Transcript Unit 4 Chemical Kinetics and Chemical Equilibrium

Unit 4
Chemical Kinetics and
Chemical Equilibrium






Reaction Rates
Rate Laws
First and Second Order Reactions
Chemical Equilibrium
Equilibrium Constants
LeChatelier’s Principle
Reaction Rates

Questions to consider:
 What makes “superglue” bond instantly
while Elmer’s glue does not?
 What factors determine how quickly food
spoils?
 Why do “glow sticks” last longer when
stored in the freezer?
 How do catalytic converters remove
various pollutants from car exhaust?
Reaction Rates

These types of questions can be answered
using chemical kinetics.
 The study of the speed or rate at which
chemical reactions occur
Reaction Rates

The rate of a chemical reaction is affected
by many factors, including:

concentration of reactants
 as concentration of reactants
increases the rate of reaction generally
increases
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
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
Reaction Rates

The rate of a chemical reaction is affected
by many factors (cont):

surface area of solid or liquid reactants
or catalysts
 as surface area increases the rate of
reaction generally increases
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 distance
time interval
= Dd
Dt
Remember, the term change always refers to
final value minus initial value.
Reaction Rates

Similarly, the rate (or speed) of a reaction can
be determined:
Rate = change in concentration (or moles) of product
time interval
Rate = D (conc. or moles)
Dt
Reaction Rates
Consider the chemical reaction:
A
Time = 0.
10. mol A
t = 20. min
5.0 mol A
5.0 mol B
B
t = 40. min
2.0 mol A
8.0 mol B
Reaction Rates
moles A or B
If the number of moles of A and B are
measured and plotted, a graph such as this
one can be obtained
12
10
8
6
4
2
0
moles A
moles B
0
20
40
time (min)
60
80
This data
can be
used to
find the
reaction
rate.
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
OR time.


the decrease in concentration (or number
of moles) of a reactant as a function of
time
Reaction Rates
In
this reaction:
Average rate of
appearance of B = change in # of moles of B
change in time
= D (mol B)
Dt
 We can calculate the average rate for any
time interval involved in the reaction.
Reaction Rates
If we consider the rate of appearance of B
over the first 20 minutes of reaction:
Average rate of
appearance of B = D (mol B)
Dt
= 5.0 mol B – 0.0 mol B
20. min – 0. min
= 0.25 mol/min
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.
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
Reaction Rates
Time
Interval
DB
Dt
0 – 20.0 min
0.25 mol
min
20.0 – 40.0 min
0.15 mol
min
Notice: DB/Dt = - DA/Dt
DA
Dt
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 = D (moles product)
Dt
= - D (moles reactant)
Dt
Reaction Rates

For most reactions, the reaction rate is
expressed as a change in concentration of
a particular reactant or product
Average Rate = D [Product] = - D [Reactant]
Dt
Dt
where [Product] = concentration of product
[Reactant] = concentration of reactant

units: M / sec or M / min
 M = molarity = moles/liter
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.
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)
Time (min)
0.0
54.0
107.0
215.0
CH3Cl (aq) + H2O (l)
[HCl] (M)
1.85
1.58
1.36
1.02
Reaction Rates
Given: [HCl]54 min = 1.58 M
[HCl]215 min = 1.02 M
Find:
avg. rate of disappearance of HCl
Avg. rate = - D [HCl]
Dt
= - (1.02 M - 1.58 M)
215 min - 54 min
= 0.0035 M / min
Reaction Rates
Example: Calculate the average reaction rate
for the reaction A  B during the first 60.0
minutes using the following data:
Time
0.0 min
20.0 min
40.0 min
60.0 min
80.0 min
[A]
1.50 M
1.00 M
0.80 M
0.75 M
0.70 M
Reaction Rates

So far, all reactions have had a one-to-one
stoichiometry.

What happens when the coefficients are
not all 1?
2A
3B
Reaction Rates
Consider the following reaction:
2 HI (g)
H2 (g) + I2 (g)
Time
(min)
mol
HI
mol
H2
mol
I2
0.0
10.0
20.0
30.0
2.00
1.50
1.00
0.75
0.0
0.0
Reaction Rates
Calculate the change in HI and H2 as a function
of time for the first 20.0 minutes of reaction:
2 HI (g)
H2 (g) + I2 (g)
Time
(min)
mol
HI
mol
H2
mol
I2
0.0
10.0
20.0
30.0
2.00
1.50
1.00
0.75
0.0
0.25
0.50
0.75
0.0
0.25
0.50
0.75
DHI
DH2
Dt
Dt
(mol/min)
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 D [HI] = D [H2]
2 Dt
Dt
Reaction Rates

In general, for a reaction:
aA + bB
cC +dD
the rate of the reaction can be found by:
Rate = - 1 D[A] = - 1 D[B] = 1 D[C] = 1 D[D]
a Dt
b Dt
c Dt
d Dt
Reaction Rates
Rate = - 1 D[A] = - 1 D[B] = 1 D[C] = 1 D[D]
a Dt
b Dt
c Dt
d Dt
This equation can be used to establish the
relationship 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!
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)
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: - D[N2O5] = 4.2 x 10-7 M /s
Dt
Reaction Rates
2 N2O5 (g)
4 NO2 (g) + O2 (g)
Rate = - 1 D[N2O5] = 1 D[NO2]
2 Dt
4 Dt
So:
D[NO2] = - 4 D[N2O5]
Dt
2 Dt
= 2 x 4.2 x 10-7 M /s = 8.4 x 10-7 M/s
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.
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!
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)
Time (min)
0.0
54.0
107.0
215.0
CH3Cl (aq) + H2O (l)
[HCl] (M)
1.85
1.58
1.36
1.02
Rate Laws

The average reaction rate decreases with
time.
 The reaction slows down as the
concentration of reactants decreases.
CH3OH (aq) + HCl (aq)
Time (min)
0.0
54.0
107.0
215.0
[HCl] (M)
1.85
1.58
1.36
1.02
CH3Cl (aq) + H2O (l)
Avg. Rate (M /min)
0.0050
0.0042
0.0031
Rate Laws

In general, the rate of any reaction depends
on the concentration of reactants.

The way in which the reaction rate varies
with the concentration of the reactants can
be expressed mathematically using a rate
law.
 An equation that shows how the reaction
rate depends on the concentration of the
reactants
Rate Laws

For a generalized chemical reaction:
wA+xB
yC+zD
the general form of the rate law is:
Rate = k[A]m [B]n
where k = rate constant
m, n = reaction order
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
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 balanced
chemical equation!
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
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
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
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
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!
Rate Laws

To determine the rate law from
experimental data,
 identify two experiments in which the
concentration of one reactant has been
changed while the concentration of the
other reactant(s) has been held constant

determine how the reaction rate changed
in response to the change in the
concentration of that reactant.
Rate Laws

To determine the rate law from
experimental data (cont)
 Repeat this process using another set of
data in which the concentration of the
first reactant is held constant while the
concentration of the other one is
changed.
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 #
1
2
3
[A] (M)
0.100
0.100
0.200
[B] (M)
0.100
0.200
0.100
Initial rate (M /s)
4.0 x 10-5
8.0 x 10-5
16.0 x 10-5
Rate Laws
Expt #
1
2
3
[A] (M)
0.100
0.100
0.200
[B] (M)
0.100
x2
0.200
0.100
Initial rate (M /s)
4.0 x 10-5
x2
-5
8.0 x 10
16.0 x 10-5
Rate = k [A]m [B]n
Compare experiments 1 and 2 to find n:
[A] = constant
Rate doubles:
[2]n = 2.0
[B] = doubles
8.0 x 10-5 = 2.0
n=1
4.0 x 10-5
Rate = k[A]m[B]1
Rate Laws
Expt #
1
2
3
[A] (M)
0.100
0.100 x 2
0.200
[B] (M)
0.100
0.200
0.100
Initial rate (M /s)
4.0 x 10-5
x4
-5
8.0 x 10
16.0 x 10-5
Rate = k [A]m [B]n
Compare experiments 1 and 3 to find m:
[A] = doubles
Rate quadruples: [2]m = 4.0
[B] = constant
16.0 x 10-5 = 4.0
n=2
4.0 x 10-5
Rate = k[A]2[B]
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
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]