Read Chapter 13 look at the problems Do the AP practice test on page AP13 AD it is in between numbered pages 547548 2 AP Test Historically the first question on the free response has ALWAYS been an equilibrium question ID: 1027632
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1. EquilibriumChapter 13
2. Homework Read Chapter 13 look at the problemsDo the AP practice test on page AP-13 A-D(it is in between numbered pages 547-548)2
3. AP TestHistorically, the first question on the free response has ALWAYS been an equilibrium question.The new style of test no longer requires that, however, I would be surprised if they didn’t have one on the test.3
4. EquilibriumA state when two competing reactions are canceling each other out.H3O+ + OH- ⇌ H2O + H2O You reach equilibrium when the rate of forward reaction is equal to the rate of backwards reaction. This does NOT mean the concentration of products and reactants are equal.The above reaction is at equilibrium when [H3O+] = [OH-] = 1x10-7 mol/L. In 1 L of water there are 55 moles.
5. Macro MicroMacroscopically, anything that is in equilibrium is stable. Meaning the concentrations of the different substances are not changing.Law of mass action- when the ratio of product to reactant is a constant value the system is at equilibrium Microscopically, atoms and molecules are constantly reacting in both reactions
6. Changes in ConcentrationN2(g) + 3H2(g) ⇌ 2NH3(g)
7. Equilibrium expression and constantK is calculated form EQUILIBRIUM CONCENTRATIONS! Its values may only be calculated experimentally. aA + bB ⇌ cC + dDequilibrium constant = equilibrium expression K = [C]c [D]d [A]a [B]b[ ] means concentration in M
8. Determine the equilibrium expressionFor the following:Br2(g) ⇌ 2Br(g)N2(g) + 3H2(g) ⇌ 2NH3(g)H2 (g) + Br2(g) ⇌ 2 HBr(g)HCN(aq) ⇌ H+(aq) + CN-(aq)
9. K values, K is always positiveIntermediate K. 0.00001<K<10000Significant concentrations of all substances are present.Very Large K. K >> 1The product concentration is very large with virtually no reactant concentration (the reaction has gone to completion).Very Small K. K<<1The reactant concentration is very large with virtually no product concentration.
10. Types of KKc = equilibrium constant in terms of concentration. This is implied if K is writtenKp = equilibrium concentration in terms of pressure.Ka = acid dissociation constantKb = base dissociation constantKw = ion-product constant for waterThese are listed on your equation sheet
11. Actual equation sheet11
12. K problemThe following equilibrium concentrations were observed for the Haber process at 127o C.[NH3] = 3.1 x 10-2 M, [N2] = 8.5 x 10-1 M, [H2] = 3.1 x10-3 MForward Reaction.N2(g) + 3H2(g) ⇌ 2NH3(g)Reverse Reaction.2NH3(g) ⇌ N2(g) + 3H2(g) Multiply by Factor n.1/2N2(g) + 3/2H2(g) ⇌ NH3(g)K is unitless
13. SummaryFor forward reaction jA + kB ⇌ lC + mD, K = [C]l [D]m [A]j [B]kFor reverse reaction jA + kB ⇌ lC + mD, K’ = K-1 = [A]j [B]k [C]l [D]mFor reaction njA + nkB ⇌ nlC + nmD K’’ = Kn = [C]nl [D]nm [A]nj [B]nk For an overall reaction of two or more steps, Koverall = K1 x K2 x K3 x ...
14. Equilibrium of Gases.PV = nRT or P = (n/V)RT n/V is mol/L, or a concentration.Concentration of a gas is C = P/RTTherefore you can determine the K of a gas from its pressure and temperature
15. Problem2NO(g) + Cl2(g) ⇌ 2NOCl(g)The reaction for the formation of nitrosyl chloride was studied at 25o C. The pressures at equilibrium were found to be PNOCl = 1.2 atm PNO = 5.0 x10-2 atm PCl2 = 3.0 x10-1 atmCalculate the value of K for the reaction at 25o C. PV = nRT R = .0821 atm L / mol K
16. KpSince we are always dividing by the pressure value by RT, there has to be a way to cancel that out.Kp is a value that looks at the pressure of the gases involved instead the concentrationsTo convert Kp to K K may be written Kc, or K with respect to conc. Kp = Kc (RT)n Where n = coefficients of products - coefficients of reactants This equation was removed from the equation sheet
17. Example of Kp and Kc N2(g) + 3H2(g) ⇌ 2NH3(g)
18. Same Problem2NO(g) + Cl2(g) ⇌ 2NOCl(g)The reaction for the formation of nitrosyl chloride was studied at 25o C. The pressures at equilibrium were found to be PNOCl = 1.2 atm PNO = 5.0 x10-2 atm PCl2 = 3.0 x10-1 atmCalculate the value of Kp for the reaction at 25o C, and then convert that Kp to K.
19. Pressure ProblemDinitrogen tetroxide in its liquid state was used as one of the fuels on the lunar lander for the NASA Apollo missions. In the gas phase it decomposes to gaseous nitrogen dioxide: N2O4(g) ⇌ 2NO2(g)Consider an experiment in which gaseous N2O4 was placed in a flask and allowed to reach equilibrium at a temperature where Kp = 0.133. At equilibrium, the pressure of N2O4 was found to be 2.71 atm. Calculate the equilibrium pressure of NO2(g).19
20. Homogeneous EquilibriaHomogeneous equilibria – involve the same phase: N2(g) + 3H2(g) ⇌ 2NH3(g) HCN(aq) ⇌ H+(aq) + CN-(aq)
21. Heterogeneous EquilibriaHeterogeneous equilibria – involve more than one phase: 2KClO3(s) ⇌ 2KCl(s) + 3O2(g) 2H2O(l) ⇌ 2H2(g) + O2(g)
22. The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.The concentrations of pure liquids and solids are constant. 2KClO3(s) ⇌ 2KCl(s) + 3O2(g)
23. Write the equilibrium expressions, K, for the following:The decomposition of solid phosphorus pentachloride to liquid phosphorus trichloride and chlorine gas. Deep blue solid copper(II) sulfate pentahydrate is heated to drive off water vapor to form solid green copper(II) sulfate.Small amounts of solid lead (II) iodide dissolve into ions.
24. Reaction Quotient, QQ is obtained by applying the law of mass action to INITIAL CONCENTRATIONS!Q is useful in determining which direction a reaction must shift to establish equilibrium.Q uses the exact same equation as K!!!!K is calculated from equilibrium concentrations or pressures.Q is calculated from concentrations or pressures we have right now (most likely not equilibrium).
25. Reaction Quotient, Q, numerical valueK = Q; The system is at equilibrium. No shift will occur.K < Q; The system shifts to the left.Consuming products and forming reactants, until equilibrium is achieved.K > Q; The system shifts to the right.Consuming reactants and forming products, to attain equilibrium.Write K then Q, the greater/less than sign points the way the reaction proceeds to reach equilibrium!
26. ShiftShift Right 2 H2O ⇌ H3O+ + OH-Shift -2y +y +yShift Left 2 H2O ⇌ H3O+ + OH-Shift +2y -y -y
27. K vs Q ProblemFor the synthesis of ammonia at 500o C, the equilibrium constant is 6.0 x 10-2. Predict the direction in which the system will shift to reach equilibrium in each of the following cases: N2(g) + 3 H2(g) ⇌ 2 NH3(g)Conc. (M) [NH3]o [N2]o [H2]oTrial 1 1.0 x 10-3 1.0 x 10-5 2.0 x10-3Trial 2 2.0 x 10-3 1.5 x 10-3 3.54 x10-1Trial 3 2.0 x 10-6 5.0 x10-1 1.0 x10-3