16-exam2

16-exam2 - CHEMISTRY 16 EXAM II Dr M Richards-Babb An optical scoring machine will grade this examination The machine is not programmed to accept

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Unformatted text preview: CHEMISTRY 16 EXAM II Dr. M. Richards-Babb July 16, 1998 An optical scoring machine will grade this examination. The machine is not programmed to accept the correct one of two sensed answers and will not sense answers which are lightly marked. Mark your answer sheet carefully with a No. 2 soft lead pencil and erase any undesired marks COMPLETELY. Avoid making any extraneous marks on the answer sheet other than the information requested below. On the answer sheet: 1. Print your name in the space for NAME (last name first, circle your last name). 2. In the space marked SUBJECT print your student number. 3. In the space marked HOUR print Summer 11 ‘98. Check to see that you have 20 examination questions, periodic table and scratch paper when the exam begins. HAND IN ONLY THE ANSWER SHEET. Useful Equations and Constants: 1atm=760 torr=760 mm Hg R=0.08206 L-atm/mol-K=8.3l4 J/mol-K Kw=1.0><10'” at 25°C x= [ -b:l: (b2 - 4ac)'/’] / 2a ln[A] = ln[A]o + -kt OR log[A] = log[A]0 + -kt/2.303 [Al/[Ale = exPH“) [A] = [Me + -kt l/[A] = l/[A]0 + kt k=Ae('E’RT) 1n k = lnA + -(EA/RT) 111(k2/k1) = 'EA/RKl/TZ) ' (1/T1)] CHEMISTRY 16 HOUR EXAM II CHOOSE THE ONE BEST ANSWER. 1. The reaction C2H6(g) -> 2 CH3(g) is first order in C2H6. If k=5.5x 10‘4 sec“, how long will it take for the initial concentration to drop to 15% of its original value? A. 3.O><102 sec B. 4.9><103 sec C. 1.3><103 sec D. 9.2><102 sec E. 3.4><103 sec 2. A plot of l/[BrO'] vs time is linear for the reaction 3 Br0'(aq) -* Br03'(aq) + 2 Br‘(aq) The order of the reaction with respect to BrO' is: A. O B. 1 C. 2 D. 3 E. -1 3. The decomposition of ozone in the stratosphere can occur by the following two-step mechanism. Br+o3 —+Br0+02 BrO+O -'Br+02 Which of the following gives a correct identification of intermediates and catalysts. I t ' t m A. Br and O BrO B. BrO Br and 03 C. BrO and O BrO and 02 D. BrO Br E. 03 and O 02 Consider. the following reactions at equilibrium. Which of the equilibria will be shifted to the right (toward formation of more products) when the pressure is increased by reducing the volume? 2 03(g) H 3 02(g) 2 H200) H 2 H2(g) + 02(g) N2(g) + 02(g) H 2 N0(g) 8 S(g) H Sa(g) None will be shifted to the right. Pressure has no effect on equilibrium. WPCPU? For the reaction below, the equilibrium constant expression in terms of concentrations Kc is: 4K02(s) + 2H20(g) H 4K0H(S) + 302(g) A. {4x[KOH]x3x[02]}/{4X[K02]X2x[H20]} B. [Of/[H20]2 C- {[KOH]4+102]’}/ {[K02]‘+[H20]2} D. [H2012/[02]3 E. {3x[02]}/{2><[H20]} The rate constants for decomposition of acetaldehyde CH3CHO to CH4 and C0 are given below at two different temperatures. The activation energy E A is: 0.011 703 4.95 865 A. 9.08X10“ J/mol B. 1.38><103 J/mol C. 1.91 X105 J/mol D. 3.17X103 J/mol E. 1.96><10'4 J/mol Consider the reaction below N2(’g) + 3 C12(g) H 2 NClz(g) Kc=3.2><10“ At equilibrium it was found that [N2]eq=0.0014 M and [C12]cq=4.3><10'4 M. What is the equilibrium concentration of NC13? A. 0.050 M B. 0.0013 M C. 1.7X10'2M D. 5.3 M E. 0.19 M 10. 11. At 127°C, Kc=2.6><10‘5 for the reaction 2 NH3(g) H N2(g) + 3 H2(g) The numerical value of Kp at 127°C is: A. 0.028 B. 4.1><107A C. 2.4><10'8 D. 8.5X10'4 E. 3.0><10'3 At 298 K the equilibrium constant Kc for the reaction below is 0.0900. H20(g) + C120(g) H 2 H0Cl(g) If the following concentrations of gases are present initially, which one will proceed to form more of the reactants in order to attain equilibrium? (HINT: Calculate QC) [HOCl]=O.10 M; [H20]=0.20 M; [C120]=2.0 M [HOC1]=0.10 M; [H20]=0.10 M; [C120]=0.10 M [HOC1]=0.75 M; [H20]=25 M; [C120]=0.25 M [HOCl]=2.0 M; [H20]=0.20 M; [C120]=4.0 M Both B and D. - P130535? Calculate the equilibrium concentration of reactant B given the table and equation below. 2A(g) + 4B(g) H C(g) + 3D(g) Initial: 0M 1.2M 5.0M 3.5 M Change: -0.7 M At Eq: 2.8 M A. 0.3 M B. 1.0 M C. 2 1 M D. 0.5 M E. 1.9 M At a particular temperature, Kc=2.50 for the reaction 302(g) + N02(g) H 503(g) + N0(g) Calculate the concentration of SO3 at equilibrium, if the initial concentrations of SO2 and N02 were 1.0 M. (HINT: Simplify algebra by taking square root of both sides of eqn.) 0.39 M A. B. 0.71 M C. 0.46 M D. 0.61 M E. 0.29 M 12. 13. 14. Hydrogen for use in ammonia production is produced by the reaction CH4(g) + H20(g) H CO(g) + 3 H2(g) AH = + (endothermic) Which of the following stresses when applied to an equilibrium mixture of the gases, will cause [H2] to decrease? meow» I. An increase in temperature. II. Removal of half of the gaseous water. III. Addition of CO(g) such that its concentration is doubled. IV. Addition of a catalyst. All will decrease [H2]. 1, II and III will decrease [H2]. II and III will decrease [H2]. Only I will decrease [H2]. I and II will decrease [H2]. Consider the reaction profile below and use it to help decide which of the statements below is TRUE? a? WP PE Reactants -' Products The transition state species is an actual product of an elementary reaction. The activation energy measures the difference in potential energy between products and reactants. The activation energy for the forward reaction and the reverse reaction are always the same. The reaction profile given indicates that the net reaction is exothermic. Addition of a catalyst speeds up the reaction by decreasing the activation energy of the reaction. A certain second order reaction has a rate constant k=3.0><105 M‘sec". How long will it take for half of the reactant to disappear (ie. what is half-fife) if its initial concentration is 5.0X10‘5 M? A. 15 sec B. 0.067 sec C. 2.3><106 sec D. 30. sec E. 0.13 sec 15. 16. 17. . 18. 19. Which of the following statements about equilibrium is TRUE? P153099? The conjugate base of boric acid H3BO3 is 15 £11,509“? Calculate [OH’] in a solution with [H3O]*= 3.5><10'3 M. mpow> At equilibrium, the reverse rate is faster than the forward rate. Large values of KS (K2103) indicate that mostly products are present at equilibrium. The numerical value of Kc depends on concentration. The numerical value of Kc is independent of temperature. All reactions go 100% to completion independent of the value of K. H2303: st . OH‘; H3O* B034; HZS H4BO3"; S'2 None of the above are correct. 5.0X10'3M 3.5><10”M 8.6X10'”M 3.5><10"7M 2.9><10"2 M The pH of7.5><10“‘M HCl is: B. C. D. E. Which one of the following solutions is basic? 91.0.0533? 7.50 10.88 3.12 . 2.82 6.15 Solution with pH=5.4 Solution with pOH=3.7 Solution with pOH=7.0 Solution with pH=8.5 Both B and (D are basic. while the conjugate acid of HS” 20. The reactant acting as a Bronsted-Lowry acid in reaction I is while the reactant acting as a Bronsted-Lowry base in reaction 11 is 1: s-2(aq) + H20 —+ Hs-(aq) + OH-(aq) 11: v HS‘(aq)+H20 —' S‘?(aq)+H3O+(aq) A. 8”; H20 - B. H20; H20 C. 8‘2; HS— D. H20; HS‘ E. These reactions are not acid-base reactions. ...
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This note was uploaded on 04/26/2008 for the course CHEM 116 taught by Professor Chigwada during the Summer '07 term at WVU.

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16-exam2 - CHEMISTRY 16 EXAM II Dr M Richards-Babb An optical scoring machine will grade this examination The machine is not programmed to accept

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