# Reaction Rates and Equilibrium

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```					KINETICS
(RATE OF REACTION)
COLLISION THEORY:
 Particles must collide in order to react
 Collisions must have sufficient energy to break
old and form new bonds
   This is called Activation Energy
   Collisions must have the appropriate orientation
for bond formation
ORIENTATION OF PARTICLES??
FACTORS THAT INCREASE THE RATE OF A
REACTION:
   Increasing Temperature
 More kinetic energy                       Double rate
 Greater number of collisions
for every
10oC
 Increasing Surface Area
increase!
 More particles in contact with one another
(generally)
   Greater chance for collisions
FACTORS THAT INCREASE THE RATE OF
A REACTION:

   Increase Concentration                       Not all reactants
 More reactant particles per volume         will affect the
 Greater chance for collisions              rate!
 Substance that is not used up in a reaction
 Provides a reaction pathway with lower activation
energy
 May be heterogeneous (different phase than
reactants) or homogeneous
POTENTIAL ENERGY DIAGRAMS
 Show whether the reaction is endothermic or
exothermic
 Represent the activiation energy necessary for
the reaction to occur
POTENTIAL ENERGY DIAGRAM FOR AN
EXOTHERMIC PROCESS
EXOTHERMIC PROCESS
MAXWELL-BOLTZMANN DISTRIBUTION
CURVE
 Shows the relative number of molecules with
enough activation energy to react
 Can show the effect of temperature

 Can show the effect of adding a catalyst
INCREASE THE TEMPERATURE!
DETERMINING THE REACTION RATE
 Change in concentration per unit time
 Not constant throughout a reaction (slows as progresses)

 Negative if based on reactants, positive if based on
products, but same value if use any species (any
reactant, any product)
CALCULATING THE AVERAGE RATE
 Simplest method of determining rate, but is not
accurate for all parts of the reaction process
D[P]
 Rate for product formation       =        Dt

_    D[P]
   Rate for reactant depletion =         Dt

See, the neg sign!
TAKING MOLAR RELATIONSHIPS INTO
ACCOUNT….
Coefficients are contained in the rate calculation:
2NO2  2NO + O2
If we determine rate for NO2 as:
_ D[NO2]
Dt
 Then…..

D[O2]          D[NO2]
 Rate for product =        Dt     =    2Dt
(divide by 2 since O2 ½ the moles of NO2, according to balance
equation)
D[NO] D[NO2]
   Rate for product NO         =          Dt = Dt
EXAMPLE 1
2NO2  2NO + O2
The initial concentration of nitrogen dioxide is
0.0100M. After 50 seconds, the concentration is
0.0079M. Determine the average rate for all
species.
ΔT = 50 sec
ave.rate=
Initial               at 50 sec         Δ[ ]             Δ[ ]/ΔT
[0.0100]NO2       [0.00799] NO2 [0.00201] NO2 -0.00201M M/s
4.02x10-5
50sec
This is the rate for conversion of NO2

This is the same ave. rate for production of NO, but the ave. rate
for O2 production is ½ (see molar ratio) = 2.01 x 10 -5 M/s
THIS AVERAGE IS THE SLOPE OF A LINE
TANGENT TO THE CURVE
Slope of the N2O5
concentration decline
rate - at one point – the
AVE RATE at this time

Curve of
the [N2O5]
decline rate
MORE DESCRIPTIVE: RATE LAWS
 Describes the rate of reaction at various
concentrations
 BASED ON EXPERIMENTAL DATA!!
 Only depends on the concentration of reactants
(and catalysts)
 Reflects the mechanism (series of reactions), not
necessarily the balanced summary equation
 Shows the effect of changing concentration of any
one reactant (or catalyst)
RATE LAWS
Rate = k [NO2]n

Where:      k = rate constant
[ ]= concentration (M)
n = order of reaction (0,1,2)
order must be determined
experimentally !!
EXAMPLE 2
 Determine the rate law expression for a reaction
from experimental data
 Understand what “order of reaction” means
B + 2A        C + D

If the concentration of reactant “A” is doubled, and the
rate is observed to be doubled, then the reaction is
said to be first order and can be written:

rate = k [A]1
and the reaction is first order in A
B + 2A       C + D
If the concentration of reactant “B” is doubled, and the
rate is observed to quadruple, then the reaction is
said to be second order and can be written:

rate = k [B]2
and the reaction is second order in B
THE OVERALL RATE LAW

B + 2A            C + D
Combining the two rate laws that each consider one
reactant, the overall rate law can be written and is
considered 3rd order overall:

rate = k [A]1 [B]2

1+2 =   3
TWO METHODS TO DETERMINE THE
RATE LAW EXPRESSION:
   Known mechanism – look at the slowest step in a
proposed reaction mechanism

   Method of Initial Rates – use experimental data
to find the “order” of each reactant (and
therefore, whole rate expression can be
determined)
METHOD 1: REACTION MECHANISMS
 Reactions occur through a series of steps
(sometimes unknown)
 Rate is not constant throughout the mechanism
(series of steps – some slower than others)
 Rate laws are developed from known mechanisms
and are used to determine the mechanism of a
reaction
 (a mechanism is “proposed” by trying to fit
various steps to support the observed data, or a
mechanism is proposed then the observed data is
used as supporting evidence)
energy   This reaction takes place in three steps

Time of reaction
Second step is rate determining
WHY ?
Ea

First step is fast
Low activation energy
Ea

Second step is slow
High activation energy
Ea

Third step is fast
Low activation energy
REACTION MECHANISMS

 Each reaction is called an elementary step .
 Anything (reactant, product) that appears in an
elementary step, but not in the balanced equation is
an intermediate .
 The rate for the slowest elementary step can be
written from its coefficients.
2CO2 + F2    2CO2F        Example 3

Rate=k [CO2] [F2] exp. determined rate law

CO2 + F2  CO2F + F                slow
F + CO2  CO2F                     fast
2CO2 + F2     2CO2F
Rate = k [CO2] [F2] from the coefficients
of the slow step--which would match the
experimental findings
2CO2 + F2    2CO2F    Example 4

CO2 + F2  CO2F + F           fast
F + CO2  CO2F                slow
2CO2 + F2     2CO2F
Rate = k [CO2]2[F2]
METHOD 2: METHOD OF INITIAL
RATES
 Use experimental data from multiple trials of an
experiment
 Compare trials so that the concentration of one
compound is constant, while the concentration of
another compound changes!
EXAMPLE 5
For the reaction        2NO2 + F2  2NO2F
the following sets of initial concentrations were used
and the rates were measured.            These two trials
[NO2]      [F2]        Rate               compare impact of
0.2 M     0.2 M         3.3 x 10-3       NO2 on the rate
0.4 M     0.2 M         6.6 x 10 -3
These two trials
0.2 M     0.6 M          2.9 x 10-2
compare impact of F   2
on the rate

Always compare the concentrations of one
compound when the concentrations of the
others remains constant
ONE MORE TO TRY – EXAMPLE 6
For the reaction
2ClO2 + F2  2ClO2F
the following sets of initial concentrations were
used and the rates were measured.
[ClO2]      [F2]        Rate
1.1 M        0.4 M        2.3 x 10-5
4.4 M        0.4 M        9.2 x 10-5
1.1 M        1.2 M         6.9 x 10-5
   Compare the rates for varying [ClO2]:
(9.2 x 10 -5 ) / (2.3 x 10 -5 )= 4 times greater rate when
[ClO2] is increased 4 times
THIS IS FIRST ORDER for [ClO2], rate =k[ClO2]

   Compare rates for varying [F2]:

(6.9 x 10-5 ) / (2.3 x 10-5) = 3 times greater rate when
is increased 3 times
THIS IS FIRST ORDER for [F2], rate = k [F2]
   Overall rate is second order
rate = k[ClO2] [F2]

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 views: 31 posted: 7/27/2012 language: pages: 37