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C h a pt e r 13 Chemical Equilibrium Chemistry 4th Edition McMurry/Fay Dr. Paul Charlesworth Michigan Technological University The Equilibrium State • 01 Chemical Equilibrium: A state achieved when the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant. • Equilibrium between phases is known as physical equilibrium. Prentice Hall ©2004 Chapter13 The Equilibrium State Prentice Hall ©2004 Chapter13 Slide 2 02 Slide 3 1 The Equilibrium State • Chemists are interested in these reversible reactions. One example is the following: Prentice Hall ©2004 Chapter13 The Equilibrium State Prentice Hall ©2004 Chapter13 The Equilibrium State • 02 Slide 4 03 Slide 5 03 Graphs of reactant and product concentrations change with time as shown below. Prentice Hall ©2004 Chapter13 Slide 6 2 The Equilibrium State Prentice Hall ©2004 04 Chapter13 Slide 7 Equilibrium Constant • The equilibrium expression compares reactant and product concentrations. [Pr oducts] = [B ] = [NO2 ] = 4.63 × 10−3 r a [Re ac tants ] [A] [N 2O4 ] p K= • 01 b 2 From this we obtain a constant (K) for the reaction, which is independent of concentration changes, but dependent on the temperature. Prentice Hall ©2004 Chapter13 Slide 8 Equilibrium Constant • 02 Homogeneous Equilibrium:When all reacting species are in the same phase, all reactants and products are included in the expression. • Amounts of components are given as molarity or partial pressure of a gas. [NO ] 2 Kc = [ 2 ] N2O4 Prentice Hall ©2004 Chapter13 2 PNO 2 Kp = PN2O4 Slide 9 3 Equilibrium Constant • 03 We can convert between Kc and Kp using an equation derived from PV = nRT: For aA æ bB Kp = Kc (0.0821T) ?n ?n = moles gas products – moles of gas reactants ?n = b – a Prentice Hall ©2004 Chapter13 Equilibrium Constant • Slide 10 04 The following pictures rerepresent mixtures of A molecules (red) and B molecules (blue), which interconvert according to the equation A æ B. If Mixture (1) is at equilibrium, which of the other mixtures is also at equilibrium? Prentice Hall ©2004 Chapter13 Equilibrium Constant • Slide 11 05 Write the Kp and Kc expressions for: 2 N 2O5(g) æ 4 NO2(g) + O2(g) • The equilibrium concentrations for the reaction between CO and Cl2 to form carbonyl chloride (phosgene gas) CO(g) + Cl2(g) æ COCl2(g) at 74°C are: [CO] = 1.2 x 10–2 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate Kc and Kp. Prentice Hall ©2004 Chapter13 Slide 12 4 Equilibrium Constant • 06 Methane (CH4) reacts with hydrogen sulfide to yield H2 and carbon disulfide, a solvent used in manufacturing. What is the value of Kp at 1000 K if the partial pressures in an equilibrium mixture at 1000 K are 0.20 atm of CH 4, 0.25 atm of H 2S, 0.52 atm of CS2, and 0.10 atm of H2? Prentice Hall ©2004 Chapter13 Equilibrium Constant Slide 13 07 • Heterogeneous Equilibrium:When reacting species are in different phases, solid and liquid phases are excluded from the expression because their concentrations “do not change.” • For CaCO3(s) æ CaO(s) + CO2(g) Kc = [CO2] because CaCO 3 and CaO are solids. Prentice Hall ©2004 Chapter13 Equilibrium Constant Prentice Hall ©2004 Chapter13 Slide 14 08 Slide 15 5 Equilibrium Constant • 09 Write the equilibrium equation for each of the following reactions: (a) CO2(g) + C(s) æ 2 CO(g) (b) Hg(l) + Hg 2+(aq) æ Hg 22+(aq) (c) 2 Fe(s) + 3 H2 O(g) æ Fe2O3(s) + 3 H2(g) (d) 2 H2 O(l) æ 2 H2(g) + O2(g) Prentice Hall ©2004 Chapter13 Using Equilibrium Constants • Slide 16 01 We can make the following generalizations concerning the composition of equilibrium mixtures: If K c > 103, products predominate over reactants. If Kc is very large, the reaction is said to proceed to completion. If K c is in the range 10–3 to 103, appreciable concentrations of both reactants and products are present. If K c < 10–3, reactants predominate over products. If Kc is very small, the reaction proceeds hardly at all. Prentice Hall ©2004 Chapter13 Using Equilibrium Constants • Slide 17 02 The reaction quotient (Q c) is obtained by substituting initial concentrations into the equilibrium constant. Predicts reaction direction. Q c > Kc System proceeds to form reactants. Q c = Kc System is at equilibrium. Q c < Kc System proceeds to form products. Prentice Hall ©2004 Chapter13 Slide 18 6 Using Equilibrium Constants • Predicting the direction of a reaction. Prentice Hall ©2004 Chapter13 Using Equilibrium Constants • 03 Slide 19 04 The equilibrium constant (Kc ) for the formation of nitrosyl chloride, from nitric oxide and chlorine gas: 2 NO(g) + Cl 2(g) æ 2 NOCl(g) is 6.5 x 104 at 35°C. In an experiment, 2.0 x 10–2 moles of NO, 8.3 x 10–3 moles of Cl2, and 6.8 moles of NOCl are mixed in a 2.0-L flask. In which direction will the system proceed to reach equilibrium? Prentice Hall ©2004 Chapter13 Using Equilibrium Constants Slide 20 05 Knowing K allows us to calculate equilibrium concentrations from initial concentrations. • We use the Initial Change Equilibrium method. • Initial (M) Change (M) Equil ibrium (M) • cis-stilbene æ trans-stilbene 0.850 0 –x +x (0.850–x) x Use Kc =24 to determine equilibrium concentrations. Prentice Hall ©2004 Chapter13 Slide 21 7 Using Equilibrium Constants • 06 A mixture of 0.500 mol H2 and 0.500 mol I2 was placed in a 1.00-L stainless steel flask at 700°C. The equilibrium constant Kc for the reaction H2(g) + I2(g) æ 2 HI(g) is 57 at this temperature. Calculate the equilibrium concentrations. If the starting concentration of HI was 0.040 M, calculate the new equilibrium concentrations. If the initial concentrations are [H2] = 0.100 M and [I 2] = 0.200 M, calculate the equilibrium concentrations. Prentice Hall ©2004 Chapter13 Le Châtelier’s Principle • 01 Le Châtelier’s principle: If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset. Prentice Hall ©2004 Chapter13 Le Châtelier’s Principle • Slide 22 Slide 23 02 Concentration Changes: The concentration stress of an added reactant or product is relieved by reaction in the direction that consumes the added substance. The concentration stress of a removed reactant or product is relieved by reaction in the direction that replenishes the removed substance. Prentice Hall ©2004 Chapter13 Slide 24 8 Le Châtelier’s Principle • 03 Haber process for synthesis of ammonia. N2(g) + 3 H2(g) æ 2 NH3(g) Kc = 0.291 at 700 K • Given an equilibrium mixture of 0.50 M N2, 3.00 M H2, and 1.98 M NH 3 at 700 K, what happens when the concentration of N2 is increased to 1.50 M? • Le Châtelier’s principle tells us the reaction will relieve the stress by converting the N2 to NH 3. Prentice Hall ©2004 Chapter13 Le Châtelier’s Principle Prentice Hall ©2004 Chapter13 Le Châtelier’s Principle Prentice Hall ©2004 Chapter13 Slide 25 04 Slide 26 05 Slide 27 9 Le Châtelier’s Principle 06 • The reaction of iron(III) oxide with carbon monoxide occurs in a blast furnace when iron ore is reduced to iron metal: • Use Le Châtelier’s principle to predict the direction of reaction when an equilibrium mixture is disturbed by: Fe2O3(s) + 3 CO(g) æ 2 Fe(l) + 3 CO2(g) •(a) Adding Fe2O 3 Prentice Hall ©2004 (b) Removing CO2 (c) Removing CO Chapter13 Slide 28 Le Châtelier’s Principle • 07 Volume and Pressure Changes: Only reactions containing gases are affected by changes in volume and pressure. Increasing pressure = Decreasing volume • PV = nRT tells us that increasing pressure or decreasing volume increases concentration. Prentice Hall ©2004 Chapter13 Slide 29 Le Châtelier’s Principle • N2(g) + 3 H2(g) æ 2 NH3(g) Prentice Hall ©2004 Chapter13 08 Kc = 0.291 at 700 K Slide 30 10 Le Châtelier’s Principle • 09 Consider the reaction: N2O4(g) æ 2 NO2(g), taking place in a cylinder with a volume = 1 unit. [NO2 ] = 2 mol/1 = 2 [N 2O4 ] = 1 mol/1 = 1 K= Prentice Hall ©2004 [NO2 ]2 [N 2 O4 ] =4 Chapter13 Slide 31 Le Châtelier’s Principle • 10 The Volume is then halved, which is equivalent to doubling the pressure. [NO2 ] = 2 mol/0.5 = 4 [N2 O4 ] = 1 mol/0.5 = 2 Q= • [NO2 ]2 =8 [N2 O4 ] Since Q > K, the [product] is too high and the reaction progresses in the reverse direction. Prentice Hall ©2004 Chapter13 Le Châtelier’s Principle • Slide 32 11 Does the number of moles of reaction products increase, decrease, or remain the same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume. 1. PCl 5(g) æ PCl 3(g) + Cl 2(g) 2. CaO(s) + CO2(g) æ CaCO3(s) 3. 3 Prentice Hall ©2004 Fe(s) + 4 H2O(g) æ Fe3O4(s) + 4 H2(g) Chapter13 Slide 33 11 Le Châtelier’s Principle • 12 The following picture represents the equilibrium mixture for the gas -phase reaction A2 æ 2A. • Draw a picture that shows how the concentrations change when the pressure is increased by decreasing the volume. Prentice Hall ©2004 Chapter13 Le Châtelier’s Principle Slide 34 13 • Temperature Changes: Changes in temperature can change the equilibrium constant. • Endothermic processes are favored when temperature increases. • Exothermic processes are favored when temperature decreases. Prentice Hall ©2004 Chapter13 Le Châtelier’s Principle • Slide 35 14 Consider the reaction N2(g) + 3 H2(g) æ 2 NH3(g) which is exothermic by 92.2 kJ. Prentice Hall ©2004 Chapter13 Slide 36 12 Le Châtelier’s Principle • 15 In the first step of the Ostwald process for synthesis of nitric acid, ammonia is oxidized to nitric oxide by the reaction: 4 NH 3(g) + 5 O2(g) æ 4 NO(g) + 6 H2O(g) ? H° = –905.6 kJ • How does the equilibrium amount vary with an increase in temperature? Prentice Hall ©2004 Chapter13 Le Châtelier’s Principle Slide 37 16 • The following pictures represent the composition of the equilibrium mixture at 400 K and 500 K for the reaction A(g) + B(g) æ AB(g). • Is the reaction endothermic or exothermic? Prentice Hall ©2004 Chapter13 Le Châtelier’s Principle • Slide 38 17 Catalysis: No effect. Prentice Hall ©2004 Chapter13 Slide 39 13