measure of the change of disorder when going from reactants to products CH 4 + 2O 2 CO 2 + 2H 2 O(g) ΔS o rxn J/K The products are more ordered than the reactants NaCl(s) Na + (aq) + Cl - (aq) ΔS o rxn 43.4 J/K The products are more disordered than the reactants more space between ions in solution 1

2 Gibb s Free Energy, ΔG, is a measure of the energy available to do work following reaction Definition: ΔG ΔH - TΔS CH 4 + 2O 2 CO 2 + 2H 2 O(g) ΔG o rxn kj NaCl(s) Na + (aq) + Cl - (aq) ΔS o rxn kj ΔG is also a measure of where the equilibrium for a reaction will lie: If ΔG rxn < 0, the reaction is product favored If ΔG rxn > 0, the reaction is reactant favored If ΔG rxn 0, the reaction is in equilibrium The equilibrium constant is related to the Gibb s Free Energy: K eq exp{-δg o rxn /RT} R gas constant J/mol K T temperature (in K) The solubility product, K sp, for a salt is a specific type of equilibrium constant Given an excess of salt, K sp for the salt will determine how much of the salt will dissolve in water: M x A y (s) x M y+ (aq) + y A x- (aq) K y+ x x y sp [M ] [A ] Many salts are only slightly soluble The solubility product is a measure of the concentration of ions in a solution saturated with the salt MA(s) M + (aq) + A - (aq) K sp [M + ][A - ] AgCl(s) Ag + (aq) + Cl - (aq) K sp [Ag + ][Cl - ]1.8x10-10 PbCl 2 (s) Pb (aq) + 2 Cl - (aq) K sp [Pb ][Cl - ] 2 1.7x10-5 AuBr 3 (s) Au 3+ (aq) + 3 Br - (aq) K sp [Au 3+ ][Br - ] 3 4.0x10-36 Knowing the K sp, we can calculate the concentration of ions in solution AgCl(s) Ag + (aq) + Cl - (aq) K sp [Ag + ][Cl - ]1.8x x x x x x x 1.3 x 10-5 M [Ag + ] [Cl - ] PbCl 2 (s) Pb (aq) + 2 Cl - (aq) K sp [Pb ][Cl - ] 2 1.7x10-5 -x x 2x x(2x) x x x 10-5 x 1.6 x 10-2 M [Pb ] 1.6 x 10-2 M [Cl - ] 3.2 x 10-2 M 2

3 AuBr 3 (s) Au 3+ (aq) + 3 Br - (aq) K sp [Au 3+ ][Br - ] 3 4.0x x x 3x x(3x) x x x x 6.2 x M [Au 3+ ] 6.2 x M [Br - ] 1.9 x 10-9 M Common ion effect How much PbI 2 will dissolve in a M solution of NaI? PbI 2 (s) Pb (aq) + 2 I - (aq) K sp 8.7 x x x 2x x(2x+.0100) x 10-9 x(4x x + 1.0x10-4 ) 8.7 x x x x10-4 x 8.7x x 8.4 x 10-5 M vs 1.3 x 10-3 M if no I - (aq) were present initially Define ion quotient, Q, as: M x A y (s) x M y+ (aq) + y A x- (aq) Q [M y+ ] x [A x- ] y Q looks just like K eq, but the system is not in equilibrium A precipitate will form only when Q exceeds K sp Q < K sp : solution is unsaturated no precipitate Q > K sp : solution is saturated precipitate forms Q K sp : solution at saturation point a solution contain M Pb and M Ag +. You want to remove one of the ions from solution by adding Cl - without precipitating the other ion. Which metal will precipitate first? K sp (AgCl) [Ag + ][Cl - ] 1.8 x K sp (PbCl 2 ) [Pb ][Cl - ] x 10-5 How much Cl - must be added before each metal begins to precipitate? Silver: [Cl - ] K sp /[Ag + ] 1.8 x / x 10-9 M Lead: [Cl - ] {K sp /[Pb ]} 1/2 {1.7 x 10-5 /0.0200} M How much Ag + will remain in solution when lead begins to precipitate? [Cl - ] M when lead begins to precipitate [Ag + ] K sp /[Cl - ] 1.8 x / x 10-9 M %Ag remaining 6.2 x 10-9 / x 100% % 3

4 Frequently, a metal may combine with one or more simple anions or neutral species to form an ion soluble in water. The resulting ion is called a complex ion: Cu + 4NH 3 (aq) Cu(NH 3 ) 4 (aq) K eq 2.3 x Copper acts as a Lewis Acid (accepts pair of electrons) and ammonia acts as a Lewis Base (donates pair of electrons) Cu electron configuration: [Ar] 4s 2 3d 6 Cu hybridizes to sp 3 d 2 which leaves unoccupied hybrid orbitals Each NH 3 has an electron lone pair on the nitrogen atom which fills the hybrid orbitals to make a complex ion: :NH 3 NH 3 H 3 N: Cu :NH 3 H 3 N-Cu-NH 3 :NH 3 NH 3 Determine the concentration of Cu in a 0.50 M NH 3 (aq) solution Cu + 4NH 3 (aq) Cu(NH 3 ) 4 (aq) because the equilibrium constant is large, the reaction will strongly favor the product let reaction go completely to right, and then allow some dissociation back to reactants Cu + 4NH 3 (aq) Cu(NH 3 ) 4 (aq) x 4x x Determine the concentration of Cu in a 0.50 M NH 3 (aq) solution [Cu(NH3 ) 4 ] 12 Keq 2.3 x 10 4 [Cu ][NH3 ] ( x) x 10 4 x (4x) rearranging gives : 256 x x10-4 x 7.3 x 10 M [Cu ] assume x isnegligible Acids and Bases Brønsted-Lowry acids and base Acid: H + ion donor Base: H + ion acceptor A Brønsted-Lowry acids are also called protic acids because they donate protons (H + ) Conjugate Acids & Bases Acids react with bases and vice versa All acids and bases come with a conjugate pair a base or acid, respectively, that is formed in conjunction with the original species HCl(aq) + H 2 O( ) H 3 O + (aq) + Cl - (aq) acid base conjugate conjugate acid base 4

5 Conjugate Acids & Bases NaOH(aq) + H 2 O( ) OH - (aq) + H 2 O( ) + Na + (aq) base acid conjugate conjugate base acid CH 3 COOH(aq) + H 2 O( ) CH 3 COO - (aq) + H 3 O + (aq) acid base conjugate conjugate base acid Strengths of Acids and Bases Strong acids donate H + ions more easily The stronger the acid, the weaker the conjugate base associated with that acid Strong bases accept H + ions more easily The stronger the base, the weaker the conjugate acid associated with that base ph is a measure of the hydronium ion content of a solution ph is defined as: ph -log[h 3 O + ] log is log base 10, not ln (natural log) [H 3 O + ] is given in molar units (M) ph of pure water ([H 3 O + ] 1.0 x 10-7 M): ph -log(1.0x10-7 ) 7.0 Neutral is defined as the ph of pure water: ph 7 Acidic solutions have ph lower than 7: ph < 7 acidic Basic solutions have ph larger than 7: ph > 7 basic We can also use poh to describe a solution poh is defined as: poh -log[oh - ] The sum of ph and poh must equal 14 ph + poh 14 assuming room temperature (25 o C) 5

6 Example Find [H 3 O + ] of a solution that has poh 9.37 Method 1: Calculate ph, then [H 3 O + ] Step 1: Determine ph ph 14 poh Step 2: Determine [H 3 O + ] [H 3 O + ] 10 -ph x 10-5 M Example (con t.) Find [H 3 O + ] of a solution that has poh 9.37 Method 2: Calculate [OH - ], then [H 3 O + ] using K w Step 1: Determine [OH - ] [OH - ] 10 -poh x M Step 2: Determine [H 3 O + ] using K w [H 3 O + ] K w /[OH - ] (1.0x10-14 )/(4.27x10-10 ) 2.34 x 10-5 M The extent of dissociation of an acid or base in H 2 O can be quantified using its ionization constant K a is a specific type of equilibrium constant HA(aq) + H 2 O( ) H 3 O + (aq) + A - (aq) acid base c. acid c. base [H3O ][A ] [H3O ][conjugate base] Ka [HA] [acid] [HA] undissociated acid in solution Acetic acid has a K a 1.8 x 10-5 Determine the ph of a 0.2 M acetic acid solution CH 3 COOH(aq) + H 2 O( ) CH 3 COO - (aq) + H 3 O + (aq) CH 3 COOH CH 3 COO - H 3 O + initial Δ -x x x equil.2 x x x Example (con t.): Determine the ph of a 0.2 M acetic acid solution [CH COO ][H O ] K 1.8 x a [CH3COOH] 2-5 x 1.8 x 10 x M.2 x ph -log[h 3 O + ] -log(.0019) 2.7 K b is a specific equilibrium constant for bases B(aq) + H 2 O( ) HB + (aq) + OH - (aq) base acid c. acid c. base [HB ][OH ] [OH ][conjugate acid] Kb [B] [base] [B] undissociated base in solution 6

7 Determine [B] in a 1.82 x 10-3 M solution of NH 3 NH 3 (aq) + H 2 O( ) NH 4+ (aq) + OH - (aq) NH 3 NH 4 + OH - initial 1.82x Δ -x x x equil 1.82x10-3 x x x NH 3 (aq) + H 2 O( ) NH 4+ (aq) + OH - (aq) [NH ][OH ] x K 1.8x10 4 b [NH ] x10 - x x 1.72 x 10-4 M [NH 4+ ] [OH - ] [NH 3 ] 1.82 x 10-3 M 1.72 x 10-4 M 1.65 x 10-3 M K a and K b are related through K w (autoionization constant of water): K a K b K w K a K w /K b K b K w /K a Acetic acid has K a 1.8 x what is K b for acetate ion (CH 3 COO - )? K b (1.0 x )/(1.8 x 10-5 ) 5.6 x