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March, 05 Chemistry 6 Cmmn ) Chlrine diide reacts in basic water t frm chlrite (ClO ) and chlrate accrding t the fllwing chemical equatin: ClO (aq) + OH (aq) ClO (aq) + ClO (aq) + H O(l) Under a certain set f cnditins, the initial rate f change f chlrine diide was determined t be -.0 0 M/s. What is the initial rate f change f the chlrite in under thse same cnditins? A) 5.75 0 M/s D).60 0 M/s B).5 0 M/s E) 9.0 0 M/s C).0 0 M/s [ ClO] [ ClO] Rate f reactin 0.5.0 M / s.50 M / s t t ) A kinetic study f the abve reactin under a certain set f cnditins yielded the data belw: Ep [ClO ] (M) [OH ] (M) -[ClO ] /t (M/s) 0.0500 0.00 5.75 0 0.00 0.00.0 0 0.00 0.0500.5 0 Frm and, dubling ClO quadruples the rate. S, the reactin is secnd rder fr this reactant. Frm and, dubling OH dubles the rate. S, the reactin is first rder fr this reactant. Which ne f the fllwing is the rate law fr this reactin? A) rate = k[clo ] [OH - ] D) rate = k[clo ] [OH - ] B) rate = k[clo ][OH - ] E) rate = k[clo ] [OH - ] C) rate = k[clo ][OH - ] ) T the right, the natural lgarithm f the cncentratin f reactant A is pltted versus time and prduces a straight line. What is the crrect rate f reactin? 0.6 Questins - A. 0.05As B. 0.05. As C. 0.05 A M s D. 0.05 A M s E. nne f the abve Fr st rder : ln A M M ln A kt 0. 0.6 k slpe 0.05 M / s 0s 0s Reactant Cnc. Ln [A] (M) 0.5 0. 0. 0. 0. 0 0 5 0 5 0 Time (sec) ) The reactin A + B prducts has the rate law, rate = k[a] 0 [B]. If the cncentratin f B is dubled while that f A is halved, by what factr will the rate f reactin increase? A) B) 9 C) 8 D) 6 E) 0 0 New rate k 0.5A B 8k B 8 times the riginal rate ( A ) 5) It takes.0 min fr the cncentratin f a reactant in a first-rder reactin t drp frm 0.5 M t 0. M at 5 C. Hw lng will it take fr the reactant cncentratin t drp frm 0.5 M t 0.05 M? A).0 min B) 86.0 min C) 7 min D) min E) 8 min ln A -ln A ln(0.5) ln(0.) Fr st rder : ln A ln A kt; k 0.008 min t min ln A -ln A ln(0.5) ln(0.05) t 8 min k 0.008 min
Feb. 6, 007 Chemistry 6 Cmmn 6) The table gives the eperimental data fr the reactin: A(g) B(g) at 800 K. What is the rder f the reactin? A. 0 rder B. 0.5 rder C. st rder D. nd rder E. rd rder 7) Suppse A k k C Rate Rate Rate k[ A] Run [A] (M) Rate (M/sec) 0.5 0.0 0.0 0.050 is an elementary reactin step which is fast, reversible, and at equilibrium. Nte, k is als written k f fr frward rate cnstant and k is als written k b fr backward rate cnstant. Which f the fllwing statements is crrect? Fr fastidius students, assume that bth cncentratins are less than M. At equilibrium, frward rate = reverse rate r k A k C. a. k must equal k. Thus, (big number) A (small number) C, implying A C. b. k is the inverse f k. Techically, A C C c. If k >> k, then [A] << [C]. (>> means much larger than and << means less than ) d. If k >> k, then [A] >> [C]. e. If k >> k, then n cnclusin is pssible abut the relative cncentratin f A and C. 0.0 k 0.5 0.8 0.8. S, =. 0.050 k 0. 8) At a certain temperature, the decmpsitin f A is first rder with a half life f 5 minutes. If the initial cncentratin f A is.5 M, what is its cncentratin after.8 hurs? A. 0.06 M B. 0.06 M C. 0.065 M D. 0.067 M E. 0.069 M.8hr 60 min # half lives / 5min. hr # half lives A A.5M 0.5 0.06M. 9) Fr the reactin A B + C, [A] is pltted versus the time in hurs. A straight line is btained whse slpe is -0.5 M hr -. What is the apprimate cncentratin f A after.5 hr if [A] = 0.90 M? A. 0.0 M B. 0. M C. 0. M D. 0.5 M E. 0. M Plt data reveal reactin is 0 rder. S, [ A] [ A] kt where k slpe0.5hr [ ] [ ] 0.9 0.5.5 0.05 A A kt M hr hr M 0) The gas-phase reactin f nitrgen mnide with chlrine prceeds t frm nitrsyl chlride. NO g Cl NOCl g The fllwing mechanism is prpsed: The rate f reactin equals (where k is k, k, r sme cmbinatin f them) A. k[no] D. k[n O ] B. k[no] E. k[n O ] [Cl ] C. k[no] [Cl ] Reactin rate is limited by slwest step. k NO g N O g slw k NOgCl g NOCl g fast
) The equilibrium cnstant epressin fr the reactin BrF 5 (g) Br (g) + 5F (g) is A) K c = [Br ] [F ] / [BrF 5 ] D) K c = [BrF 5 ] / [Br ][F ] 5 B) K c = [Br ] [F ] 5 / [BrF 5 ] E) K c = [BrF 5 ] / ([Br ] 5[F ] 5 ) C) K c = [Br ] [F ] / [BrF 5 ] 5 ) The slubility f silver brmide can be increased by disslving it in a slutin cntaining the thisulfate anin. AgBr(s) Ag + (aq) + Br - (aq) K = 7.7 0 - Overall reactin is sum f reactins Ag + - - (aq) + S O (aq) Ag(S O ) (aq) K =.7 0 t the left. S, K KK 6 What is the value f the equilibrium cnstant fr the verall reactin? AgBr(s) + S O - (aq) Ag(S O ) - (aq) + Br - (aq) A).6 0-9 B). 0 C).7 0-9 D).7 0 E) 6 ) Calculate K p (in atm) fr the reactin NOCl(g) NO(g) + Cl (g) at 00 C if K c at 00 C fr this reactin is. 0 M. A). 0 B).7 0 ml 0.08L atm P MRT; K p.0 M RT.0 67K.atm C) 0.70 L ml K n D). r K E).8 0 p Kc RT where n ).5 mles f NOCl were placed in a.50 L reactin chamber at 7ºC. After equilibrium was reached,.0 mles f NOCl remained. Calculate the equilibrium cnstant, K c, fr the reactin NOCl(g) NO(g) + Cl (g). A).0 0 B).8 0 NOCl NO Cl g.ml NOCl 0.M 0.5-; 0.0M C). 0 I 0.5 M 0 0.5L D) 5.6 0 C - + + NO ==0.06M E). 0 E 0.5- Cl 0.0M NO Cl 0.06 0.0 Kc 5.60 NOCl 0. 5) If ne starts with pure NO (g) at a pressure f 0.500 atm, the ttal pressure inside the reactin vessel when NO (g) NO(g) + O (g) reaches equilibrium is 0.67 atm. Calculate the equilibrium partial pressure f NO. A) 0.5 atm B) 0.7 atm C) 0.00 atm D) 0.6 atm E) The ttal pressure cannt be calculated because K p is nt given I C E NO NO O g 0.5 atm 0 0 - + + 0.5- NO P 0.5- + ; P 0.5 0.7atm tt P 0.5 0.5atm tt 6) Fr the fllwing reactin at equilibrium, which gives a change that will shift the psitin f equilibrium t favr frmatin f mre prducts? NOBr(g) NO(g) + Br (g), Hº rn = 0 kj/ml A) Increase the ttal pressure by decreasing the vlume. B) Add mre NO. C) Remve Br. D) Lwer the temperature. E) Remve NOBr selectively.
7) Calculate equilibrium cncentratin f hydrgen that results when 0.00 M HBr is heated in a sealed cntainer at a certain elevated temperature where K c = 0.6. The reactin is given belw. A. 8.90 - M B. 0.0 M C. 0.06 M D. 6. 0 - M E. 0.00 M I HBr g H g Br g 0.0 M 0 0 C - E 0.0 - HBr 0.0 0.0 H I 0.6 0.6 0. 0.0 0. 0.0 0.06 0.8.80.06 H 8.90 M 8) Sme sulfur diide and ygen are sealed in a cntainer and allwed t equilibrate at a particular temperature, prducing sme sulfur triide: SO g O g SO g H 0 In which f the fllwing cases des the reactin NOT prceed t right t reestablish equilibrium? A. Sme sulfur triide is remved. B. Sme ygen is added. Le Chatelier s principle is used. System mves in C. The temperature is decreased. directin t reduce stress. D. The vlume f the cntainer is reduced. E. Sme argn is added t the cntainer t increase the ttal pressure at cnstant vlume. Argn pressure des nt prduce a stress fr this reactin because it is neither a reactant nr a prduct it has n effect n Q. 9) What is the magnitude f Q and in which directin is the reactin fr the Haber prcess at 500 ºC, if P N = 55 atm, P H = 8 atm, and P NH = 5. atm. K p, is.50-5 atm - and the reactin is: A. Q =.80 - atm - ; the reactin is t the right. B. Q =.80 - atm - ; the reactin is t the left. C. Q =.90-5 atm - ; the reactin is t the right. D. Q =.90-5 atm - ; the reactin is t the left. E. Nne f the abve PNH 5.atm PN PH 55atm8atm N g H g NH g 0) Fr the reactin NOCl(g) NO(g) + Cl (g), K c = 8.0 M at a certain temperature. What cncentratin f NOCl must be put int an empty.00 L reactin vessel in rder that the equilibrium cncentratin f NOCl be.00 M? A).6 M B).5 M C).50 M D).5 M E).0 M I C E NOCl NO Cl g y 0 0 - + + y- Q.90 atm Reactin is t the left because Q K. Given : y. S, y K c NOCl NO Cl 8 5.6; y.5m
Part B (I and II are pints each) Ia-b. The ismerizatin f cyclprpane t frm prpene is a first-rder reactin. The rate cnstant at 700 K is 6.0 0 min, and the half-life at 760 K is 9.0 min. Als, at 760 K, 8.9% f a sample f cyclprpane changes t prpene in 79.0 min. (a) ( pts) Determine the rate cnstant fr this reactin at 760 K. Harder way: Easy way: ln A ln A kt; ln(0.5) ln k 79 min ; ln ln k 0.09 min ln(0.5) k 0.09 min t 9min 79min (b) ( pts) Calculate the activatin energy fr this reactin. E k Ae ; lnk ln A ; RT E a R Ea /( RT ) a i i.7 7.86.65 700 760 Ea Ea ln 6.0 ln A & ln 0.09 ln A R 700 R 760 8. J / ( ml K) Ea.7 7.86.65 69, 000J 69 kj / ml R 700K 760K.0 K Ic- d. The elementary reactin steps fr G + H L are: G + H I + J M + I K + J J + K L + M (c) ( pts) What is (are) the intermediate(s)? I, J, K (d) ( pts) What is (are) the catalyst(s)? M In each case, if nne write "nne". Intermediates are generated and used up. A catalysis is used up, but regenerated in a later step.
II. (a) PTS Calculate the slubility f calcium phsphate. Ca PO s Ca aq PO aq K.0 M 5 sp 5 7. 0 Ksp Ca PO y y y 0..0 7 y.60 ml/ L 08 (b) PTS Calculate the slubility f calcium phsphate, in 0. M ammnium phsphate, (NH ) PO. Ca PO s Ca aq PO aq K.0 M 5 sp Initial: 0 0. Change: +y +y Eqm: y 0.+y 0. 0... 0 Ksp Ca PO y y y y /.0 y.80 ml/ L. (c) PTS At what sdium phsphate cncentratin des a 0.0 M calcium chlride slutin start t precipitate calcium phsphate?. 0. 0.06. 0 Ksp Ca PO y y 0.5.0 6 y.50 M PO NaPO 0.06