Prod.C [A] t. rate = = =

Transcription

1 Concntration Concntration Practic Problms: Kintics KEY CHEM 1B 1. Basd on th data and graph blow: Ract. A Prod. B Prod.C Raction of Spcis A at 3 C Ractant A Product B Product C a) Writ th balancd quation for th raction of A: 2 A 2 B + C b) What vidnc did you us to dtrmin th cofficints in th balancd quation? Ractant A is disapparing at th sam rat as Product B is apparing, so thy must hav th sam stoichiomtric cofficint (notic that for any t, [A]t + [B]t [A] + [B]) Product B is apparing twic as fast as Product C, so B must hav a cofficint twic as larg as C (notic that for any t, [B]t [B] 2 ([C]t [C]), for xampl: look at t 5.s) c) Writ a gnral rat quation rlating th avrag rats with rspct to ach ractant and product. 1 [A] 1 [B] [C] rat 2 2 d) Calculat th avrag rat for Ractant A from 1s to 4s, and thn us th rlationship from Part c) to calculat th avrag rat for Product C ovr th sam tim intrval. [C] [A] 2. Sktch on th graph to th right: a) th initial rat b) th instantanous rat at Point C c) th avrag rat from Points B to G 1 (.39 M.16 M) (4.s 1.s) [A] A 2.2 x 1 5 mol/l s ½ ( x 1 5 mol/l s) 1.1 x 1 5 mol/l s 2 a) B C b) c) G

2 3. Basd on th following xprimntal data: Ovrall Raction: 2 I + S2O8 2 I2 + 2 SO4 2 Trial Numbr [I ]o Initial Rat (mol/l min) [S2O8 2 ]o x x x 1 6 a) Dtrmin th rat law (show work). rat k [I ] m [S2O8 2 ] n trial (2.) m m 1 trial (2.) n n 1 OR rat1 rat2 m [I ]1 [I ] x 1 5 mol/l min 6.23 x 1 6 mol/l min.8 mol/l.4 mol/l 1. rat k [I ] [S2O8 2 ] rat1 rat3 n [S2O8 2 ]1 [S2O8 2 ]3 ] 1.25 x 1 5 mol/l min 6.27 x 1 6 mol/l min.4 mol/l.2 mol/l 1. b) Using data from trial 1, dtrmin th rat constant (k) c) Dtrmin th initial rat if [I ]o.12 mol/l and [S2O8 2 ] o.15 mol/l d) Could th ovrall raction abov b an lmntary stp? Ys or No (circl on) Explain. rat k [I ] [S2O8 2 ] 1.25 x 1 5 mol/l min k (.8 mol/l) (.4 mol/l) k L/mol min.39 L/mol min rat k [I ] [S2O8 2 ] rat ( L/mol min) (.12 mol/l) (.15 mol/l) rat x 1 5 mol/l min 7. x 1 5 mol/l min Th ordrs in th xprimntally drivd rat law do not match th stoichiomtric cofficints in th ovrall raction. ) How many particls collid in th RDS, and what ar thy? Two particls collid in th RDS, on I and on S2O8 2.

3 1 / [H2O2] [H2O2] [H2O2] 4. Basd on th following xprimntal data: [H2O2] [H2O2] 1 / [H2O2] (L/mol) a) What is th ordr of this raction with rspct to th ractant H2O2, and how do you know? Th raction is 1 st ordr with rspct to H2O2, bcaus th [H2O2] vrsus tim graph is linar. b) What is th form of th intgratd rat law? [H2O2]t kt + [H2O2]o c) What is th rat constant (k)? k slop ris run s s x 1 4 s 1 k 8.3 x 1 4 s d) What would th concntration of H2O2 b at 45 sconds? [H2O2]t kt + [H2O2]o (8.333 x 1 4 s 1 ) (45 s) + (1.) 3.75 [H2O2]t mol/l.2 mol/l Not: Rviwing significant figur ruls for logs and xponnts would b usful (CHEM 1A Class Handouts, Chaptr )

4 5. Th half-lif of a first-ordr raction is found to b min. a) Using th dfinition of a half-lif, approximatly how long would it tak for 95.% of th original amount of ractant A to disappar? Giv a rang of possibl tims in multipls of t½ and in minuts. 95.% disappard 5.% rmaining 4 t½ to 5 t½ (6.25% to 3.125% rmaining) 113. min to min b) What is th rat constant (k) for this raction? t½ (2) k Not: 5 t ½ 5 (28.25 min) min which rounds DOWN according to th last digit 5, vn/odd rounding rul k (2) t½ (2) min min min 1 c) Using th intgratd rat law for a first-ordr raction, how long would it tak for 95% of th original amount of ractant A to disappar? [A]t kt + [A]o t [A]t [A]o k (.5) min 1 Not: 5.% rmaining min min 12 min d) If th initial concntration of ractant A is 2.5 M, what is th concntration aftr xactly 3 hours? [A]t kt + [A]o ( min 1 ) (18 min) + (2.5 M) [A]t [A]t mol/l.32 M ) What is th raction rat at xactly 3 hours? Assum th raction is not rvrsibl. rat k [A] rat ( min 1 ) ( mol/l) x 1 4 mol/l min rat 7.41 x 1 4 mol/l min

5 6. Without a catalyst, th activation nrgy of a raction is kj/mol. With a catalyst, th activation nrgy is kj/mol. Assuming th frquncy factor (A) and th initial concntrations of all ractants ar th sam for ach raction, how many tims fastr is th initial rat of th catalyzd raction at C? Hint: Start with th ratio of th rat laws for th two ractions, with th fastr raction in th numrator. rat k[ract.] m [ract.] n C k A ( Ea/RT) 3.3 K rat (cat) rat (uncat) k (cat) [ract.] m [ract.] n A [ract.]m [ract]n k (uncat) [ract.] m [ract] n A ( Ea uncat /RT) [ract.] m [ract] n ( Ea uncat /RT) ( kj/mol)(1 3 J/kJ) ( J/K mol) (3.3 K) ( Ea uncat /RT) ( kj/mol)(1 3 J/kJ) ( J/K mol) (3.3 K) x x x Answr: Th catalyzd raction is 3 x 1 31 tims fastr than th uncatalyzd raction