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1
Rate Equations and Rate Equations and Order of ReactionsOrder of Reactions
14.114.1 Rate Equations and Order of ReactionsRate Equations and Order of Reactions
14.214.2 Zeroth, First and Second Order ReactionsZeroth, First and Second Order Reactions
14.314.3 Determination of Simple Rate Equations from Initial Determination of Simple Rate Equations from Initial Rate MethodRate Method
14.414.4 Determination of Simple Rate Equations from DiffereDetermination of Simple Rate Equations from Differential Rate Equationsntial Rate Equations
14.514.5 Determination of Simple Rate Equations from IntegraDetermination of Simple Rate Equations from Integrated Rate Equationsted Rate Equations
1414
4
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.
5
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)
6
For the reaction aA + bB cC + dD
Rate k[A]x[B]y
x = 0 zero order w.r.t. Ax = 1 first order w.r.t. Ax = 2 second order w.r.t. A
y = 0 zero order w.r.t. By = 1 first order w.r.t. By = 2 second order w.r.t. B
7
For the reaction aA + bB cC + dD
Rate k[B]2
Describe the reaction with the following rate law.
The reaction is zero order w.r.t. A and
second order w.r.t. B.
8
Rate k[A]x[B]y
k is the rate constant
For the reaction aA + bB cC + dD
• Temperature-dependent
• Can only be determined from experiments
9
Rate k[A]x[B]y
units of k : -mol dm3 s1/(mol dm3)x+y or,mol dm3 min1 /(mol dm3)x+y
For the reaction aA + bB cC + dD
y3x3
13
yx )dm (mol)dm (mols dm mol
[B][A]rate
k
10
Rate k[A]0[B]0
units of k= mol dm3 s1/(mol dm3)0+0
= mol dm3 s1
= units of rate
For the reaction aA + bB cC + dD
12
Rate k[A][B]
units of k= mol dm3 s1/(mol dm3)1+1
= mol1 dm3 s1
For the reaction aA + bB cC + dD
The overall order of reaction can be deduced from the units of k
13
Rate k[A]x[B]y[C]z…
For the reaction
aA + bB + cC + … products
units of k : -mol dm3 s1/(mol dm3)x+y+z+…
14
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)
15
Determination of Determination of Rate Equations Rate Equations by Initial Rate by Initial Rate
MethodsMethods
16
5Cl(aq) + ClO3(aq) + 6H+(aq) 3Cl2(aq) +
3H2O(l)
Expt [Cl(aq)] / mol dm3
[ClO3(aq)]
/ mol dm3
[H+(aq)] / mol dm3
Initial rate / mol dm3 s1
1 0.15 0.08 0.20 1.0105
2 0.15 0.08 0.40 4.0105
3 0.15 0.16 0.40 8.0105
4 0.30 0.08 0.20 2.0105
rate k[Cl(aq)]x[ClO3(aq)]y[H+
(aq)]z
17
Expt [Cl(aq)] / mol dm3
[ClO3(aq)
] / mol dm3
[H+(aq)] / mol dm3
Initial rate / mol dm3 s1
1 0.15 0.08 0.20 1.0105
2 0.15 0.08 0.40 4.0105
3 0.15 0.16 0.40 8.0105
4 0.30 0.08 0.20 2.0105
zyx
zyx
5
5
(0.20)(0.08)(0.15)(0.40)(0.08)(0.15)
101.0104.0
From experiments 1 and 2,
4 = 2z
z = 2
= 2z
18
Expt [Cl(aq)] / mol dm3
[ClO3(aq)
] / mol dm3
[H+(aq)] / mol dm3
Initial rate / mol dm3 s1
1 0.15 0.08 0.20 1.0105
2 0.15 0.08 0.40 4.0105
3 0.15 0.16 0.40 8.0105
4 0.30 0.08 0.20 2.0105
zyx
zyx
5
5
(0.40)(0.08)(0.15)(0.40)(0.16)(0.15)
104.0108.0
From experiments 2 and 3,
2 = 2y
y = 1
= 2y
19
Expt [Cl(aq)] / mol dm3
[ClO3(aq)
] / mol dm3
[H+(aq)] / mol dm3
Initial rate / mol dm3 s1
1 0.15 0.08 0.20 1.0105
2 0.15 0.08 0.40 4.0105
3 0.15 0.16 0.40 8.0105
4 0.30 0.08 0.20 2.0105
zyx
zyx
5
5
(0.20)(0.08)(0.15)(0.20)(0.08)(0.30)
101.0102.0
From experiments 1 and 4,
2 = 2x
x = 1
= 2x
20
rate k[Cl(aq)][ClO3(aq)][H+
(aq)]2
Expt [Cl(aq)] / mol dm3
[ClO3(aq)
] / mol dm3
[H+(aq)] / mol dm3
Initial rate / mol dm3 s1
1 0.15 0.08 0.20 1.0105
2 0.15 0.08 0.40 4.0105
3 0.15 0.16 0.40 8.0105
4 0.30 0.08 0.20 2.0105
From experiment 1,
1.0105 k(0.15)(0.08)(0.20)2
k = 0.02 mol3 dm9 s1
21
rate k[Cl(aq)][ClO3(aq)][H+
(aq)]2
Expt [Cl(aq)] / mol dm3
[ClO3(aq)
] / mol dm3
[H+(aq)] / mol dm3
Initial rate / mol dm3 s1
1 0.15 0.08 0.20 1.0105
2 0.15 0.08 0.40 4.0105
3 0.15 0.16 0.40 8.0105
4 0.30 0.08 0.20 2.0105
From experiment 2,
4.0105 k(0.15)(0.08)(0.40)2
k = 0.02 mol3 dm9 s1
22
Q.15 2C + 3D + E P + 2Q
Expt [C] / mol dm3
[D] / mol dm3
[E] / mol dm3
Initial rate / mol dm3 s1
1 0.10 0.10 0.10 3.0103
2 0.20 0.10 0.10 2.4102
3 0.10 0.20 0.10 3.0103
4 0.10 0.10 0.30 2.7102
(a) rate k[C]x[D]y[E]z
23
Expt [C] / mol dm3
[D] / mol dm3
[E] / mol dm3
Initial rate / mol dm3 s1
1 0.10 0.10 0.10 3.0103
2 0.20 0.10 0.10 2.4102
3 0.10 0.20 0.10 3.0103
4 0.10 0.10 0.30 2.7102
(a) rate k[C]x[D]y[E]z
zyx
zyx
3
-2
(0.10)(0.10)(0.10)(0.10)(0.10)(0.20)
103.0102.4
From experiments 1 and 2,
8 = 2x
x = 3
= 2x
24
Expt [C] / mol dm3
[D] / mol dm3
[E] / mol dm3
Initial rate / mol dm3 s1
1 0.10 0.10 0.10 3.0103
2 0.20 0.10 0.10 2.4102
3 0.10 0.20 0.10 3.0103
4 0.10 0.10 0.30 2.7102
(a) rate k[C]x[D]y[E]z
zyx
zyx
3
-3
(0.10)(0.10)(0.10)(0.10)(0.20)(0.10)
103.0103.0
From experiments 1 and 3,
1 = 2y
y = 0
= 2y
25
Expt [C] / mol dm3
[D] / mol dm3
[E] / mol dm3
Initial rate / mol dm3 s1
1 0.10 0.10 0.10 3.0103
2 0.20 0.10 0.10 2.4102
3 0.10 0.20 0.10 3.0103
4 0.10 0.10 0.30 2.7102
(a) rate k[C]x[D]y[E]z
zyx
zyx
3
-2
(0.10)(0.10)(0.10)(0.30)(0.10)(0.10)
103.0102.7
From experiments 1 and 4,
9 = 3z
z = 2
= 3z
26
Expt [C] / mol dm3
[D] / mol dm3
[E] / mol dm3
Initial rate / mol dm3 s1
1 0.10 0.10 0.10 3.0103
2 0.20 0.10 0.10 2.4102
3 0.10 0.20 0.10 3.0103
4 0.10 0.10 0.30 2.7102
(a) rate k[C]3[D]0[E]2
= k[C]3[E]2
27
Expt [C] / mol dm3
[D] / mol dm3
[E] / mol dm3
Initial rate / mol dm3 s1
1 0.10 0.10 0.10 3.0103
2 0.20 0.10 0.10 2.4102
3 0.10 0.20 0.10 3.0103
4 0.10 0.10 0.30 2.7102
(b) rate k[C]3[E]2
From experiment 1,
3.0103 k(0.10)3(0.10)2
k = 300 mol4 dm12 s1
28
Q.16H+
CH3COCH3(aq) + I2(aq) CH3COCH2I(aq) + H+(aq) + I(aq)
Initial rate/ mol dm3 s1
Initial concentration/ mol dm3
[I2(aq)] [CH3COCH3(aq)] [H+(aq)]
3.5 105 2.5104 2.0101 5.0103
3.5 105 1.5104 2.0101 5.0103
1.4 104 2.5104 4.0101 1.0102
7.0 105 2.5104 4.0101 5.0103
(a) rate k[I2(aq)]x[CH3COCH3(aq)]y[H+
(aq)]z
29
Initial rate/ mol dm3 s1
Initial concentration/ mol dm3
[I2(aq)] [CH3COCH3(aq)] [H+(aq)]
3.5 105 2.5104 2.0101 5.0103
3.5 105 1.5104 2.0101 5.0103
1.4 104 2.5104 4.0101 1.0102
7.0 105 2.5104 4.0101 5.0103
(a) rate k[I2(aq)]x[CH3COCH3(aq)]y[H+
(aq)]z
z3-y1-x4-
z-3y-1x-4
5
-5
)10(5.0)10(2.0)10(1.5)10(5.0)10(2.0)10(2.5
103.5103.5
From experiments 1 and 2,
1 = 1.67x
x = 0
= 1.67x
30
Initial rate/ mol dm3 s1
Initial concentration/ mol dm3
[I2(aq)] [CH3COCH3(aq)] [H+(aq)]
3.5 105 2.5104 2.0101 5.0103
3.5 105 1.5104 2.0101 5.0103
1.4 104 2.5104 4.0101 1.0102
7.0 105 2.5104 4.0101 5.0103
(a) rate k[I2(aq)]x[CH3COCH3(aq)]y[H+
(aq)]z
z3-y1-x4-
z-3y-1x-4
5
-5
)10(5.0)10(2.0)10(2.5)10(5.0)10(4.0)10(2.5
103.5107.0
From experiments 1 and 4,
2 = 2y y = 1
= 2y
31
Initial rate/ mol dm3 s1
Initial concentration/ mol dm3
[I2(aq)] [CH3COCH3(aq)] [H+(aq)]
3.5 105 2.5104 2.0101 5.0103
3.5 105 1.5104 2.0101 5.0103
1.4 104 2.5104 4.0101 1.0102
7.0 105 2.5104 4.0101 5.0103
(a) rate k[I2(aq)]x[CH3COCH3(aq)]y[H+
(aq)]z
z3-y1-x4-
z-2y-1x-4
5
-4
)10(5.0)10(4.0)10(2.5)10(1.0)10(4.0)10(2.5
107.0101.4
From experiments 3 and 4,
2 = 2z z = 1
= 2z
32
Initial rate/ mol dm3 s1
Initial concentration/ mol dm3
[I2(aq)] [CH3COCH3(aq)] [H+(aq)]
3.5 105 2.5104 2.0101 5.0103
3.5 105 1.5104 2.0101 5.0103
1.4 104 2.5104 4.0101 1.0102
7.0 105 2.5104 4.0101 5.0103
(a) Rate = k[I2(aq)]0[CH3COCH3(aq)][H+(aq)]
= k[CH3COCH3(aq)][H+(aq)]
33
Initial rate/ mol dm3 s1
Initial concentration/ mol dm3
[I2(aq)] [CH3COCH3(aq)] [H+(aq)]
3.5 105 2.5104 2.0101 5.0103
3.5 105 1.5104 2.0101 5.0103
1.4 104 2.5104 4.0101 1.0102
7.0 105 2.5104 4.0101 5.0103
(b) Rate = k[CH3COCH3(aq)][H+(aq)]
From experiment 1,
3.5105 k(2.0101)(5.0103)
k = 0.035 mol1 dm3 s1
34
Determination of Determination of Rate Equations Rate Equations
by Graphical by Graphical MethodsMethods
36
A products
nk[A]dt
d[A] Rate
(Differential rate equation)
shows the variation of rate with [A]
Two types of plots to determine k and n
38
Examples of zero-order reactions : -
2NH3(g) N2(g) + 3H2(g)
Fe or W as catalyst
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.
2HI(g) H2(g) + I2(g) Au as catalyst
43 log10[A]
log10rate
nk[A] rate n
1010 k[A]log ratelog
slope
y-intercept
n = 0
n = 1
log10k
n = 2
[A]nlogklog 1010
slope = 1
slope = 2
slope = 0
44
nk[A]dt
d[A] (Differential rate equation)
kdtd[A]
t
0 0
A
A
t
tdtkd[A]
kt[A][A] 0t
[A]t = [A]0 – kt (Integrated rate equation)
If n = 0
Derivation not required
45
[A]t = [A]0 – kt (Integrated rate equation)
shows variation of [A] with time
time
[A]t
rate kdt
d[A]slope
[A]0
constant rate
46
nk[A]dt
d[A] (Differential rate equation)
(Integrated rate equation)
If n = 1, k[A]dt
d[A]
kdt[A]d[A]
t
0 0
[A]
[A]
t
tdtkd[A]
[A]1
loge[A]t – loge[A]0 = kt
Or [A]t [A]0 ekt loge[A]t = loge[A]0 kt
ln
47
Two types of plots to determine k and n
Or [A]t [A]0 ekt loge[A]t = loge[A]0 kt
time
loge [A]t
loge [A]0
slope = k
linear n = 1
48
Two types of plots to determine k and n
Or [A]t [A]0 ekt loge[A]t = loge[A]0 kt
time
[A]t
[A]t varies exponentially with time
constant half life n = 1
50
loge[A]t = loge[A]0 kt
21tt when 0t [A]
21
[A]
21kt[A]log[A]
21
log 0e0e
21kt[A]log[A]log
21
log 0e0ee
21kt
21
loge
21kt 2loge
k2log
t e
21
52
Q.17sucrose fructose + glucose
Rate = k[sucrose] k = 0.208 h1 at 298 K
(a) h 3.33h 0.208
ln2k
ln2t 12
1
53
Q.17sucrose fructose + glucose
Rate = k[sucrose] k = 0.208 h1 at 298 K
(b) [A]t [A]0 ekt
87.5% decomposed [A]t = 0.125[A]0
0.125 = ekt
ln0.125 = 0.208tt = 9.99 h
= e0.208t
54
nk[A]dt
d[A] (Differential rate equation)
(Integrated rate equation)
If n = 2,
kt[A]
1[A]
1
0t
tk[A]1
[A][A]
0
0t Or
56
kt[A]
1[A]
1
0t
tk[A]1
[A][A]
0
0t Or
time
[A]t
n = 2
Variable half life
n = 1
0k[A]1
t21
[A]t more rapidly with time in the early stage
57
time
[A]t
n = 1n = 2
n = 0
Plotting based on integrated rate equations
More common because [A]t and time can be obtained directly from expereiments.
58
[A]
rate n = 2 n =
1
n = 0
Plotting based on differential rate equations
Less common because rate cannot be obtained directly from expereiments.
59 log10[A]
log10rate
n = 0
n = 1
log10k
n = 2
slope = 1
slope = 2
slope = 0
Plotting based on differential rate equations
Less common because rate cannot be obtained directly from expereiments.
60
mol1 dm3 s1k against t2
s1kln[A]t against t1
mol dm3 s1k[A]t against t[A]t [A]0 – kt0
Units of kSlopeStraight line
plot
Integrated rate
equationOrder
t[A]1
kt[A]
1[A]
1
0t
kt[A][A]
ln0
t
Summary : - For reactions of the type
A Products
61
2H2O2(aq) 2H2O(l) + O2(g)
Rate = k[H2O2(aq)]
Examples of First Order ReactionsExamples of First Order Reactions
62
Examples of First Order ReactionsExamples 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
63
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
Examples of Second Order Examples of Second Order ReactionsReactions
64
Examples of Second Order Examples of Second Order ReactionsReactions
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
2. For a reaction involving one reactant only:
65
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.
66
2. For a reaction involving two reactants:
A + B products
Rate = k[A][B]
When [B] is kept constant,
excess
rate = k’[A] (where k’ = k[B]excess)
67
Rate = k[A][B]excess = k’[A]
k can be determined from k’ if [B]excess is known
Linear first order
68
2. For a reaction involving two reactants:
A + B products
Rate = k[B][A]
• When [A] is kept constant,
rate = k”[B] (where k” = k[A]excess)
excess
69
Rate = k[A]excess[B] = k’’[B]
k can be determined from k’’ if [A]excess is known
Linear first order
70
Q.18(a)
2NO2(g) 2NO(g) + O2(g)
first-order reaction, k 3.6 103 s1 at 573 K
kt[A][A]
ln0
t = (3.6103 s1)(150s)
0
t
[A][A]
= 0.58
71
Q.18(b)
2NO2(g) 2NO(g) + O2(g)
first-order reaction, k 3.6 103 s1 at 573 K
kt[A][A]
ln0
t
0
t
[A][A]
= 0.010
0
[A]0.01[A]
ln0.01 = (3.6103 s1)t
t = 1279 s
72
Q.18(c)
2NO2(g) 2NO(g) + O2(g)
first-order reaction, k 3.6 103 s1 at 573 K
RTnVP22 NONO
RT
P(g)][NO 2NO
2
(g)]RT[NORTV
nP 2
NONO
2
2
K) )(573mol K dm atm (0.082atm 1.0
113
= 0.021 mol dm3
Calculate the rate of consumption of NO2 when the partial pressure of NO2 is 1.0 atm.(Gas constant, R = 0.082 atm dm3 K1 mol1)
73
Q.18(c)
2NO2(g) 2NO(g) + O2(g)
first-order reaction, k 3.6 103 s1 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 K1 mol1)
[NO2(g)] = 0.021 mol dm3
Rate of reaction = k[NO2(g)]
= (3.6103 s1)(0.021 mol dm3)= 7.6105 mol dm3 s1
74
dt(g)]d[NO
21
reaction of rate 2
Q.18(c)
2NO2(g) 2NO(g) + O2(g)
first-order reaction, k 3.6 103 s1 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 K1 mol1)
reaction of rate2dt
(g)]d[NO2
= 2(7.6105 mol dm3 s1)
= 1.5104 mol dm3 s1
75
Q.19
ktt at Ra of masst at Ra of mass
ln[Ra][Ra]
ln00
t
= (0.104 y1)(5 y)
g 0.50 year5 after remaining Ra of mass
= 0.595
eAcRa 01
22889
22888 Half life = 6.67
years
1 y0.104 y6.67
ln2tln2
k21
Mass of Ra remaining after 5 years = 0.297 g
76
Q.20 eHePbU 01
42
20682
23892 68
Half life = 4.51 109 years
110-9 y101.54 y104.51
ln2tln2
k21
1.2311.000
ln1.231x1.000x
lnt at U of masst at U of mass
ln[U][U]
lnkt-00
t
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
= 0.208
kt = (1.541010 y1)t = 0.208 t = 1.35109 years
78
Q.21(b)
eHePbU 01
42
20682
23892 68
No. of moles of U decayed =
No. of moles of He formed8
1
= (3.20103 mol)81
= 4.00104 mol
No. of moles of Pb produced = 4.00104 mol
No. of moles of U at t0 = (4.00104 + 4.40104) mol
= 8.40104 mol
79
0
t
[U][U]
lnkt-
Q.21(b)
eHePbU 01
42
20682
23892 68
mol 108.40mol 104.40
ln 4-
-4
= 0.647
kt = 0.647
(1.541010 y1)t = 0.647
t = 4.20109 y
80
Q.22(a)
2N2O5(g) 4NO2(g) + O2(g)
1.12.74.45.97.28.910.9
13.3
/ kPa
1250
8005504003002001000t /
minute
52ONP
81
Q.22(a)
2N2O5(g) 4NO2(g) + O2(g)
1.12.74.45.97.28.910.9
13.3
/ kPa
1250
8005504003002001000t /
minute
52ONP
Constant half life 350 minutes 1st order
t against (g)]Oln[N plot line Straight t52
kt(g)]Oln[N(g)]Oln[N 052t52
82
(g)]RTO[NRTV
nP 52
ONON
52
52
RT
P(g)]O[N 52ON
52
kt(g)]Oln[N(g)]Oln[N 052t52
ktRT
Pln
RT
Pln
0
ON
t
ON 5252
ktlnRT)ln(PlnRT)ln(P 0ONtON 5252
kt)ln(P)ln(P 0ONtON 5252
plot line straight t against )ln(P tON 52
Q.22(a)
83 time
tON )ln(P52
k = slope = 1.99103 min1
linear first order
Q.22(a)/(b)
0.100.991.481.771.972.19
2.392.59
1250
8005504003002001000t /
minute
tON )ln(P52
84
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. Answer(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.
85
14.1 Rate Equations and Order of Reactions (SB p.27)
(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.)Answer
(b) k =
Unit of k =
= mol-5 dm15 min-1
23 ]C[]B][A[Rate
63-
-1-3
)dm (molmin dm mol
Back
86
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.
14.2 Zeroth, First and Second Order Reactions (SB p.29)
Answer 8.0 10-3 = k (0.20)2
k = 0.2 mol-1 dm3 s-1
Back
87
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 Initial conc.
of A (mol dm-
3)
Initial conc. of 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
88
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
89
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),
x = 1
x)02.001.0
(0010.0
5 0.000
90
14.3 Determination of Simple Rate Equations from Initial Rate Method (SB p.30)
(a) Dividing (1) by (3),
y = 2
y)04.002.0
(0010.0
5 0.000
91
14.3 Determination of Simple Rate Equations from Initial Rate Method (SB p.30)
(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
(b) Calculate the rate constant using the result of experiment 1.
(c)Write the rate equation for the reaction.
Answer
92
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:
14.3 Determination of Simple Rate Equations from Initial Rate Method (SB p.31)
Expt Initial conc. of CO(g)
(mol dm-3)
Initial conc. of NO2(g)
(mol dm-3)
Initial rate (mol dm-3 s-1)
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
93
(a) Calculate the rate constant of the reaction, and hence write the rate equation for the reaction.
14.3 Determination of Simple Rate Equations from Initial Rate Method (SB p.31)
Answer
94
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),
m = 1
m)2.01.0
(030.0
0.015
95
14.3 Determination of Simple Rate Equations from Initial Rate Method (SB p.31)
(a) Dividing (1) by (3),
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]
n)2.01.0
(030.0
0.015
96
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
97
(a) Write a chemical equation for the decomposition of hydrogen peroxide solution.
14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.34)
Answer
(a) 2H2O2(aq) 2H2O(l) + O2(g)
98
(b) Explain how you could find the rate of decomposition of hydrogen peroxide solution in the presence of a solid catalyst using suitable apparatus.
14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.34)
Answer
99
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.
100
(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)].
14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.34)
Answer
[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
102
(d) From the graph in (c), determine the order and rate constant of the reaction.
14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.34)
Answer
103
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.
k =
= 6.0 10-4 s-1
3-
-1-3
dm mol 0) - .3000(s dm mol 0) - 4-10 (1.8
104
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)
-3.74-3.82-3.98-4.23log (rate)
-0.523-0.602-0.757-1.000log [H2O2(aq)]
105
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.
106
14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.34)
(d) Slope of the graph =
=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
)8.0(5.0)02.4(71.3
Back
107
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)
Answer
108
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)
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
109
14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.36)
(i) Write the rate equation for the reaction.
Answer
(i) 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.
110
14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.36)
(ii) Determine the order of the reaction.Answer
111
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:
112
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.
k =
= 75 s-1 The rate equation for the reaction is:
Rate = 75 [N2O5(g)]
1-3-3-
-1-3
s dm mol 0) - 10 .61(s dm mol 0)(0.12
113
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)
A graph of log (rate) against log [N2O5(g)] gives a straight line of
slope m and y-intercept log k.
-0.62-0.74-0.92log (rate)
-2.50-2.62-2.80log [N2O5(g)]
114
14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.36)
Slope of the graph =
=1.0The 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)]
)62.2(50.2)74.0(62.0
115
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.Answer
116
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
117
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?
14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.39)
Answer
118
14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.39)
As radioactive decay is a first order reaction,
= 3.47 10-4 year-1
t = 4638 years
It takes 4638 years for the radioactivity of a sample of A to
dropt to 20 % of its original level.
kt
693.0
2
1
1997693.0k
kt)[A]
[A]( ln 0
t410473.)% 20% 100
( ln
Back
119
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 + O221
Answer
120
14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.40)
(a) Let the half-life of nitrogen(V) oxide be .
The half-life of nitrogen(V) oxide is 14 745 s.
2
1t
2
1
14 693.0s1047.0
t
s 745 142
1 t
121
(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.
14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.40)
Answer
122
14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.40)
(b)
Let m be the amount of CH3N=NCH3 left after 0.068
minute.
m = 0.094 g
1
2
1
min76.40min017.0
693.0693.0 t
k
0.068 40.76)1.5
( ln m
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123
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.
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
time and against time.
0.1000.1250.1670.2500.500[HI(g)] (mol dm-3)
4803602401200Time (min)
[HI(g)]1
Answer
124
14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.42)
10.000-2.3030.100480
8.000-2.0790.125360
5.988-1.7900.167240
4.000-1.3860.250120
2.000-0.6930.5000
1/[HI(g)]
(mol-1 dm3)
ln [HI(g)][HI(g)] (mol dm-3)
Time (min)
125
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,
126
14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.42)
(b) ln [HI(g)] against time,
127
14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.43)
(c) against time,)]g(HI[
1
128
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, thu
s the decomposition reaction is not of zeroth order.
Similarly, in graph (b), the plot of ln [HI(g)] against time is n
ot a straight line, thus the decomposition reaction is not of first order.
However, in graph (c), the plot of against time gives
a straight line, thus the decomposition of gaseous
hydrogen iodide is of second order.
)]g(HI[1
Back
129
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:
Determine the order of the reaction graphically.
0.056
0.063
0.072
0.083
0.100
[X] (mol dm-3)
200150100500Time (s)
Answer
130
14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.43)
17.86-2.880.056200
15.87-2.760.063150
13.89-2.630.072100
12.05-2.490.08350
10.00-2.300.1000
1 / [X] (mol-1 dm3)
ln [X][X] (mol dm-3)
Time (s)
131
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.
132
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.