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Chemical Equilibrium Reactions Go Both Ways Equilibrium Plural is equilibria (for you Latin fans) Reactions are reversible –they go forward and backward.

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Presentation on theme: "Chemical Equilibrium Reactions Go Both Ways Equilibrium Plural is equilibria (for you Latin fans) Reactions are reversible –they go forward and backward."— Presentation transcript:

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2 Chemical Equilibrium Reactions Go Both Ways

3 Equilibrium Plural is equilibria (for you Latin fans) Reactions are reversible –they go forward and backward There is always some product and reactant How much of each is there? –Depends on the reaction

4 Equilibria is when…. The rate of the forward reaction and reverse reaction are the same Sealed jar example

5 Add liquid to empty, dry jar When first added, the air has no liquid vapor. The vapor pressure of the liquid starts to populate the vapor With molecules The rate of evaporation does not change, but the rate of condensation continuously increases Until the rate of evaporation = the rate of condensation

6 Reactions Go Both Ways 2 NO 2 (g) N 2 O 4 (g) BrownColorless N 2 O 4 (g) 2 NO 2 (g) ColorlessBrown

7 In an Equilibrium The forward reaction rate is the same as the reverse reaction

8 Chemical equilibrium Different types of arrows are used in chemical equations associated with equilibria. Single arrow Assumes that the reaction proceeds to completion as written. Two single-headed arrows Used to indicate a system in equilibrium. Two single-headed arrows of different sizes. May be used to indicate when one side of an equilibrium system is favored.

9 Chemical equilibrium Homogeneous equilibriaHomogeneous equilibria - Equilibria that involve only a single phase.Examples. All species in the gas phase H 2 (g) + I 2 (g) 2HI (g) All species are in solution. HC 2 H 2 O 2 (aq) H + (aq) + C 2 H 3 O 2 - (aq)

10 Chemical equilibrium Basic steps for reaching equilibrium:Basic steps for reaching equilibrium: For the general reaction: A + B C We can view the reaction as occurring in three steps. Initial mixing Kinetic region Equilibrium region

11 Chemical equilibrium Initial mixing. When A and B are first brought together, there is no C present. The reaction proceeds as A + B C This is just at the very start of the reaction. Things change as soon as some C is produced.

12 Chemical equilibrium Kinetic region. As soon as some C has been produced, the reverse reaction is possible. A + B C Overall, we still see an increase in the net concentration of C. As we approach equilibrium, the rate of the forward reaction becomes slower.

13 Chemical equilibrium Equilibrium region. A point is finally reached where the forward and reverse reactions occur at the same rate. A + B C There is no net change in the concentration of any of the species.

14 Chemical equilibrium Concentration Time C B A Equilibrium Region Kinetic Region

15 Equilibrium A point is ultimately reached where the rates of the forward and reverse changes are the same. At this point, equilibrium is reached. Rate of FormationRate of Formation Time Reactants Products

16 Equilibrium Concentration Time KineticEquilibrium Region Products Reactants An equilibrium exists when no further change in concentration occurs. Note that the concentrations of products and reactants do not have to be equal!

17 The Equilibrium ConstantThe Equilibrium Constant Also note that the equilibrium concentrations of the reactants and products are the same regardless of the whether or not you start with only the reactants or only the products.Also note that the equilibrium concentrations of the reactants and products are the same regardless of the whether or not you start with only the reactants or only the products. No productsNo products No reactantsNo reactants

18 Equilibrium Constants The equilibrium constant is related to the rate laws of the forward and reverse reactions. Because the forward and reverse reactions are equal in rate, the rate laws must be equal to each other. k f [N 2 O 4 ] = k r [NO 2 ] 2 Isolating the concentrations and rate constants together… [NO 2 ] 2 /[N 2 O 4 ] = k f /k r = a constant (K eq ), or the equilibrium constant.

19 Equilibrium Constant The position of the equilibrium can be described mathematically We use the balanced chemical equation to develop a mathematical expression Use molarity to describe concentration –Remember, thats moles per liter of solution

20 Law of Mass Action pA + qB rC + tD K = [C] r [D] t [A] p [B] q [ ] = molarity K = equilibrium constant Products over reactants

21 Law of Mass Action 4 NH 3(g) + 7O 2 (g) 4NO 2 (g) + 6H 2 O (g) K = [C] r [D] t = [NO 2 ] 4 [H 2 O] 6 [A] p [B] q [NH 3 ] 4 [O 2 ] 7

22 Equilibrium Constant K eq = [products] x [reactants] y More products means a bigger K –Reaction lies to the right More reactants means smaller K –Reaction lies to the left Y Reactant X Product (g)

23 N 2(g) + 3H 2(g) 2NH 3(g) K = [NH 3 ] 2 [N 2 ] [H 2 ] 3 K = 6.02 x 10 -2

24 N 2(g) + 3H 2(g) 2NH 3(g) K = [NH 3 ] 2 [N 2 ] [H 2 ] 3 K = 6.02 x 10 -2

25 N 2(g) + 3H 2(g) 2NH 3(g) K = [NH 3 ] 2 [N 2 ] [H 2 ] 3 K = 6.02 x 10 -2

26 Equilibria It doesnt matter where you start The ratio of products to reactants is the same –Because the equilibria is established by the rate of the forward reaction vs the rate of the reverse reaction

27 Equilibrium Constant, K eq The equilibrium constant, K eq, is the ratio of the concentrations of the products compared to the concentrations of the reactants. K eq > 1 indicate that forming products is favored. K eq < 1 indicate that reactants dont react easily.

28 Equilibrium Constant, K eqEquilibrium Constant, K eq 3H 2 (g)+N 2 (g) 2NH 3 (g) The equilibrium constant, K eq, for the reaction above is determined by: K eq =[ NH 3 ] 2 [ H 2 ] 3 [ N 2 ] At 472 C, K eq = 0.105

29 Equilibrium Constant, K eqEquilibrium Constant, K eq 2NH 3 (g) 3H 2 (g)+N 2 (g) Reversing a reaction will result in the new equilibrium constant, K eq new, that is equal to 1 / K eq oldReversing a reaction will result in the new equilibrium constant, K eq new, that is equal to 1 / K eq old K eq new = [ H 2 ] 3 [ N 2 ] = 1 / [ NH 3 ] 2 or 9.52

30 Equilibrium Constant, K eqEquilibrium Constant, K eq NH 3 (g) 1.5H 2 (g)+0.5N 2 (g) Changing the number of moles of reactants and products will exponentially change the equilibrium constant: K eq new = (K eq old )^ nChanging the number of moles of reactants and products will exponentially change the equilibrium constant: K eq new = (K eq old )^ n K eq new = [ H 2 ] 1.5 [ N 2 ] 0.5 = (9.52)^ 0.5 [ NH 3 ] or 3.09

31 Partial pressure equilibrium constants At constant temperature, the pressure of a gas is proportional to its molarity. PV = nRT Remember, for an ideal gas: PV = nRT and molarity is: M = mole / liter or n/V so: P = MR T whereR is the gas law constant T is the temperature, K.

32 Partial pressure equilibrium constants For equilibria that involves gases, partial pressures can be used instead of concentrations. aA (g) + bB (g) eE (g) + fF (g) K p = K p is used when the partial pressures are expressed in units of atmospheres. p E e p F f p A a p B b

33 Partial pressure equilibrium constants In general, K p K c, instead K p = K c (RT) n g n g is the number of moles of gaseous products minus the number of moles of gaseous reactants. n g = (e + f) - (a + b)

34 Partial pressure equilibrium constants For the following equilibrium, K c = 1.10 x 10 7 at 700. o C. What is the K p ? atm L mol K 2H 2 (g) + S 2 (g) 2H 2 S (g) K p = K c (RT) n g T = = 973 K R= n g = ( 2 ) - ( 2 + 1) = -1

35 Partial pressure equilibrium constants K p = K c (RT) n g = 1.10 x 10 7 ( ) (973 K) = x10 5 atm L mol K [ ]

36 Equilibrium constants and expressions Heterogeneous equilibria Equilibria that involve more than one phase. CaCO 3 (s) CaO (s) + CO 2 (g) do notEquilibrium expressions for these types of systems do not include the concentrations of the pure solids (or liquids). K c = [CO 2 ]

37 Equilibrium constants and expressions Heterogeneous equilibria Heterogeneous equilibria - We dont include the pure solids and liquids because their concentrations do not vary. These values end up being included in the K value. CO 2 CaO & CaCO 3 As long as the temperature is constant and some solid is present, the amount of solid present has no effect on the equilibrium.

38 Writing an equilibrium expression Write a balanced equation for the equilibrium. Put the products in the numerator and the reactants in the denominator. Omit pure solids and liquids from the expression Omit solvents if your solutes are dilute (<0.1M). The exponent of each concentration should be the same as the coefficient for the species in the equation.

39 Writing an equilibrium expressionExample. What would be the equilibrium expression for the following? (NH 4 ) 2 CO 3 (s) 2 NH 3 (g) + CO 2 (g) + H 2 O(g) K c = [NH 3 ] 2 [CO 2 ] [H 2 O] (NH 4 ) 2 CO 3 is a pure solid so is not included K c We use [NH 3 ] 2 because the coefficient for NH 3 (g) in the equation is 2.

40 Equilibrium constant alphabet soup K cK c K pK p The equilibrium constant expressed in concentration unitsThe equilibrium constant expressed in concentration units The equilibrium constant expressed in pressure unitsThe equilibrium constant expressed in pressure units

41 Equilibrium constant alphabet soup K spK sp For the limited dissociation of insoluble solids.For the limited dissociation of insoluble solids. K sp is called the solubility product constant.solubility product constant.

42 Equilibrium constant alphabet soup K aK a K bK b K wK w For the dissociation of weak acids For the dissociation of weak bases For the dissociation of water into H + and OH -

43 Equilibrium and rate of reaction Chemical reactions tend to go to equilibrium providing that reaction takes place at a significant rate. There is no relationship between the magnitude of the equilibrium constant and the rate of a reaction. Example: Example: 2H 2 (g) + O 2 (g) 2H 2 O (g) K c = 2.9 x = [H 2 O] 2 [H 2 ] 2 [O 2 ] However, the reaction will take years to reach equilibrium at room temperature.

44 Determining equilibrium constants Equilibrium constants can be found by experiment. If you know the initial concentrations of all of the reactants, you only need to measure the concentration of a single species at equilibrium to determine the K c value. Lets consider the following equilibrium: H 2 (g) + I 2 (g) 2HI (g)

45 Determining equilibrium constants H 2 (g) + I 2 (g) 2HI (g) Assume that we started with the following initial concentrations at o C. H 2 (g) M I 2 (g) M HI (g) M At equilibrium, we determine that the concentration of iodine is M

46 Determining equilibrium constants The equilibrium expression for our system is: K c = Based on the chemical equation, we know the equilibrium concentrations of each species. I 2 (g)= the measured amount = M That means that mol I 2 reacts ( M M) to produce HI in 1.00 L of solution. [HI] 2 [H 2 ] [I 2 ]

47 Determining equilibrium constants I 2 (g) = M HI (g) = M = M H 2 (g) = M M = M Kc= = = 54 2 mol HI 1 mol I 2 [HI] 2 [H 2 ] [I 2 ] ( ) 2 ( )( )

48 Equilibrium calculations We can predict the direction of a reaction by calculating the reaction quotient. Reaction quotient, Q For the reaction: aA + bB eE + fF Q has the same form as K c with one important difference. Q can be for any set of concentrations, not just at equilibrium. Q = [E] e [F] f [A] a [B] b

49 Reaction quotient Any set of concentrations can be given and a Q calculated. By comparing Q to the K c value, we can predict the direction for the reaction. Q < K c Q < K c Net forward reaction will occur. Q = K c Q = K c No change, at equilibrium. Q > K c Q > K c Net reverse reaction will occur.

50 Reaction quotient example For an earlier example H 2 (g) + I 2 (g) 2HI (g) we determined the K c to be 54 at o C. If we had a mixture that contained the following, predict the direction of the reaction. [H 2 ] = 4.25 x M [I 2 ]= 3.97 x M [HI]= 9.83 x M

51 Reaction quotient example Q = = = 5.73 Since Q is < K c, the system is not in equilibrium and will proceed in the forward direction. [ HI ] 2 [ H 2 ] [ I 2 ] (9.83 x ) 2 (4.25 x )(3.97 x )

52 Calculating equilibrium concentrations If the stoichiometry and K c for a reaction is known, calculating the equilibrium concentrations of all species is possible. Commonly, the initial concentrations are known. One of the concentrations is expressed as the variable x. All others are then expressed in terms of x.

53 Equilibrium calculation example A sample of COCl 2 is allowed to decompose. The value of K c for the equilibrium COCl 2 (g) CO (g) + Cl 2 (g) is 2.2 x at 100 o C. If the initial concentration of COCl 2 is 0.095M, what will be the equilibrium concentrations for each of the species involved?

54 Equilibrium calculation example COCl 2 (g) CO (g) Cl 2 (g) Initial conc., M Change in conc.- X + X + X due to reaction Equilibrium Concentration, M( X) X X K c == [CO ] [Cl 2 ] [ COCl 2 ] X 2 ( X)

55 Equilibrium calculation example X 2 ( X) K c = 2.2 x = Rearrangement gives X x X x = 0 This is a quadratic equation. Fortunately, there is a straightforward equation for their solution

56 Quadratic equations An equation of the form a X 2 + b X + c = 0 Can be solved by using the following x = Only the positive root is meaningful in equilibrium problems. -b + b 2 - 4ac 2a

57 Equilibrium calculation example -b + b 2 - 4ac 2a 2.2 x x X x X x = 0 b c a b c X = x [(2.2 x ) 2 - (4)(1)( x )] 1/2 2 X = 9.1 x M

58 Equilibrium calculation example Now that we know X, we can solve for the concentration of all of the species. COCl 2 = X = M CO= X = 9.1 x M Cl 2 = X = 9.1 x M In this case, the change in the concentration of is COCl 2 negligible.

59 Summary of method of calculating equilibrium concentrations Write an equation for the equilibrium. Write an equilibrium constant expression. Express all unknown concentrations in terms of a single variable such as x. Substitute the equilibrium concentrations in terms of the single variable in the equilibrium constant expression. Solve for x. Use the value of x to calculate equilibrium concentrations.

60 You Try! A 40.0g sample of solid (NH 4 ) 2 CO 3 is placed in a closed evacuated 3L flask and heated to 400 o C. It decomposes to produce NH 3, H 2 O, and CO 2 : (NH 4 ) 2 CO 3 2 NH 3 + H 2 O + CO 2 The equilibrium constant, K p, is Write the K p expression. Calculate K c Calculate the partial pressure of NH 3 Calculate the total pressure in the flask at equilibrium. Calculate the number of grams of solid at equilibrium. What is the minimum amount of solid needed to establish equilibrium?

61 Answers K p = P NH3 2 P H2O P CO2 3.17x atm 2.08 atm 37.3g More than 2.72g

62 Predicting Shifts: LeChatelier Any stress placed on an equilibrium system will cause the system to shift to minimize the effect of the stress. How can you cause the color to change from pink to blue? s t ressYou can put s t ress on a system by adding or removing something from one side of a reaction. Co(H 2 O) 6 2+ CoCl 4 2-Co(H 2 O) Cl 1- CoCl H 2 O

63 Predicting shifts in equilibria Equilibrium concentrations are based on: –The specific equilibrium –The starting concentrations –Other factors such as: Temperature Pressure Reaction specific conditions Altering conditions will stress a system, resulting in an equilibrium shift.

64 Le Chateliers principle Co(H 2 O) 6 2+ CoCl 4 2- Co(H 2 O) Cl 1- CoCl H 2 O Note: H = + value What other stresses could be placed on this system to change the color back and forth between pink and blue? Add heatAdd heat (removes H 2 O)(removes H 2 O) Add Ag 1+Add Ag 1+ (removes Cl 1- )(removes Cl 1- ) Add iceAdd ice Add acetoneAdd acetone Add HCl (adds Cl 1- )Add HCl (adds Cl 1- )

65 Le Chateliers principle Any stress placed on an equilibrium system will cause the system to shift to minimize the effect of the stress. You can put stress on a system by adding or removing something from one side of a reaction. N 2 (g) + 3H 2 (g) 2NH 3 (g) What effect will there be if you added more ammonia? How about more nitrogen?

66 Changes in concentration Changes in concentration do not change the value of the equilibrium constant at constant temperature. When a material is added to a system in equilibrium, the equilibrium will shift away from that side of the equation. When a material is removed from a system in equilibrium, the equilibrium will shift towards that sid of the equation.

67 Le Châteliers PrincipleLe Châteliers Principle Change in Reactant or Product ConcentrationsChange 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).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.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.

68 Le Châteliers PrincipleLe Châteliers Principle Change in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations

69 Le Châteliers PrincipleLe Châteliers Principle Change in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations Adding a reactant or product shifts the equilibrium away from the increase.Adding a reactant or product shifts the equilibrium away from the increase. Removing a reactant or product shifts the equilibrium towards the decrease.Removing a reactant or product shifts the equilibrium towards the decrease. To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier). We illustrate the concept with the industrial preparation of ammonia

70 Le Châteliers PrincipleLe Châteliers Principle Change in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations N 2 and H 2 are pumped into a chamber. The pre-heated gases are passed through a heating coil to the catalyst bed. The catalyst bed is kept at C under high pressure. The product gas stream (containing N 2, H 2 and NH 3 ) is passed over a cooler to a refrigeration unit. In the refrigeration unit, ammonia liquefies but not N 2 or H 2.

71 Le Châteliers PrincipleLe Châteliers Principle Change in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations The unreacted nitrogen and hydrogen are recycled with the new N 2 and H 2 feed gas. The equilibrium amount of ammonia is optimized because the product (NH 3 ) is continually removed and the reactants (N 2 and H 2 ) are continually being added. Effects of Volume and PressureEffects of Volume and Pressure As volume is decreased pressure increases. Le Châteliers Principle: if pressure is increased the system will shift to counteract the increase by producing fewer moles of gas.Le Châteliers Principle: if pressure is increased the system will shift to counteract the increase by producing fewer moles of gas.

72 Changes in pressure Changing the pressure does not change the value of the equilibrium constant at constant temperature. Solids and liquids are not effected by pressure changes. Changing pressure by introducing an inert gas will not shift an equilibrium. Pressure changes only effect gases that are a portion of an equilibrium.

73 Changes in pressure In general, increasing the pressure by decreasing volume shifts equilibria towards the side that has the smaller number of moles of gas. H 2 (g) + I 2 (g) 2HI (g) N 2 O 2 (g) 2NO 2 (g) Unaffected by pressure Increased pressure, shift to left

74 Le Châteliers PrincipleLe Châteliers Principle Effects of Volume and PressureEffects 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.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

75 Le Châteliers PrincipleLe Châteliers Principle Effects of Volume and PressureEffects of Volume and Pressure An increase in pressure (by decreasing the volume) favors the formation of colorless N 2 O 4. The instant the pressure increases, the system is not at equilibrium and the concentration of both gases has increased.The instant the pressure increases, the system is not at equilibrium and the concentration of both gases has increased. The system moves to reduce the number moles of gas (i.e. the forward reaction is favored).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.

76 Le Châteliers PrincipleLe Châteliers Principle Effect of Temperature ChangesEffect of Temperature Changes The equilibrium constant is temperature dependent. For an endothermic reaction, H > 0 and heat can be considered as a reactant. For an exothermic reaction, H < 0 and heat can be considered as a product.For an exothermic reaction, H < 0 and heat can be considered as a product. 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.

77 Le Châteliers PrincipleLe Châteliers Principle Effect of Temperature ChangesEffect of Temperature Changes Removing heat (i.e. cooling the vessel), favors towards the decrease: if H > 0, cooling favors the reverse reaction, if H < 0, cooling favors the forward reaction. for which H > 0. –Co(H 2 O) 6 2+ is pale pink and CoCl 4 2- is blue.

78 Le Châteliers PrincipleLe Châteliers Principle Effect of Temperature ChangesEffect of Temperature Changes

79 Le Châteliers PrincipleLe Châteliers Principle Effect of Temperature ChangesEffect of Temperature Changes If a light purple room temperature equilibrium mixture is placed in a beaker of warm water, the mixture turns deep blue. Since H > 0 (endothermic), adding heat favors the forward reaction, i.e. the formation of blue CoCl If the room temperature equilibrium mixture is placed in a beaker of ice water, the mixture turns bright pink. Since H > 0, removing heat favors the reverse reaction which is the formation of pink Co(H 2 O) Co

80 Le Châteliers PrincipleLe Châteliers Principle The Effect of CatalystsThe Effect of Catalysts A catalyst lowers the activation energy barrier for the reaction. a catalyst will decrease the time taken to reach equilibriumTherefore, a catalyst will decrease the time taken to reach equilibrium. A catalyst does not effect the composition of the equilibrium mixture.A catalyst does not effect the composition of the equilibrium mixture.

81 You try! H 2(g) + S (s) H 2 S (g) H rxn = kJ/mol An amount of solid S and an amount of gaseous H 2 are placed in an evacuated container at 25 o C. At equilibrium, some solid S remains in the container. Predict and explain the following: Effect on partial pressure of H 2 S when S is added. Effect on partial pressure of H 2 when H 2 S is added Effect on mass of S when volume is increased. Effect on S when the temperature increased. Effect of adding a catalyst to initial concentrations.

82 Answers Solids have no effect on equilibrium P H2 will increase. S remains the same, equal numbers of moles of gases on both sides means no change in equilibrium. S decreases. Equilibrium is not effected by solids, but solid amounts can be changed by equilibrium. If more gases are produced, solid must go down. Catalysts make them reach equilibrium faster, but do not change equilibrium positions.


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