chem exam 01 - HOUR EXAM #1 NAME Chemistry 030.204 ' Dr. K....

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: HOUR EXAM #1 NAME Chemistry 030.204 ' Dr. K. H. Bowen ' Tuesday Feb. 25, 2003 SSN . Instruct-ions: Write your name and social security number in the blanks provided above. Be sure that yen are in the proper testing room based on your last name. This exam should be completed in ink. No regrading will be allowed for exams not completed in ink. There are 5 problems/questions on this exam. Please“ answer all of them. Percentage credit is- indicated for each. Please confirm, before you begin, that your copy of this exam contains all five problems/questions plus a periodic table. This is a closed book, closed notes exam. The use of calculators or other electronic devices is not allowed. Show all of your work on the pages provided. __—_______I___’____‘._——-————-—-—_—__——-——_—_——“——_— Problem Number Value of problem (in points) Number of points actually earned #1 20 #2 20 #3 I - I 20 #4 ‘ ' 2o #5 p _ 20 MW 5 problems 100 total pts. possible TOTAL SCORE = Problem #1 (20 points total) In basic aqueous solution the reaction I~ +‘ ocr _._>‘ (31- + 01" follows a rate law that is consistent with the following mechanism: . ‘ kl _ 0C1 (a4) + HZO(€) T HOCKaq) + OH‘qu) (fast equilibrium) Hag) + HOCKaq) 4» HOI(aq) + Cl‘(zzq) (slow) _ ’31 OH (aq) + HOKaq) —-> H200?) + 011514) (fast) What rate law is predicted by this mechanism? : The rate is determined by the slowest elementary step, the second one: ~ rate = lezfl“][HOCl] - However, the HOCl is in equilibrium with OCl‘ and OH“ due to the first step: [HOC1][OH’] = K k1 [OCl‘] 1— kiln § Solving this for [HOCl] and inserting it into the previous expression gives the prediction _ 0C1— rate = k2K1 —[I—(-gfi:]—]- Which is, in fact, the experimentally observed rate law. Problem #2 (20 pts.) Consider the following reaction mechanism: (Cl) ' H202 —+ H20 + o O + CFzClz ——> ClO + CF2Cl ClO+O3 -—-> Cl+202 Cl + CFZCI ~——> CFZCIZ (a) What is the molecularity of each elementary step? (b) Write the overall equation for the reaction. (c) Identify the reaction intermediate(s). a) The first step is unimolecular; the three subsequent steps are bimolecular. ; This is determined simply by counting the number of interacting particles on the left sides of the four equations. ,Molecularity has meaning only in reference to elementary reactions. b) The overall reaction is the sum of the steps: H202 + 03 ~—> H20 + 2 02. c) The intermediates are 0, C10, CFgCl, and Cl. These species are produced in the course of the reaction and later consumed. ,a F ’6‘) Ozone in the upper atmosphere is decomposed by nitrogen C oxide through the reaction 03+NO —-—> OZ+NOZ The experimental rate expression for this reaction is rate = 1403} [NO] Which, if any, of the following mechanisms are consistent with the observed rate expression? - (a) 03 + NO ——> O + N03. (slow) 0 + 03 —-> 2 0; (fast) N03 + NO ——> 2 N02 (fast) (b) 03 + NO ——-> 02 + NO; (slow) (c) NO + NO 2'3 N202 (fast equilibrium) W+ O; —> NO; + 2 02 (slow) Both mechanisms (a) and (b) predict the experimentally found rate expression. Mechanism (c) predicts that the reaction would be second order in N02 and first order in 03, and is not acceptable. Problem #3 ( 20 pts.) The mechanism for the decomposition of NOZCI is NOZCI No2 + (:1 N02c1 + c1 112—» No2 + Cl; By making a steady—state approximation for [Cl], express the rate of appearance of C12 in terms of the concentrations of and .\ V The reaction is the decomposition of nitryl chloride 2N0201—+ 2NO2 + C12. The mechanism involves a equilibrium breakdown of the reactant to N02 plus Cl fol— lowed by reaction of the atomic chlorine with a second molecule of N02 Cl to generate the products. The change in the concentration of Cl with time is: d 01 ' ~35] = k1[N0201] N- Mcumoz] —— k2[Cl][NOZCl] because the rate of change of the concentration of Cl equals the rate of its production minus the rate of its consumption. The steady—state approximation is that d[Cl] /dt : ; 0. If so, then: amozcu — k_1[Cl][NO2] — amumomi} = 0 and: k1[N0201] k_1[N02i + k2[N02CI] The rate of the overall reaction equals the rate of the final elementary step, which generates the two products: [01} = dmlflizmozcumu th Substitute the expression for the concentration of the C1 into this equation: rate = (4012} k1k2[N0201]2 dt k-1[N02]+k2[NOzCl] rate = Problem #4( 20 pts.) Complete and balance the following equations for nuclear reactions that are thought to take place in stars: (a) 2 lgC ——> ?+(l)1z (b) ?+{H—> 1§C+§He (C) ZgHe~>P+ZiH Use these tests for balance in nuclear equations: the sums of the left superscripts on the two sides of the equation must be equal; the sums of the left subscripts on the two sides of the equation must also be equal. a) 2% ash/1mm b) léN +1H —+ 1:0 +3He. c) 23He —» 3He+ 2iH Natural lithium consists of 7.42% 6Li and 92.58% 7Li. Much of the tritium (31H) used in experiments with fusion reactions is made by the capture of neutrons by 6Li atoms. (3) Write a balanced nuclear equation for the process. What is the other particle produced? (b) After 6Li is removed from natural lithium, the remainder is sold for other uses. Is the molar mass of the leftover lithium greater or smaller than that of natural lithium? (a) gLi +371 —-> ?H + gHe. (b) The molar mass of the lithium from which the 6Li has been depleted Will be larger because 6Li has a smaller atomic mass than the atomic mass of naturally abundant lithium. Problem #5 ( 20 ptS-) \ _ (<5 ) ' Write balanced equations that represent the following nuclear reactions. (a) Beta emission by igCl (c) Alpha emission by 2§§Ra ‘ Positron emission by fiNa (d) Electron capture by §§Sr Positron emission (loss of (116+) and electron capture by a nucleus always lower the atomic number by one; electron emission (loss of _?6—) raises the atomic number by one. a) €301 ———> i’gAr + _?e‘ + 17 b) fiNa —) igNe + (136+ + I/ C) 2333a —* 2331“ + 3H6 , d) 3331" + .016 —* 3331) + V Three atoms of element 111 were produced in 1994 by bom- d‘n 209m with 64Ni. . h 12):; {Nate a balanced equation for thlS nuclear reacuon. What ther s ecies is produced? (8 ) (b) gNrite 2i) balanced equation for the alpha decay process of this nuclide of element 111. 268 4H o - 272 1 272 u __, Une + 2 e (a) 223131 + iiiNl "’ 111Uuu + on (b) 111U11 109 PERIODIC TABLE OF THE ELEMENTS Metals Semimetals Nonmetals Transition elements ...
View Full Document

Page1 / 7

chem exam 01 - HOUR EXAM #1 NAME Chemistry 030.204 ' Dr. K....

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online