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‹#› Chapter 18 Chemical Equilibrium. solubility/chemical-stalagmite.html In this experiment sodium acetate.

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Presentation on theme: "‹#› Chapter 18 Chemical Equilibrium. solubility/chemical-stalagmite.html In this experiment sodium acetate."— Presentation transcript:

1 ‹#› Chapter 18 Chemical Equilibrium

2 http://www.chem-toddler.com/solutions-and- solubility/chemical-stalagmite.html In this experiment sodium acetate trihydrate (CH 3 COONa 3H 2 O is heated ) in the flask. Salt melting point is above room temperature, at 54°C/130°F. When the entire quantity of the salt used melted, it gave a supersaturated solution of sodium acetate in its own crystal water that was left to cool down spontaneously. A supersaturated solution contains more of the dissolved material than could be dissolved by the solvent under normal circumstances.

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5 18: The Concept of Equilibrium Typo: N 2 O 4 is a gas

6 As the N 2 O 4 is heated it begins to decompose: N 2 O 4 (g)  2NO 2 (g) A mixture of N 2 O 4 (initially present) and NO 2 (initially formed) appears light brown. When enough NO 2 is formed, it can react to form N 2 O 4 : 2NO 2 (g)  N 2 O 4 (g). At equilibrium, the forward reaction and reverse reaction are proceeding at a equal rates. The double arrow implies that the reaction proceeds in both the forward and reverse directions and is dynamic. The Concept of Equilibrium The Concept of Equilibrium

7 ‹#› The Concept of Equilibrium The Concept of Equilibrium Properties of an Equilibrium Equilibrium systems are Dynamic.Equilibrium systems are Dynamic. ReversibleReversible Can be approached from either directionCan be approached from either direction In a closed container a reversible reaction does not proceed completely.In a closed container a reversible reaction does not proceed completely.

8 Consider Forward reaction: A  B Rate = k f [A] Reverse reaction: B  A Rate = k r [B] At equilibrium k f [A] = k r [B]. Therefore, For an equilibrium we write As the reaction progresses [A] decreases to a constant, [B] increases from zero to a constant. When [A] and [B] are constant, equilibrium is achieved. The Concept of Equilibrium

9 The Concept of Equilibrium The Concept of Equilibrium Alternatively: k f [A] decreases to a constant, k r [B] increases from zero to a constant. When k f [A] = k r [B] equilibrium is achieved.

10 N 2 (g) + 3H 2 (g) → NH 3 (g) 2NH 3 (g)→ N 2 (g) + 3H 2 (g) The Equilibrium Constant The Equilibrium Constant No matter the starting composition of reactants and products, the same ratio of concentrations is achieved at equilibrium.

11 The Equilibrium Constant The Equilibrium Constant K eq is based on the molarities of reactants and products at equilibrium. We generally omit the units of the equilibrium constant. Note that the equilibrium constant expression has products over reactants.

12 For a general reaction aA + bB ↔ cC + dD the equilibrium constant expression is K eq = [C] c [D] d K eq = [C] c [D] d [A] a [B] b [A] a [B] b where K eq is the equilibrium constant. The Equilibrium Constant The Equilibrium Constant

13 The Magnitude of Equilibrium Constants The equilibrium constant, K, is the ratio of products to reactants. Therefore, the larger K the more products are present at equilibrium. Conversely, the smaller K the more reactants are present at equilibrium.

14 If K >> 1: Equilibrium lies to the right; products dominate at equilibrium. If K << 1: Equilibrium lies to the left; then reactants dominate at equilibrium. The Equilibrium Constant The Equilibrium Constant The Magnitude of Equilibrium Constants

15 What about the reverse reaction ? This reaction is strongly reactant-favored. Conc. of products is much less than that o f reactants at equilibrium. Meaning of K sp : AgCl rxn K sp = K << 1 Ag + (aq) + Cl - (aq) AgCl(s) K rev = K sp -1 = 5.6x10 9. It is strongly product-favored. [Ag + ] [Cl - ]= 1.8 x 10 -10 AgCl(s) Ag + (aq) + Cl - (aq)

16 The direction of the chemical equation and K An equilibrium can be approached from any direction. has The Equilibrium Constant The Equilibrium Constant

17 The direction of the chemical equation and K However, for reverse reaction has The equilibrium constant for a reaction in one direction is the reciprocal of the equilibrium constant of the reaction in the opposite direction. The Equilibrium Constant The Equilibrium Constant

18 Homogeneous Equilibria When all reactants and products are in one phase, the equilibrium is homogeneous. Homogeneous Equilibria If one or more reactants or products are in a different phase, the equilibrium is heterogeneous.

19 Consider: experimentally, the amount of CO 2 does not seem to depend on the amounts of CaO and CaCO 3. Why? The concentration of a solid or pure liquid is its density divided by molar mass. It can be assumed that the concentration of a solid or liquid remains constant, and therefore, is not included in the Law of Mass Action (equilibrium expression). The Equilibrium Constant The Equilibrium Constant

20 Heterogeneous Equilibria

21 Calculating Equilibrium Constants Calculating Equilibrium Constants Usually, the initial concentration of products is zero. (This is not always the case.) When in doubt, assign the change in concentration a variable.

22 Calculating Equilibrium Constants Calculating Equilibrium Constants Proceed as follows: Tabulate initial and equilibrium concentrations (or partial pressures) given. If an initial and equilibrium concentration is given for a species, calculate the change in concentration. Use stoichiometry on the change in concentration line only to calculate the changes in concentration of all species. Deduce the equilibrium concentrations of all species.

23 Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K. Solution 1. Set up a table of concentrations: [NOCl] [NO [Cl 2 ] [NOCl] [NO [Cl 2 ] 2 NOCl(g) 2 NO(g) + Cl 2 (g) Initial2.0000 Equilibrium0.66 Calculating Equilibrium Constants Calculating Equilibrium Constants

24 2 NOCl(g) 2 NO(g) + Cl 2 (g) 2 NOCl(g) 2 NO(g) + Cl 2 (g) [NOCl][NO][Cl 2 ] Initial 2.0000 Change -0.66+0.66 +0.33 Equilibrium0.66 K eq = (0.66) 2 (0.33) (1.34) 2 = 0.080 Calculating Equilibrium Constants Calculating Equilibrium Constants

25 Predicting the Direction of Reaction We define Q, the reaction quotient, for a general reaction as where [A], [B], [P], and [Q] are molarities at any time. Q = K only at equilibrium. Applications of K Constants Applications of K Constants

26 Applications of Equilibrium Constants Predicting the Direction of Reaction If Q > K then the reverse reaction must occur to reach equilibrium (i.e., products are consumed, reactants are formed, the numerator in the equilibrium constant expression decreases and Q decreases until it equals K). If Q < K then the forward reaction must occur to reach equilibrium.

27 Le Châtelier’s Principle Le Châtelier’s Principle Le Châtelier ’ s Principle: if a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance.

28 Consider the production of ammonia As the pressure increases, the amount of ammonia present at equilibrium increases. As the temperature decreases, the amount of ammonia at equilibrium increases. Can this be predicted? Le Châtelier’s Principle Le Châtelier’s Principle

29 Change in Reactant or Product Concentrations Consider the Haber process If H 2 is added while the system is at equilibrium, the system must respond to counteract the added H 2 (by Le Châtelier). That is, the system must consume the H 2 and produce products until a new equilibrium is established. Therefore, [H 2 ] and [N 2 ] will decrease and [NH 3 ] increases. Le Châtelier’s Principle Le Châtelier’s Principle

30 Effects of Volume and Pressure As volume is decreased pressure increases.As volume is decreased pressure increases. Le Châtelier ’ s Principle: if pressure is increased the system will shift to counteract the increase.Le Châtelier ’ s Principle: if pressure is increased the system will shift to counteract the increase. The shift will be in the direction where there are fewer moles of gas.The shift will be in the direction where there are fewer moles of gas.

31 Le Châtelier’s Principle Le Châtelier’s Principle Effects of Volume and Pressure That is, the system shifts to remove gases and decrease pressure. An increase in pressure favors the direction that has fewer moles of gas. In a reaction with the same number of product and reactant moles of gas, pressure has no effect. Consider :

32 Le Châtelier’s Principle Le Châtelier’s Principle Effects of Volume and Pressure The system moves to reduce the number moles of gas (i.e. the forward reaction is favored). A new equilibrium is established in which the mixture is lighter because colorless N 2 O 4 is favored.

33 Le Châtelier’s Principle Le Châtelier’s Principle Effect of Temperature Changes The equilibrium constant is temperature dependent. Adding heat (i.e. heating the vessel) favors away from the increase: if  H > 0, adding heat favors the forward reaction, if  H < 0, adding heat favors the reverse reaction.

34 Le Châtelier’s Principle Le Châtelier’s Principle Effect of Temperature Changes Consider for which ∆H > 0. for which ∆H > 0. Co(H 2 O) 6 2+ is pale pink and CoCl 4 2- is blue. Co(H 2 O) 6 2+ is pale pink and CoCl 4 2- is blue. (typo: Cr should be Co in equation) (typo: Cr should be Co in equation)

35 Le Châtelier’s Principle Le Châtelier’s Principle

36 Co(H 2 O) 6 Cl 2 (aq) Co(H 2 O) 4 Cl 2 (aq) + 2 H 2 O Pink to blue Co(H 2 O) 6 Cl 2 ---> Co(H 2 O) 4 Cl 2 + 2 H 2 O Blue to pink Co(H 2 O) 4 Cl 2 + 2 H 2 O ---> Co(H 2 O) 6 Cl 2 Le Chatelier’s Principle Le Chatelier’s Principle Properties of an Equilibrium

37 Le Châtelier’s Principle Le Châtelier’s Principle The Effect of Catalysts A catalyst lowers the activation energy barrier for the reaction. Therefore, a catalyst will decrease the time taken to reach equilibrium. A catalyst does not effect the composition of the equilibrium mixture.

38 Le Châtelier’s Principle Le Châtelier’s Principle Chromates and dichromates are salts of chromic and dichromic acid. Salts have an intense yellow or orange color, respectively. When solid potassium dichromate (K 2 Cr 2 O 7 ) is dissolved in water the resulting solution is orange. The dichromate ion in aqueous solution is in equilibrium with the chromate ion, and this can be shown with the following equation:

39 Le Châtelier’s Principle Le Châtelier’s Principle This is a dynamic equilibrium and as such is sensitive to the acidity or basicity of the solution. Shifting the equilibrium with pH changes is a classic example of Le Chatelier’s principle at work. Yellow chromate and orange dichromate are in equilibrium with each other in aqueous solution. The more acidic the solution, the more the equilibrium is shifted to the left towards the dichromate ion. As hydrochloric acid is added to the chromate solution, the yellow color turns to orange.

40 Increasing the hydrogen ion concentration is shifting the equilibrium to the left in accordance with Le Chatelier's principle, where we expect the reaction to try to remove some of the H + we have added by reacting with the CrO 4 2-, and yielding more Cr 2 O 7 2- which we observe as color change.

41 When sodium hydroxide is added to the dichromate solution, the orange color turns back to yellow, hydroxide ions react with hydrogen ions forming water, driving the equilibrium to the right (OH - removes H + ions by neutralizing them and the system acts to counteract the change) and further shifting the color. We can observe that the addition of hydroxide ions promotes the conversion of dichromate to chromate.

42 http://www.youtube.com/watch?v =wQ1VSfpZcTY&NR=1

43 http://www.chem-toddler.com/chemical- equilibrium/chromatedichromate.html


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