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8/7/2019 Reaction Rates Ppt
http://slidepdf.com/reader/full/reaction-rates-ppt 1/52
Unit 3
Chemical Kinetics andChemical Equilibrium
Reaction Rates
Rate Laws
First and Second Order Reactions
Chemical Equilibrium
Equilibrium Constants
LeChatelier¶s Principle
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Reaction Rates
Questions to consider:
What makes ³superglue´ bond instantly
while Elmer¶s glue does not?
What factors determine how quickly foodspoils?
Why do ³glow sticks´ last longer when
stored in the freezer?
How do catalytic converters removevarious pollutants from car exhaust?
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Reaction Rates
These types of questions can be answered
using chemical kinetics.
The study of the speed or rate at which
chemical reactions occur
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Reaction Rates
The rate of a chemical reaction is affected
by many factors, including:
concentration of reactantsas concentration of reactants
increases the rate of reaction generally
increases
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Reaction Rates
The rate of a chemical reaction is affected
by many factors (cont):
reaction temperature food spoils more quickly at room
temperature than in a refrigerator
bacteria grow faster at RT than at
lower temperatures
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Reaction Rates
The rate of a chemical reaction is affected
by many factors (cont):
presence of a catalyst
a substance that increases the rate of a reaction without being consumed in
the reaction
Enzymes
biological catalysts proteins that increase the rate of
biochemical reactions
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Reaction Rates
The rate of a chemical reaction is affected
by many factors (cont):
surface area of solid or liquid reactantsor catalysts
as surface area increases the rate of
reaction generally increases
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Reaction Rates
The speed of an object or event is the
change that occurs in a given time interval.
Speed of a car = change in distancetime interval
= (d
(t
Remember, the term change always refers to
final value minus initial value.
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Reaction Rates
Similarly, the rate (or speed) of a reaction can
be determined:
Rate = change in concentration (or moles) of producttime interval
Rate = ( (conc. or moles)
(t
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Reaction Rates
Consider the chemical reaction:
A B
Time = 0.
10. mol A
t = 20. min
5.0 mol A
5.0 mol B
t = 40. min
2.0 mol A
8.0 mol B
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Reaction Rates
The reaction rate for a chemical reaction
can be expressed as either:
the increase in concentration (or number of moles) of a product as a function of
time.
the decrease in concentration (or number of moles) of a reactant as a function of
time
OR
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Reaction Rates
In this reaction:
Average rate of
appearance of B = change in # of moles of Bchange in time
= ( (mol B)
( t
We can calculate the average rate for any
time interval involved in the reaction.
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Reaction Rates
If we consider the rate of appearance of B
over the first 20 minutes of reaction:
Average rate of appearance of B = ( (mol B)
( t
= 5.0 mol B ± 0.0 mol B20. min ± 0. min
= 0.25 mol/min
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Reaction Rates
The average rate of appearance of B during
the second 20 minutes of the reaction:
Avg. rate = 8.0 mol B ± 5.0 mol B
40. min ± 20. min
= 0.15 mol/min
Notice that the average rate of reaction
decreases over the course of the reaction.
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Reaction Rates
The rate of a reaction can also be
expressed as the disappearance of A as a
function of time.
For this particular reaction, when 1 mole of
B is formed, 1 mole of A must disappear.
A B
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Reaction Rates
Time (B (A
Interval (t (t
0 ± 20.0 min 0.25 mol
min
20.0 ± 40.0 min 0.15 mol
min
Notice: (B/(t = - (A/(t
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Reaction Rates
We don¶t want to report two different values
for the rate of the reaction.
For reactions with 1:1 stoichiometry:
Avg. rate = ( (moles product)
( t
= - ( (moles reactant)
( t
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Reaction Rates
For most reactions, the reaction rate is
expressed as a change in concentration of
a particular reactant or product
Average Rate = ( [Product] = - ( [Reactant]
( t ( t
where [Product] = concentration of product[Reactant] = concentration of reactant
units: M / sec or M / min
M = molarity = moles/liter
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Reaction Rates
On the exam, you will be expected to find
the average rate of reaction for a specific
time interval when given the concentration
or number of moles of either reactants or
products as a function of time.
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Reaction Rates
Example: Given the following data, what is
the average rate of the following reaction over
the time interval from 54.0 min to 215.0 min?
CH3OH (aq) + HCl (aq) CH3Cl (aq) + H2O (l)
Time (min) [HCl] (M)
0.0 1.85
54.0 1.58
107.0 1.36
215.0 1.02
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Reaction Rates
Given: [HCl]54 min = 1.58 M
[HCl]215 min = 1.02 M
Find: avg. rate of disappearance of HCl
Avg. rate = - ( [HCl]
( t
= - (1.02 M - 1.58 M)215 min - 54 min
= 0.0035 M / min
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Reaction Rates
Example: Calculate the average reaction ratefor the reaction A B during the first 60.0minutes using the following data:
Time [A]
0.0 min 1.50 M20.0 min 1.00 M
40.0 min 0.80 M60.0 min 0.75 M80.0 min 0.70 M
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Reaction Rates
So far, all reactions have had a one-to-one
stoichiometry.
What happens when the coefficients arenot all 1?
2 A 3B
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Reaction Rates
Consider the following reaction:
2 HI (g) H2 (g) + I2 (g)
Time mol mol mol
(min) HI H2 I2
0.0 2.00 0.0 0.0
10.0 1.50
20.0 1.00
30.0 0.75
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Reaction Rates
Calculate the change in HI and H2 as a functionof time for the first 20.0 minutes of reaction:
2 HI (g) H2 (g) + I2 (g)
(HI (H2
Time mol mol mol (t (t
(min) HI H2 I2 (mol/min)
0.0 2.00 0.0 0.0
10.0 1.50 0.25 0.25
20.0 1.00 0.50 0.50
30.0 0.75 0.75 0.75
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Reaction Rates
The average reaction rate must be
numerically the same, regardless of whether
you express it as the rate of appearance of
product or the rate of disappearance of
reactant.
HI disappears twice as fast as H2 appears. To
make the rates equal:
Rate = - 1 ( [HI] = ( [H2]
2 (t ( t
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Reaction Rates
In general, for a reaction:
a A + b B c C + d D
the rate of the reaction can be found by:
Rate = - 1 ([A] = - 1 ([B] = 1 ([C] = 1 ([D]
a (t b (t c (t d (t
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Reaction Rates
Rate = - 1 ([A] = - 1 ([B] = 1 ([C] = 1 ([D]
a (t b (t c (t d (t
This equation can be used to establish therelationship between rate of change of one
reactant or product to another reactant or
product.
You have to be able to do this on the
test, too!
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Reaction Rates
Example: How is the rate of disappearance of
N2O5 related to the rate of appearance of NO2
in the following reaction?
2 N2O5 (g) 4 NO2 (g) + O2 (g)
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Reaction Rates
Example: If the rate of decomposition of N2O5
in the previous example at a particular instant
is 4.2 x 10-7M /s, what is the rate of
appearance of NO2
?
2 N2O5 (g) 4 NO2 (g) + O2 (g)
Given: - ([N2O5] = 4.2 x 10-7 M /s
( t
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Reaction Rates
2 N2O5 (g) 4 NO2 (g) + O2 (g)
Rate = - 1 ([N2O5] = 1 ([NO2]
2(
t 4(
t
So:
([NO2] = - 4
([N2O5]
( t 2 ( t
= 2 x 4.2 x 10-7 M /s = 8.4 x 10-7 M/s
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Reaction Rates
Recall that the average reaction rate
changes during the course of the reaction.
Until now, we have calculated average
reaction rates.
The reaction rate at a particular time (not
time interval) is called the instantaneous
reaction rate.
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Reaction Rate
The instantaneous reaction rate is found by
determining the slope of a line tangent to
the curve at the particular time of interest.
Fortunately (for you), you won¶t have to do
this on the exam or HW!
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Rate Laws
Consider the data presented earlier for the
disappearance of HCl as a function of time
for the following reaction.
CH3OH (aq) + HCl (aq) CH3Cl (aq) + H2O (l)
Time (min) [HCl] (M)
0.0 1.8554.0 1.58
107.0 1.36
215.0 1.02
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Rate Laws
The average reaction rate decreases with
time.
The reaction slows down as the
concentration of reactants decreases.
CH3OH (aq) + HCl (aq) CH3Cl (aq) + H2O (l)
Time (min) [HCl] (M) Avg. Rate (M /min)
0.0 1.8554.0 1.58 0.0050
107.0 1.36 0.0042
215.0 1.02 0.0031
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Rate Laws
In general, the rate of any reaction depends
on the concentration of reactants.
The way in which the reaction rate varieswith the concentration of the reactants can
be expressed mathematically using a rate
law.
An equation that shows how the reactionrate depends on the concentration of the
reactants
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Rate Laws
For a generalized chemical reaction:
w A + x B y C + z D
the general form of the rate law is:
Rate = k[A]m [B]n
where k = rate constant
m, n = reaction order
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Rate Laws
Rate Constant (k) a proportionality constant that relates the
concentration of reactants to the reaction
rate
Reaction Order the power to which the concentration of
a reactant is raised in a rate law
Overall reaction order The sum of all individual reaction orders
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Rate Laws
Rate laws must be determined
experimentally.
Measure the instantaneous reaction rate
at the start of the reaction (i.e. at t = 0) for various concentrations of reactants.
You CANNOT determine the rate law by
looking at the coefficients in the balancedchemical equation!
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Rate Laws
First Order Reaction
Overall reaction order = 1
Rate = k[A]
Expt [A] (M) Rate (M/s)
1 0.50 1.00
2 1.00 2.00
3 2.00 4.00
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Rate Laws
Second Order Reaction
Overall reaction order = 2
Rate = k[A]2
Expt [A] (M) Rate (M/s)
1 0.50 0.50
2 1.00 2.00
3 1.50 4.50
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Rate Laws
Third Order Reaction
Overall reaction order = 3
Rate = k[A]3
Expt [A] (M) Rate (M/s)
1 0.50 0.25
2 1.00 2.00
3 1.50 6.75
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Rate Laws
Zero Order Reaction
Overall reaction order = 0
Rate = k[A]0 = k
Expt [A] (M) Rate (M/s)
1 0.50 2.00
2 1.00 2.00
3 1.50 2.00
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Rate Laws
REMEMBER
Rate laws must be determined
experimentally.
Determine the instantaneous reaction
rate at the start of the reaction (i.e. at t =
0) for various concentrations of
reactants.
You CANNOT determine the rate law by
looking at the coefficients in the balanced
chemical equation!
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Rate Laws
To determine the rate law from
experimental data,
identify two experiments in which the
concentration of one reactant has beenchanged while the concentration of the
other reactant(s) has been held constant
determine how the reaction rate changedin response to the change in the
concentration of that reactant.
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Rate Laws
To determine the rate law from
experimental data (cont)
Repeat this process using another set of
data in which the concentration of thefirst reactant is held constant while the
concentration of the other one is
changed.
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Rate Laws
Example: The initial reaction rate of the
reaction A + B C was measured for
several different starting concentration of A
and B. The following results were obtained.Determine the rate law for the reaction.
Expt # [A] (M) [B] (M) Initial rate (M /s)
1 0.100 0.100 4.0 x 10-5
2 0.100 0.200 8.0 x 10-5
3 0.200 0.100 16.0 x 10-5
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Rate Laws
Rate = k [A]m [B]n
Compare experiments 1 and 2 to find n:
[A] = constant[B] = doubles
Expt # [A] (M) [B] (M) Initial rate (M /s)
1 0.100 0.100 4.0 x 10-5
2 0.100 0.200 8.0 x 10-5
3 0.200 0.100 16.0 x 10-5
Rate doubles:8.0 x 10-5 = 2.0
4.0 x 10-5
x 2x 2 x 2x 2
[2]n = 2.0n = 1
Rate = k[A]m[B]1
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Rate Laws
Rate = k [A]m [B]n
Compare experiments 1 and 3 to find m:
[A] = doubles[B] = constant
Expt # [A] (M) [B] (M) Initial rate (M /s)1 0.100 0.100 4.0 x 10-5
2 0.100 0.200 8.0 x 10-5
3 0.200 0.100 16.0 x 10-5
Rate quadruples:16.0 x 10-5 = 4.0
4.0 x 10-5
x 2x 2 x 4x 4
[2]m = 4.0n = 2
Rate = k[A]2[B]
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Rate Laws
You can also solve this using algebra:
Rate = k [A]m [B]n
Compare experiments 1 and 3 to find m:
Rate 2 = k [0.200 M]m [0.100 M]n = 16.0 x 10-5 =4.0
Rate 1 k [0.100 M]m [0.100 M]n 4.0 x 10-5
[0.200 M]m = 4.0
[0.100 M]m
2m = 4.0 only if m = 2
[2.00]m = 2.0
Rate = k[A]2[B]n
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Rate Laws
You can also solve this using algebra:
Rate = k [A]m [B]n
Compare experiments 1 and 2 to find n:
Rate 2 = k [0.100 M]m [0.200 M]n = 8.0 x 10-5 = 2.0
Rate 1 k [0.100 M]m [0.100 M]n 4.0 x 10-5
[0.200 M]n = 2.0
[0.100 M]n
2n = 2.0 only if n = 1
[2.00]n = 2.0
Rate = k[A]2[B]