Chem 6 Sample exam 1 (150 points total) NAME:

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Chem 6 Sample exam 1 (150 points total) @ This is a closed book exam to which the Honor Principle applies. @ The last page contains equations and physical constants; you can detach it for easy reference. @ Please write clearly and SHOW YOUR WORK. If you need to write on the back of the exam paper, please indicate this clearly. @ Some questions are more challenging than others. Allot your time accordingly, and try to answer EVERY question. @ Please put your name both on this page and on page 2. NAME:

2 NAME: FOR GRADING USE ONLY Question 1 (15 pts)... Question 2 (12 pts)... Question 3 (10 pts)... Question 4 (14 pts)... Question 5 (16 pts)... Question 6 (15 pts)... Question 7 (20 pts)... Question 8 (14 pts)... Question 9 (14 pts)... Question 10 (20 pts)... TOTAL (150 pts)...===========

3 1. (15 pts, 5 each). Consider the reaction 2HI(g) ----> H 2 (g) + I 2 (g) At 443 C, the rate depends on [HI] as follows: [HI] (M) 0.005 0.010 0.020 Rate (Ms 1 ) 7.5x10 4 3.0x10 3 1.2x10 2 a. Find the order of the reaction and the rate law. Explain your answer. b. Find the rate constant. Don't forget the units. c. Find the rate (don't forget the units) if [HI] is 0.0020 M.

4 2. (12 pts) (a, 3 pts each) Sketch an energy profile on the axes provided for 3 different reactions with the following characteristics (all energies are in kj/mol). Be sure to clearly label the energies of the reactants (R) and products (P), as well as E and E a. (a) E = 10, E a = 25 (b) E = 10, E a = 50 (a) E = 50, E a = 50 E E E reaction coordinate reaction coordinate reaction coordinate (b, 3 pts) Assume the frequency factor A is the same for all three forward reactions (R! P). Which of the three will have the greatest rate at 298 K? Explain your answer.

5 3. (10 pts) For Michaelis-Menten kinetics, we found in class that rate = k 2[E] 0 [S] k -1 + k 2 where K M = [S] + K M Find the value of [S] (in terms of the constants k 1, k 1, k 2, and K M ) for which the rate is half the maximum possible rate. Explain your answer. k 1

6 4. (14 pts) Consider this reaction, whose proposed mechanism is shown. C 2 H 6 + I 2 C 2 H 5 I + HI proposed mechanism I 2 k 1 2I k 2 I + C 2 H 6 k 3 C 2 H 5 + I 2 C 2 H 5 + HI C 2 H 5 I + I k 1 I + I I 2 a. (4 pts, 1 each) Label each step in the box as initiation (I), propagation (P) or termination (T). b. (10 pts) Find the rate law for the disappearance of C 2 H 6. Show your work. rate = d dt [C 2 H 6 ] =

7 5. (16 pts, 8 each) A and B are in a fast equilibrium. More slowly, A reacts to form C, and B yields D. A K eq B k 2 k 3 product ratio = C D [C] [D] = d dt d dt [C] [D] a. Find the product ratio (defined above) in terms of the equilibrium constant K eq and the rate constants k 2 and k 3. Show your work. b. TRUE or FALSE. Circle your answer and explain it. If the equilibrium involving A and B lies far on the left (favoring A), then C must be the major product.

6. (15 pts) Reaction of tetrafluoroethylene (C 2 F 4, important in making Teflon) to yield C 4 F 8 is second order in C 2 F 4. At 450 K, its rate constant is k = 0.0896 M 1 s 1. If the initial concentration of C 2 F 4 is 0.100 M, find its concentration (be sure to include appropriate units) after 205 seconds. Show your work. 8

9 7. (20 pts) Consider the reaction 2N 2 O 5 (g) ----> 4NO 2 (g) + O 2 (g). Concentration vs time data was obtained at 2 different temperatures. T = 338K T = 318K Time (s) [N 2 O 5 ] (M) [N 2 O 5 ] (M) 0 1.00x10 1 1.00x10 1 100 6.14x10 2 9.54x10 2 300 2.33x10 2 8.63x10 2 600 5.41x10 3 7.43x10 2 900 1.26x10 3 6.39x10 2 Find E a for this reaction. Include appropriate units, and show your work.

8. (14 pts) The ionization energy of gold (that's the energy required to remove an electron) is 890.1 kj/mol. Is light of wavelength 225 nm capable of removing an electron from a gold atom? Circle your answer and explain it. YES NO 10

11 9. (14 pts) Consider an electron moving at an unknown velocity (it must be less than the speed of light, of course). (a, 6 pts) Find the maximum uncertainty in the velocity of the electron. (b, 8 pts) Find the minimum uncertainty in the position of the electron.

12 10. (20 pts, 4 each) Multiple choice. Circle your answers. No explanation needed, and no partial credit. 1. The half-life of 99 Mo is 67 hours. How much of a 1 gram sample remains after 268 hours? (a) 0.25 grams (b) 0.5 grams (c) 0.0625 grams (d) 0.0103 grams 2. Pick the true statement about reaction rates. a. For a first-order reaction, the rate is faster at the end than at the beginning. b. For a zero-order reaction, the rate increases steadily with time. c. For a second-order reaction, the rate doesn't depend on the initial concentrations. d. Third-order reactions are rare, because the probability of 3 molecules colliding simultaneously is very small. 3. The diagram below shows Arrhenius plots for two chemical reactions. Which reaction has the larger activation energy? Circle your choice. A B ln k A B 1/T 4. The de Broglie wavelength of an electron traveling at a speed of 10 5 m/s is (a) 7.3 nm (b) 1.2 nm (c) 4.6x10 21 nm (d)1.4x10 8 nm 5. Pick the false statement about the photoelectric effect. (a) No electrons are emitted by a given metal below a specific threshold frequency ν 0. (b) Laser light (at any frequency) is so intense that it can force the emission of electrons. (c) For light with frequency greater than hν 0, the kinetic energy of the emitted electrons increases linearly with the frequency of the light. (d) For light with frequency greater than the threshold frequency, the number of electrons emitted increases with the intensity of the light.

13 Equations, Constants, etc. rate = k[a] [A] = [A] 0 e kt ln[a] = ln[a] 0 kt t 1/2 = 0.693/k rate = k[a] 2 1 1 = + kt [A] [A] 0 t 1/2 1 = k[a] 0 k = Ae Ea/RT lnk = lna E a RT rate = k 2[E] 0 [S] k -1 + k 2 where K [S] + K M = M k 1 R = 8.314 Jmole 1 K 1 = 0.0821 Latmmole 1 K 1 T( C) + 273 = T(K) λ = h/mv = h/p pv = nrt KE = mv 2 /2 = hν hν 0 N 0 = 6.02x10 23 mole 1 E = mc 2 h = 6.626x10 34 Js x p h/2 c = 3x10 8 ms 1 h = h/2π 1 nm = 10 9 m λν = c E = hν = hc/λ electron mass = 9.11x10 31 kg p = mv 1J = 1kgm 2 /s 2