Chemical kinetics

1
CHEMICAL KINETICS
A.K.GUPTA, PGT CHEMISTRY,
KVS ZIET BHUBANESWAR
A.K.GUPTA
PGT CHEMISTRY
KVS ZIET BHUBANESWAR

2
Rate and OrderRate and Order
of Reactionsof Reactions
A.K.GUPTA, PGT CHEMISTRY, KVS ZIET BHUBANESWAR

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For the reaction aA + bB → cC + dD
Rate = k[A]n
[B]m
Rate law or rate equation:
Experimentally derived algebraic
equation which relates the rate of
reaction with the concentration of
the reactants
A.K.GUPTA, PGT CHEMISTRY, KVS
ZIET BHUBANESWAR

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Rate = k[A]n
[B]m
where n and m are the orders of reaction
with respect to A and B
n and m can be ± integers or fractional
n + m is the overall order of reaction.
ZIET BHUBANESWAR

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Rate = k[A]n
[B]m
For multi-step reactions, n & y have no direct
relation to the stoichiometric coefficients and
can ONLY be determined experimentally.
For single-step reactions (elementary
reactions),
n = a and m= b
ZIET BHUBANESWAR

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Rate = k[A]n
[B]m
n = 0 → zero order w.r.t. A
n = 1 → first order w.r.t. A
n = 2 → second order w.r.t. A
m = 0 → zero order w.r.t. B
m = 1 → first order w.r.t. B
m = 2 → second order w.r.t. B
ZIET BHUBANESWAR

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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.
ZIET BHUBANESWAR

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Rate = k[A]n
[B]m
Where k is the rate constant
(specific rate) of the reaction
• Temperature-dependent
• Can only be determined from experiments
ZIET BHUBANESWAR
k is

9
Rate = k[A]n
[B]m
units of k : -
mol L−1
s−1
/(mol L−1
)n+m
or,
mol L−1
min−1
/(mol L−1
)n+m
m1n1
11
mn
)L(mol)L(mol
sLmol
[B][A]
rate
k −−
−−
==
ZIET BHUBANESWAR

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Rate = k[A]0
[B]0
units of k
= mol L−1
s−1
/(mol L−1
)0+0
= mol L−1
s−1
= units of rate
ZIET BHUBANESWAR

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Rate = k[A][B]0
units of k
= mol L−1
s−1
/(mol L−1
)1+0
= s−1
ZIET BHUBANESWAR

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Rate = k[A][B]
units of k
= mol L−1
s−1
/(mol L−1
)1+1
= mol−1
L1
s−1
The overall order of reaction can be
deduced from the units of k
ZIET BHUBANESWAR

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Rate = k[A]n
[B]m
[C]p
…
For the reaction
aA + bB + cC + … → products
units of k : -
mol L−1
s−1
/(mol L−1
)n+m+p+…
ZIET BHUBANESWAR

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Determination of rate equations
To determine a rate equation is to find n, m, p,
z,…
Rate = k[A]n
[B]m
[C]p
…
Two approaches : -
1. Initial rate method
2. Graphical method
ZIET BHUBANESWAR

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Determination ofDetermination of
Rate LawRate Law
byby
Initial Rate MethodsInitial Rate Methods
ZIET BHUBANESWAR

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5Cl−
(aq) + ClO3
−
(aq) + 6H+
(aq) → 3Cl2(aq) + 3H2O(l)
Expt
[Cl−
(aq)]
(mol L−1
)
[ClO3
−
(aq)]
(mol L−1
)
[H+
(aq)]
(mol L−1
)
Initial rate
(mol L−1
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
Suppose the rate law for the reaction is
rate = k[Cl−
(aq)]n
[ClO3
−
(aq)]m
[H+
(aq)]p
ZIET BHUBANESWAR

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Expt
[Cl−
(aq)]
(mol L−1
)
[ClO3
−
(aq)]
(mol L−1
)
[H+
(aq)]
(mol L−1
)
Initial rate
(mol L−1
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
pmn
pmn
5
5
(0.20)(0.08)(0.15)
(0.40)(0.08)(0.15)
101.0
104.0
=
×
×
−
−
From experiments 1 and 2,
4 = 2p
⇒ p = 2
= 2p
ZIET BHUBANESWAR

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Expt
[Cl−
(aq)]
(mol L−1
)
[ClO3
−
(aq)]
(mol L−1
)
[H+
(aq)]
(mol L−1
)
Initial rate
(mol L−1
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
pmn
pmn
5
5
(0.40)(0.08)(0.15)
(0.40)(0.16)(0.15)
104.0
108.0
=
×
×
−
−
2 = 2m
⇒ m = 1
= 2m
ZIET BHUBANESWAR

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Expt
[Cl−
(aq)]
(mol L−1
)
[ClO3
−
(aq)]
(mol L−1
)
[H+
(aq)]
(mol L−1
)
Initial rate
(mol L−1
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
pmn
pmn
5
5
(0.20)(0.08)(0.15)
(0.20)(0.08)(0.30)
101.0
102.0
=
×
×
−
−
2 = 2n
⇒ n = 1
= 2n
ZIET BHUBANESWAR

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rate = k[Cl−
(aq)][ClO3
−
(aq)][H+
(aq)]2
Expt
[Cl−
(aq)]
(mol L−1
)
[ClO3
−
(aq)]
(mol L−1
)
[H+
(aq)]
(mol L−1
)
Initial rate
(mol L−1
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 experiment 1,
1.0×10−5
= k(0.15)(0.08)(0.20)2
k = 0.02 mol−3
L3
s−1
ZIET BHUBANESWAR

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rate = k[Cl−
(aq)][ClO3
−
(aq)][H+
(aq)]2
Expt
[Cl−
(aq)]
(mol L−1
)
[ClO3
−
(aq)]
(mol L−1
)
[H+
(aq)]
(mol L−1
)
Initial rate
(mol L−1
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 experiment 2,
4.0×10−5
= k(0.15)(0.08)(0.40)2
k = 0.02 mol−3
L3
s−1
ZIET BHUBANESWAR

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Question 2C + 3D + E → P + 2Q
Exp.
[C]
( mol L−1
)
[D]
( mol L−1
)
[E]
( mol L−1
)
Initial rate
( mol L−1
s-1
)
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) rate = k[C]n
[D]m
[E]p
ZIET BHUBANESWAR

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Exp.
[C]
( mol L−1
)
[D]
( mol L−1
)
[E]
( mol L−1
)
Initial rate
( mol L−1
s-1
)
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) rate = k[C]n
[D]m
[E]p
pmn
pmn
3
-2
(0.10)(0.10)(0.10)
(0.10)(0.10)(0.20)
103.0
102.4
=
×
×
−
8 = 2n
⇒ n = 3
= 2n
ZIET BHUBANESWAR

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Exp.
[C]
( mol L−1
)
[D]
( mol L−1
)
[E]
( mol L−1
)
Initial rate
( mol L−1
s-1
)
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) rate = k[C]n
[D]m
[E]p
pmn
pmn
3
-3
(0.10)(0.10)(0.10)
(0.10)(0.20)(0.10)
103.0
103.0
=
×
×
−
1 = 2m
⇒ m = 0
= 2m
ZIET BHUBANESWAR

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Exp.
[C]
( mol L−1
)
[D]
( mol L−1
)
[E]
( mol L−1
)
Initial rate
( mol L−1
s-1
)
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) rate = k[C]x
[D]y
[E]z
pmn
pmn
3
-2
(0.10)(0.10)(0.10)
(0.30)(0.10)(0.10)
103.0
102.7
=
×
×
−
9 = 3p ⇒ p = 2
= 3p
ZIET BHUBANESWAR

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Exp.
[C]
( mol L−1
)
[D]
( mol L−1
)
[E]
( mol L−1
)
Initial rate
( mol L−1
s-1
)
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) rate = k[C]3
[D]0
[E]2
= k[C]3
[E]2
ZIET BHUBANESWAR

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Exp.
[C]
( mol L−1
)
[D]
( mol L−1
)
[E]
( mol L−1
)
Initial rate
( mol L−1
s-1
)
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) rate = k[C]3
[E]2
From experiment 1,
3.0×10−3
= k(0.10)3
(0.10)2
k = 300 mol−4
L4
s−1
ZIET BHUBANESWAR

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Question
H+
CH3COCH3(aq) + I2(aq) CH3COCH2I(aq) + H+
(aq) + I−
(aq)
Initial rate
(mol L−1
s−1
)
Initial concentration
(mol L−1
)
[I2
(aq)] [CH3
COCH3
(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) Suppose,
rate = k[I2(aq)]n
[CH3COCH3(aq)]m
[H+
(aq)]p
ZIET BHUBANESWAR

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Initial rate
(mol L−1
s−1
)
(mol L−1
)
[I2
(aq)] [CH3
COCH3
(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)]n
[CH3COCH3(aq)]m
[H+
(aq)]p
p3-m1-n4-
p-3m-1n-4
5
-5
)10(5.0)10(2.0)10(1.5
)10(5.0)10(2.0)10(2.5
103.5
103.5
×××
×××
=
×
×
−
1 = 1.67n ⇒ n = 0
= 1.67n
ZIET BHUBANESWAR

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Initial rate
(mol L−1
s−1
)
(mol L−1
)
[I2
(aq)] [CH3
COCH3
(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
[CH3COCH3(aq)]m
[H+
(aq)]p
p3-m1-n4-
p-3m-1n-4
5
-5
)10(5.0)10(2.0)10(2.5
)10(5.0)10(4.0)10(2.5
103.5
107.0
×××
×××
=
×
×
−
2 = 2m ⇒m= 1
= 2m
ZIET BHUBANESWAR

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Initial rate
(mol L−1
s−1
)
(mol L−1
)
[I2
(aq)] [CH3
COCH3
(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
[CH3COCH3(aq)]m
[H+
(aq)]p
p3-m1-n4-
p-2m-1n-4
5
-4
)10(5.0)10(4.0)10(2.5
)10(1.0)10(4.0)10(2.5
107.0
101.4
×××
×××
=
×
×
−
2 = 2p ⇒ p = 1
= 2p
ZIET BHUBANESWAR

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Initial rate
(mol L−1
s−1
)
(mol L−1
)
[I2
(aq)] [CH3
COCH3
(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)]
ZIET BHUBANESWAR

33
Initial rate
(mol L−1
s−1
)
(mol L−1
)
[I2
(aq)] [CH3
COCH3
(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
L1
s−1
ZIET BHUBANESWAR

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Determination ofDetermination of
Rate LawRate Law
byby
Graphical MethodsGraphical Methods
ZIET BHUBANESWAR

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The rate of a reaction may be expressed
as : -
(1) Differential rate equation
(2) Integrated rate equation
ZIET BHUBANESWAR

36
A → products
n
k[A]
dt
d[A]
Rate =−=
(Differential rate equation)
shows the variation of rate with [A]
Two types of plots to determine k and n
ZIET BHUBANESWAR

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[A]
rate
n
k[A]
dt
d[A]
rate =−=
n = 0k
rate = k
ZIET BHUBANESWAR
Concentration of reactant A

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
ZIET BHUBANESWAR

39
[A]
rate
nk
dt
d[A]
rate [A]=−=
n = 1
slope = k
linear
ZIET BHUBANESWAR

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[A]
rate
2
k[A]
dt
d[A]
rate =−=
n = 2
k cannot be determined directly
from the graph
parabola
ZIET BHUBANESWAR

41
n
k[A]
dt
d[A]
rate =−=
[A]
rate n = 2
n = 1
n = 0
ZIET BHUBANESWAR

42 log10[A]
log10rate
n
k[A]rate =
n
1010 k[A]logratelog =
slope
y-intercept
n = 0
n = 1
log10k
n = 2
[A]nlogklog 1010 +=
slope = 1
slope = 2
slope = 0
ZIET BHUBANESWAR

43
n
k[A]
dt
d[A]
=− (Differential rate equation)
kdtd[A] −=
∫ ∫−=
t
0 0
A
A
t
t
dtkd[A]
kt[A][A] 0t −=−
[A]t = [A]0 – kt (Integrated rate
equation)
If n = 0
Derivation not required
ZIET BHUBANESWAR

44
[A]t = [A]0 – kt (Integrated rate equation)
shows variation of [A] with time
time
[A]t
ratek
dt
d[A]
slope −=−==
[A]0
constant rate
ZIET BHUBANESWAR
ConcentrationofreactantA
t
AA
k t][][ 0 −
=

45
n
k[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
t
dtkd[A]
[A]
1
loge[A]t – loge[A]0 = −kt
Or [A]t = [A]0 e−kt
loge[A]t = loge[A]0 − kt
ln
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46
time
log[A]t
log[A]0
slope = −k/2.303 k= -2.303xslpoe
linear → n = 1
ZIET BHUBANESWAR
][
][
log
303.2 0
tA
A
t
k =
intercept = log[A]0
[A]0=antilog (intercept)

47
time
[A]t
[A]t varies exponentially with time
constant half life → n = 1
ZIET BHUBANESWAR

48
seconds100tlife,half
2
1 =
ZIET BHUBANESWAR

49
2
1ttwhen = 0t [A]
2
1
[A] =
693.0301.0303.2
k
log2
303.2t
2
1 === x
ZIET BHUBANESWAR
tA
A
t
k
][
][
log
303.2 0
=

50
seconds100tlife,half
2
1 =
s100
2.303log2
k = = 6.9×10−3
s−1
ZIET BHUBANESWAR

51
Question
For hydrolysis
sucrose → fructose + glucose
Rate = k[sucrose] k = 0.208 h−1
at 298 K
a. Determine the rate constant of the
reaction.
b. Calculate the time in which 87.5%of
sucrose has decomposed
(a) h3.33
h0.208
12.303x0.30
k
2.303log2
t 12
1 === −
ZIET BHUBANESWAR

52
Q.17
sucrose → fructose + glucose
Rate = k[sucrose] k = 0.208 h−1
at 298 K
(b)
87.5% decomposed → [A]t = 0.125[A]0
On solving we get = 9.99 h
ZIET BHUBANESWAR
tA
A
t
k
][
][
log
303.2 0
=
0
0
][125.0
][
log
303.2
A
A
k
t =

53
mol−1
L1
s−1
kagainst t2
s−1−k/2.303log[A]t
against t1
mol L−1
s−1−k[A]t
against t[A]t
= [A]0
– kt0
Units of kSlope
Straight line
plot
Integrated
rate
equation
Order
t[A]
1
kt
[A]
1
[A]
1
0t
+=
kt/2.303
[A]
[A]
log
0
t
−=
Summary : - For reactions of the type
A → Products
ZIET BHUBANESWAR

54
2H2O2(aq) → 2H2O(l) + O2(g)
Rate = k[H2O2(aq)]
Examples of First Order ReactionsExamples of First Order Reactions
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55
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
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56
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 ReactionsExamples of Second Order Reactions
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57
Examples of Second Order ReactionsExamples of Second Order Reactions
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:
ZIET BHUBANESWAR

58
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.
ZIET BHUBANESWAR

59
Rate = k[A][B]
When [B] is kept constant,
excess
rate = k’[A] (where k’ = k[B]excess)
ZIET BHUBANESWAR

60
Rate = k[A][B]excess = k’[A]
k can be determined from k’ if [B]excess is known
Linear → first order
ZIET BHUBANESWAR

61
Rate = k[B][A]
• When [A] is kept constant,
rate = k”[B] (where k” = k[A]excess)
excess
ZIET BHUBANESWAR

62
Rate = k[A]excess[B] = k’’[B]
k can be determined from k’’ if [A]excess is known
Linear → first order
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63
Rate Equations and Order of Reactions
(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.
ZIET BHUBANESWAR

64
Rate Equations and Order of Reactions
(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(mol
mindmmol
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65
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.
Zeroth, First and Second Order Reactions
Answer
8.0 × 10-3
= k × (0.20)2
∴ k = 0.2 mol-1
dm3
s-1
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66
Determination of Simple Rate Equations from Initial Rate
Method
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 L-1
)
Initial conc. of
B (mol L-1
)
Initial rate
(mol L-1
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
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67
Determination of Simple Rate Equations from Initial Rate Method
(a) Calculate the order of reaction with respect to A
and that with respect to B.
Answer
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68
(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
)
02.0
01.0
(
0010.0
50.000
=
ZIET BHUBANESWAR

69
(a) Dividing (1) by (3),
∴ y = 2
y
)
04.0
02.0
(
0010.0
50.000
=
ZIET BHUBANESWAR

70
(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
(b) Calculate the rate constant using the result of
experiment 1.
(c) Write the rate equation for the reaction.
Answer
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71
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.
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 0.060A.K.GUPTA, PGT CHEMISTRY, KVS
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72
(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
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73
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
)
2.0
1.0
(
030.0
0.015
=
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74
(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
)
2.0
1.0
(
030.0
0.015
=
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75
(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
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76
(a) Write a chemical equation for the decomposition
of hydrogen peroxide solution.
Determination of Simple Rate Equations from Differential Rate Equations
Answer
(a) 2H2O2(aq) → 2H2O(l) + O2(g)
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77
(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
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78
(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.
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79
(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)].
Answer
[H2O2(aq)]
(mol L-1
)
0.100 0.175 0.250 0.300
Initial rate
(10-4
mol L-1
s-1
)
0.59 1.04 1.50 1.80
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80
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
(c)
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81
(d) From the graph in (c), determine the order and
rate constant of the reaction.
Answer
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82
(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
dmmol0)-.3000(
sdmmol0)-4-10(1.8×
ZIET BHUBANESWAR

83
(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)]
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84
(d) A graph of log (rate) against log [H2O2(aq)] gives a straight line
of slope x and y-intercept log k.
ZIET BHUBANESWAR

85
(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
−−−
−−−
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86
(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
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87
(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
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88
(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.
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89
(ii) Determine the order of the reaction.
Answer
ZIET BHUBANESWAR

90
(ii) Method 1:
A graph of the initial rates against [N2O5(g)] is shown as follows:
ZIET BHUBANESWAR

91
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
sdmmol0)-10.61(
sdmmol0)(0.12
×
−
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92
(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
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93
(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
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94
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?
Determination of Simple Rate Equations from Integrated Rate Equations
Answer
ZIET BHUBANESWAR

95
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.
k
t
693.0
2
1 =
1997
693.0
=k
kt=)
[A]
[A]
(ln 0
t4
10473.)
%20
%100
(ln −
×=
ZIET BHUBANESWAR

96
(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 + O2
2
1
Answer
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97
(a) Let the half-life of nitrogen(V) oxide be .
∴ The half-life of nitrogen(V) oxide is 14 745 s.
2
1
t
2
1
14 693.0
s1047.0
t
=× −−
s74514
2
1 =t
ZIET BHUBANESWAR

98
(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
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99
(b)
Let m be the amount of CH3N=NCH3 left after 0.068
minute.
m = 0.094 g
1
2
1
min76.40
min017.0
693.0693.0 −
===
t
k
0.06840.76)
1.5
(ln ×=
m
ZIET BHUBANESWAR

Chemical kinetics

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