Exam 1 (Ch. 14)

Exam #1 (Ch. 14) – KEY 1. (6.4 points) True or False T F Changing the concentration of reactants will change the rate constant, k. T F Changing the temperature of the reactants will change the rate constant, k. 2. (5 points) In class, teacher performed a demonstration with glow sticks and water at various temperatures. The result was the warmer the water, the brighter the light of the glow stick. Which of the following are the reasons for this observation? (Choose all that apply) ___ _x_ _x_ ___ ___ The warmer water lowered the activation energy of the reaction. The warmer water gave the reactant particles more energy. The warmer water allowed more reactant particles to collide per second. The warmer water was a catalyst in the reaction. Teacher is an evil wizard and uses magic to trick us!! 3. (5 points) Explain why 1st and 2nd order reaction rates decrease with time (assuming temperature stays constant): Both 1st and 2nd order reactions rates are dependent on the concentration of the reactants. For example, 1st order: Rate = k[A]. 2nd order: Rate = k[A]2 or [A][B]. So as the concentrations of the reactants decrease during the reactions, so do the rates. 4. (2.6 points) Do all reaction have rates that decrease with time (assuming temperature stays constant)? Explain. No, the rate law for 0th order reactions look like: Rate = k. As the reactants decrease during the reaction, it has no effect on the rate of reactions. 5. (8 points) Use the data in the table to the right to answer the following questions:  Calculate the average rate of disappearance of “A” (in M/s) from 0.0 s to 1.4 s. Rate   Calculate the average rate of disappearance of “A” (in M/s) from 0.0 s to 7.0 s. Rate   1.135M - 1.284M  0.11M/s 1.4s  0.0s 0.539M - 1.284M  0.11M/s 7.0s  0.0s time (s) conc. (M) 0 1.284 1.4 1.135 2.8 0.986 4.2 0.837 5.6 0.688 7.0 0.539 Based on your calculations, what would be the order of the reaction with respect to “A”? Since the rate of reaction does not change as the reactants decrease in concentration, it must be a 0th order reaction, Activated complex/transition state Ea ΔH Products 6. (6.5 points) Assume a certain chemical reaction releases 57kJ/mol of heat. Use the following diagram to show the energy change during the reaction. Make sure to label: products, Ea, ∆H (change in heat energy). Extra Credit (2 points): What is the chemical species at the highest point in the energy curve called? Label it on the diagram. 7. A certain chemical compound, under certain conditions, decomposes to two products by the following reaction: 2A → C + Z A student determines the value of k for the reaction to be 12.31 M-3hr-1.  (2.5 points) Based on the units of k, the reaction appears to be __4th______ order.  (4 points) Using the order you determined above, what would be the initial reaction rate (in M/min) if the initial concentration of “A” was 5.2M? Rate = k[A]4 Rate = (12.31 M-3hr-1)(5.2M)4 = 9.0x103 M/hr 8. (6 points) The following is the proposed mechanism for a reaction. Assume the reaction happens in three steps. Determine the molecularity of each elementary step.  o B+F→X molecularity: bimolecular o X + B → Z + F +Y molecularity: bimolecular o Y→C molecularity: unimolecular (2.5 points) Is there a catalyst in the reaction (circle one) ? YES NO NOT ENOUGH INFORMATION  (4 points) What is the overall chemical equation? 2B → C + Z  (6 points) What are the reaction intermediates? ___x___ and ___y___  (2.5 points) The rate law for the reaction was experimentally determined to be: Rate = k[B][F]. This is consistent with which elementary step being the “slow step” (circle one): FIRST STEP SECOND STEP THIRD STEP NOT ENOUGH INFORMATION Rate   1 Δ [reactant] 1 Δ [product]  a Δ time b Δ time  Rate   1 Δ [Br - ] 1 Δ [BrO3 ] 1 Δ [H  ] 1 Δ [Br2 ] 1 Δ [H 2 O]     5 Δ time 1 Δ time 6 Δ time 3 Δ time 3 Δ time 9. (5 points) In the following reaction: 5Br-(aq) + BrO3-(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l), the instantaneous rate of disappearance of H+(aq) is 5.63x10-2M/s determine the rate of…  …appearance of Br2(aq) _2.83x10-2 M/s_ 1 1  (-5.63x10 2 M/s)  (x) 6 3  …disappearance of Br-(aq) 4.69x10-2 M/s 1 1  (-5.63x10 2 M/s)   ( x) x=4.69x10-2M/S 6 5 x=2.83x102M/s 10. (7.5 points) A certain chemical reaction has the following rate law: Rate = k[H+]5[I-] [NH3]0  By doubling (change by 2x’s) the concentration of H+ the rate of reaction would change _32__x’s [2]5 = 32  By doubling (change by 2x’s) the conc. of NH3 the rate of reaction would change __1__x’s or 0x’s. (no change)  [2]0 = 1 By doubling (change by 2x’s) the conc. of I- the rate of reaction would change __2__x’s [2]1 = 2  Extra Credit (1.5 points) By doubling H+, I-, and NH3 simultaneously, the rate of reaction would change __64_x’s [2]5[2]1[2]0 = [2]6 = 64 or all your previous answers (32)(1)(2) = 64 11. Using the following data, determine the rate law for the equation: A + B + C → D Rate (M/s) 2.22 11.10 2.22 13.88 [A] (M) 0.10 0.50 0.10 0.10 x y [B] (M) 0.20 0.20 0.20 0.50 Trial 2  11.10 M/s   0.50 M   0.20 M   0.50 M         Trial 1  2.22 M/s   0.10 M   0.20 M   0.50 M  x y Trial 1  2.22 M/s   0.10 M   0.20 M   0.50 M         Trial 3  2.22 M/s   0.10 M   0.20 M   0.10 M  x y [C] (M) 0.50 0.50 0.10 0.10 z 5.00 = 5.00x x = 1 z Trial 4  13.88 M/s   0.10 M   0.50 M   0.10 M         Trial 3  2.22 M/s   0.10 M   0.20 M   0.10 M  1.00 = 5.00z z=0 z 6.25 = 2.5y ln(6.25) = (y)ln(2.5) 1.83=(y)(0.92) y = 1.9 ≈ 2  (5 points) Rate Law:  (2 points) Overall order of reaction: 1 + 2 + 0 = 3rd order  (2 points) Value of k = k Rate = k_[A]1[B]2[C]0 or Rate = k [A][B]2 using data from Trial 1 2.22 M/s = k(0.10)(0.20)2(0.50)0 2.22 M/s = 550 1/M2·s or 550 M-2s-1 2 0 (0.10 M)(0.20 M) (0.50 M) 12. (4 points) The k for a certain second order reaction is 5.67x10-3 M-1min-1. If the reactant starts at 3.2M, what would be the concentration after 6.0x103 sec? The unit for k tell us that the reaction is 2nd order. 1 1  kt  [A]t [A]0 [A]t  [A]t  1  1  kt  [A]0     6.0x10 3 s 1 min  1.0 x102 min 60 s 1 1   -3 -1 -1 2  (5.67x10 M min )(1.0x10 min)   3.2M   [A]t = 1.1M 13. (5 points) The half-life for 235U is 7.13x108 years. Assuming 235U decays by first-order kinetics, determine how long it will take for 75.0% of the starting amount to decay. ln [A]t  kt [A]0 ln t1  2 ln2 k 75.0%  (9.21x10 10 )(t) 100.% k ln2 7.13x10 8 yr k = 9.21x10-10 t = 3.12x108 years 14. (5 points) If you start with 1.00x109g of a reactant, what mass, in grams, would remain after:  4 half-lives? (1.00x109g)(1/2)4 = 6.15x107g  8 half-lives? (1.00x109g)(1/2)8 = 3.91x106g  Extra Credit (1.5 points) How many half-lives would it take for only 1.00x10-9g to be left? ln[(1.00x109g)(1/2)x] = ln(1.00x10-9g) 20.7 + (-0.693)(x) = -20.7 x = 59.7 15. (3 points) Give another name for the “slow step” in a multi-step reaction.  rate-determining step or rate-limiting step