General Chemistry

1. 1
Kinetics pp
The study ofThe study of reaction ratesreaction rates..
SpontaneousSpontaneous reactions are reactionsreactions are reactions
that will happen -that will happen - butbut we can’t tell howwe can’t tell how
fast.fast.
Diamond will spontaneously turn toDiamond will spontaneously turn to
graphite – eventually.graphite – eventually.
Reaction mechanismReaction mechanism - the- the stepssteps byby
which a reaction takes place.

2. 2
12.1 Reaction Rate
Rate = Conc. of A at t2 - Conc. of A at t1
t2- t1
Rate =∆[A] t
Change in concentration per unit time
For this reaction N2 + 3H2 2NH3

3. 3
N2 + 3H2 2NH3 As the reaction
progresses the concentration H2 goes down
C
o
n
c
e
n
t
r
a
t
i
o
n
Time
[H[H22]]

4. 4
As the reaction progresses the concentration N2 goes
down 1/3 as fast as for H2
3 moles H2 being consumed for every 1 mole of N2
C
o
n
c
e
n
t
r
a
t
i
o
n
Time
[H[H22]]
[N[N22]]

5. 5
As the reaction progresses the concentration
NH3 goes up “mole proportionally”
C
o
n
c
e
n
t
r
a
t
i
o
n
Time
[H[H22]]
[N[N22]]
[NH[NH33]]

6. 6
Calculating Rates
Average rates are taken over long
intervals
Instantaneous rates are 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. 7
Average slope method
C
o
n
c
e
n
t
r
a
t
i
o
n
Time
[H[H22]]
tt

8. 8
Instantaneous slope method.
C
o
n
c
e
n
t
r
a
t
i
o
n
Time
∆∆[H[H22]]
 tt

9. 9
Defining Rate pp
We can define rate in terms of theWe can define rate in terms of the
disappearancedisappearance of theof the reactantreactant oror in terms ofin terms of
the rate ofthe rate of appearanceappearance of theof the productproduct..
In our exampleIn our example
NN22 ++ 3H3H22 2NH2NH33
--∆∆[[NN22]] == --33∆∆[[HH22]] == 22∆∆[[NHNH33]]
∆∆tt ∆∆tt ∆∆tt
Since [Since [reactantsreactants]] alwaysalways decreasesdecreases
with time, any rate expression includeswith time, any rate expression includes
(-) sign(-) sign

10. 10
12.2 Rate Laws: An Introduction
Reactions areReactions are reversiblereversible..
As products accumulate they can beginAs products accumulate they can begin
to turn back into reactants.to turn back into reactants.
Early onEarly on the rate will depend onthe rate will depend on onlyonly thethe
amount ofamount of reactantsreactants present.present.
So, we want to measure the reactantsSo, we want to measure the reactants
as soon as they are mixedas soon as they are mixed..
This is called theThis is called the Initial rate methodInitial rate method..

11. 11
Two key points:Two key points:
The concentration of theThe concentration of the productsproducts dodo
not appear in the rate law because thisnot appear in the rate law because this
is anis an initialinitial rate (only the [reactants]).rate (only the [reactants]).
TheThe orderorder must be determinedmust be determined
experimentallyexperimentally..
CannotCannot be obtained from the equation’sbe obtained from the equation’s
coefficients (remember this!).coefficients (remember this!).
Rate LawsRate Laws pp

12. 12
Rate Laws
NN22 + 3H+ 3H22 ⇔⇔ 2NH2NH33
When forward & reverse reaction ratesWhen forward & reverse reaction rates
areare equalequal then there is no change inthen there is no change in
concentration of either.concentration of either.
The reaction is atThe reaction is at equilibriumequilibrium..
Rate =Rate = kk[N[N22]]mm
[H[H22]]nn
Note: “m” and “n” areNote: “m” and “n” are notnot coefficientscoefficients ofof
the balanced equation,the balanced equation, must bemust be
determined experimentallydetermined experimentally..

13. 13
You will find the rate depends only on theYou will find the rate depends only on the
concentration of theconcentration of the reactantsreactants..
Rate =Rate = kk[[NONO22]]nn
This is called aThis is called a rate law expressionrate law expression..
kk is called the rate constant.is called the rate constant.
nn is theis the orderorder of the reactant - usually aof the reactant - usually a
positive integer (1st, 2nd, etc.).positive integer (1st, 2nd, etc.).
It isIt is notnot the coefficient of the balancedthe coefficient of the balanced
reaction (e.g., it could be “1”)reaction (e.g., it could be “1”)
We canWe can insteadinstead define Rate in terms ofdefine Rate in terms of
production of Oproduction of O22 vs.vs. consumption of NOconsumption of NO22
2 NO2 NO22 22 NONO ++ OO22

14. 14
The rate ofThe rate of appearanceappearance ofof OO22 is . . .is . . .
RateRate'' == [[OO22]] == kk''[[NONO22]]nn
tt
Because there are 2 NOBecause there are 2 NO22 consumed forconsumed for
each Oeach O22 produced . . .produced . . .
Rate = k[NORate = k[NO22]]nn
==2 x Rate2 x Rate’’ = 2 x k= 2 x k''[NO[NO22]]nn
So k[NOSo k[NO22]]nn
= 2 x k= 2 x k''[NO[NO22]]nn
So k = 2 x k'So k = 2 x k'
2 NO2 NO22 22 NONO ++ OO22

15. 15
Figure 12.1 p. 562
Definition of Rate.
2NO2 → 2NO + O2

16. 16
Types of Rate Laws
Differential Rate law - describes how
rate depends on concentration.
Integrated Rate Law - Describes how
concentration depends on time.
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. 17
12.3 Determining Rate Laws
The first step is to determine the form of
the rate law (especially its order).
Must be determined from experimental
data (not from the coefficients!!).
For this reaction:
2 N2O5 (aq) 4NO2 (aq) + O2(g)
The reverse reaction won’t play a role.
Next slide shows experimental results
(like you’ll do in lab 12).

18. 18
[N[N22OO55] (mol/L)] (mol/L) Time (s)Time (s)
1.001.00 00
0.880.88 200200
0.780.78 400400
0.690.69 600600
0.610.61 800800
0.540.54 10001000
0.480.48 12001200
0.430.43 14001400
0.380.38 16001600
0.340.34 18001800
0.300.30 20002000
Now graph the data

19. 19
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
9
n"b
¢"b
×"b
b
A
u
ª
ß
b
H
To find rate we have to find the slope at
two points
We will use the tangent method.

20. 20
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
9
n"b
¢"b
×"b
b
A
u
ª
ß
b
H
At .90 M the rate is -(.98 - .76) = -0.22 = 5.5x 10 -4
M•s-1
(0-400) -400

21. 21
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
9
n"b
¢"b
×"b
b
A
u
ª
ß
b
H
At .45 M the rate is -(.52 - .31) = -0.22 = 2.7 x 10 -4
(1000-1800) -800

22. 22
Since the rate atSince the rate at twice the concentrationtwice the concentration
isis twice as fasttwice as fast, the rate law must be.., the rate law must be..
Rate = -Rate = -[N[N22OO55]] = k[N= k[N22OO55]]11
= k[N= k[N22OO55]]
tt
We say this reaction isWe say this reaction is firstfirst orderorder inin NN22OO55
TheThe onlyonly way to determine order is to runway to determine order is to run
the experiment.the experiment.
““n” isn” is notnot same number as coefficientsame number as coefficient
(only a coincidence).(only a coincidence).
1st order means doubling the [ ] doubles1st order means doubling the [ ] doubles
the ratethe rate (know this). Rate = k[A](know this). Rate = k[A]11

23. 23
A Plot of [NA Plot of [N22OO55] as f(t) for 2N] as f(t) for 2N22OO55 →→4NO4NO22 + O+ O22 pp
• Note that the reactionNote that the reaction
rate at 0.90rate at 0.90 MM is twiceis twice
that at 0.45that at 0.45 MM
• So, the reaction is 1stSo, the reaction is 1st
order becauseorder because
doubling the [ ]doubling the [ ]
doubles the ratedoubles the rate

24. 24
The method of Initial Rates pp
This method requires that a reaction beThis method requires that a reaction be
run several times (do in Lab 12).run several times (do in Lab 12).
The initial concentrations of theThe initial concentrations of the
reactants arereactants are variedvaried..
The reaction rate is measuredThe reaction rate is measured just afterjust after
the reactants are mixed.the reactants are mixed.
Eliminates the effect of the reverseEliminates the effect of the reverse
reaction and makes it easier to calculate.reaction and makes it easier to calculate.

25. 25
Here’s an AP-type QuestionHere’s an AP-type Question pp
Write general form of rate law for:Write general form of rate law for:
BrOBrO33
--
+ 5 Br+ 5 Br--
+ 6H+ 6H++
3Br3Br22 + 3 H+ 3 H22OO
Answer is . . .Answer is . . .
Rate =Rate = kk[BrO[BrO33
--
]]nn
[Br[Br--
]]mm
[H[H++
]]pp
DetermineDetermine nn,, mm,, pp by comparing ratesby comparing rates
AddAdd nn,, mm,, pp values to getvalues to get overall orderoverall order
Find value ofFind value of kk by plugging in valuesby plugging in values
from any one of the experiment rates

26. 26
An AP Test ExampleAn AP Test Example pp
For the reactionFor the reaction
BrOBrO33
--
+ 5 Br+ 5 Br--
+ 6H+ 6H++
3Br3Br22 + 3 H+ 3 H22OO
The general form of the Rate Law isThe general form of the Rate Law is
Rate = k[BrORate = k[BrO33
--
]]nn
[Br[Br--
]]mm
[H[H++
]]pp
Be able to write the above as the 1stBe able to write the above as the 1st
question, given the above reaction.question, given the above reaction.
We useWe use experimental dataexperimental data to determineto determine
the values of n,m, and pthe values of n,m, and p

27. 27
Initial concentrations (M) pp
Rate (M/s)
BrOBrO33
--
BrBr--
HH++
0.100.10 0.100.10 0.100.10 8.0 x 108.0 x 10-4-4
0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10-3-3
0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10-3-3
0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10-3-3Now we have to see how the rate changes
with concentration

28. 28
Initial concentrations (M) pp
Rate (M/s)
BrOBrO33
--
BrBr--
HH++
0.100.10 0.100.10 0.100.10 8.0 x 108.0 x 10-4-4
0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10-3-3
0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10-3-3
0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10-3-3Find n by ratio of Rate 2/Rate 1, in which
only [BrO3
-
] changes
n = 1 because (next slide)

29. 29
The math pp
Rate 2Rate 2 == 11.6 x 10.6 x 10-3-3
mol/Lsmol/Ls == kk(0.20 mol/L)(0.20 mol/L)nn
(0.10 mol/l)(0.10 mol/l)mm
(0.10 mol/L)(0.10 mol/L)pp
Rate 1Rate 1 8.0 x 108.0 x 10-4-4
mol/Lsmol/Ls kk(0.10 mol/L)(0.10 mol/L)nn
(0.10(0.10
mol/l)mol/l)mm
(0.10 mol/L)(0.10 mol/L)pp
== 2.02.0 == (0.20 mol/L ÷ 0.10 mol/L)(0.20 mol/L ÷ 0.10 mol/L)nn
= (2.0)= (2.0)nn
So,So, nn = 1= 1

30. 30
Initial concentrations (M) pp
Rate (M/s)
BrOBrO33
--
BrBr--
HH++
0.100.10 0.100.10 0.100.10 8.0 x 108.0 x 10-4-4
0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10-3-3
0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10-3-3
0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10-3-3
FindFind mm by ratio of Rate 3/Rate 2, in whichby ratio of Rate 3/Rate 2, in which
only [Bronly [Br--
] changes] changes
mm = 1= 1 (ditto on math)(ditto on math)

31. 31
Initial concentrations (M) pp
Rate (M/s)
BrOBrO33
--
BrBr--
HH++
0.100.10 0.100.10 0.100.10 8.0 x 108.0 x 10-4-4
0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10-3-3
0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10-3-3
0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10-3-3
FindFind pp by ratio of Rate 4/Rate 1, in whichby ratio of Rate 4/Rate 1, in which
only [Honly [H++
] changes] changes
pp = 2= 2 (ditto on math)(ditto on math)

32. 32
AP-type question pp
So, rate of reaction isSo, rate of reaction is firstfirst order inorder in BrOBrO33
--
andand BrBr--
andand secondsecond order inorder in HH++
Overall rateOverall rate = 1 + 1 + 2 = 4= 1 + 1 + 2 = 4
Rate lawRate law can be written as:can be written as:
Rate =Rate = kk[BrO[BrO33
--
][Br][Br--
][H][H++
]]22

33. 33
AP-type question pp
Rate = k[BrO3
-
][Br-
][H+
]2
Try calculating k from Expt. 2 BrO3
1-
Br1-
H1+
Rate (M/s) 0.20
0.10 0.10 1.6 x 10-3
The answer is . . .
k = 8.0 L3
/mol3
•s (or 8.0 M-3
s-1
)
The complete rate law is . . .
Rate = 8.0 L3
/mol3
•s [BrO3
-
][Br-
][H+
]2
Be sure to express the correct units!

34. 34
AP-type question
Units areUnits are veryvery important!!!important!!!
Let’s try Text # 27, p. 568 . . .Let’s try Text # 27, p. 568 . . .
Answer is . . .Answer is . . .
2525 a. Rate = K[NOCl]a. Rate = K[NOCl]22
b. k= 6.6 x 10b. k= 6.6 x 10-29-29
cmcm33
/molecules•s/molecules•s
c. k = 4.0 x 10c. k = 4.0 x 10-8-8
MM-1-1
ss-1-1
Do # 25 at home. (Quiz soon!)Do # 25 at home. (Quiz soon!)

35. 35
12.4 Integrated Rate Law
Expresses the reaction concentrationExpresses the reaction concentration
as a function ofas a function of timetime (vs. rate as(vs. rate as
f(conc.)).f(conc.)).
Form of the equation depends on theForm of the equation depends on the
orderorder of the rate law (from differential).of the rate law (from differential).
Changes Rate = -Changes Rate = -∆∆[A][A]nn
== kk[A][A]nn
∆∆tt
We will only work with n = 0, 1, and 2We will only work with n = 0, 1, and 2
for reactions with only afor reactions with only a single reactantsingle reactant
(decomposition)(decomposition)

36. 36
First Order
For the reaction 2NFor the reaction 2N22OO55 4NO4NO22 + O+ O22
We found the (differential) Rate =We found the (differential) Rate = kk[[NN22OO55]]11
I.e.I.e., if concentration doubles rate doubles., if concentration doubles rate doubles.
If weIf we integrateintegrate this equation with respect tothis equation with respect to
timetime we get thewe get the IntegratedIntegrated Rate LawRate Law
ln[Nln[N22OO55] = -] = - kkt + ln[Nt + ln[N22OO55]]00 (y = mx + b)(y = mx + b)
ln is the natural logln is the natural log
[N[N22OO55]]00 is theis the initialinitial concentration.concentration.

37. 37
General form is Rate = -General form is Rate = -∆∆[A] /[A] / ∆∆t = k[A]t = k[A]
ln[A] = -ln[A] = - kkt + ln[A]t + ln[A]00
In the form y = mx + bIn the form y = mx + b
y = ln[A]y = ln[A] m = -m = -kk
x = tx = t b = ln[A]b = ln[A]00
A graph ofA graph of ln[A]ln[A] vs.vs. timetime is a straightis a straight
lineline ifif it is first order (otherwise, ait is first order (otherwise, a
differentdifferent graph is a straight line, seegraph is a straight line, see
following slides).following slides).
First Order

38. 38
By getting a straight line you prove it isBy getting a straight line you prove it is
first order.first order.
Conversely, if a plot of ln[A]Conversely, if a plot of ln[A] vsvs. t (as. t (as
opposed to plotting other values) isopposed to plotting other values) is notnot
a straight line, the reaction isa straight line, the reaction is notnot firstfirst
order in [A].order in [A].
So, you have to plot several possibilitiesSo, you have to plot several possibilities
to determine which order it is.to determine which order it is.
First Order

39. 39
A Plot of [NA Plot of [N22OO55] as f(t) for 2N] as f(t) for 2N22OO55 →→4NO4NO22 + O+ O22..
• Decomposition reactionDecomposition reaction
• Note that the reactionNote that the reaction
rate at 0.90rate at 0.90 MM is twiceis twice
that at 0.45that at 0.45 MM
• So, the reaction is 1stSo, the reaction is 1st
order because doublingorder because doubling
the [ ] doubles the ratethe [ ] doubles the rate
• This is not a plot ofThis is not a plot of
ln [Nln [N22OO55], which would], which would
be a straight line if itbe a straight line if it
was first order . . .was first order . . .

40. 40
lnln[A] = -[A] = - kkt + ln[A]t + ln[A]00
By getting the straight line (plotBy getting the straight line (plot lnln[A][A] vs.vs. t)t)
you can prove it is first order.you can prove it is first order.
Often expressed in a ratioOften expressed in a ratio
First Order
[ ]
[ ]
ln
A
A
= kt
0






41. 41
Figure 12.4
A Plot of In(N2O5) Versus Time (1st order)

42. 42
Half LifeHalf Life
The time required to reachThe time required to reach halfhalf thethe
original concentration.original concentration.
If a bacterium doubles every minute,If a bacterium doubles every minute,
and takes 40 minutes to fill the jar, howand takes 40 minutes to fill the jar, how
long to fill half the jar?long to fill half the jar?
39 minutes.39 minutes.
IfIf the reaction isthe reaction is firstfirst order, then:order, then:
[A] = [A][A] = [A]00/2/2 whenwhen t = tt = t1/21/2

43. 43
Figure 12.5: A Plot of (NFigure 12.5: A Plot of (N22OO55) vs. Time for the) vs. Time for the DecompositionDecomposition
Reaction of NReaction of N22OO55

44. 44
Half Life
• The time required to reach half theThe time required to reach half the
original concentration.original concentration.
• IfIf the reaction is first orderthe reaction is first order
• [A] = [A][A] = [A]00/2 when t = t/2 when t = t1/21/2
[ ]
[ ]
ln
A
A
= kt0
0
1 2
2








ln(2) =ln(2) = kktt1/21/2
Solving for tSolving for t1/21/2::

45. 45
Half Life
tt1/21/2 = ln(2)/k = 0.693/= ln(2)/k = 0.693/kk
So, the time to reach half theSo, the time to reach half the
original concentration doesoriginal concentration does notnot
depend on the startingdepend on the starting
concentration.concentration.
This is an easy way to findThis is an easy way to find kk if youif you
know or can determine the tknow or can determine the t1/21/2 andand
the reaction is 1st order.the reaction is 1st order.

46. 46
Second Order
Rate = -Rate = -∆∆[A] /[A] / ∆∆t = k[A]t = k[A]22
integratedintegrated rate law has the form:rate law has the form:
1/[A] =1/[A] = kkt + 1/[A]t + 1/[A]00, where y = mx + b, so:, where y = mx + b, so:
y= 1/[A]y= 1/[A] m = km = k
x= tx= t b = 1/[A]b = 1/[A]00
Get a straight line ifGet a straight line if 1/[A]1/[A] vsvs tt is graphedis graphed
(instead of ln[A] = -(instead of ln[A] = - kkt + ln[A]t + ln[A]00))
KnowingKnowing kk and [A]and [A]00 you can calculate [A]you can calculate [A]
at any timeat any time tt by graphing y = mx + bby graphing y = mx + b

47. 47
Second Order Half Life
[A] = [A][A] = [A]00 /2 at t = t/2 at t = t1/21/2
1
2
0
2[ ]A
= kt +
1
[A]1
0
2
2
[ [A]
-
1
A]
= kt
0 0
1
t
k[A]1 =
1
0
2
1
[A]
= kt
0
1 2

48. 48
Z7e 545 Fig 12.6
(a) A Plot of In(C(a) A Plot of In(C44HH66) Versus) Versus tt (not straight line, so not 1st order)(not straight line, so not 1st order)
(b) A Plot of 1/(C(b) A Plot of 1/(C44HH66) Versus) Versus tt (straight line so must be 2nd order)(straight line so must be 2nd order)

49. 49
Determining Rate Laws
Since we get a straight lineSince we get a straight line
with (b) and not with (a) inwith (b) and not with (a) in
previous slide:previous slide:
The reaction must be secondThe reaction must be second
order in Corder in C44HH66

50. 50
Zero Order Rate Law
Rate = k[A]Rate = k[A]00
= k= k
Rate doesRate does notnot change with concentrationchange with concentration
Important for drug pharmacokineticsImportant for drug pharmacokinetics
IntegratedIntegrated rate law: [A] = -kt + [A]rate law: [A] = -kt + [A]00
y = mx + b formaty = mx + b format
When [A] = [A]When [A] = [A]00 /2 t = t/2 t = t1/21/2
tt1/2 =1/2 = [A][A]00 /2k/2k for zero order.for zero order.

51. 51
Z7e 546Z7e 546
Figure 12.7Figure 12.7
A Plot ofA Plot of
[A] Versus[A] Versus tt
for afor a Zero-Zero-
OrderOrder
ReactionReaction

52. 52
Occurs most often when reactionOccurs most often when reaction
happens on ahappens on a surfacesurface because thebecause the
surface area stays constant.surface area stays constant.
Also applies toAlso applies to enzyme chemistryenzyme chemistry andand
other reactions involvingother reactions involving catalystscatalysts..
Zero Order Rate Law

53. 53
Fig. 12.8 Decomposition Reaction NFig. 12.8 Decomposition Reaction N22OO →→ 2N2N22 + O+ O22
Even though [NEven though [N22O] isO] is twice as great in (b) as in (a)twice as great in (b) as in (a) thethe
reaction occurs on areaction occurs on a saturatedsaturated platinum surface, so rateplatinum surface, so rate
isis zero orderzero order

54. 54
Time
C
o
n
c
e
n
t
r
a
t
i
o
n
Zero Order Graph

55. 55
Time
C
o
n
c
e
n
t
r
a
t
i
o
n
∆[A]/∆t =
slope
∆t
-k =
∆[A]
Zero Order Graph
tt1/21/2 = [A= [A00]/2k]/2k

56. 56
More Complicated Reactions
Reactions withReactions with more than one reactantmore than one reactant..
BrOBrO33
--
+ 5 Br+ 5 Br--
+ 6H+ 6H++
3Br3Br22 + 3 H+ 3 H22OO
For this reaction we experimentallyFor this reaction we experimentally
found the (differential) rate law to befound the (differential) rate law to be
Rate = k[BrORate = k[BrO33
--
][Br][Br--
][H][H++
]]22
ToTo investigateinvestigate this reaction rate wethis reaction rate we
need to control the conditions.need to control the conditions.

57. 57
Rate = k[BrORate = k[BrO33
--
][Br][Br--
][H][H++
]]22
We set up the experiment soWe set up the experiment so twotwo of theof the
reactants are inreactants are in large excesslarge excess (Lab 12).(Lab 12).
[BrO[BrO33
--
]]00= 1.0 x 10= 1.0 x 10-3-3
MM
[Br[Br--
]]00 = 1.0= 1.0 MM (1000 x as great)(1000 x as great)
[H[H++
]]00 = 1.0= 1.0 MM (ditto)(ditto)
As the reaction proceeds [BrOAs the reaction proceeds [BrO33
--
] changes] changes
noticeably (noticeably (limiting reactantlimiting reactant).).
[Br[Br--
] and [H] and [H++
]] do notdo not (since in large excess)(since in large excess)

58. 58
This rate law can be rewritten . . .
Rate = k[BrRate = k[Br--
]]00[H[H++
]]00
22
[BrO[BrO33
--
] =] = kk’’[BrO[BrO33
--
] because,] because,
[Br[Br--
]]00 and [Hand [H++
]]00 are constant (since in largeare constant (since in large
excess). Solving forexcess). Solving for kk’’ . . .. . .
kk’’ = k[Br= k[Br--
]]00[H[H++
]]00
22
and substitutingand substituting kk’’ into theinto the
original rate law . . .original rate law . . .
Rate =Rate = kk’’[BrO[BrO33
--
]]
This is called aThis is called a pseudo first orderpseudo first order rate law.rate law.
kk ==
kk’’
Rate = k[BrORate = k[BrO33
--
][Br][Br--
][H][H++
]]22

59. 59
12.5 Summary of Rate Laws
Assume only theAssume only the forwardforward reaction isreaction is
important so can produce rate laws thatimportant so can produce rate laws that
only containonly contain reactantreactant concentrationsconcentrations
DifferentialDifferential rate lawrate law ((akaaka “rate law”) shows“rate law”) shows
howhow raterate depends ondepends on concentration.concentration.
IntegratedIntegrated rate lawrate law shows howshows how
concentrationconcentration depends ondepends on timetime..
Choice of using either method depends onChoice of using either method depends on
data being experimentally derived.data being experimentally derived.

60. 60
Summary of Rate Laws
Most common method to determine theMost common method to determine the
differentialdifferential rate law is method ofrate law is method of initialinitial
ratesrates (get(get tt as close to zero as possible)as close to zero as possible)
We will use this method in Lab 12We will use this method in Lab 12
If instead using theIf instead using the integratedintegrated rate law,rate law,
measuremeasure concentrationconcentration at various times.at various times.
You’llYou’ll graphgraph this in Lab 12 and plot thethis in Lab 12 and plot the
values to get a straight line as follows:values to get a straight line as follows:

61. 61
Summary of Rate Laws pp
Memorize the followingMemorize the following rate lawsrate laws (for AP(for AP
questions) and use the tquestions) and use the t1/21/2 for our lab 12.for our lab 12.
(rf. Table 12.6 p. 548) - use for online HW!(rf. Table 12.6 p. 548) - use for online HW!
ReactionReaction orderorder is found by whichever of theis found by whichever of the
following gives straight line:following gives straight line:
ZeroZero order whenorder when [A][A] vsvs. t. t yields straight lineyields straight line
Rate = k and tRate = k and t1/21/2 = [A]= [A]oo/2/2kk
1st1st order whenorder when lnln[A][A] vs.vs. tt yields straight lineyields straight line
Rate = k[A] and tRate = k[A] and t1/21/2 = 0.693/= 0.693/kk
2nd2nd order whenorder when 1/1/[A][A] vs.vs. tt yields straight lineyields straight line
Rate = k[A]Rate = k[A]22
and tand t1/21/2 = 1= 1/k[/k[A]A]oo

62. 62
Use for Online HW pp
Use for online HW
Hint: on one problem you will have to first determine the
order, then use the half-life equation (5th
line) to solve for
k, then use the Integrated rate law (2nd
line) to solve for
time (t).

63. 63
12.6 Reaction Mechanisms
The series of steps that actually occurThe series of steps that actually occur
in a chemical reaction.in a chemical reaction.
A balanced equation does not tell usA balanced equation does not tell us
howhow the reactants become products.the reactants become products.
Kinetics can tell us something about theKinetics can tell us something about the
mechanism.mechanism.

64. 64
Figure 12.9 A Molecular Representation of the Elementary
Steps in the Reaction of NO2 and CO
Overall: NO2 + CO → NO + CO2
Step 1: NO2 + NO2 → NO3 + NO (k1)
Step 2: NO3 + CO → NO2 + CO2 (k2)
NO3 is an intermediate

65. 65
2NO2 + F2 2NO2F
Rate = k[NO2][F2]
The proposed mechanism is:
NO2 + F2 NO2F + F (slow)
F + NO2 NO2F (fast)
F is called an intermediate. It is
formed then consumed in the reaction
Reaction Mechanisms

66. 66
Each of the two reactions is called anEach of the two reactions is called an
elementary stepelementary step..
The rate for each reaction can beThe rate for each reaction can be
written from itswritten from its molecularitymolecularity..
Molecularity is the number of piecesMolecularity is the number of pieces
that must come together (collide) tothat must come together (collide) to
produce the reaction indicated by thatproduce the reaction indicated by that
step.step.
Reaction Mechanisms

67. 67
UnimolecularUnimolecular step involves one molecule -step involves one molecule -
Rate isRate is firstfirst order.order.
BimolecularBimolecular step - requires two molecules -step - requires two molecules -
Rate isRate is secondsecond orderorder
TermolecularTermolecular step- requires three molecules -step- requires three molecules -
Rate isRate is thirdthird orderorder
TermolecularTermolecular steps are almost never heard ofsteps are almost never heard of
because the chances of three moleculesbecause the chances of three molecules
coming into contact at the same time arecoming into contact at the same time are
miniscule.miniscule.
With molecularity the coefficientsWith molecularity the coefficients cancan becomebecome
the exponents.the exponents.

68. 68
AA productsproducts
A+A productsA+A products
2A2A productsproducts
A+B productsA+B products
A+A+B ProductsA+A+B Products
2A+B Products2A+B Products
A+B+C ProductsA+B+C Products
This is also in TableThis is also in Table
12.7, p. 550. Know.12.7, p. 550. Know.
Rate = k[A]Rate = k[A]
Rate= k[A]Rate= k[A]22
Rate= k[A]Rate= k[A]22
Rate= k[A][B]Rate= k[A][B]
Rate= k[A]Rate= k[A]22
[B][B]
Rate= k[A]Rate= k[A]22
[B][B]
Rate= k[A][B][C]Rate= k[A][B][C]

69. 69
Finding the Reaction MechanismFinding the Reaction Mechanism pp
Must satisfyMust satisfy twotwo requirementsrequirements::
SumSum of theof the elementaryelementary stepssteps doesdoes givegive
thethe overalloverall balanced equation for thebalanced equation for the
reactionreaction
The proposedThe proposed mechanismmechanism mustmust agreeagree
with thewith the experimentally derivedexperimentally derived rate law.rate law.

70. 70
How to get rid of intermediatesHow to get rid of intermediates pp
Use the reactions that form them.Use the reactions that form them.
IfIf the reactions arethe reactions are fastfast andand irreversibleirreversible
the concentration of thethe concentration of the intermediateintermediate isis
based onbased on stoichiometrystoichiometry. (Use spider).. (Use spider).
IfIf it is formed by ait is formed by a reversiblereversible reactionreaction
setset thethe reversiblereversible reaction ratesreaction rates equalequal
to each other.to each other.

71. 71
Formed inFormed in reversiblereversible reactionsreactions pp
2 NO + O2 NO + O22 2 NO2 NO22 Overall reactionOverall reaction
GivenGiven: Expt-derived Rate =: Expt-derived Rate = kk[NO][NO]22
[O[O22]]
Proposed MechanismProposed Mechanism::
2 NO2 NO NN22OO22 (fast)(fast) NN22OO22 ++
OO22 2 NO2 NO22 (slow)(slow)
2nd step2nd step is rate determining (slow), so needis rate determining (slow), so need
to get this one to look like the expt.-derivedto get this one to look like the expt.-derived
rate,rate, kk[NO][NO]22
[O[O22]]
1st step1st step isis reversiblereversible so set both rates inso set both rates in thisthis
stepstep equalequal to each other:to each other:
kk11[NO][NO]22
= k= k-1-1[N[N22OO22]]

72. 72
Formed inFormed in reversiblereversible reactionsreactions pp
2 NO + O2 NO + O22 2 NO2 NO22 overall reactionoverall reaction
Given: Expt-derived Rate =Given: Expt-derived Rate = kk[NO][NO]22
[O[O22]]
Proposed MechanismProposed Mechanism
2 NO2 NO NN22OO22 (fast) N(fast) N22OO22
+ O+ O22 2 NO2 NO22 (slow)(slow)
2nd step2nd step rate comes fromrate comes from molecularitymolecularity
A+B productsA+B products Rate= k[A][B]Rate= k[A][B]
Rate = kRate = k22[N[N22OO22][O][O22]]

73. 73
Formed inFormed in reversiblereversible reactionsreactions pp
Given: Expt-derived Rate =Given: Expt-derived Rate = kk[NO][NO]22
[O[O22]]
2 NO + O2 NO + O22 2 NO2 NO22 overall reactionoverall reaction
Proposed MechanismProposed Mechanism
2 NO2 NO NN22OO22 (fast)(fast)
NN22OO22 + O+ O22 2 NO2 NO22 (slow)(slow)
Rate = kRate = k11[NO][NO]22
= k= k-1-1[N[N22OO22]] (from 1st step)(from 1st step)
Rate = kRate = k22[N[N22OO22][O][O22]] (from 2nd step)(from 2nd step)
Solve for [NSolve for [N22OO22] from (1) & sub] from (1) & sub into (2)into (2)
rate =rate = kk22 (k(k11/ k/ k-1-1)[NO])[NO]22
[O[O22]=]=k[NO]k[NO]22
[O[O22]]

74. 74
Formed in reversible reactions pp
2 NO + O2 NO + O22 2 NO2 NO22 overall reactionoverall reaction
Proposed MechanismProposed Mechanism
2 NO2 NO NN22OO22 (fast)(fast)
NN22OO22 + O+ O22 2 NO2 NO22 (slow)(slow)
Since model-derivedSince model-derived Rate =Rate = k[NO]k[NO]22
[O[O22]]
agrees with theagrees with the expt.-derivedexpt.-derived rate . . .rate . . .
And sum of the steps = overall reaction:And sum of the steps = overall reaction:
The 2-step proposed mechanismThe 2-step proposed mechanism
works!works!

75. 75
Formed inFormed in reversiblereversible reactionsreactions pp
In other words:In other words:
Since this last equation, which we derivedSince this last equation, which we derived
by eliminating the intermediateby eliminating the intermediate is theis the
samesame as theas the experimentallyexperimentally derived rate:derived rate:
AndAnd, the sum of the 2 steps add up to the, the sum of the 2 steps add up to the
overall balanced reaction:overall balanced reaction:
The proposed 2-step mechanism isThe proposed 2-step mechanism is
correct.correct.
There are AP questions that have you doThere are AP questions that have you do
this for.this for.

76. 76
Formed in fast, irreversible reactions pp
2 IBr2 IBr II22+ Br+ Br22 overall reactionoverall reaction
Experimentally-derivedExperimentally-derived Rate =Rate = kk[IBr][IBr]22
Proposed MechanismProposed Mechanism
IBrIBr I + BrI + Br (fast)(fast)
IBr + Br I + BrIBr + Br I + Br22 ((slowslow))
I + II + I II22 (fast)(fast)
Step 2 is slow, so it’s rate determiningStep 2 is slow, so it’s rate determining
Slow Rate = k[Slow Rate = k[IIBr][Br]Br][Br] butbut [Br]= [[Br]= [IIBr]Br]
So, Rate = k[So, Rate = k[IIBr][Br][IIBr] = k[Br] = k[IIBr]Br]22

77. 77
Formed in fast, irreversible reactionsFormed in fast, irreversible reactions pp
2 IBr2 IBr II22+ Br+ Br22 overall reactionoverall reaction
Proposed MechanismProposed Mechanism
IBrIBr I + BrI + Br (fast)(fast)
IBr + Br I + BrIBr + Br I + Br22 (slow)(slow) II
+ I+ I II22 (fast)(fast)
Sum of steps = overall balanced reactionSum of steps = overall balanced reaction
RateRate = k[IBr][IBr]= k[IBr][IBr] = k[IBr]= k[IBr]22
isis samesame asas
experimentally derivedexperimentally derived
So, the mechanism worksSo, the mechanism works
(But this doesn’t prove it’s the(But this doesn’t prove it’s the correctcorrect mechanism)mechanism)

78. 78
12.7 A Model for Chemical Kinetics12.7 A Model for Chemical Kinetics
Collision TheoryCollision Theory
Molecules must collide to react.Molecules must collide to react.
ConcentrationConcentration affects rates becauseaffects rates because
collisions are more likely with more [ ].collisions are more likely with more [ ].
Must collideMust collide hard enoughhard enough..
TemperatureTemperature andand raterate are related.are related.
Only a small number of collisionsOnly a small number of collisions
produce reactions (produce reactions (orientationorientation,, forceforce).).

79. 79
TermsTerms
Activation energyActivation energy - the minimum energy- the minimum energy
needed to make a reaction happen.needed to make a reaction happen.
Activated Complex or Transition StateActivated Complex or Transition State --
The arrangement of atoms at the top ofThe arrangement of atoms at the top of
the energy barrier.the energy barrier.
Write rate laws forWrite rate laws for CHCH33NCNC(g)(g) →→ CHCH33CNCN(g)(g)
OO3(g)3(g) + O+ O(g)(g) →→ 2O2O2(g)2(g)
Try #53 p. 571 ( similar to next slide):Try #53 p. 571 ( similar to next slide):

80. 80
P
o
t
e
n
t
i
a
l
E
n
e
r
g
y
Reaction Coordinate
Reactants
Products

81. 81
P
o
t
e
n
t
i
a
l
E
n
e
r
g
y
Reaction Coordinate
Reactants
Products
Activation
Energy Ea

82. 82
P
o
t
e
n
t
i
a
l
E
n
e
r
g
y
Reaction Coordinate
Reactants
Products
Activated
complex

83. 83
P
o
t
e
n
t
i
a
l
E
n
e
r
g
y
Reaction Coordinate
Reactants
Products
∆E
}

84. 84
P
o
t
e
n
t
i
a
l
E
n
e
r
g
y
Reaction Coordinate
2BrNO
2NO + Br
Br---NO
Br---NO
2
Transition
State

85. 85
ArrheniusArrhenius
Said thatSaid that reaction ratereaction rate shouldshould increaseincrease withwith
temperaturetemperature..
AtAt high temperaturehigh temperature more molecules have themore molecules have the
energy required to get over the barrierenergy required to get over the barrier (i.e(i.e..
the activation energy).the activation energy).
So aSo a higher Ehigher Eaa means slower reactionmeans slower reaction at aat a
given temperature.given temperature.
The number of collisions with the necessaryThe number of collisions with the necessary
energy increasesenergy increases exponentiallyexponentially with higherwith higher
temperature.temperature.

86. 86
Arrhenius
Number of collisions with the requiredNumber of collisions with the required
energy = zeenergy = ze-E-Eaa/RT/RT
z = total collisionsz = total collisions
e is Euler’s number (opposite of ln)e is Euler’s number (opposite of ln)
EEaa = activation energy= activation energy
R = ideal gas constant = 8.314R = ideal gas constant = 8.314 JJ/K mol/K mol
T is temperature in KelvinT is temperature in Kelvin

87. 87
ProblemsProblems
ObservedObserved rate israte is lessless than the numberthan the number
of collisions that have the minimumof collisions that have the minimum
energy.energy.
This is due to MolecularThis is due to Molecular orientationorientation
Written into equation asWritten into equation as pp (the(the stericsteric
factorfactor).).
k = zk = zppee-E-Eaa/RT/RT

88. 88
Figure 12.13 Several Possible Orientations for aFigure 12.13 Several Possible Orientations for a
Collision Between Two BrNO Molecules.Collision Between Two BrNO Molecules.
(a) & (b) can lead to a collision, (c) cannot.(a) & (b) can lead to a collision, (c) cannot.

89. 89
O
N
Br
O
N
Br
O
N
Br
O
N
Br
O N Br ONBr ONBr
O NBr
O N BrONBr No
Reaction

90. 90
Arrhenius Equation
kk = zpe= zpe-E-Eaa/RT/RT = Ae= Ae-E-Eaa/RT/RT
A is called the frequency factor = zpA is called the frequency factor = zp
lnln kk = -(E= -(Eaa/R)(1/T) + ln A/R)(1/T) + ln A
Another line !!!! (y = mx + b)Another line !!!! (y = mx + b)
That is,That is, lnln kk vsvs 1/T1/T is a straight lineis a straight line
Remember, T is temperature in Kelvin!Remember, T is temperature in Kelvin!
Another form: ln(kAnother form: ln(k22/k/k11) = (E) = (Eaa/R)(1/T/R)(1/T11 - 1/T- 1/T22))
You will do this in Lab 12You will do this in Lab 12

91. 92
Figure 12.14Figure 12.14
Plot of In(Plot of In(kk)) vs.vs. 1/1/TT forfor
2N2N22OO5(5(gg)) → 4ΝΟ→ 4ΝΟ22
((gg))
+ O+ O2(2(gg).).
Can get ECan get Eaa fromfrom
slope = -Eslope = -Eaa/R/R
Do in lab 12.Do in lab 12.

92. 93
Activation Energy and Rates
The final saga

93. 94
Mechanisms and ratesMechanisms and rates
There is an activation energy forThere is an activation energy for eacheach
elementaryelementary step.step.
Activation energy determinesActivation energy determines kk..
kk = Ae= Ae- (E- (E
aa/RT)/RT)
kk determinesdetermines raterate
So,So, slowestslowest stepstep (rate determining)(rate determining)
must have themust have the highesthighest activation
energy.

94. 95
 This reaction takes place in three steps

95. 96

Ea
First step is fast
Low activation energy

96. 97
Second step is slow
High activation energy

Ea

97. 98

Ea
Third step is fast
Low activation energy

98. 99
Second step is rate determining

99. 100
Intermediates are present

100. 101
Activated Complexes or
Transition States

101. 102
12.8 Catalysts
Speed up a reactionSpeed up a reaction without being usedwithout being used
upup in the reaction.in the reaction.
Enzymes are biological catalysts.Enzymes are biological catalysts.
Homogenous CatalystsHomogenous Catalysts are in theare in the samesame
phasephase as the reactants.as the reactants.
Heterogeneous CatalystsHeterogeneous Catalysts are in aare in a
differentdifferent phase as the reactants.phase as the reactants.

102. 103
How Catalysts WorkHow Catalysts Work
Catalysts allow reactions to proceed byCatalysts allow reactions to proceed by
aa different mechanismdifferent mechanism - a new pathway.- a new pathway.
NewNew pathway has apathway has a lowerlower activationactivation
energy (so, has a faster rate).energy (so, has a faster rate).
MoreMore molecules will have this activationmolecules will have this activation
energy.energy.
DoesDoes notnot change overallchange overall EE

103. 104
Figure 12.15
Energy Plots
for a Catalyzed
and an
Uncatalyzed
Pathway for a
Given
Reaction.
∆E is the same
in both cases.

104. 105
Figure 12.16Figure 12.16
Effect of a CatalystEffect of a Catalyst
on theon the NumberNumber ofof
Reaction-ProducingReaction-Producing
CollisionsCollisions
(increases even(increases even
though nothough no
temperaturetemperature
increase).increase).

105. 106
Pt surface
HH
HH
HH
HH
Hydrogen bonds to
surface of metal.
Break H-H bonds
Heterogenous Catalysts

106. 107
Pt surface
HH
HH
Heterogenous Catalysts
C HH C
HH

107. 108
Pt surface
HH
HH
Heterogenous Catalysts
C HH C
HH
The double bond breaks and bonds to the
catalyst.

108. 109
Pt surface
HH
HH
Heterogenous Catalysts
C HH C
HH
The hydrogen atoms bond with the
carbon

109. 110
Pt surface
H
Heterogenous Catalysts
C HH C
HH
H HH

110. 111
Heterogenous CatalystsHeterogenous Catalysts
Summary: usually in 4 steps:Summary: usually in 4 steps:
Adsorption & activationAdsorption & activation of reactantsof reactants
onto the catalytic surface.onto the catalytic surface.
MigrationMigration of the adsorbedof the adsorbed reactantsreactants
across the surface.across the surface.
ReactionReaction of the adsorbed reactants.of the adsorbed reactants.
EscapeEscape (desorption) of the(desorption) of the productsproducts..

111. 112
Homogenous Catalysts
ChlorofluorocarbonsChlorofluorocarbons catalyze thecatalyze the
decomposition of ozone.decomposition of ozone.
Both are in same phase (gas)Both are in same phase (gas)
Another example:Another example: EnzymesEnzymes regulatingregulating
the body processes. (Protein catalysts)the body processes. (Protein catalysts)

112. 113
Catalysts and rateCatalysts and rate
Catalysts will speed up a reactionCatalysts will speed up a reaction butbut
only to a certain pointonly to a certain point..
Past a certain point adding morePast a certain point adding more
reactants won’t change the rate.reactants won’t change the rate.
BecomesBecomes Zero OrderZero Order (saturation kinetics)(saturation kinetics)
See saturation kinetics with renal diseaseSee saturation kinetics with renal disease
& some drugs (aminoglycosides).& some drugs (aminoglycosides).

113. 114
Catalysts and rate.
Concentration of reactants
R
a
t
e
Rate increases until the active
sites of catalyst are filled.
Then rate is independent of
concentration

114. 115

115. 116