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Chapter Fourteen Chemical Equilibrium.

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Presentation on theme: "Chapter Fourteen Chemical Equilibrium."— Presentation transcript:

1 Chapter Fourteen Chemical Equilibrium

2 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 Dynamic Equilibrium Illustrated
NaCl containing radioactive Na+ is added to a saturated NaCl solution. After a time, the solution contains radioactive Na+ … NaCl dissolves and recrystallizes continuously. … and the added salt now contains some stable Na+.

4 Concentration vs. Time Beginning with 1 M H2 and 1 M I2, the [HI] increases and both [H2] and [I2] decrease. If we begin with only 1 M HI, the [HI] decreases and both [H2] and [I2] increase. Beginning with 1 M each of H2, I2, and HI, the [HI] increases and both [H2] and [I2] decrease.

5 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 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 [G]g[H]h Kc = [A]a[B]b Products in numerator. Reactants in denominator.

7 The Equilibrium Constant
The equilibrium constant is constant regardless of the initial concentrations of reactants and products. This constant is denoted by the symbol Kc 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 Example 14.1 If the equilibrium concentrations of COCl2 and Cl2 are the same at 395 °C, find the equilibrium concentration of CO in the reaction: CO(g) + Cl2(g) COCl2(g) Kc = 1.2 x 103 at 395 °C

9 The Condition of Equilibrium
The kinetics view: Kc = (forward rate)/(reverse rate) = kf/kr The thermodynamics view: The equilibrium constant can be related to other fundamental thermodynamic properties and is called the thermodynamic equilibrium constant, Keq. The thermodynamic equilibrium constant expression uses dimensionless quantities known as activities in place of molar concentrations.

10 Modifying the Chemical Equation
Consider the reaction: 2 NO(g) + O2(g) NO2(g) [NO2]2 Kc = ––––––––– = 4.67 x 1013 (at 298 K) [NO]2 [O2] Now consider the reaction: 2 NO2(g) NO(g) + O2(g) What will be the equilibrium constant K'c for the new reaction? [NO]2 [O2] K'c = ––––––––– = ––––––––––– = [NO2] [NO2]2 ––––––––– [NO]2 [O2] –– = ––––––––– = 2.14 x 10–14 Kc x 1013

11 Modifying the Chemical Equation (cont’d)
Consider the reaction: 2 NO(g) + O2(g) NO2(g) [NO2]2 Kc = ––––––––– = 4.67 x 1013 (at 298 K) [NO]2 [O2] Now consider the reaction: NO2(g) NO(g) + ½ O2(g) What will be the equilibrium constant K"c for the new reaction? [NO] [O2]1/ /2 K"c = ––––––––– = ––– [NO2] Kc = x 10–14 = x 10–7

12 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 Kc 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 Kc.

13 The equilibrium constant for the reaction:
Example 14.2 The equilibrium constant for the reaction: ½ H2(g) + ½ I2(g) HI(g) at 718 K is 7.07. (a) What is the value of Kc at 718 K for the reaction HI(g) ½ H2(g) + ½ I2(g) (b) What is the value of Kc at 718 K for the reaction H2(g) + I2(g) HI(g)

14 The Equilibrium Constant for an Overall Reaction
Suppose we need: N2O(g) + 3/2 O2(g) NO2(g) Kc(1) = ?? and we’re given: N2O(g) + ½ O2(g) NO(g) Kc(2) = 1.7 x 10–13 2 NO(g) + O2(g) NO2(g) Kc(3) = 4.67 x 1013 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 …)

15 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, Kp, is used. (PG)g(PH)h Kp = (PA)a(PB)b Kc and Kp are related by: Kp = Kc (RT)Δn(gas) where Dn(gas) is the change in number of moles of gas as the reaction occurs in the forward direction. Dn(gas) = mol gaseous products – mol gaseous reactants

16 Example 14.3 Consider the equilibrium between dinitrogen tetroxide and nitrogen dioxide: N2O4(g) NO2(g) Kp = at 319 K (a) What is the value of Kc for this reaction? (b) What is the value of Kp for the reaction 2 NO2(g) N2O4(g)? (c) If the equilibrium partial pressure of NO2(g) is atm, what is the equilibrium partial pressure of N2O4(g)?

17 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. Example: CaCO3(s) CaO(s) + CO2(g) [CaO] [CO2] Kc = –––––––––– [CaCO3] Kc = [CO2]

18 C(s) + H2O(g) CO(g) + H2(g)
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) + H2O(g) CO(g) + H2(g) Write the equilibrium constant expression for Kc for this reaction.

19 Equilibrium Constants: When Do We Need Them and When Do We Not?
A very large numerical value of Kc or Kp signifies that a reaction goes (essentially) to completion. A very small numerical value of Kc or Kp 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 Example 14.5 Is the reaction CaO(s) + CO2(g) CaCO3(s) likely to occur to any appreciable extent at 298 K?

21 The Reaction Quotient, Q
For nonequilibrium conditions, the expression having the same form as Kc or Kp is called the reaction quotient, Qc or Qp. 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 Qc to that of Kc.

22 The Reaction Quotient, Q
When Q is smaller than K, the denominator of Q is too big; we have “too much reactants.” When Q = K, equilibrium has been reached. When Q is larger than K, the numerator of Q is too big; we have “too much products.”

23 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 Changing the Amounts of Reacting Species
At equilibrium, Q = Kc. If the concentration of one of the reactants is increased, the denominator of the reaction quotient increases. Q is now less than Kc. This condition is only temporary, however, because the concentrations of all species must change in such a way so as to make Q = Kc again. In order to do this, the concentrations of the products increase; the equilibrium is shifted to the right.

25 … 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. When acetic acid (a reactant) is added to the equilibrium mixture …

26 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. CH3(CH2)6CH2OH(soln) + CH3COOH(soln) CH3(CH2)6CH2OCOCH3(soln) + H2O(soln) H+

27 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 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: H2(g) + I2(g) HI equilibrium is unaffected by pressure changes.

29 When pressure is increased …
Initial When pressure is increased … … to give one molecule of N2O4, reducing the pressure increase. … two molecules of NO2 combine …

30 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 Example 14.8 An equilibrium mixture of O2(g), SO2(g), and SO3(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 SO3(g) SO2(g) + O2(g) Example 14.9 Is the amount of NO(g) formed from given amounts of N2(g) and O2(g), N2(g) + O2(g) NO(g) ΔH° = kJ greater at high or low temperatures?

32 Example 14.10 A Conceptual Example
Flask A, pictured below, initially contains an equilibrium mixture of the reactants and products of the reaction CO(g) + H2O(g) CO2(g) + H2(g) ΔH = –41 kJ; Kc = 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, H2, CO2, and H2O 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 CO2; (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 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 H2(g) + I2(g) 2 HI(g) Kc = 54.3 at 698 K
Example 14.11 In a 10.0-L vessel at 1000 K, mol SO2 and mol O2 react to form mol SO3 at equilibrium. What is Kc, at 1000 K, for the reaction that is shown here? 2 SO2(g) + O2(g) SO3(g) Example 14.12 Consider the reaction H2(g) + I2(g) HI(g) Kc = 54.3 at 698 K If we start with mol I2(g) and mol H2(g) in a 5.25-L vessel at 698 K, how many moles of each gas will be present at equilibrium?

35 Calculating Equilibrium Quantities from Kc and Kp 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 Example 14.13 Suppose that in the reaction of Example 14.12, the initial amounts are mol H2 and mol I2. What will be the amounts of reactants and products when equilibrium is attained? Example 14.14 Carbon monoxide and chlorine react to form phosgene, COCl2, which is used in the manufacture of pesticides, herbicides, and plastics: COCl2(g) CO(g) + Cl2(g) Kc = 1.2 x 103 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 Cl2, and mol COCl2 in a 10.0-L flask?

37 Example 14.15 A sample of phosgene, COCl2(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) + Cl2(g) COCl2(g) Kp = 22.5 what will be the partial pressure of each gas and the total gas pressure?

38 Cumulative Example A mixture of H2S(g) and CH4(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 H2S. The CS2 present at equilibrium was converted, first to H2SO4 and then to BaSO4, with 1.42 x 10–3 mol BaSO4 being obtained. Use these data to determine Kp at 700 °C for the reaction 2 H2S(g) + CH4(g) CS2(g) + 4 H2(g)

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