FORM A is EXAM II, VERSION 1 (v1) Name

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1 FORM A is EXAM II, VERSION 1 (v1) Name 1. DO NOT TURN THIS PAGE UNTIL DIRECTED TO DO SO. 2. These tests are machine graded; therefore, be sure to use a No. 1 or 2 pencil for marking the answer sheets. 3. Completely blacken the answer circle. If you change an answer, erase completely the previous mark. 4. You may remove you answer sheet from this booklet. If you have a pink test, test form A should be darkened (B for Blue), if it is not, notify your instructor immediately. 5. Fill in your last name, first name, and initial. Blacken the corresponding letters below. 6. Fill in your ID number. CAREFULLY, blacken the corresponding numbers below this section. 7. Fill in the Dept. Course No. and Section. The Dept. = CHEM, the Course No. = 102, and your section refers to your lab section. 8. If you what your scores posted by a portion of your ID #, mark A under the option column. 9. READ THE TEST CAREFULLY. The time limit on this test is 50 minutes. 10. Use the test for scratch paper. 11. Mark your answers in this booklet as well as on the answer sheet so you can check your score with the key after the test. 12. There are 17 questions. Each counts 6 points for a total of 102 points. NO GRADES OVER 100 WILL BE RECORDED. 13. Your score will be calculated from the number of correct answers. There is no penalty for guessing. 14. Turn in your scan sheet, show your ID, and have your calculator checked. You will also turn in your test. 15. A key will be on the electronic class bulletin board at 2:15PM. This is accessed through the class homepage IMPORTANT INFORMATION: S for liquid water is J/g degree C c = 3.00 x 10 8 m/s R = L atmos/ K mol or R = J/K mol F = 1.8 C + 32 S liquid water = 4.184J/g C q = ms t Kp = Kc(RT) Δn G = H -T S S liquid water = 4.184J/g C ln[a] = -akt + ln [A] 0 [A] = -akt + [A] 0 lnk = -Ea/R (1/T) + ln A [A] o = t 1/2 1 = t 1/ = t 1/ ak ak[a] o ak [A] = akt + [A] o ln [ k 2 ] = Ea [ 1-1 ] [ k 1 ] R [ T 1 T 2 ] 102 Sample Exam 2-1 page1

2 1. Consider: 2A + 3B à 2C Experiment Initial [A] Initial [B] Initial Rate (M/S) Assume that the reaction described above is first order in both reactants A and B. Calculate the value of the reaction rate constant using data for Experiment 1. A B C. 5.6 D. 37 E In the reaction below, which average rate relationship is INCORRECT? 2NaOH + H 2 SO 4 à 3 Na 2 SO 4 + 2H 2 O A. - [NaOH]/2 t = - [H 2 SO 4 ]/ t B. [Na 2 SO 4 ]/3 t = [H 2 O]/2 t C. - [NaOH]/2 t = [Na 2 SO 4 ]/3 t D. - [H 2 SO 4 ]/ t = [H 2 O]/2 t E. 2 [NaOH]/ t = [H 2 SO 4 ]/ t 3. Given the initial-rate data below for the reaction mm + zz à products, the correct rate law is rate = Initial Conc. (M) Reactants Expt. [M] [Z] Rate (mol /L s) x x x 10-5 A. k[m][z] B. k[m][z] 2 C. k[m] 4 [Z] 2 D. k[m] 2 [Z] 2 E. k[m] 2 [Z] 4. Which one of the following graphs shows the correct relationship between concentration and time for a second order reaction? l n [A] [A] A. ti me B. time 1/ [A] [A] time time C. D. E. None of these plots should give a straight-line graph for 2 nd order. 102 Sample Exam 2-1 page2

3 5. Give the instantaneous rate for the reaction 2A --> B at 25 seconds. A. 5.0 x 10-5 M/s B. 3.0 x 10-4 M/s C. 2.0 x 10-4 M/s D. 1.0 x 10-4 M/s E. 1.5 x 10-4 M/s 6. A certain second order reaction has a half- life of 18 seconds when the initial concentration of reactant is 0.71 M. What is the reaction rate constant ( in L/mol s) for this reaction? A B. 1.3 C D E Choose the one with the largest number of TRUE statements: (1) the rate constant, k, changes with temperature (2) increasing the temperature will decrease the rate of an endothermic reaction (3) there are no units on the rate constant, k (4) a catalyst increases the rate of the reaction (5) the rate of the forward and reverse reactions are not equal at equilibrium, but are unchanging (6) the rate of a reaction will increase with increased concentration of the reactants (7) the rate increases with increased temperature because there are more collisions with the minimum energy to react A. 1,3,4,6 B. 2,3,4,6 C. 2,3,5 D. 1,5,6,7 E. 1,4,6,7 102 Sample Exam 2-1 page3

4 8. A possible mechanism for the reaction Br NO à 2NOBr is shown below: K 1 2NO N 2 O 2 K -1 fast, equilibrium N 2 O 2 + Br 2 K 2 2NOBr slow The proper form of its rate law should be rate = A. (k 2 k 1 /k -1 )[NO] 2 [Br 2 ] B. k 1 [Br 2 ] 0.5 C. k 1 [NO] 0.5 D. (k 1 /k -1 ) 2 [NO] 2 E. (k 2 k 1 /k -1 )[NO][Br 2 ] 2 9. Consider the proposed mechanism: A + F ---> 2C slow C ---> E + F fast C + 2A ---> E fast Which statements are true? I. The overall equation is: 3A ---> 2E II. F functions as an intermediate III. C functions as a catalyst IV. The rate law for the overall reaction is k[a][f] A. Only I B. Only II C. Both I and IV D. Both II and III E. the correct answer is not given 10. Use the following energy profile for the uncatalyzed and catalyzed forward reaction. Give the E for the forward catalyzed reaction. A. A B. B C. C D. D E. 102 Sample Exam 2-1 page4

5 11. Which of the following situations would be the best choice to increase the rate of a chemical reaction? A. Reducing the surface area of the reactants and increasing the temperature. B. Using a catalyst and increasing the concentration of a reactant. C. Using a catalyst and decreasing the temperature of the reactants. D. Increasing the temperature and reducing the concentration of a reactant. E. None of these 12. At equilibrium, which is/are true? A. The particles that are products will remain as products until a stress is applied to the system. B. The concentration of the products equals the concentration of the reactants. C. All chemical processes have ceased. D. Only products are present in the container. E. The rate of reactants becoming products equals the rate of products becoming reactants. 13. Choose the proper Kc expression for the following reaction: 4CuO(s) + CH 4 (g) <--> CO 2 (g) + 4Cu(s) + 2H 2 O(g) A. [CH 4 ]/[CO 2 ][H 2 O] 2 B. [CO 2 ][H 2 O] 2 /[CH 4 ] C. [Cu] 4 /[CuO] 4 D. [CuO] 4 /[Cu] 4 E. [CuO]/[Cu] 14. The value of Kc for the following reaction is 1.6. C(s) + CO 2 (g) 2 CO(g) What is the equilibrium concentration of CO if the equilibrium concentration of CO 2 is 0.50 M? A B C D E The reaction 2 SO 2 (g) + O 2 (g) <=> 2 SO 3 (g) was carried out in a 2.0 liter container. Initially, 8.0 moles of SO 2 and 7.0 moles of O 2 were placed in the container. At equilibrium 2.0 moles of SO 2 were found to be present. What is the value of K for the reaction? A B. 1.5 C. 4.5 D. 3.0 E Kp at 1000K for the following reaction is 18 2H 2 (g) + O 2 (g) <=> 2 H 2 O(g) What is the value of Kc? A B C D E. 160 NOTE: See sample exam 2-2 for an Arrhenius equation problem. 102 Sample Exam 2-1 page5

6 17. The reaction A (aq)----> B (aq) has an equilibrium constant of 2. Below are drawings of the 1 liter box containing the solution at equilibrium. Choose the box that best shows this equilibrium. Note that the water molecules are not shown. A. B. C. D. E. 18. The following information is available. CoO(s) + H 2 (g) <=> Co(s) + H 2 O(g) K = 67 CoO(s) + CO(g) <=> Co(s) + CO 2 (g) K = 490 Use the provided information to calculate Kc for the following reaction: CO 2 (g) + H 2 (g) <=> CO(g) + H 2 O(g) A. 3.3 x 10 4 B C. 7.3 D. 557 E. 423 Key question 6 pts each (unless noted) 1 C 2 E 3 E 4 C 5 E (B = 3 pts) 6 A 7 E (A, D = 3pts) 8 A 9 C (A=3) 102 Sample Exam 2-1 page6

7 10 C 11 B 12 E 13 B 14 D 15 C (B =3) 16 B (D=3) 17 A 18 B 102 Sample Exam 2-1 page7