Chapter 6 General Concepts of Chemical Equilibrium

Transcription

1 Chapter 6 General Concepts of Chemical Equilibrium

2 Introduction Chemical Equilibrium Ø The state where the concentrations of all reactants and products remain constant with time. Ø Equilibrium is not static, but is a highly dynamic situation.

3 Chemical Reactions: The rate Concept Ø A + B Ø C + D C + D ( forward) A + B (reverse) Ø Initially there is only A and B so only the forward reaction is possible Ø As C and D build up, the reverse reaction speeds up while the forward reaction slows down. Ø Eventually the rates are equal

4 Reaction Rate Forward Reaction Rate forwards = Rate reverse Reverse reaction Equilibrium Dynamic? Time

5 The equilibrium constant Law of Mass Action Ø For any reaction Ø aa + bb cc + dd Ø K = [C] c [D] d [A] a [B] b Ø K is called the equilibrium constant. Ø is how a reversible reaction is identified

6 v Comments on Law of mass action v K is constant regardless of the amounts of materials mixed initially v Equilibrium concentrations will not always be the same but K is the same v Each set of equilibrium concentrations in an equilibrium system is called equilibrium position v There is only one K value for a given system but infinite number of equilibrium positions v The law of mass action applies to solution and gaseous equilibria

7 Changing the chemical equation of an equilibrium system: Reciprocal rule Ø If we write the reaction in reverse. Ø cc + dd aa + bb Ø Then the new equilibrium constant is Ø K reversible = [A] [C] a c [B] [D] b d 1 = Reciprocal rule K forward

8 Multiplying the equation by a coefficient: Coefficient Rule Ø If we multiply the equation by a constant Ø naa + nbb ncc + ndd Ø Then the equilibrium constant is Ø K = [C] nc [D] nd = ([C] c [D] d ) n = K n [A] na [B] nb ([A] a [B] b ) n

9 Rules of Multiple Equlibria Ø Reaction 3 = Reaction 1 + Reaction 2 A 2 B A + AB K 1 = 2.2 AB A + B; K 2 = 4.0 [ A][ AB] K 1 = [ A B] 2 [ A][ B] K 2 = [ AB] A2B K 3 = [ A] [ A 2 2 2A + B [ B] B] = K 1 XK 2 K 3 = K 1 XK 2 K 3 =2.2 X4.0 = 8.8 Ø K (Reaction 3) =K (reaction 1) X K (reaction 2)

10 Notes on Equilibrium Expressions v The Equilibrium Expression for a reaction is the reciprocal of that for the reaction written in reverse. v When the equation for a reaction is multiplied by n, the equilib expression changes as follows: v (Equilib Expression) final = (Equilib Expression initial )n v Usually K is written without units

11 Calculation of K Ø N 2 + 3H 2 2NH 3 Ø Initial At Equilibrium Ø [N 2 ] 0 =1.000 M [N 2 ] = 0.921M Ø [H 2 ] 0 =1.000 M [H 2 ] = 0.763M Ø [NH 3 ] 0 =0 M [NH 3 ] = 0.157M K = [ NH 3] [ N ][ H ] 3 = 9.47X10-3

12 Calculation of K Ø N 2 + 3H 2 2NH 3 Ø Initial Ø [N 2 ] 0 = 0 M Ø [H 2 ] 0 = 0 M At Equilibrium [N 2 ] = M [H 2 ] = M Ø [NH 3 ] 0 = M [NH 3 ] = 0.157M Ø K is the same no matter what the amount of starting materials

13 Homogeneous Equilibria Ø All reactants and products are in one phase, gases for example Ø K can be used in terms of either concentration or pressure.

14 Heterogeneous Equilibria Ø If the reaction involves pure solids or pure liquids as well as gases, the concentration of the solid or the liquid doesn t change. Ø As long as they are not used up they are left out of the equilibrium expression. Ø Thus, there is no term for L or S in K expression. Ø However, the presence of L or S is a must for equilibrium to occur.

15 Example: Equilibrium expression for heterogeneous equilibria Ø H 2 (g) + I 2 (s) 2HI(g) Ø Ø But the concentration of I 2 does not change. Ø K ' = [ HI] [ H ][ I 2 2 [ HI] K '[ I2] = = K [ H ] ]

16 Applications of the equilibrium Constant Ø The magnitude of K helps prediction of the feasibility (extent or direction but not the speed) of the reaction Ø K> 1; the reaction system consists mostly products (equilibrium mostly lies to the right) Ø Systems with very large K go mostly to completion Ø Systems with very small values of K do not occur to any significant extent Ø There is no relation between the value of K and the time to reach equilibrium (the rate of reaction) Ø Time to reach equilibrium depends on E a for reactants and products

17 The Reaction Quotient, Q (Quantitative prediction of direction of reaction) Ø Q Tells how the direction of a reaction will go to reach equilibrium Ø Q s are calculated the same as K s, but for a system not at equilibrium Ø [Products] coefficient Q = [Reactants] coefficient aa(g) + bb(g) cc(g) + dd(g) c d ( PC ) X ( PD ) Q = a b ( PA ) X ( PB ) Ø Compare value of Q to that of K

18 What Q tells us? Ø If Q<K Ø Not enough products Ø Equilibrium shifts to right; forward reaction is predominant Ø If Q>K Ø Too many products Ø Equilibrium shifts to left; reverse reaction is predominant Ø If Q=K system is at equilibrium; there is no further change

19 Le Chatelier s Principle Ø if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change. Ø If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress. Ø There are 3 Types of stress

20 External conditions that cause a disturbance to a chemical equilibrium Ø Adding or removing reactants or products Ø Changing the volume (or pressure) of the system Ø Changing the temperature

21 The effect of a change in concentration of reactants and/or products Ø The system will shift away from the added component Ø Adding product makes Q>K Ø Removing reactant makes Q>K Ø Adding reactant makes Q<K Ø Removing product makes Q<K Ø knowing the effect on Q, will tell you the direction of the shift Ø Adding or removing liquids or solids does not affect the equilibrium

22 The effect of a Change in Pressure Ø The pressure changes as a result of: Ø Adding or removing gaseous reactant or product Ø Changing the volume of the container Ø By reducing the volume of the container, the system will move in the direction that reduces its volume.

23 Changes in volume

24 Changes in Temperature Ø Affects the rates of both the forward and reverse reactions. Ø changes the equilibrium constant. Ø The direction of the shift depends on whether it is exo- or endothermic

25 Ø DH<0 Ø Releases heat Exothermic Ø heat is a product Ø Raising temperature shifts the reaction direction toward reactants (to the left)

26 Ø DH>0 Endothermic Ø Heat is added to the system Ø Heat as a reactant Ø Raising temperature shifts the direction of reaction toward products (to the right)

27 Example Ø Consider: A + B C + D At 25 o C, K is Calculate the equilibrium concentrations of all species if 0.20 mol of A is reacted with 0.5 mol of B and dissolved in 1.00 L flask A + B C + D Initial(mol) [ ] (mol/l) Change(mol/L -1 ) -x -x +x +x Equilib 0.2-x 0.5-x +x +x

28 ( x)( x) ( x) 2 K = (0.20 x)(0.5 x) = (1 x) 2 = 0.3 K = (0.20 ( x)( x) x)(0.5 x) = ( x) (1 2 x) 2 = x X = 0 x = b ± b 2 4ac 2a X = 0.11 M

29 Example Ø Consider: A + B C + D At 25 o C, K is 2.0x Calculate the equilibrium concentrations of all species if 0.20 mol of A is reacted with 0.5 mol of B and dissolved in 1.00 L flask Since K is very large, the reaction of A with B will be virtually complete to the right Thus, only traces of A will be left at equilibrium. Let x represent the equilibrium concentration of A.

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31 Example Assume 0.10 mol of A is reacted with 0.20 mol of B in a volume of 1000 ml; K = 1.0 X What are the equilibrium concentrations of A, B, and C

32 Dissociation Equilibria Calculate the equilibrium concentrations of A and B in a 0.10 M solution of a weak electrolyte AB with an equilibrium constant of 3.0 X 10-6.

33 The Common Ion Effect-Shifting the Equilibrium Calculate the equilibrium concentrations of A and B in a 0.10 M solution of a weak electrolyte AB with an equilibrium constant of 3.0 X 10-6 and 0.20 molar B

34 Systematic approach to equilibrium calcualations This approach involves writing expressions for mass balance of species and one for charge balance of Species Mass Balance

35 Example

36 Charge balance equations principle of electroneutrality All solutions arc electrically neutral that is, there is no solution containing a detectable excess of positive or negative charge The sum of the positive charges equals the sum or negative charges A single charge balance equation lor a given set of equilibria.

37 Example The total charge concentrations from all sources is always equal to the net equilibrium concentration of the species multiplied by its charge

38 Example Write a charge balance expression for a solution containing KCI, AI 2 (SO 4 ) 3 and KNO 3. Neglect the dissociation of water.

39 Example Write the charge balance equation for a solution of [Ag(NH 3 ) 2 ]Cl

40 Example Calculate the equilibrium concentrations of A and B In a 0.10 M solution of a weak electrolyte AB with an equilibrium constant of 3.0 X Use the mass charge balance approach

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43 Example Calculate the equilibrium concentrations of A and B In a 0.10 M solution of a weak electrolyte AB with an equilibrium constant of 3.0 X Use the mass charge balance approach. Assume the charge A is + I, the charge on B is - I, and that the extra B (0.20M) comes from MB; MB is completely dissociated.

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