Chapter 15 REVIEW. Part 1. Part 2

Transcription

1 () Yes, the evidence from many systems shows that the rate at which reactant particles are colliding to form products is equal to the rate at which products are colliding to form reactants. (3) When a system reaches equilibrium, the reactant amounts will have decreased to some specific extent. If the remaining reactants are present to much less than 1%, the reaction is considered quantitative. In reactions where significant (detectable) amounts of reactants remain, equilibrium theory must be taken into account. (4) Yes, if the equilibrium constant is known, the chemical amount of product that should form can be accurately predicted. Chapter 15 REVIEW Part 1 (Pages ) 1. B. C D 5. A 6. B 7. C 8. B 9. A Solutions 4. mol 3. p 100% 70.3% 6.00 mol Kreverse K.4 10 forward 10. [CH 4(g)] mol/l mol/l mol/l 4 [H (g)] mol/l mol/l 1 [H (g)] mol/l = mol/l Part eq 4 (Pages ) 11. Chemical equilibrium is a state of a closed system in which all macroscopic properties are constant. 1. Chemical equilibrium is explained as a balance between forward and reverse processes occurring simultaneously, at the same rate. 13. Reactants favoured refers to an equilibrium where the reactant to product ratio is quite high. If the ratio is greater than 99%, the reaction may be considered nonspontaneous or the reaction extent described as negligible. 14. (a) When a soft drink bottle has just been opened, it is in a non-equilibrium state. Carbon dioxide gas escapes from the solution as the rate of decomposition of carbonic acid into carbon dioxide exceeds the rate for formation of carbonic acid from carbon dioxide gas and water. Copyright 007 Thomson Nelson Unit 8 Solutions Manual 617

2 (b) When the bottle is sealed and at a constant temperature, it is in an equilibrium state. Carbon dioxide gas and water are in equilibrium with carbonic acid. 15. A catalyst does not affect the state of equilibrium. It affects both forward and reverse reactions, and decreases the time required to reach equilibrium. 10% [NH 16. (a) N (g) + 3 H (g) 3(g)] NH 3 (g) K c = [N (g)] [H (g)] 3 (b) H (g) + O (g) H O(g) (c) CO(g) + H O(g) 675 CO (g) + H (g) K c K c [HO(g)] = [H (g)] [O (g)] [CO (g)] [H (g)] = [CO(g)] [H O(g)] 17. At equilibrium the concentration of the product is very much greater than the concentration of the reactants. The extremely large equilibrium constant shows that the reaction is quantitative. 18. O (g) O (aq) [O (aq)] K c = [O (g)] g O(aq) 1 mol 1 L 3.00 g mol/l 1 mol 0. O (g) mol/l 4.8 L [O (aq)] mol/l 1. Kc = [O (g)] mol/l = No. The value of the equilibrium constant will not change. The only factor that affects the value of the equilibrium constant is temperature. 3. When the gas above the water is pure O (g), the concentration of reactant (oxygen gas) in the equilibrium is much higher than it is when the gas above the water is air. This shifts the equilibrium towards the product side, resulting in more O dissolving into the water. 4. If climate change leads to an increase in the average temperature of water, the solubility of O (g) in such areas will, according to the solubility rules for gases, decrease. Aquatic organisms that require oxygen will therefore have less dissolved oxygen available to them and will have lower survival rates. 5. Both increasing the concentration of the reactant (by adding more) and decreasing the concentration of the product (by removing some) will increase the yield of the product. Ideally, both these changes can be done continuously, in which case the reaction will be constantly shifting toward products and never reaching equilibrium thus producing the most product in the least time. 6. (a) Concentration C H (g) (mol/l) H (g) (mol/l) C H 4 (g) (mol/l) Initial Change Equilibrium [CH 4(g)] K c = = = [C H (g)] [H (g)] (0.94) 618 Unit 8 Solutions Manual Copyright 007 Thomson Nelson

3 (b) 7. (a) [H (g)] [Br (g)] [HBr(g)] Concentration (mol/l) (mol/l) (mol/l) Initial Change x x x Equilibrium 4.00 x 4.00 x x K c [HBr(g)] ( x) = = = 1.0 [H (g)] [ Br (g)] ( x) x x 1.0 x = 3.46 ( x) x =.54 mol/l At equilibrium, [H (g)] = [Br (g)] = ( ) mol/l = 1.46 mol/l, [HBr(g)] = x.54 mol/l = 5.07 mol/l (unrounded value of x used) (b) [H (g)] [Br (g)] [HBr(g)] Concentration (mol/l) (mol/l) (mol/l) Initial Change x x x Equilibrium 6.00 x 6.00 x x [HBr(g)] ( x) Kc = = [H (g)] [ Br (g)] ( x) = 1.0 x x x = 3.46 ( x) x = 3.80 mol/l At equilibrium, [H (g)] = [Br (g)] = ( ) mol/l =.0 mol/l [HBr(g)] = 3.80 mol/l = 7.60 mol/l Copyright 007 Thomson Nelson Unit 8 Solutions Manual 619

4 mol/l CO (g) H (g) Kc 4.00 CO(g) H O(g) CO(g).00 CO(g) 9. When a chemical system at equilibrium is disturbed by a change in some property of the system, the system adjusts, if possible, in a way that opposes the change. 30. Temperature, container volume, and individual concentrations are commonly manipulated to shift a chemical equilibrium system. 31. Decreasing the volume increases the overall system pressure, and increasing the volume decreases the overall system pressure. [NO 4(g)] 3. (a) 1.15 = t = 55 C [NO (g)] (b) [N O 4 (g)] = 1.15 [NO (g)] = 1.15 (0.050) = mol/l (units assumed) (c) According to Le Châtelier s principle, if the concentration of nitrogen dioxide is increased, then the equilibrium would shift to the right. The system shifts in such a way as to reduce the concentration of nitrogen dioxide (some reacts to produce more dinitrogen tetraoxide). 33. (a) The equilibrium system will shift to the left, partially counteracting the applied heat by undergoing an endothermic reaction. (b) The equilibrium system will shift to the left, partially counteracting the decreased total pressure by producing more gas phase molecules. (c) The equilibrium system will shift to the right, partially counteracting the increased concentration of oxygen by reacting it with HCl(g). (d) The equilibrium position will remain unchanged. Catalysts have no effect on the position of an equilibrium system. 34. (a) high concentration of C H 6 (g), low concentration of C H 4 (g) and H (g), high temperature, low overall pressure (b) high concentration of CO(g) and H (g), low concentration of CH 3 OH(g), low temperature, high overall pressure 35. (a) The equilibrium system will shift to the left, partially counteracting the applied heat by undergoing an endothermic reaction. (b) The equilibrium system will shift to the left, partially counteracting the decreased total pressure by producing more gas phase molecules. (c) The equilibrium system will shift to the right, partially counteracting the increased concentration of oxygen by reacting more oxygen with carbon monoxide. (d) No system shift will occur. Catalysts have no effect on the position of an equilibrium system. (e) The equilibrium system will shift to the right, partially counteracting the decreased concentration of carbon dioxide by reacting oxygen with carbon monoxide to produce carbon dioxide. 36. (a) The equilibrium system will shift to the right. Dissolving CuSO 4 (s) increases the concentration of Cu + (aq) ions, which initially increases the forward reaction rate, while the reverse reaction rate is not changed. (b) The equilibrium system will shift to the left. Decreasing the temperature slows down the forward (endothermic) reaction rate more than it slows the reverse (exothermic) reaction rate. 60 Unit 8 Solutions Manual Copyright 007 Thomson Nelson

5 (c) The equilibrium system will not shift. Both the forward and reverse rates are increased by the same factor as the solid surface area increases. (Adding more sodium carbonate does not change its concentration, since pure substances in the solid state have concentrations that are essentially fixed, i.e., the number of moles per unit volume of each substance is a constant.) (d) The equilibrium system will shift to the left. Decreasing the volume of the vessel compresses the CO (g) and thereby increases its concentration, which increases the reverse reaction rate, while the forward reaction rate is not changed. (e) The equilibrium system will shift to the right. Decreasing the concentration of Cl (aq) decreases the reverse reaction rate, while the forward reaction rate is not changed. (Added AgNO 3 (s) will dissolve, introducing Ag + (aq), which will react with Cl (aq) to form AgCl(s).) (f) The equilibrium system will not shift. Increasing the vessel volume allows all the gases in the system to expand and thereby decreases their concentrations. Since the concentration of each gas is changed by the same factor and the forward and reverse reactions involve the same number of moles of gases, both the forward and the reverse reaction rates are decreased equally. (g) The equilibrium system will shift to the right. Adding soluble Fe(NO 3 ) 3 (s) increases the concentration of Fe 3+ (aq), which increases the forward reaction rate, while the reverse reaction rate is not changed. 37. The forward reaction is favoured (i.e., increased) by the changes in (b) and (c). (These are the only exothermic changes.) 38. (a) The structural formula for diethanolamine is: (b) Since the forward reaction is exothermic (negative r H ), keeping the temperature low will favour the forward reaction in both equilibria; the system will shift right to try to compensate for the removal of heat. High pressure ensures that more hydrogen sulfide remains dissolved, favouring the forward progress of the solubility equilibrium for H S(g). 39. To regenerate the scrubber solution, reverse the conditions. Heat the reaction vessel to shift both equilibria to the left, and decrease the pressure to encourage the most H S to come out of the solution. 40. CO (g) + H O(l) H CO 3 (aq) (negative r H ) 41. H CO 3 (aq) + (C H 4 OH) NH(aq) (C H 4 OH) NH + (aq) + HCO 3 (aq) (negative r H ) Because this equation is very similar to the one representing the removal of H S(g), the process conditions for removing H S(g) heating the reaction vessel should work to remove CO (g) as well. Extension 4. In an industrial amine scrubber system: 1. Sour gas enters the contactor tower and rises through the descending amine solution.. Purified gas flows from the top of the tower. 3. The amine solution, carrying absorbed acid gases, leaves the tower for the heat exchanger or optional flash tank. 4. Rich amine is heated by hot, regenerated lean amine in the heat exchanger. Copyright 007 Thomson Nelson Unit 8 Solutions Manual 61

6 5. Rich amine is further heated in the regeneration still column, by heat supplied from the reboiler. The steam rising through the still liberates H S and CO, regenerating the amine. The equilibrium shifts to favour the reverse reaction. 6. Steam and acid gases, separated from the rich amine, are condensed and cooled, respectively, in the reflux condenser. 7. Condensed steam is separated in the reflux accumulator and is then returned to the still. Acid gases may be vented or directed to a sulfur recovery system. 8. Hot, regenerated lean amine is cooled in a solvent aerial cooler and circulated to the contactor tower, completing the cycle. Both the dissolving of the hydrogen sulfide gas and its reaction with diethanolamine are equilibrium processes. The conditions are manipulated to first maximize the quantity of hydrogen sulfide that is removed from the gas mixture, and then to regenerate the hydrogen sulfide after it has been separated from the gas mixture. Chemical equilibrium is crucial to this scrubber system. If the main steps were not equilibrium processes, then it would not be possible to shift the equilibrium in the desired direction. The same description applies to the removal of carbon dioxide. Refineries in Alberta that use this process to reduce sulfur content include Shell (Fort Saskatchewan), PetroCanada (Edmonton), and Imperial Oil (Edmonton). 43. The tungsten and iodine reaction to produce tungsten(ii) iodide is exothermic. A high temperature of the system forces the equilibrium to shift to the left, depositing metallic tungsten on the filament. This deposition reverses the tendency of the tungsten to be lost gradually from the filament. The presence of the halogen (iodine) helps to establish an equilibrium that, at high temperature, restores the filament. 44. At high altitude, the concentration of oxygen in air is significantly reduced. The rate at which oxygen will dissolve (react with hemoglobin) in blood depends on the concentration of gaseous oxygen in the lungs. Thus, at high altitudes the rate at which humans absorb oxygen from the air they breathe is considerably reduced. The theory of dynamic equilibrium explains that the rate of dissolving of oxygen is reduced, while the rate of oxygen escaping from solution is unchanged, by a decrease in concentration of gaseous oxygen. Le Châtelier s principle leads to the conclusion that more oxygen will escape from solution in blood as the equilibrium shifts to counteract the decreased gaseous oxygen concentration in the air. People who are born and live their lives at high altitudes tend to have larger than average lung capacities, so that the amount of oxygen absorbed is increased by increasing the total amount that is inhaled with each breath. 6 Unit 8 Solutions Manual Copyright 007 Thomson Nelson