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Chapter 15 Chemical Equilibrium Lecture Presentation © 2012 Pearson Education, Inc.

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Equilibrium The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. © 2012 Pearson Education, Inc.

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Equilibrium The Concept of Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate. © 2012 Pearson Education, Inc. Once equilibrium is achieved, the amount of each reactant and product remains constant

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Equilibrium The Equilibrium Constant Consider the generalized reaction The equilibrium expression for this reaction would be K c = [C] c [D] d [A] a [B] b aA + bBcC + dD © 2012 Pearson Education, Inc.

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Equilibrium The Equilibrium Constant © 2012 Pearson Education, Inc. N2O4(g)N2O4(g)2NO 2 (g) K eq = [NO 2 ] 2 [N 2 O 4 ]

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Equilibrium The Equilibrium Constant Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written K p = (P C c ) (P D d ) (P A a ) (P B b ) © 2012 Pearson Education, Inc.

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Equilibrium The equilibrium constant K p for the reaction is 158 at 1000K. What is the equilibrium pressure of O 2 if the P NO = 0.400 atm and P NO = 0.270 atm? 2 2NO 2 (g) 2NO (g) + O 2 (g) 14.2 K p = 2 P NO P O 2 P NO 2 2 POPO 2 = K p P NO 2 2 2 POPO 2 = 158 x (0.400) 2 /(0.270) 2 = 347 atm

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Equilibrium Relationship Between K c and K p where K p = K c (RT) n n = (moles of gaseous product) (moles of gaseous reactant) © 2012 Pearson Education, Inc.

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Equilibrium The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 74 0 C are [CO] = 0.012 M, [Cl 2 ] = 0.054 M, and [COCl 2 ] = 0.14 M. Calculate the equilibrium constants K c and K p. CO (g) + Cl 2 (g) COCl 2 (g) Kc =Kc = [COCl 2 ] [CO][Cl 2 ] = 0.14 0.012 x 0.054 = 220 K p = K c (RT) n n = 1 – 2 = -1 R = 0.0821T = 273 + 74 = 347 K K p = 220 x (0.0821 x 347) -1 = 7.7 14.2

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Equilibrium Equilibrium Can Be Reached from Either Direction It doesn’t matter whether we start with N 2 and H 2 or whether we start with NH 3, we will have the same proportions of all three substances at equilibrium. © 2012 Pearson Education, Inc.

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Equilibrium What Does the Value of K Mean? If K>>1, the reaction is product-favored; product predominates at equilibrium. If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium. © 2012 Pearson Education, Inc.

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Equilibrium Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction: K c = = 0.212 at 100 C [NO 2 ] 2 [N 2 O 4 ] K c = = 4.72 at 100 C [N 2 O 4 ] [NO 2 ] 2 N2O4(g)N2O4(g)2NO 2 (g) N2O4(g)N2O4(g) © 2012 Pearson Education, Inc.

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Equilibrium Manipulating Equilibrium Constants The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number: K c = = 0.212 at 100 C [NO 2 ] 2 [N 2 O 4 ] K c = = (0.212) 2 at 100 C [NO 2 ] 4 [N 2 O 4 ] 2 4NO 2 (g)2N 2 O 4 (g) N2O4(g)N2O4(g)2NO 2 (g) © 2012 Pearson Education, Inc.

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Equilibrium A + B C + D C + D E + F A + B E + F K c = ‘ [C][D] [A][B] K c = ‘ ‘ [E][F] [C][D] [E][F] [A][B] K c = KcKc ‘ KcKc ‘‘ KcKc KcKc ‘‘ KcKc ‘ x If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 14.2

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Equilibrium Sample Exercise 15.5 Combining Equilibrium Expressions Given the reactions determine the value of K c for the reaction Solution Analyze We are given two equilibrium equations and the corresponding equilibrium constants and are asked to determine the equilibrium constant for a third equation, which is related to the first two. Plan We cannot simply add the first two equations to get the third. Instead, we need to determine how to manipulate the equations to come up with the steps that will add to give us the desired equation.

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Equilibrium Sample Exercise 15.5 Combining Equilibrium Expressions Continued Solve If we multiply the first equation by 2 and make the corresponding change to its equilibrium constant (raising to the power 2), we get Reversing the second equation and again making the corresponding change to its equilibrium constant (taking the reciprocal) gives Now we have two equations that sum to give the net equation, and we can multiply the individual K c values to get the desired equilibrium constant.

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Equilibrium Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO 3 (s) CaO (s) + CO 2 (g) K c = ‘ [CaO][CO 2 ] [CaCO 3 ] [CaCO 3 ] = constant [CaO] = constant K c = [CO 2 ] = K c x ‘ [CaCO 3 ] [CaO] K p = P CO 2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant. 14.2

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Equilibrium K c = [Pb 2+ ] [Cl ] 2 PbCl 2 (s) Pb 2+ (aq) + 2Cl (aq) © 2012 Pearson Education, Inc. Example

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Equilibrium At 1280 0 C the equilibrium constant (K c ) for the reaction Is 1.1 x 10 -3. If the initial concentrations are [Br 2 ] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. Br 2 (g) 2Br (g) Let x be the change in concentration of Br 2 Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x [Br] 2 [Br 2 ] K c = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 Solve for x 14.4

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Equilibrium K c = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 4x 2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x 2 + 0.0491x + 0.0000747 = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac 2a2a x = Br 2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x x = -0.00178x = -0.0105 At equilibrium, [Br] = 0.012 + 2x = -0.009 Mor 0.00844 M At equilibrium, [Br 2 ] = 0.062 – x = 0.0648 M 14.4

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Equilibrium Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations Solution Analyze We are given the initial concentrations of H 2 and l 2 and the equilibrium concentration of HI. We are asked to calculate the equilibrium constant K c for Solve First, we tabulate the initial and equilibrium concentrations of as many species as we can. We also provide space in our table for listing the changes in concentrations. As shown, it is convenient to use the chemical equation as the heading for the table. Second, we calculate the change in HI concentration, which is the difference Between the equilibrium and initial values: A closed system initially containing 1.000 10 3 M H 2 and 2.000 10 3 M I 2 at 448 C is allowed to reach equilibrium, and at equilibrium the HI concentration is 1.87 10 3 M. Calculate K c at 448 C for the reaction taking place, which is Plan We construct a table to find equilibrium concentrations of all species and then use the equilibrium concentrations to calculate the equilibrium constant. Change in [HI] = 1.87 10 3 M 0 = 1.87 10 3 M

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Equilibrium Third, we use the coefficients in the balanced equation to relate the change in [HI] to the changes in [H 2 ] and [I 2 ]: Fourth, we calculate the equilibrium concentrations of H 2 and I 2, using initial concentrations and changes in concentration. The equilibrium concentration equals the initial concentration minus that consumed: Our table now looks like this (with equilibrium concentrations in blue for emphasis): Continued [H 2 ] = 1.000 10 3 M 0.935 10 3 M = 0.065 10 3 M [I 2 ] = 2.000 10 3 M 0.935 10 3 M = 1.065 10 3 M Notice that the entries for the changes are negative when a reactant is consumed and positive when a product is formed. Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations

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Equilibrium Finally, we use the equilibrium-constant expression to calculate the equilibrium constant: Continued Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations

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Equilibrium The Reaction Quotient (Q) Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium. To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. © 2012 Pearson Education, Inc. Q = [C initial ] c [D initial ] d [A initial ] a [B initial ] b aA + bBcC + dD

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Equilibrium Q c > K c system proceeds from right to left to reach equilibrium Q c = K c the system is at equilibrium Q c < K c system proceeds from left to right to reach equilibrium 14.4

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Equilibrium Sample Exercise 15.10 Predicting the Direction of Approach to Equilibrium Solution Analyze We are given a volume and initial molar amounts of the species in a reaction and asked to determine in which direction the reaction must proceed to achieve equilibrium. Plan We can determine the starting concentration of each species in the reaction mixture. We can then substitute the starting concentrations into the equilibrium-constant expression to calculate the reaction quotient, Q c. Comparing the magnitudes of the equilibrium constant, which is given, and the reaction quotient will tell us in which direction the reaction will proceed. Solve The initial concentrations are The reaction quotient is therefore At 448 C the equilibrium constant K c for the reaction is 50.5. Predict in which direction the reaction proceeds to reach equilibrium if we start with 2.0 10 2 mol of HI, 1.0 10 2 mol of H 2, and 3.0 10 2 of I 2 in a 2.00-L container. Because Q c < K c, the concentration of HI must increase and the concentrations of H 2 and I 2 must decrease to reach equilibrium; the reaction as written proceeds left to right to attain equilibrium.

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Equilibrium Le Châtelier’s Principle “If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.” © 2012 Pearson Education, Inc.

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Equilibrium Le Châtelier’s Principle Changes in Concentration continued ChangeShifts the Equilibrium Increase concentration of product(s)left Decrease concentration of product(s)right Decrease concentration of reactant(s) Increase concentration of reactant(s)right left 14.5 aA + bB cC + dD Add Remove

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Equilibrium Le Châtelier’s Principle Changes in Volume and Pressure A (g) + B (g) C (g) ChangeShifts the Equilibrium Increase pressureSide with fewest moles of gas Decrease pressureSide with most moles of gas Decrease volume Increase volumeSide with most moles of gas Side with fewest moles of gas 14.5

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Equilibrium uncatalyzedcatalyzed 14.5 Catalyst does not change equilibrium constant or shift equilibrium. Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner Le Châtelier’s Principle

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Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 1 Chapter 7 Chemical Quantities 7.5 Mole Relationships in Chemical Equations.

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 1 Chapter 7 Chemical Quantities 7.5 Mole Relationships in Chemical Equations.

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