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UT CH 302 - Practice Exam 3 Answer Key
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Spring 2009 CH302 Practice Exam 3 Answer Key 1. Consider the reaction below: C6H12O6(s) + 12 O2(g) Æ 6 CO2(g) + 6 H2O(g) Which of the following is an incorrect expression of the rate? 1. rate = (Δ[H2O] / 6·Δt) 2. rate = -(Δ[O2] / 12·Δt) 3. rate = (Δ[CO2] / 6·Δt) 4. rate = -(Δ[C6H12O6] / Δt) Correct Explanation: rate = -(Δ[O2] / 12·Δt) = (Δ[CO2] / 6·Δt) = (Δ[H2O] / 6·Δt) 2. The overall reaction Br2(g) + 2 NO2(g) Æ2 BrNO(g) + O2(g) Has an empirically determined rate law, rate = k·[NO2]2·[Br2]·[O2]-1. If k = 3.0 x 104 M-1·s-1, [NO2] = 0.01 M, [Br2] = 0.02 M and [O2] = 0.01 M, what is the observed rate? 1. 0.3 M·s-1 2. 0.0006 M·s-1 3. 300 M·s-1 4. 6.0 M·s-1 Correct Explanation: rate = k·[NO2]2·[Br2]·[O2]-1 = (3.0 x 104 M-1·s-1)·(0.01 M)2·(0.02 M)·(0.01 M)-1 = 6.0 M·s-1 3. Consider the rate constants below: I. k = 7.45 x 10-2 M-2·s-1 II. k = 1.79 x 10-2 M3·s-1 III. k = 4.77 x 10-2 M1·s-1 Which response arranges them from lowest to highest order. 1. III, II, I 2. I, II, III 3. I, III, II 4. II, I, III 5. II, III, I Correct 6. III, I, II Explanation: k = 1.79 x 10-2 M3·s-1 would correspond to a negative 2nd order reaction. k = 4.77 x 10-2 M1·s-1 would correspond to a 0 order reaction. k = 7.45 x 10-2 M-2·s-1 would correspond to a 3rd order reaction. 4. Consider the data below: Experiment number [A] M [B] M [C] M [D] M initial rate M·s-1 1 0.42 0.5 1.12 2.01 1.06 x 10-6 2 0.84 0.5 1.12 2.01 2.12 x 10-6 3 0.75 0.25 1.12 2.01 1.89 x 10-6 4 1.23 0.93 0.97 2.01 3.58 x 10-6 5 0.21 1.35 0.56 5.53 8.02 x 10-6 What is the overall order of this reaction? 1. 12. 2 Correct 3. 3 4. 4 Explanation: The most obvious order to solve for first is the order A. This is because A is the only species for which two experiments were performed in which it was the only species whose concentration was changed. The order of A is 1, as shown below. (rate1 / rate2) = ([A]1 / [A]2)a (1.06 x 10-6 / 2.12 x 10-6) = (0.42 / 0.84)a (1/2) = (1/2)a a = 1 Next we can solve for the order of B, by making use of knowing the order of A. (rate1 / rate3) = ([A]1 / [A]3) ([B]1 / [B]3)b (1.06 x 10-6 / 1.89 x 10-6) = (0.42 / 0.75) (0.5 / 0.25)b 0.56 = (0.56) (0.5 / 0.25)b b = 0 Now we know we can ignore B for the rest of the work. (rate1 / rate5) = ([A]1 / [A]5) ([C]1 / [C]5)-1([C]1 / [C]5)-1 (1.06 x 10-6 / 3.58 x 10-6) = (0.42 / 0.21) (1.12 / 0.56)-1 0.296 = (0.341) (1.12 / 0.97)c 0.867 = (1.15)c log 0.867 = c log 1.15 -0.06 = c 0.06 c = -1 Lastly we can solve for the order of D. (rate1 / rate5) = ([A]1 / [A]5) ([C]1 / [C]5)-1 ([D]1 / [D]5)d (1.06 x 10-6 / 8.02 x 10-6) = (0.42 / 0.21) (1.12 / 0.56)-1 (2.01 / 5.53)d 0.132 = (2) (2)-1 (0.363)d log 0.132 = d log 0.363 -0.88 = d -0.44 d = 2 The overall order is 2. 5. Consider the elementary reaction: CH4(g) + 2 O2(g) Æ CO2(g) + 2 H2O(g) If k = 9.7 x 106 M-1·hr-1, and there is initially 0.014 M H2O, how long will it take for the H2O concentration to reach 7.95 M? 1. 36 ms 2. 22 ms 3. 13 ms Correct 4. 5 ms Explanation: (1/[H2O]) = (1/[H2O]o) - 2kt (1/7.95) = (1/0.014) - 2(9.7 x 106 M-1·hr-1)(t) t = 0.0000036 hours = 13 ms 6. Consider the elementary reaction: SO2(aq) + H2O(l) Æ H2SO3(aq) If k = 1.21 x 10-4 M-1·s-1, and there is initially 2.3 M of SO2, what is the half life of the reaction? 1. 1.0 hr Correct 2. 1.6 hr 3. 2.6 hr 4. not enough information Explanation: t1/2 = 1 / [SO2]k = 1 / 1.19 x 10-4 M-1·s-1 · 2.3 M = 3,593 sec7. A student studying the kinetics of a reaction finds that the natural log of some concentration data produces a straight line when plotted as a function of time. What is the order of the reaction? 1. 0th order 2. 1st order Correct 3. 2nd order 4. not enough information Explanation: First order reactions exhibit a log-linear relationship between elapsed time and the concentration of reactants or products. 8. Collision theory predicts that 1. raising a system's temperature will accelerate any reactions. Correct 2. reaction intermediates are short-lived. 3. activation energy has no effect on reaction rate. 4. all collisions are productive. Explanation: Collision theory states that molecules must collide with sufficient energy and correct orientation in order to react and that both the frequency and energy of the collisions is directly proportional to system's temperature. 9. Transition state theory assumes that formation of the transition state is (reversible/irreversible) and (does/doesn't) require a minimum amount of energy. 1. irreversible, does 2. reversible, doesn't 3. reversible, does Correct 4. reversible, doesn't Explanation: TST states that a reactant (or set of reactants) must achieve their transition state (also called activated complex) before forming products. The transition state is inherently short-lived as it is in quasi-equilibrium with the reactant(s) ground state and either rapidly decays back to the ground state or goes on to form products (this is treated as irreversible for the purposes of TST). 10. What is the activation energy for a reaction that has a rate constant (k) of magnitude 4.03 x 105 and a pre-exponential factor (A) of 106? 1. 2.25 kJ·mol-1 Correct 2. 2.25 J·mol-1 3. 2,251 kJ·mol-1 4. not enough information Explanation: k = A·e(-Ea/R·T) Ea = -ln(k/A)·R·T = -ln(4.03 x 105/106)·R·T = 2,251.66 J·mol-1 11. What is a reaction's activation energy of raising the temperature from 100 °C to 1000 °C causes the rate to increase by a factor of 5? 1. Ea = 1.2 kJ·mol-1 2. Ea = 1.5 kJ·mol-1 3. Ea = 3.9 kJ·mol-1 4. Ea = 7.1 kJ·mol-1 Correct Explanation: ln(k2/k1) = (Ea/R)·(1/T1 - 1/T2) Ea = R·ln(k2/k1) / (1/T1 - 1/T2) We are told the rate increased by a factor of 5, therefore k2/k1 = 5 Ea = 8.314·ln(5) / (1/373 - 1/1273) Ea = 7,059 J·mol-1 = 7.1 kJ·mol-1 12. Consider the reaction mechanism below: step 1: H2O2 Æ H2O• + O• step 2: CO + O• Æ CO2• step 3: CO2• + H2O• Æ H2O + CO2overall: H2O2 + CO Æ H2O + CO2 Which step must be the slow step if the reaction is experimentally determined to be 2nd order overall? 1. step 1 2. step 2 3. step 3 Correct 4. Any step. Explanation: If steps 1 or 2 were the rate-determining step, the overall order of the reaction would be


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UT CH 302 - Practice Exam 3 Answer Key

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