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Chapter Fourteen 1 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chemical Equilibrium Chapter Fourteen.

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Presentation on theme: "Chapter Fourteen 1 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chemical Equilibrium Chapter Fourteen."— Presentation transcript:

1 Chapter Fourteen 1 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chemical Equilibrium Chapter Fourteen

2 2 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Dynamic Nature of Equilibrium When a system reaches equilibrium, the forward and reverse reactions continue to occur … but at equal rates. We are usually concerned with the situation after equilibrium is reached. After equilibrium the concentrations of reactants and products remain constant.

3 Chapter Fourteen 3 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Dynamic Equilibrium Illustrated NaCl dissolves and recrystallizes continuously. After a time, the solution contains radioactive Na + … NaCl containing radioactive Na + is added to a saturated NaCl solution. … and the added salt now contains some stable Na +.

4 Chapter Fourteen 4 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Concentration vs. Time If we begin with only 1 M HI, the [HI] decreases and both [H 2 ] and [I 2 ] increase. Beginning with 1 M H 2 and 1 M I 2, the [HI] increases and both [H 2 ] and [I 2 ] decrease. Beginning with 1 M each of H 2, I 2, and HI, the [HI] increases and both [H 2 ] and [I 2 ] decrease.

5 Chapter Fourteen 5 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Regardless of the starting concentrations; once equilibrium is reached … … the expression with products in numerator, reactants in denominator, where each concentration is raised to the power of its coefficient, appears to give a constant.

6 Chapter Fourteen 6 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 The Equilibrium Constant Expression For the general reaction: aA + bB  gG + hH The equilibrium expression is: Each concentration is simply raised to the power of its coefficient Products in numerator. Reactants in denominator. [G] g [H] h K c = [A] a [B] b

7 Chapter Fourteen 7 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 The Equilibrium Constant The equilibrium constant is constant regardless of the initial concentrations of reactants and products. This constant is denoted by the symbol K c and is called the concentration equilibrium constant. Concentrations of the products appear in the numerator and concentrations of the reactants appear in the denominator. The exponents of the concentrations are identical to the stoichiometric coefficients in the chemical equation.

8 Chapter Fourteen 8 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example 14.1 If the equilibrium concentrations of COCl 2 and Cl 2 are the same at 395 °C, find the equilibrium concentration of CO in the reaction: CO(g) + Cl 2 (g) COCl 2 (g) K c = 1.2 x 10 3 at 395 °C

9 Chapter Fourteen 9 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 The Condition of Equilibrium The kinetics view: K c = (forward rate)/(reverse rate) = k f /k r The thermodynamics view: –The equilibrium constant can be related to other fundamental thermodynamic properties and is called the thermodynamic equilibrium constant, K eq. –The thermodynamic equilibrium constant expression uses dimensionless quantities known as activities in place of molar concentrations.

10 Chapter Fourteen 10 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Modifying the Chemical Equation What will be the equilibrium constant K' c for the new reaction? [NO 2 ] 2 K c = ––––––––– = 4.67 x (at 298 K) [NO] 2 [O 2 ] Now consider the reaction: 2 NO 2 (g) 2 NO(g) + O 2 (g) [NO] 2 [O 2 ] 1 K' c = ––––––––– = ––––––––––– = [NO 2 ] 2 [NO 2 ] 2 ––––––––– [NO] 2 [O 2 ] 1 1 –– = ––––––––– = 2.14 x 10 –14 K c 4.67 x Consider the reaction:2 NO(g) + O 2 (g) 2 NO 2 (g)

11 Chapter Fourteen 11 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Modifying the Chemical Equation (cont’d) What will be the equilibrium constant K" c for the new reaction? Consider the reaction:2 NO(g) + O 2 (g) 2 NO 2 (g) [NO 2 ] 2 K c = ––––––––– = 4.67 x (at 298 K) [NO] 2 [O 2 ] Now consider the reaction: NO 2 (g) NO(g) + ½ O 2 (g) [NO] [O 2 ] 1/2 1 1/2 K" c = ––––––––– = ––– [NO 2 ] 2 K c = 2.14 x 10 –14 = 1.46 x 10 –7

12 Chapter Fourteen 12 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Modifying the Chemical Equation (cont’d) For the reverse reaction, K is the reciprocal of K for the forward reaction. When an equation is divided by two, K for the new reaction is the square root of K for the original reaction. General rule: –When the coefficients of an equation are multiplied by a common factor n to produce a new equation, we raise the original K c value to the power n to obtain the new equilibrium constant. It should be clear that we must write a balanced chemical equation when citing a value for K c.

13 Chapter Fourteen 13 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example 14.2 The equilibrium constant for the reaction: ½ H 2 (g) + ½ I 2 (g) HI(g) at 718 K is (a)What is the value of K c at 718 K for the reaction HI(g) ½ H 2 (g) + ½ I 2 (g) (b) What is the value of K c at 718 K for the reaction H 2 (g) + I 2 (g) 2 HI(g)

14 Chapter Fourteen 14 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 The Equilibrium Constant for an Overall Reaction Adding the given equations gives the desired equation. Multiplying the given values of K gives the equilibrium constant for the overall reaction. (To see why this is so, write the equilibrium constant expressions for the two given equations, and multiply them together. Examine the result …) and we’re given: Suppose we need: N 2 O(g) + 3/2 O 2 (g) 2 NO 2 (g) K c (1) = ?? 2 NO(g) + O 2 (g) 2 NO 2 (g) K c (3) = 4.67 x N 2 O(g) + ½ O 2 (g) 2 NO(g) K c (2) = 1.7 x 10 –13

15 Chapter Fourteen 15 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Equilibria Involving Gases In reactions involving gases, it is often convenient to measure partial pressures rather than molarities. In these cases, a partial pressure equilibrium constant, K p, is used. K c and K p are related by: K p = K c (RT) Δn(gas) where  n(gas) is the change in number of moles of gas as the reaction occurs in the forward direction. (P G ) g (P H ) h K p = (P A ) a (P B ) b  n(gas) = mol gaseous products – mol gaseous reactants

16 Chapter Fourteen 16 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example 14.3 Consider the equilibrium between dinitrogen tetroxide and nitrogen dioxide: N 2 O 4 (g) 2 NO 2 (g) K p = at 319 K (a) What is the value of K c for this reaction? (b) What is the value of K p for the reaction 2 NO 2 (g) N 2 O 4 (g)? (c) If the equilibrium partial pressure of NO 2 (g) is atm, what is the equilibrium partial pressure of N 2 O 4 (g)?

17 Chapter Fourteen 17 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Equilibria Involving Pure Solids and Liquids The equilibrium constant expression does not include terms for pure solid and liquid phases because their concentrations do not change in a reaction. Although the amounts of pure solid and liquid phases change during a reaction, these phases remain pure and their concentrations do not change. [CaO] [CO 2 ] K c = –––––––––– [CaCO 3 ] K c = [CO 2 ] Example: CaCO 3 (s) CaO(s) + CO 2 (g)

18 Chapter Fourteen 18 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example 14.4 The reaction of steam and coke (a form of carbon) produces a mixture of carbon monoxide and hydrogen, called water-gas. This reaction has long been used to make combustible gases from coal: C(s) + H 2 O(g) CO(g) + H 2 (g) Write the equilibrium constant expression for K c for this reaction.

19 Chapter Fourteen 19 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Equilibrium Constants: When Do We Need Them and When Do We Not? A very large numerical value of K c or K p signifies that a reaction goes (essentially) to completion. A very small numerical value of K c or K p signifies that the forward reaction, as written, occurs only to a slight extent. An equilibrium constant expression applies only to a reversible reaction at equilibrium. Although a reaction may be thermodynamically favored, it may be kinetically controlled … Thermodynamics tells us “it’s possible (or not)” Kinetics tells us “it’s practical (or not)”

20 Chapter Fourteen 20 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example 14.5 Is the reaction CaO(s) + CO 2 (g) CaCO 3 (s) likely to occur to any appreciable extent at 298 K?

21 Chapter Fourteen 21 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 The Reaction Quotient, Q For nonequilibrium conditions, the expression having the same form as K c or K p is called the reaction quotient, Q c or Q p. The reaction quotient is not constant for a reaction, but is useful for predicting the direction in which a net change must occur to establish equilibrium. To determine the direction of net change, we compare the magnitude of Q c to that of K c.

22 Chapter Fourteen 22 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 The Reaction Quotient, Q When Q is larger than K, the numerator of Q is too big; we have “too much products.” When Q = K, equilibrium has been reached. When Q is smaller than K, the denominator of Q is too big; we have “too much reactants.”

23 Chapter Fourteen 23 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Le Châtelier’s Principle When any change in concentration, temperature, pressure, or volume is imposed on a system at equilibrium, the system responds by attaining a new equilibrium condition that minimizes the impact of the imposed change. Analogy: Begin with 100 men and 100 women at a dance. Assume that there are 70 couples dancing, though not always the same couples (dynamic equilibrium). If 30 more men arrive, what happens? The equilibrium will shift, and shortly, more couples will be dancing … but probably not 30 more couples.

24 Chapter Fourteen 24 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Changing the Amounts of Reacting Species At equilibrium, Q = K c. If the concentration of one of the reactants is increased, the denominator of the reaction quotient increases. Q is now less than K c. This condition is only temporary, however, because the concentrations of all species must change in such a way so as to make Q = K c again. In order to do this, the concentrations of the products increase; the equilibrium is shifted to the right.

25 Chapter Fourteen 25 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 When acetic acid (a reactant) is added to the equilibrium mixture … … the acetic acid concentration first increases … … then the concentrations of both reactants decrease … … and the concentrations of both products increase, until a new equilibrium is established.

26 Chapter Fourteen 26 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example 14.7 Water can be removed from an equilibrium mixture in the reaction of 1-octanol and acetic acid, for example, by using a solid drying agent that is insoluble in the reaction mixture. Describe how the removal of a small quantity of water affects the equilibrium. CH 3 (CH 2 ) 6 CH 2 OH(soln) + CH 3 COOH(soln) CH 3 (CH 2 ) 6 CH 2 OCOCH 3 (soln) + H 2 O(soln) H+H+

27 Chapter Fourteen 27 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Heterogeneous Equilibria and Le Chatelier’s Principle Addition or removal of pure solids or pure liquids from a system at equilibrium does not affect the equilibrium.

28 Chapter Fourteen 28 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Changing External Pressure or Volume in Gaseous Equilibria When the external pressure is increased (or system volume is reduced), an equilibrium shifts in the direction producing the smaller number of moles of gas. When the external pressure is decreased (or the system volume is increased), an equilibrium shifts in the direction producing the larger number of moles of gas. If there is no change in the number of moles of gas in a reaction, changes in external pressure (or system volume) have no effect on an equilibrium. Example: H 2 (g) + I 2 (g) 2 HI equilibrium is unaffected by pressure changes.

29 Chapter Fourteen 29 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Initial When pressure is increased … … two molecules of NO 2 combine … … to give one molecule of N 2 O 4, reducing the pressure increase.

30 Chapter Fourteen 30 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Temperature Changes and Catalysis Raising the temperature of an equilibrium mixture shifts equilibrium in the direction of the endothermic reaction; lowering the temperature shifts equilibrium in the direction of the exothermic reaction. –Consider heat as though it is a product of an exothermic reaction or as a reactant of an endothermic reaction, and apply Le Châtelier’s principle. A catalyst lowers the activation energy … of both the forward and the reverse reaction. Adding a catalyst does not affect an equilibrium state. A catalyst merely causes equilibrium to be achieved faster.

31 Chapter Fourteen 31 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example 14.8 An equilibrium mixture of O 2 (g), SO 2 (g), and SO 3 (g) is transferred from a 1.00-L flask to a 2.00-L flask. In which direction does a net reaction proceed to restore equilibrium? The balanced equation for the reaction is 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) Example 14.9 Is the amount of NO(g) formed from given amounts of N 2 (g) and O 2 (g), N 2 (g) + O 2 (g) 2 NO(g) ΔH° = kJ greater at high or low temperatures?

32 Chapter Fourteen 32 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example A Conceptual Example Flask A, pictured below, initially contains an equilibrium mixture of the reactants and products of the reaction CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) ΔH = –41 kJ; K c = 9.03 at 698 K It is isolated from flask B by a closed valve. When the valve is opened, a new equilibrium is established as the contents of the two flasks mix. Describe, qualitatively, how the amounts of CO, H 2, CO 2, and H 2 O in the new equilibrium compare with the amounts in the initial equilibrium if (a) flask B initially contains Ar(g) at 1 atm pressure; (b) flask B initially contains 1.0 mol CO 2 ; (c) flask B initially contains 1.0 mol CO and the temperature of the A– B mixture is raised by 100 °C. If you are uncertain of the result in any of the three cases, explain why.

33 Chapter Fourteen 33 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Determining Values of Equilibrium Constants Experimentally When initial amounts of one or more species, and equilibrium amounts of one or more species, are given, the amounts of the remaining species in the equilibrium state and, therefore, the equilibrium concentrations often can be established. A useful general approach is to tabulate under the chemical equation: –the concentrations of substances present initially –changes in these concentrations that occur in reaching equilibrium –the equilibrium concentrations. This sort of table is sometimes called an “ICE” table: Initial/Change/Equilibrium.

34 Chapter Fourteen 34 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example In a 10.0-L vessel at 1000 K, mol SO 2 and mol O 2 react to form mol SO 3 at equilibrium. What is K c, at 1000 K, for the reaction that is shown here? 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) Example Consider the reaction H 2 (g) + I 2 (g) 2 HI(g) K c = 54.3 at 698 K If we start with mol I 2 (g) and mol H 2 (g) in a 5.25-L vessel at 698 K, how many moles of each gas will be present at equilibrium?

35 Chapter Fourteen 35 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Calculating Equilibrium Quantities from K c and K p Values When starting with initial reactants and no products and with the known value of the equilibrium constant, these data are used to calculate the amount of substances present at equilibrium. Typically, an ICE table is constructed, and the symbol x is used to identify one of the changes in concentration that occurs in establishing equilibrium. Then, all the other concentration changes are related to x, the appropriate terms are substituted into the equilibrium constant expression, and the equation solved for x.

36 Chapter Fourteen 36 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example Suppose that in the reaction of Example 14.12, the initial amounts are mol H 2 and mol I 2. What will be the amounts of reactants and products when equilibrium is attained? Example Carbon monoxide and chlorine react to form phosgene, COCl 2, which is used in the manufacture of pesticides, herbicides, and plastics: COCl 2 (g) CO(g) + Cl 2 (g) K c = 1.2 x 10 3 at 668 K How much of each substance, in moles, will there be at equilibrium in a reaction mixture that initially has mol CO, mol Cl 2, and mol COCl 2 in a 10.0-L flask?

37 Chapter Fourteen 37 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example A sample of phosgene, COCl 2 (g), is introduced into a constant-volume vessel at 395 °C and observed to exert an initial pressure of atm. When equilibrium is established for the reaction CO(g) + Cl 2 (g) COCl 2 (g) K p = 22.5 what will be the partial pressure of each gas and the total gas pressure?

38 Chapter Fourteen 38 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Cumulative Example A mixture of H 2 S(g) and CH 4 (g) in the mole ratio 2:1 was brought to equilibrium at 700 °C and a total pressure of 1.00 atm. The equilibrium mixture was analyzed and found to contain 9.54 x 10 –3 mol H 2 S. The CS 2 present at equilibrium was converted, first to H 2 SO 4 and then to BaSO 4, with 1.42 x 10 –3 mol BaSO 4 being obtained. Use these data to determine K p at 700 °C for the reaction 2 H 2 S(g) + CH 4 (g) CS 2 (g) + 4 H 2 (g)


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