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AP Notes Chapter 16 Equilibrium Dynamic chemical system in which two reactions, equal and opposite, occur simultaneously

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Properties 1. Appear from outside to be inert or not functioning 2. Can be initiated in both directions

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Pink to blue Co(H 2 O) 6 Cl 2 Co(H 2 O) 4 Cl H 2 O Blue to pink Co(H 2 O) 4 Cl H 2 O Co(H 2 O) 6 Cl 2

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Equilibrium achieved Product conc. increases and then becomes constant at equilibrium Reactant conc. declines and then becomes constant at equilibrium

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At any point in the reaction H 2 + I 2 2 HI

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Equilibrium achieved In the equilibrium region

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Kinetics Definition R f = R r

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At equilibrium, the rates of the forward and reverse reactions are equal.

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aA + bB cC R f (eq) = k f [A] a [B] b R r (eq) = k r [C] c R f = R r k f [A] a [B] b = k r [C] c

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By convention

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K is a concentration quotient for a system at equilibrium. K = Q

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For a system NOT at equilibrium Q K

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if Q > K The reverse reaction will occur until equilibrium is achieved.

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if Q < K The forward reaction will occur until equilibrium is achieved.

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Achieving equilibrium is a driving force in chemical systems and will occur when possible. It cannot be stopped (spontaneous)

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1.For the equilibrium system, 2NO 2 (g) N 2 O 4 (g), the equilibrium constant, K C, is 8.8 at 25 0 C. If analysis shows that 2.0 x mole of NO 2 and 1.5 x mole of N 2 O 4 are present in a 10.0 L flask, is the reaction at equilibrium?

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Types of Reactions 1. one way (goes to completion) NaOH (s) Na + (aq) + OH - (aq)

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Types of Reactions 2. Equilibrium (two opposite reactions at same time) a. dimerization 2NO 2 (g) N 2 O 4 (g)

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b. dissociation of a weak electrolyte CH 3 COOH + H 2 O CH 3 COO - + H 3 O +

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c. saturated aqueous solutions AgCl (s) Ag + (aq) + Cl - (aq) C 6 H 12 O 6 (s) C 6 H 12 O 6 (aq)

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By convention EQUILIBRIUM CONSTANT K eq

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if K eq > 1 [products] coeff > [reactants] coeff the forward reaction proceeded to a greater extent than the reverse reaction to achieve equilibrium (i.e. the products predominate at equilibrium)

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if K eq < 1 [products] coeff < [reactants] coeff the forward reaction proceeded to a lesser extent than the reverse reaction to achieve equilibrium (i.e. the reactants predominate at equilibrium)

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How are k f and k r related to temperature? k f and k r are temperature dependent thus, K eq is temperature dependent

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N 2 O 4 + heat 2 NO 2 (colorless) (brown) H o = kJ K c (273 K) = K c (298 K) =

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N 2 O 4 (g) 2NO 2 (g) Examples of Equilibrium Expressions

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CH 3 COOH + H 2 O CH 3 COO - +H 3 O +

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AgCl (s) Ag + (aq) + Cl - (aq)

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Concentrations of pure liquids and solids are NOT included in equilibrium expressions, as their concentrations are themselves constants.

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The value of K eq may appear to change based on way equation is balanced.

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A value that is mathematically related to another (eg. temp) is NOT considered a new value

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Multiple Equilibria H 3 PO H 2 O PO H 3 O + H 3 PO 4 + H 2 O H 2 PO H 3 O + H 2 PO H 2 O HPO H 3 O + HPO H 2 O PO H 3 O +

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K eq = K 1. K 2. K 3 for the complete dissociation of phosphoric acid

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So far, K eq has been studied as a function of concentration, or expressed with appropriate notation, K c

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But, what about equilibrium systems where all components are gases? Partial pressures mole distribution

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where V = a container parameter (constant for all gases) T = constant for given values of K R = constant

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aA(g) + bB(g) cC(g)

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Substituting for a gas

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Let c - (a + b) = n where n is the change in # of moles of gas (product - reactant) for the forward reaction.

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If we express the equilibrium constant as a function of partial pressures

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Thus K C = K P (RT) - n or K P = K c (RT) n

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2.When 2.0 moles of HI(g) are placed in a 1.0 L container and allowed to come to equilibrium with its elements, it is found that 20% of the HI decomposes. What is K C and K P ?

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Applications of the Equilibrium Constant & LeChateliers Principle

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mol of n-butane is placed in a 0.50 L container and allowed to come to equilibrium with its isomer isobutane. K C at 25 0 C is 2.5. What are the equilibrium concentrations of the two isomers?

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Set up an ICE table Initial [ ] of components Change in [ ] Equilibrium [ ]

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n-butane isobutane I C -x +x E x x

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solve

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4.2.0 mols Br 2 are placed in a 2.0 L flask at 1756 K, which is of sufficient energy to split apart some of the molecules. If K C = 4.0 x at 1756 K, what are the equilibrium concentrations of the bromine molecules and atoms?

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Br 2 (g) 2 Br(g) I C -x +2x E x 2x

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solve

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if K <<< [A] 0, then can assume amount that dissociated to reach equilibrium is VERY small, thus

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solve

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5.Calculate [OH - ] at equilibrium of a solution that is initially M nicotine.

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2H 2 O + Nic NicH OH - I C -x +x +2x E x x 2x

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K C = K 1. K 2 K C = (7.0 x )(1.1 x ) K C = 7.7 x

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LeChatliers Principle When a stress is placed on a system at equilibrium, the system will adjust so as to relieve that stress.

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Stress Factors 1. Change in concentration of reactants or products 2. Change in volume or pressure (for gases) 3. Change in temperature

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Responses to Stress 3 H 2 (g) + N 2 (g) 2 NH 3 (g) 1. Change concentration a. add either H 2 or N 2 b. remove NH 3

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Responses to Stress 3H 2 (g) + N 2 (g) 2 NH 3 (g) 2. Change in volume or pressure a. increase volume b. Increase pressure c. add He

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Responses to Stress 3H 2 (g) + N 2 (g) 2 NH 3 (g) H = -92 kJ 3. Change in temperature a. increase temperature

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AgCl(s) Ag + (aq) + Cl - (aq) a. add AgCl b. add H 2 O c. add NaCl d. add NH 3 (aq)

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K eq is a temperature dependent constant, similar to the rate constant, k f or k r slope of line is different

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1/T ln K eq m = - H R /R

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Change T – change in K – therefore change in P or concentrations at equilibrium Use a catalyst: reaction comes more quickly to equilibrium. K not changed. Add or take away reactant or product: –K does not change –Reaction adjusts to new equilibrium position Le Chateliers Principle

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