Name______________________________________________period____________AP Unit 9: equilibrium 1. What is equilibrium? Rate of the forward reaction equals the rate of the reverse reaction 2. How can you tell when equilibrium has been reached? The concentrations of the reactants and products don’t change. 3. What is the only factor that can change the value of Kc? temperature 4. Write the equilibrium expression for the following reactions. a. H2(g) + Cl2(g) ↔2HCl(g) [HCl]2 / [H2][Cl2] c. 2NO(g) + O2(g) ↔2NO2(g) [NO2]2 / [NO]2[O2] b. 4NH3(g) + 7O2(g) ↔ 4NO2(g) + 6H2O (g) [H2O]6 [NO2]4/ [O2]7[NH3]4 d. 2N2O5(g) ↔ 4NO2(g) +O2(g) [NO2]4[O2] / [N2O5]2 f. FeO(s) + H2(g) ↔ Fe(s) + H2O (g) g. 4HCl (g) + O2(g) ↔2H2O(l) + 2Cl(g) [H2O]/ [H2] [Cl]2/[HCl]4[O2] 5. When the following reactions come to equilibrium, does the equilibrium mixture contain mostly reactants or mostly products? a. N2 + O2  2NO; Kc= 1.5 x 10-10 reactants b. 2SO2 + O2  H2O + 2SO3; Kc =2.5 x 109 products 6. If Kc = 0.042 for PCl3 (g) + Cl2 (g) ↔PCl5 (g) at 500 K, what is the value of Kp for this reaction at this temperature? 1.0 x 10-3 7. The equilibrium constant for this reaction 2SO3(g) ↔ 2SO2(g) + O2 (g) is Kc = 2.4 x 10 -3 at 200 ⁰C. a. Calculate Kc for 2SO2(g) + O2 (g)↔ 2SO3(g) 420 b. Does the equilibrium favor SO2 and O2, or does it favor SO3 at this temperature? SO3 8. When does Kp = Kc? when the number of moles of gas are the same on both sides of the equation 9. The reaction AB4C(g) ⇄ 2 B2 (g) + AC (g) reached equilibrium at 900 K in a 5.00 L vessel. At equilibrium, 0.084 mol of AB4C, 0.035 mol of B2, and 0.059 mol of AC were detected. What is the equilibrium constant (Kc) at this temperature for this reaction? Is this reaction product favored or reactant favored at equilibrium? 3.5 x 10-5 Reaction favored 10. Carbonyl bromide, COBr2, decomposes to CO and Br2 at 73⁰C. If you begin with 0.10 moles of COBr2 in a 2.5 Liter flask and find that there are 0.015 moles of COBr2 at equilibrium, calculate the equilibrium concentrations of the products. What is KC for this reaction? Is this reaction reactant-favored or product-favored at equilibrium? COBr2 (g) ↔ CO (g) + Br2 (g) 0.19 M Reactant favored 11. Hydrogen fluoride gas can be formed from hydrogen and fluorine gases according to the balanced equation shown below. At a specific temperature, the equilibrium constant KP for this reaction is 37.6. What are the partial pressures of all of the gases at equilibrium when the initial partial pressures of hydrogen gas and fluorine gas are 0.600 atm? H2 (g) + F2 (g) ↔ 2 HF (g) 0.150 atm for H2; 0.150 atm for F2; 0.900 atm for HF 12. Nitrogen gas reacts with oxygen gas at high temperature to make nitrogen oxides. This is a major source of air pollution from auto exhaust. At 1500 K, the equilibrium constant for the formation of nitrogen monoxide from its elements is 1.0 x 10-5. Calculate the equilibrium concentration of NO when the initial concentration of N2 is 0.60 M and O2 is 0.30M. 1.3 x 10 -3 M 13. The equilibrium constant, Kc, for the following reaction is 5.9 x 10 -3 at 250⁰C, Suppose 0.17 moles of N2O4(g) is placed in a 5.00 L flask. What is the equilibrium concentration of NO2(g)? N2O4(g) ↔ 2 NO2(g) 0.014 M 14. Define Reaction Quotient (Q). What is the difference between the reaction quotient and the equilibrium constant (K)? reaction quotient can have values at any time in the reaction and equilibrium constant must use values at equilibrium. Q is equal to [products]/[reactants], but not necessarily at equilibrium. Use the following information for questions 16 and 17: The formation of ammonia is an extremely important reaction worldwide for the production of fertilizers and explosives. At 200°C, the equilibrium constant, Kc, is 0.65. N2 (g) + 3 H2 (g) ⇄ 2 NH3 (g) 15. For the reaction above, Q is found to be = 3.5 at 200°C. This means that… a. More NH3(g) must form in order to reach equilibrium. b. More N2(g) must form in order to reach equilibrium. c. More H2(g) must form in order to reach equilibrium. d. The reaction is at equilibrium. e. Cannot tell from the information given. 16. After this reaction runs for a while, the following concentrations are measured: [N2] = 0.07 M, [H2] = 0.009 M, and [NH3] = 0.0002 M. Is the system at equilibrium? If not, in which direction will the reaction run in order to establish equilibrium? Q = 0.78 so the system is not at equilibrium; reverse reaction would run in order to reach equilibrium 17. At a certain temperature, Kc for the reaction shown is 0.18. PCl3 (g) + Cl2 (g) ⇄ PCl5 (g) Suppose a reaction vessel at this temperature contained these three gases at the following concentrations: [PCl3] = 0.0420 M, [Cl2] = 0.0240 M, [PCl5] = 0.0050 M. Is the system at equilibrium? If not, which direction will the reaction have to proceed to get to equilibrium? Q = 4.9 so reaction is not at equilibrium; reaction will go in reverse to reach equilibrium 18. Consider the following equilibrium, for which ∆H < 0: 2SO2(g) + O2(g) ↔ 2SO3(g) How will each of the following affect an equilibrium mixture of the three gases? a. Oxygen gas is added to the system more products b. The reaction mixture is heated more reactants c. The volume of the reaction vessel is doubled more reactants d. A catalyst is added to the mixture no change e. A noble gas is added no change f. Sulfur dioxide gas is removed from the system more reactants 19. For the following reaction, ∆H = 2816 kJ: 6CO2(g) + 6H2O(l) ↔ C6H12O6(S) + 6O2(g) How is the equilibrium yield of C6H12O6 affected by a. Increasing the pressure of carbon dioxide no change b. Increasing the temperature more c. Decreasing the total pressure no change d. Removing carbon dioxide less e. Adding a catalyst no change 20. How do the following changes affect the value of the equilibrium constant for an exothermic reaction: a. Removal of a reactant or product no change b. Decrease in the volume no change c. Decrease in the temperature increases d. Addition of a catalyst no change 21. Predict the effect of decreasing the temperature on the position (forward or reverse) of the following equilibrium. a. 2NH3(g) ↔ N2(g) + 3H2(g) Δ H = 37.2 kJ reverse b. CO(g) + H2O(g) ↔ CO2(g) + H2(g) Δ H = -27.6 kJ forward 22. For the reaction below, which change would cause the equilibrium to shift toward the product? CH4(g) + 2H2S(g) + heat ↔ CS2(g) + 4H2(g) a. Decrease the concentration of dihydrogen sulfide. b. Increase the temperature of the system. c. Increase the concentration of carbon disulfide. d. Decrease the concentration of methane (CH4). CO2(g) + H2(g)  H2O(g) + CO(g) 23. When H2(g) is mixed with CO2(g) at 2,000 K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured. [H2] = 0.20 mol/L [CO2] = 0.30 mol/L [H2O] = [CO]= 0.55 mol/L (a) Using the equilibrium concentrations given above, calculate the value of Kc, the equilibrium constant for the reaction. 5.0 (b) Determine Kp in terms of Kc for this system. 5.0 (c) In a different experiment, 0.50 mole of H2(g) is mixed with 0.50 mole of CO2(g) in a 3.0-liter reaction vessel at 2,000 K. Calculate the equilibrium concentration, in moles per liter, of CO(g) at this temperature. 0.11 M 2 H2S(g)  2 H2(g) + S2(g) 24. When heated, hydrogen sulfide gas decomposes according to the equation above. A 3.40 g sample of H2S(g) is introduced into an evacuated rigid 1.25 L container. The sealed container is heated to 483 K, and 3.7210–2 mol of S2(g) is present at equilibrium. (a) Write the expression for the equilibrium constant, Kc, for the decomposition reaction represented above. K = [H2]2[S2]/[H2S]2 (b) Calculate the equilibrium concentration, in molL-1, of the following gases in the container at 483 K. (i) H2(g) 0.0595 M (ii) H2S(g) 0.0203 M (c) Calculate the value of the equilibrium constant, Kc, for the decomposition reaction at 483 K. 0.256 25. Write the Ksp expression for these compounds. a. AgI b. Al(OH)3 Ksp = [Ag][I] Ksp = [Al][OH]3 c. ZnCO3 Ksp = [Zn][CO3] d. Zn(OH)2 Ksp = [Zn][OH]2 26. A saturated solution of silver chloride contains 1.94 x 10-4 g of AgCl dissolved in 100 mL of water at 20°C. Write the equilibrium constant expression for this reaction and calculate the equilibrium constant (Ksp) for AgCl. Ksp = [Ag][Cl]; 1.83 x 10-10 27. If the molar solubility of CaF2 at 35⁰C is 1.24 x 10-3 mol/L, what is the Ksp at this temperature? 7.00 x 10-9 28. The Ksp for calcium fluoride (CaF2) is Ksp = 4.2 x 10-11 at 25°C. What is the solubility of CaF2 at this temperature? (In other words, what is the molarity of CaF2 in a saturated solution?) 0.00022 M 29. A 1.00 L solution saturated at 25⁰C with calcium oxalate is evaporated to dryness, giving a 0.0061 g residue of calcium oxalate. Calculate Ksp for this salt at 25 ⁰C. 2.3 x 10-9 30. Calculate the molar solubility of AgBr in (Ksp = 5.0 x 10-13) a) pure water b) 0.03 M AgNO3 solution c) 0.10 M NaBr solution -7 -11 7.1 x 10 M 1.7 x 10 M 5.0 x 10-12 M 31. Calculate the molar solubility of CaF2 in (Ksp = 3.9 x 10 -11) a) pure water b) 0.15 M KF solution 2.1 x 10-4 M 1.7 x 10-9 M c) 0.080M Ca(NO3)2 solution 1.1 x 10-5 M 32. Will silver sulfate precipitate when 100 mL of 0.050 M silver nitrate is mixed with 10 mL of 5.0 x 10-2 M sodium sulfate solution? (Ksp for silver sulfate = 1.5 x 10-5) Q = 9.1x10-6 Q < Ksp so no precipitate 33. A solution contains 2.0 x 10 -4 M Ag+ and 1.5 x 10 -3 M Pb2+. If NaI is added, will AgI (Ksp = 8.3 x 10 -17) or PbI2 (Ksp = 7.9 x 10 -9) precipitate first? Specify the concentration of I- needed to begin precipitation. AgI will precipitate first at [I-] = 4.2 x 10-13 M 34. a. The Ksp for AgCl is 1.8 x 10-10. If Ag+and Cl-are both in solution and in equilibrium with AgCl. What is [Ag+] If [Cl-] =0.020 M? 9 x 10 -9 M 35. Will a precipitate of lead (II) sulfate form if 100 mL of a 1.0x10-3 mol/L solution of lead (II) nitrate is added to 100 mL of 2.0x10-3 mol/L magnesium sulfate solution? (Ksp for lead sulfate = 6.3 x 10-7) Q = 5.0 x 10-7 Q < Ksp so no precipitate 36. a) Ksp = [Mg2+][F¯]2 = (1.21 x 10¯3) (2 x 1.21 x 10¯3)2 = 7.09 x 10¯9 b) Ksp = [Mg2+] (2x + 0.100)2 since 2x is much less than 0.100= 7.09 x 10¯9 = [Mg2+] (0.010)2 [Mg2+] = (7.09 x 10¯9) / (10¯2) = 7.09 x 10¯7 M Note: OK if 0.102 is used for [F¯], then Ksp = 6.76 x 10¯7 c) [Mg2+]:100.0 x 3.00 x 10¯3 = 300.0 x [Mg2+] [Mg2+] = 1.00 x 10¯3 M [F¯]: 200.0 x 2.00 x 10¯3 = 300.0 x [F¯] [F¯] = 1.33 x 10¯3 M Q = Ion Product = [Mg2+] [F¯]2 = (1.00 x 10¯3) (1.33 x 10¯3)2 = 1.77 x 10¯9 Since Q < Ksp, no precipitate will form. d) Solubility of MgF2 decreases with the increasing temperature, thus dissolution process is exothermic. MgF2 ---> Mg2+ + 2F¯ + Q (or H) Reason: i) Increased temperature puts a stree on the system (LeChâtelier). The system will reduce the stree by shifting the equilibrium in the endothermic (left) direction. OR ii) A data supported argument such as comparing ion concentrations, calculating second Ksp and giving proper interpretations Review: 36. Of the following 100 mL solutions: 0.10 M NaF, 0.10 M MgCl2, 0.10M C2H5OH, 0.10 M CH3COOH a. Which solution has the lowest electrical conductivity? Explain 0.10M C2H5OH; this is a nonelectrolyte there are no ions in solution 37. Consider the molecules CF4 and SF4 a. Draw lewis structures for each molecule b. In terms of molecular geometry, account for the fact that the CF4 molecule is nonpolar, whereas the SF4 molecule is polar CF4 is tetrahedral ;all of the dipoles cancel out and the molecule is nonpolar SF4 is see saw shape; all of the dipoles do not cancel out and the molecule is polar 38. Write a balanced net ionic equation for the following a. Liquid bromine is shaken with a 0.5M sodium iodide solution Br2 + 2I-  2Br- + I2 b. Nitric acid is added to a sodium carbonate solution 2H+ + Na2CO3 2Na+ + CO2 + H2O