Rates of Reaction

1
Rates of Reaction

• Chapter 13

2
Kinetics

• Kinetics is the study of the rates of chemical
  reactions.

? concentration
Rate
? time
3
Lets Consider an Example

• Nitrogen Dioxide decomposes into nitric oxide and
  oxygen.

2NO2 (g) 2NO (g) O2 (g)
4
Lets Consider an Example

• Nitrogen Dioxide decomposes into nitric oxide and
  oxygen.

2NO2 (g) 2NO (g) O2 (g)
Concentrations
5
Concentrate on the Rate of NO2 Loss with Time

• Nitrogen Dioxide decomposes into nitric oxide and
  oxygen.

2NO2 (g) 2NO (g) O2 (g)
Concentrations
6
Rates of Formation

• Instantaneous rate
• The rate during a brief portion of a reaction (a
  tangent to the curve)
• Average rate
• The average of the rate over the entire reaction

7
Rate Laws

• Consider a reaction 2N2O5 (soln)
  4NO2 (soln) O2 (g)
• O2 escapes, so the reverse reaction is
  negligible.
• Differential rate law rate of the reaction
  depends on concentrations

8
Rate Laws

• Rate k N2O5n
• k is the rate constant, n is the order of the
  reactant.
• n does NOT necessarily correspond to a
  coefficient in the balanced chemical equation!
• The values for n and k must be determined
  experimentally.

9
Determining the Form of the Rate Law

• What is n for
  2N2O5 (soln) 4NO2 (soln) O2 (g)
• Lets consider some data
  N2O5 i Rate (mol/ L s) i
  0.90 M 5.4 x10-4
  0.45 M 2.7 x
  10-4
• Rate 2 2.7 x 10-4 kN2O5n
  Rate 1 5.4
  x10-4 kN2O5n

10
Looking at the Ratio of the Rates

• Rate 2 2.7 x 10-4 kN2O5n
  Rate 1 5.4
  x10-4 kN2O5n
• 0.5 (0.5)n
• Our n value is 1, first order reaction in
  N2O5
• Rate kN2O5

11
Rate Laws

• Nitric oxide, NO, reacts with hydrogen to give
  nitrous oxide, N2O, and water.
• 2NO(g) H2(g) N2O(g) H2O(g)

12
Rate Laws

• In a series of experiments, the following initial
  rates of disappearance of NO were obtained
• Initial concentrations Initial rate of
  NO H2
  reaction of NO with H2

1) 6.4 x 10-3 M 2.2 x 10-3 M 2.6 x 10-5
M/s
2) 12.8 x 10-3 M 2.2 x 10-3 M 1.0 x 10-4
M/s
3) 6.4 x 10-3 M 4.5 x 10-3 M 5.1 x
10-5 M/s
13
Rate Laws

• Find the rate law and the value of the rate
  constant for the reaction of NO.
• Doubling NO quadruples the rate, so the
  reaction is second order in NO.
• Doubling H2 doubles the rate, so the reaction
  is first order in H2.
• Rate kNO2H2

14
Rate Laws
Rate kNO2H2
Substituting values for the rate and
concentrations yields a value for k k

rate
2.6 x 10-5 M/s


NO2H2
6.4 x 10-3 M22.2 x 10-3 M
2.9 x 102 M-2s-1
15
Integrated Rate Laws

• Integrating the rate law allows us to determine
  the concentration at any given time.

16
First Order

• A gt product
• rate -?A/?t
• or
• rate kA

?A
kA
?t
17
First Order Plots

• ln A -kt ln A0
• or
• A A0 exp(-kt)

A
lnA
time
time
18
Half Life

• The amount of time for half of the reactants to
  be consumed.

at t1/2 we define A A0/2 so t1/2 1/k ln
A0
A0/2
(ln 2) /k
19
Second Order

• Using the same integrating procedure we can show
  that rate kA2

20
Second Order

• Using the same integrating procedure we can show
  that rate kA2
• Becomes

1
1
1
kt

A
A
A0
where t1/2 1/kA0
time
21
A Problem

• The recombination of iodine atoms to form
  molecular iodine in the gas phase follows
  second-order kinetics with a rate constant of 7.0
  x 109 M-1 s-1 at 23 oC. If the initial
  concentration of I was 0.086 M, calculate the
  concentration after 2.0 minutes.

I (g) I (g) I2 (g)
22
The Solution
Second Order
1
1
kt

A
A0
1
1/0.086M 7.0 x 109 (120 s)
A
A
1.2 x 10-12 M
23
A Question of Half-life

• If a first order reaction proceeds to 75
  completion in 32 minutes, what is the half-life
  of the reaction?
• t1/2 ln2/k 0.693/k
• But, what is k?

24
Use Integrated Rate Law
lnA lnA0 - kt
or
A
ln
-kt
A0
0.25A0
ln
-k(32 min. x 60 sec/min.)
A0
25
Use Integrated Rate Law
lnA lnA0 - kt
or
A
ln
-kt
A0
0.25A0
ln
-k(32 min. x 60 sec/min.)
A0
k 7.2 x 10-4 s-1
26
A Question of Half-life

• If a first order reaction proceeds to 25
  completion in 32 minutes, what is the half-life
  of the reaction?
• t1/2 ln2/k 0.693/k
• 0.693/7.2 x 10-4 s-1
• 960 s

27
Change of Concentration with Time Half-life

• Sulfuryl chloride, SO2Cl2, decomposes when
  heated.
• SO2Cl2(g) SO2(g) Cl2(g)
  
• In an experiment, the initial concentration of
• sulfuryl chloride was 0.0248 mol/L. If the
• rate constant is 2.2 x 10-5/s, what is the
• concentration of SO2Cl2 after 4.5 hrs? The
• reaction is first order.

28
Change of Concentration with Time Half-life

• Let SO2Cl2 0.0248 M and SO2Cl2t the
• concentration after 4.5 hrs. Substituting these
• and k 2.2 x 10-5/s into the first-order rate
• equation gives

3600s
)
SO2Cl2t
-(2.2 x 10-5/s)
(4.5 hrs x
1 hr
log

0.0248 M
2.303
-0.1547
29
Change of Concentration with Time Half-life

• Taking the antilogs of both sides gives

SO2Cl2t

0.7003
0.0248 M
SO2Cl2 0.7003 x 0.0248 M
1.73 x 10-2 1.7 x 10-2 M
30
Another Experiment

• What is the effect of temperature on the rate of
  reaction?

31
Collision Theory of Reactions

• If molecules in a solution are going to react,
  they must collide with one another.
• Factors affecting rate
• Rate of collision
• Orientation of collision
• Energy of collision

32
Temperature Dependence

• The Arrhenius Equation k A exp -Ea/RT
• Where
• k is the rate constant
• A is the frequency factor
• Ea is the activation energy
• R and T are as defined previously

33
Activation Barriers

• k A exp -Ea/RT

Ea
Energy
?H
Reaction Coordinate
34
Transition StatesA Temporary Step on the
Reaction Path

• The collisional encounter generates new species
  on the reaction path.

Ea
Energy
?H
Reaction Coordinate
35
Molecularity of Intermediates

• Unimolecular Reaction
• One molecule reacts alone to form product
• Bimolecular Reaction
• Two molecules combine in the transition state
• Termolecular Reaction
• Three or more molecules combine in the transition
  state

36
Reaction Mechanisms

• A sequential series of simple reactions which
  combine to form a larger, balanced chemical
  equation.

37
Lets Consider

• 2NO2 (g) F2 (g) 2NO2F (g)
• What is the mechanism?
• The rate law is RATE kNO2F2
• What is the intermediate?
• Experimentally, we can see evidence for NO2F
  being formed!

38
What If...?

• Experimentally, we can see evidence for NO2F
  being formed. We might speculate two elementary
  reactions occur.

NO2 F2
NO2F F
Then
F NO2
NO2F
39
What If...?

• Experimentally, we can see evidence for NO2F
  being formed. We might speculate two elementary
  reactions occur.

k1
NO2 F2
NO2F F
Slow
k2
Fast
F NO2
NO2F
40
A Reaction Mechanism MUST!

• Sum to the balanced chemical equation
• Must agree with the experimentally determined
  rate law

41
What If...?

• Experimentally, we can see evidence for NO2F
  being formed. We might speculate two elementary
  reactions occur.

k1
NO2 F2
NO2F F
Slow
k2
Fast
F NO2
NO2F
2NO2 F2 F
2NO2F F
42
What If...?

• Experimentally, we can see evidence for NO2F
  being formed. We might speculate two elementary
  reactions occur.

k1
NO2 F2
NO2F F
Slow
k2
Fast
F NO2
NO2F
2NO2 F2
2NO2F
rate kNO2F2 according to the elementary
reaction
43
Catalysis

• In order to facilitate beneficial reactions that
  are very slow, we often use catalysts to decrease
  the activation barrier for a reaction.

Ea
Energy
?H
Reaction Coordinate
44
Types of Catalysis

• Homogeneous Catalysis
• The interaction of reactants and catalysts in the
  same phase.
• e.g., CFCs (gas/gas)
• Heterogeneous Catalysis
• The interaction of reactants and catalysts in
  different phases.
• e.g., catalytic converters (solid/gas)