Chapter 11 Student Notes PHW

Chapter 11 Student Notes PHW - Chapter 11 Objectives 11.1....

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Unformatted text preview: Chapter 11 Objectives 11.1. Define equilibrium and explain on a microscopic level why equilibrium is observed. 11.2. Introduce the equilibrium constant as a way of defining the extent of equilibrium. Include a discussion of the use of pressure, concentration, and activity when expressing equilibrium. 11.3. Describe the relationships which exist among equilibrium constants for heterogeneous reactions. 11.4. Provide examples of how to solving equilibrium problems. 11.5. Outline the fundamentals of Le Chatelier's Principle 10/6/2009 Zumdahl Chapter 2 H2O (g) + CO (g) 10/6/2009 1 H2 (g) + CO2 (g) Zumdahl Chapter 2 2 1 Chemical Reactions and Equilibrium 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. Four fundamental characteristics of equilibrium states in isolated isolated systems: 1. They display no macroscopic evidence of change. 2. They are reached through spontaneous processes. 3. They show a dynamic balance of forward and backward processes. 4. 10/6/2009 Zumdahl Chapter 2 3 The Form of Equilibrium Expressions Consider the equilibrium reaction where 2 molecules of NO2 combine reversibly to form N2O4 : 2 NO2 ↔ N2O4 (1) Rate of forward reaction (2 NO2 → N2O4) = Rate of reverse reaction (N2O4 → 2 NO2) = At equilibrium, the concentrations have adjusted to make the rate rate of the forward reaction equal to the rate of the reverse reaction: reaction: (2) Rearranging eqt. (2) allows us to define a Pseudo Equilibrium eqt. Constant (K′) as follows: K′ = 10/6/2009 Zumdahl Chapter 2 = cN2O4 / cNO22 4 2 “Legitimate” Equilibrium Constants are Unitless 2 NO2 ↔ N2O4 = K′ cref K′ K = We can also define an equilibrium constant in terms of the partial pressures of partial reactants and products: KP = Typically, the reference concentration (pressure) is chosen to be 1 of whatever be unit is being used, i.e., 1 mol L-1, 1 atm, 1 bar, … It is often inferred. atm, inferred. An equilibrium constant is a unitless number that has no meaning unless the balanced chemical equation and the reference pressure or concentration are also specified 10/6/2009 Zumdahl Chapter 2 5 Law of Mass Action 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 if pressures if concentrations (PC )c (PD )d (PA )a (PB )b [C]c [D]d [A]a [B]b = KP =K Reference pressuresZumdahl Chapter 2 or concentrations are inferred 10/6/2009 6 3 From Text, p. 506: “The law of mass action is widely applicable. It correctly describes describes equilibrium behavior of all chemical reaction systems whether they occur in they solution or in the gas phase. Although, as we will see later, corrections for nonnonideal behavior must be applied in certain cases, such as for concentrated cases, aqueous solutions and for gases at high pressures, the law of mass action mass provides a remarkably accurate description of all types of chemical equilibria. chemical equilibria. When dealing with non-ideal behavior, activity replaces concentration nonor partial pressure. Gas phase activity is often referred to as fugacity. fugacity. 10/6/2009 Zumdahl Chapter 2 7 Sample Problem Write equilibrium expressions for the reactions defined by the following equations. Reference pressures can be inferred. inferred. 3 H2(g) + SO2(g) ↔ H2S(g) + 2 H2O(g) = KP 2 C2F5Cl(g) + 4 O2(g) ↔ Cl2(g) + 4 CO2(g) + 5 F2(g) = KP 10/6/2009 Zumdahl Chapter 2 8 4 Heterogeneous Equilibria Solids CaCO3(s) ↔ CaO(s) + CO2(g) Liquids H2O(l) ↔ H2O(g) KP is independent of the amounts of CaCO3(s) or CaO(s) CaO(s) Dissolved I2(s) ↔ I2(aq) species 10/6/2009 Zumdahl Chapter 2 9 Law of Mass Action Extended to Heterogeneous Equilibria 1. Gases enter equilibrium expressions as partial pressures, e.g., PCO2 2. Dissolved species enter as concentrations, 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., I2(s) ↔ I2(aq) [I2(aq) ] = K aq) aq) I 2 ( s ) ↔ I 2 ( aq ) K = 10/6/2009 [ I 2 ( aq )] [ I 2 ( aq )] = = [ I 2 ( aq )] [ I 2 ( s )] 1 Zumdahl Chapter 2 10 5 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 forward reaction. #1 2 H2(g) + O2(g) ↔ 2 H2O(g) O(g (PH2O)2 (PH2)2(PO2) = K1 #2 2 H2O(g) ↔ 2 H2(g) + O2(g) O(g (PH2)2(PO2) (PH2O)2 = K2 10/6/2009 Zumdahl Chapter 2 11 Relationships Among the K’s of Related Reactions # 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. #1 #3 2 H2(g) + O2(g) ↔ 2 H2O(g) Rxn 1 O(g (PH2O)2 (PH2)2(PO2) = K1 H2(g) + ½ O2(g) ↔ H2O(g) Rxn 3 = Rxn 1 times 1/2 O(g (PH2O) = K3 (PH2)(PO2)½ 10/6/2009 Zumdahl Chapter 2 12 6 Relationships Among the K’s of Related Reactions # 3: When chemical equations are added to give a new equation, their equilibrium constants are multiplied to give the equilibrium constant constant associated with the new equation. (PBr2)(PCl2) 2 BrCl (g) ↔ Br2 (g) + Cl2 (g) 2 BrCl (g) + I2 (g) ↔ 2 IBr (g) + Cl2(g) (PBr2)(PCl2) (PBrCl)2 X = K1 = 0.45 @ 25oC (PBrCl)2 Br2 (g) + I2 (g) ↔ 2 IBr (g) (PIBr)2 o (PBr2) (PI2) = K2 = 0.051 @ 25 C (PIBr)2 (PBr2) (PI2) = = (0.45)(0.051) = 0.023 @ 25oC 10/6/2009 Zumdahl Chapter 2 13 The Reaction Quotient aA + bB (PC )c (PD )d (PA )a (PB )b forward ⎯⎯ ⎯→ cC + dD reverse ←⎯ ⎯ ⎯ (PC )c (PD )d (PA )a (PB )b =K P = QP Note that KP (the Equilibrium Constant) uses equilibrium partial pressures Note that QP (the reaction quotient) uses prevailing partial pressures, not necessarily at equilibrium 10/6/2009 Zumdahl Chapter 2 14 7 The Reaction Quotient (con’t) forward ⎯⎯ ⎯→ aA + bB reverse cC + dD ←⎯ ⎯ ⎯ (PC )c (PD )d (PA )a (PB )b =Q P (PC )c (PD )d (PA )a (PB )b =K P If Q < K, reaction proceeds in a K, If Q > K, reaction proceeds in a K, If Q = K, the reaction is at equilibrium K, 10/6/2009 Zumdahl Chapter 2 15 The equilibrium constant for the reaction P4(g) ↔ 2 P2(g) is 1.39 at 400 Suppose that 2.75 mol of P4(g) and 1.08 mol of P2(g) are mixed in a closed 25.0 L container at 400oC. Compute Q(init) (the Q at the moment of mixing) and state the direction in which the reaction proceeds. oC . ( PP 2 )2 = K = 1.39 ( PP 4 )1 P PV = nRT KP = 1.39 @ 400oC nP4t=0 = 2.75 mol nP2t=0 = 1.08 mol PP4t=0 = nP4t=0 RT/V = [(2.75 mol)(0.0821 atm L mol-1 K-1)(673 K)] ÷ (25.0 L) = 6.08 atm PP2t=0 = nP2t=0 RT/V = [(1.08 mol)(0.0821 atm L mol-1 K-1)(673 K)] ÷ (25.0 L) = 2.39 atm QP = (2.39)2 / (6.08) = 0.939 (assuming Pref = 1 atm) atm) 10/6/2009 Zumdahl Chapter 2 16 8 Sample Problem Consider the equilibrium 4 NO2(g) ↔ 2 N2O(g) + 3 O2(g) O(g The three gases are introduced into a container at partial pressures of 3.6 atm pressures (for NO2), 5.1 atm (for N2O), and 8.0 atm (for O2) and react to reach equilibrium at a fixed temperature. The equilibrium partial pressure of the NO2 is measured to be 2.4 atm. Evaluate the equilibrium constant of the reaction at this temperature, assuming that no competing reactions occur. 4 NO2(g) ↔ 2 N2O(g) + 3 O2(g) O(g initial partial pressure (atm) change in partial pressure (atm) equilibrium partial pressure (atm) 10/6/2009 Zumdahl Chapter 2 17 Evaluate the equilibrium constant of the reaction at this temperature, assuming temperature, that no competing reactions occur. 4 NO2(g) ↔ 2 N2O(g) + 3 O2(g) O(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 KP = 10/6/2009 (PN2O)2 (PO2)3 (PNO2)4 = Zumdahl Chapter 2 5.1 + 2x 8.0 + 3x = 690 = KP 18 9 Sample Problem The compound GeWO4(g) forms at high temperature in the reaction 2 GeO(g) + W2O6(g) ↔ 2 GeWO4(g) GeO(g Some GeO(g) and W2O6(g) are mixed. Before they start to react, their partial GeO(g pressures both equal 1.000 atm. After their reaction at constant temperature constant and volume, the equilibrium partial pressure of GeWO4(g) is 0.980 atm. Assuming that this is the only reaction that takes place, (a) determine the equilibrium partial pressures of GeO and W2O6, and (b) determine the equilibrium constant for the reaction. reaction. 2 GeO (g) + W2O6 (g) ↔ 2 GeWO4(g) (g 1.000 1.000 0 initial partial pressure (atm) change in partial pressure (atm) – 2x –x equilibrium partial pressure (atm) 1.000 – 2x 1.000 – x 10/6/2009 +2x +2x 0.980 Zumdahl Chapter 2 19 (a) determine the equilibrium partial pressures of GeO and W2O6, and (b) determine the equilibrium constant for the reaction. 2 GeO(g) + W2O6(g) ↔ 2 GeWO4(g) GeO(g 1.000 1.000 0 initial partial pressure (atm) change in partial pressure (atm) –2x equilibrium partial pressure (atm) 1.000 – 2x 0 + 2x = 0.980 atm GeWO4 x = 0.490 atm KP = (PGeWO4)2 (PGeO)2 (PW2O6) 10/6/2009 = –x 1.000 – x +2x 0.980 1.000 – 2(0.490) = 0.020 atm GeO 1.000 – 0.490 = 0.510 atm W2O6 (0.980)2 = 4700 = KP (0.020)2 (0.510) Zumdahl Chapter 2 20 10 Sample Problem A vessel holds pure CO(g) at a pressure of 1.282 atm and a temperature of CO(g) 354K. A quantity of nickel is added, and the partial pressure of CO(g) drops to of CO(g) 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. 354K. Construct an “ICE” table ICE” PCO (atm) PNi(CO)4 (atm) initial partial pressure (atm) change in partial pressure (atm) equilibrium partial pressure (atm) => At equil., Pco = equil., KP = PNi(CO) 1.282 -4x 0.709 4 (P CO ) 4 [Ni(s)] 10/6/2009 = PNi(CO) 4 (P CO ) 4 (1) 0 +1x +1x x x = = KP = 0.143 / (0.709)4 = Zumdahl Chapter 2 21 Sample Problem At a particular temperature, K = 2.0 x 10-6 mol/L for the reaction 2CO2(g) 2CO(g) + O2(g) 2CO(g If 2.0 mol CO2 is initially placed into a 5.0-L vessel, calculate the equilibrium 5.0concentrations of all species. 2CO2(g) 2CO(g) + O2(g) 2CO(g initial partial pressure (mol/L) 0.4 change in partial pressure (mol/L) – 2x equilibrium partial pressure (mol/L) 0.4 -2x Kp = K 0 +2x +2x 2x 0 +1x 1x [CO]2 [O 2 ] (2 x)2 ( x) = 2.0x10-6 = 2 [CO 2 ] (0.40 − 2 x) 2 check assumption 10-3 assume 2x << 0.40 <――――――――> 2x = 2(4.3x10-3)) x 100 = 2.2% ――――――――> 0.40 0.40 assumpiton valid, less than 5% assumption (2x) 2 (x) 2.0x10-6 = ⇒ x = 4.3x10-3M (0.40) 2 10/6/2009 Zumdahl Chapter 2 (cref = 1 M) 22 11 At a particular temperature, K = 2.0 x 10-6 mol/L for the reaction 2CO2 (g) 2CO (g) + O2 (g) (g If 2.0 mol CO2 is initially placed into a 5.0-L vessel, calculate the equilibrium 5.0concentrations of all species. 2CO2 (g) initial partial pressure (mol/L) 0.4 change in partial pressure (mol/L) – 2x equilibrium partial pressure (mol/L) 0.4 -2x Kp = = 2CO (g) + O2 (g) 0 +2x +2x 2x 0 +1x 1x [CO]2 [O 2 ] = 2.0x10-6 [CO 2 ]2 (2 x) 2 ( x) (0.40 − 2 x) 2 x = 4.3x10-3 M [O 2 ] = x = 4.3 x 10-3M 10/6/2009 Zumdahl Chapter 2 23 The Person Behind the Science 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 labor- Moments in a Life – Le Chatelier was named "chevalier" (knight) of the Légion d'honneur in 1887, decoration established by Napoléon Napolé Bonaparte in 1802. Zumdahl Chapter 2 10/6/2009 24 12 Effects of External Stresses on Equilibria: Le Châtelier’s Principle Le Châtelier’s Principle provides a way to predict the response of Châtelier’ an equilibrium system to an external perturbation, such as… as… 1. Effects of Adding or Removing Reactants or Products 2. Effects of Changing the Volume (or Pressure) of the System 3. Effects of Changing the Temperature 10/6/2009 Zumdahl Chapter 2 25 Effect of Adding or Removing Reactants or Products PCl5(g) ↔ PCl3(g) + Cl2(g) Q = [PCl3] [Cl2] / [PCl5] K = [PCl3]eq [Cl2]eq / [PCl5]eq = Qeq Suppose that a state of equilibrium initially exists, which is then perturbed by then addition or removal of a reactant or product. How does the system respond? add extra PCl5(g) add extra PCl3(g) remove some Cl2(g) 10/6/2009 Zumdahl Chapter 2 26 13 Effect of Changing the System Pressure/Volume Volume Decreased Volume Increased (Pressure Increased) (Pressure Decreased) Preactants > Pproducts Equilibrium shift right (toward products) Equilibrium Shifts left (toward reactants) Preactants < Pproducts Equilibrium Shifts left (toward reactants) Equilibrium shift right (toward products) Preactants = Pproducts Equilibrium not affected Equilibrium not affected Boyles Law: PV = Const. 2 P2(g) ↔ P4(g) PCl5(g) ↔ PCl3(g) + Cl2(g) CO(g) + H2O(g) ↔ CO2(g) + H2(g) 10/6/2009 Zumdahl Chapter 2 27 Effect of Changing the Temperature If a reaction is endothermic in the forward direction and we increase the temperature, the equilibrium will shift toward products because some heat is used temperature, up to convert reactants to products, thus partially counteracting the temperature counteracting increase. If a reaction is endothermic in the forward direction and we decrease the temperature, the equilibrium will shift toward reactants because some heat is temperature, liberated when products are converted to reactants, thus partially counteracting partially the temperature decrease. 10/6/2009 Zumdahl Chapter 2 28 14 Driving Reactions to Completion Industrial Synthesis of Ammonia (Haber Process) N2(g) + 3H2(g) ↔ 2NH3(g) (Forward reaction is exothermic) Forward exothermic What conditions should be adopted to increase the yield of the product ammonia? Volume Decreased (Pressure Increased) Volume Increased (Pressure Decreased) Equilibrium Shifts left (toward reactants) Temperature Raised Temperature Lowered Equilibrium Shifts left (toward reactants) 10/6/2009 Zumdahl Chapter 2 29 Equilibria Involving Real Gases N2(g) + 3 H2(g) ↔ 2 NH3(g) KP = PNH32 / (PN2 PH23) KP is an accurate equilibrium constant only in the limit where all gases behave ideally. At high pressures, such as those employed in the industrial synthesis industrial of ammonia, non-ideal behavior of gases must be taken into account: non- The φ’s are called fugacity coefficients and represent a measure of deviations from ideal gas behavior for each gas. They are readily determinable determinable experimentally for pure gases, but are very difficult to determine in gas determine mixtures because they must be evaluated at the equilibrium composition which, typically is not known if the φ’s are not known. The liquid phase analog of φ is known. the activity coefficient, γ. coefficient, 10/6/2009 Zumdahl Chapter 2 30 15 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 10/6/2009 Zumdahl Chapter 2 31 16 ...
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This note was uploaded on 10/06/2009 for the course CHEM 1101 taught by Professor Bottomley during the Fall '08 term at Georgia Tech.

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