Chapter 14 - Chemical Kinetics

So far, we have only been concerned with the fact that a reaction does occur, and not how long it takes for
the reaction to occur.

2A  B combination reaction
Practical examples of how long!
H2(g) + ½ O2  H2O (l)  Very Slow
N2O(g)  N2(g) ½ O2(g)  SLOW
combustion  fast process
TNT exploding  very fast reaction
1.
2.
3.
4.

chemical kinetics. 
concerned with determining the speed or rate at which a reaction occurs.
How is the reaction rate affected by
temperature?
states of reactants?
amount of reactants?
catalyst?
surface area of the reacting species?
a)
b)
c)
d)
e)
Example
Br2(aq) + HCOOH(aq)  2 Br-(aq) + 2 H+(aq) + CO2(g)

Br2 (aq) is orange-red in aqueous solution. The concentration of bromine can be followed with a
spectrophotometer.

Define  average rate = -

NOTE: “-“ placed before equation since 1[Br2] . Reaction rates are always positive.
[ Br 2 ] fin - [ Br 2 ]in
 [ Br 2 ]
=t
t f - ti
look at the following calculations of average rate for this reaction. (Figure 1).
0.0 - 50.0s
-
[Br 2]
[0.0101] - [0.0120]
=
t
50.0s - 0.0s
moles
= 3.80 x 10-5 M / S
L
[Br 2]
[0.00710] - [0.00846]
100.0 - 150.0s =
t
150.0s - 100.0s
5
 2.72 x10 M / s
remember [
] 
1
200.0 - 250.0s
-
[Br 2]
[0.00420] - [0.00500]
=
t
300.0s - 250.0s
5
 160
. x10 M / s

the average rate is the sum of the rates over the specified time period. Note - each time we used different
50.0 s intervals, the average rate changed!

we want to use the instantaneous rate, i.e. the rate that is obtained over a very small time interval.

plot the instantaneous rate vs. the [Br2] at that time interval and see what we get. (Figure 2).

note that the rate (instantaneous) is [Br2]
i.e.
rate (t) = k[Br2 (t)]
k = 3.50 x 10-3/s (note units)

the constant k is called the rate constant

The rate of the reaction is dependent on reactant concentration but the RATE CONSTANT IS
INDEPENDENT OF THE REACTANT CONCENTRATION.
Reaction Rates and Reaction Stoichiometry
look at the reaction
O3(g) + NO(g)  NO2(g) + O2(g)
rate = 
O3
 [ NO ]
[ NO2 ]
[O ]
= = +
= + 2
t
t
t
t
another example.
2NOCl(g)  2NO(g) + Cl2 (g)
rate = -
 [NOCl(g)]
1  [NO(g)]  [Cl2 (g)]
=+
=
2t
2
t
t
WHY? 2 moles of NOCl disappear for every 1 mole Cl2 formed.

for a general reaction
aA + bB  cC + dD
2
rate =
-1 [A]
-1 [B]
1 [C]
1 [D]
=
=
=
a t
b t
C t
d t

Why do we define our rate in this way?

removes ambiguity in the measurement of reaction rates in that we now obtain a single rate for the entire
equation, not just for the change in a single reactant or product.
Notes for Determining Rate Laws
(1)
When we determine rate laws, we will look at the changes in the initial rates as a function of changing
concentration of reactants. The initial rate is the rate we obtain immediately after mixing the reactant (as
close to t = 0 as possible!)
(2)
Generally, initial rates are larger and are more easily measured. Finally, side reactions are usually minimal
initially.
(3)
We will use the isolation method. The concentration of all but one reactant is fixed and we study its effect
on the measured initial rate. We then apply the same procedure to all other reactants.
O3 (g) + NO(g)  NO2 (g) + O2 (g)
x
y
rate = k[O3 (g) ] [NO(g) ]

rate law  relates rate of the reaction to the form to concentration of reactants and rate constant.

For a general reaction
aA + bB + cC  dD + eE
rate = k[A]x[B]y[C]z


THE ONLY WAY THAT WE CAN DETERMINE THE SUPERSCRIPTS (k, x, y, and z) FOR NONELEMENTARY CHEMICAL REACTION IS BY EXPERIMENTATION.
 x + y + z = reaction order
e.g. x = 1; y = 1; z = 0
2nd order reaction (x + y + z = 2)
x = 0; y = 0; z = 1 (1st order reaction)
x = 2; y = 0; z = 0 (2nd order)
3
Example

Use the isolation method to determine the rate law for the following reaction.
NO (g) + O3 (g)  NO2 (g) + O2 (g)
Exp. #
1
2
3
4
5
[O3]o /M
2.10 x 10-6
4.20 x 10-6
6.30 x 10-6
6.30 x 10-6
6.30 x 10-6
[NO]o/M
2.10 x 10-6
2.10 x 10-6
2.10 x 10-6
4.20 x 10-6
6.30 x 10-6
Initial Rate/(M/s)
1.60 x 10-5
3.21 x 10-5
4.79 x 10-5
9.62 x 10-5
14.5 x 10-5
Examine the experimental data for set 1 - set 2
Set 1  Set 2
[O3] doubles  rate doubles
[NO] = constant
Set 2  Set 3 again[O3] x 2  rate doubles
Look at where [NO] constant
ratio (Rate)2/(Rate)1

2 = 2x

([O3]2/[O3]1)y
(Rate 2/Rate 1) = 2 = ([4.2 * 10-6]/[2.1 * 10-6])x
x=1
Set 3  Set 4
[O3] constant
rate doubles
[NO] x 2
y=1
[O3] constant
rate doubles
[NO] x 2
y=1
check for Set 4 - Set 5
 rate law = k * [O3]1[NO]1
Calculate the rate constant – use any one of the experiments to calculate k – for example experiment 1
4
rate  k NO  O 3   rate 1  k  NO    O 3  
1
1
1.60  10 5 Ms 1  k 2.10  10 6  2.10  10 6 
1.60  10 5 Ms 1
k 
 3.63  106 M 1s 1
2
6
2.10  10 M


The average of all five experiments yield k = 3.63 x 106 M-1 s-1

This illustrates the isolation method of determining reaction rates i.e., fix the concentration of one of the
reactants, and examine the effect that changing the concentration of the other reactant has on the rate of
reaction.
Another example
2 NO(g) + 2H2(g)  N2(g) + 2H2O(g)
Exp. #
(PH )/kPa
(PNO)/kPa)
Initial Rate/(kPa/s)
53.3
53.3
38.5
19.6
40.0
20.3
53.3
53.3
0.142
0.038
0.213
0.105
2
1
2
3
4
Ans:
Rate Law: Rate = k[NO]x[H2]y
exp (1)  (2) (PH2)0 is constant.
 (Rate)2/(Rate)1 = ({PNO}2 )x / ({PNO}1)x

0.033/0.137 = (20.3/40.0)x
4 = (2)x  x = 2
exp (3)  (4)
(PNO)0 constant

 (Rate)3/(Rate)4 = ({PH2}3)y / ({PH2}4)y
0.105/0.213 = (19.6/38.5)y
(1/2)y = 1/2  y = 1
 rate law
rate = k * [NO]2[H2]1  reaction order = 2 + 1 = 3
Types of Reactions: Variation of [Reactants] with Time
5
Again, to calculate the rate constant – use any one of the experiments in the table above. For experiment 1
rate  k PNO  PH  rate 1  k
2
2
0.137kPa s
k 
1
P   P 
2
NO
1
H2
1
 k  40.0 kPa   53.3kPa 
2
0.137kPa s 1
 1.61  10 6 KPa 2s 1
3
85280 kPa
The average of the four experiments yields k = 1.81 x 106 KPa-2 s-1
First Order Reaction

A  product rate = -[A]/t = k[A]
note ln [A]/[A]0 = ln[A] - ln[A]0
 [A ] 
ln  t   ln[A]t  ln[A]0   kt
[A]0 
k has units of s-1
look at the following situation
Example

A certain first order reaction has a rate constant of k = 5.1 x 10-3 /s. How long does it take for ½ of the
reactant A to disappear?
ln([A] t /[A] o) = - k t
We want the time for ½ the concentration of A to disappear.
i.e. when [A ]t = ½ [A]0 we designate the time as t = t1/2 = the half life of the reaction.
ln {[A]0/2[A]0} = -k t {note using pressures ln{P0/P(t)} = kt}
ln{1/2} = - kt
ln (1) - ln (2) = -kt = 0 - 0.693 = -kt
0.693/k = t  t = 0.693/5.1 x 10-3  s-1
t1/2 = 136 seconds

general equation for A FIRST ORDER REACTION

Radioactive Samples decay according to first outer kinetics. This is the half-life of samples of e.g. 14C, 239Pu,
99Tc.
t1/2 = 0.693/k
6
Example
For the first order chemical reaction
A  products
The rate constant is 3.07 x 10-4 s-1. Given that the initial concentration of A is 0.177 M, calculate the time
required for the [A] to reach 0.116 mol/L.
A 
0.116 
ln    kt  ln 
 3.07  10 4 s 1 t
 A o
0.177 
0.4226  3.07  104 s 1 t
0.4226
t 
 1376 s  1.38  103 s
4 1
3.07  10 s
2nd Order Reaction

A + B  products

Reaction 1 is 1st order in A and B and 2nd order overall

Reaction 2 is 2nd order in A
i.e.
or A  products
Rate = k[A][B]
Rate = k[A]2
look at the rate equations

for the second case
rate = -A]/t = k[A]2

We find that the concentration time profile is as follows

1/[A] t = 1/[A] o + kt
ONLY

half-life for this type of second order reaction.

[A] at t = t½ = ½ [A]0 2
FOR A SECOND ORDER REACTION WITH THE RATE LAW RATE = k[A]2
1
1
2
1

 kt 1/2 
 kt 1/2
[A ]0 / 2 [A ]o
[A ]o [A ]o
1
[A ] 0
 kt 1/2  t 1/2 
1
k [A ]0
7

note the dependence of the half-life on the initial concentration unlike the first order case.
Example
2NO(g)  N2(g) + O2(g)
The reaction is found to be second order in NO(g)
(i.e. rate = k*[NO]2). [A]0 = 2.8 x 10-3 mol/L
1.
Find the value of the rate constant, given that the time it took for ½ of the original concentration to disappear
was 5030s (t½ = 5030s).
2.
How long will it take for C to decrease to 2.0 x10-3 mol/L.

t½ = 1/ (k [A]0) 

k = 1/ (t½ [A]0)
k = 2.8*10-3 mol/L*5030s
= 7.1*10-2/MS
1/[A] = 1/ [A]o + kt
2.0 x 10-3M = 2.8 x10-3 M + 7.1*10-2/MS t
500 / M - 357 / M = 143/M = k t  t = 143/M * 1/7.1*10-2/MS

t = 2014 s = 2.0 x 10 3 s

Note that we can also have the following
Example hydration of methyl iodide

CH3I(aq) + H2O(l)  CH3OH(aq) + H+(aq) + I-(aq)
reaction is first order in H2O(l) and CH3I
rate = k*[CH3I][H2O(l)]
since we carry out the reaction in aqueous solution
[H2O] >>>> [CH3I]  [H2O] doesn’t change by a lot
this reaction is essentially, then, a first order reaction in CH3I only
i.e.
rate
= k*[CH3I]*[H2O]
but since the concentration of H2O is essentially constant
= k*[CH3I] where k = k*[H2O]
 the reaction is pseudo - first order since it appears to be first order, but it is actually a second order
process.
8
Collision Theory of Kinetics

with few exceptions, the reaction rate  with temperature
i.e. rate 

number of collisions / unit time
assume that the reactions take place due to collisions between molecules
consider A2 + B2  product
rate = k[A2][B2]
“collision of A2 and B2 causes a reaction”

but not all collisions lead to products. There is a minimum amount of energy (a threshold collision energy)
below which we don’t expect a reaction to occur.

if the molecules don’t have this energy, then molecules remain unchanged (no reaction).

the minimum amount of energy need for initiation of a chemical reaction is the activation energy (Ea). In
order for a reaction to occur, the minimum kinetic energy of the molecules must be  activation energy or Ea.

note  the species temporarily formed by the reactant molecules as a result of collisions before they turn to
products is the activated complex.
(Slide of reaction profile in text)

note Ea, E’s
note only a small # of molecules usually have a K.E. > Ea
Arrhenuis  showed how the rate constant depended on temperature.
ln k = ln A - Ea/RT
A is called the frequency factor which represents an estimate of the number of collisions in the system.
note the dependence of k
ln k  -1/T
ln k  ln A
(k  as T )
(k  as A )
NOTE however, if the activation energy is large, the rate constant is small (reaction is SLOW)
if we have rate constant data at different temperatures, we can calculate Ea
ln k 1  ln A 
Ea
RT1
ln k 2  ln A 
Ea
RT 2
Assuming the Arrhenius frequency factor is constant at the two temperatures (a reasonable assumption for many
reactions)
9
ln k 1 
Ea
E
 ln A  ln k 2  a
RT1
RT 2
1 1
  
 T1 T 2 
k 
E 1 1
 ln  2    a   
R  T 2 T1 
 k1 
 ln k 2  ln k 1 
Ea
R
Example
The hydration of sucrose yields glucose and fructose
C12H22O11(aq) + H2O(l) C6H12O6(aq) + C6H12O6(aq)
glucose
fructose
k has been measured at 300 K and 310 K and found to be k(300 K) = 2.12*10-4/Ms
4/Ms
E
 
ln  k 2    a
R
k1
1 1
  
 T 2 T1 
 8.46  10 -4 M 1s 1 
Ea
ln 



1 1
-4
8.314 J
 2.12  10 M s 
K mol
ln(0.251) 
k(310 K) = 8.46*10-
1 
 1

 310K 300K 


Ea
3.225x 10 3 K 1  3.333x 103 K 1 

8.314 J
K mol
1.382x 8.314 J
K mol
 Ea
3 1
1.08x 10 K
solving for Ea  Ea = 1.07x105 J/mol = 107 kJ/mol
NOTE  Arrhenius Equation is best suited for studying reactions between simple species (atoms, diatomic
molecules)

The orientation of reactant (how they collide) becomes very important when the species get bigger.
Catalysts  so far, we have considered one way of speeding up a reaction (i.e. increasing T usually
increases k). Another way is by the use of a catalyst.
Definition
reaction.
- a substance that speeds up the rate of the reaction without being consumed in the overall
10
one way  the catalyst may form an intermediate complex that decomposes to form the product and
regenerating the catalyst.

look at A+B

A+B
C
C
rate constant k
rate constant in the presence of the catalyst is kc
 NOTE RATE WITH CATALYST > RATE WITHOUT CATALYST
3 general types
1. Homogeneous
2. Heterogeneous
3. Enzyme
Homogeneous Catalysis

the catalyst and the reactants are in the same phase
e.g. Oxidation of SO2 to SO3
2 SO2(g) + O2(g)  2 SO3(g)

rate is SLOW
Presence of NO(g), the following occurs.
(1) NO(g) + O2(g)  NO2(g)
(2) NO2(g) + SO2(g)  SO3(g) + NO(g)
FAST

SO2 (g) is a potent acid rain gas

note the rate of NO2(g) oxidizing SO2(g) to SO3(g) is faster than the direct oxidation. NOx(g) are produced
from burning fossil fuels such as gas, coal, oil
HETEROGENEOUS Catalysis



the catalyst and the reactants are in different phases
adsorption the binding of molecules to the surface to a surface. Adsorption on the surface occurs on
active sites; places where reacting molecules are adsorbed and physically bond to the metal surface.
e.g. the hydrogenation of ethene (C2H4 (g)) to ethane
C2H4 (g) + H2(g)  C2H6 (g)

this reaction is energetically favourable, rxnH -136.98 kJ/mole of ethane , but it is very slow.
11

In the presence of a finely divided metal such as Ni (s), Pt(s), or Pd(s), the reaction goes very quickly .
(Slide 14.18 in textbook pg. 520)
the H2 and C2H4 the following five steps
1. approach the surface;
2. adsorb on the surface;
3. H2 dissociates to form H (g) on the surface; the adsorbed H atoms migrate to the adsorbed C2H4 and react to
form the product (C2H6) on the surface (note that the products is also in an adsorbed state);
4. the product desorbs from the surface and diffuses back to the gas phase
Two other important heterogeneous catalysis processes
(1) petroleum cracking (refining crude oil)
(2) catalytic converters  very efficient in reducing exhaust emission when hot; cold is another story
Enzyme Catalysis
enzymes - biological catalyst

enzymes are proteins (M > 10000 g/mol)

high degree of specificity (i.e., they will react with one substance and one substance primarily

living cell > 3000 different enzymes

large molecules, usually floppy. According to Fischer, active site is rigid, the substrate moleucle’s structure
exactly fit’s the “lock” (the “key”). This is a very simple, but not altogether correct, picture, of enzyme
catalysis.
Simplified
E + S  ES
E  enzyme; S  substrate; P  product
ES
P+E
rate = k*[ES}
Rate Laws for Multistep Process
note: given that a reaction may proceed through a large number of elementary steps, the experimentally
established rate law must reflect the reaction rate of the slowest elementary step.  the rate determining
step (rds)
What do we mean by an rds?
Let’s look at the following situation: A commuter goes through a two step process to get to work in Halifax.
(1) highway  MacKay Bridge Toll booth
12
(2) toll booth  downtown
MacKay Bridge Toll Booth
Overall reaction
highway  downtown
Situation 1.

highway clogged, toll booth is fast.
Situation 2.

fast highway, clogged toll booth.

for the first situation, the clogged highway is the slowest step in the commuting process  rate-determining
step

for the second situation, the clogged toll-booth is the slowest step in the commuting process  ratedetermining step

note that the overall process (highway  downtown) depends on now fast step 1 and 2 occur or which step
is the rate-determining step.
13
Look at specific examples
reaction of NO2 with CO  NO + CO2
Step 1: 2 NO2(g)  NO3(g) + NO(g)
Step 2: NO3 (g) + CO(g)  NO2(g) + CO2(g)
Overall NO2(g) + CO(g)  NO(g) + CO2(g)

both steps are elementary steps (of course)
Step 1 rate = k1[NO2]2
Step 2 rate = k2[NO3][CO]

bimolecular
bimolecular
the experimentally determined rate law is
rate = k[NO2]2
examine two possibilities
Step 1 is the slow step (k1 <<<< k2)
 what we would observe is that
rate = k[NO2]2
which is what is found experimentally
 Step 1 is rate determining step (or rds)
if step 2 is slow (i.e., k2 <<<<< k1)
rate = k [NO3][CO]  different than observed experimentally  ruled out
Mechanisms where the initial step is fast

slightly more complex
H2(g) + H2(g)  2 HI(g)

for many years, chemists thought the reaction was elementary since its rate law was found experimentally to
be
rate = k [H2][I2]

in the 1960’s evidence mounted for the existence of an initial fast step.
Mechanism
Step 1 I2(g) 2 I(g)
Step 2 2 I(g) + H2(g)  2 HI(g) k2
14
k1
How does the mechanism account for the experimentally rate law?

Step 1 is fast  k1 >>>> k2
I2(g)  2 I(g) is fast
2 I(g) + H2(g)  2 HI(g) is slow
note that the rate law for the rds is written
rate = k2 [I]2[H2]

I(g) is what we call an intermediate  a generally unstable species with a low unknown concentration. We
have a problem in that our rate law now depends on the concentration of some that are intermediates.
What happens to I(g)?
2 I(g) + H2(g)  2 HI(g) react
but
I(g) + I(g)  I2(g)
may also occur
 reactions involving our intermediate are isolated
I2(g)  2 I(g)
2 I(g) + H2(g)  2 HI(g) k2
2 I(s)  I2(g)
k-1  the rate constant for the reverse reaction

look at the production of I(g) in Step 1 to find out how much I(g) is obtained in the first step.

Say that the amount of I(g) produced is limited by how fast it reacts back to form I2(g) i.e.
 [I]2 = (k1/k-1) [I2]
Subst. into Step 2
rate

= k2[H2] x (k1/k-1) [I2]
= k2(k1/k-1) [H2][I2]
= k [H2][I2]
{experimental rate law}
in general, whenever a fast step proceeds a slow step, we solve for the concentration of the intermediate by
assuming the rate of production of the intermediate is equal to the rate of the intermediate recombining to the
initial reactant.
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