(C) Pavel Sedach and Prep PDF Free Download

Chemistry 209 F15 Version A Page 6 of 9 SECTION II: Long Answers: To be graded manually (Total value 21) Answers must be written in non-erasable ink to be considered for re-grading. For full marks show all your work. Question 1 [Total value: 10 points] The plot below describes experimental results for two different measurements of the reaction A products. Use this plot to answer parts a) through d). Assume 2 significant figures when reading values from the plot. 0.35 0.3 Data For Question 1 Experiment 1 Experiment 2 a. [2 points] Calculate the average rate of consumption of A in the time period 0-10 s for Experiment 2. [A], mol/l 0.25 0.2 0.15 b. [2 points] Calculate the initial rate of reaction for Experiment 1. 0.1 0.05 0 0 5 10 15 20 time (s) c. [2 points] Calculate the initial rate of reaction for Experiment 2. d. [4 points] Using the [A] 0 values from the graph and your results from parts b. and c., write the rate law for this reaction, including the value of k and its units. Assume a whole-number order for all components.

Chemistry 209 F15 Version A Page 7 of 9 Question 2 [Total value: 4 points] Calcium carbide (CaC 2 ), a solid, reacts with water to produce acetylene gas (C 2 H 2 ) and aqueous calcium hydroxide (Ca(OH) 2 ). a. Write the balanced equation for this reaction. [1 point] b. If 100.0 g of calcium carbide is consumed in this reaction and the acetylene produced is captured in a 30.00 L container at 20.0 C, at what pressure will the container be after the reaction is complete? [3 points] Question 3 [Total value: 4 points] One step in coal gasification relies upon the following endothermic process: C(s) + H 2 O (g) CO(g) + H 2 (g) K P = 0.797 at 700.0 0 C. a. If 1.00 g of carbon is added to a reaction vessel containing 1.00 M each of H 2 O, CO, and H 2 gases, predict the direction in which the reaction will proceed. Justify your answer. [2 points] [Question 3 continued next page]

Chemistry 209 F15 Version A Page 8 of 9 b. Based on the fact that the reaction is endothermic, compare the yield of the reaction at room temperature to the yield at 700 0 C. Use 1 to 3 grammatically correct sentences to explain your answer. [2 points] Question 4 [Total value: 4 points] You prepare a 0.134 M solution of chlorous acid (HClO 2 ). Chlorous acid has a pk a of 1.921. Find the ph of this solution.

Chemistry 209 F15 Version A Page 9 of 9 1 1A 1 H 1.008 3 Li 6.941 11 Na 22.99 19 K 39.10 37 Rb 85.47 55 Cs 132.9 87 Fr (223) 2 2A 4 Be 9.012 12 Mg 24.31 20 Ca 40.08 38 Sr 87.62 56 Ba 137.3 88 Ra 226.0 Legend: Data Sheet CHEM 209 1 H 1.008 Periodic Table 18 8A Atomic number (Z) 13 Atomic symbol 3A Atomic mass (amu) 3 4 5 6 7 8 9 10 11 12 21 Sc 44.96 39 Y 88.91 57* La 138.9 89** Ac (227) 22 Ti 47.88 40 Zr 91.22 72 Hf 178.5 104 Rf (261) 23 V 50.94 41 Nb 92.91 73 Ta 180.9 105 Ha (262) 24 Cr 52.00 42 Mo 95.94 74 W 183.9 106 Sg (263) 25 Mn 54.94 43 Tc (98) 75 Re 186.2 107 Ns (262) 26 Fe 55.85 44 Ru 101.1 76 Os 190.2 108 Hs (265) 27 Co 58.93 45 Rh 102.9 77 Ir 192.2 109 Mt (266) 28 Ni 58.69 46 Pd 106.4 78 Pt 195.1 110 Uun (269) 29 Cu 63.55 47 Ag 107.9 79 Au 197.0 111 Uuu (272) 30 Zn 65.38 48 Cd 112.4 80 Hg 200.6 5 B 10.81 13 Al 26.98 31 Ga 69.72 49 In 114.8 81 Tl 204.4 14 4A 6 C 12.01 14 Si 28.09 32 Ge 72.59 50 Sn 118.7 82 Pb 207.2 15 5A 7 N 14.01 15 P 30.97 33 As 74.92 51 Sb 121.8 83 Bi 209.0 16 6A 8 O 16.00 16 S 32.07 34 Se 78.96 52 Te 127.6 84 Po (209) 17 7A 9 F 19.00 17 Cl 35.45 35 Br 79.90 53 I 126.9 85 At (210) 2 He 4.003 10 Ne 20.18 18 Ar 39.95 36 Kr 83.80 54 Xe 131.3 86 Rn (222) Lanthanides * Actinides ** 58 Ce 140.1 90 Th 232.0 59 Pr 140.9 91 Pa 231.0 60 Nd 144.2 92 U 238.0 61 Pm (145) 93 Np 237.0 62 Sm 150.4 94 Pu (244) 63 Eu 152.0 95 Am (243) 64 Gd 157.3 96 Cm (247) 65 Tb 158.9 97 Bk (247) 66 Dy 162.5 98 Cf (251) 67 Ho 164.9 99 Es (252) 68 Er 167.3 100 Fm (257) 69 Tm 168.9 101 Md (258) 70 Yb 173.0 102 No (259) 71 Lu 175.0 103 Lr (260) Strong Acids: HCl, HBr, HI, HNO3, H2SO4, HClO4 Strong Bases: Hydroxides of Group 1 (Li to Cs) and Group 2 (Ca, Sr, Ba) Constants: Conversion Factors: Gas Constant: R = 0.08205 L atm mol -1 K-1 1 J = 1 kg m 2 s2 = 8.314 L kpa mol -1 K-1 = 8.314 J mol -1 K-1 T (K) = T ( C) + 273.15 = 0.08314 L bar mol -1 K-1 1 Pa = 1 kg m -1 s-2 = 10-5 bar Avogadro s number: NA = 6.022 10 23 mol -1 1 L atm = 101.3 J Faraday s Constant: F = 96 485 C/mol electrons 1 atm = 760.0 torr = 101.3 kpa = 760.0 mm Hg = 1.013 bar Planck s Constant: h = 6.626 10-34 J s 1 L = 10-3 m 3 Speed of Light: c = 2.998 10 8 m s -1 1 C = 1 J/V Rydberg Constant: R = 1.096776 10 7 m -1 STP conditions: 0 C, 100 kpa Factoring Rydberg Constant: RH = R h c = 2.18 10-18 J Electrochemical standard state: 1 atm, 1 M, 25 C ln ln ln 1 1 log 0.0592 log 2 0.693 0 ln 1 1 1 1 4 2 0.0592 log Δ ln 1 1 ln 1 1 log ln

(C) Pavel Sedach and Prep101 1