Unit IV Chemical Equilibrium Focusing on Acid-Base Systems - PowerPoint Presentation

1. Unit IVChemical Equilibrium Focusing on Acid-Base Systems

2. Chemical System: A system is described in terms of empirical properties such as temperature, pressure, volume and amount of substance present.By definition, all chemical systems are fast, quantitative, stoichiometric and spontaneous! However, this assumption is NOT always true.

3. A chemical system that is separate from it’s surroundings where no matter can enter or leave is called a closed system.When a bottle of carbonated beverage is opened, the pressure on the system changes and dissolved gas is allowed to leave the system. The equilibrium has been disturbed.

4. Closed Systems at EquilibriumEvidence from many chemical reactions occurring in a closed system has shown us that after some reactions appear to have stopped, there is a mixture of reactants and products present. Na2SO4(aq) + CaCl2(aq) CaSO4(s) + 2 NaCl(aq)forwardreverseWe assume that any closed system with no observable changes occurring is in a state of dynamic equilibrium.The forward reaction (collisions between reactants to form products) and the reverse reaction (collisions between products to form reactants) are occurring simultaneously and at the same rate.

5. phase equilibriumsolubility equilibriumEquilibrium (constant macroscopic properties) can be classified as one of three types.H20 (l)  H2O(g)CuSO4(s)  Cu2+(aq) + SO4 2-(aq)where the rate of heating = rate of coolingWhere rate of crystallization = rate of dissolving in a closed system

6. Chemical Reaction EquilibriumH2(g) + I2(g)  2HI(g), t = 448°C

7. 1 – start with reactants only 2 – start with a mixture of reactants and product 3 – start with products only No matter the starting conditions, the system reaches a state of dynamic equilibrium each time.H2(g) + I2(g)  2HI(g), t = 448°C

8. The rate of the forward reaction decreases as the number of reactant molecules decreases (fewer collisions).The rate of the reverse reaction in increases as the number of product molecules increases (more collisions).Dynamic equilibrium is reached when the rate of the forward reaction is equal to the rate of the reverse reaction.H2(g) + I2(g)  2HI(g), t = 448°C

9. Dynamic Equilibrium and the Collision Reaction TheoryDynamic equilibrium occurs when there is a balance between two opposing processes at the same rateIt is attributed to the collision reaction theory which states that particles are in constant motion and as a result continuously collide, rearrange and reorient themselves. As a result, there is equal competition between collisions of reactants to form products and products to form reactants.

10. Percent YieldPercent yield provides a way to refer to the chemicals present in equilibrium systems.The maximum possible yield is calculated using stoichiometry, assuming all of the reactant molecules are used up to form products. In this case only 78% of the theoretical amount of the products was produced.78%H2(g) + I2(g)  2HI(g), t = 448°C

11. Based on percent yield, equilibrium systems fall under one of the above classifications.Percent Reaction = output x 100% inputNOTE: A negligible % yield indicates that there is NO apparent reaction taking place (non-spontaneous)

12. Sample ProblemFor each of the following, write the chemical reaction equation with appropriate equilibrium arrows1a. pH measurements indicate that acetic acid in vinegar is approximately 1% ionized into hydrogen and acetate ions.1b. Quantitative analysis of the reaction between sodium sulfate and calcium chloride solutions show that products are favored.1c. An aluminum sulfate solution reacts quantitatively with a sodium hydroxide solution. 1%CH3COOH(aq)  H+(aq) + CH3COO-(aq) >50%Na2SO4(aq) + CaCl2(aq)  2 NaCl(aq) + CaSO4(s)Al2(SO4)3(aq) + 6 NaOH(aq)  3 Na2SO4(aq) + 2 Al(OH)3(s)

13. Sample ProblemWhen 1.5 mol of chlorine gas was mixed with excess carbon monoxide in a 1.00L container, 0.80 mol of COCl2(g) was produced at equilibrium. Determine the % reaction.

14. Equilibrium Law and the Equilibrium Constant (Kc)According to equilibrium law, when a chemical system is at equilibrium, the value for this equilibrium is constant.An equilibrium law expression is based on a balanced equation for the reaction system. Balance the equation using whole number co-efficients and IGNORE the concentrations of PURE solids and liquids!!!The equilibrium constant is a unitless numerical value that is mathematically equal to the concentration of all product species, divided by the concentration of all reactant species, raised to the power of the co-efficient of that species

15. The Equilibrium Constant, KcConsider the following generic reaction equation for a system at equilibrium:a, b, c, d – coefficients A, B, C, D – chemical formulasThe equilibrium law expression allows us to calculate the value of the equilibrium constant, Kc.If Kc > 1, then the products are favoured at equilibrium.If Kc < 1, then the reactants are favoured at equilibrium.productsreactantsThe greater the value of the equilibrium constant, the more the products are favoured at equilibrium. a A + b B  c C + d D

16. Guidelines for Determining KcPure SOLIDS are NOT included in the equilibrium constant formula since the concentration (density) does not vary.2 NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)

17. Guidelines for Determining KcPure LIQUIDS are NOT included in the equilibrium constant since the concentration (density) does not vary. A pure liquid occurs when two different elements combine to produce a liquid; otherwise the liquid is classified as a homogeneous mixture.2 H2(g) + O2(g)  2 H2O(l)

18. Guidelines for Determining KcAqueous IONS and GASES in solution ARE included since the concentration of aqueous ions and gases can be altered by varying the volume of solvent. N2H4(g) + H2O(l)  N2H5+(aq) + OH-(aq)

19. Determining Equilibrium using Kc less than 1Reactants are favored(more reactant than product at equilibrium)<50%greater than 1Products are favored(more product than reactant at equilibrium)>50%equal to 1Neither is favoredEQUILIBRIUM YIELD IS EQUAL TO 50%

20. Note:An equilibrium constant value is dependent upon:The systems temperatureAn equilibrium constant value is independent upon:The reagent concentrationAny catalyst presentThe time taken to reach equilibrium

21. First, write the balanced equation with whole-number coefficients.2 NO(g) + O2(g)  2 NO2(g)

22. formationdecompositionreciprocalKc = 40 (given)Kc = 0.025reciprocal2 HI(g) H2(g) + I2(g)H2(g) + I2(g)  2 HI(g)

23. Equilibrium law expressions do NOT include solids or liquids because their concentrations are fixed – the chemical amount (number of moles) per unit volume is a constant value.NH4Cl(s)  NH3(g) + HCl(g)

24. Ions in solution must be represented as single entities. Equilibrium constant expressions are always written from the net ionic form of reaction equations. Spectator ions are not included.Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq)

25. Solutions of carbolic acid, commonly known as phenol (HC6H5(aq)) are widely used as disinfectants. One such solution has a concentration of 6.44 x 10-2 mol/L and a pH of 5.60. Carboxylic acid dissociates in water according to the equationHC6H5O(aq) + H2O(l)  C6H5O-(aq) + H3O+(aq)The Kc expression for the reaction is Kc = [C6H5O-(aq)][H3O+(aq)] b. Kc = [HC6H5O(aq)][H2O(l)] [HC6H5O(aq)][H2O(l)] [C6H5O-(aq)][H3O+(aq)] c. Kc = [C6H5O-(aq)][H3O+(aq)] d. Kc = [HC6H5O(aq)] [HC6H5O(aq)] [C6H5O-(aq)][H3O+(aq)] Sample Diploma Question

26. Sample Diploma QuestionFor the steam-hydrocarbon reforming process, the equilibrium law expression isKc = [CO2(g)][H2(g)]4 [CH4(g)][H2O(g)]2The reaction described by this equilibrium is:CH4(g) + 2 H2O(g)  CO2(g) + 4 H2(g)CO2(g) + 4 H2(g)  CH4(g) + 2 H2O(g)CH4(g) + H2O(g)  CO2(g) + H2(g)CO2(g) + H2(g)  CH4(g) + H2O(g)

27. Sample Diploma QuestionCoal and natural gas contain trace amounts of sulfur compounds which, when burned, may lead to acid rain pollution. A reaction that is related to acid rain is as follows:2 H2S(g) + 3 O2(g)  2 H2O(g) + 2 SO2(g)The equilibrium law expression for this reaction is: Kc = [H2O(g)]2 +[SO2(g)]2 b. Kc = [H2S(g)]2 + [O2(g)]3 [H2S(g)]2 +[O2(g)]3 [H2O(g)]2 + [SO2(g)]2 c. Kc = [H2O(g)]2 [SO2(g)]2 d. Kc = [H2S(g)]2 [O2(g)]3 [H2S(g)]2 [O2(g)]3 [H2O(g)]2 [SO2(g)]2

28. Sample Diploma QuestionNitrogen fixation occurs slowly in the atmosphere. The equation for this reaction is: N2(g) + O2(g) + 180.4 kJ  2 NO(g) Kc = 4.0 x 10 -31The equilibrium, expression for this reaction is:Kc = 2 [NO(g)] b. Kc = [NO(g)]2 [N2(g)][O2(g)] [N2(g)][O2(g)] Kc = [N2(g)][O2(g)] d. Kc = [N2(g)][O2(g)] 2 [NO(g)] [NO(g)]2

29. Sample Diploma Question continued…N2(g) + O2(g) + 180.4 kJ  2 NO(g) Kc = 4.0 x 10 -31At equilibrium, if the [O2(g)]= [N2(g)], then:[NO(g)] = [N2(g)][NO(g)] > [N2(g)][NO(g)] = 2 [N2(g)][NO(g)] < [N2(g)]

30. Finding concentrations at equilibriumIn order to determine the equilibrium constant for a given reaction, equilibrium concentrations are required. If they are not given, an ICE table can be used to determine the equilibrium concentrationI: Initial concentrationC: Change in concentration (can be either an increase or a decrease)E: Equilibrium concentration

31. Creating ICE tablesDetermine the reaction that is occurring at equilibriumSet up an ICE tableConcentration[reactant][product]InitialChangeEquilibrium3. Fill in all information given in the problem4. Calculate the change for one chemical species5. Using stoichiometry, (mol-mol) determine the change in concentration for each chemical species6. Calculate the equilibrium concentration

32. NOTESince all substances in the reaction are gases, stoichiometric calculations can involve concentration directly. This is because the volume will be the same for every gaseous substance in the container.This is also true when every substance in a reaction is an aqueous entity dissolved in the same volume of solvent.

33. ICE TablesI – initialC – changeE – equilibriumICE tables are used for quantitative calculations involving chemical equilibrium systems that are not quantitative (i.e. < 99.9% yield).

34. Predicting Final Equilibrium Concentrations

35. Graphing Equilibrium Reactions

36. Sample Diploma QuestionThe Haber-Bosch process for the industrial production of ammonia involves the equilibriumN2(g) + 3 H2(g)  2 NH3(g) + 92.2 kJIn a lab experiment a chemical engineer injects 0.20 mol of N2(g) and 0.60 mol of H2(g) into a 1.0L flask at 500oC. She records her analysis of the contents of the flask at 5 second intervals in the following table.Time (s)Concentration (mol/L)N2(g)H2(g)NH3(g)00.200.600.0050.140.420.12100.110.330.18150.100.300.20200.100.300.20250.100.300.20

37. Sample diploma continued… Analyze the data. Your response should include:A plot of the concentration of N2(g), H2(g) and NH3(g), versus time on the graph paper providedThe time required to establish equilibriumThe equilibrium constant for the reaction

38.

39. Sample Diploma Question

40. Read pgs. 676 – 687pg. 682 Practice #’s 1 – 3 Lab exercise 15.A on p.683Lab exercise 15.B on p.686pgs. 688 – 689 Section 15.1 Questions #’s 1, 3, 5 – 10 Homework: