1. R = k [A]m [B]n If A + B  C DETERMINING THE RATE LAW
Then the rate law would ultimately be written as
R = k [A]m [B]n

2. 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

3. NH4+ + NO2-  N2 + 2H2O
exp. var
control
Dep. var

4. R = k [NH4+]m [NO2]n
exp. var
control
Dep. var 2x

5. R = k [NH4+]1 [NO2]n
exp. var
control
Dep. var 2x

6. R = k [NH4+]1 [NO2]n
exp. var
control
Dep. var 2x

7. R = k [NH4+]1 [NO2]
exp. var
control
Dep. var 2x 1

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

9. R = k [0.02M]1 [NO2-]1

10. R = k [0.02M]1[0.200M]1

11. 10.8 x 10-7M/s = k [0.02M]1[0.200M]1

12. 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

13. 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

14. 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)
1st order in Br2
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 2 2
(0.0320)x 1 =
.5x 4
x = 2
R = k[NO]2 [Br2]1

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

16. H3C N C: H3C C N:

17. 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
R = - [A] = k[A] t
ln[A]t - ln[A]0 = ln
[A]t
[A]0
= - kt
ln [A]t = - k t + ln [A]0
y = mx + b

18. H3C N C: H3C C N:

19. H3C N C: H3C C N: CH3NC1 1 mm Hg 150 140 130 118 90 70 55 35 1/ CH3NC3
.0067
.0071
.0076
.0084
.0111
.0142
.0182
.0286
R = - [A] = k[A]2 t 1
= kt +
[A]t
[A]0
y = mx + b
3 r = 0.92

20. H3C N C: H3C C N: CH3NC1 Time (sec) 1 mm Hg 1/ CH3NC3 .0067 .0071
.0076
.0084
.0111
.0142
.0182
.0286
ln CH3NC2
5.02
4.94
4.86
4.77
4.49
4.24
4.00
3.55
2 r = 0.98
3 r = 0.92
150
140
130
118 90 70 55 35
900
2,500
5,000
10,000
15,000
20,000
30,000

21. 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 yr 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] t
ln[A]t - ln[A]0 = ln
[A]t
= - kt
[A]0
ln [A]t = - k t + ln [A]0
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

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

23. 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

24. Examining the Relationship Between the Rate Constant
and Temperature

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

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

27. 1010 collisions/sec occur
1 in collisions succeeds

28. ...And What Do the Mathematicians Have to Say About
Collision Theory?
k = Ae-Ea/RT

29. Rewriting the Arrhenius Equation
k = Ae-Ea/RT
ln k1 = -Ea/RT1 + ln Ae
ln k2 = -Ea/RT2 + ln Ae
lnk1 - ln k2 = (-Ea/RT1 + ln Ae) - (Ea/RT2 + ln Ae) ln k1 k2 Ea R 1 T2 T1 - =

30. But there is a simpler way
Getting a Line on the Arrhenius Equation
But there is a simpler way
to think about the
Arrhenius Equation
Slope = -Ea/R*
k = Ae-Ea/RT
ln k
ln k1 = -Ea/RT1 + ln Ae
y = m x b
1/T
* R (ideal gas constant) = 8.31 J/K-mol

31. Mastering the Arrhenius Equation with your TI
The following table shows the rate constants for the rearrangement of CH3CN
(methyl isonitrile) at various temperatures
Temperature (°C) k (s-1)
189.7
198.9
230.3
251.2
2.52 x 10-5
5.25 x 10-5
6.30 x 10-4
3.16 x 10-3
From these data calculate the activation energy of this reaction

32. Mastering the Arrhenius Equation with your TI
k = Ae-Ea/RT
ln k1 = -Ea/RT1 + ln Ae
L L L L L5
T(°C) L /L k ln L4
y = m x b
Slope = -Ea/R*
It’s as easy as L1,
L2, L3
ln k (L5)
1/T (L3)

33. Mastering the Arrhenius Equation with your TI
The following table shows the rate constants for the following reaction at varying
temperatures: CO(g) + NO2(g)  CO2(g) + NO(g)
Temperature (°C) k (M-1 s-1)
600
650
700
750
800
0.028
0.22
1.3 6 23
From these data calculate the activation energy for this reaction
Now, you try it!

34. Reaction Mechanisms: The Shortest Distance
Between Two Points Often Requires More Steps
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)

35. Reaction Mechanisms: The Shortest Distance
Between Two Points Often Requires More Steps
NO2(g) +NO2(g) NO3(g) + NO(g)
Rate Determining step
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
Yea, but what if the second step
is the rate determining step?

36. Reaction Mechanisms: The Shortest Distance
Between Two Points Often Requires More Steps
OK.
Let’s try this one experimentally:
2NO(g) + Br2(g)  2NOBr(g)
what we get is R= k[NO]2[Br2]

37. Reaction Mechanisms: The Shortest Distance
Between Two Points Often Requires More Steps k1
NO(g) +Br2(g) NOBr2(g)
Fast step
k-1
NOBr2(g) + NO(g)  2NOBr(g) k2
Rate Determining step
Slow step
2NO(g) + Br2(g)  2NOBr(g)
1. We assume that reaction 1 is at equilibrium and that Rf = Rr
2. If the above is true, then k1[NO][Br2] = k-1 [NOBr2]
3. 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.
4. We can substitute for NOBr2 using step two, however. Doing so would allow us to use NO(g) and Br2(g) which are not intermediates

38. Reaction Mechanisms: The Shortest Distance
Between Two Points Often Requires More Steps k1
k-1
Let [NOBr2] = [NO][Br2]
If R = k [NOBr2][Br2].
Then by substitution, R = k [NO][Br2] [Br2]. k1
k-1 k1
k-1
We then let k = k2
Finally, we get the rate law R = k[NOBr2][Br2].

39. Reaction Mechanisms: The Shortest Distance
Between Two Points Often Requires More Steps
Prove that the following mechanism is consistent with R = k[NO]2[Br2], the
rate law which was derived experimentally: k1
NO(g) +NO(g)  N2O2(g)
k-1
N2O2(g) + Br2(g)  2NOBr(g) k2
2NO(g) + Br2(g)  2NOBr(g)
Now, you try it

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

41. Catalyst:a substance that changes the speed of a chemical
reaction without it self undergoing a permanent chemical
change in the process
Ethylene
Ethane
This metallic substrate is a heterogeneous catalyst