# Reaction

```Chapter VIII Phenomenological chemical
kinetics (empirical/classical)
-- experimental aspects
3 The rate equation of reaction with simple order
The purposes of the rate law:
1) Given the composition of the reaction mixture,
the rate law can predict the reaction rate.
2) Gives clues to the mechanism and testify the
proposed mechanism.
3) To classify reactions on the reaction order.
It was found that reactions with same reaction order are
usually of same kinetic characteristics, therefore, reactions
are usually classified on the basis of reaction order.
1
1
r

k
[H
]
[I
]
H2 + I2 = 2 HI
2
2
1
0.5
r

k
[H
]
[Cl
]
H2 + Cl2 = 2 HCl
2
2
H2 + Br2 = 2 HBr
Overall
reactions
[H 2 ][Br2 ]0.5
rk
[HBr]
1 k '
[Br2 ]
Reaction with
definite order
Reaction without
definite order
Reaction with
simple order
Reaction with simple order:
The reaction whose rate only depends on the
concentration of reactants, and both the partial order
and the reaction order is zero or plus integer is called
reaction with simple order.
r = kcn
n
kinds
0
zeroth-order reaction
1
first-order reaction
2
second-order reaction
3
third-order reaction
5.1 the first-order reaction:
Reaction: A  P
at t = 0
c0
at t = t
c
Differential rate equation:
can be rearranged into:
dc

 k1c
dt
dc

 k1dt
c
Which can be integrated directly
c0
ln  k1t
c
c  c0 exp(k1t )
c  c0 exp(k1t )
1.0
C / mol dm-3
c~t curve of firstorder reaction
0.8
0.6
0.4
0.2
0.0
0
1000
2000
3000
t/s
Half-life
c0
c
2
c0
ln  k1t
c
4000
5000
Only when t  , can
c  0, which suggests
that, the first-order
reaction can not
complete.
ln 2 0.6932
t1 

k1
k1
2
ln(C/mol dm-3)
0
lnc ~ t curve of the
first-order reaction
-1
-2
-3
-4
-5
0
1000
2000
3000
4000
5000
t/s
ln c  ln c0  k1t
The slope of the lnc ~ t curve is the k1
Characteristics of the first-order reaction
1) Unit of k is s-1
2) lnc is in linear proportion to t
3) can not complete
4) Half-life does not depend on c0
Example:
1) Decay of isotopes
226
88
Ra 
226
86
Rn  42 He
2) Decomposition
1
N 2 O5  N 2 O 4  O 2
2
3) Isomerization
Example:
The half-life of the first-order decay
The natural abundance of
14C
isotope
is 1.1  10-13 mol% in living matter.
Willard F. Libby
1960 Noble Prize
obtained
in
USA
excavation shows that the
1908/12/17 ~1980/09/08
content
Application of 14C for age
determinations
Calculate the age of the object.
is
0.89
an

archeological
14C
10-14
isotope
mol%.
isotope
Half-life
isotope
Half-life
234Th--205Pb
24 day
235U--207Pb
7.13108 yr
210Pb--205Pb
22 yr
40K--40Ar
1.5 109 yr
14C--14N
5692 yr
238U--208Pb
4.5 109 yr
230Th--208Pb
7.52 104 yr
234Th--208Pb
1.39 1010 yr
87Rb--87Sr
5.0 1010 yr
5.2 Second-order reaction
2A  P;
A + B  P
A +
B  P
a
b
cA= ax cB =bx
Differential rate equation:
dx
 k 2 (a  x)(b  x)
dt
dCA

 k 2 CA C B
dt
dx
 k2 dt
(a  x)(b  x)

x
0

x
0
t
dx
  k 2 dt
(a  x)(b  x) 0
x
t
dx
dx

  k 2 dt
0
(a  b)(b  x)
(a  b)( a  x) 0
ln( b  x)
ln b
ln( a  x)
ln a




 k 2t
( a  b) ( a  b) ( a  b) ( a  b)
1
b( a  x )
ln
 k 2t
(a  b) a(b  x)
When a = b
dc

 k2 c 2
dt
dc
  2  k2 dt
C0 c
C
1
1 
 
  k2t
 c c0 
C / mol dm-3
1.0
0.8
0.6
0.4
0.2
0.0
0
1000
2000
3000
t/s
4000
5000
c~t curve of
second-order
reaction
When c  0, t  , which suggests that, the pure secondorder reaction can not complete, either.
Half-life
t1/ 2
1

k2 c0
1/C/ mol dm-3
1/c ~ t curve of
second-order
reaction
50
40
30
20
10
0
0
1000
2000
3000
4000
5000
t/s
For pure second-order reaction
1 dc

 k2 c 2
2 dt
1
1 
 
  2k2t
 c c0 
t1/ 2
1

2k2 c0
Characteristics of second-order reaction
1) Unit of k is mol-1dm3s-1
2) 1/c is in linear proportion to t
3) can not complete
4) Half-life
1
t1 
c0
2
Increasing the initial concentration of the reactant
will shorten the reaction time.
Example:
1) dimerization
2) Decomposition
2 HI = H2 + I2
3) recombination
2CH3 = C2 H6
4) esterification
CH 3 COOH+C 2 H 5 OH →
CH 3 COOC 2 H 5 +H 2 O
5) hydrolysis
C12H22O11 + H2O 
C12H22O11 + C12H22O11
C12H22O11 + H2O  C6H12O6 + C6H12O6
In 1850, experiment done by Wilhelmy suggested
the rate equation of the reaction is:
r  k[C12 H22O11 ]
r  k[C12 H 22 O11 ][H 2O]v
Because the amount of water keeps nearly unchanged
during the reaction, [H2O] keeps nearly constant, and the
rate equation can be then simplified as
r  k '[C12 H 22O11 ]
Pseudo first-order reaction
r  k[C12 H 22 O11 ][H 2 O]6 [H 3O  ]
5.3 third-order reaction
3A  P
A + B + C  P
2A + B  P
3A  P
1 1 1 
 2  2   3k3t
2  c c0 
1 dc

 k3 c 3
3 dt
1 1
 2  6k3t
2
c c0
1
t1 
2
2
k
c
3 0
2
For A + B + C  P
with same initial concentration
Differential rate equation
dc

 k3 c 3
dt
Integrated rate equation
1 1
 2  2k3t
2
c c0
1 1 1 
 2  2   k3t
2  c c0 
3
t1 
2
2
k
c
3 0
2
Only five third-order gaseous reactions
have been observed.
2NO + X2  N2O + X2O; X = H, D
2NO + O2  2NO2;
2NO + X2  2NOX; X = Br, Cl
Are these true third order reactions ?
r = k [C6H5CHO]2[CN-]
r = k [C2H4O][H+][Br-]
Hydrolysis of sucrose
r  k[C12 H 22O11 ][H 2O][H  ]
During reaction, both [H2O] and [H+] keep nearly constant
and the reaction behaves as a first-order reaction.
5.4 zeroth-order reaction
A  P
Differential rate equation
c
t
c0
0
  dc   k0 dt
c0  c  k0t
dc
  k0
dt
t1/ 2
c0

2k0
When c = 0, the reaction completes, the reaction time is:
c0
t 
k0
The zero-order reaction can complete.
C / mol dm-3
1.0
0.8
0.6
0.4
0.2
0.0
0
1000
2000
3000
4000
5000
t/s
c ~ t curve for zero-order reaction
Characteristics of zeroth-order reaction
1) Unit of k is mol dm-3s-1
2) c is in linear proportion to t
3) can complete
4) When c increases, reaction time will
be prolonged.
C0
t 
k0
Examples:
Decomposition over catalysts:
1) 2N2O  2N2 + O2 over Pt wire
2) 2NH3  N2 + 3H2 over W wire
Photochemical reaction:
r=kI
5.5 for nth-order reaction
r  kc
n
1
1
1
{
 n 1 }  k n t
n 1
n  1 (a  x)
a
t1/ 2
2n 1  1
A

 n 1
n 1
(n  1)kc0
c0
For n  1
```