Chemical Kinetics and

Transcription

1 Chemical Kinetics and Equilibrium Part 2: Chemical Equilibrium David A. Katz Department of Chemistry Pima Community College Tucson, AZ USA

2 The Concept of Equilibrium Kinetics applies to the speed of a reaction, the concentration of product appearing (or of reactant disappearing) per unit time Equilibrium i applies to the extent of a reaction, the concentration of product that has appeared given unlimited time, or when no further macroscopic change occurs. Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

3 The Cobalt(II) Chloride Equilibrium Co(H 2 O) 6 2+ (aq) +4 Cl (aq) CoCl 4 (aq) +6 H 2 O (l)

4 The equilibrium system N 2 O 4(g) 2 NO 2 (g) Initially, N 2 O 4 (colorless) is added to the reaction container As the system moves toward an equilibrium condition, the color changes as more NO 2 (red-brown) is formed When equilibrium is reached, the numbers of N 2 O 4 and NO 2 molecules is constant and the color remains unchanged

5 The Concept of Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate.

6 A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant.

7 Depicting Equilibrium A system at equilibrium is dynamic on the molecular level, l that t is, both the forward and the reverse reactions are still taking place at the same rate. rate forward = rate backward No net change is observed because changes in one direction are balanced by changes in the other. To show an equilibrium system, we write its equation with a double arrow N 2 O 4(g) 2NO 2(g)

8 Chemical Equilibrium Iron(III) and thiocyanate Fe(H 2 O) 3+ 6 (aq) + SCN - Fe(SCN)(H 2 O) 2+ (aq) 5 (aq) + H 2 O (l) +

9 The Equilibrium Constant

10 The Equilibrium Constant Forward reaction: Rate law: Reverse reaction: Rate law: N 2 O 4 (g) 2 NO 2 (g) Rate = k f [N 2 O 4 ] 2 NO 2 (g) N 2 O 4 (g) Rate = k r [NO 2 ] 2

11 The Equilibrium Constant At equilibrium Rate f = Rate r Substitute t the rate equations k f [N 2 O 4 ] = k r [NO 2 ] 2 Rewriting this, it becomes k f [NO 2 ] 2 = k [N 2 2O r 4 ]

12 The Equilibrium Constant The ratio of the rate constants is a constant at that temperature, so the constants are combined into a single constant, and the expression becomes k f [NO = 2 ] 2 K eq = kr [N 2 O 4 ]

13 The Equilibrium Constant To generalize this expression, consider the reaction aa + bb cc + dd The equilibrium expression for this reaction would be K eq = [C]c [D] d [A] a [B] b This equation is known as the Law of Mass Action

14 The Equilibrium Constant For any reaction of the form: aa+bb + cc +dd [C]c [D] d The equilibrium expression is K eq = [A]a [B] b The values of a, b, c, and d are those of the coefficients in the balanced chemical equation. Note that this is equilibrium, not kinetics. Equilibrium is a State Function, that is, the value of K c depends on the concentrations of the reactants and the products.

15 The Equilibrium Constant Because pressure is proportional p to concentration for gases in a closed system, the equilibrium expression for 4 gases, A, B, C, and D, can also be written K = (P C) c (P D ) d p (P A ) a (P B ) b

16 Relationship between K and K c p From the ideal gas law we know that PV = nrt Rearranging to solve for P, we get n P = RT V Note that n/v is moles/liter or Molarity, so P =MRT

17 Relationship between K and K c p Substituting for P into the expression for K p for each substance, the relationship between K c and K p becomes K p = (M C RT) c (M D RT) d (M A RT) a (M B RT) b Factor out the RT terms to get K n p = K c (RT) Where: n = (moles of gaseous s product) (moles of gaseous s reactant)

18 Equilibrium Can Be Reached from Either Direction Examining the data in the table, above, the equilibrium ratio of [NO 2 ] 2 to [N 2 O 4 ] remains constant at this temperature no matter what the initial concentrations of NO 2 and N 2 O 4 are. Note: There must be sufficient quantities of compounds to reach an equilibrium condition

19 Equilibrium Can Be Reached from Either Direction This is the data from the last two trials from the table on the previous slide.

20 Equilibrium Can Be Reached from Either Direction It does not 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 at the specified temperature.

21 What Does the Value of K Mean? If K >> 1, the reaction is product-favored; d product predominates at equilibrium.

22 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.

23 Manipulating Equilibrium Constants t The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. N 2 O 4 (g) 2 NO 2 (g) K c = [NO 2 ] 2 [N 2 O 4 ] = at 100 C 2NO 2 (g) N 2O 4 (g) [N 2O 4 K c = 2 4] [NO 2 ] 2 = = 4.72 at 100 C

24 Manipulating Equilibrium Constants t The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number. N 2 O 4 (g) [NO 2] 2NO 2 2 (g) K c = = at 100 C [N 2 O 4 ] 2 N 2 O 4 (g) 4 NO 2 (g) K c = [NO 2 ] 4 [N 2 O 4 ] 2 = (0.212) 2 at 100 C

25 Manipulating Equilibrium Constants t The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps.

26 Heterogeneous Equilibria

27 The Concentrations of Solids and Liquids Are Considered to be Constant Both can be obtained by dividing the density Both can be obtained by dividing the density of the substance by its molar mass and both of these are constants at constant temperature.

28 The Concentrations of Solids and Liquids Are Considered to be Constant Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression PbCl 2 (s) Pb 2+ (aq) + 2 Cl (aq) K c = [Pb 2+ ] [Cl ] 2

29 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.

30 What Are the Equilibrium Expressions for These Equilibria? SnO 2(s) + 2 CO (g) Sn (s) + 2 CO 2 (g) CaCO 3 (s) CaO (s) + CO 2 (g) Zn (s) + Cu 2+ (aq) Cu (s) + Zn 2+ (aq)

31 Equilibrium i Calculations

32 Equilibrium Calculations A closed system initially containing x 10 3 M H and x 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(g) 2HI (g)

33 What Do We Know? (This is our initial data) [H 2 ], M [I 2 ], M [HI], M Initially x x Change At equilibrium 1.87 x 10-3 Note: We assume there is no HI present initially

34 Determine changes: [HI] Increases by 1.87 x 10-3 M [H 2 ], M [I 2 ], M [HI], M Initially x x Change x 10-3 At equilibrium 1.87 x 10-3

35 Stoichiometry tells us [H 2 2] and [I 2 ] decrease by half as much H 2 (g) + I 2 (g) 2 HI (g) [H 2 ], M [I 2 ], M [HI], M Initially x x Change x x x 10-3 At equilibrium 1.87 x 10-3

36 We can now calculate the equilibrium concentrations of all three compounds [H 2 ], M [I 2 ], M [HI], M Initially x x Change x x x 10-3 At equilibrium 6.5 x x x 10-3

37 and substitute the concentrations into the equilibrium constant expression [HI] 2 K c = [H2 ] [I 2 ] = (1.87 x 10-3 ) 2 (6.5 x 10-5 )(1.065 x 10-3 ) = 51 Note that the equilibrium value has no units

38 The Reaction Quotient (Q) The reaction quotient is used when we are given data (i.e., initial iti concentrations) ti for a reaction but we do not know if the system is in equilibrium To calculate Q, one substitutes the initial concentrations of reactants and products into the equilibrium expression. Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

39 If Q = K, the system is at equilibrium.

40 If Q > K, there is too much product and the equilibrium shifts to the left.

41 If Q < K, there is too much reactant, and the equilibrium shifts to the right.

42 Le Châtelier s Principle

43 Le Châtelier s Principle Henri Louis Le Châtelier ( ) In 1884, Le Chatelier stated: Any system in stable chemical equilibrium, subjected to the influence of an external cause which tends to change either its temperature or its condensation (pressure, concentration, number of molecules in unit volume), either as a whole or in some of its parts, can only undergo such internal modifications as would, if produced alone, bring about a change of temperature t or of condensation of opposite sign to that resulting from the external cause. In 1888, he restated this as: Every change of one of the factors of an equilibrium occasions a rearrangement of the system in such a direction that the factor in question experiences a change in a sense opposite to the original change.

44 Le Châtelier s Principle Our modern statement is: If a system at equilibrium is disturbed by a change in temperature, t 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.

45 Le Châtelier s Principle Effect of a change in T: change in K therefore change in P or concentrations at equilibrium Change in concentration (Add or take away reactant or product): K does not change Reaction adjusts to new equilibrium position Use a catalyst: reaction comes more quickly to equilibrium. K not changed.

46 The Effect of Changes in Temperature Co(H 2 O) 6 2+ (aq) + 4 Cl - (aq) + HEAT CoCl 4 2- (aq) + 6 H 2 O (l)

47 Predicting the Effect of a Change in Concentration on the Equilibrium Position PROBLEM: To improve air quality and obtain a useful product, sulfur is often removed from coal and natural gas by treating the fuel contaminant hydrogen sulfide with O 2 ; 2H 2 S(g) + O 2 (g) 2S(s) + 2H 2 O(g) What happens to (a) [H 2 O] if O 2 is added? (c) [O 2 ]ifh 2 S is removed? (b) [H 2 S] if O 2 is added? (d) [H 2 S] if sulfur is added? SOLUTION: Write an expression for Q and compare it to K when the system is disturbed to see in which direction the reaction will progress. Q = [H 2 O] 2 [H 2 S] 2 [O 2 ] (a) When O is added the denominator increases so Q decreases (a) When O 2 is added, the denominator increases, so Q decreases. The reaction must progress to the right to come back to K. Therefore [H 2 O] increases.

48 Predicting the Effect of a Change in Concentration on the Equilibrium Position Q = [H 2O] 2 [H 2 S] 2 [O 2 ] (b) When O 2 is added, the denominator increases and Q decreases. The reaction must progress to the right to come back to K. Therefore [H 2 S] decreases. (c) When H 2 S is removed, the denominator decreases and Q increases. The reaction must progress to the left to come back to K. Therefore [O 2 ] increases. (d) Sulfur is not part of the Q (K) expression because it is a solid. Therefore, as long as some sulfur is present the reaction is unaffected. [H 2 S] is unchanged.

49 The effect of pressure (volume) on an equilibrium system + lower P (higher V) less moles per liter of gas shift to the left higher P (lower V) More moles per liter of gas shift to the right

50 Catalysts increase the rate of both the forward and reverse reactions.

51 Equilibrium is achieved faster, but the equilibrium composition remains unaltered.

52 The Effect of Various Changes on an Equilibrium System

53 The Haber-Bosch Process The Haber-Bosch process is the transformation of atmospheric nitrogen and hydrogen into ammonia (NH 3 ) for the production of ammonia-based fertilizers. Initially developed by Fritz Haber ( ) in 1905 by passing a mixture of N 2 and H 2 over an iron catalyst at 1000 C. Later, Haber modified the process by increasing the pressure to atm. over a catalyst at 500 C. In 1908, BASF acquired the process and assigned Carl Bosch ( ) the task of scaling the process up to industrial quantities. Bosch ss modifications of the Haber process provided ammonium sulphate for use as a fertilizer for the soil. N 2(g) + 3 H 2(g) 2 NH 3(g) + 92 kj. In 1914, Germany s supplies of sodium and potassium nitrates for making explosives were blocked off by the Allied forces. Using the Haber-Bosch process, they were able to produce explosives, prolonging World War I. Fritz Haber Carl Bosch

54 The Haber-Bosch Process K = 3.5 x 10 8 at 298 K If H 2 is added to the system, N 2 will be consumed and the two reagents will form more NH 3.

55 The Haber-Bosch Process This apparatus helps push the equilibrium i to the right by removing the ammonia (NH 3 ) from the system as a liquid.

56 Effect of Temperature on K c for Ammonia Synthesis T (K) K c x x x x x x x 10-2

57 Percent yield of ammonia vs. temperature ( C) at five different operating pressures