1. Iodide ion is oxidized by H SO
1. Iodide ion is oxidized by H2SO3 to triiodide ion, which can be detected by the blue color of a starch-triiodide compound: 6 I–(aq) + H2SO3(aq) + 4 H+(aq) a. S(s) + 2 I3–(aq) + 3 H2O(ℓ) Estimate the instantaneous rate of formation of I3– at 450 s from the graph. (4 pts.) [H 2SO3 ] (0.445 0.529) M 8.40 x 104 M/s t (500 400) s mole H 2SO3 2 mole I3 8.40 x 104 x 1.68 x 10–3 M I3–/s Ls 1 mole H 2SO3 rate b. Initial rate data for the reaction are tabulated below: [I–]i, M 0.0300 0.0600 0.0300 0.0300 0.0400 Exp# 1 2 3 4 5 [H2SO3]i, M 0.0200 0.0200 0.0400 0.0200 0.0300 [H+]i, M 0.0300 0.0300 0.0300 0.0600 0.0500 Initial rate, M/s 1.05 x 10–4 8.40 x 10–4 2.10 x 10–4 4.20 x 10–4 ? Write the rate law for the reaction. Show work. (4 pts.) 1-2 1-3 1-4 2x [I–] 2x [H2SO3] 2x [H+] = 8x rate = 2x rate = 4x rate [I–]3 [H2SO3]1 [H+]2 rate = k[I–]3[H2SO3][H+]2 c. What is the order of the reaction in hydrogen ion? (3 pts.) 2nd order d. Calculate the numerical value of the rate constant. (3 pts.) k e. 1.05 x 104 M/s 2.16 x 105 M–5s–1 (0.0300 M)3 (0.0200 M)(0.0300 M) 2 Calculate the rate of experiment #5. (4 pts.) rate = 2.16 x 105 M–5s–1 (0.0400 M)3 (0.0300 M) (0.0500 M)2 = 1.04 x 10–3 M/s f. What function of [H2SO3], when plotted versus time, would give a straight line? What information can be obtained from the slope of this line? (3 pts.) ln[H2SO3] versus t, slope = –k The rate data of the following reaction were collected at different temperatures and functions of the data plotted to determine the activation energy. CH3Br(aq) + OH–(aq) T (ºC) 17.0 35.7 46.4 57.6 0.0030 -10 k (M–1·s–1) 1.07 x 10–7 1.65 x 10–6 6.71 x 10–6 2.47 x 10–5 CH3OH(aq) + Br–(aq) 0.0031 -11 0.0032 0.0033 0.0034 0.0035 y = (-1.29 x 104)x + 28.4 -12 y = lnk 2. -13 -14 -15 -16 -17 -1 x = 1/T (K ) a. Clearly label the x and y axes (above) with the proper function. (2 pts.) lnk versus 1/T b. Determine Ea of the reaction. (4 pts.) slope = –(Ea/R); Ea = –(–1.29 x 104 K) x 8.314 J/mole·K = 107 kJ/mole c. Sketch a reaction profile, to scale, if ΔHreaction = –120 kJ/mole. (3 pts.) 3. Barium peroxide (BaO2) is a powerful oxidizer that can produce Cl2 from Cl–: 1) BaO2(ℓ) + 4 HCl(g) K = 223.7 at 527 oC a. 2 BaCl2(s) + 4 H2O(g) + 2 Cl2(g) 2 1 1 K2 = = 223.7 K1 2 BaO2(ℓ) + 8 HCl(g) 2 = 1.998 x 10–5 Write the K equilibrium expression for reaction 1. (3 pts.) K = c. ΔH = –376 kJ Calculate the value of K at 527 oC, for reaction 2: (3 pts.) 2) b. BaCl2(s) + 2 H2O(g) + Cl2(g) [H 2O]2 [Cl 2 ] [HCl]4 Calculate the value of Kp for reaction 1 at 527 oC. (4 pts.) Kp = K(RT)Δn = 223.7(0.08206 x 800.15)3–4 = 3.41 d. BaO2, HCl, BaCl2, H2O and Cl2 are placed in a cylinder with a movable piston, heated to 527 oC and allowed to reach equilibrium. Will the amount of Cl2 in the cylinder increase, decrease, or not change in response to the following. Explain briefly. Use reaction 1 as reference. (8 pts.) i. H2O(g) is added to the flask. decrease adding a product (H2O) shifts to remove product; toward reactant ii. The piston is pulled up. decrease pressure decreases – reactions shifts in direction of more moles of gas to increase pressure; 4 moles reactants versus 3 moles product iii. BaCl2 is added to the flask. not change not part of the equilibrium expression; cannot change the concentration of a solid iv. The temperature is decreased. increase goes in the direction to produce heat; to the right in this case 4. A flask is filled with 0.500 M N2 and 0.800 M H2. The following takes place at 1100 oC: N2(g) + 3 H2(g) 2 NH3(g) At equilibrium, the concentration of NH3 is 0.160 M. Calculate K. (6 pts.) Initial Equil [] [N2] 0.500 M –x 0.500 – x 0.500 –0.080 = 0.420 M [H2] 0.800 M –3x 0.800 – 3x 0.800 – 3(0.080) = 0.560 M [NH3] 0M +2x 2x = 0.160 M 0.160 M 0.160 [NH3 ] = 0.347 = 3 3 [N 2 ][H 2 ] 0.420 0.560 2 K= 5. Cyanogen, C2N2, is known as a pseudohalogen because it undergoes reactions similar to the halogen, such as forming “inter”halogen compounds.. 2 ICN(solv) I2(solv) + C2N2(solv) K = 2.08 (at 50.0 °C) (solv) means dissolved in CHCl3 0.300 mole ICN and 0.450 mole of both I2 and C2N2 are placed in 1.00 L of solution. a. Which direction does the reaction go (to the left or to the right) to reach equilibrium? Show a calculation. (4 pts.) Q = [I2 ]i [C2 N 2 ]i (0.450)(0.450) = = 2.25; Q > K, reaction goes to left. 2 [ICN]i (0.300)2 b. Calculate the molarity of all substances at equilibrium. (8 pts.) initial equilibrium [ICN] 0.300 +2x 0.300 + 2x [I2] 0.450 –x 0.450 – x [C2N2] 0.450 –x 0.450 – x [I2 ][C2 N 2 ] (0.450 x)(0.450 x) (0.450 x) 2 K = = = = 2.08 [ICN]2 (0.300 + 2x)2 (0.300 + 2x) 2 (0.450 x) = (0.300 + 2x) 2.08 ; 0.4327 + 2.8844x = 0.450 – x 3.8844x = 0.01733; x = 0.00446 M [ICN] = 0.300 + 2(0.00446) = 0.309 M [I2] = 0.450 – 0.00446 = 0.446 M [C2N2] = 0.450 – 0.00446 = 0.446 M Check: K = [I2 ][C2 N 2 ] (0.446)(0.446) = = 2.08 2 [ICN] (0.309) 2 6. Calculate the pH of the following aqueous solutions. To receive full credit, you must show an appropriate chemical equation for each. Constants in the handout. (30 pts.) a. 0.20 M HC8H8O2 (phenylacetic acid) C8H8O2–(aq) + H3O+(aq) HC8H8O2(aq) + H2O(ℓ) initial equilbrium [HC8H8O2] 0.20 M –x 0.20 – x [H3O+] ~0 +x x H2O – – – [C8H8O2–] 0 +x x [H3O+ ][C8 H8O2 ] x2 x2 5 Ka = = 5.2 x 10 = [HC8 H8O2 ] 0.20 x 0.20 x = 3.2 x 10–3 M = [H3O+]; pH = –log(3.2 x 10–3) = 2.49 b. 0.10 M NaC7H14O2 (sodium heptanoate) NaC7H14O2(aq) Na+(aq) + C7H14O2–(aq) C7H14O2–(aq) + H2O(ℓ) initial equilbrium Kb = [C7H14O2–] 0.10 M –x 0.10 – x HC7H14O2(aq) + OH–(aq) H2O – – – [OH–] ~0 +x x [HC7H14O2] 0 +x x [OH ][HC7 H14O2 ] Kw 1.0 x 1014 x2 10 = = = 7.7 x 10 [C7 H14O2 ] K a HC7 H14O2 1.3 x 105 0.10 x = 8.8 x 10–6 M = [OH–]; pOH = –log(8.8 x 10–6) = 5.06; pH = 14.00 – pOH = 8.94 c. 0.0015 M HClO4 (perchloric acid) HClO4(aq) + H2O(ℓ) [H3O+] = [HClO4]i = 0.0015 M pH = –log(0.0015) = 2.82 H3O+(aq) + ClO4–(aq) d. 0.25 M Li2O (lithium oxide) 2 Li+(aq) + O2–(aq) Li2O(aq) 2 OH–(aq) O2–(aq) + H2O(ℓ) [OH–] = 2 [O2–] = 0.50 M; pOH = 0.30; pH = 14.00 – pOH = 13.70 e. 0.34 M KBr (potassium bromide) K+(aq) + Br–(aq) KBr(aq) Br–(aq) + H2O(ℓ) HBr(aq) + OH–(aq) Br– is the conjugate base of a strong acid, therefore a VERY weak base. K+ is not an acid or a base. The only species remaining that affects the pH is H2O. The pH of pure water is 7.00. f. 0.15 M (C5H5NH)NO3 (pyridinium nitrate) (C5H5NH)NO3(aq) C5H5NH+(aq) + NO3–(aq) C5H5NH+(aq) + H2O(ℓ) initial equilbrium [C5H5NH+] 0.15 M –x 0.15 – x C5H5N(aq) + H3O+(aq) H2O – – – [H3O+] ~0 +x x [C5H5N] 0 +x x [H3O+ ][C5 H5 N] Kw 1.0 x 1014 Ka = = = = 5.9 x 106 + 9 [C5 H5 NH ] K b C5H5 N 1.7 x 10 5.9 x 106 x2 0.15 x = 9.4 x 10–4 M = [H3O+]; pH = –log(9.4 x 10–4) = 3.03
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