TMK CHAPTER 8.0 REACTION KINETICS

CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

TUTORIAL 8

8.1: Reaction Rate

1. (a) Reaction rate is the change in the concentration of a reactant or a product with time.

Rate law is an equation or expression that relates the reaction rate to the rate
constant and the concentration of the reactants.

(b) i. rate =  d[I]   d[OCl]   d[Cl]   d[OI]
dt dt dt dt

ii. rate =  1 d[NH3]   1 d[O2]   1 d[NO]   1 d[H2O]
4 dt 5 dt 4 dt 6 dt

2. (a) rate   1 dH2    1 dNH3 

3 dt 2 dt
Rate of consumption of hydrogen is 3/2 rate of formation of ammonia.

(b) Given:

 dH2  = 0.074 M s–1

dt

i. rate =  1 dH2   1 dNH3 

3 dt 2 dt

 1 0.074 M s1  1 dNH3 

3 2 dt

dNH3  = 0.049 M s–1

dt
the rate of ammonia formation is 0.049 M s–1

ii. rate =  1 dH2    dN2 
3 dt dt

 1 0.074 M s1  dN2 

3 dt

dN2  = 0.025 M s–1

dt
the rate of nitrogen depletion is 0.025 M s–1

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CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

3. (a) Given: T = 25oC = 298.15 K
P = 1.5 atm
Question:
[O2] = ?

Assume that oxygen gas collected behave ideally.

PV = nRT

P = [O2]RT
(1.5 atm) = [O2](0.08206 atm L mol−1 K−1)(298.15 K)

[O2] = 0.0613 M

(b) Reaction rate = O2 gas collected

t

0.0613 M

=
45 s

= 1.36 × 10−3 M s–1

4. General rate law: rate = k[A]m[B]n

Exp. 1: 3 × rate = k (3[A])m[B]n

Exp. 2: 8 × rate = k [A]m(2[B])n

To find m, compare general rate law with exp. 1.

rate  k Am Bn

3 rate k 3Am Bn

1  Am
3 3m Am

3 = 3m
m=1
order of reaction with respect to A = 1.

To find n, compare general rate law with exp. 2.

rate  k Am Bn

8 rate k Am 2Bn

1  Bn
8 2n Bn

8 = 2n
n=3
order of reaction with respect to B = 3.

Rate = k[A][B]3

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CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

dC

5. (a) Rate of reaction =
dt

 Exp. 1: rate =
2.5103 M = 8.33×10−5 M min−1

30 min

 Exp. 2: rate =
1.0 102 M = 3.33×10−4 M min−1

30 min

 Exp. 3: rate =
5.0 103 M = 4.17×10−5 M min−1

120 min

(b) rate = k[C]m[D]n
Exp. 1: 8.33×10−5 = k(0.1)m(1.0)n
Exp. 2: 3.33×10−4 = k(0.1)m(2.0)n
Exp. 3: 4.17×10−5 = k(0.05)m(1.0)n

To find m, compare exp. 1 with exp. 3.

 8.33 105 k 0.1m 1.0n
 

4.17  105 k 0.05m 1.0n

2 = 2m
m=1
order of reaction with respect to C = 1.

To find n, compare exp. 1 with exp. 2.

 8.33105 k 0.1m 1.0n
 

3.33104 k 0.1m 2.0n

0.25 = 0.5n
n=2
order of reaction with respect to D = 2.

rate = k[C][D]2

(c) From exp. 1

rate = k[C][D]2

k= rate = 8.33105 M min1 = 8.33×10−4 M-2 min-1

CD2 0.11.02 M3

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CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

(d) rate  k C D2

x  rate k C 2D2

 1  D2

x 22 D2

x=4
rate of reaction will increase 4×.

6. General rate law: rate  k MnO4 p ClO3 q H r

Given: Exp.1: 5.2 103  k 0.1p 0.1q 0.1r
Exp.2: 3.3102  k 0.25p 0.1q 0.1r
Exp.3:1.6 102  k 0.1p 0.3q 0.1r
Exp.4: 7.4 103  k 0.1p 0.1q 0.2r

To find p, compare Exp. 1 and Exp. 2.

5.2 103  k 0.1p 0.1q 0.1r
3.3 102 k 0.25p 0.1q 0.1r

0.158 = (0.4)p
p=2
order of reaction with respect to MnO4 = 2.

To find q, compare Exp. 1 and Exp. 3.

5.2 103  k 0.1p 0.1q 0.1r
1.6 102 k 0.1p 0.3q 0.1r

0.325 = (0.33)q
q=1
order of reaction with respect to ClO3 = 1.

To find r, compare Exp. 1 and Exp. 4.

5.2  103  k 0.1p 0.1q 0.1r
7.4  103 k 0.1p 0.1q 0.2r

0.7 = (0.5)r
r = 0.5
order of reaction with respect to H = 0.5.

rate  k MnO4 2 ClO3  H 0.5

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CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

From exp. 1,

rate = 5.2 103 M s1 = 16.4 M−2.5 s−1
k = MnO4 2 ClO3  H 0.5
0.12 0.1 0.1 0.5 M3.5

7. Given:
k = 5.1×10-4 s-1 (First order reaction)

a) t1/ 2  ln 2 = 5.1 ln 2 s1 =1359 s
k 104

b) Given: Question:
[N2O5]o = 0.25 M [N2O5]192 = ?
t = 3.2 min = 192 s

ln  N2O5   kt
 N2O5 o


t

 0.25 M 192 s

 ln 

N2O5 192
5.1104 s1

[N2O5]192 = 0.23 M

c) Given: Question:
[N2O5]o = 0.35 M t=?

i. if [N2O5]t = 0.08 M

ln  N2O5   kt
 N2O5 o


t

 ln  0.35 M 
 0.08 M   5.1104 s1 t

t = 2894 s

ii. if [N2O5]t = 38% of 0.35 M = 0.133 M

ln  N2O5   kt
 N2O5 o


t

 0.35 M s 1

ln 0.133 M 
5.1104 t

t = 1897 s

13

CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

8. 0 1000 1800 3000 4000
13.3 8.54 6.67 5.00 4.17
Time/s -4.32 -4.76 -5.01 -5.30 -5.48
[C4H6]/x 10-3 M 75.2 117 150 200 239

ln[C4H6]
1/[C4H6] M-1

1/[C4H6] M- 300 1/[C4H6]vs time
250
200 500 1000 1500 2000 2500 3000 3500 4000 4500
150 Time, s
100
50

0
0

a) Second order
b) The rate constant,k = gradient of the line

= (200 – 75.2)/ (3000 – 0)
= 0.0416 M-1s-1
c) t1/2 = 1/k[C4H6]
= 1/(0.0416x13.3x10-3)
=1807 s

8.2: Collision Theory and Transition State Theory & 8.3: Factors Affecting Reaction Rate

1. a) two requirement for effective collision to occur:
 Molecules of reactants must possess a minimum kinetic energy equal to or
greater than the activation energy.
 The molecules must collide in the correct orientation.

b) N2 (g)  H2 (g)  NH3 (g)
 When N2 and H2 molecules possess kinetic energy which is equal to or greater
than the activation energy and collide with each other in the correct orientation,
effective collisions occur, thus form NH3 as the product.

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CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

2. a) Function of platinum:
act as catalyst to increase the rate of reaction by providing an alternative pathway with
lower activation energy.

effect on the rate of formation of H2 and I2.
 Platinum provide alternative pathway with lower activation energy, Ea.
 More molecules of HI able to collide with minimum or higher energy than the
Ea.
 Thus, effective collision increase and rate of reaction increase.

b)

Activated complex

Potential Energy 2HI(g) Activated complex Ea without Pt
Ea with Pt

kJ

H2(g)+I2(g)
Reaction Progress

c) Concentration – If the concentration is increased,
 number of molecules per unit volume increase the
number of collisions would also increase .
 increasing the frequency of effective collisions.
 rate of reaction increases.

Temperature – The higher the temperature,
 the number of molecules possessing higher kinetic

energy would increase.
 More molecules would possess kinetic energy higher

than the activation energy.
 increasing the frequency of effective collisions.
 thus rate of reaction increases.

3. a)

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CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

FractionNuofmbmeorleocfulmeoslewcitulheesnergy, E Maxwell-Boltzmann distribution

T1 T2T>2TT>21>T1T1
NoN. Noofomfoomflmeocloeulcleeucsluewlseiwsthwitshitpheeneeendreg≥ryguy≥1 ≥aEtaETaa1taTt 1T1

T2 NoN. Noofoomfoomflmeocloeulcleeucsluewlseiwsthwitshitpheeneeendreg≥ryguy≥1 ≥aEtaETaa2taTt 2T2

Ea KinEetniceregnye,rEgy, kJFractionNoufmmboelreocfulemsolewictuhleesnergy, E

b) Explanation:
 The above diagram shows Maxwell-Boltzmann distribution for a gas at
temperature T1 and temperature T2, where T2 is higher than T1.
 The shaded areas represent the fraction of molecules possessing kinetic energy
equals or greater than activation energy Ea.
 At higher temperature, T2 more molecules moving at high speed thus have kinetic
energy equals to or greater than activation energy. Frequency of effective collisions
increase. Thus, rate of reaction increases.

c) If a catalyst is added, the Ea value will decreased.

Maxwell-Boltzmann distribution
T1 T2T>2 >T1T1
NoN.oofomf moloelceucluelseswwithithspeeneedrg≥yu≥1 Eata Tat1 T1
NoN.oofomf moloelceucluelseswwithithspeeneedrg≥yu≥1 Eata Tat2 T2
T2

Ea with Ea KEinneetrigcye,nEergy, kJ
catalyst

Explanation:

 catalyst to increase the reaction rate by providing an alternative pathway with

lower activation energy.

 Therefore, more molecules moving at high speed thus have kinetic energy

equals to or greater than activation energy.

 Frequency of effective collisions increase.

 Thus, rate of reaction increases.

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CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

4. a) Activation energy is the minimum amount of energy required to initiate a

chemical reaction.

b) Given: Question:
k1 = 3.46×10−2 s−1 k2 = ?

T1 = 298 K

T2 = 350 K
Ea = 50.2 kJ mol1

ln k1  Ea 1  1 
k2 R  T1 
 T2 

     ln
3.46 102 s1  50200 J mol1  1 K  1 K 
k2 8.314 J mol1K1  350 298 

k2 = 0.702 s1 Question:
Ea = ?
c) Given:
k1 = 2.52×10−5 s−1

T1 = 463 K
k2 = 6.30×10−4 s−1

T2 = 503 K

ln k1  Ea 1  1 
k2 R  T1 
 T2 

   ln
2.52 105 s1  Ea  1 K  1 K 
6.30 104 s1 8.314 J mol1K1  503 463 

Ea = 156 kJ mol1

5. Decomposition of P→Q,

Given:

[P]o = 2.50 M
k1 = 3.95×10−3 M─1 s−1 (Second Order Reaction)
T1 = 700oC = 973.15 K

a) Questions:

Derive half-life

t1/2 = ?

P

At t1/2, [P]1/2 = o
2

1   1  kt t1/ 2  k 1

P P o Po

2  1  kt1/2 1
3.95 103
P  P  t1/2  2.5 = 101.3 s
oo

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CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

t1/ 2  k 1

P o

1  kt1/2

P o

b) Given: Questions:
T2 = 800oC = 1073.15 K k2 = ?
Ea = 34.8 kJ mol1

ln k1  Ea 1  1 
k2 R  T1 
 T2 

 3.95 103 M1 s1   34.8103 J mol1 1  1 
ln  k2  8.314 J mol1 K1  1073.15 973.15 
 K K


k2 = 5.90 x 10-3 M-1 s-1

6. a) k = AeEa/RT
Arrhenius equation is an equation that relate the rate constant with temperature.

b) Based on Arrhenius equation, k = AeEa/RT in which eEa/RT is determine by the fraction
of molecules that sufficient energy to react.
 At high temperature, more molecules possess kinetic energy equals to or
greater than Ea.
 Thus magnitude of Ea will decreased while rate constant increase.
 This lead to the increase in the rate of reaction.

c) i.

ln k 1/T (K1)
1.44 5.00103
1.20 3.33103
1.08 2.50103
1.01 2.00103
0.95 1.67103
0.92 1.43103
0.89 1.25103

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CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

ii. From the graph,

Gradient =  Ea
R

 Ea 1.30  (0.93)
R = 3.9103 1.5103

0.37
= 2.4103

= - 154

Ea = 154×8.314
= 1282 kJ mol1

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CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

MEKA 8 Differential rate equation
1. a)
Rate =  1 d[NH3]   1 d[O2]   1 d[NO]   1 d[H2O]
b) 4 dt 5 dt 4 dt 6 dt

2. a) Given:
 d[NH3] = 0.58 M s−1

dt

i. Rate of disappeatance of O2.

 d[O2 ]  5   d[NH3 ] 
dt 4  dt 

 = 5 0.58Ms1
4
= 0.725 M s−1

ii. Rate of formation of NO.

 1 d[NH3]   1 d[NO]
4 dt 4 dt

 d [ NO]  4   d[NH3 ] 
dt 4  dt 

 = 4 0.58Ms1
4
= 0.58 M s−1

General rate law: rate  kNOp H2 q

   Given: Exp.1: 1.25105  k 5.00 103 p 2.00 103 q
   Exp.2: 5.00105  k 10.00103 p 2.00103 q
   Exp.3: 10.00105  k 10.00103 p 4.00 103 q

To find p, compare Exp. 1 and Exp. 2.

   1.25 105  k 5.00 103 p 2.00 103 q
   5.00  105 k 10.00 103 p 2.00 103 q

0.25 = (0.5)p
p=2
order of reaction with respect to NO = 2.

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CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

To find q, compare Exp. 2 and Exp. 3.

10.00 103 p 2.00 103 q
   5.00 105
k

   10.00  105 k 10.00 103 p 4.00 103 q

0.5 = (0.5)q
q=1
order of reaction with respect to H2 = 1.

∴ rate law: rate  kNO2 H2 

b) Rate constant

   1.25105 Ms1  k 5.00103 M 2 2.00103 M

k = 2.5×102 M−2 s−2

3. Given:
k = 30.0 M−1 min−1 (second order reaction)

a) Time needed
Given:
[HI]i = 0.10 M
[HI]t = 0.01 M

1    1   kt
HI
HI o

 1 1

0.01M 0.1M
    30.0M 1 min1 t

t = 3.0 min

b) Energy profile diagram

Activated complex

Potential Energy H2(g)+I2(g)

2HI(g) vekJ
Reaction Progress

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CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

4. [H2O2] vs Time

2.5

2

[H2O2], M 1.5

1

0.5

0

0 500 1000 1500 2000 2500 3000

Time, s

a) First order.
b) Half-life, t1/2 = 960 s
c) t1/2 = ln 2/k

k = ln 2/960
= 7.22 x 10-4

5. In the reaction of Y→ Z,
Given:
[Y]i = 4.0 M
k1 = 2.95 M−1 s−1 (Second order reaction)
T1 = 298 K
T2 = 350 K
Ea = 70.8 kJ mol−1

Half-life

t1  k 1 
2
Y o

1

 = 2.95102 M 1s1 4M 

= 8.47 s

Rate constant at 350 K, k2

ln k1  Ea 1 1 
k2 R  T1 
 T2 

 2.95 102 M1 s1   70.8103 J mol1 1  1 
ln  k2  8.314 J mol1 K1  350 298 
 K K


k2 = 2.06 M-1 s-1

22

6. a) Arrhenius equation CHEMISTRY SK025
TMK Chapter 8
k = AeEa RT
REACTION KINETICS
b) graph
1/T (K1)
ln k 2.00103
19.40 1.67103
11.92 1.43103
6.77 1.25103
2.56

From the graph,

Gradient =  Ea
R

 Ea = 19.40  (2.56)
R 2103  1.25103

= 16.84
7.5  104

Ea = 2.24104×8.314
= 186102 kJ mol1

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CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

KUMBE 8

1. Given:
 d[Br2 ] = 0.039 M s−1
dt

a)  1 d[H2O]   1 d[Br2 ]
3 dt 3 dt

 d[H2O]  3   d[Br2 ] 
dt 3  dt 

= 1 (0.039 M s−1)

= 0.039 M s−1

b)  1 d[Br ]   1 d[Br2 ]
5 dt 3 dt

d[Br  ] 5  d[Br2 ] 
dt 3  dt 
  

= 5 (0.039 M s−1)
3

= 0.065 M s−1

c)  1 d[H  ]   1 d[Br2 ]
6 dt 3 dt

d[H  ] 6  d[Br2 ] 
dt 3  dt 
  

= 2 (0.039 M s−1)
= 0.078 M s−1

2. a) Order of reaction

General rate law: rate  k CH3  CBr p OH q
3

Given: Exp.1: 0.0050  k 0.50p 0.05q

Exp.2: 0.010  k 1.0p 0.05q

Exp.3: 0.015  k 1.50p 0.05q

Exp.4: 0.010  k 1.0p 0.10q

Exp.5: 0.010  k 1.0p 0.20q

24

CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

To find p, compare Exp. 3 and Exp. 1.

0.015  k 1.5p 0.05q
0.005 k 0.5p 0.05q

3 = (3)p
p=1
order of reaction with respect to (CH3)3CBr = 1.

To find q, compare Exp. 5 and Exp. 2.

0.010  k 1.0p 0.20q
0.010 k 1.0p 0.05q

1 = (4)q
q=0
order of reaction with respect to OH− = 0.

b) rate law
= [( ) ]

c) rate constant,
0.005 M s−1 = k(0.5M)
k = . −

3. a) Given:
Second order reaction
t = 0.20 min, [NOCl]t = 85% of 0.30 M
[NOCl]o = 0.30 M

85
[NOCl]t = 85 % = 100 × 0.3

= 0.255 M

Rate constant

1 - 1 = kt
[NOCl] [NOCl]

1 M - 1 M = k (0.2 min)
0.255 0.30

k = 2.94 M-1 min-1

b) Given:

obtained straight line for graph 1/[C2F4] versus time
with slope = k = 0.04 M-1 min-1
∴ the reaction is follow second order reaction.

Rate law,
Rate = k [C2F4]2

25

CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

4. Given:
First order reaction

a) Rate law
rate = k [NH3]

b) Given:
[NH3]o = 0.67 M
[NH3]t = 0.26 M
t = 19 s

rate constant,

ln [NH3]o - ln [NH3] = kt
ln (0.67 M) – ln (0.26 M) = k (19 s)
k = 0.05 s-1

c) half-life,

1 = ln 2

2

ln 2
= 0.05 s−1

= 13.91 s

5. a) At 380 K, 100 kPa

Number of molecules

280 K

380 K

Ea energy

b) At 280 K, 100 kPa, catalyst

26

CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

Number of molecules
280 K

Ea catalyst Ea energy

∴ Conclusion: Shaded area show the number of molecules having kinetic energy greater or
 equal to activation energy, Ea.
Number of molecules having kinetic energy greater or equal to activation
 energy is increase at high temperature and in the presence of catalyst.
Thus, the frequency of effective collision increase and increase the rate of
 reaction.

6. a) i. Graph of ln k vs 1/T.

In k -12.09 -10.35 -8.74 -7.18
3.53 3.41 3.30
1/T (x 10-3 K-1) 3.66

0 Graph In k vs 1/T 3.6 3.7
3.2
1/T (x 10-3 K-1 )
-2 3.3 3.4 3.5
-4
In k -6
-8
-10
-12
-14

27

CHEMISTRY SK025
TMK Chapter 8

REACTION KINETICS

ii. activation energy, Ea

gradient = -1.361×104 = -ERa
Ea

= - 8.314

Ea = 113.15 kJ/mol

iii. Half-life
from the graph, k = 3.165×10−3

t1 = ln 2
k
2

ln 2
= 3.165×10−3

= 219 s

b) Modify statement
i. A catalyst is a substance that speeds up a chemical reaction without being
consumed in the reaction.
ii. The function of a catalyst is provides an alternative pathway to lower the
activation energy for a chemical reaction.

c)

Energy

2HI(g) Ea without catalyst
Ea with catalyst

H2 (g) + I2 (g)
Reaction Progress

28