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Chapter 11 Chemical Equilibrium 11.1 The Equilibrium Condition 11.2 The Equilibrium Constant 11.3 Equilibrium Expressions Involving Pressures 11.4 The.

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Presentation on theme: "Chapter 11 Chemical Equilibrium 11.1 The Equilibrium Condition 11.2 The Equilibrium Constant 11.3 Equilibrium Expressions Involving Pressures 11.4 The."— Presentation transcript:

1 Chapter 11 Chemical Equilibrium 11.1 The Equilibrium Condition 11.2 The Equilibrium Constant 11.3 Equilibrium Expressions Involving Pressures 11.4 The Concept of Activity 11.5 Heterogeneous Equilibria 11.6 Applications of the Equilibrium Constant 11.7 Solving Equilibrium Problems 11.8 Le Chatelier's Principle 11.9 Equilibria Involving Real Gases

2 The Equilibrium Condition (General) Thermal equilibrium indicates two systems in thermal contact with each do not exchange energy by heat. If two bricks are in thermal equilibrium their temperatures are the same. Chemical equilibrium indicates no unbalanced potentials (or driving force). A system in equilibrium experiences no change over time, even infinite time. The opposite of equilibrium systems are non-equilibrium systems that are off balance and change with time. Example 1 atm O 2 + 2 atm H 2 at 298K

3 aA + bB cC + dD The same equilibrium state is achieved whether starting with pure reactants or pure products. The equilibrium state can change with temperature. The Equilibrium Condition (Chem Rxn)

4 The Equilibrium State (Chem Rxn) H 2 O (g) + CO (g) H 2 (g) + CO 2 (g) Change [CO] to P CO [H 2 O] to P H2O etc

5 As the equilibrium state is approached, the forward and backward rates of reaction approach equality. At equilibrium the rates are equal, and no further net change occurs in the partial pressures of reactants or products. 1. No macroscopic evidence of change. 2. Reached through spontaneous processes. 3. Show a dynamic balance of forward and backward processes. 4. Same regardless of the direction from which they are approached. Fundamental characteristics of equilibrium states: Chemical Reactions and Equilibrium 5. No change over time.

6 ↔ Use this in an equilibrium expression. Use this to indicate resonance. Arrows: Chemical Symbolism

7 Chemical Reactions and Equilibrium The equilibrium condition for every reaction can be described in a single equation in which a number, the equilibrium constant (K) of the reaction, equals an equilibrium expression, a function of properties of the reactants and products. H 2 O(l) H 2 O(g) @ 25 o C Temperature ( o C) Vapor Pressure (atm) 15.00.01683 17.00.01912 19.00.02168 21.00.02454 23.00.02772 25.00.03126 30.00.04187 50.00.1217 H 2 O(l) H 2 O(g) @ 30 o C K = 0.03126 K = 0.04187

8 Partial pressures and concentrations of products appear in the numerator and those of the reactants in the denominator. Each is raised to a power equal to its coefficient in the balanced chemical equation. aA + bB cC + dD Law of Mass Action (1)

9 1. Gases enter equilibrium expressions as partial pressures, in atmospheres. E.g., P CO2 2. Dissolved species enter as concentrations, in molarity (M) moles per liter. E.g., [Na + ] 3. Pure solids and pure liquids are represented in equilibrium expressions by the number 1 (unity); a solvent taking part in a chemical reaction is represented by unity, provided that the solution is dilute. E.g., I 2 (s) ↔ I 2 (aq) [I 2 (aq) ] = K Law of Mass Action (2)

10 P H 2 O P ref = K P ref is numerically equal to 1 The convention is to express all pressures in atmospheres and to omit factors of P ref because their value is unity. An equilibrium constant K is a pure number. H 2 O (l) H 2 O (g) K p = P H 2 O The concept of Activity (i-th component) = a i = P i / P reference @ 25 o C K p = 0.03126 atm K = 0.03126 Activities

11 H 2 O (g) + CO (g) H 2 (g) + CO 2 (g) The Equilibrium State

12 The Equilibrium Expressions In a chemical reaction in which a moles of species A and b moles of species B react to form c moles of species C and d moles of species D, The partial pressures at equilibrium are related through K = P c C P d D /P a A P b B aA + bB cC + dD

13 Write equilibrium expressions for the following reactions 3 H 2 (g) + SO 2 (g) H 2 S(g) + 2 H 2 O(g) 2 C 2 F 5 Cl(g) + 4 O 2 (g) Cl 2 (g) + 4 CO 2 (g) + 5 F 2 (g)

14 Gases and Solids CaCO 3 (s) CaO(s) + CO 2 (g) K=P CO 2 K is independent of the amounts of CaCO 3 (s) or CaO(s) Heterogeneous Equilibrium

15 Liquids Solutions H 2 O(l) H 2 O(g) K=P H 2 O I 2 (s) I 2 (aq) K=[I 2 ] Heterogeneous Equilibrium

16 Relationships Among the K’s of Related Reactions #1: The equilibrium constant for a reverse reaction is always the reciprocal of the equilibrium constant for the corresponding forward reaction. 2 H 2 (g) + O 2 (g) 2 H 2 O (g) (P H 2 O ) 2 (P H 2 ) 2 (P O 2 ) = K 1 2 H 2 O (g) 2 H 2 (g) + O 2 (g) (P H 2 ) 2 (P O 2 ) (P H 2 O ) 2 = K 2 K 1 = 1/K 2 #1 #2 aA + bB cC + dDcC + dD aA + bB versus

17 # 2: When the coefficients in a balanced chemical equation are all multiplied by a constant factor, the corresponding equilibrium constant is raised to a power equal to that factor. (P H 2 O ) (P H 2 )(P O 2 ) ½ = K 3 K 3 = K 1 ½ Relationships Among the K’s of Related Reactions 2 H 2 (g) + O 2 (g) 2 H 2 O (g) Rxn 1 #1 H 2 (g) + ½ O 2 (g) H 2 O (g) Rxn 3 = Rxn 1 times 1/2#3 (P H 2 O ) 2 (P H 2 ) 2 (P O 2 ) = K 1

18 # 3: when chemical equations are added to give a new equation, their equilibrium constants are multiplied to give the equilibrium constant associated with the new equation. 2 BrCl (g) ↔ Br 2 (g) + Cl 2 (g) (P Br 2 )(P Cl 2 ) (P BrCl ) 2 Br 2 (g) + I 2 (g) ↔ 2 IBr (g) (P IBr ) 2 (P Br 2 ) (P I 2 ) 2 BrCl (g) + I 2 (g) ↔ 2 IBr (g) + Cl 2 (g) = K 1 K 2 = (0.45)(0.051) =0.023 @ 25 o C Relationships Among the K’s of Related Reactions (P Br 2 )(P Cl 2 ) (P BrCl ) 2 = K 1 = 0.45 @ 25 o C X (P IBr ) 2 (P Br 2 ) (P I 2 ) = K 2 = 0.051 @ 25 o C = K 1 K 2 = K 3 wrong arrow

19 Calculating Equilibrium Constants Consider the equilibrium 4 NO 2 (g)↔ 2 N 2 O(g) + 3 O 2 (g) The three gases are introduced into a container at partial pressures of 3.6 atm (for NO 2 ), 5.1 atm (for N 2 O), and 8.0 atm (for O 2 ) and react to reach equilibrium at a fixed temperature. The equilibrium partial pressure of the NO 2 is measured to be 2.4 atm. Calculate the equilibrium constant of the reaction at this temperature, assuming that no competing reactions occur. 4 NO 2 (g) ↔ 2 N 2 O(g) + 3 O 2 (g) initial partial pressure (atm) change in partial pressure (atm) equilibrium partial pressure (atm)

20 4 NO 2 (g) ↔ 2 N 2 O(g) + 3 O 2 (g) initial partial pressure (atm) 3.6 5.1 8.0 change in partial pressure (atm) – 4x +2x +3x equilibrium partial pressure (atm) 2.4 5.1 + 2x 8.0 + 3x 3.6 – 4x = 2.4 atm NO 2 ; x = 0.3 atm 5.1 + 2(0.3 atm) = 5.7 atm N 2 O 8.0 + 3(0.3 atm) = 8.9 atm O 2 K = (P N 2 O ) 2 (P O 2 ) 3 (P NO 2 ) 4 = Calculate the equilibrium constant of the reaction at this temperature, assuming that no competing reactions occur.

21 The compound GeWO 4 (g) forms at high temperature in the reaction 2 GeO (g) + W 2 O 6 (g) ↔ 2 GeWO 4 (g) Some GeO (g) and W 2 O 6 (g) are mixed. Before they start to react, their partial pressures both equal 1.000 atm. After their reaction at constant temperature and volume, the equilibrium partial pressure of GeWO 4 (g) is 0.980 atm. Assuming that this is the only reaction that takes place, (a) determine the equilibrium partial pressures of GeO and W 2 O 6, and (b) determine the equilibrium constant for the reaction. 2 GeO (g) + W 2 O 6 (g) ↔ 2 GeWO 4 (g) initial partial pressure (atm)1.000 1.000 0 change in partial pressure (atm) – 2x – x +2x equilibrium partial pressure (atm)

22 0 + 2x = 0.980 atm GeWO 4 ; x = 0.490 atm 1.000 – 2(0.490) = 0.020 atm GeO 1.000 – 0.490 = 0.510 atm W 2 O 6 K = (P GeWO 4 ) 2 (P GeO ) 2 (P W 2 O 6 ) = 2 GeO(g) + W 2 O 6 (g) ↔ 2 GeWO 4 (g) initial partial pressure (atm)1.000 1.000 0 change in partial pressure (atm) –2x –x +2x equilibrium partial pressure (atm) 1.000 – 2x 1.000 – x 0.980 (a)determine the equilibrium partial pressures of GeO and W 2 O 6, and (b)determine the equilibrium constant for the reaction.

23 Skip Solving quadratic equations Will utilize approximation method –Systems that have small equilibrium constants. –Assume “x” (the change in concentration) is small (less than 5%) of the initial concentration.

24 A vessel holds pure CO (g) at a pressure of 1.282 atm and a temperature of 354K. A quantity of nickel is added, and the partial pressure of CO (g) drops to an equilibrium value of 0.709 atm because of the reaction Ni (s) + 4CO (g) ↔ Ni(CO) 4 (g) Compute the equilibrium constant for this reaction at 354K. initial partial pressure (atm) 1.282 0 change in partial pressure (atm) -4x +1x equilibrium partial pressure (atm) 0.709 x P CO (atm) P Ni(CO)4 (atm) At equil. P co =x = P Ni(CO)4 = Construct an “ICE” table Ni (s) + 4CO (g) ↔ Ni(CO) 4 (g)

25 Equilibrium Calculations At a particular temperature, K = 2.0 x 10 -6 mol/L for the reaction 2CO 2 (g) 2CO (g) + O 2 (g) If 2.0 mol CO 2 is initially placed into a 5.0-L vessel, calculate the equilibrium concentrations of all species. initial partial pressure (mol/L) 0.4 0 0 change in partial pressure ( mol/L ) – 2x +2x+1x equilibrium partial pressure ( mol/L ) 0.4 -2x 2x 1x 2CO 2 (g) 2CO (g) + O 2 (g)

26 At a particular temperature, K = 2.0 x 10 -6 mol/L for the reaction 2CO 2 (g) 2CO (g) + O 2 (g) If 2.0 mol CO 2 is initially placed into a 5.0-L vessel, calculate the equilibrium concentrations of all species. initial partial pressure ( mol/L ) 0.4 0 0 change in partial pressure ( mol/L ) – 2x +2x+1x equilibrium partial pressure ( mol/L ) 0.4 -2x 2x 1x 2CO 2 (g) 2CO (g) + O 2 (g)

27 At a particular temperature, K = 2.0 x 10 -6 mol/L for the reaction 2CO 2 (g) 2CO (g) + O 2 (g) If 2.0 mol CO 2 is initially placed into a 5.0-L vessel, calculate the equilibrium concentrations of all species. initial partial pressure ( mol/L ) 0.4 0 0 change in partial pressure ( mol/L ) – 2x +2x+1x equilibrium partial pressure ( mol/L ) 0.4 -2x 2x 1x 2CO 2 (g) 2CO (g) + O 2 (g)

28 K (the Equilibrium Constant) uses equilibrium partial pressures Q (the reaction quotient) uses prevailing partial pressures, not necessarily at equilibrium wrong arrow Non-Equilibrium Conditions: The Reaction Quotient (1)

29 If Q < K, reaction proceeds in a forward direction (toward products); If Q > K, reaction proceeds in a backward direction (toward reactants); If Q = K, the reaction is in equilibrium. wrong arrow The Reaction Quotient (2)

30 The equilibrium constant for the reaction P 4 (g) ↔ 2 P 2 (g) is 1.39 at 400 o C. Suppose that 2.75 mol of P 4 (g) and 1.08 mol of P 2 (g) are mixed in a closed 25.0 L container at 400 o C. Compute Q (init) (the Q at the moment of mixing) and state the direction in which the reaction proceeds. K = 1.39 @ 400 o C; n P 4 (init) = 2.75 mol; n P 2 (init) = 1.08 mol P P 4 (init) = n P 4 (init) RT/V P P 2 (init) = n P 2 (init) RT/V = [(2.75mol)(0.08206 atm L mol -1 K -1 )(273.15+400 o C)]/(25.0L) = [(1.08mol)(0.08206 atm L mol -1 K -1 )(273.15+400 o C)]/(25.0L) = 6.08 atm = 2.39 atm Q =

31 Henri Louis Le Châtelier (1850-1936) Highlights –1884 Le Chatelier's Principle: A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress –If a chemical system at equilibrium experiences a change in concentration, temperature or total pressure the equilibrium will shift in order to minimize that change. –Industrial chemist involved with industrial efficiency and labor-management relations Moments in a Life –Le Chatelier was named "chevalier" (knight) of the Légion d'honneur in 1887, decoration established by Napoléon Bonaparte in 1802.

32 Effects of External Stresses on Equilibria: Le Châtelier’s Principle A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. 2. Effects of Changing the Volume (or Pressure) of the System 3. Effects of Changing the Temperature 1. Effects of Adding or Removing Reactants or Products Le Châtelier’s Principle provides a way to predict the response of an equilibrium system to an external perturbation, such as…

33 Effects of Adding or Removing Reactants or Products PCl 5 (g) PCl 3 (g) + Cl 2 (g) K = 11.5 @ 300 o C = Q add extra PCl 5 (g) add extra PCl 3 (g) remove some PCl 5 (g) remove some PCl 3 (g) A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. In this case adding or removing reactants or products

34 Effects of Changing the Volume of the System PCl 5 (g) PCl 3 (g) + Cl 2 (g) Let’s decrease the volume of the reaction container Let’s increase the volume of the reaction container Less room :: less amount (fewer moles) Shifts reaction to restore equilibrium More room :: more amount (greater moles) Shifts reaction to restore equilibrium 1 mole1+1 = 2 moles A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. In this case a change in volume

35 Volume DecreasedVolume Increased (Pressure Increased)(Pressure Decreased) V reactants > V products Equilibrium shift right (toward products) Equilibrium Shifts left (toward reactants) V reactants < V products Equilibrium Shifts left (toward reactants) Equilibrium shift right (toward products) V reactants = V productsEquilibrium not affected 2 P 2 (g) P 4 (g) PCl 5 (g) PCl 3 (g) + Cl 2 (g) CO (g) +H 2 O (g) CO 2 (g) + H 2 (g) Boyles Law: PV = Constant A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. In this case a change in volume (or pressure)

36 Effects of Changing the Temperature Endothermic: heat is aborbed by a reaction Reactants + heat gives Products Exothermic: heat is liberated by a reaction Reactants gives Products + heat A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. In this case a change in temperature

37 Effects of Changing the Temperature Let’s decrease the temperature of the reaction Let’s increase the temperature of the reaction, what direction does the equilibrium reaction shift Endothermic: absorption of heat by a reaction Reactants + heat gives Products

38 Let’s decrease the temperature of the reaction Let’s increase the temperature of the reaction Effects of Changing the Temperature Exothermic: heat liberated by a reaction Reactants  Products + heat A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. In this case a change in temperature

39 Temperature RaisedTemperature Lowered Endothermic Reaction (absorb heat) Equilibrium shift right (toward products) Equilibrium Shifts left (toward reactants) Exothermic Reaction (liberate heat) Equilibrium Shifts left (toward reactants) Equilibrium shift right (toward products) If a forward reaction is exothermic, Then the reverse reaction must be endothermic

40 Driving Reactions to Completion/ Increasing Yield Industrial Synthesis of Ammonia (Haber) N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) Forward reaction is exothermic What conditions do we need to increase the yield, i.e., produce more ammonia? Volume DecreasedVolume Increased (Pressure Increased)(Pressure Decreased) V reactants > V products Equilibrium shift right (toward products) Equilibrium Shifts left (toward reactants) Temperature RaisedTemperature Lowered Exothermic Reaction (liberate heat) Equilibrium Shifts left (toward reactants) Equilibrium shift right (toward products)


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