2Equilibrium Plural is equilibria (for you Latin fans) Reactions are reversiblethey go forward and backwardThere is always some product and reactantHow much of each is there?Depends on the reaction
3Equilibria is when….The rate of the forward reaction and reverse reaction are the sameSealed jar example
4Add liquid to empty, dry jar The rate of evaporation does not change,but the rate of condensation continuously increasesUntil the rate of evaporation = the rate of condensationWhen first added, the air has no liquid vapor.The vapor pressure of the liquid starts to populate the vaporWith molecules
5Reactions Go Both Ways 2 NO2 (g) N2O4 (g) Brown Colorless Colorless Brown
6The forward reaction rate is the same as the reverse reaction In an EquilibriumThe forward reaction rate is the same asthe reverse reaction
7Chemical equilibriumDifferent types of arrows are used in chemical equations associated with equilibria.Single arrowAssumes that the reaction proceeds to completion as written.Two single-headed arrowsUsed 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.
8Chemical equilibriumHomogeneous equilibria - Equilibria that involve only a single phase.Examples.All species in the gas phaseH2 (g) + I2 (g) HI (g)All species are in solution.HC2H2O2 (aq) H+ (aq) + C2H3O2- (aq)
9Chemical equilibrium A + B C Basic steps for reaching equilibrium: For the general reaction:A + B CWe can view the reaction as occurring in three steps.Initial mixingKinetic regionEquilibrium region
10Chemical equilibrium A + B C Initial mixing. When A and B are first brought together, there is no C present.The reaction proceeds asA + B CThis is just at the very start of the reaction. Things change as soon as some C is produced.
11Chemical equilibrium A + B C Kinetic region. As soon as some C has been produced, the reverse reaction is possible.A + B COverall, we still see an increase in the net concentration of C.As we approach equilibrium, the rate of the forward reaction becomes slower.
12Chemical equilibrium A + B C Equilibrium region. A point is finally reached where the forward and reverse reactions occur at the same rate.A + B CThere is no net change in the concentration of any of the species.
13Chemical equilibrium Concentration Time C B A Equilibrium Region Kinetic
14Equilibrium A point is ultimately reached where the rates of the forwardand reverse changesare the same.At this point,equilibriumis reached.ReactantsRate of FormationProductsTime
15EquilibriumAn equilibrium exists when no further change in concentration occurs. Note that the concentrations of products and reactants do not have to be equal!ProductsKinetic EquilibriumRegion RegionConcentrationReactantsTime
16The 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.No productsNo reactants
17Equilibrium 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.kf [N2O4] = kr[NO2]2Isolating the concentrations and rate constants together…[NO2]2/[N2O4] = kf/kr = a constant (Keq), or the equilibrium constant.
18Equilibrium ConstantThe position of the equilibrium can be described mathematicallyWe use the balanced chemical equation to develop a mathematical expressionUse molarity to describe concentrationRemember, that’s moles per liter of solution
19Law of Mass Action pA + qB rC + tD K = [C]r • [D]t [A]p • [B]q [ ] = molarityK = equilibrium constantProducts “over”reactants
20Law of Mass Action 4 NH3(g) + 7O2 (g) 4NO2 (g) + 6H2O (g) K = [C]r • [D]t = [NO2]4 • [H2O]6 [A]p • [B]q [NH3]4 • [O2]7
21Equilibrium Constant More products means a bigger K Y Reactant X Product (g)Keq = [products]x [reactants]yMore products means a bigger KReaction lies to the rightMore reactants means smaller KReaction lies to the left
23N2(g) + 3H2(g) 2NH3(g) K = [NH3]2 [N2] • [H2]3 K = 6.02 x 10-2
24N2(g) + 3H2(g) 2NH3(g) K = [NH3]2 [N2] • [H2]3 K = 6.02 x 10-2
25Equilibria It doesn’t matter where you start The “ratio” of products to reactants is the sameBecause the equilibria is established by the rate of the forward reaction vs the rate of the reverse reaction
26Equilibrium Constant, Keq The equilibrium constant, Keq, is the ratio of the concentrations of the products compared to the concentrations of the reactants.Keq > 1 indicate that forming products is favored. Keq < 1 indicate that reactants don’t react easily.
27Equilibrium Constant, Keq 3H2(g)+N2(g) NH3(g)The equilibrium constant, Keq, for the reaction above is determined by:Keq = [ NH3 ] 2[ H2 ] 3 [ N2 ]At 472C, Keq = 0.105
28Equilibrium Constant, Keq 2NH3(g) H2(g)+N2(g)Reversing a reaction will result in the new equilibrium constant, Keq new, that is equal to 1 / Keq oldKeq new = [ H2 ] 3 [ N2 ] = 1 / 0.105[ NH3 ] or 9.52
29Equilibrium Constant, Keq NH3(g) H2(g)+0.5N2(g)Changing the number of moles of reactants and products will exponentially change the equilibrium constant: Keq new= (Keq old)^nKeq new = [ H2 ]1.5 [ N2 ]0.5 = (9.52)^0.5[ NH3 ] or 3.09
30Partial pressure equilibrium constants At constant temperature, the pressure of a gas is proportional to its molarity.Remember, for an ideal gas: PV = nRTand molarity is: M = mole / liter or n/Vso: P = MR Twhere R is the gas law constantT is the temperature, K.
31Partial pressure equilibrium constants For equilibria that involves gases, partial pressures can be used instead of concentrations.aA (g) + bB (g) eE (g) + fF (g)Kp =Kp is used when the partial pressures are expressed in units of atmospheres.pEe pFfpAa pBb
32Partial pressure equilibrium constants In general, Kp ≠ Kc, insteadKp = Kc (RT)DngDng is the number of moles of gaseous products minus the number of moles of gaseous reactants.Dng = (e + f) - (a + b)
33Partial pressure equilibrium constants For the following equilibrium, Kc = 1.10 x 107 at 700. oC. What is the Kp?2H2 (g) + S2 (g) H2S (g)Kp = Kc (RT)DngT = = 973 KR =Dng = ( 2 ) - ( 2 + 1) = -1atm Lmol K
34Partial pressure equilibrium constants Kp = Kc (RT)Dng= 1.10 x ( ) (973 K)= x105-1][atm Lmol K
35Equilibrium constants and expressions Heterogeneous equilibriaEquilibria that involve more than one phase.CaCO3 (s) CaO (s) + CO2 (g)Equilibrium expressions for these types of systems do not include the concentrations of the pure solids (or liquids).Kc = [CO2]
36Equilibrium constants and expressions Heterogeneous equilibria - We don’t include the pure solids and liquids because their concentrations do not vary. These values end up being included in the K value.CO2CaO &CaCO3As long as the temperature is constant and some solid is present, the amount of solid present has no effect on the equilibrium.
37Writing 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 expressionOmit 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.
38Writing an equilibrium expression Example.What would be the equilibrium expression for the following?(NH4)2CO3(s) NH3(g) + CO2(g) + H2O(g)Kc = [NH3]2 [CO2] [H2O](NH4)2CO3 is a pure solid so is not included KcWe use [NH3]2 because the coefficient for NH3(g) in the equation is 2.
39Equilibrium constant alphabet soup The equilibrium constant expressed in concentration unitsKcThe equilibrium constant expressed in pressure unitsKp
40Equilibrium constant alphabet soup For the limited dissociation of insoluble solids.KspKsp is called thesolubility product constant.
41Equilibrium constant alphabet soup KaFor the dissociation of weak acidsKbFor the dissociation of weak basesFor the dissociation of water into H+ and OH-Kw
42Equilibrium 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: 2H2 (g) + O2 (g) H2O (g)Kc = 2.9 x = [H2O]2[H2]2 [O2]However, the reaction will take years to reach equilibrium at room temperature.
43Determining 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 Kc value.Lets consider the following equilibrium:H2 (g) + I2 (g) HI (g)
44Determining equilibrium constants H2 (g) + I2 (g) HI (g)Assume that we started with the following initial concentrations at 425.4oC.H2 (g) MI2 (g) MHI (g) MAt equilibrium, we determine that the concentration of iodine is M
45Determining equilibrium constants The equilibrium expression for our system is:Kc =Based on the chemical equation, we know the equilibrium concentrations of each species.I2 (g) = the measured amount= MThat means that mol I2 reacts ( M M) to produce HI in 1.00 L of solution.[HI]2[H2] [I2]
47Equilibrium calculations We can predict the direction of a reaction by calculating the reaction quotient.Reaction quotient, QFor the reaction: aA + bB eE + fFQ has the same form as Kc 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
48Reaction quotientAny set of concentrations can be given and a Q calculated. By comparing Q to the Kc value, we can predict the direction for the reaction.Q < Kc Net forward reaction will occur.Q = Kc No change, at equilibrium.Q > Kc Net reverse reaction will occur.
49Reaction quotient example For an earlier exampleH2 (g) + I2 (g) HI (g)we determined the Kc to be 54 at oC.If we had a mixture that contained the following, predict the direction of the reaction.[H2] = 4.25 x 10-3 M[I2] = 3.97 x 10-1 M[HI] = 9.83 x 10-2 M
50Reaction quotient example ==Since Q is < Kc, the system is not in equilibrium and will proceed in the forward direction.[ HI ]2[ H2 ] [ I2 ](9.83 x 10-2)2(4.25 x 10-3)(3.97 x 10-1)
51Calculating equilibrium concentrations If the stoichiometry and Kc 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.
52Equilibrium calculation example A sample of COCl2 is allowed to decompose. The value of Kc for the equilibriumCOCl2 (g) CO (g) + Cl2 (g)is 2.2 x at 100 oC.If the initial concentration of COCl2 is 0.095M, what will be the equilibrium concentrations for each of the species involved?
53Equilibrium calculation example COCl2 (g) CO (g) Cl2 (g)Initial conc., MChange in conc. - X + X + Xdue to reactionEquilibriumConcentration, M ( X) X XKc = =[CO ] [Cl2 ][ COCl2 ]X2( X)
54Equilibrium calculation example Kc = 2.2 x =Rearrangement givesX x X x = 0This is a quadratic equation. Fortunately, thereis a straightforward equation for their solution
55Quadratic equations An equation of the form a X2 + b X + c = 0 Can be solved by using the followingx =Only the positive root is meaningful in equilibrium problems.-b b2 - 4ac2a
56Equilibrium calculation example X x X x = 0a b c-b b2 - 4ac2aX =- 2.2 x [(2.2 x 10-10)2 - (4)(1)( x 10-11)]1/22X = x 10-6 M
57Equilibrium calculation example Now that we know X, we can solve for the concentration of all of the species.COCl2 = X = MCO = X = 9.1 x 10-6 MCl2 = X = 9.1 x 10-6 MIn this case, the change in the concentration of is COCl2 negligible.
58Summary 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.
59You Try!A 40.0g sample of solid (NH4)2CO3 is placed in a closed evacuated 3L flask and heated to 400oC. It decomposes to produce NH3, H2O, and CO2 :(NH4)2CO3 ↔ 2 NH3 + H2O + CO2The equilibrium constant, Kp , is 0.295Write the Kp expression.Calculate KcCalculate the partial pressure of NH3Calculate 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?
60Answers Kp = PNH32 PH2O PCO2 3.17x10-8 1.04 atm 2.08 atm 37.3g More than 2.72g
61Predicting Shifts: LeChatelier 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.Co(H2O) Cl CoCl H2OHow can you cause the color to change from pink to blue?
62Predicting shifts in equilibria Equilibrium concentrations are based on:The specific equilibriumThe starting concentrationsOther factors such as:TemperaturePressureReaction specific conditionsAltering conditions will stress a system, resulting in an equilibrium shift.
63Le Chatelier’s principle Co(H2O) Cl CoCl H2ONote: ∆H = + valueAdd HCl (adds Cl1-)Add heatAdd acetoneAdd Ag1+(removes Cl1- )(removes H2O)Add iceWhat other stresses could be placed on this system to change the color back and forth between pink and blue?
64Le Chatelier’s 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.N2(g) + 3H2 (g) NH3 (g)What effect will there be if you added moreammonia? How about more nitrogen?
65Changes 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.
66Le Châtelier’s Principle Change in Reactant or Product ConcentrationsConsider the Haber processIf H2 is added while the system is at equilibrium, the system must respond to counteract the added H2 (by Le Châtelier).That is, the system must consume the H2 and produce products until a new equilibrium is established.Therefore, [H2] and [N2] will decrease and [NH3] increases.
67Le Châtelier’s Principle Change in Reactant or Product Concentrations
68Le Châtelier’s Principle Change in Reactant or Product ConcentrationsAdding a reactant or product shifts the equilibrium away from the increase.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
69Le Châtelier’s Principle Change in Reactant or Product ConcentrationsThe catalyst bed is kept at C under high pressure.N2 and H2 are pumped into a chamber.The pre-heated gases are passed through a heating coil to the catalyst bed.In the refrigeration unit, ammonia liquefies but not N2 or H2.The product gas stream (containing N2, H2 and NH3) is passed over a cooler to a refrigeration unit.
70Le Châtelier’s Principle Change in Reactant or Product ConcentrationsThe unreacted nitrogen and hydrogen are recycled with the new N2 and H2 feed gas.The equilibrium amount of ammonia is optimized because the product (NH3) is continually removed and the reactants (N2 and H2) are continually being added.Effects of Volume and PressureAs volume is decreased pressure increases.Le Châtelier’s Principle: if pressure is increased the system will shift to counteract the increase by producing fewer moles of gas.
71Changes in pressureChanging 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.
72Changes in pressureIn general, increasing the pressure by decreasing volume shifts equilibria towards the side that has the smaller number of moles of gas.H2 (g) + I2 (g) HI (g)N2O2 (g) NO2 (g)Unaffected by pressureIncreased pressure, shift to left
73Le Châtelier’s Principle Effects of Volume and PressureThat 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
74Le Châtelier’s Principle Effects of Volume and Pressure An increase in pressure (by decreasing the volume) favors the formation of colorless N2O4.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).A new equilibrium is established in which the mixture is lighter because colorless N2O4 is favored.
75Le Châtelier’s Principle Effect of Temperature ChangesThe 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.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.
76Le Châtelier’s Principle Effect of Temperature ChangesRemoving 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 DH > 0.Co(H2O)62+ is pale pink and CoCl42- is blue.
77Le Châtelier’s Principle Effect of Temperature Changes
78Le Châtelier’s Principle Effect 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 CoCl42-.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(H2O)62+.Co
79A catalyst does not effect the composition of the equilibrium mixture. Le Châtelier’s PrincipleThe Effect of CatalystsA 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.
80You try! H2(g) + S(s) → H2S(g) ∆Hrxn = -20.17kJ/mol An amount of solid S and an amount of gaseous H2 are placed in an evacuated container at 25oC. At equilibrium, some solid S remains in the container. Predict and explain the following:Effect on partial pressure of H2S when S is added.Effect on partial pressure of H2 when H2S is addedEffect on mass of S when volume is increased.Effect on S when the temperature increased.Effect of adding a catalyst to initial concentrations.
81Answers 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.