Chemistry 4th Edition McMurry/Fay

Transcription

1 13 Ch a pt e r 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. Chapter 13 Slide 2 The Equilibrium State 02 Chapter 13 Slide 3 1

2 The Equilibrium State 02 Chemists are interested in these reversible reactions. One example is the following: Chapter 13 Slide 4 The Equilibrium State 03 Chapter 13 Slide 5 The Equilibrium State 03 Graphs of reactant and product concentrations change with time as shown below. Chapter 13 Slide 6 2

3 The Equilibrium State 04 Chapter 13 Slide 7 Equilibrium Constant 01 The equilibrium expression compares reactant and product concentrations. [ Pr oducts K = ]p Reactants [ [ ] r = B] b [ [ A] a = NO 2 ] 2 [ N 2 O 4 ] = From this we obtain a constant (K) for the reaction, which is independent of concentration changes, but dependent on the temperature. Chapter 13 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. [ K c = NO 2 ] 2 [ N 2 O 4 ] K p = P 2 NO 2 P N2O4 Chapter 13 Slide 9 3

4 Equilibrium Constant 03 We can convert between K c and K p using an equation derived from PV = nrt: For aa æ bb K p = K c (0.0821T)?n?n = moles gas products moles of gas reactants?n = b a Chapter 13 Slide 10 Equilibrium Constant 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? Chapter 13 Slide 11 Equilibrium Constant 05 Write the K p and K c expressions for: 2 N 2 O 5 æ 4 NO 2 + O 2 The equilibrium concentrations for the reaction between CO and Cl 2 to form carbonyl chloride (phosgene gas) CO + Cl 2 æ COCl 2 at 74 C are: [CO] = 1.2 x 10 2 M, [Cl 2 ] = M, and [COCl 2 ] = 0.14 M. Calculate K c and K p. Chapter 13 Slide 12 4

5 Equilibrium Constant 06 Methane (CH 4 ) reacts with hydrogen sulfide to yield H 2 and carbon disulfide, a solvent used in manufacturing. What is the value of K p 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 2 S, 0.52 atm of CS 2, and 0.10 atm of H 2? Chapter 13 Slide 13 Equilibrium Constant 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 CaCO 3 (s) æ CaO(s) + CO 2 K c = [CO 2 ] because CaCO 3 and CaO are solids. Chapter 13 Slide 14 Equilibrium Constant 08 Chapter 13 Slide 15 5

6 Equilibrium Constant 09 Write the equilibrium equation for each of the following reactions: (a) CO 2 + C(s) æ 2 CO (b) Hg(l) + Hg 2+ (aq) æ Hg 2 2+ (aq) (c) 2 Fe(s) + 3 H 2 O æ Fe 2 O 3 (s) + 3 H 2 (d) 2 H 2 O(l) æ 2 H 2 + O 2 Chapter 13 Slide 16 Using Equilibrium Constants 01 We can make the following generalizations concerning the composition of equilibrium mixtures: If K c > 10 3, products predominate over reactants. If K c is very large, the reaction is said to proceed to completion. If K c is in the range 10 3 to 10 3, appreciable concentrations of both reactants and products are present. If K c < 10 3, reactants predominate over products. If K c is very small, the reaction proceeds hardly at all. Chapter 13 Slide 17 Using Equilibrium Constants 02 The reaction quotient (Q c ) is obtained by substituting initial concentrations into the equilibrium constant. Predicts reaction direction. Q c > K c Q c = K c Q c < K c System proceeds to form reactants. System is at equilibrium. System proceeds to form products. Chapter 13 Slide 18 6

7 Using Equilibrium Constants 03 Predicting the direction of a reaction. Chapter 13 Slide 19 Using Equilibrium Constants 04 The equilibrium constant (K c ) for the formation of nitrosyl chloride, from nitric oxide and chlorine gas: 2 NO + Cl 2 æ 2 NOCl is 6.5 x 10 4 at 35 C. In an experiment, 2.0 x 10 2 moles of NO, 8.3 x 10 3 moles of Cl 2, and 6.8 moles of NOCl are mixed in a 2.0-L flask. In which direction will the system proceed to reach equilibrium? Chapter 13 Slide 20 Using Equilibrium Constants 05 Knowing K allows us to calculate equilibrium concentrations from initial concentrations. We use the Initial Change Equilibrium method. cis-stilbene æ trans-stilbene Initial (M) Change (M) x +x Equil ibrium (M) (0.850 x) x Use K c =24 to determine equilibrium concentrations. Chapter 13 Slide 21 7

8 Using Equilibrium Constants 06 A mixture of mol H 2 and mol I 2 was placed in a 1.00-L stainless steel flask at 700 C. The equilibrium constant K c for the reaction H 2 + I 2 æ 2 HI is 57 at this temperature. Calculate the equilibrium concentrations. If the starting concentration of HI was M, calculate the new equilibrium concentrations. If the initial concentrations are [H 2 ] = M and [I 2 ] = M, calculate the equilibrium concentrations. Chapter 13 Slide 22 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. Chapter 13 Slide 23 Le Châtelier s Principle 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. Chapter 13 Slide 24 8

9 Le Châtelier s Principle 03 Haber process for synthesis of ammonia. N H 2 æ 2 NH 3 K c = at 700 K Given an equilibrium mixture of 0.50 M N 2, 3.00 M H 2, and 1.98 M NH 3 at 700 K, what happens when the concentration of N 2 is increased to 1.50 M? Le Châtelier s principle tells us the reaction will relieve the stress by converting the N 2 to NH 3. Chapter 13 Slide 25 Le Châtelier s Principle 04 Chapter 13 Slide 26 Le Châtelier s Principle 05 Chapter 13 Slide 27 9

10 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: Fe 2 O 3 (s) + 3 CO æ 2 Fe(l) + 3 CO 2 Use Le Châtelier s principle to predict the direction of reaction when an equilibrium mixture is disturbed by: (a) Adding Fe 2 O 3 (b) Removing CO 2 (c) Removing CO Chapter 13 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. Chapter 13 Slide 29 Le Châtelier s Principle 08 N H 2 æ 2 NH 3 K c = at 700 K Chapter 13 Slide 30 10

11 Le Châtelier s Principle 09 Consider the reaction: N 2 O 4 æ 2 NO 2, taking place in a cylinder with a volume = 1 unit. [NO 2 ] = 2 mol/1 = 2 [N 2 O 4 ] = 1 mol/1 = 1 K = [NO 2 ]2 [N 2 O 4 ] = 4 Chapter 13 Slide 31 Le Châtelier s Principle 10 The Volume is then halved, which is equivalent to doubling the pressure. [NO 2 ] = 2 mol/0.5 = 4 [N2O4] = 1 mol/0.5 = 2 Q = [NO 2 ] 2 = 8 [N 2 O 4 ] Since Q > K, the [product] is too high and the reaction progresses in the reverse direction. Chapter 13 Slide 32 Le Châtelier s Principle 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 æ PCl 3 + Cl 2 2. CaO(s) + CO 2 æ CaCO 3 (s) 3. 3 Fe(s) + 4 H 2 O æ Fe 3 O 4 (s) + 4 H 2 Chapter 13 Slide 33 11

12 Le Châtelier s Principle 12 The following picture represents the equilibrium mixture for the gas-phase reaction A 2 æ 2A. Draw a picture that shows how the concentrations change when the pressure is increased by decreasing the volume. Chapter 13 Slide 34 Le Châtelier s Principle 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. Chapter 13 Slide 35 Le Châtelier s Principle 14 Consider the reaction N H 2 æ 2 NH 3 which is exothermic by 92.2 kj. Chapter 13 Slide 36 12

13 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 O 2 æ 4 NO + 6 H 2 O?H = kj How does the equilibrium amount vary with an increase in temperature? Chapter 13 Slide 37 Le Châtelier s Principle 16 The following pictures represent the composition of the equilibrium mixture at 400 K and 500 K for the reaction A + B æ AB. Is the reaction endothermic or exothermic? Chapter 13 Slide 38 Le Châtelier s Principle 17 Catalysis: No effect. Chapter 13 Slide 39 13