1. Chapter 12
Chemical Kinetics

2. Spontaneity – inherent tendency for a process to occur (but doesn’t imply speed) * some can be slow, they can occur in weeks or even years

3. Reaction mechanism – steps by which a reaction takes place
Reaction rate – change in concentration of a reactant or product per unit of time

4. Rate = conc. of A at time t2 – conc. of A at time t1
= - D [A]
D t
Instantaneous rate – can be obtained by computing the slope of a line tangent to the curve of that point
= - D [A]
D t
(in terms of reactants)

5. Find Rate in terms of Products
- must account for coefficients in balanced equations Rate Law – show how the rate depends on concentration of reactants Rate = k[A]n k = rate constant n = order

6. Types of Rate Laws
1. Differential rate law (rate law) – expresses how the rate depends on concentration 2. integrated rate law – express how the concentration depends over time initial rate – is the instantaneous rate determined just after the reaction begins (t=0)

7. Several experiments are conducted using different concentrations and an initial rate is determined for each
Overall reaction order – the sum of the orders of each reactant (n, m……)

8. Determine Rate Law
Exp initial conc. NH initial conc. NO initial rate (mol/L*s) M M X M M X M M X NH4+ (aq) + NO2- (aq)  N2 (g) + 2H2O (l) Rate = D[NH4+ ] = k[NH4+ ]n [NO2-]m D t

9. EXP2 = 2.70X10-7 mol/L*s = k[0.100 mol/L]n [0.010 mol/L]m
2.00
2.00 = [0.010 mol/L]m
[ mol/L]m
= (2.0)m
m = 1 reaction is first order for NO2
EXP3 = 5.40X10-7 mol/L*s = k[0.200 mol/L]n[ mol/L]m
2.00 = (0.200/0.100)n
2.00 = (2.0)n n =1 reaction is first order for NH4+

10. Overall reaction order 1 + 1 = 2
Overall reaction is second order

12. BrO3- (aq) + 5 Br- (aq) + 6H+ (aq)  3Br2 (l) + 3H2O (l)
EXAMPLE: Given the results of the four experiments, determine the order of all three reactants, the over all order, and the value of the rate constant (k).
BrO3- (aq) + 5 Br- (aq) + 6H+ (aq)  3Br2 (l) + 3H2O (l)
Ex. Initial con Initial con Initial con Initial rate
BrO Br H+
X10-4
X10-3
X10-3
X10-3
Rate = k[BrO3- ]n [Br-]m [H+]p

13. Ex 2 = 1.6X10-3 = k(0.20)n (0.10)m (0.10)p
Ex 1 = 8.0X10-4 = k(0.10)n (0.10)m (0.10)p
2.0 = (0.20/0.10)n
2.0 = 2.0n
n = 1
Ex3 = 3.2X10-3 = k(0.20)n (0.20)m (0.10)p
2.0 = (0.20/0.10)m
2.0 = 2.0m
m = 1

14. Ex4 = 3.2X10-3 = k(0.10)n (0.10)m (0.20)p
Ex 1 = 8.0X10-4 = k(0.10)n (0.10)m (0.10)p
4.0 = (0.20/0.10)p
4.0 = 2.0p
p = 2
Rate of the reaction is first order for n and m, but second order for p
Rate law
Rate = k[BrO3- ] [Br-] [H+]2
overall reaction order is = = 4

15. To solve for k (use any of the experiments given)
8.0X10-4 mol/L*s = k(0.10 mol/L) (0.10 mol/L) (0.10 mol/L)2
8.0X10-4 mol/L*s = k (1.0 X10-4 mol4/L4)
1.0 X10-4 mol4/L X10-4 mol4/L4
8.0X10-4 mol/L*s = k
1.0 X10-4 mol4/L4
8.0 L3/mol3*s = k

17. Integrated Rate Law
FIRST ORDER
2 N2O5 (soln)  4 NO2 (soln) + O2 (g)
Rate law
Rate = D[N2O5] = k[N2O5]
D t
Since rate of reaction depends on the concentration of N2O5 to the first power, therefore if the concentration of N2O5 doubles than the production of NO2 ad O2 would also double

18. Integrated rate =
ln [N2O5] = -kt + ln[N2O5] (ln = natural log)
Rate law = D[A] = k[A] Dt
Plotting natural log of concentration vs. time ALWAYS gives a straight line (first order)
Equation is in the form of y=mx+b, where m= slope and b= intercept
ln [N2O5] = -kt + ln[N2O5]0
y = ln [N2O5] x = t m = -k b = ln[N2O5]0

20. Half Life of First Order Reaction
Ln ([A]0 / [A] ) = kt [A] = [A]0 2
Combine them:
ln [A] = kt1/2
[A]0/2
Ln(2) = kt1/2

21. Example: A certain first order reaction has a half life of 20
Example: A certain first order reaction has a half life of 20.0 minutes.
a) calculate the rate constant (k) for the reaction
b) how much time is required for the reaction to
be 75% complete
Using exp pg. 567
k = = = X10-2 min-1
t1/
b) Ln ([A]0 / [A] ) = kt
[A]0 / [A] = 1 / .25 = 4.0

22. ln (4.0) = kt
ln (4.0) = X t
min
ln (4.0)
3.47X = t
40 min = t

24. SECOND ORDER
Rate = -D[A] = k[A]2 Dt
Integrated second
1 = kt + 1
[A] [A]0
* 1/[A] vs time will produce a straight line with slope equal to k

26. Half life for second order
Half life = [A] = [A]0/2
= kt
[A]0/ [A]0
= kt1/2
[A] [A]0
1 = kt1/2 therefore t1/2 =
[A] k[A]0

27. Determine Rate Laws
Butadiene reacts to forms its dimer according to the equation
2 C4H6 (g)  C8H12 (g)
The following data was collected for this reaction at a given temperature:
[C4H6] (mol/L) Time (+- 1 s)

28. Is the reaction first or second order?
What is the value of the rate constant (k) for the reaction?
What is the half life for the reaction under the initial conditions of this experiment?
SOLUTION:
Must see whether the plot of ln[C4H6] vs time is a straight line (1st order) or plot of 1/[C4H6] vs time is a straight line (2nd order)

29. T (s) 1/[C4H6] ln[C4H6]

30. use points t=0 and t =6200 to find the rate constant (k)
Therefore the reaction is SECOND ORDER so that the rate can be written as:
Rate = D [C4H6] = k[C4H6]2
D t
Slope = k because y= mx +b, y = 1/[C4H6] and x = t therefore the slope = Dy/Dx = D 1
[C4H6] Dt
use points t=0 and t =6200 to find the rate constant (k)
for the reaction
k = slope = ( ) L/mol = 381 L/mol =
(6200-0) s s
6.14 X 10-2 L/mol*s

31. c) Expression of half life of second order reaction is:
k[A]0
k = 6.14 X10-2 L/mol*s
[A]0 = [C4H6]0 = M (concentration of t=0)
T ½ = =
(6.14 X10-2 L/mol*s) ( mol/L)
1.63 X 10-3 s so the initial concentration
C4H6 of is halved in 1630 s

33. ZERO ORDER Rate Law Rate = k[A]0 = k(1) = k
Rate is constant, doesn’t change with concentration as a first and second order reaction
Integrated rate law
[A] = -kt + [A]0
* Plotting [A] vs. t gives a straight line of slope -k

35. Half Life can be obtained from integrated rate law
[A] = [A]0/2 when t = t1/2 so 2
[A]0 = - kt1/2 + [A]0 or kt1/2 = [A]0 2k
solving for t1/2 gives
t1/2 = [A]0
* Zero order reactions often occur when a substance such as a metal surface or an enzyme is required for the reaction to occur for example

36. 2 N2O (g)  2 N2 (g) + O2 (g)
Occurs on a hot platinum surface
When platinum surface is completely covered with N2O, and increase in concentration of N2O has NO effect on the rate, since only N2O molecules on surface can react therefor rate is a constant

37. Integrated rate laws of reactions with more than one reactant
BrO3- (aq) + 5Br-(aq) + 6H+(aq)  3Br2(l) + 3H2O(l)
Rate law is by experimental data:
rate = D[BrO3-] = k[BrO3-][Br-][H+]2 Dt
*run under conditions where Br- ion and H+ ions concentrations are much larger than BrO3- ion concentration, assume that
[Br-] = [Br-]0 and [H+] = [H+]0
Then the rate law can be written as:

38. Rate = k[Br-]0[H+]02[BrO3-] = k’[BrO3-]
Where [Br-]0 and [H+]]0 are constants
k’ = k[Br-]0[H+]02
Rate law:
rate = k’[BrO3-]
This is a pseudo-first order
Rate law – because law is obtained by simplifying a more complicated one

39. To solve for k rearrange to:
k = k’
[Br-]0[H+]02

41. REACTION MECHANISMS
Reactions mechanisms – reaction occur by a series of steps
Consider:
NO2(g) + CO(g)  NO(g) + CO2(g)
Rate law
rate = k[NO2]2

42. The mechanism:
Where k1 and k2 are rate constants of individual reactions
NO3 is an intermediate – a species that is neither a reactant nor a product but is formed and consumed during the reaction sequence
elementary step – each of the two reactions (rate law can be written from its molecularity)

43. Molecularity – the number of species that must collide to produce the reaction indicated by the step
Unimolecular step – reaction with only one step
Bimolecular – reaction with two steps
Ternomolecular – reaction with 3 steps (rare)

44. Reaction mechanism can now be define more precisely and its series of elementary steps must meet two requirements:
1. sum of the elementary steps must give the
overall balanced equation from the reaction
2. mechanism must agree with the
experimentally determined rate law

45. First note the sum of two steps gives the overall balanced equation
Target equation: NO2(g) + CO(g)  NO(g) + CO2(g)
NO2(g) + NO2(g)  NO3(g) + NO(g)
NO2(g) + CO(g)  NO2(g) + CO2(g)
NO2(g) + NO2(g) + NO3(g) + CO(g)  NO3(g) + NO(g) + NO2(g) + CO2(g)
Overall reaction: NO2(g) + CO(g)  NO(g) + CO2(g)
First requirement met

46. NO2(g) + NO2(g)  NO3(g) + NO(g) slow (rate determining step)
Rate determining step – in multistep reactions often one step is much slower than the others
Reactants become products only as fast as they can get through slowest step
Overall reaction can not be faster than slowest (rate determining step)
Assuming the first step is the rate determining and the second step is relatively fast
NO2(g) + NO2(g)  NO3(g) + NO(g) slow (rate determining step)
NO2(g) + CO(g)  NO2(g) + CO2(g) fast

47. Therefore the rate of CO2 production is controlled by the rate of formation of NO3 in the first step
The bimolecular first step has a rate of:
rate of formation = D[NO3] = k1[NO2]2
of NO Dt
Since the overall reaction rate can be no faster than slowest step
overall rate = k1[NO2]2

48. SOLVE REACTION MECHANISM
The balanced equation for a reaction of the gases is:
2 NO2(g) + F2(g)  2 NO2F(g)
The experimentally determined rate law is:
rate = k[NO2][F2]
Suggested mechanism for reaction is:
NO2 + F NO2F + F slow
F + NO NO2F fast
Is this an acceptable mechanism? Does it satisfy two requirements? k1 k2

49. SOLUTION:
First step – the sum of the steps should give the balance equation
NO2 + F NO2F + F
F + NO NO2F
NO2 + F2 + F + NO2  NO2F + F + NO2F
Overall reaction: 2 NO2 + F2  2 NO2F
Step met

50. Second step: mechanism must agree experimentally determined rate law
rate = k1[NO2][F2]
Proposed mechanism is acceptable

52. MODEL for CHEMICAL KINETICS
Experimentally shown that virtually all rate constants show an experimental increase with absolute temperature
Collision model – model used to account for observed characteristics of reaction rates (build around central idea that molecules must collide to react)
Kinetic molecular theory – predicts that an increase in temperature raises molecular velocities and so an increase in the frequency of collisions between molecules

53. Rate of reaction is much smaller than calculated collision frequency in a collection of gas particles which means that only a small fraction of the collisions produces a reaction
Activation energy (proposed by Svante Arrhenius) thresh-hold energy that must be overcome to produce a chemical reaction

54. 2 BrNO(g)  2 NO(g) + Br2(g)
Two Br-N bonds must be broken and one Br-Br must be formed
Br-N bond requires considerable energy (243 KJ/mol) which must come from somewhere
Energy comes from kinetic energies possessed by reacting molecules before collision
Kinetic energy is change to potential enegy as molecules are distorted during collision to break bonds and re-arrange the atoms into product molecules

56. Activated complex (transition state) – arrangement of atoms found at the top of the potential energy “hill,” or barrier
Conversion of BrNO to NO and Br2 is exothermic, cause the products have a lower potential energy than the reactants
DE has no effect on rate of the reaction, it rather depends on the size of the activation energy (Ea)
Only collisions with energy greater than activation energy are able to react (get over barrier)

57. # of collisions with = (total # of collisions)e –Ea/RT
activation energy
Where Ea = activation energy
R = universal gas constant
T = Kelvin temperature
The e –Ea/RT factor represent the fraction of collision with energy Ea or greater at temperature T

58. Experiments show that a considerably smaller than the rate collision with enough energy to surmount the barrier still do not produce a reaction. Why?
Molecular orientation – during collisions, some collision orientations can lead to a reaction and others can not, therefore collision factor to allow for collision with non-productive molecular orientation

59. Two requirements must be satisfied for reactants to collide successfully (to re-arrange to form products):
Collision must involve enough energy to produce reaction: collision energy must equal or exceed activation energy
relative orientation must allow formation of any new bonds necessary to produce products
Can represent the rate constant as :
k = zpe-Ea/RT
Where z = collision frequency
p = (called steric factor) always > 1 and reflects the
fraction of collisions with effective orientation
e-Ea/RT = represent fraction of collisions with sufficient
energy to produce reaction

60. Often written as (Arrhenius equation) k = Ae-Ea/RT
Where A replaces zp called the frequency factor
Ln (k) = -Ea ln(A)
R T
Iinear equation similar to y = mx + b
Where
Y = ln(k) m = -Ea/R = slope x = 1/T b = ln(A) = intercept
* Most rate constants obey the Arrhemius equation to a good approximation indicated that the collision model for chemical reactions is reasonable

61. DETERMINE ACTIVATION ENERGY I Reaction: 2 N2O5(g)  4NO2(g) + O2(g)
Was studied at several temperatures and the following vales for k were obtained:
K (s-1) T (oC)
2.0 X
7.3 X
2.7 X
9.1 X
2.9 X
Calculate the Ea for this reaction.

62. SOLUTION:
Obtains value for Ea, construct a plot of ln(k) versus 1/T. First calculate values of ln(k) and 1/T
T (oC) T(k) /T (k) k (s-1) ln (k)
X X
X X
X X
X X
X X
Plot ln(k) vs 1/T where the slope is
D ln(k) = -1.2X10-4 K
D (1/T)
Value of Ea can be solved by
slope = -Ea R

63. Ea = -R (slope) = -(8.3145 J/K*mol) (-1.2X104 K) =
1.0X105 J/mol
Value for activation energy is 1.0X105 J/mol
Alternate equation to find Ea
ln k2 = Ea
k R T T2

65. DETERMINE ACTIVATION ENERGY II A gas phase reaction
CH4(g) + 2S2(g)  CS2(g) + 2H2S(g)
At 550oC the rate constant is 1.1 L/mol*s and at 6250C the rate constant is 6.4 L/mol*s. Using these values calculate Ea for the reaction.
SOLUTION:
K (L/mol*s) T(oC) T (K)
1.1 = k = T1
6.4 = k = T2

66. ln k2 = Ea
k R T T2
ln = Ea
J/k*mol K K
Solve for Ea
Ea = ( J/K*mol) ln 6.4
1.1
823 K K
1.4 X105 J/mol

68. Catalysis
Sometimes increasing the temperature to accelerate a reaction is not feasible (EX: living cells)
Enzymes – substances in the body which increase rates of reactions
Catalyst – substances that speed up the reaction without being consumed itself

69. * reaction proceeds rapidly at a relative low
temperature
* works by providing a new pathway for the
reaction, one at a lower activation energy,
a much larger faction of collisions are
effective at a given temperature and
reaction rate is increased
* they lower the activation energy , Ea, but
does not affect the energy difference DE
between the products and the reactants

71. Classification of Catalysts
Homogeneous catalyst – one that is present in the same phase as the reacting molecules
* found in gas and liquid phase
* example: catalytic behavior of nitric oxide
towards ozone (O3)
Heterogenous catalyst – one that is existing in a different phase (usually a solid)

72. Heterogenous Catalyst
Most often involves gaseous reactants being adsorbed on surface of the solid catalyst
adsorption – collection of one substance on the
surface of another substance
absorption – penetration of one substance into
another
Examples: catalysis occur in the hydration of unsaturated hydrocarbons where the C=C are converted to C-C bonds

74. Heterogenous catalyst involves 4 steps
Adsorption and activation of the reactants
Migration of the adsorbed reactant on surface
Reaction of adsorbed substances
Escape, or desorption of the products