Chemical Kinetics CHAPTER 14. Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson, Brady, & Hyslop. CHAPTER 14 Chemical Kinetics
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1 Chemical Kietics CHAPTER 14 Chemistry: The Molecular Nature of Matter, 6 th editio By Jesperso, Brady, & Hyslop CHAPTER 14 Chemical Kietics Learig Objectives: Factors Affectig Reactio Rate: o Cocetratio o State o Surface Area o Temperature o Catalyst Collisio Theory of Reactios ad Effective Collisios Determiig Reactio Order ad Rate Law from Data Itegrated Rate Laws Rate Law " Cocetratio vs Rate Itegrated Rate Law " Cocetratio vs Time Uits of Rate Costat ad Overall Reactio Order Half Life vs Rate Costat (1 st Order) Arrheius Equatio Mechaisms ad Rate Laws
2 CHAPTER 14 Chemical Kietics Lecture Road Map: Factors that affect reactio rates Measurig rates of reactios Rate Laws Collisio Theory Trasitio State Theory & Activatio Eergies Mechaisms Catalysts A) CHAPTER 14 Chemical Kietics Factors that Affect Reactio Rates B)
3 Kietics The Speed at Which Reactios Occur Kietics: Study of factors that gover o How rapidly reactios occur ad o How reactats chage ito products Rate of Reactio: o Speed with which reactio occurs o How quickly reactats disappear ad products form C) Kietics The Speed at Which Reactios Occur A B Reactio rate is measured by the amout of product produced or reactats cosumed per uit time. o [B] cocetratio of products will icrease over time o [A] cocetratio of reactats will decrease over time >)
4 Kietics Factors Affectig Reactio Rates 1. Chemical ature of reactats o What elemets, compouds, salts are ivolved? o What bods must be formed, broke? o What are fudametal differeces i chemical reactivity? D) Kietics Factors Affectig Reactio Rates 2. Ability of reactats to come i cotact (Reactats must meet i order to react) The gas or solutio phase facilitates this o Reactats mix ad collide with each other easily o Homogeeous reactio o All reactats i same phase o Occurs rapidly o Heterogeeous reactio o Reactats i differet phases o Reactats meet oly at iterface betwee phases o Surface area determies reactio rate o Icrease area, icrease rate; decrease area, decrease rate E)
5 Kietics Factors Affectig Reactio Rates 3. Cocetratios of reactats o Rates of both homogeeous ad heterogeeous reactios affected by [X ] o Collisio rate betwee A ad B icrease if we icrease [A] or icrease [B ]. o Ofte (but ot always) reactio rate icreases as [X ] icreases F) Kietics Factors Affectig Reactio Rates 4. Temperature o Rates are ofte very sesitive to temperature o Raisig temperature usually makes reactio faster for two reasos: o Faster molecules collide more ofte ad collisios have more eergy o Most reactios, eve exothermic reactios, require eergy to occur o Rule of thumb: Rate doubles if temperature icreases by 10 C (10 K) GH)
6 Kietics Factors Affectig Reactio Rates 5. Presece of Catalysts o Substaces that icrease rates of chemical reactios without beig used up o Rate-acceleratig agets o Speed up rate dramatically o Rate ehacemets of 10 6 ot ucommo o Chemicals that participate i mechaism but are regeerated at the ed GG) CHAPTER 14 Chemical Kietics Measurig Reactio Rates G@)
7 Rates Measurig Rate of Reactio o Rate = ratio with time uit i deomiator o Rate of Chemical Reactio o Chage i cocetratio per uit time. "#%&'()#%" = ["#%#(%] %&*" o Always with respect to a give reactat or product o [reactats] decrease with time o [products] icrease with time GA) Rates Measurig Rate of Reactio "#%X = [X ] [X ] t t' & % = % "[X ] t & t ' "t o Cocetratio i M uits o Time i s uits o Uits o rate: mol/l s = mol L s = M s o [product] icreases by 0.50 mol/l per secod rate = 0.50 M/s o [reactat] decreases by 0.20 mol/l per secod rate = 0.20 M/s GB)
8 Rates Rate of Reactio Always positive whether somethig is icreasig or decreasig i [X ] o Reactats o Reactat cosumed o So "[X ] is egative o Need mius sig to make rate positive Rate = "[reactat] "t o Products o Produced as reactio goes alog o So "[X ] is positive o Thus rate already positive Rate = [product] t GC) Rates Measurig Rate of Reactio Coefficiets idicate the relative rates at which reactats are cosumed ad products are formed o Related by coefficiets i balaced chemical equatio o Kow rate with respect to oe product or reactat o Ca use equatio to determie rates with respect to all other products ad reactats. Rate = 1 #A + B %& 'C + (D "A "B "C "D "t = 1 " "t = 1 # "t = 1 "t G>)
9 Rates Rate of Reactio: Example C 3 H 8 (g) + 5O 2 (g) %& 3CO 2 (g) + 4H 2 O(g) O 2 reacts 5 times as fast as C 3 H 8 Rate = "[O ] 2 = 5 "[C H ] 3 8 "t "t CO 2 forms 3 times faster tha C 3 H 8 cosumed Rate = [CO 2 ] t H 2 O forms 4/5 as fast as O 2 cosumed [H 2 O] t = " 4 5 = "3 [C 3 H 8 ] t [O 2 ] t GD) Group Problem 1/&%&I)J/"+92)4#)#&,4:3)2-&92/&%45"0)K5)#2&:/,)'"L"%) J")34I",)M452)+94,#()N/4O")L4'"P+%Q)J"9+:#")45)<&%3#) 92/&%4'")P+#6)) ) R)S);+1/) ) Q)4#)<&%3",)+5)+)%+5")&<)) C0H)I)GH UB) 3&/VW#)M2+5)%+5")4#).1/)9&'#:3",X) ) GE)
10 Rates Chage of Reactio Rate with Time Geerally reactio rate chages durig reactio, it is t costat Why? o Ofte iitially fast whe lots of reactat preset o Slower ad slower as reactats are depleted o Rate depeds o the cocetratio of the reactats o Reactats beig used up, so the cocetratio of the reactats are decreasig ad therefore the rate decreases Measured i 3 ways: o istataeous rate, average rate, iitial rate GF) Rates Istataeous & Iitial Reactio Rate Istataeous rate o Slope of taget to curve at some specific time Iitial rate o Determied at time = NO2 appearace [N O Tim e
11 Rates Average Rate of Reactio Average Rate: Slope of lie coectig startig ad edig coordiates for specified time frame NO2 appearace [NO2] Time (s) [Product] = time Rates Example Reportig Differet Types of Rates Cocetratio vs. Time Curve for 0.005M pheolphthalei reactig with 0.61 M NaOH at room temperature Rate at ay time t = egative slope (or taget lie) of curve at that poit "#%&'()* = y x = +,' +-.
12 Rates Example Reportig Differet Types of Rates [P] (mol/l) Time (s) rate = Iitial rate = Average rate betwee first two data poits ( ) M (10.5 0) s = ( M ) 10.5 s = 4.76 "10 5 M / Rates Example Reportig Differet Types of Rates Istataeous Rate at 51.1 s (90,0.0028) Rate = slope = "y "x = rise ru ( )M = (160 90) s = 0.001M =1.4 #10 5 M / s 70 s
13 Rates Example Reportig Differet Types of Rates Average Rate betwee 0 ad s Rate = slope = "y "x = rise ru ( )M = ( ) s = 0.003M s = 2.7#105 M / Group Problem A reactio was of NO 2 decompositio was studied. The cocetratio of NO 2 was foud to be M at 5 miutes ad at 10 miutes the cocetratio was M. What is the average rate of the reactio betwee 5 mi ad 10 mi? A. 310 M/mi B M/mi C M/mi D
14 CHAPTER 14 Chemical Kietics Rate Rate Laws Rates Based o All Reactats #A + B %& 'C + (D Rate = "[A] "t = k[a] m [B] o Rate Law or Rate expressio o k is the rate costat o Depedet o Temperature & Solvet o m ad = expoets foud experimetally o No ecessary coectio betwee stoichiometric coefficiets (#, ) ad rate expoets (m, ) o Usually small itegers o Sometimes simple fractios (", #) or
15 Rate Laws Rates Based o All Reactats Below is the rate law for the reactio 2A +B 3C rate= M 1 s 1 [A][B] If the cocetratio of A is 0.2 M ad that of B is 0.3 M, ad the reactio is 1 st order (m & = 1) what will be the reactio rate? rate=0.045 M 1 s 1 [0.2][0.3] rate= M/s Rate Laws Order of Reactios Rate = k[a] m [B] Expoets specify the order of reactio with respect to each reactat Order of Reactio o m = 1 [A] 1 1 st order i [A] o m = 2 [A] 2 2 d order i [A] o m = 3 [A] 3 3 rd order i [A] o m = 0 [A] 0 0 th order i [A] [A] 0 = 1 meas A does't affect rate Overall order of reactio = sum of orders (m ad ) of each reactat i rate law AH)
16 Rate Laws Order of Reactios: Example 5Br + BrO 3 + 6H + %& 3Br 2 + 3H 2 O "[BrO ] 3 = k[bro "t 3 ] x [Br ] y [H + ] z x = 1 y = 1 z = 2 o 1 st order i [BrO 3 ] o 1 st order i [Br ] o 2 d order i [H + ] o Overall order = = 4 "# = k[%& ' ] ( [% ] ( [) + ] * AG) Rate Laws Order of Reactio & Uits for k A@)
17 Group Problem The followig rate law has bee observed: Rate = k[h 2 SeO][I ] 3 [H + ] 2. The rate with respect to I ad the overall reactio rate is: A. 6, 2 B. 2, 3 C. 1, 6 D. 3, 6 AA) Rate Laws Calculatig k If we kow rate ad cocetratios, ca use rate law to calculate k From Text Example of decompositio of HI at 508 C Rate= M/s [HI] = M rate = "[HI] "t = k[hi] 2 k = rate [HI] 2 = "4 M / s ( M ) 2 = M -1 s -1 AB)
18 Rate Laws Determiig Expoets i Rate Law Experimetal Determiatio of Expoets o Method of iitial rates o If reactio is sufficietly slow o or have very fast techique o Ca measure [A] vs. time at very begiig of reactio o Before it slows very much, the " iitial rate = [A] [A] % 1 0 # t 1 t ' 0 & o Set up series of experimets, where iitial cocetratios vary AC) Rate Laws Determiig Rate Law Expoets: Example 3A + 2B %& products Rate = k[a] m [B] Expt. # [A] 0, M [B] 0, M Iitial Rate, M/s ) ) ) 10 4 # Coveiet to set up experimets so # The cocetratio of oe species is doubled or tripled # Ad the cocetratio of all other species are held costat # Tells us effect of [varied species] o iitial rate A>)
19 Rate Laws Determiig Rate Law Expoets If reactio is 1 st order i [X], Doublig [X] 1 & 2 1 Doubles the rate If reactio is 2 d order i [X], Doublig [X] 2 & 2 2 Quadruples the rate If reactio is 0 th order i [X], Doublig [X] 0 & 2 0 Rate does't chage If reactio is th order i [X] Doublig [X] & 2 times the iitial rate AD) Rate Laws Determiig Rate Law Expoets: Example Expt. # [A] 0, M [B] 0, M Iitial Rate, M/s ) ) ) 10-4 Comparig Expt. 1 ad 2 o Doublig [A] o Quadruples rate Rate 2 Rate = = 4 " 4 o Reactio 2 d order i A = [A] " 4 4 = Rate 2 Rate 1 = k A # " 2 k A # " 1 " m B2 " m B1 2 m = 4 or m = 2 # # = k 0.20 # m " 0.10 # " k " 0.10# m " 0.10 # = 0.20 # " " 0.10 # m = m 2m AE)
20 Rate Laws Determiig Rate Law Expoets: Example Expt. # [A] 0, M [B] 0, M Iitial Rate, M/s ) ) ) 10-4 "#%&'()*+,-.**/*%(0*1* o Y&:J/4'P)Z[) o \+5"),&"#)'&5)92+'P") o \"+9]&')2-3 *&%,"%)4'))^)Z[ H)^)G) 1 = Rate 3 Rate 2 = k A # " 3 k A # " 2 2 = 1 or = 0 " m B3 " m B2 Rate 3 Rate = " 4 " 4 # # = k 0.20 # m " 0.20 # " k " 0.20# m " 0.10 # = 0.20 # " " 0.10 # = 1 = 2 AF) Rate Laws Determiig Rate Law Expoets: Example Expt. # [A] 0, M [B] 0, M Iitial Rate, M/s ) ) ) 10 4 o Coclusio: rate = k[a] 2 o Ca use data from ay experimet to determie k o Let s choose first experimet k = "# " A # = &'% % &( ) *M +, % (.&(*M ( ) % = &.% % &( &% M &&, && Jesperso, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 40
21 Rate Laws Determiig Rate Law Expoets: Ex 2 2 SO 2 + O 2 %& 2 SO 3 Rate = k[so 2 ] m [O 2 ] Expt # [SO 2 ] M [O 2 ] M Iitial Rate of SO 3 Formatio, M s ) ) ) ) 10-2 BG) Rate Laws Determiig Rate Law Expoets: Ex 2 Rate 2 Rate 1 = k SO # " 2 k SO # " 2 m O2 # 2 " 2 m O2 # 1 " 1 4 = k " 0.50 # m " 0.30 # k " 0.25# m " 0.30 # 4 = " 0.50 # " 0.25 # m = m 2m 4 = 2 m or m = 2 B@)
22 Rate Laws Determiig Rate Law Expoets: Ex 2 Rate 4 Rate 2 = k SO # " 2 k SO # " 2 m O2 # 4 " 4 m O2 # 2 " 2 3 = k " 0.50 # m " 0.90 # k " 0.50# m " 0.30 # 3 = " 0.90 # " 0.30 # = 3 3 = 3 or = 1 BA) Rate Laws Determiig Rate Law Expoets: Ex 2 Rate = k[so 2 ] 2 [O 2 ] 1 1 st order i [O 2 ] 2 d order i [SO 2 ] 3 rd order overall Ca use ay experimet to fid k k = "# [%& ' ] ' [& ' ] ( = ).* (*"' +M /, (*.-*+M ) ' (*..*+M ) = *.()+M ', "( BB)
23 Group Problem Usig the followig experimetal data, determie the order with respect to NO ad O 2. A. 2, 0 B. 3,1 C. 2, 1 D. 1, 1 Expt # [NO] M [O 2 ] M Iitial Rate M s ) ) ) 10 2 BC)
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