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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(H2O)6Cl2 Co(H2O)4Cl H2O Blue to pink Co(H2O)4Cl H2O Co(H2O)6Cl2
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Product conc. increases and then becomes constant at equilibrium
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
H2 + I2 2 HI
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In the equilibrium region
Equilibrium achieved In the equilibrium region
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Kinetics Definition Rf = Rr
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At equilibrium, the rates of the forward and reverse reactions are equal.
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aA + bB cC Rf(eq) = kf [A]a [B]b Rr(eq) = kr [C]c Rf = Rr
kf [A]a [B]b = kr [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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The reverse reaction will occur until equilibrium is achieved.
if Q > K The reverse reaction will occur until equilibrium is achieved.
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The forward reaction will occur until equilibrium is achieved.
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, 2NO2 (g) N2O4 (g) , the equilibrium constant, KC , is 8.8 at 250C. If analysis shows that 2.0 x 10-3 mole of NO2 and 1.5 x 10-3 mole of N2O4 are present in a 10.0 L flask, is the reaction at equilibrium?
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2NO2 (g) N2O4 (g) , the equilibrium constant, KC , is 8. 8 at 250C
2NO2 (g) N2O4 (g) , the equilibrium constant, KC , is 8.8 at 250C. If analysis shows that 2.0 x 10-3 mole of NO2 and 1.5 x 10-3 mole of N2O4 are present in a 10.0 L flask, is the reaction at equilibrium? 2NO2 (g) N2O4 (g) KC = 8.8 so… Q = [product]m = [reactant]n
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2NO2 (g) N2O4 (g) , the equilibrium constant, KC , is 8. 8 at 250C
2NO2 (g) N2O4 (g) , the equilibrium constant, KC , is 8.8 at 250C. If analysis shows that 2.0 x 10-3 mole of NO2 and 1.5 x 10-3 mole of N2O4 are present in a 10.0 L flask, is the reaction at equilibrium? 2NO2 (g) N2O4 (g) KC = 8.8 so… Q = [product]m = (2.0 x 10-3 mole NO2/ 10.0L) [reactant]n (1.5 x 10-3 mole of N2O4 /10.0L)2 =
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2NO2 (g) N2O4 (g) , the equilibrium constant, KC , is 8. 8 at 250C
2NO2 (g) N2O4 (g) , the equilibrium constant, KC , is 8.8 at 250C. If analysis shows that 2.0 x 10-3 mole of NO2 and 1.5 x 10-3 mole of N2O4 are present in a 10.0 L flask, is the reaction at equilibrium? 2NO2 (g) N2O4 (g) KC = 8.8 so… Q = [product]m = (2.0 x 10-3 mole NO2/ 10.0L) [reactant]n (1.5 x 10-3 mole of N2O4 /10.0L)2 = 8888 KC = 8.8 so… Q = 8888 Is K = Q? Concentrations and orders must be used!! NO Q > K so [product] is BIGGER than is should be to be at Equilibrium with given K value so products will convert to reactants to reach 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 2NO2(g) N2O4(g)
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b. dissociation of a weak electrolyte
CH3COOH + H2O CH3COO- + H3O+
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c. saturated aqueous solutions
AgCl(s) Ag+(aq) + Cl-(aq) C6H12O6(s) C6H12O6(aq)
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EQUILIBRIUM CONSTANT Keq
By convention
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[products]coeff > [reactants]coeff
if Keq > 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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[products]coeff < [reactants]coeff
if Keq < 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 kf and kr related to temperature?
kf and kr are temperature dependent thus, Keq is temperature dependent
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N2O4 + heat 2 NO2 (colorless) (brown) ∆Ho = + 57.2 kJ
Kc (273 K) = Kc (298 K) =
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Examples of Equilibrium
Expressions N2O4(g) 2NO2(g)
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CH3COOH + H2O CH3COO- +H3O+
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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 Keq may appear to change based on way equation is balanced.
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A value that is mathematically related to another (eg
A value that is mathematically related to another (eg. temp) is NOT considered a new value
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Multiple Equilibria H3PO4 + 3 H2O PO43- + 3 H3O+
H3PO4 + H2O H2PO4- + H3O+ H2PO4- + H2O HPO42- + H3O+ HPO42- + H2O PO43- + H3O+
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for the complete dissociation of
Keq = K1. K2. K3 for the complete dissociation of phosphoric acid
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So far, Keq has been studied as a function of concentration, or expressed with appropriate notation, Kc
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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 KC = KP(RT)-n or KP = Kc(RT)Dn
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2. When 2. 0 moles of HI(g) are placed in a 1
2. When 2.0 moles of HI(g) are placed in a 1.0 L container and allowed to come to equilibrium with it’s elements, it is found that 20% of the HI decomposes. What is KC and KP?
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Applications of the Equilibrium Constant & LeChatelier’s Principle
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3. 017 mol of n-butane is placed in a 0
mol of n-butane is placed in a 0.50 L container and allowed to come to equilibrium with its isomer isobutane. KC at 250C is 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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mols Br2 are placed in a 2.0 L flask at 1756 K, which is of sufficient energy to split apart some of the molecules. If KC = 4.0 x 10-4 at 1756 K, what are the equilibrium concentrations of the bromine molecules and atoms?
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Br2(g) 2 Br(g) I C x x E x x
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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 0
5. Calculate [OH-] at equilibrium of a solution that is initially M nicotine.
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2H2O + Nic NicH OH- I C x x x E x x x
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KC = K1 . K2 KC = (7.0 x 10-7)(1.1 x 10-10) KC = 7.7 x 10-17
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LeChatlier’s 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 1. Change concentration a. add either H2 or N2
3 H2(g) + N2(g) 2 NH3(g) 1. Change concentration a. add either H2 or N2 b. remove NH3
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Responses to Stress 2. Change in volume or pressure a. increase volume
3H2(g) + N2(g) 2 NH3(g) 2. Change in volume or pressure a. increase volume b. Increase pressure c. add He
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Responses to Stress 3. Change in temperature a. increase temperature
3H2(g) + N2(g) 2 NH3(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 H2O c. add NaCl d. add NH3(aq)
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slope of line is different
Keq is a temperature dependent constant, similar to the rate constant, kf or kr slope of line is different
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m = -HR/R ln Keq 1/T
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Le Chatelier’s Principle
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”
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