M. R. Naimi-Jamal
Faculty of Chemistry
Iran University of Science & Technology

Contents
14-1 The Rate of a Chemical Reaction
14-2 Measuring Reaction Rates
14-3 Effect of Concentration on Reaction
Rates: The Rate Law
14-4 Zero-Order Reactions
14-5 First-Order Reactions
14-6 Second-Order Reactions
14-7 Reaction Kinetics: A Summary

Contents
14-8 Theoretical Models for Chemical Kinetics
14-9 The Effect of Temperature on Reaction Rates
14-10 Reaction Mechanisms
14-11 Catalysis
Focus On Combustion and Explosions

همدقم
.
اهنآ تعرس لرتنک یاههار و ییایمیش یاهشنکاو تعرس هعلاطم ینعی کیتنیس
لص ف رد
.
دنوش یم یدنب هقبط نگمهان و نگمه تروص هب ییایمیش یاهشنکاو
نگمهان یاهشنکاو و دنریگ یم تروص زاف کی رد اهنت نگمه یاهشنکاو
.
اهزاف کرتشم
•
•
•
H

( aq )

2 NO ( g )

OH
O
2
(
 g
(
) aq )

H
2
O ( l )

2 NO
2
( g )
2 Mg ( s )
Zn ( s )


2
O
2
( g )
H

( aq )


2 MgO ( s )
Zn
2

( aq )

H
2
( g )

تعرس هلداعم
.
تسا طبترم نآ داوم تظلغ اب شنکاو تعرس
ای هیلوا داوم نتفر نیب زا تعرس ، شنکاو تعرس یضایر ظاحل هب
.
تسا نامز دحاو رد لصاح داوم دیلوت تعرس
.
دنهد یم شیامن
[ ]
اب ار رلاوم تظلغ و
R
اب ار شنکاو تعرس
: تشون ناوت یم لااب فیرعت قباطم
•
•
•
•
A

B
R
  d [ A ] dt
  d [ B ] dt

14-1 The Rate of a Chemical Reaction
•
Rate of change of concentration with time.
2 Fe 3+ (aq) + Sn 2+ → 2 Fe 2+ (aq) + Sn 4+ (aq) t = 38.5 s [Fe 2+ ] = 0.0010 M
Δt = 38.5 s Δ[Fe 2+ ] = (0.0010 – 0) M
Δ[Fe 2+ ]
Rate of formation of Fe 2+ = =
Δt
0.0010 M
38.5 s
= 2.6 x 10 -5 M s -1

Rates of Chemical Reaction
2 Fe 3+ (aq) + Sn 2+ → 2 Fe 2+ (aq) + Sn 4+ (aq)
Δ[Sn 4+ ]
Δt
=
1
2
Δ[Fe 2+ ]
= -
Δt
1
2
Δ[Fe 3+ ]
Δt

General Rate of Reaction a A + b B → c C + d D
Rate of reaction = rate of disappearance of reactants
= -
1 a
Δ[A]
Δt
= -
1 b
Δ[B]
Δt
= rate of appearance of products
=
1 c
Δ[C]
Δt
=
1 d
Δ[D]
Δt

14-2 Measuring Reaction Rates
H
2
O
2
(aq) → H
2
O(l) + ½ O
2
(g)
2 MnO
4
(aq) + 5 H
2
O
2
(aq) + 6 H + →
2 Mn 2+ + 8 H
2
O(l) + 5 O
2
(g)
Experimental set-up for determining the rate of decomposition of H
2
O
2
. Oxygen gas given off by the reaction mixture is trapped, and its volume is measured in the gas buret. The amount of H
2
O
2 remaining concentration of H
2 the measured volume of O
2
(g).
O
2 consumed and the can be calculated from

Example:
Determining and Using an Initial Rate of Reaction.
H
2
O
2
(aq) → H
2
O(l) + ½ O
2
(g) Rate =
-Δ[H
2
O
2
]
Δt
Initial rate:
-(-2.32 M / 1360 s) = 1.7 x 10 -3 M s -1

Example:
What is the concentration at 100s?
[H
2
O
2
] i
= 2.32 M Rate = 1.7 x
10 -3 M s -1
=
- Δ[H
2
O
2
]
Δt
-Δ[H
2
O
2
] = -([H
2
O
2
] f
- [H
2
O
2
] i
) = 1.7 x
10 -3 M s -1 x
Δt
[H
2
O
2
]
100 s
– 2.32 M = -1.7 x
10 -3 M s -1 x
100 s
[H
2
O
2
]
100 s
= 2.32 M - 0.17 M
= 2.17 M

14-3 Effect of Concentration on Reaction
Rates: The Rate Law a A + b B
…. → g G + h H
….
Rate of reaction = k [A] m [B] n
….
Rate constant = k
Overall order of reaction = m + n +
….

شنکاو هبترم
A B
A

B
R
 d [ A ]
 k dt
R
 d [ A ]
 k [ A ] dt
R
 d [ A ]
 k [ A ]
2 dt
R
 d [ A ]
 k [ A ]
3 dt
رفص هبترم یاهشنکاو
لوا هبترم یاهشنکاو
مود هبترم یاهشنکاو
موس هبترم یاهشنکاو
•
•
•
•

لوا هبترم یاهشنکاو
 d [ A ] dt
 d [ A ]
 kdt
[ A ]
 k [ A ]
[
[

A ]
A ]
0
 d [ A ]
[ A ]
  t
0 kdt ln
[ A ]
0
[ A ]
 kt بسح رب اهتظلغ تبسن یمتیراگل رادومن
.
تسا یطخ نامز

Example:
Establishing the Order of a reaction by the Method of Initial
Rates.
Use the data provided establish the order of the reaction with respect to HgCl reaction.
2 and C
2
O
2
2and also the overall order of the

Example:
Notice that concentration changes between reactions are by a factor of 2.
Write and take ratios of rate laws taking this into account.

Example:
R
3
= k [HgCl
2
]
3 m [C
2
O
4
2]
3 n
R
2
= k[HgCl
2
]
2 m [C
2
O
4
2]
2 n
R
2
R
3
= k (0.105) m [C
2
O
4
2]
2 n k (0.052) m [C
2
O
4
2]
3 n
R
2
R
3
= 2 m =
7.1 x 10 -5
3.5 x 10 -5
2 m = 2.0 therefore m = 1.0

Example:
R
2
= k[HgCl
2
]
2
1 [C
2
O
4
2]
2 n = k(0.105)( 0.30
) n
R
1
= k[HgCl
2
]
1
1 [C
2
O
4
2]
1 n = k(0.105)( 0.15
) n
R
2
R
1
= k(0.105)(0.30) k(0.105)(0.15) n n
R
2
R
1
=
(0.30) n
(0.15) n
= 2 n =
7.1
x
10 -5
1.8
x
10 -5
= 3.94
2 n = 3.98 therefore n = 2.0

Example:
First order
R = k [HgCl
2
] [C
2
O
4
2]
2

15-4 Zero-Order Reactions
A → products
R rxn
= k [A] 0
R rxn
= k
[k] = mol L -1 s -1

Integrated Rate Law
-Δ[A]
= k
Δt
Move to the infinitesimal
-d[A]
= k dt
And integrate from 0 to time t
[A] t
-∫
[A]
0 d[A]
0 t dt
-[A] t
+ [A]
0
= kt
[A] t
= [A]
0
- kt

15-5 First-Order Reactions
H
2
O
2
(aq) → H
2
O(l) + ½ O
2
(g) d[H
2
O
2
] dt
= -k [H
2
O
2
] ;
∫
[A] t
[A]
0 d[H
2
O
2
]
[H
2
O
2
]
0 t
[k] = s -1 ln
[A] t
[A]
0
= -kt ln[A] t
= -kt + ln[A]
0

First-Order Reactions

Half-Life
• t
½ is the time taken for one-half of a reactant to be consumed.
For a first order reaction: ln
[A] t
[A]
0
= -kt ln
½[A]
0
[A]
0
= -kt
½ ln 2 = kt
½ t
½
= ln 2 k
=
0.693
k

Half-Life
Bu t OOBu t (g) → 2 CH
3
CO(g) + C
2
H
4
(g)

Some Typical First-Order Processes
Some typical first-order processes

15-6 Second-Order Reactions
•
Rate law where sum of exponents m + n +
…
= 2
A → products d[A]
= -k[A] 2 ; dt t [A] t
∫
[A]
0 d[A]
[A] 2 dt
0
1
[A] t
1
= kt +
[A]
0
[k] = M -1 s -1 = L mol -1 s -1

Second-Order Reaction
1
[A] t
1
= kt +
[A]
0

Pseudo First-Order Reactions
•
Simplify the kinetics of complex reactions
•
Rate laws become easier to work with
CH
3
CO
2
C
2
H
5
+ H
2
O → CH
3
CO
2
H + C
2
H
5
OH
•
If the concentration of water does not change appreciably during the reaction.
–
Rate law appears to be first order
•
Typically hold one or more reactants constant by using high concentrations and low concentrations of the reactants under study.

Testing for a Rate Law
Plot [A] vs t.
Plot ln[A] vs t.
Plot 1/[A] vs t. 2nd order

15-7 Reaction Kinetics: A Summary
•
Calculate the rate of a reaction from a known rate law using:
Rate of reaction = k [A] m [B] n
….
•
Determine the instantaneous rate of the reaction by:
Finding the slope of the tangent line of [A] vs t or,
Evaluate –Δ[A]/Δt, with a short Δt interval.

Summary of Kinetics
• Determine the order of reaction by:
Using the method of initial rates
Find the graph that yields a straight line
Test for the half-life to find first order reactions
Substitute data into integrated rate laws to find the rate law that gives a consistent value of k.

Summary of Kinetics
• Find the rate constant k by:
Determining the slope of a straight line graph.
Evaluating k with the integrated rate law.
Measuring the half life of first-order reactions.
•
Find reactant concentrations or times for certain conditions using the integrated rate law after determining k.

Activation Energy
• For a reaction to occur there must be a redistribution of energy sufficient to break certain bonds in the reacting molecule(s).
•
Activation Energy is:
–
The minimum energy above the average kinetic energy that molecules must bring to their collisions for a chemical reaction to occur.

Activation Energy

Kinetic Energy

Collision Theory
•
If activation barrier is high, only a few molecules have sufficient kinetic energy and the reaction is slower.
•
As temperature increases, reaction rate increases.
•
Orientation of molecules may be important.

Collision Theory

Transition State Theory
•
The activated complex is a hypothetical species lying between reactants and products at a point on the reaction profile called the transition state.

15-9 Effect of Temperature on
Reaction Rates
•
Svante Arrhenius demonstrated that many rate constants vary with temperature according to the equation: k = A e -Ea/RT
Ea 1 ln k = + ln A
R T

Arrhenius Plot
N
2
O
5
(CCl
4
) → N
2
O
4
(CCl
4
) + ½ O
2
(g)
E a
= -1.2
x
10 4 K
R
E a
= 1.0
x
10 2 kJ mol -1

Arrhenius Equation k = A e -Ea/RT
E a
1 ln k = + ln A
R T ln k
2
– ln k
1
E a
1
= + ln A -
R T
2
E a
R
1
T
1
- ln A k
2
E ln = k
1
R
1
T
1
1
T
2 k log = k
1
2
E
2.3 R
1
T
1
1
T
2

A Rate Determining Step

11-5 Catalysis
• Alternative reaction pathway of lower energy.
•
Homogeneous catalysis.
– All species in the reaction are in solution.
•
Heterogeneous catalysis.
– The catalyst is in the solid state.
– Reactants from gas or solution phase are adsorbed.
– Active sites on the catalytic surface are important .

11-5 Catalysis