New Jersey Center for Teaching and Learning Slide 1 / 27 Progressive Science Initiative This material is made freely available at www.njctl.org and is intended for the non-commercial use of students and teachers. These materials may not be used for any commercial purpose without the written permission of the owners. NJCTL maintains its website for the convenience of teachers who wish to make their work available to other teachers, participate in a virtual professional learning community, and/or provide access to course materials to parents, students and others. Click to go to website: www.njctl.org Slide 2 / 27 Equilibrium PART A: Introduction and Ice Tables www.njctl.org EQUILIBRIUM Slide 3 / 27 When a battery reaches equilibrium, the electrons are drawn equally to both terminals of the battery and no more useful work can be obtained.
Equilibrium Chemical equilibrium is reached when the rate of conversion to product is equal to the rate of conversion to reactants. Slide 4 / 27 2H 2O(g) <--> 2H 2(g) + O 2(g) Since the products and reactants will be made at the same rate they are used, the concentrations will remain constant. Time (min) [H2O] [H2] [O2] 0 10 0 0 5 4 6 3 10 4 6 3 Equilibrium was reached somewhere between 0 and 5 minutes. Equilibrium The time at which a reaction reaches equilibrium can be seen easily on a graph. Slide 5 / 27 2H 2O(g) <--> 2H 2(g) + O 2(g) Time (min) [H2O] [H2] [O2] 10 0 10 0 0 5 4 6 3 10 4 6 3 Conc. 5 time (min) 5 10 Equilibrium is reached after roughly 2.5 minutes. Equilibrium The equilibrium constant can be derived by examining the relationship of the forward and reverse rates at equilibrium. Slide 6 / 27 2H 2O(g) <--> 2H 2(g) + O 2(g) forward rate = kf*[h 2O] 2 = reverse rate = kr*[h 2] 2 [O 2] kf*[h 2O] 2 = kr*[h 2] 2 [O 2] Since the ratio of the rates will be constant at equilibrium... K (equilibrium constant) = kf = [H 2] 2 [O 2] kr [H 2O] 2 K = [Product] x [Reactant] y
Equilibrium Slide 7 / 27 The favorability of a reaction (degree to which it creates product) can be determined by examining the equilibrium constant. 2H 2O(g) <--> 2H 2(g) + O 2(g) Time (min) [H2O] [H2] [O2] 0 10 0 0 5 4 6 3 10 4 6 3 K = [H 2] 2 [O 2] [H 2O] 2 = [6] 2 [3] [4] 2 = 6.75 Since K > 1, the reaction is considered relatively favorable. The greater the value, the more favorable it is. If K <1, the reaction would be deemed relatively unfavorable as reactants are preferred. Equilibrium Expressions Slide 8 / 27 An equilibrium expression shows the ratio of products to reactants for a given reaction. Write the equilibrium expression for the following: CH 4(g) + 2O 2(g) --> CO 2(g) + 2H 2O(g) K = [CO 2][H 2O] 2 [CH 4][O 2] 2 Write the equilibrium expression for the following: Ag 3PO 4(s) --> 3Ag + (aq) + PO 4 3- (aq) Pure solids and liquids are considered unchanging compared to aqueous ions or gases so they are not included... K = [Ag + ] 3 [PO 4 3- ] 1 Which of the following is NOT necessarily true at equilibrium? Slide 9 / 27 A The concentration of products will equal the concentration of reactants B The rates of product formation and product degradation will be the same C The concentrations of products and reactants will remain constant D The equilibrium constant will be equal to 1 E Both A and D
1 Which of the following is NOT necessarily true at equilibrium? A The concentration of products will equal the concentration of reactants B The rates of product formation and product degradation will be the same E Slide 9 () / 27 C The concentrations of products and reactants will remain constant D The equilibrium constant will be equal to 1 E Both A and D 2 At what time does the reaction below reach equilibrium? Slide 10 / 27 A Around 4 seconds B Around 10 seconds C Around 15 seconds D Around 20 seconds Conc. 2CO(g) <--> O2(g) + C(s) O2 CO 5 10 15 20 25 Time(sec) E Around 25 seconds 2 At what time does the reaction below reach equilibrium? Slide 10 () / 27 A Around 4 seconds B Around 10 seconds C Around 15 seconds D Around 20 seconds Conc. 2CO(g) <--> O2(g) + C(s) O2 CO C 5 10 15 20 25 Time(sec) E Around 25 seconds
3 What would be the value of the equilibrium constant for the following reaction? Slide 11 / 27 A 0.06 B 0.3 C 16.7 D 0.667 E 3.33 2CO(g) <--> O 2(g) + C(s) 8 CO Conc. 4 O 2 5 10 15 20 25 Time(sec) 3 What would be the value of the equilibrium constant for the following reaction? Slide 11 () / 27 A 0.06 B 0.3 C 16.7 D 0.667 E 3.33 2CO(g) <--> O 2(g) + C(s) 8 CO Conc. 4 O 2 A 5 10 15 20 25 Time(sec) 4 Which of the following ionic compounds would produce the fewest ions in saturated solutions of each? Slide 12 / 27 A AgCl Ksp = 1.7 x 10-10 B PbCl 2 Ksp = 1.6 x 10-5 C CuS Ksp = 8 x 10-37 D Ag 3PO 4 Ksp = 1.8 x 10-18 The equilibrium constant here is expressed as Ksp which stands for "solubility product". This version of K is used to describe the dissolving process of ionic compounds.
4 Which of the following ionic compounds would produce the fewest ions in saturated solutions of each? Slide 12 () / 27 A AgCl Ksp = 1.7 x 10-10 B PbCl 2 Ksp = 1.6 x 10-5 C CuS Ksp = 8 x 10-37 D Ag 3PO 4 Ksp = 1.8 x 10-18 C The equilibrium constant here is expressed as Ksp which stands for "solubility product". This version of K is used to describe the dissolving process of ionic compounds. 5 What is the proper equilibrium expression for the following reaction? Slide 13 / 27 OCl-(aq) + H 2O(l) --> HOCl(aq) + OH-(aq) A K = [HOCl][OH-] / [OCl-][H2O] B K = [OCl-][H2O] / [HOCl][OH-] C K = [HOCl] / [OCl-] D K = [HOCl][OH-] / [OCl-] 5 What is the proper equilibrium expression for the following reaction? Slide 13 () / 27 OCl-(aq) + H 2O(l) --> HOCl(aq) + OH-(aq) A K = [HOCl][OH-] / [OCl-][H2O] B K = [OCl-][H2O] / [HOCl][OH-] C K = [HOCl] / [OCl-] D D K = [HOCl][OH-] / [OCl-]
6 What is the value of K for the following reaction given that the equilibrium concentrations of [OH - ] = 2 x 10-4 and [Cu 2+ ] = 1 x 10-4 M respectively? Slide 14 / 27 Cu(OH) 2(s) <--> Cu 2+ (aq) + 2OH - (aq) 6 What is the value of K for the following reaction given that the equilibrium concentrations of [OH - ] = 2 x 10-4 and [Cu 2+ ] = 1 x 10-4 M respectively? Slide 14 () / 27 Cu(OH) 2(s) <--> Cu 2+ (aq) + 2OH - (aq) 4x10-12 Gaseous Equilibria Slide 15 / 27 Equilibrium constants for reactions occurring exclusively in the gas phase can be expressed as either Kp or Kc values. Kp = expresses equilibrium amounts in pressures Kc = expresses equilibrium amounts in concentrations Given the following reaction, write the expressions for Kp and for Kc: CH 4(g) + 2O 2(g) --> CO 2(g) + 2H 2O(g) Kp = (P CO2)(P H2O) 2 (P CH4)(P O2) 2 Kc = [CO 2][H 2O]2 [CH 4][O 2] 2
Gaseous Equilibria Slide 16 / 27 Kp and Kc are related by the equation Kp = Kc(RT) (dmol), where dmol = mol gaseous product - mol gaseous reactant. Calculate Kc for the reaction below @ 10 C: 3H 2(g) + N 2(g) --> 2NH 3(g) Kp = 0.045 Kc = Kp / (RT) dmol Kc = 0.045 / (23.2) -2 Kc = 24.2 Note: If the change in moles is zero = Kp = Kc. This will occur when there are equal numbers of gaseous moles of product and reactant. Ice Tables Slide 17 / 27 The changing amounts of reactants and products in a reaction are often best kept track of in an "ICE" table. Given the following reaction, what are the concentrations of all materials at equilibrium given a [HCl] initial=0.5 M and [HCl] eq=0.2 M? 2HCl(g) --> H 2(g) + Cl 2(g) Initial (I) 0.5 - - Change (C) -0.3 +0.15 +0.15 Equilibrium (E) 0.2 0.15 0.15 The equilibrium amount of HCl was used to find the change in [HCl]. The changes in H 2 and Cl 2 were found using the stoichiometrical coefficients. Exactly 1/2 the amount of hydrogen gas is produced for a given amount of HCl that reacts. Ice Tables Slide 18 / 27 The equilibrium constant can be used to determine the concentrations of substances at equilibrium Given that there are initially 0.2 atm of HCl, what would be the pressure of all gases at equilibrium given the reaction below: 2HCl(g) --> H 2(g) + Cl 2(g) Kp = 0.67 Initial (I) 0.2 - - Change(C) -2x +x +x Equilibrium(E) 0.2-2x x x Kp = 0.67 = x 2 / (0.2-2x) 2 Take square root of both sides = 0.82 = x / 0.2-2x Solve for x = 0.062 atm. P HCl = 0.12 atm P H2 = P Cl2 = 0.062 atm
Tips for Equilibrium Problems Slide 19 / 27 There are three steps to solving most equilibrium problems. Step One: Write a balanced reaction, including states. Step Two: Write a proper ice table, being mindful of stoichiometrical coefficients Step Three: Write a proper expression and solve for what is required/ 7 For which of the following reactions would Kc = Kp? Slide 20 / 27 A 2H 2O(g) --> 2H 2(g) + O 2(g) B PCl 5(g) --> PCl 3(g) + Cl 2(g) C 2HBr(g) --> H 2(g) + Br 2(l) D CaCO 3(s) --> CaO(s) + CO 2(g) E Mg(s) + 2H + (aq) --> H 2(g) + Mg 2+ (aq) 7 For which of the following reactions would Kc = Kp? Slide 20 () / 27 A 2H 2O(g) --> 2H 2(g) + O 2(g) B PCl 5(g) --> PCl 3(g) + Cl 2(g) C 2HBr(g) --> H 2(g) + Br 2(l) C D CaCO 3(s) --> CaO(s) + CO 2(g) E Mg(s) + 2H + (aq) --> H 2(g) + Mg 2+ (aq)
8 What would be the value of Kp for the reaction below @25 C? Slide 21 / 27 2KClO 3(s) --> 2KCl(s) + 3O 2(g) Kc = 0.034 8 What would be the value of Kp for the reaction below @25 C? Slide 21 () / 27 2KClO 3(s) --> 2KCl(s) + 3O 2(g) Kc = 0.034 498 9 What would be the value of Kp for the reaction below where the initial pressure of methane and oxygen were 0.4 atm and the equilibrium pressure of carbon dioxide is 0.1 atm? CH 4(g) + 2O 2(g) --> CO 2(g) + 2H 2O(g) Slide 22 / 27
9 What would be the value of Kp for the reaction below where the initial pressure of methane and oxygen were 0.4 atm and the equilibrium pressure of carbon dioxide is 0.1 atm? CH 4(g) + 2O 2(g) --> CO 2(g) + 2H 2O(g) Slide 22 () / 27 0.33 10 What would be the equilibrium concentration of hydrogen gas given a initial concentrations of 0.2 M for both the CO and H 2O? Slide 23 / 27 CO(g) + H 2O(g) --> CO 2(g) + H 2(g) K = 1.56 10 What would be the equilibrium concentration of hydrogen gas given a initial concentrations of 0.2 M for both the CO and H 2O? Slide 23 () / 27 CO(g) + H 2O(g) --> CO 2(g) + H 2(g) K = 1.56 0.12
11 Which of the following would be TRUE regarding the following reaction? Slide 24 / 27 PbCl 2(s) --> Pb 2+ (aq) + 2Cl - (aq) K sp = 1.8 x 10-5 A The reaction is highly favorable B The product concentrations will be much higher than the reactant concentrations at equilibrium C The [Cl-] will be 1/2 that of [Pb2+] at equilibrium D The [Cl-] will be 2x that of [Pb2+] at equilibrium E None of these 11 Which of the following would be TRUE regarding the following reaction? Slide 24 () / 27 PbCl 2(s) --> Pb 2+ (aq) + 2Cl - (aq) K sp = 1.8 x 10-5 A The reaction is highly favorable B The product concentrations will be much higher than the reactant concentrations at equilibrium C The [Cl-] will be 1/2 that of [Pb2+] at equilibrium D The [Cl-] will be 2x that of [Pb2+] at equilibrium E None of these D 12 What is the concentration of hydroxide ions in a saturated solution of Ca(OH) 2 given the following reaction and Ksp? (Note: no initial amount of Ca (OH) 2 is given as it is a solid and does not change appreciably) Slide 25 / 27 Ca(OH) 2(s) --> Ca 2+ (aq) + 2OH - (aq) Ksp = 5.02 x 10-6
12 What is the concentration of hydroxide ions in a saturated solution of Ca(OH) 2 given the following reaction and Ksp? (Note: no initial amount of Ca (OH) 2 is given as it is a solid and does not change appreciably) Slide 25 () / 27 Ca(OH) 2(s) --> Ca 2+ (aq) + 2OH - (aq) Ksp = 5.02 x 10-6 0.0108 M Slide 26 / 27 See you in part "b" where we will deal with Le-Chatelier's principle and how the equilibrium of a reaction can be shifted... Slide 27 / 27