Title: Chemical Kinetics Rates and Mechanisms of Chemical Reactions
1Chemical KineticsRates and Mechanisms of
Chemical Reactions
KINETICS - the study of REACTION RATES
MECHANISM- the way the reaction proceeds
2Recap
- Integrated rate laws
- Half Lives
- First Order t1/2 ln 2 / k
- Second Order t1/2 1 / (k Ao)
- Reaction Mechanisms
- Rate Determining Step the slowest step in a
mechanism - Can be deduced from the rate laws
3COLLISION THEORY
- Molecules/ ions must collide to react
- - Orientation and energy factors slow down
the reaction. - - Rates go up faster than collision frequency
increases as Temperature increases - Not all collisions consummate a reaction
- - Orientation does not depend on T.
- Not all properly oriented collisions Consummate a
reaction - - Energy levels may be inadequate
4Chemical Reaction and Molecular Collisions
Molecular collisions lead to chemical reactions.
Thus, the reaction constant, k is determined by
several factors. k Z f p Z collision
frequencyp, the fraction with proper
orientationf, fraction of collision having
sufficient energy for reactionf is related to
the potential energy barrier called activation
energy, Ea. f ? e Ea / RT or exp ( Ea / R
T) Thus, k A e Ea / RT
z p and A are constants
5Energy in chemical reactions
Potential energy diagram for an endothermic (?H lt
0) reaction.
6Arrhenius Equation
- k A e -Ea/RT
- The activation energy, Ea, of a reaction is the
minimum amount of energy that the reacting
molecules must possess if the reaction is to be
successful - The Arrhenius equation describes the temperature
dependence of the rate constant that is
exponentially related to the activation energy (A
is the pre-exponential factor, or the A factor
also called the frequency factor)
7The Arrhenius Equation
The temperature dependence of the rate constant k
is best described by the Arrhenius equation k
A e Ea / R Tor ln k ln A Ea / R T If k1
and k2 are the rate constants at T1 and T2
respectively, then k1 Ea 1 1
ln k2 R T1 T2
8Application of Arrhenius Equation
The reaction 2 NO2(g) ? 2NO(g) O2(g) The
rate constant k 1.0 x 10-10 s-1 at 300 K and
the activation energy Ea 111 kJ mol-1. What
are A, k at 273 K and T when k 1x
10-11? Method Make use of, and rearrange k A
e Ea / R T
9ANSWERS Blanked out of the online
version, these will be completed in the lecture.
10How A works
- Measure of what fraction of the possible
collision geometries can result in a reaction
11How Ea works
Kinetic energy distributions for a reaction
mixture at two different temperatures
12How Ea works
The greater Ea, the more energy reactants need
to collide together with
13 Determining Ea k A e -Ea/RT
- If the rate constant is measured at two or more
different temperatures, the activation energy may
then be determined. - For example
- at T1 Ln k1 Ln A - Ea / RT1
- and at T2 Ln k2 Ln A - Ea / RT2
- subtracting
- Ln k1 - Ln k2 Ln (k1/ k2) - Ea
(1/T1 - 1/T2) - R
14 Determining Ea k A e -Ea/RT
- Example
- If a reaction has a rate constant of 2.0 x 10-5
s-1 at 20 ?C and 7.3 x 10-5 s-1 at 30 ?C, what
is the activation energy ? -
- Ln (k1/ k2) - Ea (1/T1
- 1/T2) - R
- Ln (2.0 x 10-5)/ (7.3 x 10-5) - (Ea/8.314)
1/293 - 1/303 - Ea 96 kJ mol-1
15Ideally you want a lot more than two T values
16Question
- The rate constant of a particular reaction
triples when the temperature is increased from 25
?C to 35 ?C. Calculate the activation energy,
Ea, for this reaction. -
- Ln (k1/ k2) - Ea (1/T1 - 1/T2)
- R
- Ln (1/3) - (Ea / 8.314)(1/298 - 1/308)
- -1.099 - Ea(1.310 x 10-5)
- Ea 83 800 J mol-1 or 83.8 kJ mol-1
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18Example questions
- TEXT PROBLEMS
- Practice Exercise 14.9, 14.10
- Problems 14.56, 14.80, 14.102
- CONCEPTS
- Collision Theory
- Steady State Approximation
- CALCULATIONS
- Half-life
- Arrhenius Equation