- 1.Kineticspp The study ofreaction rates . Spontaneousreactions
are reactions that will happen -butwe cant tell how fast. Diamond
will spontaneously turn to graphite eventually. Reaction mechanism-
thestepsbywhich a reaction takes place.
2. 12.1 Reaction Rate Rate =Conc. of A at t 2- Conc. of A at t 1
t 2 - t 1 Rate = [A] t Change in concentration per unit time For
this reaction N 2+ 3H 2 2NH 3 3. N 2+ 3H 2 2NH 3 As the reaction
progresses the concentration H 2goes down Concentration Time [H 2 ]
4. As the reaction progresses the concentration N 2goes down 1/3 as
fast as for H 2 3 moles H 2being consumed for every 1 mole of N 2
Concentration Time [H 2 ] [N 2 ] 5. As the reaction progresses the
concentration NH 3goes up mole proportionally Concentration Time [H
2 ] [N 2 ] [NH 3 ] 6. Calculating Rates Average ratesare taken over
long intervals Instantaneous ratesare determined by finding the
slope of a line tangent to the curve at any given point because the
rate can change over time We take the derivative. 7. Average slope
method Concentration Time [H 2 ] t 8. Instantaneous slope method.
Concentration Time [H 2 ] t 9. Defining Ratepp We can define rate
in terms of thedisappearanceof thereactant orin terms of the rate
ofappearanceof theproduct . In our example N 2+3H 2 2NH 3 - [ N 2
]=- 3 [ H 2 ]=2 [ NH 3 ] t t t Since [ reactants ]alwaysdecreases
with time, any rate expression includes (-) sign 10. 12.2 Rate
Laws:An Introduction Reactions arereversible . As products
accumulate they can begin to turn back into reactants. Early onthe
rate will depend ononlythe amount ofreactantspresent. So, we want
to measure the reactantsas soon as they are mixed . This is called
theInitial rate method . 11. Two key points: The concentration of
theproductsdo not appear in the rate law because this is
aninitialrate (only the [reactants]). Theordermust be
determinedexperimentally . Cannotbe obtained from the equations
coefficients (remember this!). Rate Lawspp 12. Rate Laws N 2+ 3H 2
2NH 3 When forward & reverse reaction rates areequalthen there
is no change in concentration of either. The reaction is
atequilibrium . Rate =k [N 2 ] m [H 2 ] n Note:m and n arenot
coefficientsof the balanced equation,must be determined
experimentally . 13. You will find the rate depends only on the
concentration of thereactants . Rate =k [ NO 2 ] n This is called
arate law expression . kis called the rate constant. nis theorderof
the reactant - usually a positive integer (1st, 2nd, etc.). It
isnotthe coefficient of the balanced reaction (e.g., it could be 1)
We caninsteaddefine Rate in terms ofproduction of O 2
vs.consumption of NO 2 2 NO 22 NO+O 2 14. The rate ofappearanceofO
2is . . . Rate '= [ O 2 ] =k ' [ NO 2 ] n t Because there are 2 NO
2consumed for each O 2produced . . . Rate = k[NO 2 ] n = 2 x Rate =
2 x k ' [NO 2 ] n Sok[NO 2 ] n = 2 x k ' [NO 2 ] n So k = 2 x k'2
NO 22 NO+O 2 15. Figure 12.1 p. 562 Definition of Rate. 2NO 2 2NO +
O 2 16. Types of Rate Laws DifferentialRate law - describes
howratedepends onconcentration . IntegratedRate Law - Describes
howconcentrationdepends ontime . For each type of differential rate
law there is an integrated rate law and vice versa. Rate laws can
help us better understand reaction mechanisms. 17. 12.3 Determining
Rate Laws Thefirststep is to determine theformof the rate law
(especially itsorder ). Mustbe determined from experimental data (
notfrom the coefficients!!). For this reaction: 2 N 2 O 5(aq)4NO
2(aq) + O 2 (g) The reverse reaction wont play a role. Next slide
shows experimental results (like youll do in lab 12). 18. Now graph
the data [N 2 O 5 ] (mol/L)Time (s) 1.00 0 0.88 200 0.78 400 0.69
600 0.61 800 0.54 1000 0.48 1200 0.43 1400 0.38 1600 0.34 1800 0.30
2000 19. To find rate we have to find theslopeat two points We will
use thetangentmethod. 20. At.90M the rate is - (.98 - .76)=
-0.22=5.5x 10-4 M s -1 (0-400)-400 21. At.45 Mthe rate is - (.52 -
.31)= -0.22= 2.7 x 10-4 (1000-1800)-800 22. Since the rate attwice
the concentrationistwice as fast , the rate law must be.. Rate = -
[N 2 O 5 ] = k[N 2 O 5 ] 1= k[N 2 O 5 ] tWe say this reaction
isfirstorderinN 2 O 5 Theonlyway to determine order is to run the
experiment. n isnotsame number as coefficient (only a coincidence).
1st order means doubling the [] doubles the rate(know this).Rate =
k[A] 1 23. A Plot of [N 2 O 5 ] as f(t) for 2N 2 O 5 4NO 2+ O 2 pp
Note that the reaction rate at 0.90Mis twice that at 0.45M So, the
reaction is 1st order because doubling the [ ] doubles the rate 24.
The method of Initial Ratespp This method requires that a reaction
be run several times (do in Lab 12). The initial concentrations of
the reactants arevaried . The reaction rate is measuredjust
afterthe reactants are mixed. Eliminates the effect of the reverse
reaction and makes it easier to calculate. 25. Heres an AP-type
Questionpp Write general form of rate law for: BrO 3 -+ 5 Br -+ 6H
+ 3Br 2+ 3 H 2 O Answer is . . . Rate =k [BrO 3 - ] n [Br - ] m [H
+ ] p Determinen ,m ,pby comparing rates Add n ,m ,pvalues to
getoverall order Find value ofkby plugging in valuesfromany oneof
the experiment rates 26. An AP Test Examplepp For the reactionBrO 3
-+ 5 Br -+ 6H + 3Br 2+ 3 H 2 O The general form of the Rate Law is
Rate = k[BrO 3 - ] n [Br - ] m [H + ] p Be able to write the above
as the 1st question, given the above reaction. We useexperimental
datato determine the values of n,m, and p 27. Now we have to see
how theratechanges withconcentration Initial concentrations (M)pp
Rate (M/s) BrO 3 - Br - H + 0.10 0.10 0.10 8.0 x 10 -4 0.20 0.10
0.10 1.6 x 10 -3 0.20 0.20 0.10 3.2 x 10 -3 0.10 0.10 0.20 3.2 x 10
-3 28. Findnby ratio of Rate 2/Rate 1, in which only [BrO 3 - ]
changes n= 1 because (next slide) Initial concentrations ( M )pp
Rate (M/s) BrO 3 - Br - H + 0.10 0.10 0.10 8.0 x 10 -4 0.20 0.10
0.10 1.6 x 10 -3 0.20 0.20 0.10 3.2 x 10 -3 0.10 0.10 0.20 3.2 x 10
-3 29. The mathpp Rate 2=1 .6 x 10 -3mol/Ls=k (0.20 mol/L) n (0.10
mol/l) m (0.10 mol/L) p Rate 18.0 x 10 -4mol/Ls k (0.10 mol/L) n
(0.10 mol/l) m (0.10 mol/L) p =2.0=(0.20 mol/L0.10 mol/L) n= (2.0)
n So,n= 1 30. Findmby ratio of Rate 3/Rate 2, in which only [Br - ]
changes m= 1(ditto on math) Initial concentrations ( M )pp Rate
(M/s) BrO 3 - Br - H + 0.10 0.10 0.10 8.0 x 10 -4 0.20 0.10 0.10
1.6 x 10 -3 0.20 0.20 0.10 3.2 x 10 -3 0.10 0.10 0.20 3.2 x 10 -3
31. Findpby ratio of Rate 4/Rate 1, in which only [H + ] changes p=
2(ditto on math) Initial concentrations ( M )pp Rate (M/s) BrO 3 -
Br - H + 0.10 0.10 0.10 8.0 x 10 -4 0.20 0.10 0.10 1.6 x 10 -3 0.20
0.20 0.10 3.2 x 10 -3 0.10 0.10 0.20 3.2 x 10 -3 32. AP-type
questionpp So, rate of reaction isfirstorder inBrO 3 - and Br
-andsecondorder inH + Overall rate= 1 + 1 + 2 = 4 Rate lawcan be
written as: Rate =k [BrO 3 - ][Br - ][H + ] 2 33. AP-type
questionpp Rate =k [BrO 3 - ][Br - ][H + ] 2 Try calculatingkfrom
Expt. 2BrO 3 1- Br 1- H 1+ Rate ( M /s )0.20 0.10 0.10 1.6 x 10 -3
The answer is . . . k = 8.0 L 3 /mol 3 s(or 8.0M -3 s -1 ) The
complete rate law is . . . Rate =8.0 L 3 /mol 3 s[BrO 3 - ][Br -
][H + ] 2 Be sure to express the correct units! 34. AP-type
question Units areveryimportant!!! Lets try Text # 27, p. 568 . . .
Answer is . . . 25 a. Rate = K[NOCl] 2 b. k= 6.6 x 10 -29cm 3
/moleculess c. k = 4.0 x 10 -8 M -1 s -1 Do # 25 at home.(Quiz
soon!) 35. 12.4 Integrated Rate Law Expresses the reaction
concentration as a function oftime(vs. rate as f(conc.)). Form of
the equation depends on theorderof the rate law (from
differential). ChangesRate = - [A] n =k [A] n t We will only work
with n = 0, 1, and 2 for reactions with only asingle
reactant(decomposition) 36. First Order For the reaction2N 2 O 5
4NO 2+ O 2 We found the (differential) Rate =k [ N 2 O 5 ] 1 I.e. ,
if concentration doubles rate doubles. If weintegratethis equation
with respect totimewe get theIntegratedRate Law ln[N 2 O 5 ] = -k t
+ ln[N 2 O 5 ] 0 (y = mx + b) ln is the natural log [N 2 O 5 ] 0is
theinitialconcentration. 37. General formis Rate = - [A] / t = k[A]
ln[A] = -k t + ln[A] 0 In the form y = mx + b y = ln[A] m = - k x =
t b = ln[A] 0 A graph ofln[A]vs.timeis a straight lineifit is first
order (otherwise, adifferentgraph is a straight line, see following
slides). FirstOrder 38. By getting a straight line you prove it is
first order. Conversely, if a plot of ln[A]vs . t (as opposed to
plotting other values) isnota straight line, the reaction
isnotfirst order in [A]. So, you have to plot several possibilities
to determine which order it is. First Order 39. A Plot of [N 2 O 5
] as f(t) for 2N 2 O 5 4NO 2+ O 2 . Decomposition reaction Note
that the reaction rate at 0.90Mis twice that at 0.45M So, the
reaction is 1st order because doubling the [ ] doubles the rate
This is not a plot ofln [N 2 O 5 ], which would be a straight line
if it was first order . . . 40. ln [A] = -k t + ln[A] 0 By getting
the straight line (plotln [A]vs.t) you can prove it is first order.
Often expressed in a ratio First Order 41. Figure 12.4 A Plot ofIn
(N 2 O 5 ) Versus Time (1st order) 42. Half Life The time required
to reachhalfthe original concentration. If a bacterium doubles
every minute, and takes 40 minutes to fill the jar, how long to
fill half the jar? 39 minutes. Ifthe reaction isfirstorder, then:
[A] = [A] 0 /2whent = t 1/2 43. Figure 12.5:A Plot of (N 2 O 5 )
vs. Time for theDecompositionReaction of N 2 O 5 44. Half Life
ln(2) =k t 1/2 Solving for t 1/2 : The time required to reach half
the original concentration. Ifthe reaction is first order [A] = [A]
0 /2when t = t 1/2 45. Half Life t 1/2= ln(2)/k = 0.693/ k So, the
time to reach half the original concentration doesnotdepend on the
starting concentration. This is an easy way to findkif you know or
can determine the t 1/2and the reaction is 1st order. 46. Second
Order Rate = - [A] / t= k[A] 2 integratedrate law has the form:
1/[A] =k t + 1/[A] 0 , where y = mx + b, so: y= 1/[A]m = k x= t b =
1/[A] 0 Get a straight line if1/[A]vstis graphed (instead of ln[A]
= -k t + ln[A] 0 ) Knowingkand [A] 0you can calculate [A] at any
timetby graphing y = mx + b 47. Second Order Half Life [A] = [A]
0/2 att = t 1/2 48. Z7e545 Fig 12.6 (a) A Plot of In(C 4 H 6 )
Versust(not straight line, so not 1st order) (b) A Plot of 1/(C 4 H
6 ) Versust(straight line so must be 2nd order) 49. Determining
Rate Laws Since we get a straight line with (b) and not with (a) in
previous slide:The reaction must be second order in C 4 H 6 50.
Zero Order Rate Law Rate = k[A] 0= k Rate doesnotchange with
concentration Important for drug pharmacokinetics Integratedrate
law:[A] = -kt + [A] 0 y = mx + b format When [A] = [A] 0/2t = t
1/2t 1/2=[A] 0/2kfor zero order. 51. Z7e 546 Figure 12.7 A Plot
of[A] Versust for aZero-OrderReaction 52. Occurs most often when
reaction happens on asurfacebecause the surface area stays
constant. Also applies toenzyme chemistryand other reactions
involvingcatalysts . Zero Order Rate Law 53. Fig. 12.8
Decomposition Reaction N 2 O2N 2+ O 2 Even though [N 2 O] is twice
as great in (b) as in (a) the reaction occurs on a saturated
platinum surface, so rate is zero order 54. Zero Order Graph Time
Concentration 55. Zero Order Graph Time Concentration A]/ t = slope
t -k = A] t 1/2= [A 0 ]/2k 56. More Complicated Reactions Reactions
withmore than one reactant . BrO 3 -+ 5 Br -+ 6H + 3Br 2+ 3 H 2 O
For this reaction we experimentally found the (differential) rate
law to be Rate = k[BrO 3 - ][Br - ][H + ] 2 Toinvestigatethis
reaction rate we need to control the conditions. 57. Rate = k[BrO 3
- ][Br - ][H + ] 2 We set up the experiment sotwoof the reactants
are inlarge excess(Lab 12). [BrO 3 - ] 0 = 1.0 x 10 -3 M [Br - ] 0=
1.0M (1000 x as great) [H + ] 0= 1.0M (ditto) As the reaction
proceeds [BrO 3 - ] changes noticeably ( limiting reactant ).[Br -
] and [H + ]do not (since in large excess) 58. This rate law can be
rewritten . . . Rate = k[Br - ] 0 [H + ] 0 2 [BrO 3 - ] =k [BrO 3 -
] because, [Br - ] 0and [H + ] 0are constant (since in large
excess).Solving fork . . . k = k[Br - ] 0 [H + ] 0 2and
substitutingk into the original rate law . . . Rate =k [BrO 3 - ]
This is called apseudo first orderrate law. k= k [Br - ] 0 [H + ] 0
2 Rate = k[BrO 3 - ][Br - ][H + ] 2 59. 12.5 Summary of Rate Laws
Assume only theforwardreaction is important so can produce rate
laws that only containreactantconcentrations Differentialrate law(
akarate law) shows howratedepends onconcentration. Integratedrate
lawshows howconcentrationdepends ontime . Choice of using either
method depends on data being experimentally derived. 60. Summary of
Rate Laws Most common method to determine thedifferentialrate law
is method ofinitial rates(gettas close to zero as possible) We will
use this method in Lab 12 If instead using theintegratedrate law,
measureconcentrationat various times. Youllgraphthis in Lab 12 and
plot the values to get a straight line as follows: 61. Summary of
Rate Lawspp Memorize the followingrate laws(for AP questions) and
use the t 1/2for our lab 12. (rf. Table 12.6 p. 548) - use for
online HW! Reactionorderis found by whichever of the following
gives straight line: Zeroorder when[A]vs . tyields straight line
Rate = k and t 1/2= [A] o /2 k 1storder whenln [A]vs.tyields
straight lineRate = k[A] and t 1/2= 0.693/ k 2ndorder when1/
[A]vs.tyields straight lineRate = k[A] 2and t 1/2= 1 /k[ A] o 62.
Use for Online HWpp Use for online HW Hint:on one problem you will
have to first determine theorder , then use thehalf-life equation(5
thline) to solve for k, then use theIntegrated rate law(2 ndline)
to solve for time (t). 63. 12.6 Reaction Mechanisms The series of
steps that actually occur in a chemical reaction. A balanced
equation does not tell ushowthe reactants become products. Kinetics
can tell us something about the mechanism. 64. Figure 12.9 A
Molecular Representation of the Elementary Steps in the Reaction of
NO 2and CO Overall:NO 2+ CONO + CO 2 Step 1:NO 2+ NO 2 NO 3+ NO(k 1
) Step 2:NO 3+ CONO 2+ CO 2 (k 2 ) NO 3is anintermediate 65. 2NO 2+
F 2 2NO 2 FRate = k[NO 2 ][F 2 ] Theproposedmechanism is: NO 2+ F 2
NO 2 F+ F(slow) F+ NO 2 NO 2 F (fast)F is called anintermediate. It
is formedthenconsumed in the reaction Reaction Mechanisms 66. Each
of the two reactions is called anelementary step . The rate for
each reaction can be written from itsmolecularity . Molecularity is
the number of pieces that must come together (collide) to produce
the reaction indicated by that step. Reaction Mechanisms 67.
Unimolecularstep involves one molecule - Rate isfirstorder.
Bimolecularstep - requires two molecules - Rate issecondorder
Termolecularstep- requires three molecules - Rate
isthirdorderTermolecularsteps are almost never heard of because the
chances of three molecules coming into contact at the same time are
miniscule. With molecularity the coefficientscanbecome the
exponents. 68. Aproducts A+Aproducts 2Aproducts A+Bproducts
A+A+BProducts 2A+BProducts A+B+CProducts This is also in Table
12.7, p. 550.Know. Rate = k[A]Rate= k[A] 2 Rate= k[A] 2 Rate=
k[A][B] Rate= k[A] 2 [B] Rate= k[A] 2 [B] Rate= k[A][B][C] 69.
Finding the Reaction Mechanismpp Must satisfytworequirements :
Sumof theelementarystepsdoesgive theoverallbalanced equation for
the reaction The proposedmechanismmustagreewith theexperimentally
derivedrate law. 70. How to get rid of intermediatespp Use the
reactions that form them. Ifthe reactions arefastandirreversible
the concentration of theintermediateis based onstoichiometry .(Use
spider). Ifit is formed by areversiblereactionsetthereversible
reaction ratesequalto each other. 71. Formed
inreversiblereactionspp 2 NO + O 2 2 NO 2 Overall reaction Given
:Expt-derived Rate =k [NO] 2 [O 2 ] Proposed Mechanism :2 NON 2 O 2
(fast) N 2 O 2+ O 22 NO 2(slow) 2nd stepis rate determining (slow),
so need to get this one to look like the expt.-derived rate,k [NO]
2 [O 2 ] 1st stepisreversibleso set both rates inthisstepequalto
each other: k 1 [NO] 2= k -1 [N 2 O 2 ] 72. Formed
inreversiblereactionspp 2 NO + O 2 2 NO 2 overall reaction
Given:Expt-derived Rate =k [NO] 2 [O 2 ] Proposed Mechanism2 NON 2
O 2 (fast)N 2 O 2+ O 22 NO 2(slow) 2nd steprate comes
frommolecularity A+Bproducts Rate= k[A][B] Rate = k 2 [N 2 O 2 ][O
2 ] 73. Formed inreversiblereactionspp Given:Expt-derived Rate =k
[NO] 2 [O 2 ] 2 NO + O 2 2 NO 2 overall reaction Proposed
Mechanism2 NON 2 O 2 (fast) N 2 O 2+ O 22 NO 2(slow)Rate =k 1 [NO]
2= k -1 [N 2 O 2 ](from 1st step)Rate = k 2 [N 2 O 2 ][O 2 ](from
2nd step) Solve for [N 2 O 2 ]from (1) & subinto (2) rate =k
2(k 1 / k -1 )[NO] 2 [O 2 ]= k[NO] 2 [O 2 ] 74. Formed
inreversiblereactionspp 2 NO + O 2 2 NO 2 overall reaction Proposed
Mechanism2 NON 2 O 2 (fast) N 2 O 2+ O 22 NO 2(slow) Since
model-derivedRate = k[NO] 2 [O 2 ]agrees with theexpt.-derivedrate
. . . And sum of the steps = overall reaction: The 2-step proposed
mechanism works! 75. Formed inreversiblereactionspp Inother words:
Since this last equation, which we derived by eliminating the
intermediateis the sameas theexperimentallyderived rate: And , the
sum of the 2 steps add up to the overall balanced reaction:The
proposed 2-step mechanism is correct. There are AP questions that
have you do this for. 76. Formed infast, irreversiblereactionspp 2
IBrI 2 + Br 2 overall reaction Experimentally-derivedRate =k [IBr]
2 Proposed Mechanism IBr I + Br(fast) IBr + BrI + Br 2 ( slow ) I +
I I 2 (fast) Step 2 is slow, so its rate determining Slow Rate = k[
I Br][Br]but[Br]= [ I Br] So, Rate = k[ I Br][ I Br] = k[ I Br] 2
77. Formed in fast, irreversible reactionspp 2 IBrI 2 + Br 2
overall reaction Proposed Mechanism IBr I + Br(fast) IBr + BrI + Br
2 (slow) I + I I 2 (fast) Sum of steps = overall balanced reaction
Rate= k[IBr][IBr]= k[IBr] 2issame as experimentally derived So, the
mechanism works (But this doesnt prove its thecorrectmechanism) 78.
12.7 A Model for Chemical Kinetics Collision Theory Molecules must
collide to react. Concentrationaffects rates because collisions are
more likely with more []. Must collidehard enough .
Temperatureandrateare related. Only a small number of collisions
produce reactions ( orientation ,force ). 79. Terms Activation
energy- the minimum energy needed to make a reaction happen.
Activated Complex or Transition State- The arrangement of atoms at
the top of the energy barrier. Write rate laws forCH 3 NC (g) CH 3
CN (g) O 3(g)+ O (g) 2O 2(g) Try #53 p. 571 ( similar to next
slide): 80. Potential Energy Reaction Coordinate Reactants Products
81. Potential Energy Reaction Coordinate Reactants Products
Activation Energy E a 82. Potential Energy Reaction Coordinate
Reactants Products Activated complex 83. Potential Energy Reaction
Coordinate Reactants Products E } 84. Potential Energy Reaction
Coordinate 2BrNO 2NO + Br Br---NO Br---NO 2 Transition State 85.
Arrhenius Said thatreaction rateshouldincreasewithtemperature .
Athigh temperaturemore molecules have the energy required to get
over the barrier(i.e . the activation energy). So ahigher E ameans
slower reactionat a given temperature. The number of collisions
with the necessary energy increasesexponentiallywith higher
temperature. 86. Arrhenius Number of collisions with the
requiredenergy = ze -E a /RT z = total collisions e is Eulers
number (opposite of ln) E a= activation energy R = ideal gas
constant = 8.314J /K mol T is temperature in Kelvin 87. Problems
Observedrate islessthan the number of collisions that have the
minimum energy. This is due to Molecularorientation Written into
equation asp(thesteric factor ). k = z p e -E a /RT 88. Figure
12.13 Several Possible Orientations for a Collision Between Two
BrNO Molecules. (a) & (b) can lead to a collision, (c) cannot.
89. No Reaction O N Br O N Br O N Br O N Br O N Br O N Br O N Br O
N Br O N Br O N Br 90. Arrhenius Equation k= zpe -E a /RT= Ae -E a
/RT A is called the frequency factor = zp lnk= -(E a /R)(1/T) + ln
A Another line !!!! (y = mx + b) That is,lnkvs1/Tis a straight line
Remember, T is temperature in Kelvin! Another form:ln(k 2 /k 1 ) =
(E a /R)(1/T 1- 1/T 2 )You will do this in Lab 12 91. Figure
12.14pp Plot of In( k )vs.1/ Tfor 2N 2 O 5( g ) g )+ O 2( g ). Can
get E afrom slope = -E a /R Do in lab 12. 92. Figure 12.14 Plot of
In( k )vs.1/ Tfor 2N 2 O 5( g ) g )+ O 2( g ). Can get E afrom
slope = -E a /R Do in lab 12. 93. Activation Energy and Rates The
final saga 94. Mechanismsand rates There is an activation energy
foreach elementarystep. Activation energy determinesk . k= Ae - (E
a /RT)kdeterminesrate So,sloweststep(rate determining) must have
thehighestactivation energy. 95. This reaction takes place in three
steps 96. First step is fast Low activation energy E a 97. Second
step is slow High activation energy E a 98. E a Third step is fast
Low activation energy 99. Second step is rate determining 100.
Intermediates are present 101. Activated Complexes or Transition
States 102. 12.8 Catalysts Speed up a reactionwithout being used
upin the reaction. Enzymes are biological catalysts. Homogenous
Catalystsare in thesame phaseas the reactants. Heterogeneous
Catalystsare in adifferentphase as the reactants. 103. How
Catalysts Work Catalysts allow reactions to proceed by adifferent
mechanism- a new pathway. Newpathway has aloweractivation energy
(so, has a faster rate). Moremolecules will have this activation
energy. Doesnotchange overall E 104. Figure 12.15 Energy Plots for
a Catalyzed and an Uncatalyzed Pathway for a Given Reaction. Eis
the same in both cases. 105. Figure 12.16 Effect of a Catalyst on
theNumberof Reaction-ProducingCollisions (increases even though no
temperature increase). 106. Hydrogen bonds to surface of metal.
Break H-H bonds Heterogenous Catalysts Pt surface H H H H H H H H
107. Heterogenous Catalysts Pt surface H H H H C H H C H H 108.
Heterogenous Catalysts The double bond breaks and bonds to the
catalyst. Pt surface H H H H C H H C H H 109. Heterogenous
Catalysts The hydrogen atoms bond with the carbon Pt surface H H H
H C H H C H H 110. Heterogenous Catalysts Pt surface H C H H C H H
H H H 111. Heterogenous Catalysts Summary:usually in 4 steps:
Adsorption & activationof reactants onto the catalytic surface.
Migrationof the adsorbedreactantsacross the surface. Reactionof the
adsorbed reactants. Escape(desorption) of theproducts . 112.
Homogenous Catalysts Chlorofluorocarbonscatalyze the decomposition
of ozone. Both are in same phase (gas) Another
example:Enzymesregulating the body processes. (Protein catalysts)
113. Catalysts and rate Catalysts will speed up a reactionbut only
to a certain point . Past a certain point adding more reactants
wont change the rate. BecomesZero Order(saturation kinetics) See
saturation kinetics with renal disease & some drugs
(aminoglycosides). 114. Catalysts and rate. Rate increases until
the active sites of catalyst are filled. Then rate isindependentof
concentration Concentration of reactants Rate
