Transcript AP Notes Chapter 15
AP Notes Chapter 15
Principles of Reactivity: Chemical Kinetics
Chemical Kinetics The study of how fast chemical reactions occur. How fast species are appearing and disappearing.
It has nothing to do with the extent to which a reaction will occur – that is thermodynamics.
There is a difference between the probability that molecules will react when they meet each other and once they do react how stable is the product.
First, probability of reaction has to do with overcoming an energy barrier or activation energy.
Second, has to do with the differences between reactants and products.
Thermodynamics: Spontaneity deals with how likely a reaction will take place.
Spontaneous reactions are reactions that will happen - but we can’t tell how fast.
Diamond will spontaneously turn to graphite – eventually.
Factors Affecting Rates 1. Surface area
Powders
2. Temperature
Dye
3. Pressure 4. Catalyst /Inhibitor 5. Concentration
H 2 0 2 & I 2
Reaction Rate
The change in concentration of a reactant or product per unit time For a reaction A B Average Rate = Change in Moles of B Change in time Rate = D D [B] = D t D [A] t
Reaction Rate
Rate = Conc. of A at t 2 t 2 - t 1 Rate = D [A] D t -Conc. of A at t 1 Change in concentration per unit time Changes in Pressure concentration to changes in
N 2 + 3H 2 2NH 3 As the reaction progresses the concentration H 2 goes down
C t i o n o n c e t n r a
[H 2 ]
Time
N 2 + 3H 2 2NH 3 As the reaction progresses the concentration N 2 goes down 1/3 as fast
t a i o n C o n c t r e n
[H 2 ] [N 2 ]
Time
N 2 + 3H 2 2NH 3 As the reaction progresses the concentration NH 3 goes up.
t a i o n C o n c t r e n
[H 2 [NH ] 3 ] [N 2 ]
Time
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 Derivative and Integrated Rates.
Average slope method
t a i o n C o n c t r e n
D [H 2 ] D t
Time
Instantaneous slope method.
t a i o n C o n c t r e n
D [H 2 ] D t
Time
Defining Rate
We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product.
In our example N 2 + 3H 2 2NH 3 D [N 2 ] D t = - 1 D [H 2 ] = 1 D [NH 3 ] 3 D t 2 D t
RATE LAW Mathematical model that contains specific information about what reactants determine the rate and how the rate is controlled
Rate Laws
Reactions are reversible.
As products accumulate they can begin to turn back into reactants.
Early on the rate will depend on only the amount of reactants present.
We want to measure the reactants as soon as they are mixed.
This is called the Initial rate method .
Rate Laws
Two key points The concentration of the products do not appear in the rate law because this is an initial rate.
The order must be determined experimentally, can’t be obtained from the equation
2 NO
2
2 NO + O
2 You will find that the rate will only depend on the concentration of the reactants.
Rate = k [NO 2 ] n This is called a rate law expression .
k is called the rate constant.
n is the order of the reactant -usually a positive integer.
2 NO
2
2 NO + O
2 The rate of appearance of O 2 can be said to be.
Rate' = D [O 2 ] = D [NO 2 ] = k'[NO 2 ] D t 2 D t Because there are 2 NO 2 So k = 2 x k' for each O 2 Rate = 2 x Rate' So k[NO 2 ] n = 2 x k'[NO 2 ] n
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.
Determining Rate Laws
The first step is to determine the form of the rate law (especially its order).
Must be determined from experimental data.
For this reaction 2 N 2 O 5 (aq) 4NO 2 (aq) + O 2 (g) The reverse reaction won’t play a role
[N 2 O 5 ] (mol/L) 1.00
0.88
0.78
0.69
0.61
0.54
Now graph the data
Time (s) 0 200 400 600 800 1000 0.48
0.43
0.38
0.34
0.30
1200 1400 1600 1800 2000
1.2
1
To find rate we have to find the slope at two points
0.8
0.6
0.4
0.2
We will use the tangent method.
0
1.2
1 0.8
0.6
0.4
0.2
0 At .90 M the rate is (.98 - .76) = 0.22 =- 5.5x 10 -4 (0-400) -400
1.2
1 0.8
0.6
0.4
0.2
0 At .40 M the rate is (.52 - .31) = 0.22 =- 2.7 x 10 -4 (1000-1800) -800
Since the rate at twice the concentration is twice as fast the rate law must be..
Rate = D [N 2 O 5 ] = k[N D t 2 O 5 ] 1 = k[N 2 O 5 ] We say this reaction is first order in N 2 O 5 The only way to determine order is to run the experiment.
The method of Initial Rates
This method requires that a reaction be run several times.
The initial concentrations of the reactants are varied.
The reaction rate is measured bust after the reactants are mixed.
Eliminates the effect of the reverse reaction.
Integrated Rate Law
Expresses the reaction concentration as a function of time.
Form of the equation depends on the order of the rate law (differential).
Rate = D [A] n D t We will only work with n=0, 1, and 2 although negative and fractional exist
General Rate Law
aA + bB
cC Rate
-[A] or -[B] r = k [A] m [B] n
Where: m = order with respect to A n = order wrt B and n + m = order of reaction Notice we are primarily discussing the reactants
For example: aA + bB Zero Order: r = k cC + dD First Order in A: r = k [A] First Order in B: r = k [B] Second Order in A: r = k [A] 2 Second Order in B: r = k [B] 2 First Order in A, first Order in B, for a second order overall: r = k [A][B]
Initial Rate Method to establish the exact rate law R 0 = k[A 0 ] m [B 0 ] n but experimental data must provide orders Use ratio method to solve for m and n R 0 = k[A 0 ] m [B 0 ] n
R 0 = k[A 0 ] m [B 0 ] n Exp Rate [A 0 ] [B 0 ] 1 0.05 0.50 0.20
2 0.05 0.75 0.20
3 0.10 0.50 0.40
Exp
1 Exp 2 : 0 .
05 0 .
05
k k
0 .
50 0 .
75 0 .
20 0 .
20
n n
1 0 .
50 0 .
75
m
m
0
Exp
1 Exp 3 : 0 .
05 0 .
10
k k
0 .
50 0 .
50 0 .
20 0 .
40
n n
1 2 0 .
20 0 .
40
n
n
1
Thus, rate = k[A] 0 [B] 1 or rate = k[B]
Where k is the rate constant , with appropriate units.
rate = k[B] Using Exp 1: 0 .
05
mol L
s
k
0 .
20
mol L
k
0 .
25 s 1
INTEGRATED RATE LAWS Each “order” has a separate “law” based on the integration of the rate expression.
You NEED TO KNOW the rate expressions & half-life formulas for 0, 1 st , and 2 nd order reactions.
Third order and fractional order reactions exist, but are rare.
Nuclear reactions are typically 1 st order, therefore, use the first-order rate law and half-life to solve problems of this type
aA B Rate Law Zero Order First Order r=k[A] o r=k[A] 1 Second Order r=k[A] 2 d[A] = k[A]m y=mx +b Plot for a linear graph slope of linear plot half-life t1/2 when [A]=-kt +[A] o ln[A]=-kt+ln[A] o [A] vs t -k t 1/2 = [A] o 2k t ln[A] vs t -k 1/2 = 0.693
k 1 = kt + 1 [A] [A] o 1 vs t [A] k t 1/2 = 1 k[A] o
Zero Order Rate Law
Most often when reaction happens on a surface because the surface area stays constant.
Also applies to enzyme chemistry.
i o n C o n c e n t r a t Time
i o n C o n c e n t r a t
D[
A]
D
t Time k =
D[
A]/
D
t
Zero Order Rate Law
Rate = k[A] 0 = k Rate does not change with concentration.
Integrated [A] = -kt + [A] 0 When [A] = [A] 0 /2 t = t 1/2 t 1/2 = [A] 0 /2k
For a zero-order reaction
, the rate is independent of the concentration. Expressed in terms of calculus, a zero - order reaction would be expressed as:
rate
-
or
dB dB dt
k kB dt
0
B
B
0
dB
k
0
t dt
B
B
0
k
t
0
or B
kt
B
0
m = -k t
Time for the concentration to become one half of its initial value
if B 1 2
B
0
then
substitute into
B
k
t
B
0 1 2 B 0 k t 1 2 B 0 t 1 2 B 0 2 k Half-life 0-Order Reaction
First Order
For the reaction 2N 2 O 5 4NO 2 We found the Rate = k[ N 2 O 5 ] 1 + O If concentration doubles rate doubles.
2 If we integrate this equation with respect to time we get the Integrated Rate Law ln[N 2 O 5 ] = - kt + ln[N 2 O 5 ] 0 ln is the natural log [N 2 O 5 ] 0 is the initial concentration.
First Order
General form Rate = D [A] / D t = k[A] ln[A] = - kt + ln[A] 0 In the form y = mx + b y = ln[A] m = -k x = t b = ln[A] 0 A graph of ln[A] vs time is a straight line.
First Order
By getting the straight line you can prove it is first order Often expressed in a ratio
First Order
By getting the straight line you can prove it is first order Often expressed in a ratio ln [ 0 = kt
First Order Nuclear Decay and Half life
The time required to reach half the original concentration.
If the reaction is first order [A] = [A] 0 /2 when t = t 1/2
• The time required to reach half the original concentration.
• If the reaction is first order • [A] = [A] 0 /2 when t = t 1/2 ln 0 0 2 = kt 1 2 ln(2) = kt 1/2
t 1/2 = 0.693/k The time to reach half the original concentration does not depend on the starting concentration.
An easy way to find k
For a 1st-order reaction
, the rate is dependent of the concentration. Expressed in terms of calculus, a 1st - order reaction would be expressed as: dA dt
kA
1
or dA A
k dt
A A
0
dA A
k
0
t dt
t
ln
A A
0
k
t
0 ln A kt ln A 0
m = -k
ln
A A
0
k
t ln 1 2
A A
0 0
k
t 1 2 ln 2 k t 1 2
or
t 1 2 ln 2 k 0 .
693
k
t 1 2 0 .
693
k
Half-life 1 st -order reaction
Second Order
Rate = D [A] / D t = k[A] 2 integrated rate law 1/[A] = kt + 1/[A] 0 y= 1/[A] Knowing k and [A] 0 at any time t m = k x= t b = 1/[A] 0 A straight line if 1/[A] vs t is graphed you can calculate [A]
Second Order Half Life
[A] = [A] 0 1 /2 at t = t 1/2 A 0 = kt 1 2 + 2 2 1 [ A] 0 - [ A] 0 1 [A] 0 = kt 1 2 1 [ A] 0 = kt 1 2 t 1 k[A] 0
For a 2nd-order reaction
, the rate is dependent of the concentration. Expressed in terms of calculus, a 2nd - order reaction would be expressed as:
dB dt
kB
2
or
dB B 2
k
dt B B 0 dB B 2 k 0 t dt
1/[B] m = k t
1
B
1
B
0
k
1
or
B
kt
1
B
0 1 B 1 2 1 k t B 0 1 k t B 0 1 2 1 B 0 t 1 2 1 k B 0
Half-life for 2nd - order
General Rate Law
aA + bB + cC dD
rate D A D t
or rate
k[A] m rate rate D B D t
or rate
k[B] n D
C
D t
or rate
k[C] p
A + B + C
products
Exp 1 2 3 4 5 [A] [B] [C] 1.25
2.50
1.25
1.25
3.01
1.25
1.25
3.02
3.02
1.00
1.25
1.25
1.25
3.75
1.15
Initial Rate 8.7
17.4
50.8
457 ??
Exp 1 2 [A] 1.25
2.50
[B] 1.25
1.25
[C] 1.25
1.25
Initial Rate 8.7
17.4
•If [B] and [C] are held constant, then any change in rate is caused by [A].
•[A] doubles from 1.25 to 2.50 the rate double from 8.7 to 17.4
therefore order for A is 1 st order
Exp 3 4 [A] 1.25
1.25
[B] 3.02
3.02
[C] 1.25
3.75
Initial Rate 50.8
457
•If [A] and [B] are held constant, then any change in rate is caused by [C].
•[C] doubles from 1.25 to 3.75 the rate quadruples or a squaring of the change from 50.8 to 475 •therefore order for C is 2nd order
Exp 1 3 [A] 1.25
1.25
[B] 1.25
3.02
[C] 1.25
1.25
Initial Rate 8.7
50.8
•If [A] and [C] are held constant, then any change in rate is caused by [B].
•[B] doubles from 1.25 to 3.02 the rate quadruples or a squaring of the change from 8.7 to 50.8
•therefore order for B is 2nd order
aA + bB + cC
A + B + C
dD
products rate = k[A]
1
[B]
2
[C]
2 if we make substitutions with data from experiment 1 to solve for k
8.7 = k[1.25]
1
[1.25]
2
[1.25]
2
k=2.85
Exp 5 [A] [B] 3.01
1.00
[C] 1.15
rate = 2.85[3.01] 1 [1.00] 2 [1.15] 2 Initial Rate ??
rate = 11.35
Another Complex Reaction For the reaction BrO 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 We use experimental data to determine the values of n,m,and p
BrO 3 + 5 Br + 6H + 3Br 2 + 3 H 2 O For this reaction we found the rate law to be Rate = k[BrO 3 ][Br ][H + ] 2 To investigate this reaction rate we need to control the conditions
Rate = k[BrO 3 ][Br ][H + ] 2 We set up the experiment so that two of the reactants are in large excess.
[BrO 3 ] 0 = 1.0 x 10 -3 M [Br ] 0 = 1.0 M [H + ] 0 = 1.0 M As the reaction proceeds [BrO 3 ] changes noticably [Br ] and [H + ] don’t
Rate = k[BrO 3 ][Br ][H + ] 2 This rate law can be rewritten Rate = k[BrO Rate = k[Br ] 3 0 ][Br [H + ] 0 ] 2 0 [H + [BrO ] 3 0 2 ] Rate = k’[BrO 3 ] This is called a pseudo first order rate law.
k = k’ [Br ] 0 [H + ] 0 2
Initial concentrations (M) Rate (M/s)
BrO 3 0.10
0.20
0.20
0.10
Br 0.10
0.10
0.20
0.10
H + 0.10
0.10
0.10
0.20
8.0 x 10 -4 1.6 x 10 -3 3.2 x 10 -3 3.2 x 10 -3
Now we have to see how the rate changes with concentration
Reaction Mechanisms
The series of steps that actually occur in a chemical reaction.
Kinetics can tell us something about the mechanism A balanced equation does not tell us how the reactants become products.
Reaction Mechanisms
2NO 2 + F 2 Rate = k[NO 2 ][F 2 ] 2NO 2 F The proposed mechanism is NO 2 + F 2 F + NO 2 NO NO 2 2 F + F F (slow) (fast) F is called an intermediate It is formed then consumed in the reaction
Reaction Mechanisms
Each of the two reactions is called an elementary step .
The rate for a reaction can be written from its molecularity .
Molecularity is the number of pieces that must come together.
Mechanism must pass two tests to be accepted 1. Sum of mechanism steps produces the net equation.
2. Rate laws of all rate determining steps must be combined to produce experimental rate law.
Unimolecular rirst order.
step involves one molecule - Rate is Bimolecular step - requires two molecules - Rate is second order Termolecular step- requires three molecules Rate is third order Termolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule.
A products A+A products 2A products A+B products Rate = k[A] Rate= k[A] 2 Rate= k[A] 2 Rate= k[A][B] A+A+B Products Rate= k[A] 2 [B] 2A+B Products Rate= k[A] 2 [B] A+B+C Products Rate= k[A][B][C]
How to get rid of intermediates
Use the reactions that form them If the reactions are fast and irreversible - the concentration of the intermediate is based on stoichiometry.
If it is formed by a reversible reaction set the rates equal to each other.
Formed in reversible reactions
2 NO + O 2 2 NO 2 Mechanism 2 NO N 2 O 2 + O 2 N 2 O 2 2 NO 2 (fast) (slow) rate = k 2 [N 2 O 2 ][O 2 ] k 1 [NO] 2 rate = k 2 = k -1 [N 2 O 2 ] (k 1 / k -1 )[NO] 2 [O 2 ]=k[NO] 2 [O 2 ]
Formed in fast reactions
2 IBr Mechanism IBr I 2 + Br I + Br 2 IBr + Br I + Br 2 I + I I 2 Rate = k[IBr][Br] but [Br]= [IBr] Rate = k[IBr][IBr] = k[IBr] 2 (fast) (slow) (fast)
Collision theory
Molecules must collide to react.
Concentration affects rates because collisions are more likely.
Must collide hard enough.
Temperature and rate are related.
Only a small number of collisions produce reactions.
E n e r g y P o t e n t i a l Reactants
Reaction Coordinate
Products
E n e r g y P o t e n t i a l Reactants
Reaction Coordinate
Activation Energy E a Products
E n e r g y P o t e n t i a l Reactants Activated complex
Reaction Coordinate
Products
E n e r g y P o t e n t i a l Reactants
Reaction Coordinate
Products
} D
E
Br---NO E n e r g y P o t e n t i a l 2BrNO Br---NO Transition State
Reaction Coordinate
2NO + Br 2
Misconception about catalysts Catalysts do not take part in reactions A + B --> A-B* --> AB A-B* A + B E a (cat)
D
H AB Reaction Coordinate
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.
Arrhenius
Said the reaction rate should increase with temperature.
At high temperature more molecules have the energy required to get over the barrier.
The number of collisions with the necessary energy increases exponentially.
Arrhenius Equation
Number of collisions with the required k = Ae ^(-E a /RT) k = rate constant A = total collisions ≈ frequency factor e is Euler’s number (opposite of ln) E a = activation energy R = Universal gas constant T = temperature in Kelvin Linearized Arrhenius Equation lnk = lnA – (E a /RT)
Using the linear equation we can find a number of things Plot lnk vs 1/T the slope will give you the activation energy If we know Ea and k at one temperature then you can find k at a different temperature ln k2 = E a 1 - 1 k1 R T1 T2
Problems
Observed rate is less than the number of collisions that have the minimum energy.
Due to Molecular orientation written into equation as p the steric factor.
O N Br O N Br O N Br Br N O Br N O O N Br O N Br Br Br O N Br N O O N No Reaction
Arrhenius Equation
k = zpe -E a /RT = Ae -E a /RT A is called the frequency factor = zp ln k = -(E a /R)(1/T) + ln A Another line !!!!
ln k vs t is a straight line
Mechanisms and rates
There is an activation energy for each elementary step.
Activation energy determines k.
k = Ae - (E a /RT) k determines rate Slowest step (rate determining) must have the highest activation energy.
This reaction takes place in three steps
E a First step is fast Low activation energy
E a Second step is slow High activation energy
Third step is fast Low activation energy E a
Second step is rate determining
Intermediates are present
Activated Complexes or Transition States
Possible Mechanisms a. CO + NO 2
CO 2 + NO rate
k
[
NO 2
[
b. 2NO 2 N 2 O 4
N 2 O 4 + 2CO
2CO 2 + 2NO (slow) rate rate
k ' (
[
forward eq ) ( r everse eq ) NO 2
2
k ' '
[
N 2 k O k ' ' 4
'
[ [
NO 2 N 2 O 4
2
rate ( slo )
k ' ' '
[
N 2 rate O 4
[
2 k
[
NO 2
2
[
2
c. 2NO 2 NO 3
NO 3 + CO
+ NO (slow) NO 2 + CO 2
rate
(fast)
k
[
NO
2 2
d. 2NO 2
2NO + O 2 2 CO + O 2
2CO 2
rate
(slow) (fast)
k [ NO 2 2
Transition State Theory
transition state = activated complex A + B ---> A--B* ---> AB
A + B --> A-B* --> AB activated complex A-B* A + B E a activation energy
D
H AB Heat of reaction Reaction Coordinate
A + B --> A-B* --> AB A-B* AB E a
D
H A + B Reaction Coordinate
In General
rate(R)
[A] R = k[A] k = rate constant
If one examines the relationship between the rate constant, k, and temperature, we find that ln k
1/T
m = -E
a
/R 1/T
ln k
E a R
1 T
ln A Arrhenius Equation
k
Ae
E a RT
A = frequency factor R = 8.314 J/mol
.
K
ln k 1
E a R
1 T 1
ln A ln k 2
E a R
1 T 2
ln A
ln k k 2 1
E R a
1 T 2
1 T 1
Catalysts
Speed up a reaction without being used up in the reaction.
Enzymes are biological catalysts.
Homogenous Catalysts phase as the reactants.
are in the same Heterogeneous Catalysts phase as the reactants.
are in a different
How Catalysts Work
Catalysts allow reactions to proceed by a different mechanism - a new pathway.
New pathway has a lower activation energy.
More molecules will have this activation energy.
Do not change D E
Heterogenous Catalysts
H H Hydrogen bonds to surface of metal.
Break H-H bonds H H Pt surface H H H H
Heterogenous Catalysts
H H C H C H H H H H Pt surface
Heterogenous Catalysts
The double bond breaks and bonds to the catalyst.
H H H H C H C Pt surface H H H
Heterogenous Catalysts
The hydrogen atoms bond with the carbon H H H H C H C Pt surface H H H
H
Heterogenous Catalysts
H H C H H C H H H Pt surface
Homogenous Catalysts
Chlorofluorocarbons catalyze the decomposition of ozone.
Enzymes regulating the body processes. (Protein catalysts)
Catalysts and rate
Catalysts will speed up a reaction but only to a certain point.
Past a certain point adding more reactants won’t change the rate.
Zero Order
Catalysts and rate.
t e R a Rate increases until the active sites of catalyst are filled.
Then rate is independent of concentration Concentration of reactants