Chapter 15 Chemical Equilibrium Test Bank

March 26, 2018 | Author: kitty_krumley | Category: Chemical Equilibrium, Chemical Reactions, Hydrogen, Chemistry, Oxygen


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Chemistry, 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 15.1 Multiple-Choice and Bimodal Questions 1) The value of K eq for the equilibrium ˆˆ ˆ H 2 (g) + I 2 (g) ˆˆ 2 HI (g) ˆ is 794 at 25 °C. What is the value of K eq for the equilibrium below? ˆ‡ ˆ 1/ 2H 2 (g) + 1/ 2I 2 (g) ‡ ˆ HI (g) ‡ A) 397 B) 0.035 C) 28 D) 1588 E) 0.0013 Answer: C Diff: 3 Page Ref: Sec. 15.3 2) The value of K eq for the equilibrium ˆ‡ ˆ H 2 (g) + I 2 (g) ‡ ˆ 2 HI (g) ‡ is 794 at 25 °C. At this temperature, what is the value of K eq for the equilibrium below? ˆ‡ ˆ HI (g) ‡ ˆ 1/ 2H 2 (g) + 1/ 2I 2 (g) ‡ A) 1588 B) 28 C) 397 D) 0.035 E) 0.0013 Answer: D Diff: 3 Page Ref: Sec. 15.3 Chemistry, 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 3) The value of K eq for the equilibrium ˆ‡ ˆ H 2 (g) + I 2 (g) ‡ ˆ 2 HI (g) ‡ is 54.0 at 427 °C. What is the value of K eq for the equilibrium below? ˆ‡ ˆ HI (g) ‡ ˆ 1/ 2H 2 (g) + 1/ 2I 2 (g) ‡ A) 27 B) 7.35 C) 0.136 D) 2.92 × 103 E) 3.43 × 10−4 Answer: C Diff: 3 Page Ref: Sec. 15.3 4) Consider the following chemical reaction: ˆ‡ ˆ H 2 (g) + I 2 (g) ‡ ˆ 2 HI (g) ‡ At equilibrium in a particular experiment, the concentrations of H 2 , I 2 , and HI were 0.15M 0.033M and 0.55M respectively. The value of K eq for this reaction is __________. A) 23 B) 111 C) 9.0 × 10−3 D) 6.1 E) 61 Answer: E Diff: 3 Page Ref: Sec. 15.5 Chemistry, 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 5) A reaction vessel is charged with hydrogen iodide, which partially decomposes to molecular hydrogen and iodine: ˆ‡ ˆ 2HI (g) ‡ ˆ H 2 (g) + I 2 (g) ‡ When the system comes to equilibrium at 425 °C, PHI = 0.708atm , and PH 2 = PI 2 = 0.0960 atm. The value of Kp at this temperature is __________. A) 6.80 × 10−2 B) 1.30 × 10 −2 C) Kp cannot be calculated for this gas reaction when the volume of the reaction vessel is not given. D) 54.3 E) 1.84 × 10−2 Answer: E Diff: 3 Page Ref: Sec. 15.5 6) Acetic acid is a weak acid that dissociates into the acetate ion and a proton in aqueous solution: ˆ‡ ˆ HC 2 H 3 O 2 (aq) ‡ ˆ C 2 H 3 O 2 − (aq) + H + (aq) ‡ At equilibrium at 25 °C a 0.100 M solution of acetic acid has the following concentrations: [HC 2 H 3 O 2 ] = 0.0990M [C 2 H 3 O 2 − ] = 1.33 × 10−3 M and H + = 1.33 × 10−3 M The equilibrium constant, K eq , for the ionization of acetic acid at 25 °C is __________. A) 5.71 × B) 0.100 C) 1.75 × D) 1.79 × E) 5.71 × 104 10−7 10−5 106 Answer: D Diff: 3 Page Ref: Sec. 15.5 At equilibrium at 373 K.0055 mol of N 2 O 4 remains.Chemistry. 0. 15. A) 0. 15.566 D) 0.682 mol of H 2 and 0.5 8) Dinitrogentetraoxide partially decomposes according to the following equilibrium: ˆ‡ ˆ N 2 O 4 (g) ‡ ˆ 2NO 2 (g) ‡ A 1.000 B) 0.22 D) 0. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 7) At elevated temperatures.022 E) 0. At equilibrium at 700 K.324 Answer: E Diff: 2 Page Ref: Sec.5 . A) 2.00-L flask is charged with 0. molecular hydrogen and molecular bromine react to partially form hydrogen bromide: ˆ‡ ˆ H 2 (g) + Br 2 (g) ‡ ˆ 2HBr (g) ‡ A mixture of 0.2 × 10−4 B) 13 C) 0.00 L.566 mol of H 2 present. K eq for this reaction is __________. At equilibrium.0400mol of N 2 O 4 . there are __________ mol of Br2 present in the reaction vessel. there are 0.232 E) 0.87 Answer: E Diff: 3 Page Ref: Sec.440 C) 0.440 mol of Br2 is combined in a reaction vessel with a volume of 2. 15.0 Answer: D Diff: 4 Page Ref: Sec. there were 0. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 9) At 22 °C. assuming that some solid NH 4 HS remains. 0.9 × 10−3 E) 3.070 C) 0. carbon monoxide is converted to carbon dioxide via the following reaction: ˆ‡ ˆ CO (g) + H 2 O (g) ‡ ˆ CO 2 (g) + H 2 (g) ‡ In an experiment.78 D) 0. Calculate the equilibrium partial pressure (atm) of ammonia.56 E) 1.52 D) 4. A) 5.Chemistry.35 mol of CO and 0.5 . 15.5 10) In the coal-gasification process.070 for the equilibrium: ˆ‡ ˆ NH 4 HS (s) ‡ ˆ NH 3 (g) + H 2 S (g) ‡ A sample of solid NH 4 HS is placed in a closed vessel and allowed to equilibrate. A) 0.00-L reaction vessel.26 B) 0. K eq at the temperature of the experiment is __________.19 mol of CO remaining.40 mol of H 2 O were placed in a 1. At equilibrium. Kp = 0.75 C) 1.8 Answer: A Diff: 4 Page Ref: Sec.47 B) 0. 15.8 Answer: C Diff: 3 Page Ref: Sec. A) 11 B) 4.123 atm.0782 B) 0.0908 D) 0. The value of K eq is __________. the partial pressure of PCl 3 is __________ atm.500 mol of Br 2 . At equilibrium. giving an initial pressure of 0.0121: ˆ‡ ˆ PCl 5 (g) ‡ ˆ PCl 3 (g) + Cl 2 (g) ‡ A vessel is charged with PCl 5 .0455 C) 0. A) 0.6 .0 L flask is charged with 0. the flask contains 0.0 C) 110 D) 6. An equilibrium reaction ensues: ˆ‡ ˆ I 2 (g) + Br 2 (g) ‡ ˆ 2IBr (g) ‡ When the container contents achieve equilibrium.84 mol of IBr.5 12) The equilibrium constant (Kp) for the interconversion of PCl 5 and PCl 3 is 0.Chemistry.123 Answer: D Diff: 4 Page Ref: Sec.500 mol of I 2 and 0. 15.1 E) 2.0330 E) 0. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 11) A sealed 1. the partial pressures of H 2 and I 2 at equilibrium are 0. 15.50 × 10−2 Answer: C Diff: 4 Page Ref: Sec.389 Answer: C Diff: 3 Page Ref: Sec.40 × 103 .6 .7 C) 17. At equilibrium. A) 294 B) 35.1 atm of NO. 15.64 D) 0. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 13) K P = 0.888 atm. the partial pressure of O 2 is __________ atm. The partial pressure of HI is __________ atm.0198 at 721 K for the reaction ˆ‡ ˆ 2HI (g) ‡ ˆ H 2 (g) + I 2 (g) ‡ In a particular experiment. ˆ‡ ˆ 2NO (g) ‡ ˆ N 2 (g) + O 2 (g) ‡ A closed vessel is charged with 36.87 B) 1. A) 7.9 D) 6.710 and 0.00 E) 1.6 14) At 200 °C.Chemistry.98 C) 5. respectively. the equilibrium constant (Kp) for the reaction below is 2.125 E) 0. Answer: D Diff: 1 Page Ref: Sec.Chemistry.2 3) In what year was Fritz Haber awarded the Nobel Prize in chemistry for his development of a process for synthesizing ammonia directly from nitrogen and hydrogen? A) 1954 B) 1933 C) 1918 D) 1900 E) 1912 Answer: C Diff: 1 Page Ref: Sec. 15. C) Haber was working in his lab with his instructor at the time he worked out the process.2 .1 2) What role did Karl Bosch play in development of the Haber-Bosch process? A) He discovered the reaction conditions necessary for formation of ammonia. E) He was the German industrialist who financed the research done by Haber. 15. B) He originally isolated ammonia from camel dung and found a method for purifying it. A) all chemical reactions have ceased B) the rates of the forward and reverse reactions are equal C) the rate constants of the forward and reverse reactions are equal D) the value of the equilibrium constant is 1 E) the limiting reagent has been consumed Answer: B Diff: 1 Page Ref: Sec. D) He developed the equipment necessary for industrial production of ammonia. __________. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 15.2 Multiple-Choice Questions 1) At equilibrium. 15. D) It is an industrial synthesis of sodium chloride that was discovered by Karl Haber. B) It is a process used for the synthesis of ammonia.Chemistry. 15. Answer: B Diff: 2 Page Ref: Sec.2 5) Which one of the following will change the value of an equilibrium constant? A) changing temperature B) adding other substances that do not react with any of the species involved in the equilibrium C) varying the initial concentrations of reactants D) varying the initial concentrations of products E) changing the volume of the reaction vessel Answer: A Diff: 2 Page Ref: Sec. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 4) Which one of the following is true concerning the Haber process? A) It is a process used for shifting equilibrium positions to the right for more economical chemical synthesis of a variety of substances.2 . E) It is a process for the synthesis of elemental chlorine. 15. C) It is another way of stating LeChatelier's principle. 2 7) The equilibrium-constant expression depends on the __________ of the reaction. 15.2 . 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 6) Which of the following expressions is the correct equilibrium-constant expression for the equilibrium between dinitrogen tetroxide and nitrogen dioxide? ˆ‡ ˆ N 2 O 4 (g) ‡ ˆ 2NO 2 (g) ‡ A) [NO 2 ] [N 2 O 4 ] [NO 2 ]2 B) [N 2 O 4 ] C) [NO 2 ] [N 2 O 4 ]2 D) [NO 2 ] [N 2 O 4 ] E) [NO 2 ]2 [N 2 O 4 ] Answer: B Diff: 2 Page Ref: Sec. A) stoichiometry B) mechanism C) stoichiometry and mechanism D) the quantities of reactants and products initially present E) temperature Answer: A Diff: 2 Page Ref: Sec.Chemistry. 15. K c = __________.2 .70 x 10−4 E) 723 Answer: D Diff: 3 Page Ref: Sec. A) 0.23 D) 2.44 x 105 D) 2. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 5 8) Given the following reaction at equilibrium.44 x 10 at 230.2 9) Given the following reaction at equilibrium at 450.0821 C) 7.Chemistry. K p = __________. if K c = 6.0 °C.66 x 106 E) 2. 15.0 °C: ˆ‡ ˆ CaCO 3 (s) ‡ ˆ CaO (s) + CO 2 (g) ‡ If pCO 2 = 0.67 x 10−2 B) 1.0160 atm. ˆ‡ ˆ 2NO (g) + O 2 (g) ‡ ˆ 2NO 2 (g) ‡ A) 3.67 x 107 Answer: B Diff: 3 Page Ref: Sec. 15.0160 B) 0.56 x 104 C) 6. 99 x 10−4Ź C) 1.0 Answer: B Diff: 3 Page Ref: Sec.2 11) Given the following reaction at equilibrium at 300. K p = __________.05 D) 42. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 10) Given the following reaction at equilibrium.Chemistry.66 x 10−3 Answer: A Diff: 3 Page Ref: Sec.111 atm .90 x 10−6 B) 2. 15.9 E) 45. A) 1.2 .11 x 10−1 D) 8.23 x 10−2 B) 4. K c = __________.05 at 250.45 x 10−2 C) 1. 15.0 °C. ˆ‡ ˆ PCl 5 (g) ‡ ˆ PCl 3 (g) + Cl 2 (g) ‡ A) 3.12 x 10−2 E) 5.0 K: ˆ‡ ˆ NH 4 HS (s) ‡ ˆ NH 3 (g) + H 2 S (g) ‡ If pNH 3 = pH 2 S = 0. if K p = 1. 4 . 15.2. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 12) Which of the following expressions is the correct equilibrium-constant expression for the reaction below? ˆ‡ ˆ (NH 4 ) 2 Se (s) ‡ ˆ 2NH 3 (g) + H 2 Se (g) ‡ A) [NH 3 ][H 2 Se] / (NH 4 )2 Se B) (NH 4 )2 Se / [NH 3 ]2 [H 2 Se] C) 1/[(NH 4 ) 2 Se] D) [NH 3 ]2 [H 2 Se] E) [NH 3 ]2 [H 2 Se] / [(NH 4 )2 Se] Answer: D Diff: 3 Page Ref: Sec.Chemistry.2. 15.4 13) Which of the following expressions is the correct equilibrium-constant expression for the reaction below? ˆ‡ ˆ HF (aq) + H 2 O (l) ‡ ˆ H 3 O + (aq) + F− (aq) ‡ A) [HF][H 2O] / {H 3O + ][F− ] B) 1 / HF C) [H 3O+ ][F − ] / [HF][H 2O] D) [H 3O+ ][F − ] / [HF] E) [F − ] / [HF] Answer: D Diff: 3 Page Ref: Sec. 15. 15. At equilibrium. A) products predominate B) reactants predominate C) roughly equal amounts of products and reactants are present D) only products are present E) only reactants are present Answer: A Diff: 1 Page Ref: Sec. __________.3 15) The equilibrium constant for the gas phase reaction ˆ‡ ˆ 2NH 3 (g) ‡ ˆ N 2 (g) + 3H 2 (g) ‡ is Keq = 230 at 300 °C.Chemistry. At equilibrium.34 × 10−3 at 300 °C. 15. __________. 15.3 . A) products predominate B) reactants predominate C) roughly equal amounts of products and reactants are present D) only products are present E) only reactants are present Answer: B Diff: 1 Page Ref: Sec. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 14) The equilibrium constant for the gas phase reaction ˆ‡ ˆ N 2 (g) + 3H 2 (g) ‡ ˆ 2NH 3 (g) ‡ is Keq = 4. 3 . 15.063 C) 0.50 B) 0.25: ˆ‡ ˆ SO 2 (g) + NO 2 (g) ‡ ˆ SO 3 (g) + NO (g) ‡ The value of K eq at the same temperature for the reaction below is __________. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 16) The equilibrium constant for reaction 1 is K. ˆ‡ ˆ 2SO 2 (g) + 2NO 2 (g) ‡ ˆ 2SO 3 (g) + 2NO (g) ‡ A) 0.25 E) 16 Answer: B Diff: 2 Page Ref: Sec.12 D) 0. 15. The equilibrium constant for reaction 2 is __________.3 17) The value of K eq for the following reaction is 0. ˆ‡ ˆ ‡ (1) SO 2 (g) + (1/ 2) O 2 (g) ‡ ˆ SO 3 (g) ˆ‡ ˆ ‡ (2) 2SO 3 (g) ‡ ˆ 2SO 2 (g) + O 2 (g) A) K 2 B) 2K C) 1/2K D) 1 / K 2 E) -K 2 Answer: D Diff: 4 Page Ref: Sec.Chemistry. ˆ‡ ˆ 2O 3 (g) ‡ ˆ 3O 2 (g) ‡ A) B) C) D) 3PO 2 2PO 3 2PO 3 3PO 2 3PO 3 2PO 2 PO 32 PO 2 2 PO 23 E) PO 32 Answer: E Diff: 3 Page Ref: Sec.3 19) The K eq for the equilibrium below is 7. ˆ‡ ˆ 2Cl 2 (g) + 2H 2 O (g) ‡ ˆ 4HCl (g) + O 2 (g) ‡ What is the value of K eq at this temperature for the following reaction? 1 ˆ‡ ˆ Cl 2 (g) + H 2 O (g) ‡ ˆ 2HCl (g) + O 2 (g) ‡ 2 A) 0. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 18) The equilibrium expression for Kp for the reaction below is __________.3 .0752 B) 5. 15.66 × 10−3 C) 0.274 D) 0.150 Answer: C Diff: 2 Page Ref: Sec.0 °C.52 × 10−2 at 480.Chemistry. 15.0376 E) 0. ˆ‡ ˆ 2Cl 2 (g) + 2H 2 O (g) ‡ ˆ 4HCl (g) + O 2 (g) ‡ What is the value of K eq at this temperature for the following reaction? 1 ˆ‡ ˆ 2HCl (g) + O 2 (g) ‡ ˆ Cl 2 (g) + H 2 O (g) ‡ 2 A) 13. 15.0376 D) 5.52 × 10−2 at 480.52 × 10−2 at 480.3 21) The K eq for the equilibrium below is 7.66 × 10−3 E) 0.274 Answer: B Diff: 2 Page Ref: Sec.0 °C.Chemistry. ˆ‡ ˆ 2Cl 2 (g) + 2H 2 O (g) ‡ ˆ 4HCl (g) + O 2 (g) ‡ What is the value of K eq at this temperature for the following reaction? ˆ‡ ˆ 4HCl (g) + O 2 (g) ‡ ˆ 2Cl 2 (g) + 2H 2 O (g) ‡ A) 0.3 .0752 B) -0.3 B) 3.3 D) 5. 15.65 C) -0.150 Answer: C Diff: 2 Page Ref: Sec. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 20) The K eq for the equilibrium below is 7.0752 C) 13.0 °C.66 × 10−3 E) 0. 1 ˆ‡ ˆ SO 2 (g) + O 2 (g) ‡ ˆ SO 3 (g) ‡ 2 What is the value of K eq at this temperature for the following reaction? 1 ˆ‡ ˆ SO 3 (g) ‡ ˆ SO 2 (g) + O 2 (g) ‡ 2 A) 0.112 at 700.112 at 700.93 E) -0.0 ° C.112 Answer: C Diff: 2 Page Ref: Sec.3 23) The K eq for the equilibrium below is 0.0560 E) 0. 15.112 D) 8.Chemistry.224 B) 0. 15.335 C) 0.0125 C) 0.3 .112 Answer: D Diff: 2 Page Ref: Sec.224 B) 0.0 °C.0125 D) 0. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 22) The K eq for the equilibrium below is 0. 1 ˆ‡ ˆ SO 2 (g) + O 2 (g) ‡ ˆ SO 3 (g) ‡ 2 What is the value of K eq at this temperature for the following reaction? ˆ‡ ˆ 2SO 2 (g) + O 2 (g) ‡ ˆ 2SO 3 (g) ‡ A) 0. 3 .99 C) 17. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 24) The K eq for the equilibrium below is 0. 15.0 K. Calculate Kp for the reverse reaction. the equilibrium constant for the reaction ˆ‡ ˆ 2NO (g) + Br 2 (g) ‡ ˆ 2NOBr (g) ‡ is Kp = 0.0 °C. ˆ‡ ˆ 2NOBr (g) ‡ ˆ 2NO (g) + Br 2 (g) ‡ . 1 ˆ‡ ˆ SO 2 (g) + O 2 (g) ‡ ˆ SO 3 (g) ‡ 2 What is the value of K eq at this temperature for the following reaction? ˆ‡ ˆ 2SO 3 (g) ‡ ˆ 2SO 2 (g) + O 2 (g) ‡ A) 79.013 . 15.6Ź × 10−4 C) 77 D) 0.013 B) 1.93 Answer: A Diff: 2 Page Ref: Sec.Chemistry.1 Answer: C Diff: 2 Page Ref: Sec.86 D) 4.46 E) 8.7 B) 2. A) 0.112 at 700.3 25) At 1000.99 E) 1. Chemistry.0 mol SO 3 (g) Answer: B Diff: 3 Page Ref: Sec.25 mol of SO 3 (g) D) 0.3.3 27) The expression for K p for the reaction below is __________. 15.25 mol of SO 2 (g) and 0. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 26) Consider the following equilibrium. 15. ˆ‡ ˆ 2 SO 2 (g) + O 2 (g) ‡ ˆ 2 SO 3 (g) ‡ The equilibrium cannot be established when __________ is/are placed in a 1.25 mol SO 2 (g) and 0.50 mol SO 3 (g) E) 1.4 . 15.25 mol O 2 (g) B) 0. ˆ‡ ˆ 4CuO (s) + CH 4 (g) ‡ ˆ CO 2 (g) + 4Cu (s) + 2H 2 O(g) ‡ A) B) PCH 4 PCO 2 PH 2 2 [Cu]PCO 2 PH 2 O 2 [CuO] 4 PCH 4 PCO 2 PH 2 O 2 C) PCH 4 D) E) PCO 2 PH 2 O 2 PCuO PCH 4 PH 2 O 2 PCO 2 Answer: C Diff: 2 Page Ref: Sec.50 mol O 2 (g) and 0.0-L container.75 mol SO 2 (g) C) 0. A) 0. A) The equilibrium partial pressures of Br 2 . Answer: D .00 atm of Cl 2 (g) . D) The equilibrium partial pressure of BrCl(g) will be greater than 2.00 atm. Cl 2 . and 2.00 atm. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 28) The equilibrium-constant expression for the reaction ˆ‡ ˆ Ti (s) + 2Cl 2 (g) ‡ ˆ TiCl 4 (l) ‡ is given by A) [TiCl4 (l)] [Ti (s)][Cl 2 (g)] [Ti (s)][Cl 2 (g)]2 B) [TiCl4(l)] [TiCl4 (l)] C) [Cl 2 (g)]2 D) [Cl2 (g)]−2 E) [TiCl4 (l)] [Ti(s)][Cl 2 (g)]2 Answer: D Diff: 2 Page Ref: Sec. 1.00 atm of Br 2 (g) .00 atm of BrCl(g) . Use Q to determine which of the statements below is true. B) The equilibrium partial pressure of Br 2 will be greater than 1. the equilibrium constant for the reaction ˆ‡ ˆ Br 2 (g) + Cl 2 (g) ‡ ˆ 2BrCl(g) ‡ is K P = 7. C) At equilibrium. 15. and BrCl will be the same as the initial values. 15. the total pressure in the vessel will be less than the initial total pressure.4 29) At 400 K.0 p . A closed vessel at 400 K is charged with 1.Chemistry. E) The reaction will go to completion since there are equal amounts of Br 2 and Cl 2 .3. 6 .Chemistry. 15. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium Diff: 3 Page Ref: Sec. 15. C) The reaction is at equilibrium when Q < K eq . E) Q is the same as K eq when a reaction is at equilibrium. C) K does not depend on the concentrations or partial pressures of reaction components. Answer: E Diff: 3 Page Ref: Sec.Chemistry. D) The reaction is at equilibrium when Q > K eq . D) Q does not depend on the concentrations or partial pressures of reaction components. whereas Q is temperature dependent. the reaction quotient is undefined. E) The reaction is at equilibrium when Q = K eq .6 . B) K eq does not change with temperature. B) At equilibrium. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 30) Which of the following statements is true? A) Q does not change with temperature. 15.6 31) How is the reaction quotient used to determine whether a system is at equilibrium? A) The reaction quotient must be satisfied for equilibrium to be achieved. Answer: E Diff: 3 Page Ref: Sec. are in the flask at equilibrium? A) 0.6 33) Of the following equilibria.46.64 mol) was placed in a 1.00-L flask containing no NO or Br2 . only __________ will shift to the left in response to a decrease in volume. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 32) Nitrosyl bromide decomposes according to the following equation. 0.18.23 C) 0.Chemistry.090 D) 0. At equilibrium the flask contained 0. 0. respectively. 0.360 E) 0.18. ˆ‡ ˆ 2NOBr (g) ‡ ˆ 2NO (g) + Br2 (g) ‡ A sample of NOBr (0.46. ˆ‡ ˆ ‡ A) H 2 (g) + Cl2 (g) ‡ ˆ 2 HCl (g) ˆ‡ ˆ ‡ B) 2 SO 3 (g) ‡ ˆ 2 SO2 (g) + O2 (g) ˆ‡ ˆ ‡ C) N 2 (g) + 3H 2 (g) ‡ ˆ 2 NH3 (g) ˆ‡ ˆ ‡ D) 4 Fe (s) + 3O 2 (g) ‡ ˆ 2Fe2 O3 (s) ˆ‡ ˆ ‡ E) 2HI (g) ‡ ˆ H 2 (g) + I 2 (g) Answer: B Diff: 3 Page Ref: Sec.46 Answer: C Diff: 4 Page Ref: Sec. 15.18. 0.46 mol of NOBr.7 .18 B) 0. 15. How many moles of NO and Br2 . 0. 15.7 .Chemistry. A) increasing the temperature B) decreasing the temperature C) increasing the pressure D) removing some of the CaCO3 (s) E) none of the above Answer: A Diff: 3 Page Ref: Sec.7 35) For the endothermic reaction ˆ‡ ˆ CaCO3 (s) ‡ ˆ CaO (s) + CO2 (g) ‡ Le Chatelier's principle predicts that __________ will result in an increase in the number of moles of CO 2 . 15. A) increasing the pressure B) decreasing the pressure C) increasing the temperature D) removing some oxygen E) increasing the volume of the container Answer: A Diff: 3 Page Ref: Sec. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 34) The reaction below is exothermic: ˆ‡ ˆ 2SO 2 (g) + O2 (g) ‡ ˆ 2SO3 (g) ‡ Le Chatelier's Principle predicts that __________ will result in an increase in the number of moles of SO3 (g) in the reaction container. based on Le Chatelier's principle? ˆ‡ ˆ ‡ A) N 2 (g) + 3H2 (g) ‡ ˆ 2NH3 (g) ˆ‡ ˆ ‡ B) N 2O4 (g) ‡ ˆ 2NO2 (g) ˆ‡ ˆ ‡ C) N 2 (g) + 2O2 (g) ‡ ˆ 2NO2 (g) ˆ‡ ˆ ‡ D) 2N 2 (g) + O2 (g) ‡ ˆ 2N2 O(g) ˆ‡ ˆ ‡ E) N 2 (g) + O2 (g) ‡ ˆ 2NO(g) Answer: E Diff: 4 Page Ref: Sec.7 .Chemistry. 15.4 kJ Le Chatelier's principle predicts that adding N 2 (g) to the system at equilibrium will result in __________. 15.7 37) Consider the following reaction at equilibrium: ˆ‡ ˆ 2NH 3 (g) ‡ ˆ N 2 (g) + 3H 2 (g) ‡ ΔH° = +92. A) a decrease in the concentration of NH 3 (g) B) a decrease in the concentration of H 2 (g) C) an increase in the value of the equilibrium constant D) a lower partial pressure of N 2 E) removal of all of the H 2 (g) Answer: B Diff: 3 Page Ref: Sec. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 36) In which of the following reactions would increasing pressure at constant temperature not change the concentrations of reactants and products. A) increase the partial pressure of CO(g) at equilibrium B) decrease the partial pressure of CO 2 (g) at equilibrium C) increase the value of the equilibrium constant D) increase the partial pressure of CO 2 (g) at equilibrium E) decrease the value of the equilibrium constant Answer: D Diff: 3 Page Ref: Sec.7 39) Consider the following reaction at equilibrium: ˆ‡ ˆ 2CO 2 (g) ‡ ˆ 2CO(g) + O2 (g) ‡ ΔH° = -514 kJ Le Chaelier's principle predicts that adding O 2 (g) to the reaction container will __________. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 38) Consider the following reaction at equilibrium: ˆ‡ ˆ 2NH 3 (g) ‡ ˆ N 2 (g) + 3H 2 (g) ‡ Le Chatelier's principle predicts that the moles of H 2 in the reaction container will increase with __________. 15.7 .Chemistry. 15. A) some removal of NH 3 from the reaction vessel (V and T constant) B) a decrease in the total pressure (T constant) C) addition of some N 2 to the reaction vessel (V and T constant) D) a decrease in the total volume of the reaction vessel (T constant) E) an increase in total pressure by the addition of helium gas (V and T constant) Answer: B Diff: 3 Page Ref: Sec. Chemistry. 15. 15. A) increase the partial pressure of O 2 (g) B) decrease the partial pressure of CO 2 (g) C) decrease the value of the equilibrium constant D) increase the value of the equilibrium constant E) increase the partial pressure of CO Answer: C Diff: 3 Page Ref: Sec. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 40) Consider the following reaction at equilibrium: ˆ‡ ˆ 2CO 2 (g) ‡ ˆ 2CO (g) + O2 (g) ‡ ΔH° = -514 kJ Le Chatelier's principle predicts that an increase in temperature will __________.7 41) Consider the following reaction at equilibrium: ˆ‡ ˆ C(s) + H 2 O(g) ‡ ˆ CO(g) + H2 (g) ‡ Which of the following conditions will increase the partial pressure of CO? A) decreasing the partial pressure of H 2O(g) B) removing H 2O(g) from the system C) decreasing the volume of the reaction vessel D) decreasing the pressure in the reaction vessel E) increasing the amount of carbon in the system Answer: D Diff: 5 Page Ref: Sec.7 . 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 42) Consider the following reaction at equilibrium. ˆ‡ ˆ 2CO 2 (g) ‡ ˆ 2CO (g) + O2 (g) ‡ ΔH° = -514 kJ Le Chatelier's principle predicts that the equilibrium partial pressure of CO (g) can be maximized by carrying out the reaction __________.7 43) Consider the following reaction at equilibrium: ˆ‡ ˆ 2SO 2 (g) + O2 (g) ‡ ˆ 2SO3 (g) ‡ ΔH° = -99 kJ Le Chatelier's principle predicts that an increase in temperature will result in __________.7 . 15.Chemistry. A) at high temperature and high pressure B) at high temperature and low pressure C) at low temperature and low pressure D) at low temperature and high pressure E) in the presence of solid carbon Answer: C Diff: 3 Page Ref: Sec. A) a decrease in the partial pressure of SO3 B) a decrease in the partial pressure of SO2 C) an increase in K eq D) no changes in equilibrium partial pressures E) the partial pressure of O 2 will decrease Answer: A Diff: 3 Page Ref: Sec. 15. 15.3 Short Answer Questions. then the equilibrium mixture contains mostly __________.3 2) If Reaction A + Reaction B = Reaction C. 15. then K c Reaction C = __________. Answer: reciprocal Diff: 1 Page Ref: Sec. 15. A) increase the rate of the forward reaction only B) increase the equilibrium constant so that products are favored C) slow the reverse reaction only D) increase the rate at which equilibrium is achieved without changing the composition of the equilibrium mixture E) shift the equilibrium to the right Answer: D Diff: 3 Page Ref: Sec.Chemistry. 1) The equilibrium-constant expression for a reaction written in one direction is the __________ of the one for the reaction written for the reverse direction. 15.7 15. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 44) The effect of a catalyst on an equilibrium is to __________. Answer: K c Reaction A x K c Reaction B Diff: 3 Page Ref: Sec.3 .3 3) If the value for the equilibrium constant is much greater than 1. Answer: products Diff: 2 Page Ref: Sec. Answer: reaction quotient Diff: 2 Page Ref: Sec.6 .0-L container and allowed to reach equilibrium described by the equation ˆ‡ ˆ N 2 O4 (g) ‡ ˆ 2NO2 (g) ‡ If at equilibrium the N 2O 4 is 25% dissociated. liquids Diff: 1 Page Ref: Sec. 15. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 4) Pure __________ and pure __________ are excluded from equilibrium-constant expressions.58 Diff: 4 Page Ref: Sec. 15.6 7) If the reaction quotient Q for a reaction is less than the value of the equilibrium constant K for that reaction at a given temperature. 15.5 6) The number obtained by substituting starting reactant and product concentrations into an equilibrium-constant expression is known as the __________. Answer: reactants. 15.4 5) Exactly 3. Answer: solids.Chemistry. what is the value of the equilibrium constant for the reaction? Answer: 0.5 moles if N 2O 4 is placed in an empty 2. products Diff: 3 Page Ref: Sec. __________ must be converted to __________ for the system to reach equilibrium. Answer: True Diff: 2 Page Ref: Sec. effectively removing it and shifting the equilibrium to the right.2 2) The effect of a catalyst on a chemical reaction is to react with product. 15. Answer: False Diff: 2 Page Ref: Sec. __________ the reaction temperature results in an increase in K. increasing the reaction temperature results in a(an) __________ in K. 15. 15. __________ must be converted to __________ for the system to reach equilibrium. 15. Answer: decrease Diff: 3 Page Ref: Sec.4 True/False Questions 1) The relationship between the concentrations of reactants and products of a system at equilibrium is given by the law of mass action. reactants Diff: 3 Page Ref: Sec. Answer: products. 15. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 8) If the reaction quotient Q for a reaction is greater than the value of the equilibrium constant K for that reaction at a given temperature.Chemistry.7 10) If a reaction is endothermic.7 .6 9) For an exothermic reaction. Answer: increasing Diff: 3 Page Ref: Sec.7 15. reducing the volume of a gaseous equilibrium mixture causes the reaction to shift in the direction that increases the number of moles of gas in the system. Answer: True Diff: 2 Page Ref: Sec. Answer: False Diff: 2 Page Ref: Sec. 15.7 4) In an exothermic equilibrium reaction. Answer: True Diff: 3 Page Ref: Sec. 15. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 3) At constant temperature.7 .Chemistry. 15.7 5) Le Chatelier's principle states that if a system at equilibrium is disturbed. the equilibrium will shift to minimize these disturbance. increasing the reaction temperature favors the formation of reactants. 36 × 10−2 mol of N 2 O 4 remains.202 atm PCl2 = 0. K eq for this reaction is __________.50 × 10−2 C) 1.5 Algorithmic Questions 1) Phosphorous trichloride and phosphorous pentachloride equilibrate in the presence of molecular chlorine according to the reaction: PCl3 (g) + Cl2 (g) → PCl5 (g) An equilibrium mixture at 450 K contains PPCl3 = 0.What is the value of Kp at this temperature? A) 66.212 D) 6.45 atm . 15.Chemistry.92 × 10−4 Answer: D Diff: 3 Page Ref: Sec.54 Answer: A Diff: 3 Page Ref: Sec. 2. At equilibrium.5 2) Dinitrogen tetroxide partially decomposes according to the following equilibrium: N 2 O 4 (g) → 2NO 2 (g) A 1.723 B) 0.78 × 10−1 D) 2. A) 0. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 15. 15.99 E) 7.00 × 10−2 mol of N 2 O 4 .256 atm . and s PPCl5 = 3.391 C) 0.000-L flask is charged with 3.7 B) 1.5 .94 × 10−3 E) 1. 15.10 × 102Ź B) 2. the equilibrium constant (K p ) for the following reaction is 0. 2SO 2 +O 2 (g) → 2SO 3 (g) At equilibrium.9 atm.01 × 1014 C) 4.72 × 1011Ź Answer: A Diff: 3 Page Ref: Sec. 11e(Brown/LeMay/Bursten/Murphy) Chapter 15 Chemical Equilibrium 3) The K p for the reaction below is 1.60 × 10−4 atm .0 K.0 B) 4. The partial pressure of the product.28 × 105Ź E) 1. is __________ atm.21× 10−3 C) 192 D) 6. A) 1. A) 82. The partial pressure of SO 3 is __________ atm.20 × 10−4 E) 40. PCO = PCl 2 = 8.2 Answer: A Diff: 3 Page Ref: Sec.0 °C: CO (g) + Cl 2 (g) → COCl 2 (g) In an equilibrium mixture of the three gases.345.49 × 108 at 100.0 atm and that of O 2 is 15.96 × 10−15 D) 1. phosgene (COCl 2 ) .Chemistry.5 4) At 900. 15.5 . the partial pressure of SO 2 is 35. 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