1. Chemical Kinetics How fast can a reaction go
Four things that affect rate
a. concentration
b. temperature
c. catalyst
d. surface area

2. Average rate
Instantaneous rate

3. Rate Laws
NH4+ + NO2-  N2 + 2H2O
R = k [NH4+]m [NO2-]n
k = rate proportionality constant
m = reaction order for the
ammonium ion
n = reaction order for the nitrite
ion

4. How do the three chemicals compare?
Rate
N2 + 3H2 → 2NH3
R or rate = -Δ[N2] = M/s Δt
How do the three chemicals compare?

5. NH4+ + NO2-  N2 + 2H2O
indep. var
control
Dep. var

6. R = k [NH4+]m [NO2]n
Skeleton rate law
Dep. var
indep. var
control 2x

7. R = k [NH4+]1 [NO2]1

8. 10.8 x 10-7M/s = k [0.02M]1[0.200M]1

9. 10.8 x 10-7M/s = k[0.02M]1 [0.200M]1
k= 10.8 x 10-7 M/s /[0.02M]1[0.200M]1
k = 10.8 x 10-7 M/s / 0.004M2
k = 2.7 x /M • s
M = ? M1 M1
s ?

10. Suppose the concentrations for
NH4+ and NO2- are
0.15 M and 0.20 M, respectively
Since R = k[NH4+]1 [NO2]1
where k = 2.7 x /M - s at 25 C
then R =
2.7 x /M - s [0.15 M] [0.20 M]
R = 8.1 x 10-6 M/s

11. 2NO(g) + Br2(g)  2NOBr
x 10-4
x 10-4
x 10-4
Initial [NO] Initial [Br2] Rate of
appearance
NOBr
(M/s)
(M/L)
R = k[NO]m [Br2]n
3.24 x 10-4
k(0.0160)x (0.0120)1 =
6.42 x 10-4
k(0.0320)x (0.0060)1 1
(0.0160)x 1 2 =
.5x
x = 2 = 2
(0.0320)x 4
R = k[NO]2 [Br2]1

12. Using the calculator to find order
H3C N C: H3C C N:
Using the calculator to find order
CH3NC1
Time (sec)
1 mm Hg
150
140
130
118 90 70 55 35
900
2,500
5,000
10,000
15,000
20,000
30,000

13. H3C N C: H3C C N:

14. y = m x + b ln [A]t = - k t + ln [A]0 H3C N C: H3C C N: CH3NC1
1 mm Hg
150
140
130
118 90 70 55 35
Ln CH3NC2
5.02
4.94
4.86
4.77
4.49
4.24
4.00
3.55
2 r = 0.98
ln [A]t = - k t + ln [A]0
y = m x + b

15. H3C N C: H3C C N: 1 = kt + [A]t y = mx + b [A]0 CH3NC1 1/ CH3NC3 150
140
130
118 90 70 55 35
.0067
.0071
.0076
.0084
.0111
.0142
.0182
.0286 1
= kt +
[A]t
[A]0
y = mx b
3 r = 0.92
1 mm Hg

16. H3C N C: H3C C N: CH3NC1 ln CH3NC2 1/ CH3NC3 Time (sec) 150 140 130
118 90 70 55 35
5.02
4.94
4.86
4.77
4.49
4.24
4.00
3.55
.0067
.0071
.0076
.0084
.0111
.0142
.0182
.0286
900
2,500
5,000
10,000
15,000
20,000
30,000

17. zero order Rate = k[A]o 1st order Rate = k[A]1 t
plot concentration [A] vs. time (t) – linear, negative slope
integrated rate law [A]t = -kt + [A]0
half life – t1/2 = [A]0/2k [A]
1st order Rate = k[A] t
plot of ln conc. vs. time is linear, negative slope
integrated rate law ln[A]t = -kt + ln[A]0
half life – t1/2 = .693/k ln[A]
2nd order Rate = k[A] t
plot inverse of conc. vs. time is linear, positive slope
integrated rate law /[A]t = kt + 1/[A]0
half life – t1/2 = 1/k[A] /[A] t

18. ln [A]t = - (1.45 yr.-1)(1.00 yr.) + ln( 5.0 x 10-7 g/cm3)
Using Rate Law Equations to Determine Change
in Concentration Over Time
The first order rate constant for the decomposition of a certain insecticide in water at 12 C° is 1.45 yr1-. A quantity of insecticide is washed into a lake in June leading to a concentration of 5.0 x 10-7 g/cm3. Assume that the effective temperature of the lake is 12 C°. (a)What is the concentration of the insecticide in June of the following year? (b) How long will it take for the concentration of the insecticide to drop to
3.0 x 10-7 g/cm3?
R = - [A] = k[A]
ln [A]t = - k t + ln [A]0 t
ln [A]t = - (1.45 yr.-1)(1.00 yr.) + ln( 5.0 x 10-7 g/cm3)
ln [A]t = , [A]t = e
[A]t = 1.2 x 10-7 g/cm3

19. Using Rate Law Equations to Determine the Half-Life of a Reaction
Let’s begin with ln[A]t - ln[A]0 = ln
= - kt
[A]0
[A]t
given ln
let [A]t = ½ [A]0
[A]0
½ [A]0
then ln
= - kt,
and ln ½ = -kt ½
[A]0
t ½ = -ln ½ = 0.693
= - kt, k k

20. Using Rate Law Equations to Determine the Half-Life of a Reaction
The concentration of an insecticide accidentally spilled into a lake was measured at 1.2 x 10-7 g/cm3. Records from the initial accident show that the concentration of the insecticide was 5.0 x 10-7 g/cm3. Calculate the 1/2 life of the insecticide.
1.2 x 10-7 g/cm3 ln
= - k (1.00 yr.)
5.0 x 10-7 g/cm3
k = 1.43 yr-1
t ½ = = yr.
1.43 yr-1

21. Examining the Relationship Between the Rate Constant
and Temperature

22. ..it’s because of Collision Theory
Why does Temperature Have an Effect on Rate?
..it’s because of
Collision Theory

23. It’s All About Activation Energy and Getting over
the HUMP!!

24. 1010 collisions/sec occur
1 in collisions succeeds

25. Reaction Mechanisms NO(g) + O3(g)  NO2(g) + O2(g)
A Reaction Mechanism is the process which describes in great detail the order in which bonds are broken and reformed, changes in orientation and the energies involved during those rebondings, and changes in orientations
NO(g) + O3(g)  NO2(g) + O2(g)
A single reaction event is called an elementary reaction
NO2(g) + CO(g)  NO(g) + CO2(g)
While this reaction looks like an elementary reaction, it
actually takes place in a series of steps
NO2(g) +NO2(g) NO3(g) + NO(g)
NO3(g) + CO(g)  NO2(g) + CO2(g)
NO2(g) + CO(g)  NO(g) + CO2(g)

26. Reaction Mechanisms NO2(g) +NO2(g) NO3(g) + NO(g)
Rate Determining step
NO2(g) +NO2(g) NO3(g) + NO(g)
Slow step
NO3(g) + CO(g)  NO2(g) + CO2(g)
Fast step
NO2(g) + CO(g)  NO(g) + CO2(g)
The rate law is constructed from the rate
determining step
Rate = k1[NO2]2

27. Reaction Mechanisms: Catalysts
NO + O3 → NO2 + O2
O + NO2 → NO + O2
O + O3 → 2O2

28. Fast step NO(g) +Br2(g) NOBr2(g) NOBr2(g) + NO(g)  2NOBr(g)
Rate Determining step
Slow step
2NO(g) + Br2(g)  2NOBr(g)
If step 2 is the rate determining step, then R = k[NOBr2][Br2]. Keep in mind, however, that one cannot include an intermediate in the rate determining step.
We can substitute for NOBr2 using the previous equation NO(g) and Br2(g) which are not intermediates
rate = k [NO]2[Br2]

29. rate law which was derived experimentally:
Prove that the following mechanism is consistent with R = k[NO]2[Br2], the
rate law which was derived experimentally:
NO(g) +NO(g)  N2O2(g)
N2O2(g) + Br2(g)  2NOBr(g)
2NO(g) + Br2(g)  2NOBr(g)

30. Catalyst: a substance that changes the speed of a
chemical reaction without it self undergoing a permanent chemical change in the process or it is reconstituted at the end.
Br - is a homogeneous catalyst

31. Heterogeneous Catalysts
Ethylene
Ethane
This metallic substrate is a heterogeneous catalyst