Chm 152 – Equilbrium

Name: ____KEY_________________ CHM 152 – Equilbrium (Ch. 13) Equilibrium Constant, Kc and Kp 1. Once a system has reached equilibrium, are the following true or false? a. The reaction is finished, no more products are forming. __ false __ b. The concentrations of the reactants and the products are equal. false __ c. The concentrations are no longer changing. _ true _ d. The reaction is not over, but will continue forever if isolated. _ true _ e. The speed at which products are made equals the speed at which reactants form. true 2. What is equal at equilibrium? ____forward and reverse rates_____ 3. What general information can be gathered by observing the magnitude of the equilibrium constant? Whether a reaction is reactant- or product-favored. 4. Answer the following for the reaction of NO gas with chlorine gas to produce NOCl gas. Write out the balanced reaction, and the Kc and Kp expressions. 2 PNOCl [ NOCl]2 2 NO(g) + Cl2(g)  2 NOCl(g) Kc = K p= 2 [NO]2 [Cl 2 ] (PNO )(PCl 2 ) 5. Write the equilibrium expression Kc for the reaction between potassium phosphate and calcium nitrate in water. Show the balanced reaction. 2 K3PO4(aq) + 3 Ca(NO3)2(aq)  6 KNO3(aq) + Ca3(PO4)2(s) Kc = [KNO 3 ]6 [K 3 PO 4 ]2 [Ca(NO 3 ) 2 ]3 Kp = NA 6. Write the equilibrium expression Kc for the reaction between sodium carbonate and calcium hydroxide in water. Na2CO3(aq) + Ca(OH)2(aq)  2 NaOH(aq) + CaCO3(s) Kc = [NaOH]2 / [Na2CO3][ Ca(OH)2] 7 .Write the expression for Kc for the reaction: Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) Kc = [Ag(NH3)2+] / [Ag+][(NH3)2] 8. Write the expression for Kp for the reaction : H2(g) + Br2(l)  2 HBr(g) Kp = PHBr2 / (PH2) 9. Write the expression for Kp for the reaction: Kp = 1 / PCO2 CO2(g) + CaO(s)  CaCO3(s) Name: ____KEY_________________ 10. Write Kc expressions for the following reactions: a) 3 O2 (g)  2 O3 (g) Kc = [(O3)2] / [(O2)3] b) N2 (g) + 3 H2 (g)  2 NH3 (g) Kc = [(NH3)2] / [N2][(H2)3] c) H2 (g) + I2 (g)  2 HI (g) Kc = [(HI)2] / [H2][I2] d) PCl5 (g)  PCl3 (g) + Cl2 (g) Kc = [PCl3][Cl2] / [PCl5] e) SO2 (g) + ½ O2 (g)  SO3 (g) Kc = [SO3] / [SO2][(O2)1/2] 11. Write Kp expressions for each of the following reactions: a) Ni(s) + 4CO(g)  Ni(CO)4(g) Kp = PNi(CO)4 / (PCO)4 b) 5CO(g) + I2O5(s)  I2(g) + 5CO2(g) Kp = (PI2)(PCO2)5 / (PCO)5 c) Ca(HCO3)2(aq)  CaCO3(s) + H2O(l) + CO2(g) d) AgCl(s)  Ag+(aq) + Cl(aq) Kp = PCO2 Kp = 1 12. Arrange the reactions in order of their increasing tendency to proceed toward completion: _B_ _C_ _D_ _A_ (a) 4NH3(g) + 3O2(g)  2N2(g) + 6H2O(g) Kp = 1 x 10228 atm (b) N2(g) + O2(g)  2NO(g) Kp = 5 x 10-31 (c) 2HF(g)  H2(g) + F2(g) Kp = 1 x 10-13 (d) 2NOCl(g)  2NO(g) + Cl2(g) Kp = 4.7 x 10-4 atm 13. Nitrogen dioxide dimerizes to form dinitrogen tetraoxide: 2 NO2(g)  N2O4(g) Calculate the value of Kc, given that the gas phase equilibrium constant, Kp, for the reaction is 1.3 × 103 at 273 K. (R = 0.08206 L·atm/mol·K) n = -1; Kp = Kc(RT)n = 2.9 x 104 Name: ____KEY_________________ Calculating K, Q, ICE tables 14. Answer the following for the reaction of NO gas with chlorine gas to produce NOCl gas. Calculate Kp and Kc at 25.0oC given the pressure of NO gas is 1.56 atm, chlorine gas is 0.887 atm, and NOCl gas is 3.45 atm. Kp = = 5.51398 = 5.51 Kc = = 135 15. Calculate the equilibrium constant for this reaction: 2 PO2Br (aq)  2 PO2 (aq) + Br2 (aq) Given: [PO2Br] = 0.0255M, [PO2] = 0.155M, and [Br2] = 0.00351M at equilibrium. K = [PO2]2 [Br2] / [PO2Br]2 = (0.155)2(0.00351) / (0.0255)2 = 0.130 16. Calculate the equilibrium constant Kc for this unbalanced reaction: 2 SO3 (g)  2 SO2 (g) + O2 (g) Given: [SO3] = 0.0255M, [SO2] = 1.08M, and [O2] = 1.45M at equilibrium. Kc = = 2.60 x 103 = 17. Use the equilibrium reaction: H2 (g) + I2 (g)  2 HI (g) to calculate Kc for each of the sets of equilibrium concentrations below. a. [H2] = 0.0505 M [I2] = 0.0498 M [HI] = 0.349 M Kc = 48.4 b. [H2] = 0.00560 M [I2] = 0.000590 M [HI] = 0.0127 M Kc = 48.8 c. [H2] = 0.00460 M [I2] = 0.000970 M [HI] = 0.0147 M Kc = 48.4 What do you notice about the 3 Kc values? They’re all about the same value (or should be close). 18. For the combination reaction: H2 (g) + I2 (g)  2 HI (g), calculate all three equilibrium concentrations when [H2]o = [I2]o = 0.200 M and Kc = 64.0. ICE table: x = 0.160 M [H2] = 0.200 - x = 0.040 M [I2] = 0.200 - x = 0.040 M [HI] = 2 (x) = 0.320 M Name: ____KEY_________________ 19. For the combination reaction, PCl3 (g) + Cl2 (g)  PCl5 (g), calculate all three equilibrium concentrations when Kc = 16.0 and [PCl5]o = 1.00 M. Bad problem – no solution. You can set up the ICE table and equilibrium expressions, but the solution doesn’t work out with these numbers. I’ll fix it later…. ICE table: Kc = (1.00 - x) / x2 = 16.0 20. For the decomposition reaction, COCl2 (g)  CO (g) + Cl2 (g), calculate all three equilibrium concentrations when Kc = 0.680 with [CO]o = 0.500 and [Cl2]o = 1.00 M. ICE table: Kc = (0.500 – x)(1.00 – x) / x = 0.680 x = (2.18 ± 1.66) / 2 = 0.26048 [COCl2] eq = 0.260 M, [CO] eq = 0.500 – 0.260 = 0.240 M, [Cl2] eq = 1.00 – 0.260 = 0.740 M 21. We place 0.064 mol N2O4 (g) in a 4.00 L flask at 200 K. After reaching equilibrium, the concentration of NO2(g) is 0.0030 M. What is Kc for the reaction N2O4(g)  2 NO2(g)? x = 0.0015 [N2O4]eq = 0.0145, [NO2]eq = 0.0030 Kc = 6.2 x 10-4 22. Carbonyl bromide decomposes to carbon monoxide and bromine: COBr2(g)  CO(g) + Br2(g) Kc is 0.190 at 73oC. If an initial concentration of 0.330 M COBr2 is allowed to equilibrate, what are the equilibrium concentrations of COBr2, CO, and Br2? Kc = x2 / (0.330 – x) = 0.190 x = 0.1728 [COBr2]eq = 0.330 – 0.1728 = 0.157 M; [CO]eq = [Br2]eq = 0.173 M 23. H2(g) + CO2(g)  H2O(g) + CO(g) a) It is found at 986oC that there are 11.2 atm each of CO and water vapor and 8.8atm each of H2 and CO2 at equilibrium. Calculate the equilibrium constant. Kp = (11.2)(11.2) / (8.8)(8.8) = 1.6198 = 1.62 b) If there were 8.8 moles of H2 and CO2 in a 500.0mL container at equilibrium, how many moles of CO(g) and H2O(g) would be present? Kc = 1.62 = x2 / y2 y = 8.8 moles / 0.5000 L = 17.6 24. Consider the equilibrium: 2N2O(g) + O2(g) 4NO(g) 1.62 = x2 / 309.76 x = 11.2 moles Name: ____KEY_________________ 3.00 moles of NO(g) are introduced into a 1.00-Liter evacuated flask. When the system comes to equilibrium, 1.00 mole of N2O(g) has formed. Determine the equilibrium concentrations of each substance. Calculate the Kc for the reaction based on these data. 2N2O(g) 0 +2x 2x + O2(g) 0 +x x 4NO(g) 3.00 M -4x 3.00 – 4x Kc = (3.00 – 4x)4 / (2x)2(x) 1.00 mole of N2O in 1.00 L = 1.00 M; Therefore, 2x = 1.00 M and x = 0.50 M Kc = (3.00 – 4x)4 / (2x)2(x) = (3.00 – 4*0.5)4 / (2*0.5)2(0.5) = 2 [NO]eq = [N2O]eq = 1.00 M; [O2]eq = 0.50 M 25. For the reaction: SiH4(g) + O2(g)  SiO2(g) + H2O(g) a. Balance the equation. SiH4(g) + 2 O2(g)  SiO2(g) + 2 H2O(g) b. Write the equilibrium expression for the forward reaction: Kc = [SiO2][H2O]2 / [SiH4][O2]2 c. Write the equilibrium expression for the reverse reaction: Kc = [SiH4][O2]2 / [SiO2][H2O]2 d. What is the equilibrium constant in the forward direction if [SiH4] = 0.45M; [O2] = 0.25M; [SiO2] = 0.15M; and [H2O] = 0.10M at equilibrium? Kf = (0.15)(0.10)2 / (0.45)(0.25)2 = 0.053 e. What is the equilibrium constant in the reverse reaction? Kf = (0.45)(0.25)2 / (0.15)(0.10)2 = 19 f. If [SiH4] = 0.34M; [O2] = 0.22M; [SiO2] = 0.35M; and [H2O] = 0.20M, what would be the reaction quotient (Q) in the forward direction? Q = (0.35)(0.20)2 / (0.34)(0.22)2 = 0.85 g. Which direction will the reaction in part f go? (Toward products or reactants?) Toward reactants 26. Phosphorus pentachloride decomposes into phosphorus trichloride and chlorine gas. What is the initial concentration of phosphorus pentachloride if at equilibrium the concentration of chlorine gas is 0.500M? Given: Kc = 10.00 (Hint: ICE table) I C E PCl5  x -0.500 (x-0.500) PCl3 + Cl2 0 0 +0.500 +0.500 (0.500) (0.500) 10.00 = (0.500)2 / (x-0.500) 10.00 x – 5 = 0.25 x = 0.525 M = [PCl5] Name: ____KEY_________________ 27. A 1.000 L flask is initially filled with 1.000 mole of hydrogen gas and 2.000 moles of iodine gas at 448oC. At this temperature Kc is 50.5. Calculate the equilibrium concentrations for all the chemical species in the reaction, which is hydrogen gas and iodine gas produce HI gas. H2(g) + I2(g)  2 HI(g) I 1.000M 2.000M 0 C -x -x +2x E (1.000-x) (2.000-x) (2x) Kc = = = 50.5 50.0(1.000-x)(2.000-x) = 50.5(2.000-3.000+x2) = (2x)2 101-151.5x + 50.5x2 = 4x2 46.5x2 – 151.5x + 101.0 = 0 Plug into quadratic x = {+151.5 +/} / 2(46.5) X = 2.3231 or 0.93498 (the first gives negative concentrations so is wrong) [H2] = 1.000-x = 0.065M [I2] = 2.000-x = 1.065M [HI] = 2x = 1.870M 28. Answer the following three questions for this reaction: the formation of hydrogen chloride gas. a. Write the Kc expression. H2(g) + Cl2(g)  2 HCl(g) Kc = [HCl]2 / [H2][Cl2] b. Given Kc = 3.56 x 105 calculate Kp at 25.0oC. Kp = (3.56 x 105)(RT)0 = 3.56 x 105 c. If the original concentrations are [H2] = 1.55M and [Cl2] = 1.55M, calculate all the equilibrium concentrations for each species in the reaction in moles/L. I C E H2(g) + Cl2(g)  2 HCl(g) 1.55 1.55 0 -x -x +2x (1.55-x) (1.55-x) (2x) 3.56 x 105 = (2x)2 / (1.55-x)2 Take square root both sides, solve for x = 1.54 [H2] = [Cl2] = 0.01M and [HCl] = 3.08M 29. A 0.0240 mol sample of N2O4(g) is allowed to reach equilibrium with NO2(g) in a 0.372 L flask at 25.0oC. Calculate the concentration of N2O4(g) at equilibrium. N2O4(g)  2 NO2(g) Kc = 4.61 x 10-3 I 0.06452M 0 C -x +2x E (0.06452-x) (2x) Name: ____KEY_________________ 4.61 x 10-3 = 4x2 / (0.06452 – x) 4x2 = -4.61 x 10-3x + 2.974 x 10-4 4x2 + 4.61 x 10-3x - 2.974 x 10-4 = 0 quadratic formula with a = 4, b = +4.61 x 10-3 and c = - 2.974 x 10-4 x = 8.065 x 10-3 (the other solution is negative and can’t have negative concentration) thus [N2O4] = 0.0565M 30. For the reaction 2 HI(g)  H2(g) + I2(g) Kc = 12.3. If [H2] = [I2] = [HI] = 3.21 x 10-3 M, which one of the following statements is true? a. The concentrations of H2 and HI will decrease as the system approaches equilibrium b. The concentrations of H2 and I2 will increase as the system approaches equilibrium c. The system is at equilibrium, so the concentrations will not change d. The concentration of HI will rise as the system approaches equilibrium e. We need pressures to solve this problem Q = (3.21 x 10-3) (3.21 x 10-3) / (3.21 x 10-3)2 = 1 Q < K so reaction goes forward so there will be more products Le Chatelier’s Principle 31. a. b. c. d. Consider this endothermic reaction: 3 O2(g)  2 O3(g). To shift this reaction to the reactants: You could ___ decrease ___ the pressure. ( increase or decrease) You could ____ increase ___ the volume. ( increase or decrease) You could ____ remove ____ oxygen gas. ( add or remove) You could ___ decrease __ the temperature. ( increase or decrease) 32. Which of the following, if increasing, will change the value of the equilibrium constant? ________ a. Pressure b. Volume c. [Product] d. Temperature e. [Reactant] 33. a. b. c. Consider this reaction: 2 SO2(g) + O2(g)  2 SO3(g). To shift this reaction towards the products: You could ___increase___ the pressure. (increase or decrease) You could ___decrease____ the volume. (increase or decrease) You could ___add____ oxygen gas. (add or remove) 36. For each system described below, indicate in which direction the equilibrium will shift when each stress is added or removed. Also explain how the system will react to alleviate the stress. a) N2 (g) + 3 H2 (g)  2 NH3 (g): more H2 is added to this reaction at equilibrium. Reaction will shift toward products to offset the additional H2 b) Using the same reaction, some NH3 is removed from the reaction when it is at equilibrium. Reaction will shift toward products to produce more NH3 c) 2 SO2 (g) + O2 (g)  2 SO3 (g) + heat: the system temperature goes up (heat is added). Toward reactants; heat is a product, adding heat shifts reaction left. d) Using the same reaction, heat is removed (that is, the temperature goes down). Toward products: heat is a product; removing some shifts reaction right. Name: ____KEY_________________ e) PCl3 (g) + Cl2 (g)  PCl5 (g): volume is reduced by half. Toward products: reducing volume shifts rxn to side with fewer moles. f) Using the same system as above, a catalyst is added to the system. No change; catalysts do not affect equilibrium. g) H2 (g) + Cl2 (g)  2 HCl (g): volume is doubled. No change; changing volume or pressure will not affect this system; same # moles on both sides. h) Using the same system as above, some neon is added to the system. No change; neon is an inert gas; it won’t react with or affect the system. 37. Explain how the following changes in reaction conditions will affect the position of the equilibrium below, and explain your reasoning. A(g) + B(aq)  C(s) ΔHrxn= -453 kJ/mol 1) The pressure of A in the reaction chamber is increased. The reaction is pushed toward products. 2) The temperature of the reaction is increased by 200 C. Because heat can be thought of as being a product, the reaction will be pushed toward reactants. 3) A catalyst is added to the system. No change. A catalyst doesn’t change the equilibrium position, it only changes how quickly equilibrium is reached. 4) As the reaction progresses, more of compound B is steadily added to the reaction chamber. The reaction is pushed toward products. 5) An inhibitor is added to the reaction chamber. No change, though the reaction will move more slowly. 6) Argon gas is added to the reaction chamber, doubling the pressure. No change. If the partial pressure of gaseous comopunds is changed, the equilibrium will shift position. However, adding argon gas doesn’t change the partial pressures of A, so the equilibrium position is unaffected.