Question? Leave a message!




Chemical Equilibrium

Chemical Equilibrium
Chemical Equilibrium www.ThesisScientist.comEquilibrium vs. Disequilibrium • When systems are at equilibrium with their surroundings, their conditions are the same as the surroundings and they stay that way. • When systems are in disequilibrium with their surroundings, their conditions are not the same as the surroundings. • Systems that are in disequilibrium tend to change until they reach equilibrium with their surroundings. • Living things are in controlled disequilibrium with their environment—they are not at the same conditions as the environment and do not tend to change toward those conditions. www.ThesisScientist.comReaction Rates • Some chemical reactions proceed rapidly.  Like the precipitation reactions studied in Chapter 7 where the products form practically the instant the two solutions are mixed. • Other reactions proceed slowly.  Like the decomposition of dye molecules of a sofa placed in front of a window. • The rate of a reaction is measured in the amount of reactant that changes into product in a given period of time.  Generally moles of reactant used per second. Like miles per hour. • Chemists study ways of controlling reaction rates. www.ThesisScientist.comCollision Theory • In order for a reaction to take place, the reacting molecules must collide with each other. • Once molecules collide they may react together or they may not, depending on two factors: 1. Whether the collision has enough energy to ―start to break the bonds holding reactant molecules together." 2. Whether the reacting molecules collide in the proper orientation for new bonds to form. www.ThesisScientist.comEffective Collisions • Collisions in which these two conditions are met (and therefore the reaction occurs) are called effective collisions. • The higher the frequency of effective collisions, the faster the reaction rate. • There is a minimum energy needed for a collision to be effective. We call this the activation energy. The lower the activation energy, the faster the reaction will be. www.ThesisScientist.comEffective Collisions: Kinetic Energy Factor For a collision to lead to overcoming the energy barrier, the reacting molecules must have sufficient kinetic energy so that when they collide, it can form the activated complex. www.ThesisScientist.comEffective Collisions: Orientation Effect www.ThesisScientist.comReaction Energy Diagram www.ThesisScientist.comFactors Effecting Reaction Rate: Reactant Concentration • The higher the concentration of reactant molecules, this increases the frequency of reactant molecule collisions the faster the reaction will generally go. • Since reactants are consumed as the reaction proceeds, the speed of a reaction generally slows over time. www.ThesisScientist.comEffect of Concentration on Rate Low concentrations of reactant molecules leads to fewer effective collisions, therefore a slower reaction rate. High concentrations of reactant molecules lead to more effective collisions, therefore a faster reaction rate. www.ThesisScientist.comFactors Effecting Reaction Rate: Temperature • Increasing the temperature increases the energy so that their collisions can overcome the activation energy. • And, increasing the temperature also increases the frequency of collisions. • Both these mean that increasing temperature increases the reaction rate. www.ThesisScientist.comEffect of Temperature on Rate Low temperatures lead to fewer molecules with enough energy to overcome the activation energy, and less frequent reactant collisions, therefore a slower reaction rate High temperatures lead to more molecules with enough energy to overcome the activation energy, and more frequent reactant collisions, therefore, a faster www.ThesisScientist.com reaction rate.Chemical Equilibrium • When a reaction reaches equilibrium, the production of both reactants and products is constant. • The forward and reverse reactions still continue. • Because they go at the same rate, the amounts of materials does not appear to change. www.ThesisScientist.comEquilibrium As the reaction proceeds, the forward reaction Initially, we only have reactant molecules in the slows down as the reactants get used up. At the mixture. The reaction can only proceed in the same time, the reverse reaction speeds up as forward direction, making products. product concentration increases. Once equilibrium is established, the concentrations of the reactants and products in Eventually, the forward and reverse rates are equal. the final mixture do not change, (unless At this time equilibrium is established. www.ThesisScientist.com conditions are changed). Equilibrium  Equal • The rates of the forward and reverse reactions are equal at equilibrium. • But that does not mean the concentrations of reactants and products are equal. www.ThesisScientist.comEquilibrium, Continued Initially, only the forward B As Onc ecthe ae use e fquil or the wa ibrium rre d ar ce tan a c is ti t co e on pr stnc abli eoc ntra shed, eeti ds on Eventually, the reaction proceeds reaction takes place. de it th make c ere for ase wa s produc s, rd the f and re or ts wa aver nd use rd se rere as re ca ticon ti ac ons tants. slows. in the reverse direction as fast as As proc the ee pr d a oduc t the ts same accumulate rate, so t , the he it proceeds in the forward direction. re conce verse ntra reati cons tion spe of all eds mater up. ials At this time equilibrium is established. stay constant. Rate forward Rate reverse Time www.ThesisScientist.com Rate• What occurs in a reaction at equilibrium. • When the number of people moving up is the same as the number of people moving down, the number of people on each floor remains constant, and the two populations are in equilibrium. • Equilibrium occurs when the forward and reverse reactions are the same. www.ThesisScientist.co mAn Analogy: Population Changes When Narnians feel However, as time passes, overcrowded, some will emigration will occur in both emigrate to Middle Earth. directions at the same rate, leading to populations in Narnia and Middle Earth that are constant, though not necessarily equal. www.ThesisScientist.comEquilibrium Constant • Even though the concentrations of reactants and products are not equal at equilibrium, there is a relationship between them. • For the reaction H (g) + I (g)  2HI(g) at 2 2 equilibrium, the ratio of the concentrations raised to the power of their coefficients is constant. 2 HI K eq HI 2 2 www.ThesisScientist.comEquilibrium Constant, Continued • For the general equation aA + bB  cC + dD, the relationship is given below: The lowercase letters represent the coefficients of the balanced chemical equation. Always products over reactants. • The constant is called the equilibrium constant, K . eq c d CD K eq a b AB www.ThesisScientist.comWriting Equilibrium Constant Expressions • For aA + bB  cC + dD, c d CD the equilibrium constant K eq a b AB expression is: • So for the reaction 4  NO O 2 N O 4 NO + O , the 2 2 2 5 2 2 K eq 2  equilibrium constant N O 2 5 expression is: www.ThesisScientist.comEquilibrium Constants for Heterogeneous Equilibria • Pure substances in the solid and liquid state have constant concentrations.  Adding or removing some does not change the concentration because they do not expand to fill the container or spread throughout a solution. • Therefore, these substances are not included in the equilibrium constant expression. For the reaction CaCO (s) + 2 HCl(aq)  CaCl (aq) + CO (g) + H O(l): 3 2 2 2 CaClCO 2 2 K eq 2 HCl www.ThesisScientist.comWrite the Equilibrium Constant Expressions, K , for Each of the Following: eq • 2 CO (g)  2 CO(g) + O (g) 2 2 +2 2 • BaSO (s)  Ba (aq) + SO (aq) 4 4 • CH (g) + 2 O (g)  CO (g) + 2 H O(l) 4 2 2 2 www.ThesisScientist.comWrite the Equilibrium Constant Expressions, K , for Each of the Following, eq Continued: • 2 CO (g)  2 CO(g) + O (g) 2 2 2 CO •O 2 K = 2 eq CO 2 +2 2 • BaSO (s)  Ba (aq) + SO (aq) 4 4 +2 2 K =Ba •SO eq 4 • CH (g) + 2 O (g)  CO (g) + 2 H O(l) 4 2 2 2 CO 2 K = 2 eq CH •O www.ThesisScientist.com 4 2What Does the Value of K Imply eq • When the value of K 1, we know that when the eq reaction reaches equilibrium, there will be many more product molecules present than reactant molecules.  The position of equilibrium favors products. • When the value of K 1, we know that when the eq reaction reaches equilibrium, there will be many more reactant molecules present than product molecules.  The position of equilibrium favors reactants. www.ThesisScientist.comA Large Equilibrium Constant www.ThesisScientist.comA Small Equilibrium Constant www.ThesisScientist.comWrite the Equilibrium Constant Expressions, K , and Predict the Position of eq Equilibrium for Each of the Following: 95 • 2 HF(g)  H (g) + F (g) K = 1 x 10 2 2 eq 25 • 2 SO (g) + O (g)  2 SO (g) K = 8 x 10 2 2 3 eq 17 • N (g) + 2 O (g)  2 NO (g) K = 3 x 10 2 2 2 eq www.ThesisScientist.comWrite the Equilibrium Constant Expressions, K , and Predict the Position of Equilibrium for eq Each of the Following, Continued: 95 • 2 HF(g)  H (g) + F (g) K = 1 x 10 2 2 eq HF 2 2 Favors reactants. K eq 2 HF 25 • 2 SO (g) + O (g)  2 SO (g) K = 8 x 10 2 2 3 eq 2 SO 3 K Favors products. eq 2 SOO 2 2 17 • N (g) + 2 O (g)  2 NO (g) K = 3 x 10 2 2 2 eq 2  NO 2 Favors reactants. K www.ThesisScientist.com eq 2 NO 2 2Calculating K eq • The value of the equilibrium constant may be determined by measuring the concentrations of all the reactants and products in the mixture after the reaction reaches equilibrium, then substituting in the expression for K . eq • Although you may have different amounts of reactants and products in the equilibrium mixture, the value of K will always be the same. eq  The value of K depends only on the temperature. eq  The value of K does not depend on the amounts of eq reactants or products with which you start. www.ThesisScientist.comInitial and Equilibrium Concentrations for H (g) + I (g)  2HI(g) 2 2 Equilibrium Equili Initial brium Constant 2 HI K eq H I HI H I HI 2 2 2 2 H I 2 2 2 0.78  50 0.50 0.50 0.0 0.11 0.11 0.78 0.110.11 2 0.39  50 0.0 0.0 0.50 0.055 0.055 0.39 0.0550.055 2 1.17  50 0.50 0.50 0.50 0.165 0.165 1.17 0.1650.165 2 0.934  50 1.0 0.5 0.0 0.53 0.033 0.934 0.530.033 www.ThesisScientist.comExample 15.3—Find the Value of K for the Reaction eq from the Given Concentrations: 2 CH (g)  C H (g) + 3 H (g). 4 2 2 2 Given: CH = 0.0203 M, C H = 0.0451 M, H = 0.112 M 4 2 2 2 Find: K eq Solution Map: CH , C H , H K 4 2 2 2 eq 3 C H H 2 2 2 K eq 2 Relationships: CH 4 Solve: 3 3 C H H 0.04510.112 2 2 2 K 0.154 eq 2 2 CH  0.0203 4 Check: K is unitless. eq www.ThesisScientist.comPractice—Calculate K for the Reaction eq 2 NO (g)  N O (g) 2 2 4 at 100 C if the Equilibrium Concentrations Are NO = 0.0172 M and N O = 0.0014 M. 2 2 4 www.ThesisScientist.comPractice—Find the Value of K for the Reaction from the eq Given Concentrations: 2 NO (g)  N O (g). 2 2 4 Given: NO = 0.0172 M, N O = 0.0014 M 2 2 4 Find: K eq Solution Map: NO , N O K 2 2 4 eq N O 2 4 K eq 2 NO Relationships: 2 Solve: N O 0.0014  2 4 K 4.7 eq 2 2 NO 0.0172  2 Check: K is unitless. eq www.ThesisScientist.comExample 15.4—Find the Value of HI for the Reaction at Equilibrium from the Given Concentrations and K : eq H (g) + I (g) 2HI(g) 2 2 Given: I = 0.020 M, H = 0.020 M, K = 69 2 2 eq Find: HI Solution Map: I , H , K HI 2 2 eq 2 HI K eq Relationships: H I 2 2 Solve: 2 2 HI HI 2 KH K  KH I  I  HI H HII eq eq 2 22 2 2 2 eq H I H I 2 2 2 2 690.0200.020HI 0.17 M www.ThesisScientist.comDisturbing and ReEstablishing Equilibrium • Once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same. • However, if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is reestablished. • The new concentrations will be different, but the equilibrium constant will be the same.  Unless you change the temperature. www.ThesisScientist.comLe Châtelier’s Principle • Le Châtelier’s principle guides us in predicting the effect on the position of equilibrium when conditions change. • ―When a chemical system at equilibrium is disturbed, the system shifts in a direction that will minimize the disturbance.‖ www.ThesisScientist.comAn Analogy: Population Changes, due to a Gold Rush When the populations of Narnia and When an influx of population enters Middle Earth are in equilibrium, the Middle Earth from somewhere outside emigration rates between the two Narnia, it disturbs the equilibrium states are equal so the populations established between Narnia and stay constant. Middle Earth. The result will be people moving from Middle Earth into Narnia faster than people moving from Narnia into Middle Earth. This will continue until a new equilibrium between the populations is established, However, the new populations will have different numbers of www.ThesisScientist.com people than the old ones.The Effect of Concentration Changes on Equilibrium • Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found. That has the same K . eq • Removing a product will increase the amounts of the other products and decrease the amounts of the reactants. You can use to this to drive a reaction to completion • Remember: Adding more of a solid or liquid does not change its concentration and, therefore, has no effect on the equilibrium. www.ThesisScientist.comThe Effect of Concentration Changes on Equilibrium, Continued When NO 2 is added, some of it combines to make more N O . 2 4 www.ThesisScientist.comThe Effect of Concentration Changes on Equilibrium, Continued When N O 2 4 is added, some of it decomposes to make more NO . 2 www.ThesisScientist.comPractice—Predict the Effect on the Equilibrium When the Underlined Substance Is Added to the Following Systems: • 2 CO (g)  2 CO(g) + O (g) 2 2 2+ 2 • BaSO (s)  Ba (aq) + SO (aq) 4 4 • CH (g) + 2 O (g)  CO (g) + 2 H O(l) 4 2 2 2 www.ThesisScientist.comPractice—Predict the Effect on the Equilibrium When the Underlined Substance Is Added to the Following Systems, Continued: • 2 CO (g)  2 CO(g) + O (g) 2 2 Shift right, removing some of the added CO and 2 increasing the concentrations of CO and O . 2 2+ 2 • BaSO (s)  Ba (aq) + SO (aq) 4 4 2+ Shift left, removing some of the added Ba and 2 reducing the concentration of SO . 4 • CH (g) + 2 O (g)  CO (g) + 2 H O(l) 4 2 2 2 Shift right, removing some of the added CO and decreasing the O , 2 2 www.ThesisScientist.com while increasing the concentration of CO . 2Effect of Volume Change on Equilibrium • For solids, liquids, or solutions, changing the size of the container has no effect on the concentration. • Changing the volume of a container changes the concentration of a gas. Same number of moles, but different number of liters, resulting in a different molarity. www.ThesisScientist.comEffect of Volume Change on Equilibrium, Continued • Decreasing the size of the container increases the concentration of all the gases in the container.  This increases their partial pressures. • If their partial pressures increase, then the total pressure in the container will increase. • According to Le Châtelier’s principle, the equilibrium should shift to remove that pressure. • The way to reduce the pressure is to reduce the number of molecules in the container. • When the volume decreases, the equilibrium shifts to the side with fewer molecules. www.ThesisScientist.comThe Effect of Volume Change on Equilibrium, Continued WS he ince n the ther pre e a ssure re more is de gc are s ased molec by incule reas sing on the the r volume, the eactants side posi of ti the r on of e e ac quil tion, ibrium when the shifts towa prerssure d the is i side w ncreit ah the sed the gr eater posi num tion be of r e of quil mol ibrium ecules shifts —the towarrd the eactan pr t si oduc de.ts. www.ThesisScientist.comThe Effect of Temperature Changes on Equilibrium • Exothermic reactions release energy and endothermic reactions absorb energy. • If we write heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s principle to predict the effect of temperature changes. However, heat is not matter and not written in a proper equation. www.ThesisScientist.comThe Effect of Temperature Changes on Equilibrium for Exothermic Reactions • For an exothermic reaction, heat is a product. • Increasing the temperature is like adding heat. • According to Le Châtelier’s principle, the equilibrium will shift away from the added heat. • The concentrations of C and D will decrease and the concentrations of A and B will increase. • The value of K will decrease. eq • How will decreasing the temperature effect the system c d CD aA + bB  cC + dD + heat K eq a b AB www.ThesisScientist.comThe Effect of Temperature Changes on Equilibrium for Endothermic Reactions • For an endothermic reaction, heat is a reactant. • Increasing the temperature is like adding heat. • According to Le Châtelier’s principle, the equilibrium will shift away from the added heat. • The concentrations of C and D will increase and the concentrations of A and B will decrease. • The value of K will increase. eq • How will decreasing the temperature effect the c d system CD K Heat + aA + bB  cC + dD eq a b AB www.ThesisScientist.comThe Effect of Temperature Changes on Equilibrium www.ThesisScientist.comPractice—Predict the Effect on the Equilibrium When the Temperature Is Reduced. • Heat + 2 CO (g)  2 CO(g) + O (g) 2 2 2+ 2 • BaSO (s)  Ba (aq) + SO (aq) (endothermic) 4 4 • CH (g) + 2 O (g)  CO (g) + 2 H O(l) (exothermic) 4 2 2 2 www.ThesisScientist.comPractice—Predict the Effect on the Equilibrium When the Temperature Is Reduced, Continued. • Heat + 2 CO (g)  2 CO(g) + O (g) 2 2 Shift left, reducing the value of K . eq 2+ 2 • Heat + BaSO (s)  Ba (aq) + SO (aq) 4 4 Shift left, reducing the value of K . eq • CH (g) + 2 O (g)  CO (g) + 2 H O(l) + Heat 4 2 2 2 Shift right, increasing the value of K . eq www.ThesisScientist.comSolubility and Solubility Product • Even ―insoluble‖ salts dissolve somewhat in water.  Insoluble = less than 0.1 g per 100 g H O. 2 • The solubility of insoluble salts is described in terms of equilibrium between undissolved solid and aqueous ions produced. + A Y (s)  n A (aq) + m Y (aq) n m • Equilibrium constant for this process is called the solubility product constant, K . sp + n m K = A Y sp • If there is undissolved solid in equilibrium with the solution, the solution is saturated. • Larger K = more soluble. sp  For salts that produce the same number of ions. www.ThesisScientist.comExample—Determine the K of PbBr if its sp 2 2 Solubility Is 1.44 x 10 M. 2+ – PbBr (s)  Pb (aq) + 2 Br (aq) 2 init 0 0 equil 0.0144 0.0288 2+ – 2 2 5 K = Pb Br = (0.0144)(0.0288) = 1.19 x 10 sp www.ThesisScientist.comExample 15.9— Calculating Molar Solubility from K sp www.ThesisScientist.comExample 15.9: • Calculate the molar solubility of BaSO . 4 10 K = 1.07 x 10 at 25 °C sp www.ThesisScientist.comExample: Information: 10 Calculate the molar solubility Given: K = 1.07 x 10 sp of BaSO . 2+ 2 4 Find: BaSO , M = Ba = SO 4 4 10 K = 1.07 x 10 at 25 °C sp 2+ 2 Equation: K = Ba SO sp 4 2+ Solution Map: K → Ba sp • Apply the solution map: 2 2 KBaSO sp 4 2 10 2 2 2 1.07  10BaBaBa 10 2 1.0710Ba 5 2 1.03  10Ba www.ThesisScientist.comActivation Energy • The energy barrier that prevents any collision between molecules from being an effective collision is called the activation energy. • The larger the activation energy of a reaction, the slower it will be. At a given temperature. www.ThesisScientist.comExothermic Reaction Activation energy, large Activation energy, Reactants small DH reaction Products Progress of reaction www.ThesisScientist.com Relative potential energyEndothermic Reaction Activation energy Products DH Reactants reaction Progress of reaction www.ThesisScientist.com Relative potential energyCatalysts • A catalyst is a substance that increases the rate of a reaction, but is not consumed in the reaction. • Catalysts lower the activation energy of a reaction. • Catalysts work by providing an easier pathway for the reaction. www.ThesisScientist.comCatalyst Effect on Activation Energy www.ThesisScientist.comCatalyst Effect on Activation Energy, Continued www.ThesisScientist.comEnzymes • Enzymes are protein molecules produced by living organisms that catalyze chemical reactions. • The enzyme molecules have an active site to which organic molecules bind. • When the organic molecule is bound to the active site, certain bonds are weakened. • This allows a particular chemical change to occur with greater ease and speed.  i.e., the activation energy is lowered. www.ThesisScientist.comSucrase www.ThesisScientist.com
sharer
Presentations
Free
Document Information
Category:
Presentations
User Name:
Dr.SamuelHunt
User Type:
Teacher
Country:
United Arab Emirates
Uploaded Date:
21-07-2017