15.1 The Concept of Equilibrium

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1 Lecture Presentation Chapter 15 Chemical Yonsei University 15.1 The Concept of N 2 O 4 (g) 2NO 2 (g) 2 Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

2 The Concept of As a system approaches equilibrium, both the forward and reverse reactions are occurring. Once equilibrium is achieved, the amount of each reactant and product remains constant. 3 4 Depicting Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow: N 2 O 4 (g) 2NO 2 (g) ----The double arrow implies that the process is dynamic., the reaction is reversible At equilibrium K eq = Rate f = Rate r k f [N 2 O 4 ] = k r [NO 2 ] 2 The ratio of the rate constants is a constant at that temperature, and the expression becomes k f [NO 2 ] = 2 k r [N 2 O 4 ]

3 The Constant the law of mass action Consider the generalized reaction aa + bb cc + dd The equilibrium constant expression for this reaction would be 5 K c = [C]c [D] d [A] a [B] b K c, equilibrium constant K c : equilibrium constant. The subscript c indicates that molar concentrations were used to evaluate the constant. The equilibrium expression depends on stoichiometry. It does not depend on the reaction mechanism. The value of K c varies with temperature. We generally omit the units of the equilibrium constant. 6

4 Sample Exercise 15.1 Writing -Constant Expressions Write the equilibrium expression for K c for the following reactions: (a) (b) (c) Solution (a) (b) (c) Practice Exercise Write the equilibrium-constant expression K c for (a), (b). Answer: (a) (b) 7 K p, Constant in Terms of Pressure Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written K p = (P C c ) (P Dd ) (P Aa ) (P Bb ) From the ideal-gas law we know that PV = nrt P = n RT V K p = K c (RT) n 8 n = (moles of gaseous product) (moles of gaseous reactant)

5 Constants and Units The constants are defined in terms of activities rather than concentrations or partial pressures. The activity of a substance in an ideal mixture is the ratio of the concentration or pressure of the substance either to a reference concentration (1 M) or a reference pressure (1 atm). Activities have no units. For pure solids and pure liquids, the activities equal 1. real systems Here activities also have no units. The activities are not exactly equal to concentration but we will ignore the differences. Thermodynamic equilibrium constants derived from these activities also have no units. 9 Sample Exercise 15.2 Converting between K c and K p For the Haber process, K c = 9.60 at 300 C. Calculate K p for this reaction at this temperature. Solution Practice Exercise For the equilibrium, Kc is at 1000 K. Calculate the value for K p. Answer:

6 15.3 Understanding and Working with Constants Can Be Reached from Either Direction : It doesn t matter whether we start with N 2 and H 2 or whether we start with NH 3, we will have the same proportions of all three substances at equilibrium. 11 What Does the Value of K Mean? If K>>1, the reaction is product-favored; product predominates at equilibrium. If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium. 12

7 product favored reaction Meaning of K reactant favored reaction 2SO O 2 SO K HSO H HSO K SO O 2 SO K SO2 O2 2 SO3 K HSO 2 3 H HSO3 K Manipulating Constants Reversing an : Reciprocal Rule N 2 O 4 (g) 2NO 2 (g) K c = [NO 2 ] 2 [N 2 O 4 ] = at 100 C 2NO 2 (g) N 2 O 4 (g) K c = [N 2 O 4 ] [NO 2 ] 2 = 4.72 at 100 C 14

8 N 2 O 4 (g) Manipulating Constants Multiples of an Reaction : Coefficient Rule 2NO 2 (g) K c = [NO 2 ] 2 [N 2 O 4 ] = at 100 C 2N 2 O 4 (g) 4NO 2 (g) K c = [NO 2 ] 4 [N 2 O 4 ] 2 = (0.212) 2 at 100 C Manipulating Constants The Sum of Reactions : Rule of multiple equilibria 15 K=K 1 K 2 K 3. Sample Exercise 15.4 Evaluating an Constant When an Equation is Reversed For the reaction that is run at 25 C, K c = Use this information to write the equilibrium-constant expression and calculate the equilibrium constant for the reaction Solution Practice Exercise For reaction? Answer: , K p = at 300 C.What is the value of K p for the reverse 16

9 Sample Exercise 15.5 Combining Expressions Given the reactions determine the value of K c for the reaction Solution Practice Exercise Given that, at 700 K, Kp = 54.0 for the reaction and K p = for the reaction, determine the value of K p for the reaction 6 HI(g) + N 2 (g) at 700 K. Answer: Heterogeneous The concentrations of solids and liquids do not appear in the equilibrium expression. PbCl 2 (s) K c = [Pb 2+ ] [Cl ] 2 Pb 2+ (aq) + 2Cl (aq) Note: Although the concentrations of these species are not included in the equilibrium expression, they do participate in the reaction and must be present for an equilibrium to be established! 18

10 CaCO 3 (s) CO 2 (g) + CaO(s) As long as some CaCO 3 or CaO remain in the system, the amount of CO 2 above the solid will remain the same. K c = [CO 2 ] and K p = P CO 2 19 Solvents in Solutions H 2 O(l) + CO 2 3 (aq) OH (aq) + HCO 3 (aq) K c = [OH ][HCO 3 ] / [CO 2 3 ] Here the concentration of water is essentially constant and we can think of it as a pure liquid. 20

11 Sample Exercise 15.6 Writing -Constant Expressions for Heterogeneous Reactions Write the equilibrium-constant expression K c for (a) (b) Solution (a). (b). Practice Exercise Write the following equilibrium-constant expressions: (a) (b) Answer: (a) (b) Calculating Constants A closed system initially containing x 10 3 M H 2 and x 10 3 M I 2 at 448 C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10 3 M. Calculate K c at 448 C for the reaction taking place, which is H 2 (g) + I 2 (s) 2HI(g) 22

12 What Do We Know? H 2 (g) + I 2 (s) 2HI(g) [H 2 ], M [I 2 ], M [HI], M Initially x x Change At equilibrium 1.87 x [HI] Increases by 1.87 x 10 3 M H 2 (g) + I 2 (s) 2HI(g) [H 2 ], M [I 2 ], M [HI], M Initially x x Change At equilibrium x x

13 Stoichiometry tells us [H 2 ] and [I 2 ] decrease by half as much. H 2 (g) + I 2 (s) 2HI(g) [H 2 ], M [I 2 ], M [HI], M Initially x x Change 9.35 x x x 10 3 At equilibrium 1.87 x We can now calculate the equilibrium concentrations of all three compounds H 2 (g) + I 2 (s) 2HI(g) [H 2 ], M [I 2 ], M [HI], M Initially x x Change 9.35 x x x 10 3 At equilibrium 6.5 x x x

14 and, therefore, the equilibrium constant: K c = = [HI] 2 [H 2 ] [I 2 ] (1.87 x 10 3 ) 2 (6.5 x 10 5 )(1.065 x 10 3 ) = Practice Exercise Sulfur trioxide decomposes at high temperature in a sealed container: 2SO 3 (g) 2SO 2 (g)+o 2 (g). Initially, the vessel is charged at 1000 K with SO 3 (g) at a partial pressure of atm. At equilibrium the SO 3 partial pressure is atm. Calculate the value of K p at 1000 K. Answer:

15 15.6 Applications of Constants The Reaction Quotient (Q) Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium. To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. 29 If Q < K, there is too much reactant, and the equilibrium shifts to the right. If Q = K, the system is at equilibrium. If Q > K, there is too much product, and the equilibrium shifts to the left. 30

16 Sample Exercise Predicting the Direction of Approach to At 448 C the equilibrium constant K c for the reaction is Predict in which direction the reaction proceeds to reach equilibrium if we start with mol of HI, mol of H 2, and of I 2 in a 2.00-L container. Solution The initial concentrations are Because Q c < K c, the concentration of HI must increase and the concentrations of H 2 and I 2 must decrease to reach equilibrium; the reaction as written proceeds left to right to attain equilibrium. Practice Exercise At 1000 K the value of K p for the reaction is Calculate the value for Q p, And predict the direction in which the reaction proceeds toward equilibrium if the initial partial pressures are P SO3 = 0.16 atm; P SO2 = 0.41 atm; P O2 = 2.5 atm. Answer: Q p = 16; Q p > K p, and so the reaction will proceed from right to left, forming more SO Applications of K Calculating Concentrations The same steps used to calculate equilibrium constants are used to calculate equilibrium concentrations. Generally, we do not have a number for the change in concentration. Therefore, we need to assume that x mol/l of a species is produced (or used). The equilibrium concentrations are given as algebraic expressions. 32

17 Small K Value Ascorbic acid is a weak acid with K = The acid dissociates into hydrogen ion and ascorbate ion as follows: HC 6 H 7 O 6 (aq) H + (aq) + C 6 H 7 O 6 (aq) Starting with 0.50 M HC 6 H 7 O 6, what will be the equilibrium concentrations of H + and C 6 H 7 O 6 ions? Constructing the ICEA table, HC 6 H 7 O 6 (aq) H + (aq) + C 6 H 7 O 6 (aq) I 0.50 M 0 M 0 M C x +x +x E 0.50x +x +x A 0.50 x x 33 Cont. (Small K Value) K = = + - H C 6H7O6 [HC H O ] xx 0.50 x= x = x = H = C 6H7O 6 = M 34

18 5% Rule x 100% 5% M M 100% = 1.3% 5% 0.50 M 35 Slightly Soluble Solids Small Value of K To determine the concentration of calcium and fluoride ions in a saturated solution of calcium fluoride, let x = amount of CaF 2 that dissolves: CaF 2 (s) Ca 2+ (aq) + 2 F (aq) I 0 0 C +x +2x E x 2x K=[Ca 2+ ][F ] 2 ==

19 Slightly Soluble Solids Small Value of K x 2x 3 12 K = [Ca ][F ] x = x M x = Ca [F ] 2x = M 2 37 Large Value of K Consider the hydrogen iodide equilibrium: H 2 (g) + I 2 (g) 2 HI(g) K = at 298 K. If a reaction was started with 1.00 M hydrogen and 0.30 M iodine, calculate the equilibrium concentrations of H 2, I 2 and HI. Since K is large, the reaction is largely product favored. Let x represent the small amount that does not react to form product. H 2 + I 2 2HI 38 I C x x x E 0.70+x x x A 0.70 x 0.60

20 Large Value of K = 2 K = H 2 I HI x -4 x = M = I H = 0.70 M HI = 0.60 M 2 39 Intermediate Value of K Consider the acid dissociation: HNO 2 (aq) H + (aq)+ NO 2 (aq) which has an intermediate K = Calculate the equilibrium concentration of H + and NO 2- in a 0.50 M solution of HNO 2 Since the value of K is intermediate, no approximation can be made. Instead use the quadratic formula. HNO 2 (aq) H + (aq) + NO 2 (aq) I C x x x E 0.50-x x x 40

21 Intermediate Value of K xx 0.50-x 2 b b ac 4 x = 2a H NO K = = HNO F M or = x x M [H + ] =[NO 2- ] = M cf) approximation 5% rule 41 Sample Exercise Calculating Concentration from Initial Concentrations A L flask is filled with mol of H 2 (g) and mol of I 2 (g) at 447 C. The value of the equilibrium constant K c for the reaction at 448 C is 50.5.What are the equilibrium concentrations of H 2, I 2, and HI in moles per liter? Solution 42

22 15.7 Le Châtelier s Principle If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance. 43 Le Châtelier s Principle (Comprression) (Expansion) 44

23 The Haber Process N 2 + 3H 2 2NH 3 H = kj/mol. cf) ammonia-based fertilizers 45 Change in Reactant or Product Concentration The Haber Process If H 2 is added to the system, N 2 will be consumed and the two reagents will form more NH 3. 46

24 The Haber Process This apparatus helps push the equilibrium to the right by removing the ammonia (NH 3 ) from the system as a liquid. 47 Effects of Volume and Pressure Changes If the equilibrium involves gaseous products or reactants, the concentration of these species will be changed if we change the volume of the container. For example, if we decrease the volume of the container, the partial pressures of each gaseous species will increase. Le Châtelier s principle predicts that if pressure is increased, the system will shift to counteract the increase. That is, the system shifts to remove gases and decrease pressure. An increase in pressure favors the direction that has fewer moles of gas. cf) n r =n p ; inert gas addition 48

25 Sample Exercise Using Le Châtelier s Principle to Predict Shifts in Consider the equilibrium In which direction will the equilibrium shift when (a) N 2 O 4 is added, (b) NO 2 is removed, (c) the pressure is increased by addition of N 2 (g), (d) the volume is increased, (e) the temperature is decreased? Solution (a) The system will adjust to decrease the concentration of the added N 2 O 4, so the equilibrium shifts to the right, in the direction of product. (b) The system will adjust to the removal of NO 2 by shifting to the side that produces more NO 2 ; thus, the equilibrium shifts to the right. (c) Adding N 2 will increase the total pressure of the system, but N 2 is not involved in the reaction. The partial pressures of NO 2 and N 2 O 4 are therefore unchanged, and there is no shift in the position of the equilibrium. (d) If the volume is increased, the system will shift in the direction that occupies a larger volume (more gas molecules); thus, the equilibrium shifts to the right. (e) The reaction is endothermic, so we can imagine heat as a reagent on the reactant side of the equation. Decreasing the temperature will shift the equilibrium in the direction that produces heat, so the equilibrium shifts to the left, toward the formation of more N 2 O 4. Note that only this last change also affects the value of the equilibrium constant, K. 49 Effect of Temperature Changes If we increase the temperature of an endothermic reaction, K increases. If we increase the temperature of an exothermic reaction, K decreases. 50

26 Changes in Temperature Co(H 2 O) 6 2+ (aq) + 4Cl(aq) CoCl 4 (aq) + 6H 2 O(l) 51 Sample Exercise Predicting the Effect of Temperature on K (a) Using the standard heat of formation data in Appendix C, determine the standard enthalpy change for the reaction (b) Determine how the equilibrium constant for this reaction should change with temperature. Solution (a) At 25 C, for NH 3 (g) is kj/mol. The values for H 2 (g) and N 2 (g) are zero by definition because the enthalpies of formation of the elements in their normal states at 25 C are defined as zero. (See section 5.7). Because 2 mol of NH 3 is formed, the total enthalpy change is (2 mol)(46.19 kj/mol) 0 = kj (b) Because the reaction in the forward direction is exothermic, we can consider heat a product of the reaction. An increase in temperature causes the reaction to shift in the direction of less NH 3 and more N 2 and H 2. This effect is seen in the values for K p presented in Table Notice that K p changes markedly with changes in temperature and that it is larger at lower temperatures. 52

27 Catalysts Catalysts increase the rate of both the forward and reverse reactions. 53 When one uses a catalyst, equilibrium is achieved faster, but the equilibrium composition remains unaltered. Problems 12, 26, 40, 50, 64, 84, 94, 98 54

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