1. 12/01/2011
Course Title: General and Inorganic
Chemistry
TOPIC 3: CHEMICAL KINETICS
Lecturer and contacts
Mr. Vincent Madadi
Department of Chemistry, University of Nairobi
P. O. Box 30197-00100,
Nairobi, Kenya
Chemistry Dept. Rm 114
Tel: 4446138 ext 2185
Email: vmadadi@uonbi.ac.ke, madadivin2002@yahoo.com
Website: http://www.uonbi.ac.ke/staff/vmadadi
1/12/2011 mov 1
Introduction
• Kinetics is the study of rates of chemical reactions and
the mechanisms by which they occur.
• The reaction rate is the increase in concentration of a
product per unit time or decrease in concentration of a
reactant per unit time.
• A reaction mechanism is the series of molecular steps
by which a reaction occurs.
1/12/2011 mov 2
1

2. 12/01/2011
Thermodynamic vs kinetics of reaction
• Thermodynamics determines if a reaction can occur.
Kinetics determines how quickly a reaction occurs
• Some reactions that are thermodynamically feasible
are kinetically so slow as to be imperceptible
• The Rate of a Reaction
Cdiamond + O2(g)→ CO2(g) ΔG°= -396kJ
Very Slow
H+(aq) + OH-(aq) → H2O(l) ΔG°= -79kJ
1/12/2011 Very Fastmov 3
Rate of chemical reaction
• 2 N2O5 → 4 NO2 + O2
• 2 moles of N2O5 disappear for every 4 moles of NO2
and 1 mole of O2 formed.
• Reaction rates are the rates at which reactants
disappear or products appear.
• This movie is an illustration of a reaction rate.
1/12/2011 mov 4
2

3. 12/01/2011
Reaction rate
• Reaction rate is the change of concentration of a
reactant or product per unit time
aA + bB → cC + Dd
Reaction rate = ΔConcentration
Δtime
• Rate is expressed either as rate of appearance of product
or rate of disappearance of reactant
• E.g.
aA + bB → cC + dD
1/12/2011 mov 5
Reaction rate
2 NO2(g) → 2 NO(g) + O2(g)
1/12/2011 mov 6
3

4. 12/01/2011
Reaction rate
• Mathematically, the rate of a reaction can be written as:
• Square brackets [ ] are often used to express molarity
(i.e.[HCl] means Molarity of HCl)
• The relative rates of consumption of reactants and
formation of products depend on the reaction stoichiometry
1/12/2011 mov 7
Reaction rate
• The relative rates of consumption of reactants and formation
of products depend on the reaction stoichiometry
• For the reaction
2HBr (g) → H2 (g) + Br2 (g)
• two moles of HBr are consumed for every one mole of H2
which is formed
1/12/2011 mov 8
4

5. 12/01/2011
Reaction rate
• Experimental Rate Law: the rate of a reaction is
proportional to the product of the concentrations of
the reactants raised to some power.
• For a reaction
aA + bB → products,
the rate law is the equation
rate = k[A]x[B]y
• Relationships Between Rate and Concentration
1/12/2011 mov 9
Reaction rate
• x and y are the orders of the reaction in [A] and [B]
respectively
• The overall order of the reaction is x + y
• x and y are usually small integers, but may be zero,
negative, or fractions
• k is the specific rate constant
1/12/2011 mov 10
5

6. 12/01/2011
Rate constant k
• Units depend on overall reaction order
• Value does not change with concentration
• Value does not change with time
• Valid for a specific temperature
• Dependent on presence or absence of a catalyst
• Value must be determined experimentally
1/12/2011 mov 11
Factors that affect rate of reaction
• They are six key factors that affect rate of
reaction:
1) Nature of reactants and products
2) Concentration
3) Temperature
4) Catalyst
5) Surface area
6) Light radiation
1/12/2011 mov 12
6

7. 12/01/2011
Factors that affect rate of reaction
1. Nature of reactants and products
• Chemical reactions involve rearrangement of bonds: Bonds in
reactants are broken and new bonds are formed in products
• In organic or molecular reactions, large number of bonds are
broken in reactants and formed in products-hence reactions are
slow
• E.g. Hydrolysis of cane sugar:
C12H22O11 + H2O C6H12O6 + C6H12O6
Cane sugar glucose fructose
1/12/2011 mov 13
Factors that affect rate of reaction...
• Inorganic reactions involve ions, hence no bonds to be broken in
reactants . The reactions are faster.
• E.g.
Ag+(aq) + NO3-(aq) + Na+Cl-(aq) AgCl(aq) + NaNO3(aq)
• Concentration
• Based on the rate law of mass action, the rate of reaction is
directly proportional to the product of the concentration of the
reactants at a particular temperature
• For reaction aA + bB cC + dC
1/12/2011
Rate = K[A]a[B]b
mov 14
7

8. 12/01/2011
Factors that affect rate of reaction ...
3) Temperature
• In most reactions, the rate of reaction doubles with 10 K increase
in temperature
• But with 10 k increase in temperature, collision frequency (Z)
increases by a factor of 1.016
I.e. Z α T½
Z2/Z1 = (T2/T1)1/2 = (310/300)1/2 = 1.016
• However, the rate increases by almost 100%
• This is because increasing temperature by 10 K increases the
number of active molecules (molecules with E > Ea) which
increases the rate drastically mov
1/12/2011 15
Factors that affect rate of reaction ...
Effect of increase in temperature by 10 K
1/12/2011 mov 16
8

9. 12/01/2011
Factors that affect rate of reaction...
• 4) Catalyst
• A catalyst is a substance that alters the rate of reaction without
itself getting consumed
• There are two types of catalysts:
• Positive catalyst: It increases the rate of reaction e.g.
MnO2
2KClO3 2KCl + 3 O2
300 ⁰C
• MnO2 acts as a positive catalyst since uncatalysed reaction
takes place at 700 ⁰C and is slower
1/12/2011 mov 17
Factors that affect rate of reaction...
• Negative catalyst: This is the catalyst which retards
the rate of reaction e.g. Oxidation of chloroform is
retarded by ethanol
1% ethanol
4CHCl3 + 3 O2 4COCl2 + 2Cl2 + H2O
Mechanism of catalysis
• A catalyst alters the rate of reaction by providing a
path with lower or higher activation energy
1/12/2011 mov 18
9

10. 12/01/2011
Factors that affect rate of reaction...
5) Surface area
• The rate of homogeneous reaction is influenced by
the surface area of the reactants
• Particle size decreases, surface area increases for the
same mass because of creation of new surfaces.
Hence the rate of reaction increases
• E.g. Powdered zinc reacts faster with dilute HCl than a
block of zinc
1/12/2011 mov 19
Factors that affect rate of reaction...
6) Light radiation
• The rate of photochemical reaction is affected by light
radiation
• Photons (E = hv) supply the necessary energy of
activation to the reactant molecules to form products
λ = 400 nm
• E.g. H2 + Cl2 2HCl
1/12/2011 mov 20
10

11. 12/01/2011
Order of reaction
• This is the sum of the powers of the concentration
term in an experimentally established rate law
• Example
aA + bB → cC + dD
• The theoretical rate law is:
Rate = K[A]a[B]b
• But experimentally determined rate expression is:
Rate = K[A]α[B]β
1/12/2011 mov 21
Order of reaction cont.
• Where α and β are the actual moles of A and B and
may not be necessarily be equal to “a” and “b”
• Overall order of reaction is given by the sum of the
individual orders i.e.
• Overall order = α + β
• Order of reaction can be zero, +ve, -ve or fraction.
But higher orders of reaction are rare
1/12/2011 mov 22
11

12. 12/01/2011
Order of reaction cont.
• In complex reactions the order of reaction is
determined by the slowest step of the reaction which
is also called the rate determining step
1/12/2011 mov 23
Molecularity of reaction
• This is the total number of molecules in the step
leading to chemical reaction
• For any reaction, the least number of molecules is one
• Thus molecularity cannot be zero or fraction
• Types of molecularity:
1) Unimolecular reactions:
•1/12/2011
Reactions involving one molecule of the reactants
mov 24
12

13. 12/01/2011
Molecularity of reaction
• Example:
PCl5 PCl3 + Cl2
2) Bimolecular reactions
CH3COOC2H5 + H2O CH3COOH + C2H5OH
3) Thermolecular reactions
2NO + O2 2NO2
1/12/2011 mov 25
Pseudo uni-molecular reactions
• The reaction in which the order is one but molecularity
is 2
• E.g.
CH3COOC2H5 + H2O CH3COOH + C2H5OH
• Order is one because water is in excess, hence
concentration does not change during the reaction
• Thus, the rate is independent of the conc. Of water but
only dependent on conc. of ester
1/12/2011 mov 26
13

14. 12/01/2011
Rate law
• Exercise
1) The rate of reaction 2NO + O2 2NO2
follows the rate law Rate = K[NO]2[O2]
If K = 2x 10-6 mol-1L2, What is the rate of the reaction
when [NO] = 0.04 molL-1 and [O2] = 0.2 molL-1
[ Ans = 6.4 x 10-10 molL-1s-1]
1/12/2011 mov 27
Rate law
2. The rate of the reaction
2NO + O2 2NO2
is doubled when the concentration of O2 is doubled,
but increases by factor of 8 when the concentration
of both reactants is doubled.
Determine the:
a) Order of reaction wrt NO and O2
b) Overall order of reaction
1/12/2011 mov 28
14

15. 12/01/2011
Activation energy
• This is the additional amount of energy that reactant
molecules must acquire in order to react and form products
• It is defined as the amount of energy that the reactants must
absorb to pass over the activation energy barrier to form
products
• Activation energy diagram
Ea
Reactant
ER
E Th
Product
1/12/2011 mov 29
Activation Energy cont.
• E = Eth –ER
= Threshold Energy –Energy possessed by molecules
• The activation energy is related to the rate constant K and
temperature T according to Arrhenius equation
K = Ae-Ea/RT
Where A = Frequency factor
K = rate constant
T = temperature in Kelvin
R = Gas constant
E = Activation energy
e = base of natural logarithm
1/12/2011 mov 30
15

16. 12/01/2011
Derivation of Arrhenius equation
• Arrhenius equation is derived from the Vant’
Hoffs reacton isochore
dlnKc/dT = ΔE/RT2
and the reaction dynamic equilibrium
A+B=C+D
• Kc for the reaction,
Kc = Kf/Kb = [C][D]/[A][B]
Kf[A][B] = Kb[C][D]
• Hence, Kc = Kf/Kb
1/12/2011 mov 31
Derivation of Arrhenius equation cont.
• If ΔE is written as Ef –Eb, then from equation 1 and 2
then,
• dlnKf/dT – dlnKb/dT = Ef/RT2 – Eb/RT2
• or dlnk/dT = E/RT2
dlnk = EdT/RT2
• Integrating the equation gives,
lnk = -E/RT + C
1/12/2011 mov 32
16

17. 12/01/2011
Derivation of Arrhenius equation cont.
• Provided E is a constant, the equation can be written
as:
k = Ae-E/RT
• This is the Arrhenius equation
• Application of activation energy
1) To determine activation energy
1/12/2011 mov 33
Determination of activation Energy
• There are two methods for determining of activation
energy Ea
1) Graphical method
2) Rate constant method
1) Graphical method
• From Arrhenius equation, K = Ae-E/RT
• Applying lo to both sides of the equation,
1/12/2011 mov 34
17

18. 12/01/2011
Determination of activation Energy cont.
• lnK = lnA – Ea/RT
• >2.303logK = 2.303logA – Ea/RT
• >logK = logA – Ea/2.303RT
• >logK = -Ea/2.303RT + logA
Ξ y = mx + C
• Plotting logK against 1/T give a straight line and Ea can
be calculated
1/12/2011 mov 35
Graphical Determination of activation
energy
• Graph Log A
Slope = -Ea/2.303RT
Log K
Ea = slope x -2.303 R
1/12/2011 mov 36
18

19. 12/01/2011
Graphical Determination of activation
energy...
• Slope = -Ea/2.303xR
• Ea = -2.303 x R x Slope
• Log A = Intercept
A = Antilog (Intercept)
1/12/2011 mov 37
2. Determination of activation energy
from rate constant method
• From K = Ae-Ea/RT
• lnK = lnA – Ea/RT
• Let at Temperature T1 and rate constant K1; and at T2
and rate constant K2
• For small change in temperature, the change Ea and
A do not significantly change
1/12/2011 mov 38
19

20. 12/01/2011
Determination of activation energy from rate
constant method ...
• Thus,
lnK1 = lnA – Ea/RT1 3
lnK2 = lnA – Ea/RT2 4
• Subtracting 3 from 4,
• LnK2 – lnk1 = Ea/RT1 – Ea/RT2 5
• logK2/K1 = Ea [1/T1 – 1/T2]
2.303
• logK2/K1 = Ea [(T2-T1)/T1T2]
2.303R
1/12/2011 mov 39
Determination of activation energy from rate
constant method...
• Log K2/K1 = Ea [(T2-T1)/T1T2] Eq. 6
2.303R
Since K1, K2, T1 and T2 are known, Ea can be calculated
from the equation 6
1/12/2011 mov 40
20

21. 12/01/2011
Rate Laws
• A rate law shows the relationship between the reaction rate and
the concentrations of reactants. Exponents tell the order of the
reaction with respect to each reactant.
• This reaction is
First-order in [NH4+]
First-order in [NO2−]
• The overall reaction order can be found by adding the
exponents on the reactants in the rate law.
• This reaction is second-order overall.
• For gas-phase reactants use PA instead of [A]
1/12/2011 mov 41
Integrated Rate Laws
• These are expressions which relate the concentration of
the reactants with time
• Application:
1) Used to predict amount of reactant or product at a
particular time
2) Predict how long the reaction will take
3) To predict when a toxic chemical can be disposed
• Can be classified into Zero, first, second and third
order reactions
1/12/2011 mov 42
21

22. 12/01/2011
Integrated Rate Laws cont.
1) Zero order reactions
• This is a reaction where the rate of reaction is
independent of the concentration of the reactants
• Derivation:
• Let A Product
• Initial conc. (mol/l) t = 0, a 0
• At time t = t, a-x x
• Where x is the concentration of the reactant (A)
undergoing decomposition
1/12/2011 mov 43
Zero order reactions cont.
• Rate expression for zero order reaction:
• dx/dt α (a-x)0 ........................................................Eq.1
• Or dx/dt = Ko(a-x)o = ko .......................................Eq. 2
• K0 is the rate constant for zero order reaction
• Integrating the equation,
• ∫dx = ∫Kodt = Ko ∫dt
x = Kot + C, .............................................................Eq. 3
• Where C is the integration constant
1/12/2011 mov 44
22

23. 12/01/2011
Zero order reactions cont.
• But when t = 0 and x = 0
• Thus,
0 = Ko x 0 + C .............................................Eq. 4
• Hence C = 0
• Substituting the value of C into equation 3 gives,
x = Kot
• Therefore, Ko = x/t ...........................................Eq. 5
• Eq.5 is the rate constant equation for zero order
reaction
1/12/2011 mov 45
Zero order reactions cont.
• Units for rate constant:
• Ko = x/t = conc/time = molL-1/s = molL-1s-1
• Half-life –This is the time duration in which half the
concentration of reactants is transformed into products
• Thus at t = t½, x = a/2,
• Substituting these values into equation 5
K0 = a/2t½ t ½ = a/2k0 t½ α a
• This shows that half-life of a zero order reaction is
directly proportional to the initial concentration
1/12/2011 mov 46
23

24. 12/01/2011
Zero order reactions cont.
• Graph for zero order reaction
The rate of reaction is independent
of the concentration of reactants
Rate
Concentration of reactants
1/12/2011 mov 47
First Order Reactions
• First Order Reactions
• These are reactions where the sum of the powers of
concentration of the exponential term in an
experimentally established rate law is one
• Thus the rate of reaction is dependent on the single
power of the concentration term of the reactants
• Example:
• For the reaction, A Product
1/12/2011 mov 48
24

25. 12/01/2011
First Order Reactions cont.
• Initial conc (molL-1) at t=0, a o ......Eq. 1
• Conc (molL-1) at t=t a-x x .......Eq. 2
• Where x moles of A have decomposed into products in
time “t”
• Deferential, -d(a-x)/dt or dx/dt α(a-x) ...............EQ. 3
• Or dx/dt = k1(a-x) ...............................................Eq. 4
• Where, K1 is the rate constant for first order reaction.
1/12/2011 mov 49
First Order Reactions cont.
• Rearrangement,
• dx/(a-x) = k1dt ....................................................Eq. 5
• Integrating, [identity ∫dx/x = ln x]
∫dx/(a-x) = k1 ∫dt
-ln(a-x) = k1t + C ..............................................Eq. 6
• Where C = constant of integration
• But at t=0, x = o
• Substituting the values into equation 6,
1/12/2011 mov 50
25

26. 12/01/2011
First Order Reactions cont.
• -lna = C .............................................................Eq. 7
• Substitute the value of C into equation c,
• -ln(a-x)= k1t –lna ................................................Eq. 8
• Rearranging
• k1t = lna – ln(a-x)
• K1 = ln[a/(a-x)]x 1/t
• Change ln to log10 [lnx = 2.303logx]
• K1 = (2.303/t)log(a/a-x) .......................................Eq.8
• This is the expression for rate constant for first order
reaction
1/12/2011 mov 51
First Order Reactions cont.
• Unit of the rate constant
• From Eq. 8
• K1 = (2.303/t)log(a/a-x) = conc/(time xconc)
k1 = 1/time = s-1 ...................................................................Eq. 9
Graph for first order reaction
• From equation 8
• -ln(a-x)= k1t –lna
• Multiply though by -1
•1/12/2011
ln(a-x)= -k1t + lna mov 52
26

27. 12/01/2011
First Order Reactions cont.
• Introduce log10
• 2.303log(a-x) = -k1t + 2.303 x log(a)
• Or log(a-x) = -k1 x t + log(a) ...........................Eq. 10
2.303
Ξ y =mx +c
Log(a)
Slope = -k1/2.303
Log(a-x)
1/12/2011 t mov 53
Second order reactions cont.
• From the graph,
• K1 = -2.303 x slope
• Half-life of first order reaction
• From Eq. 8, K1 = 2.303 Log (a/a-x)
t
• At t = t ½ , x = a/2
• Substituting the values into the equation
• K1 = 2.303 Log (a/a - a/2) = 2.303 log 2
t½ t½
1/12/2011 mov 54
27

28. 12/01/2011
First Order Reactions cont.
• Making t ½ the subject,
• t½ = 2.303 log 2 = 2.303 x 0.301 = 0.693
k1 k1 k1
• Thus, t½ = 0.693/k1
• This means that for 1st order reactions, half-life is
independent of the initial concentration of reactants
1/12/2011 mov 55
3. Second order reaction
• These are reactions in which the sum of the powers of
the of the concentration term in an experimentally
established rate law is 2
• There are two cases of second order reaction
• Case 1: 2A Product
• Thus, Rate law = dx/dt = K[A]2 Order = 2
• Case 2: A+B Product
• Rate law = dx/dt = k[A][B] order = 2
1/12/2011 mov 56
28

29. 12/01/2011
Second order reaction cont.
• Case 1: Occurs when the concentration of both reactants
is the same
• Case 2: Occurs when the concentration of both reactants
is NOT the same
• 1) Case 1: Concentration of both reactants is the same
• Let the reaction, A + A Product
• Initial conc. t=0 a a 0 ..Eq. 1
• Conc. At t = t a-x a-x x ...Eq. 2
1/12/2011 mov 57
Second order reaction cont.
• Conc. is in mol L-1
• Thus, x molL-1 of the reactant A decomposes in time t
• Deferential,
• dx/dt α (a-x)(a-x) or dx/dt α (a-x)2 ..................Eq. 3
• Introduce rate constant k2
• dx/dt = k2 (a-x)2 ......................................Eq. 4
• Where k2 is the rateconstant for the second order
reaction
1/12/2011 mov 58
29

30. 12/01/2011
Second order reaction cont.
• Rearranging,
• dx/ (a-x)2 = k2 dt .....................................................Eq. 5
• Integrating,
• ∫dx/(a-x)2 = k2∫dt note [∫xndx = xn+1/n+1]
• Hence,
• 1/(a-x) = k2t + C ...................................................Eq. 6
• Where C = constant of integration
• But when t= 0, x= 0
1/12/2011 mov 59
Second order reaction cont.
• Substituting the values into the equation,
• 1/a = C
• Substitute the value of C into equation 6
• 1/(a-x) = k2t + 1/a .............................Eq. 7
• Thus, K2t = 1/(a-x) - 1/a
• Hence, K2 = (1/at) x [x/(a-x)] ...................Eq. 8
• Eq.8 is the expression for rate constant for second order reaction
where initial conc. of the reactants is the same
1/12/2011 mov 60
30

31. 12/01/2011
Second order reaction cont.
• Case 2: When concentration of both reactants is
different
• Let the reaction, A + B Product
• Initial conc. t=0 a b 0 ..Eq. 1
• Conc. At t = t a-x b-x x ...Eq. 2
• Where x mol/L of A and B decomposed in time t to
form product
1/12/2011 mov 61
Second order reaction cont.
• Differential
• dx/dt α (a-x)(b-x)
• dx/dt = K2(a-x)(b-x)
• Where K2 is the rate constant for second order
reaction
• Separating the variables,
• dx/[(a-x)(b-x)] = K2 dt ..........................................Eq. 3
1/12/2011 mov 62
31

32. 12/01/2011
Second order reaction cont.
• Integration by parts
• ∫ dx = ∫ A dx + ∫B dx = ∫k2dt ...Eq.4
(a-x)(b-x) (a-x) (b-x)
• 1 = A + B ......................Eq.5
(a-x)(b-x) (a-x) (b-x)
• Multiply through by (a-x)(b-x)
• 1 = A(b-x) + B(a-x) ..............................................Eq. 6
1/12/2011 mov 63
Second order reaction cont.
• When x = a, 1 = A(b-a)
Thus, A = 1/(b-a)
• When x = b, 1 = B(a-b)
Thus, B = 1/(a-b)
Or
• ∫ dx = 1 ∫dx + 1 ∫dx = k2 ∫ dt ..Eq. 7
(a-x)(b-x) (b-a) (a-x) (a-b) (b-x)
• Factorise the middle term by 1/(a-b)
• = ∫ dx = 1 [ ∫dx - ∫dx ] = k2 ∫ dt ...Eq. 8
(a-x)(b-x) (a-b) (b-x) (a-x)
1/12/2011 mov 64
32

33. 12/01/2011
Second order reaction cont.
• ∫ dx = 1 [ -ln(b-x) + ln(a-x)] = k2 ∫ dt ..Eq. 9
(a-x)(b-x) (a-b)
Therefore,
• = 1 [ ln (a-x) ] = k2t + C ......................Eq. 10
(a-b) (b-x)
When t = 0, x = 0,
Substituting into the equation 14,
1/12/2011 mov 65
Second order reaction cont.
• = 1 [ ln (a ] = C .....................................Eq. 11
(a-b) (b)
• Substitute the value of C in eq. 15 into eq. 14
• = 1 [ ln (a-x) ] = k2t + 1 [ ln (a ]..........Eq. 12
(a-b) (b-x) (a-b) (b)
Thus
• K2 = 1 [ ln b(a-x) ] ............................Eq. 13
(a-b)t a(b-x)
1/12/2011 mov 66
33

34. 12/01/2011
Second order reaction cont.
• Applying log10
• K2 = 2.303 log b(a-x) .............................................Eq. 14
(a-b)t a(b-x)
Characteristics:
Unit : K2 = 1 x x
at (a-x)
= 1/(conc x time)
K2 = [conc. X time]-1 .............................................................Eq. 15
1/12/2011 mov 67
Second order reaction cont.
• K2 = [mol-1s]-1 = mol-1Ls-1
• When one of the reactants is in excess,
• i.e. Let a>> b
• a-b = a, a-x =a
• Thus b and x can be neglected in comparison to a
• Hence, K2 = 2.303 logb(a-x) ................................Eq. 16
(a-b)t a(b-x)
• k2 = = 2.303 log(b/b-x)) .....................................Eq. 17
at
1/12/2011 mov 68
34

35. 12/01/2011
Second order reaction cont.
• k2a = = 2.303 log(b/b-x)) ................................Eq. 18
t
• k’ = = 2.303 log(b/b-x)) .................................Eq. 20
t
Where K2a = k’ = rate constant for the first order
reaction
1/12/2011 mov 69
Second order reaction cont.
• Graph for second order reactions
1 lnb(a-x)
(a-b) a(b-x)
t
1/12/2011 mov 70
35

36. 12/01/2011
Second order reaction cont.
• Half life
• From equation 8, case 1
• K2 = 1 x x
at (a-x)
• t = t 1/2 , x = a/2
• K2 = 1 x a/2
at ½ (a – a/2)
• t ½ = 1/k2.a .....................................................Eq. 21
1/12/2011 mov 71
Second order reaction cont.
• Thus t ½ is directly proportional to 1/a
1/12/2011 mov 72
36