Transcript Chemical Kinetics - Part 2

14
1
Rate Equations and
Order of Reactions
14.1
Rate Equations and Order of Reactions
14.2
Zeroth, First and Second Order Reactions
14.3
Determination of Simple Rate Equations from Initial
Rate Method
14.4
Determination of Simple Rate Equations from
Differential Rate Equations
14.5
Determination of Simple Rate Equations from
Integrated Rate Equations
Rate Equations
and Order of
Reactions
2
For the reaction
aA + bB  cC + dD
Rate  k[A]x[B]y
rate law or rate equation
3
For the reaction
aA + bB  cC + dD
Rate  k[A]x[B]y
where x and y are the orders of reaction
with respect to A and B
x and y can be  integers or fractional
x  y is the overall order of reaction.
4
For the reaction
aA + bB  cC + dD
Rate  k[A]x[B]y
For multi-step reactions,
x, y have no direct relation to the
stoichiometric coefficients and can ONLY be
determined experimentally.
For single-step reactions (elementary reactions),
x = a and y = b (refer to p.35)
5
For the reaction
aA + bB  cC + dD
Rate  k[A]x[B]y
x = 0  zero order w.r.t. A
x = 1  first order w.r.t. A
x = 2  second order w.r.t. A
y = 0  zero order w.r.t. B
y = 1  first order w.r.t. B
y = 2  second order w.r.t. B
6
For the reaction
aA + bB  cC + dD
Describe the reaction with the following rate law.
Rate  k[B]2
The reaction is zero order w.r.t. A and
second order w.r.t. B.
7
For the reaction
aA + bB  cC + dD
Rate  k[A]x[B]y
k is the rate constant
• Temperature-dependent
• Can only be determined from experiments
8
For the reaction
aA + bB  cC + dD
Rate  k[A]x[B]y
rate
mol dm s
k

y
x
3 x
3 y
[A] [B]
(mol dm ) (mol dm )
3
1
units of k : mol dm3 s1/(mol dm3)x+y or,
mol dm3 min1 /(mol dm3)x+y
9
For the reaction
aA + bB  cC + dD
Rate  k[A]0[B]0
units of k
= mol dm3 s1/(mol dm3)0+0
= mol dm3 s1
= units of rate
10
For the reaction
aA + bB  cC + dD
Rate  k[A][B]0
units of k
= mol dm3 s1/(mol dm3)1+0
= s1
11
For the reaction
aA + bB  cC + dD
Rate  k[A][B]
units of k
= mol dm3 s1/(mol dm3)1+1
= mol1 dm3 s1
The overall order of reaction can be
deduced from the units of k
12
For the reaction
aA + bB + cC + …  products
Rate  k[A]x[B]y[C]z…
units of k : mol dm3 s1/(mol dm3)x+y+z+…
13
Determination of rate equations
To determine a rate equation is to find k, x, y,
z,…
Rate  k[A]x[B]y[C]z…
Two approaches : -
1. Initial rate method (pp.17-18)
2. Graphical method (pp.19-26)
14
Determination of
Rate Equations by
Initial Rate Methods
15
5Cl(aq) + ClO3(aq) + 6H+(aq)  3Cl2(aq) + 3H2O(l)
Expt
[Cl(aq)]
[ClO3(aq)] [H+(aq)]
Initial rate
/ mol dm3 / mol dm3 / mol dm3 / mol dm3 s1
1
0.15
0.08
0.20
1.0105
2
0.15
0.08
0.40
4.0105
3
0.15
0.16
0.40
8.0105
4
0.30
0.08
0.20
2.0105
rate  k[Cl(aq)]x[ClO3(aq)]y[H+(aq)]z
16
Expt
[Cl(aq)]
[ClO3(aq)] [H+(aq)]
Initial rate
/ mol dm3 / mol dm3 / mol dm3 / mol dm3 s1
1
0.15
0.08
0.20
1.0105
2
0.15
0.08
0.40
4.0105
3
0.15
0.16
0.40
8.0105
4
0.30
0.08
0.20
2.0105
From experiments 1 and 2,
4.0  105 (0.15)x (0.08)y (0.40)z
z

=
2
y
1.0  105 (0.15)x (0.08) (0.20)z
4 = 2z  z = 2
17
Expt
[Cl(aq)]
[ClO3(aq)] [H+(aq)]
Initial rate
/ mol dm3 / mol dm3 / mol dm3 / mol dm3 s1
1
0.15
0.08
0.20
1.0105
2
0.15
0.08
0.40
4.0105
3
0.15
0.16
0.40
8.0105
4
0.30
0.08
0.20
2.0105
From experiments 2 and 3,
8.0  105 (0.15)x (0.16)y (0.40)z
y

=
2
y
4.0  105 (0.15)x (0.08) (0.40)z
2 = 2y  y = 1
18
Expt
[Cl(aq)]
[ClO3(aq)] [H+(aq)]
Initial rate
/ mol dm3 / mol dm3 / mol dm3 / mol dm3 s1
1
0.15
0.08
0.20
1.0105
2
0.15
0.08
0.40
4.0105
3
0.15
0.16
0.40
8.0105
4
0.30
0.08
0.20
2.0105
From experiments 1 and 4,
2.0  105 (0.30)x (0.08)y (0.20)z
x

=
2
y
1.0  105 (0.15)x (0.08) (0.20)z
2 = 2x  x = 1
19
Expt
[Cl(aq)]
[ClO3(aq)] [H+(aq)]
Initial rate
/ mol dm3 / mol dm3 / mol dm3 / mol dm3 s1
1
0.15
0.08
0.20
1.0105
2
0.15
0.08
0.40
4.0105
3
0.15
0.16
0.40
8.0105
4
0.30
0.08
0.20
2.0105
rate  k[Cl(aq)][ClO3(aq)][H+(aq)]2
From experiment 1,
1.0105  k(0.15)(0.08)(0.20)2
20
k = 0.02 mol3 dm9 s1
Expt
[Cl(aq)]
[ClO3(aq)] [H+(aq)]
Initial rate
/ mol dm3 / mol dm3 / mol dm3 / mol dm3 s1
1
0.15
0.08
0.20
1.0105
2
0.15
0.08
0.40
4.0105
3
0.15
0.16
0.40
8.0105
4
0.30
0.08
0.20
2.0105
rate  k[Cl(aq)][ClO3(aq)][H+(aq)]2
From experiment 2,
4.0105  k(0.15)(0.08)(0.40)2
21
k = 0.02 mol3 dm9 s1
Q.15
2C + 3D + E  P + 2Q
Expt
[C]
/ mol dm3
1
0.10
0.10
0.10
3.0103
2
0.20
0.10
0.10
2.4102
3
0.10
0.20
0.10
3.0103
4
0.10
0.10
0.30
2.7102
(a)
22
[D]
[E]
Initial rate
/ mol dm3 / mol dm3 / mol dm3 s1
rate  k[C]x[D]y[E]z
Expt
[C]
/ mol dm3
1
0.10
0.10
0.10
3.0103
2
0.20
0.10
0.10
2.4102
3
0.10
0.20
0.10
3.0103
4
0.10
0.10
0.30
2.7102
(a)
[D]
[E]
Initial rate
/ mol dm3 / mol dm3 / mol dm3 s1
rate  k[C]x[D]y[E]z
From experiments 1 and 2,
y
2.4  10
(0.20) (0.10) (0.10)z
x

=
2
3.0  103 (0.10)x (0.10)y (0.10)z
-2
23
8 = 2x  x = 3
x
Expt
[C]
/ mol dm3
1
0.10
0.10
0.10
3.0103
2
0.20
0.10
0.10
2.4102
3
0.10
0.20
0.10
3.0103
4
0.10
0.10
0.30
2.7102
(a)
[D]
[E]
Initial rate
/ mol dm3 / mol dm3 / mol dm3 s1
rate  k[C]x[D]y[E]z
From experiments 1 and 3,
y
3.0  10
(0.10) (0.20) (0.10)z
y

=
2
3.0  103 (0.10)x (0.10)y (0.10)z
-3
24
1 = 2y  y = 0
x
Expt
[C]
/ mol dm3
1
0.10
0.10
0.10
3.0103
2
0.20
0.10
0.10
2.4102
3
0.10
0.20
0.10
3.0103
4
0.10
0.10
0.30
2.7102
(a)
[D]
[E]
Initial rate
/ mol dm3 / mol dm3 / mol dm3 s1
rate  k[C]x[D]y[E]z
From experiments 1 and 4,
y
2.7  10
(0.10) (0.10) (0.30)z
z

=
3
3.0  103 (0.10)x (0.10)y (0.10)z
-2
25
9 = 3z  z = 2
x
Expt
[C]
/ mol dm3
1
0.10
0.10
0.10
3.0103
2
0.20
0.10
0.10
2.4102
3
0.10
0.20
0.10
3.0103
4
0.10
0.10
0.30
2.7102
(a)
26
[D]
[E]
Initial rate
/ mol dm3 / mol dm3 / mol dm3 s1
rate  k[C]3[D]0[E]2 = k[C]3[E]2
Expt
[C]
/ mol dm3
1
0.10
0.10
0.10
3.0103
2
0.20
0.10
0.10
2.4102
3
0.10
0.20
0.10
3.0103
4
0.10
0.10
0.30
2.7102
(b)
[D]
[E]
Initial rate
/ mol dm3 / mol dm3 / mol dm3 s1
rate  k[C]3[E]2
From experiment 1,
3.0103  k(0.10)3(0.10)2
k = 300 mol4 dm12 s1
27
Q.16
CH3COCH3(aq) + I2(aq)
Initial rate
/ mol dm3 s1
(a)
28
H+
CH3COCH2I(aq) + H+(aq) + I(aq)
Initial concentration
/ mol dm3
[I2(aq)]
[CH3COCH3(aq)]
[H+(aq)]
3.5 105
2.5104
2.0101
5.0103
3.5 105
1.5104
2.0101
5.0103
1.4 104
2.5104
4.0101
1.0102
7.0 105
2.5104
4.0101
5.0103
rate  k[I2(aq)]x[CH3COCH3(aq)]y[H+(aq)]z
Initial rate
/ mol dm3 s1
(a)
Initial concentration
/ mol dm3
[I2(aq)]
[CH3COCH3(aq)]
[H+(aq)]
3.5 105
2.5104
2.0101
5.0103
3.5 105
1.5104
2.0101
5.0103
1.4 104
2.5104
4.0101
1.0102
7.0 105
2.5104
4.0101
5.0103
rate  k[I2(aq)]x[CH3COCH3(aq)]y[H+(aq)]z
From experiments 1 and 2,
-1 y
3.5  10
(2.5  10 ) (2.0  10 ) (5.0  10-3)z
x

=
1.67
3.5  105 (1.5  10-4 )x (2.0  10-1 ) y (5.0  10-3 )z
-5
29
-4 x
1 = 1.67x  x = 0
Initial rate
/ mol dm3 s1
(a)
Initial concentration
/ mol dm3
[I2(aq)]
[CH3COCH3(aq)]
[H+(aq)]
3.5 105
2.5104
2.0101
5.0103
3.5 105
1.5104
2.0101
5.0103
1.4 104
2.5104
4.0101
1.0102
7.0 105
2.5104
4.0101
5.0103
rate  k[I2(aq)]x[CH3COCH3(aq)]y[H+(aq)]z
From experiments 1 and 4,
-1 y
7.0  10
(2.5  10 ) (4.0  10 ) (5.0  10-3)z
y

=
2
3.5  105 (2.5  10-4 )x (2.0  10-1 ) y (5.0  10-3 )z
-5
30
-4 x
2 = 2y  y = 1
Initial rate
/ mol dm3 s1
(a)
Initial concentration
/ mol dm3
[I2(aq)]
[CH3COCH3(aq)]
[H+(aq)]
3.5 105
2.5104
2.0101
5.0103
3.5 105
1.5104
2.0101
5.0103
1.4 104
2.5104
4.0101
1.0102
7.0 105
2.5104
4.0101
5.0103
rate  k[I2(aq)]x[CH3COCH3(aq)]y[H+(aq)]z
From experiments 3 and 4,
-1 y
1.4  10
(2.5  10 ) (4.0  10 ) (1.0  10-2)z
z

=
2
7.0  105 (2.5  10-4 )x (4.0  10-1 ) y (5.0  10-3 )z
-4
31
-4 x
2 = 2z  z = 1
Initial rate
/ mol dm3 s1
Initial concentration
/ mol dm3
[I2(aq)]
[CH3COCH3(aq)]
[H+(aq)]
3.5 105
2.5104
2.0101
5.0103
3.5 105
1.5104
2.0101
5.0103
1.4 104
2.5104
4.0101
1.0102
7.0 105
2.5104
4.0101
5.0103
(a)
Rate = k[I2(aq)]0[CH3COCH3(aq)][H+(aq)]
= k[CH3COCH3(aq)][H+(aq)]
32
Initial rate
/ mol dm3 s1
Initial concentration
/ mol dm3
[I2(aq)]
[CH3COCH3(aq)]
[H+(aq)]
3.5 105
2.5104
2.0101
5.0103
3.5 105
1.5104
2.0101
5.0103
1.4 104
2.5104
4.0101
1.0102
7.0 105
2.5104
4.0101
5.0103
(b)
Rate = k[CH3COCH3(aq)][H+(aq)]
From experiment 1,
3.5105  k(2.0101)(5.0103)
k = 0.035 mol1 dm3 s1
33
Determination of
Rate Equations by
Graphical Methods
34
Two types of rate equations : (1) Differential rate equation
(2) Integrated rate equation
35
A  products
d[A]
n
Rate  
 k[A]
dt
(Differential rate equation)
shows the variation of rate with [A]
Two types of plots to determine k and n
36
rate
k
d[A]
rate  
 k[A] n
dt
n=0
rate = k
[A]
37
Examples of zero-order reactions : 2NH3(g)
2HI(g)
Fe or W as catalyst
Au as catalyst
N2(g) + 3H2(g)
H2(g) + I2(g)
Decomposition of NH3/HI can take place only
on the surface of the catalyst.
Once the surface is covered completely
(saturated) with NH3/HI molecules at a given
concentration of NH3/HI, further increase in
[NH3]/[HI] has no effect on the rate of
reaction.
38
rate
k
d[A]
rate  
 k[A] n
dt
n=0
rate = k
[A]
39
rate
d[A]
rate  
 k[A]
dt
n=1
linear
slope = k
[A]
40
rate
d[A]
rate  
 k[A]2
dt
n=2
parabola
k cannot be determined directly
from the graph
[A]
41
d[A]
rate  
 k[A]n
dt
rate
n=2
n=1
n=0
[A]
42
slope
rate  k[A]n
log10rate  log10k[A]n log10k  nlog10[A]
log10rate
n=2
slope = 2
y-intercept
n=1
slope = 1
log10k
slope = 0
43
n=0
log10[A]
d[A]
n

 k[A]
dt
(Differential rate equation)
If n = 0
d[A]  kdt
At

A0
Derivation not required
t
d[A]  k  dt
t0
[A]t  [A]0  kt
[A]t = [A]0 – kt (Integrated rate equation)
44
[A]t = [A]0 – kt (Integrated rate equation)
shows variation of [A] with time
[A]t
[A]0
d[A]
slope 
 k   rate
dt
constant rate
time
45
d[A]
n

 k[A]
dt
If n = 1,
ln
(Differential rate equation)
d[A]

 k[A]
dt
d[A]
 kdt
[A]
t
1
[A]0 [A] d[A]  k t0dt
[A] t
loge[A]t – loge[A]0 = kt
loge[A]t = loge[A]0  kt
Or [A]t  [A]0 ekt
(Integrated rate equation)
46
loge[A]t = loge[A]0  kt
Or [A]t  [A]0 ekt
Two types of plots to determine k and n
loge [A]t
loge [A]0
linear  n = 1
slope = k
time
47
loge[A]t = loge[A]0  kt
Or [A]t  [A]0 ekt
Two types of plots to determine k and n
[A]t
[A]t varies exponentially with time
constant half life  n = 1
time
48
half life,t1  100 seconds
2
49
loge[A]t = loge[A]0  kt
1
[A]t  [A]0
2
when t  t1
2
1
loge [A]0  loge [A]0  kt 1
2
2
1
loge  loge [A]0  loge [A]0  kt 1
2
2
1
loge  kt 1
loge 2
2
2
t1 
2
k
 loge 2  kt1
2
50
half life,t1  100 seconds
2
ln2
k
100 s
51
= 6.9103 s1
Q.17
sucrose  fructose + glucose
Rate = k[sucrose]
(a)
52
k = 0.208 h1 at 298 K
ln2
ln2
t1 

 3.33h

1
2
k
0.208h
Q.17
sucrose  fructose + glucose
Rate = k[sucrose]
(b)
k = 0.208 h1 at 298 K
[A]t  [A]0 ekt
87.5% decomposed  [A]t = 0.125[A]0
0.125 = ekt = e0.208t
ln0.125 = 0.208t
53
t = 9.99 h
d[A]
n

 k[A]
dt
(Differential rate equation)
If n = 2,
1
1

 kt
[A]t [A]0
Or
[A]0
[A]t 
1  k[A]0 t
(Integrated rate equation)
54
1
1

 kt
[A]t [A]0
Or
[A]0
[A]t 
1  k[A]0 t
1
[A]t
Linear  n = 2
1
[A]0
Slope = k
time
55
1
1

 kt
[A]t [A]0
Or
[A]0
[A]t 
1  k[A]0 t
[A]t
[A]t  more rapidly with time in the
early stage
1
Variable half life t1 
2
k[A]0
n=1
n=2
56
time
Plotting based on integrated rate equations
[A]t
More common because [A]t and time can be
obtained directly from expereiments.
n=0
n=1
n=2
time
57
Plotting based on differential rate equations
rate
n=2
n=1
n=0
Less common because rate
cannot be obtained directly
from expereiments.
[A]
58
Plotting based on differential rate equations
log10rate
n=2
slope = 2
slope = 1
log10k
59
slope = 0
n=1
Less common because rate
cannot be obtained directly
from expereiments.
n=0
log10[A]
Summary : - For reactions of the type
A  Products
Order
Integrated
rate
equation
Straight line
plot
Slope
Units of k
0
[A]t  [A]0 – kt
[A]t against t
k
mol dm3 s1
1
[A]t
ln
 kt ln[A]t against t
[A]0
k
s1
2
1
1
1

 kt
against t
[A]
[A]t [A]0
t
k
mol1 dm3 s1
60
Examples of First Order Reactions
2H2O2(aq)  2H2O(l) + O2(g)
Rate = k[H2O2(aq)]
61
Examples of First Order Reactions
Reaction
Rate equation
2N2O5(g)  4NO2(g) + O2(g)
Rate = k[N2O5(g)]
SO2Cl2(l)  SO2(g) + Cl2(g)
Rate = k[SO2Cl2(l)]
(CH3)3CCl(l) + OH-(aq)
 (CH3)3COH(l) + Cl-(aq)
Rate = k[(CH3)3CCl(l)]
(SN1)
All radioactive decays
e.g. Rate = k[Ra]
SN1 : 1st order Nucleophilic Substitution Reaction
62
Examples of Second Order Reactions
1.
For a reaction involving one reactant only:
2NOCl(g)  2NO(g) + Cl2(g)
Rate = k[NOCl(g)]2
2NO2(g)  2NO(g) + O2(g)
Rate = k[NO2(g)]2
63
Examples of Second Order Reactions
2. For a reaction involving one reactant only:
Reaction
Rate equation
H2(g) + I2(g)  2HI(g)
Rate = k[H2(g)][I2(g)]
CH3Br(l) + OH(aq)
 CH3OH(l) + Br(aq)
Rate = k[CH3Br(l)][OH(aq)] (SN2)
CH3COOC2H5(l) + OH(aq)
 CH3COO(aq) + C2H5OH(l)
Rate = k[CH3COOC2H5(l)][OH(aq)]
SN2 : 2nd order Nucleophilic Substitution Reaction
64
2. For a reaction involving two reactants:
A + B  products
Rate = k[A][B]
To determine the rate equation, the concentration
of one of the reactants must be kept constant (in
large excess) such that the order of reaction w.r.t.
the other reactant can be determined.
65
2. For a reaction involving two reactants:
A + B  products
Rate = k[A][B]excess
When [B] is kept constant,
rate = k’[A] (where k’ = k[B]excess)
66
Rate = k[A][B]excess = k’[A]
k can be determined from k’ if [B]excess is known
Linear  first order
67
2. For a reaction involving two reactants:
A + B  products
Rate = k[B][A]excess
•
When [A] is kept constant,
rate = k”[B] (where k” = k[A]excess)
68
Rate = k[A]excess[B] = k’’[B]
k can be determined from k’’ if [A]excess is known
Linear  first order
69
Q.18(a)
2NO2(g)  2NO(g) + O2(g)
first-order reaction, k  3.6  103 s1 at 573 K
[A]t
ln
 kt = (3.6103 s1)(150s)
[A]0
[A]t
= 0.58
[A]0
70
Q.18(b)
2NO2(g)  2NO(g) + O2(g)
first-order reaction, k  3.6  103 s1 at 573 K
[A]t 0.01[A]0

= 0.01
[A]0
[A]0
[A]t
ln
 kt
[A]0
ln0.01 = (3.6103 s1)t
t = 1279 s
71
Q.18(c)
2NO2(g)  2NO(g) + O2(g)
first-order reaction, k  3.6  103 s1 at 573 K
Calculate the rate of consumption of NO2 when the
partial pressure of NO2 is 1.0 atm.
(Gas constant, R = 0.082 atm dm3 K1 mol1)
PNO2V  nNO2RT
PNO2 
nNO2
V
RT  [NO2 (g)]RT
PNO2
1.0 atm
[NO2 (g)] 

RT
(0.082atm dm3 K 1 mol1 )(573K)
= 0.021 mol dm3
72
Q.18(c)
2NO2(g)  2NO(g) + O2(g)
first-order reaction, k  3.6  103 s1 at 573 K
Calculate the rate of consumption of NO2 when the
partial pressure of NO2 is 1.0 atm.
(Gas constant, R = 0.082 atm dm3 K1 mol1)
[NO2(g)] = 0.021 mol dm3
Rate of reaction = k[NO2(g)]
= (3.6103 s1)(0.021 mol dm3)
= 7.6105 mol dm3 s1
73
Q.18(c)
2NO2(g)  2NO(g) + O2(g)
first-order reaction, k  3.6  103 s1 at 573 K
Calculate the rate of consumption of NO2 when the
partial pressure of NO2 is 1.0 atm.
(Gas constant, R = 0.082 atm dm3 K1 mol1)
1 d[NO2 (g)]
rate of reaction  
2
dt
d[NO2 (g)]
 2  rate of reaction
dt
= 2(7.6105 mol dm3 s1)
= 1.5104 mol dm3 s1
74
Q.19
228
88
Ra Ac e
228
89
0
1
Half life = 6.67 years
ln2
ln2
k

 0.104 y 1
t1
6.67 y
2
[Ra]t
mass of Ra at t
ln
 ln
 kt
[Ra]0
mass of Ra at t0
= (0.104 y1)(5 y)
mass of Ra remainingafter 5 year
= 0.595
0.50 g
Mass of Ra remaining after 5 years = 0.297 g
75
Q.20
4
0
U 206
Pb

8
He

6
82
2
1 e
238
92
Half life = 4.51  109 years
k
ln2
ln2
-10
1


1.54

10
y
t1
4.51  109 y
2
Let 1.000x be the mass of U left behind at time t
 Mass of Pb produced at time t = 0.231x
 Mass of U consumed at time t = 0.231x
 Mass of U at t0 = 1.231x
- kt  ln
[U]t
mass of U at t
1.000x
1.000
 ln
 ln
 ln
= 0.208
[U]0
mass of U at t0
1.231x
1.231
kt = (1.541010 y1)t = 0.208
76
t = 1.35109 years
Q.21(a)
U  Pb 8 He 6 e
238
92
206
82
4
2
0
1
ln2
ln2
9
t1 


4.50

10
y
10
1
2
k
1.54  10 y
77
Q.21(b)
U  Pb 8 He 6 e
238
92
206
82
4
2
0
1
1
No. of moles of U decayed = No. of moles of He formed
8
1
= (3.20103 mol)
8
= 4.00104 mol
No. of moles of Pb produced = 4.00104 mol
No. of moles of U at t0 = (4.00104 + 4.40104) mol
= 8.40104 mol
78
Q.21(b)
U  Pb 8 He 6 e
238
92
206
82
4
2
0
1
-4
[U]t
4.40 10 mol
- kt  ln
 ln
= 0.647
-4
8.40 10 mol
[U]0
kt = 0.647
(1.541010 y1)t = 0.647
t = 4.20109 y
79
Q.22(a)
t/
minute
2N2O5(g)  4NO2(g) + O2(g)
0
100 200 300
PN2O5 / kPa 13.3 10.9 8.9
80
7.2
400
550
800 1250
5.9
4.4
2.7
1.1
Q.22(a)
t/
minute
2N2O5(g)  4NO2(g) + O2(g)
0
100 200 300
PN2O5 / kPa 13.3 10.9 8.9
7.2
400
550
800 1250
5.9
4.4
2.7
1.1
Constant half life  350 minutes  1st order
Straightlineplot  ln[N2O5 (g)]t againstt
ln[N2O5 (g)]t  ln[N2O5 (g)]0  kt
81
Q.22(a)
PN2O5 
nN2O5
V
RT  [N2O5 (g)]RT
[N2O5 (g)] 
PN2O5
RT
ln[N2O5 (g)]t  ln[N2O5 (g)]0  kt
 PN2O5
ln
 RT

 PN2O5 
  ln
  kt
t
 RT  0
ln(PN2O5 )t  lnRT  ln(PN2O5 )0  lnRT  kt
ln(PN2O5 )t  ln(PN2O5 )0  kt
ln(PN2O5 )t againstt  straight lineplot
82
Q.22(a)/(b)
t/
minute
0
100 200 300
400
550
800 1250
ln(PN2O5 )t 2.59 2.39 2.19 1.97 1.77 1.48 0.99 0.10
ln(PN2O5 )t
linear  first order
k =  slope = 1.99103 min1
83
time
14.1 Rate Equations and Order of Reactions (SB p.27)
(a) The reaction between tyrosine (an amino acid) and
iodine obeys the rate law: rate = k [Tyr] [I2].
Write the orders of the reaction with respect to tyrosine
and iodine respectively, and hence the overall order.
(a) The order of the reaction with respect to
tyrosine is 1, and the order of the reaction
with respect to iodine is also 1. Therefore, the
overall order of the reaction is 2.
84
Answer
14.1 Rate Equations and Order of Reactions (SB p.27)
Back
(b) Determine the unit of the rate constant (k) of the
following rate equation:
Rate = k [A] [B]3 [C]2
(Assume that all concentrations are measured in
mol dm–3 and time is measured in minutes.)
Rate
[ A ][B]3 [C] 2
-3
-1
mol
dm
min
 Unit of k =
(mol dm - 3 )6
(b) 
k=
= mol-5 dm15 min-1
85
Answer
14.2 Zeroth, First and Second Order Reactions (SB p.29)
The initial rate of a second order reaction is 8.0 × 10–3 mol
dm–3 s–1. The initial concentrations of the two reactants,
A and B, are 0.20 mol dm–3. Calculate the rate constant of
the reaction and state its unit.
8.0 

=k
k = 0.2 mol-1 dm3 s-1
10-3
(0.20)2
Back
86
Answer
14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.30)
For a reaction between two substances A and B,
experiments with different initial concentrations of A and
B were carried out. The results were shown as follows:
Expt
87
Initial conc. of A Initial conc. of
(mol dm-3)
B (mol dm-3)
Initial rate
(mol dm-3 s-1)
1
0.01
0.02
0.0005
2
0.02
0.02
0.001 0
3
0.01
0.04
0.002 0
14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.30)
(a) Calculate the order of reaction with respect to A
and that with respect to B.
Answer
88
14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.30)
(a) Let x be the order of reaction with respect to A, and y be the
order of reaction with respect to B. Then, the rate equation for
the reaction can be expressed as:
Rate = k [A]x [B]y
Therefore,
0.0005 = k (0.01)x (0.02)y .......................... (1)
0.0010 = k (0.02)x (0.02)y .......................... (2)
0.002 0 = k (0.01)x (0.04)y .......................... (3)
Dividing (1) by (2),
0.000 5
0.01 x
(
)
0.0010
0.02

x=1
89
14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.30)
(a) Dividing (1) by (3),
0.000 5
0.02 y
(
)
0.0010
0.04

y=2
90
14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.30)
(b) Calculate the rate constant using the result of
experiment 1.
(c) Write the rate equation for the reaction.
(b) Using the result of experiment (1),
Rate = k [A] [B]2
0.000 5 = k  0.01  0.022
k = 125 mol-2 dm6 s-1
(c) Rate = 125 [A] [B]2
Back
91
Answer
14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.31)
In the kinetic study of the reaction,
CO(g) + NO2(g)  CO2(g) + NO(g)
four experiments were carried out to determine the initial
reaction rates using different initial concentrations of
reactants. The results were as follows:
Expt
Initial conc. Initial conc. Initial rate
of CO(g)
of NO2(g) (mol dm-3 s-1)
(mol dm-3)
(mol dm-3)
92
1
0.1
0.1
0.015
2
0.2
0.1
0.030
3
0.1
0.2
0.030
4
0.4
0.1
0.060
14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.31)
(a) Calculate the rate constant of the reaction, and hence
write the rate equation for the reaction.
Answer
93
14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.31)
(a) Let m be the order of reaction with respect to CO, and n be the
order of reaction with respect to NO2. Then, the rate equation for
the reaction can be expressed as:
Rate = k [CO]m [NO2]n
Therefore,
0.015 = k (0.1)m (0.1)n .......................... (1)
0.030 = k (0.2)m (0.1)n .......................... (2)
0.030 = k (0.1)m (0.2)n .......................... (3)
Dividing (1) by (2),
0.015
0.1 m
(
)
0.030 0.2

m=1
94
14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.31)
(a) Dividing (1) by (3),
0.015
0.1 n
(
)
0.030 0.2

n=1
 Rate = k [CO] [NO2]
Using the result of experiment (1),
0.015 = k (0.1)2
k = 1.5 mol-1 dm3 s-1
 Rate = 1.5 [CO] [NO2]
95
14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.31)
(b) Determine the initial rate of the reaction when the initial
concentrations of both CO( g) and NO2( g) are 0.3 mol
dm–3.
Answer
(b) Initial rate = 1.5  0.3  0.3
= 0.135 mol dm-3 s-1
Back
96
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
(a) Write a chemical equation for the decomposition
of hydrogen peroxide solution.
Answer
(a) 2H2O2(aq)  2H2O(l) + O2(g)
97
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
(b) Explain how you could find the rate of
decomposition of hydrogen peroxide solution in
the presence of a solid catalyst using suitable
apparatus.
Answer
98
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
(b) In the presence of a suitable catalyst such as manganese(IV)
oxide, hydrogen peroxide decomposes readily to give oxygen
gas which is hardly soluble in water. A gas syringe can be used
to collect the gas evolved. To minimize any gas leakage, all
apparatus should be sealed properly. A stopwatch is used to
measure the time. The volume of gas evolved per unit time (i.e.
the rate of evolution of the gas) can then be determined.
99
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
(c) The table below shows the initial rates of
decomposition of hydrogen peroxide solution of
different concentrations. Plot a graph of the initial
rate against [H2O2(aq)].
[H2O2(aq)]
(mol dm-3)
0.100
0.175
0.250
0.300
Initial rate
(10-4 mol dm-3 s-1)
0.59
1.04
1.50
1.80
Answer
100
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
(c)
101
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
(d) From the graph in (c), determine the order and
rate constant of the reaction.
Answer
102
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
(d) There are two methods to determine the order and rate constant
of the reaction.
Method 1:
When the concentration of hydrogen peroxide solution
increases from 0.1 mol dm–3 to 0.2 mol dm–3, the reaction rate
increases from 0.59 × 10–4 mol dm–3 s–1 to about 1.20 × 10–4
mol dm–3 s–1.
∴ Rate  [H2O2(aq)]
Therefore, the reaction is of first order.
The rate constant (k) is equal to the slope of the graph.
(1.8  10 - 4 - 0) mol dm -3 s -1
k=
(0.300 - 0) mol dm - 3
103
= 6.0  10-4 s-1
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
(d) Method 2:
The rate equation can be expressed as:
Rate = k [H2O2(aq)]x
where k is the rate constant and x is the order of reaction.
Taking logarithms on both sides of the rate equation,
log (rate) = log k + x log [H2O2(aq)] ................. (1)
104
log
[H2O2(aq)]
-1.000
-0.757
-0.602
-0.523
log (rate)
-4.23
-3.98
-3.82
-3.74
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
(d) A graph of log (rate) against log [H2O2(aq)] gives a straight line
of slope x and y-intercept log k.
105
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
(d) Slope of the graph =
 3.71  ( 4.02)
 0.5  ( 0.8)
=1.0
 The reaction is of first order.
Substitute the slope and one set of value into equation (1):
-4.23 = log k + (1.0) (-1.000)
log k = -3.23
k = 5.89  10-4 s-1
Back
106
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
(a) Decide which curve in the following graph corresponds
to
(i) a zeroth order reaction;
(ii) a first order reaction.
(a) (i) (3)
(ii) (2)
107
Answer
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
(b) The following results were obtained for the
decomposition of nitrogen(V) oxide.
2N2O5(g)  4NO2(g) + O2(g)
108
Concentration of N2O5
(mol dm-3)
Initial rate (mol dm-3 s-1)
1.6  10-3
0.12
2.4  10-3
0.18
3.2  10-3
0.24
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
(i) Write the rate equation for the reaction.
(i)
109
The rate equation for the reaction can be
expressed as:
Rate = k [N2O5(g)]m
where k is the rate constant and m is the
order of reaction.
Answer
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
(ii) Determine the order of the reaction.
110
Answer
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
(ii) Method 1:
A graph of the initial rates against [N2O5(g)] is shown as follows:
111
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
As shown in the graph, when the concentration of N2O5 increases
from 1.0  10–3 mol dm–3 to 2.0  10–3 mol dm–3, the rate of the
reaction increases from 0.075 mol dm–3 s–1 to 0.15 mol dm–3 s–1.

Rate  [N2O5(g)]

The reaction is of first order.
Then, the rate constant k is equal to the slope of the graph.
(0.12  0) mol dm -3 s -1
k=
(1.6  10 - 3 - 0) mol dm - 3 s -1
= 75 s-1
 The rate equation for the reaction is:
Rate = 75 [N2O5(g)]
112
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
Method 2:
Taking logarithms on both sides of the rate equation, we obtain:
log (rate) = log k + m log [N2O5(g)] .......... (1)
log [N2O5(g)]
-2.80
-2.62
-2.50
log (rate)
-0.92
-0.74
-0.62
A graph of log (rate) against log [N2O5(g)] gives a straight line of
slope m and y-intercept log k.
113
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
Slope of the graph =  0.62  ( 0.74)
 2.50  ( 2.62)
=1.0
The reaction is of first order.
Substitute the slope and one set
of value into equation (1):
-0.92 = log k + (1.0) (-2.80)
log k = 1.88
k = 75.86 s-1
The rate equation for the
reaction is:
Rate = 75.86 [N2O5(g)]
114
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
(iii) Determine the initial rate of reaction when the
concentration of nitrogen(V) oxide is:
(1) 2.0 × 10–3 mol dm–3.
(2) 2.4 × 10–2 mol dm–3.
115
Answer
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
(iii) The rate equation, rate = 75 [N2O5(g)], is used for the
following calculation.
(1) Rate = 75 [N2O5(g)]
= 75 s–1  2.0  10–3 mol dm–3
= 0.15 mol dm–3 s–1
(2) Rate = 75 [N2O5(g)]
= 75 s–1  2.4  10–2 mol dm–3
= 1.8 mol dm–3 s–1
Back
116
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.39)
The half-life of a radioactive isotope A is 1 997
years. How long does it take for the radioactivity of
a sample of A to drop to 20% of its original level?
Answer
117
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.39)
Back
As radioactive decay is a first order reaction,
0.693
k
2
0.693
k
1997
= 3.47  10-4 year-1
t1 
[A]0
)  kt
[A]
100 %
ln (
)  3.47  10  4 t
20 %

t = 4638 years
It takes 4638 years for the radioactivity of a sample of A to
dropt to 20 % of its original level.
 ln (

118
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.40)
(a) At 298 K, the rate constant for the first order
decomposition of nitrogen(V) oxide is 0.47 × 10–4 s–1.
Determine the half-life of nitrogen(V) oxide at 298 K.
N2O5  2NO2 + 1 O2
2
Answer
119
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.40)
(a) Let the half-life of nitrogen(V) oxide be t 1 .
2
0.693
0.47  10 s 
t1
4
1
2
t 1  14 745 s
2
 The half-life of nitrogen(V) oxide is 14 745 s.
120
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.40)
(b) The decomposition of CH3N = NCH3 to form N2 and
C2H6 follows first order kinetics and has a half-life of
0.017 minute at 573 K. Determine the amount of
CH3N = NCH3 left if 1.5 g of CH3N = NCH3 was
decomposed for 0.068 minute at 573 K.
Answer
121
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.40)
(b)
k
0.693
0.693

 40.76 min 1
t1
0.017 min
2
Let m be the amount of CH3N=NCH3 left after 0.068
minute.
1.5
ln ( )  40.76  0.068
m
m = 0.094 g
122
Back
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.42)
In the decomposition of gaseous hydrogen iodide,
the following experimental data were obtained.
Time
(min)
[HI(g)]
(mol
dm-3)
123
0
120
240
360
480
0.500
0.250
0.167
0.125
0.100
Determine the order of decomposition of gaseous
hydrogen iodide graphically. You may try to plot
graphs of [HI(g)] against time, ln[HI(g)] against
1
time and [HI(g)] against time.
Answer
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.42)
124
Time (min)
[HI(g)]
(mol dm-3)
ln [HI(g)]
1/[HI(g)]
(mol-1 dm3)
0
0.500
-0.693
2.000
120
0.250
-1.386
4.000
240
0.167
-1.790
5.988
360
0.125
-2.079
8.000
480
0.100
-2.303
10.000
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.42)
The order of decomposition can be determined by plotting:
(a) [HI(g)] against time,
125
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.42)
(b) ln [HI(g)] against time,
126
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.43)
(c)
127
against time,
1
[HI(g)]
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.43)
In graph (a), the plot of [HI(g)] against time is not a straight line, thus
the decomposition reaction is not of zeroth order.
Similarly, in graph (b), the plot of ln [HI(g)] against time is
not a straight line, thus the decomposition reaction is not of first
order.
1 against time gives
However, in graph (c), the plot of
[HI(g)]
a straight line, thus the decomposition of gaseous
hydrogen iodide is of second order.
Back
128
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.43)
The change in concentration of substance X as it
decomposed at 698 K was recorded in the following
table:
Time
(s)
[X]
(mol
dm-3)
0
50
100
150
200
0.100
0.083
0.072
0.063
0.056
Determine the order of the reaction graphically.
129
Answer
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.43)
130
Time (s)
[X]
(mol dm-3)
ln [X]
1 / [X]
(mol-1 dm3)
0
0.100
-2.30
10.00
50
0.083
-2.49
12.05
100
0.072
-2.63
13.89
150
0.063
-2.76
15.87
200
0.056
-2.88
17.86
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.43)
As the graph of [X]
against time is not a
straight line, the
reaction is not of
zeroth order.
131
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.43)
Similarly, the plot of ln [X}
against time is not a
straight line, thus the
reaction is not of first
order.
132
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.43)
Back
1
The plot of
[ X]
Against time gives a
straight line, therefore the
reaction is of second
order.
133