2. The units in which the rate of a chemical reaction in solution is measured are (could be); 4rate. sec L.sec

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1 Kineic Pblem Fm Ramnd F. X. Williams. Accding he equain, NO(g + B (g NOB(g In a ceain eacin miue he ae f fmain f NOB(g was fund be ml L - s -. Wha is he ae f cnsumpin f B (g, als in ml L - s -? a ml L - s - b ml L - s - ½(- NO = (- B = ½ ( NOB c ml L - s - d ml L - s - = (- B = ½ ( NOB = ½ ( = ml L - s - e ml L - s -. The unis in which he ae f a chemical eacin in sluin is measued ae (culd be; a. L ml - s - b. ml L - s - c. s - d. ml s L - e. sec L - ml - ae M ml ml. L. sec sec L.sec. If a eacin invlving a single eacan is fis de wih a ae cnsan f s -, hw much ime is equied f 75.0% f he iniial quani f eacan be used up? a. 6.7 secnds b. 0.8 secnds c.. secnds d. 5. secnds e..6 secnds k , ans, ( sec 4. A eacin has he ae law, ae = k[a][b]. Which ne f he fllwing will cause he geaes incease in he eacin ae? a. deceasing he empeaue wihu changing he cncenains b. dubling he cncenain f B c. quadupling he cncenain f A d. ipling he cncenain f B e. dubling he cncenain f A a. b. c. d. e. T, ae Rae [B] Rae [4A] 4ae Rae [B] 4ae 9ae Rae [A] ae 5. The eacin, NO(g + O (g NO (g, was fund be fis de in each f he w eacans and secnd de veall. The ae law shuld heefe be wien as; a. ae = k[no] b. ae = k([no][o ] c. ae = k[no ] [NO] - [O ] -½ d. ae = k[no] [O ] e. ae = k([no][o ] Rae = k [NO] [O ]

2 6. A eacin has he ae law, ae = k[a][b]. Wha is he veall de f he eacin? a. b. 4 Rae k[ A][ B] c. d. n e F he eacin, XO + O XO, The ae law is heefe; a. ae = k[xo] [O ] b. ae = k[xo][o ] c. ae = k[xo][o ] d. ae = k[xo] [O ] e. ae = k[xo] /[O ] un # [XO] [O ] ae, mml L - s [ O ( [ O ] ] [ XO] ( [ XO] ae k[ XO] [ O ],, ( 9 ( 8. The epeimenal ae law is ae = k[no][b ]. In a ceain eacin miue he ae f fmain f NOB(g was fund be ml L - s -. Which uni belw is he cec uni f he ae cnsan in his case? NO(g + B (g NOB(g a. ml L - s - b. s - n=+= c. ml L - s - ae M L d. ml - L s - k ml. L. sec.sec.sec e. ml - L s - sec. M M ml 9. Given he eacin, C aa bb dd ee whee C is a caals and ae = k[a] q [B] [C] s, which ne f he saemens belw is false? a. The epnens q,, and s ae fen ineges. b. The epnen s mus be deemined epeimenall. c. The epnens q and ae equal he cefficiens a and b, especivel. d. The veall de f he eacin is q + + s. e. The smbl k epesens he ae cnsan. 0. The half-life f a chemical eacin was fund be independen f he quani f maeial which he eseache empled. The eacin is heefe; a. pssibl fis de b. definiel fis de c. ze de d. pssibl secnd de e. definiel secnd de

3 . In a fis de eacin wih nl ne eagen, he eacin was saed wih a cncenain f eacan equal mla. Afe eacl w hus, he cncenain had fallen mla. Wha is he mlai afe eacl hee hus? a M 0.69 b M k c M d M e M 0.69 shif,, ans,,0.08 ans 0.08M. A fis de eacin A B wih a ae cnsan f min -. If [A] = mla, wha will he cncenain be 50 minues lae? a M b M k c M d M e M shif,, ans 0.05 ans. The ae cnsan f a fis de decmpsiin eacin is 0.0 min -. Wha is /? a. min b. 6.4 min c sec 6.4 min k 0.0 d..5 min e. 7. min 0.048M 4. Given a eacin, A + B P, f which he bseved ae law is ae = k[a]. Which ne f he fllwing is ue? a. [A] = /k b. [A] = k/ c. /[A] = k d. he half-life is 0.69/k e. e [A] = -k 5. In a fis de eacin, wha facin f he maeial will emain afe 4 half-lives? a. /6 b. /8 c. /9 d. /4 e. / 4 n ( 4 6. The iniial cncenain f a eacan in a fis de eacin is 0.60 mla. Wha will be is cncenain afe half-lives? a M b. 0.0 M c M d. 0.0 M e M n ( M

4 7. F he eacin, A B + C, he ae law is k[a]. If i akes 80.0 secnds f 70.0% f a 0.0 gam sample f A be ansfmed in pducs, wha is he value f he ae cnsan? a s - b s - k c s - 00 k80 d s e s ,, ans,, 80 k 0.050s 8. A eacin is fis de veall. F a given sample, if is iniial ae is ml L - s - and 5.0 das lae is ae dpped ml L - s -, wha is is half-life? a. 5.0 das b das c..5 das d. 5.0 das e. 7.5 das k ,, ans,0.695 ans 5das 9. F a ne sep eacin, he acivain eneg f he fwad eacin is 40.0 kj ml -, and he enhalp f eacin is -0.0 kj ml -. Which saemen belw is ue? a. The acivain eneg f he fwad eacin wuld be affeced a geae een han he acivain eneg f he evese eacin b addiin f a caals. b. The value f he enhalp f eacin wuld be deceased b addiin f a caals. c. The eacin is endhemic. d. The evese eacin is slwe han he fwad eacin (smalle ae cnsan. e. The eacin ae wuld be deceased b an incease in empeaue. 0. F a ne sep eacin, he acivain eneg f he fwad eacin is 40.0 kj ml -, and he enhalp f eacin is -0.0 kj ml -. Calculae he acivain eneg f he evese eacin. a kj ml - b kj ml - c. -00 kj kj ml - d kj kj ml - e. +00 kj kj ml - 4

5 . The acivain eneg f a eacin can be fund b finding he slpe f a pl f l(k vs T - and a. adding his slpe -R b. mulipling his slpe b.0 c. dividing his slpe b -R d. mulipling his slpe b.0r e. mulipling his slpe b R Ea k A R T b m k,, T m( R E a m E R a. F a chemical eacin, he ae cnsan a 50.0 C is s -, and he acivain eneg is.40 kiljules. Calculae he value f he ae cnsan a 5.0 C. a s - b s - c s - d s - k Ea T T k R T T k e s -.4( 85 ( , shif,, ans ans k sec. The ae cnsan f a ceain chemical eacin is L ml - s - a 5.0 C and 0.05 L ml - s - a 50.0 C. Wha is he acivain eneg f he eacin, epessed in kj? a. 5. kj b. 5.6 kj c. 7.6 kj d. 45. kj k Ea T T k R T T Ea e. 60. kj ,, ans ( ( 5 E 5.6kJ 5. Which ne f he fllwing saemens cncening he ae f a chemical eacin is false? a. I will be ve apid if he acivain eneg is lage. b. I will be slw if ne me f he seps is slw. c. I ma be inhibied smeimes b ceain caalic agens. d. I is dependen n empeaue. e. I fen inceases when he cncenains f ne f he eacans is inceased. 6. A vaiable which has n effec n he ae f a chemical eacin unde an cicumsances is; a. eneg f acivain b. caals c. cncenain f he eacans d. empeaue e. sandad enhalp f eacin f he ssem a 5

6 7. A caals ales he ae f a chemical eacin b a. pviding an alenae pahwa which has a diffeen acivain eneg b. changing he pducs fmed in he eacin c. changing he fequenc f cllisins beween mlecules d. alwas pviding a suface n which mlecules eac e. changing he enhalp f eacin f he eacin 8. Cnside he fllwing eacin: A B The aveage ae f appeaance f B is given b B. Cmpaing he ae f appeaance f B and he ae f disappeaance f A, we ge [B] ( [A] A -/ B +/ C -/ D + ( NH ( B ( B ( A 9. Which subsance in he eacin belw eihe appeas disappeas he fases? 4NH +7O 4NO +6H O A NH B O C NO D H O ( NH ( O ( NO ( ( O ( NH ( NO ( H O ( fases O H O 0. The veall de f a eacin is. The unis f he ae cnsan f he eacin ae. A M s B M s C s ae k M sec. M M.sec M. sec D M 6

7 A flask is chaged wih 0.4 ml f A and allwed eac fm B accding he eacin A(g B(g. The fllwing daa ae bained f [A] as he eacin pceeds in L flask: (-. The aveage ae f disappeaance f A beween 0 s and 0 s is ml/s. A. 0 B.0 C D 454 ae The aveage ae f appeaance f B beween 0 s and 0 s is ml/s. A.50 B C.50 D 7.0 ae Hw man mles f B ae pesen a 0 s? A 0.0 B 0.0 C 0.0 D [ B] 0 ae A he sa f an epeimen, hee ae 0.00 ml f eacan and 0 ml f pduc in he eacin vessel (L. Afe 5 min, 0.08 ml f eacan (CH NC emain. Thee ae ml f pduc (CH CN in he eacin vessel. CH NC(g CH CN(g A 0.0 B C 0.00 D 0.08 E 0.09 [ CH CN] M 7

8 5. If he ae law f he eacin; A B pducs is fis de in A and secnd de in B, hen he ae law is ae =. A k A B B k[a] [B] C k[a][b] D k[a] [B] ae k 6. The kineics f he eacin belw wee sudied and i was deemined ha he eacin ae inceased b a fac f 9 when he cncenain f B was ipled. The eacin is de in B. A B P Rae =k[b] 9ae=[B] X= A ze B fis C secnd D hid 7. A eacin was fund be hid de in A. Inceasing he cncenain f A b a fac f will cause he eacin ae. Rae =k[b] ae=[b] ae = imes = 7 A emain cnsan B incease b a fac f 7 C incease b a fac f 9 D iple The daa in he able belw wee bained f he eacin: (8-40 A B P 8. The de f he eacin in A is. A B C D 4 (, ( 8

9 9. The de f he eacin in B is. A B C D 4 E 0 (, ( The veall de f he eacin is. n= +0= A B C D 4 4. F a fis-de eacin, a pl f vesus is linea. A [A], B A, C [A], D A, k b m k k m k 4. The eacin belw is fis de in[ho ]: HO (l HO(l O (g A sluin iginall a M HO is fund be M afe 54 min. The half-life f his eacin is min. A 6.8 B 8 C 4 D k ,, ans, ans 4. Of he fllwing, uni f a eacin ae is. 8das A ml / L B M / s C L/ml s D s/ml.l ae M ml ml. L. sec sec L.sec 9

10 44. The daa in he able belw wee bained f he eacin A B P The ae law f his eacin is ae =. A k[a][b] (, B k[p] C ka B D ka B E ka ( ae k[ A], ( 9 ( A cmpund decmpses b a fis-de pcess. If 5.0 % f he cmpund decmpses in 60.0 minues, he half-life f he cmpund is k A 65 minues B 0 minues C 45 minues 60 D 80 minues ` ,, ans, ans 46. Of he fllwing, will lwe he acivain eneg f a eacin. A inceasing he cncenains f eacans B aising he empeaue f he eacin C adding a caals f he eacin D emving pducs as he eacin pceeds 45 min ues 47. A paicula fis-de eacin has a ae cnsan f.50 s a 5.0 C. Wha is he magniude f k a 95.0 C if E a 55.5 kj/ml? A B C 576 D k k.50 k Ea T T R T T ( 70 ( , shif,, ans.50 ans k sec 0

Sections 3.1 and 3.4 Exponential Functions (Growth and Decay)

Sections 3.1 and 3.4 Exponential Functions (Growth and Decay) Secions 3.1 and 3.4 Eponenial Funcions (Gowh and Decay) Chape 3. Secions 1 and 4 Page 1 of 5 Wha Would You Rahe Have... $1million, o double you money evey day fo 31 days saing wih 1cen? Day Cens Day Cens