I. Equilibrium a dynamic condition in which a forward reaction proceeds at the same rate as the reverse reaction; no changes will be observed

Transcription

1 UNIT IX EQUILIBRIUM 1 I. Equilibrium a dynamic condition in which a forward reaction proceeds at the same rate as the reverse reaction; no changes will be observed II. Chemical Equilibrium * physical equilibrium equilibrium in physical processes (example: vapor-liquid equilibrium * chemical equilibrium chemical changes H (g I (g HI(g [ HI] [ H ][ I ] eq Understanding Equilibrium The concept of equilibrium is based on the understanding that a reaction will always proceed in a direction that gets it to the desired equilibrium state, regardless of the starting amounts of reactant or product. Assuming the conditions don t change, the equilibrium state (represented by the numerical value of eq will be the same for the same reaction. A B A B [ A B] [ A] [ B] eq * in this example, eq 1 A B A B starting conditions final conditions ( A B Q ( A ( B (0 0 (10 (5 Q <, so rxns goes right (more product ( A B Q ( A ( B ( (6 ( Q <, so rxns goes right (more product ( A B Q ( A ( B ( 1 ( (1 Q, so rxn is at equilibrium

2 UNIT IX EQUILIBRIUM starting conditions ( 5 Q ( 0 ( 0 Q >, so rxns goes left (more reactant ( Q 1 ( (1 Q, so rxn is at equilibrium * mass-action statement * example: aa bb cc dd eq c [ C] [ D] a [ A] [ B] d b S. Give the mass-action statement for: 1 A B C D A B A B N O (g NO (g eq [ C][ D] [ A] [ B] eq [ A] [ B ] [ A B] For the following equilibrium: SO (g O (g SO (g eq [ NO ] [ N O ] If at 50 o C, the equilibrium concentrations are: [SO ] 0.75 M, [O ] 0.0 M, [SO ] 0.15 M, what is eq the numerical value of? [ SO ] [ SO ] [ O ] [ 0.15] [ 0.75] [ 0.0] 0.1 < 1, so reactants are favored * shows relationship between amount of products and reactants at equilibrium * If >1, equilibrium favors products. If <1, equilibrium favors reactants. III. Ways of expressing Equilibrium Constants A. Homogeneous Equilibrium all ecies are in the same phase * c equilibrium constant using concentrations of ecies (in M * equilibrium constant using partial pressures of gas ecies For the following equilibrium, write the mass-action statement for the c and N O (g NO (g c [ NO ] [ N O ] NO N O

3 UNIT IX EQUILIBRIUM 1 For the following equilibrium: Cl 5 (g Cl (g Cl (g 1.05 at 50 o C. If the equilibrium partial pressures of Cl 5 and Cl are atm and 0.6atm, reectively, what is the equilibrium partial pressure of Cl at this temperature? Cl Cl (0.6 x (1.05( x 1. 98atm Cl 5 We can relate the c to the with the following equation: ( RT c n The c for the reaction: N O (g temperature? NO (g is.6 x 10 - at 5 o C. What is the value for at this n mol NO (.6x10 (g 1mol N O (g ( B. Heterogeneous Equilibria ecies are in two or more different phases * We do not include the concentration of pure solids and liquids because they will not change appreciably in a reaction. 1 Write the c and for the following equilibria: a (NH Se(s NH (g H Se(g b AgCl(s Ag (aq Cl - (aq c ( NH ( H Se c ( Ag ( Cl 1 NH H Se c (s 6 Cl (g Cl (l 1 1 c 6 [ Cl ] 6 Cl

4 UNIT IX EQUILIBRIUM Consider the following equilibrium: CaCO (s CaO(s CO (g. At 800 o C the pressure of CO is 0.6 atm. Calculate a and (b c at that temperature. a b c CO c.69x ( RT n 0.6 c ( both 's are < 1, so favors reactants 1 C. Multiple Equilibria if a reaction can be expressed as the sum of two or more reactions, the for the overall reaction ( is: 1 D. Other Rules 1 If the reaction is reversed: ` 1/ o If the reaction is modified by a multiple(n: ` o n At a certain temperature the following reactions have the constants shown S (s O (g SO (g c1. x 10 1 S (s O (g SO (g c 9.8 x 10 8 Calculate the equilibrium constant c for the following reaction at room temperature switch and double top rxn : bottom rxn stays same : sum :(S and O SO (g O (g SO (g cancel : SO S O SO O S O SO SO ' ' c c1 1.x10 9.8x10 8 " (5.7x x10 (9.8x x10 15 IV. The Relationship between inetics and Equilbrium Consider the following equilibrium: A B AB k r Then: rate fwd k f [A][B] and rate rev k r [AB ] At equilibrium: rate fwd rate rev so k f [A][B] k r [AB ] k f Well: k k f r [ AB ] [ A][ B] c therefore: c k k f r

5 UNIT IX EQUILIBRIUM 5 V. What does tell us? A. redicting the direction of a reaction * If we know the for an equilibrium, and we are given some initial conditions, we can calculate a quotient (Q for the initial conditions - If Q >, the reaction will proceed in the reverse direction (left this makes Q smaller until it is equal to - If Q, the reaction is already at equilibrium - If Q <, the reaction will proceed in the forward direction (right At the start of a reaction, the concentrations of N, H, and NH are 7.11 x 10 - M, 9.17 x 10 - M, and 1.8 x 10 - M in a.50 L reaction vessel at 00 o C. If the c 0.65 at this temperature, is this system at equilibrium? If not, predict the direction the reaction will proceed. N (g H (g NH (g Q (NH (N (H (1.8x10 (7.11x10 (9.17x Q <, so reaction proceeds to the right (products B. Calculating Equilibrium Concentrations; for the sample problems 1-, use the following equilibrium: H (g I (g HI(g 1 If you begin the reaction in a.00 L container with mol of hydrogen, and 0.00 mol of iodine, and the reaction proceeds until it reaches an equilibrium with 0.00 mol of hydrogen iodide present, calculate c. Initial :.500mol Change : Final :.00mol 0 x 0.00 x 0.00 n H f HI x.00mol mol n mol 0.00mol ( H f 0.150M ( I f M ( HI.00L ( HI (0.00 c 5. ( H ( I (0.150( H I 0 I f Q 0, so rxn Change must relate by stoichiometric ratio f 0.00M A mixture of mol H and mol I was placed in a 1.00 L container at 0 o C. Calculate the concentrations of H, I, and HI at equilibrium. The c for this equilibrium is 5. at this temperature. H Initial :.500mol Change : Final :.5.500mol.5 x square root of both sides : 7. x 7.( x x.67 7.x 9.x.67 x 0.9 ( HI x 0.786M ( H I HI 0 x 0 x (x (.5 ( I M 5.

6 UNIT IX EQUILIBRIUM 6 Suppose the initial concentrations at 0 o C are [H ] 0.006M, [I ] 0.001M and [HI] 0.0 M, what will their concentrations be at equilibrium? Initial :.006mol Change : Final :.006 (.0 x 5. 5.(.006 (.001 (.0 x (.006 ( (.58x10 50.x H x x.65x 8.98x10.001mol.001 (H M I HI.0 x 0.0 x 5.0x x x 0 x.011 (too big, (.0 Q 19.5 : Q < so rxn (0.006(.001 ignore OR (I.0058M (HI.055M FOIL Suppose at 0 o C, the initial concentration of HI is 0. M, what will the concentrations of all ecies be at equilibrium? Initial : Change : x Final : (. - x.059.9x x 0 50.x ( H H 0 x I 0 x x 5.x HI.. 5.x.9x.059 x.05 (. ( x( x 5. ( I x.05m ( HI..180M 5 Suppose at a temperature of 550,. x If you begin the reaction with hydrogen and iodine gases at partial pressures of 1. atm and atm reectively, what will the final pressures of the three gases be at equilibrium? H Initial : 1.mol Change : Final : 1. (x.x10 (1.(.798 I.798mol.798 x x 0 x If the value of is 5% or less than the value of you may ignore the x's in the denominator 6 HI 0.x10 (x.x10 (1. (.798 the numbers next to x (1. &.798, x 9.0x10 HI x 1.8x10 atm H 1. 1.atm N atm 6 6

7 UNIT IX EQUILIBRIUM 7 VI. Stresses on an Equilibrium * Le Chatlier s rinciple - when a system at equilibrium undergoes a stress, the equilibrium will shift in a direction which relieves that stress Example: A B C D * If reaction leans to the right, [C] & [D] go up; [A] and [B] go down * If reaction leans to the left, [C] & [D] go down; [A] and [B] go up For each of the following stresses, indicate the direction the equilibrium will shift, then indicate whether the concentrations of the other three participants will increase, decrease or stay the same: A. Changes in Concentration A B C D * if concentration of a ecies increases, equilibrium leans AWAY from that ecies * if concentration of a ecies decreases, equilibrium leans TOWARDS that ecies * these changes DO NOT affect the equilibrium constant ( T. 1 increase [A] decrease [C] (B decreases; (C&(D increase (A&(B decrease; (D increases S 1 decrease [B] increase [D] (A increases; (C&(D decrease (A&(B increase; (C decreases ** Note that none of these changes affect the rate constant (k B. Changes in Temperature - depends on whether reaction is endothermic or exothermic T. A B C H -6.5 kj/mol Same as: A B C heat * If T goes up, reaction leans away from heat, If T goes down, reaction goes toward heat * If product increases, increases * if product decreases, decreases 1 increase temperature decrease temperature (A&(B increase; (C& k decrease (A&(B decrease; (C&k increase S. A B C D H 5. kj/mol Same as: heat A B C D 1 increase temperature decrease temperature (A&(B decrease; (C,(D,&k increase (A&(B increase (C, (D, & k decrease

8 UNIT IX EQUILIBRIUM 8 C. Changes in ressure - for gas equilibria only starting conditions NO N O NO (g N O (g * Note that as the pressure increases, the volume decreases, so the side of the reaction that contains less moles of gas (product side is favored, therefore the concentration of N O increases and NO decreases * If pressure increases, volume decreases, equilibrium leans to side with less moles * If pressure decreases, volume increases, equilibrium leans to side with more moles * Also, this could apply if volume is changed with no change in pressure * These changes also affect in a manner similar to temperature T. A(g B(g AB(g * Note 1 mole of gas on the right, on the left 1 increase pressure decrease pressure (A&(B decrease; (AB&k increase (A&(B increase; (AB&k decrease S. NH (g H (g N (g * moles of gas on the right, on the left 1 increase volume decrease volume (NH decreases; (NH increases; (H,(N,k increase (H,(N,k decrease D. Effect of a Cayalyst * A catalyst has no effect on the balance of an equilibrium, it only shortens the time it takes to get from the start of the reaction to the equilibrium state E. Addition of an Inert Gas this includes the Noble Gases * Addition of an inert gas has no effect on the balance of the equilibrium

9 UNIT IX EQUILIBRIUM 9 VII. Solubility roduct - a way to measure solubility of a compound * When an ionic solid dissolves in water, it separates into its cations and anions - dissociation equation : MX(s M (aq X - (aq * When the water is saturated with the ions, an equilibrium is established between the solid compound and the ions: [ M ][ X ] - solubility expression : mass action: * if compound is soluble, equilibrium leans to the right, is large * if compound is insoluble, equilibrium leans to the left, is small 1 [M ][X - ] Give the dissociation equation and write the solubility expression for the following ionic solids: 1 AgCl(s Ag Cl - Ca (O (s Ca O - * Note that in order for ions to exist in isolation, they must be in aqueous solution; therefore phase notation is rarely used * How this value relates to saturation: assume the starting concentrations of some ions are [M ] o and [X - ] o ; [M ] o [X - ] o Q 1 If Q<, the solution is unsaturated If Q, the solution is saturated If Q>, the solution is supersaturated, a precipitate will form * molar solubility solubility expressed in mol/l (M * This quantity represents the number of moles of salt dissolving per liter of solution * solubility solubility expressed in g/l 1 The molar solubility of silver sulfate is M. Calculate the solubility product of the salt. Ag SO (s Ag - SO For every Ag SO that lits, you get one sulfate ion, and two silver ions, therefore (SO - molar solubility 0.015M and then (Ag is twice that of sulfate, so (Ag 0.00M (Ag (SO - (0.00 ( x 10-5 It is found experimentally that the solubility of calcium sulfate is 0.67 g/l. Calculate the value of for calcium sulfate. CaSO (s Ca SO - (Ca (SO - (.9x10 -. x g CaSO 1mol CaSO.9x10 L 16.g CaSO mol L

10 UNIT IX EQUILIBRIUM 10 Calculate the molar solubility of copper(ii hydroxide (. x Cu( OH.x10 5.5x10 ( Cu ( x(x 0 1 x 1.8x10 Cu x x x 7 ( OH M OH dissociation equation x solubility expression Calculate the solubility of calcium phohate ( 1. x Ca ( O 1.x10 ( Ca (x 6 x.6x10 Ca x ( O (x 108x 6 5 O x.6x10 M L 6 mol 10g 8.0x10 1mol g / L 5 Exactly 00 ml of M BaCl are added to exactly 600 ml of M SO. Will a precipitate form? The for barium sulfate is 1.1 x BaSO total volume of (0.00L(0.000M 8.0x10 (0.600L(0.0080M.8x10 Q ( Ba Ba ( SO SO ( Ba solution 0.00L 0.600L 0.800L mol Ba mol SO ( SO - (0.0010( x10 Q >, so equilibrium shifts left (to reactants 0.800L M 0.800L M since reactant is the precipitate, the precipitate will form 6 ( Ba ( SO

11 UNIT IX EQUILIBRIUM 11 VIII. The Common Ion Effect and Solubility 1 Calculate the solubility of silver chloride (in g/l in pure water. 1.6 x AgCl Ag Cl x x ( Ag ( Cl 1.6x10 10 x 1.x10 x L 5 mol 1g 1mol 1.8x10 g / L Calculate the solubility of silver chloride (in g/l in a M silver nitrate solution M ( Ag AgCl Ag Cl x x ( Ag ( Cl 1.6x10 1.6x x10 x L ( NO ( x( x i (0.0065( x 5% rule 8 mol 1g.5x10 1mol i only Ag is in equilibrium, so ignore 6 g / L * Note that the solubility of AgCl in a silver nitrate solution is 500 times less than that of AgCl in pure water. This is the essence of the common ion effect. Having ions already present in solution reduces the solubility of a salt containing that ion. What is the molar solubility of lead(ii iodide in a 0.050M solution of sodium iodide? 1. x M ( Na bi b I x x ( b ( I 1.x10 1.x x 5.6x10 ( x(0.050 x ( x( i mol / L ( I i 5% rule only I is in equilibrium, so ignore Na NO

12 UNIT IX EQUILIBRIUM 1 Calculate the final concentrations of (aq, C O - (aq, Ba (aq, and Br -1 (aq in a solution prepared by adding 0.100L of 0.00M C O to 0.150L of 0.50M BaBr. For BaC O,. x 10-8 BaC total volume of (0.150L(0.50M 0.075mol Ba BaC.x10.x10 and Br O ( Ba 8 8 x.x10 Ba - solution 0.100L 0.150L 0.50L don' t form the ppt, so they are ectators, you can calculate their final concentrations now : ( x( x (0.0700( x 5% rule 7 Ba x ( C O C M ( C O O x (0.100L(0.00M 0.000mol C O ( Ba mol mol Br -1 x as many 's ( ( BaBr 0.000mol mol ( 0.160M ( Br 0.00M 0.50L 0.50L LR Ba BaCO C O 0.075mol 0mol 0.000mol mol ( Ba 0.000mol 0.000mol 0.50L 0.000mol mol NA( solid ~ 0mol O C O ( C O Ba Br M C O C O