chapter 14 kinetics...box 1: rate = k(5)(5)2 = 125k box 2 contains 7 red spheres and 3 purple...
TRANSCRIPT
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Chapter 14
Chemical Kinetics
James F. Kirby
Quinnipiac University
Hamden, CT
Lecture Presentation
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Chemical Kinetics• In chemical kinetics we study the rate (or
speed) at which a chemical process occurs.
• Besides information about the speed at
which reactions occur, kinetics also sheds
light on the reaction mechanism, a
molecular-level view of the path from
reactants to products.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
14.1 Factors that Affect
Reaction Rates
1) Physical state of the reactants
2) Reactant concentrations
3) Reaction temperature
4) Presence of a catalyst
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Physical State of the Reactants
• The more readily the reactants collide,
the more rapidly they react.
• Homogeneous reactions are often
faster.
• Heterogeneous reactions that involve
solids are faster if the surface area is
increased; i.e., a fine powder reacts
faster than a pellet or tablet.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Reactant Concentrations
• Increasing reactant
concentration generally
increases reaction rate.
• Since there are more
molecules, more
collisions occur.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Temperature
• Reaction rate generally increases with
increased temperature.
• Kinetic energy of molecules is related to
temperature.
• At higher temperatures, molecules
move more quickly, increasing numbers
of collisions and the energy the
molecules possess during the collisions.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Presence of a Catalyst
• Catalysts affect rate without being in the
overall balanced equation.
• Catalysts affect the kinds of collisions,
changing the mechanism (individual
reactions that are part of the pathway
from reactants to products).
• Catalysts are critical in many biological
reactions.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
14.2 Reaction Rate• Rate is a change in concentration over
a time period: Δ[ ]/Δt.
• Δ means “change in.”
• [ ] means molar concentration.
• t represents time.
• Types of rate measured:
average rate
instantaneous rate
initial rate
Chemical
Kinetics
© 2015 Pearson Education, Inc.
From the data in Figure 14.3, calculate the
average rate at which A disappears over the
time interval from 20 s to 40 s.
Sample Exercise 14.1 Calculating an Average Rate of Reaction (pg. 2 of notes)
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solution
Analyze We are given the concentration of A at 20 s (0.54 M) and at 40 s (0.30 M) and asked to calculate
the average rate of reaction over this time interval.
Plan The average rate is given by the change in concentration, Δ[A], divided by the change in time, Δt.
Because A is a reactant, a minus sign is used in the calculation to make the rate a positive quantity.
Solve
From the data in Figure 14.3, calculate the average rate at which A disappears over the time interval from
20 s to 40 s.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Following Reaction Rates
Rate of a reaction
is measured using
the concentration
of a reactant or a
product over time.
In this example,
[C4H9Cl] is
followed.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Following Reaction Rates
The average rate
is calculated by the
–(change in
[C4H9Cl]) ÷
(change in time).
The table shows
the average rate
for a variety of time
intervals.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Plotting Rate Data• A plot of the data gives more
information about rate.
• The slope of the curve at one
point in time gives the
instantaneous rate.
• The instantaneous rate at
time zero is called the initial
rate; this is often the rate of
interest to chemists.
• This figure shows
instantaneous and initial rate
of the earlier example.
Note: Reactions
typically slow down
over time!
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Using Figure
14.4, calculate the
instantaneous
rate of
disappearance of
C4H9Cl at
t = 0 s (the initial
rate).
Sample Exercise 14.2 Calculating an Instantaneous Rate of Reaction (pg.2)
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solution
Analyze We are asked to determine an
instantaneous rate from a graph of reactant
concentration versus time.
Using Figure 14.4, calculate the instantaneous rate
of disappearance of C4H9Cl at
t = 0 s (the initial rate).
Plan To obtain the instantaneous rate at t = 0 s, we must
determine the slope of the curve at t = 0. The tangent is
drawn on the graph as the hypotenuse of the tan triangle.
The slope of this straight line equals the change in the
vertical axis divided by the corresponding change in the
horizontal axis (which, in the case of this example, is the
change in molarity over change in time).
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solve The tangent line falls
from [C4H9Cl] = 0.100 M to
0.060 M in the time change
from 0 s to 210 s. Thus, the
initial rate is
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Relative Rates
• As was said, rates are followed using a reactant or
a product. Does this give the same rate for each
reactant and product?
• Rate is dependent on stoichiometry.
• If we followed use of C4H9Cl and compared it to
production of C4H9OH, the values would be the
same. Note that the change would have opposite
signs—one goes down in value, the other goes up.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Relative Rates and Stoichiometry
• What if the equation is not 1:1?
• What will the relative rates be for:
2 O3 (g) → 3 O2 (g)
Chemical
Kinetics
© 2015 Pearson Education, Inc.
(a) How is the rate at which ozone
disappears related to the rate at which
oxygen appears in the reaction
2 O3(g) → 3O2(g)?
(b) If the rate at which O2 appears,
Δ[O2]/Δt, is 6.0 × 10–5 M/s at a particular
instant, at what rate is O3 disappearing at
this same time, –Δ[O3]/Δt?
Sample Exercise 14.3 Relating Rates at Which Products Appear and Reactants Disappear (pg. 3)
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solution
Analyze We are given a balanced chemical
equation and asked to relate the rate of appearance
of the product to the rate of disappearance of the
reactant.
Plan We can use the coefficients in the chemical
equation as shown in Equation 14.4 to express the
relative rates
of reactions.
(a) How is the rate at which ozone disappears related to the rate at
which oxygen appears in the reaction
2 O3(g) → 3O2(g)?
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solution
Solve
(a) Using the coefficients in the balanced equation and the
relationship given by Equation 14.4, we have:
(a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the
reaction
2 O3(g) → 3O2(g)?
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Check We can apply a stoichiometric factor to convert the
O2 formation rate to the O3 disappearance rate:
(b)If the rate at which O2 appears, Δ[O2]/Δt, is 6.0 × 10–5
M/s at a particular instant, at what rate is O3 disappearing
at this same time, –Δ[O3]/Δt? Solving the equation from
part (a) for the rate at which O3 disappears, –Δ[O3]/Δt, we
have:
Chemical
Kinetics
© 2015 Pearson Education, Inc.
14.3 Determining
Concentration Effect on Rate• How do we determine what effect the
concentration of each reactant has on
the rate of the reaction?
• We keep every concentration constant
except for one reactant and see what
happens to the rate. Then, we change a
different reactant. We do this until we
have seen how each reactant has
affected the rate.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
An Example of How
Concentration Affects Rate• Experiments 1–3 show how [NH4
+] affects rate.
• Experiments 4–6 show how [NO2−] affects rate.
• Result: The rate law, which shows the
relationship between rate and concentration for all
reactants:
Rate = k [NH4+] [NO2
−]
Chemical
Kinetics
© 2015 Pearson Education, Inc.
An Example of How Concentration Affects Rate
rate law Rate = k [NH4+] [NO2
−]
Chemical
Kinetics
© 2015 Pearson Education, Inc.
More about Rate Law
• The exponents tell the order of the reaction with
respect to each reactant.
• In our example from the last slide:
Rate = k [NH4+] [NO2
−]
• The order with respect to each reactant is 1. (It is
first order in NH4+ and NO2
−.)
• The reaction is second order (1 + 1 = 2; we just
add up all of the reactants’ orders to get the
reaction’s order).
• What is k? It is the rate constant. It is a
temperature-dependent quantity.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Consider a reaction A + B → C for which rate =
k[A][B]2. Each of the following boxes represents
a reaction mixture in which A is shown as red
spheres and B as purple ones. Rank these
mixtures in order of increasing rate of reaction.
Sample Exercise 14.4 Relating a Rate Law to the Effect of Concentration on Rate (pg.5)
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solution
Analyze We are given three boxes containing
different numbers of spheres representing mixtures
containing different reactant concentrations. We are
asked to use the given rate law and the
compositions of the boxes to rank the mixtures in
order of increasing reaction rates.
Plan Because all three boxes have the same
volume, we can put the number of spheres of each
kind into the rate law and calculate the rate for each
box.
Consider a reaction A + B → C for which rate = k[A][B]2. Each of the following boxes represents
a reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures
in order of increasing rate of reaction.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solution
Solve Box 1 contains 5 red spheres and 5 purple spheres,
giving the following rate:
Box 1: Rate = k(5)(5)2 = 125k
Box 2 contains 7 red spheres and 3 purple spheres:
Box 2: Rate = k(7)(3)2 = 63k
Box 3 contains 3 red spheres and 7 purple spheres:
Box 3: Rate = k(3)(7)2 = 147k
The slowest rate is 63k (Box 2), and the highest is 147k (Box
3). Thus, the rates vary in the order 2 < 1 < 3.
Consider a reaction A + B → C for which rate = k[A][B]2. Each of the following boxes represents
a reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures
in order of increasing rate of reaction.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Check Each box contains 10 spheres. The rate law
indicates that in this case [B] has a greater
influence on rate than [A] because B has a larger
reaction order. Hence, the mixture with the highest
concentration of B (most purple spheres) should
react fastest. This analysis confirms the order 2 < 1
< 3.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
2N2O5(g) → 4NO2(g) + O2(g) [14.9]
Rate = k [N2O5]
H2(g) + I2(g) → 2HI2(g) [14.10]
Rate = k [H2] [I2]
CHCl3(g) + Cl2(g) → CCl4(g) + HCl(g) [14.11]
Rate = k [CHCl3] [Cl2]1/2
The following are some additional examples of
experimentally determined rate laws:
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solution
Analyze We are given two rate laws and asked to express
(a) the overall reaction order for each and (b) the units for
the rate constant for the first reaction.
Plan The overall reaction order is the sum of the exponents
in the rate law. The units for the rate constant, k, are found
by using the normal units for rate (M/s) and concentration
(M) in the rate law and applying algebra to solve for k.
(a)What are the overall reaction orders for the reactions described in
Equations 14.9 and 14.11?
(b)What are the units of the rate constant for the rate law in Equation
14.9?
Sample Exercise 14.5 Determining Reaction Orders and Units for Rate Constants (pg. 5)
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solution
Solve
(a)The rate of the reaction in Equation 14.9 is first
order in N2O5 and first order overall. The reaction
in Equation 14.11 is first order in CHCl3 and one-
half order in Cl2. The overall reaction order is
three halves.
(a) What are the overall reaction orders for the reactions described in Equations 14.9 and 14.11?
Sample Exercise 14.5 Determining Reaction Orders and Units for Rate Constants (pg. 5)
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solution
Solve
(b) For the rate law for Equation 14.9, we have
Units of rate = (units of rate constant)(units of concentration)
so…
Notice that the units of the rate constant change as
the overall order of the reaction changes.
(b)What are the units of the rate constant for the rate law in Equation 14.9?
Sample Exercise 14.5 Determining Reaction Orders and Units for Rate Constants (pg. 5)
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solution
Analyze We are given a table of data that relates concentrations of reactants with initial rates of reaction
and asked to determine (a) the rate law, (b) the rate constant, and (c) the rate of reaction for a set of
concentrations not listed in the table.
Plan (a) We assume that the rate law has the following form: Rate = k[A]m[B]n. We will use the given data to
deduce the reaction orders m and n by determining how changes in the concentration change the rate. (b)
Once we know m and n, we can use the rate law and one of the sets of data to determine the rate constant
k. (c) Upon determining both the rate constant and the reaction orders, we can use the rate law with the
given concentrations to calculate rate.
The initial rate of a reaction A + B → C was measured for several different starting concentrations of A
and B, and the results are as follows:
Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the
reaction when [A] = 0.050M and [B] = 0.100 M.
Sample Exercise 14.6 Determining a Rate Law from Initial
Rate Data
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solve
(a) If we compare experiments 1 and 2, we see that [A] is held constant and [B] is doubled. Thus, this pair
of experiments shows how [B] affects the rate, allowing us to deduce the order of the rate law with
respect to B. Because the rate remains the same when [B] is doubled, the concentration of B has no
effect on the reaction rate. The rate law is therefore zero order in B (that is, n = 0).
In experiments 1 and 3, [B] is held constant, so these
data show how [A] affects rate. Holding [B] constant
while doubling [A] increases the rate fourfold. This
result indicates that rate is proportional to [A]2 (that is,
the reaction is second order in A). Hence, the rate law is
(b) Using the rate law and the data from experiment 1,
we have
(c) Using the rate law from part (a) and the
rate constant from part (b), we have
Because [B] is not part of the rate law, it is irrelevant to the rate if there is at least some B present to react
with A.
Check A good way to check our rate law is to use the concentrations in experiment 2 or 3 and see if we
can correctly calculate the rate. Using data from experiment 3, we have
Rate = k[A]2 = (4.0 × 10–3 M –1 s–1)(0.200 M)2 = 1.6 × 10–4 M/s
Thus, the rate law correctly reproduces the data, giving both the correct number and the correct units for
the rate.
Continued
Sample Exercise 14.6 Determining a Rate Law from Initial
Rate Data
Chemical
Kinetics
© 2015 Pearson Education, Inc.
First Order Reactions
• Some rates depend only on one
reactant to the first power.
• These are first order reactions.
• The rate law becomes:
Rate = k [A]
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Relating k to [A] in a
First Order Reaction• rate = k [A]
• rate = −Δ [A] / Δt
• So: k [A] = −Δ [A] / Δt
• Rearrange to: Δ [A] / [A] = − k Δt
• Integrate: ln ([A] / [A]o) = − k t
• Rearrange: ln [A] = − k t + ln [A]o
• Note: this follows the equation of a line:
y = m x + b
• So, a plot of ln [A] vs. t is linear.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
An Example: Conversion of Methyl
Isonitrile to Acetonitrile• The equation for the reaction:
CH3NC → CH3CN
• It is first order.
Rate = k [CH3NC]
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Finding the Rate Constant, k
• Besides using the rate law, we can find
the rate constant from the plot of ln [A]
vs. t.
• Remember the integrated rate law:
ln [A] = − k t + ln [A]o
• The plot will give a line. Its slope will
equal −k.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
The decomposition of a certain insecticide in water at
12 °C follows first-order kinetics with a rate constant
of 1.45 yr–1. A quantity of this insecticide is washed
into a lake on June 1, leading to a concentration of
5.0 × 10–7 g/cm3. Assume that the temperature of
the lake is constant (so that there are no effects of
temperature variation on the rate). (a) What is the
concentration of the insecticide on June 1 of the
following year? (b) How long will it take for the
insecticide concentration to decrease to 3.0 × 10–7
g/cm3?
Sample Exercise 14.7 Using the Integrated First-Order Rate Law (pg.7 notes)
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Analyze We are given the rate constant for a reaction that
obeys first order kinetics, as well as information about
concentrations and times, and asked to calculate how much
reactant (insecticide) remains after 1 yr. We must also
determine the time interval needed to reach a particular
insecticide concentration. Because the exercise gives time
in (a) and asks for time in (b), we will find it most useful to
use the integrated rate law, Equation 14.13.
Plan
(a)We are given k = 1.45 yr–1, t = 1.00 yr, and [insecticide]0 =
5.0 × 10–7 g/cm3, and so Equation 14.13 can be solved for
[insecticide]t .
The decomposition of a certain insecticide in water at 12 °C follows first-order kinetics with a rate
constant of 1.45 yr–1. A quantity of this insecticide is washed into a lake on June 1, leading to a
concentration of 5.0 × 10–7 g/cm3. Assume that the temperature of the lake is constant (so that there are
no effects of temperature variation on the rate). (a) What is the concentration of the insecticide on June 1
of the following year? (b) How long will it take for the insecticide concentration to decrease to 3.0 × 10–7
g/cm3?
Sample Exercise 14.7 Using the Integrated First-Order Rate Law
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solve
(a)Substituting the known quantities into Equation 14.13, we have
We use the ln function on a calculator to evaluate the
second term on the right [that is, ln(5.0 × 10–7)], giving
To obtain [insecticide]t = 1 yr, we use the inverse
natural logarithm, or ex, function on the calculator:
Note that the concentration units for [A]t and [A]0 must be the
same.
ln[insecticide]t = 1 yr = –(1.45 yr–1)(1.00 yr) + ln(5.0 × 10–7)
ln[insecticide]t = 1 yr = –1.45 + (–14.51) = –15.96
[insecticide]t = 1 yr = e–15.96 = 1.2 × 10–7g/cm3
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solution
Analyze We are given the rate constant for a reaction that
obeys first order kinetics, as well as information about
concentrations and times, and asked to calculate how much
reactant (insecticide) remains after 1 yr. We must also
determine the time interval needed to reach a particular
insecticide concentration. Because the exercise gives time
in (a) and asks for time in (b), we will find it most useful to
use the integrated rate law, Equation 14.13.
Plan
(b)We have k = 1.45 yr–1, [insecticide]0 = 5.0 × 10–7 g/cm3,
and [insecticide]t = 3.0 × 10–7 g/cm3, and so we can solve
Equation 14.13 for time, t.
The decomposition of a certain insecticide in water at 12 °C follows first-order kinetics with a rate
constant of
1.45 yr–1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 ×10–7 g/cm3. Assume that the temperature of the lake is constant (so that there are no effects of temperature
variation on the rate). (a) What is the concentration of the insecticide on June 1 of the following year? (b)
How long will it take for the insecticide concentration to decrease to 3.0 × 10–7 g/cm3?
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solve
(b) Again substituting into Equation 14.13, with
[insecticide]t = 3.0 × 10–7 g/cm3, gives
Solving for t gives
Check In part (a) the concentration remaining after 1.00 yr (that is, 1.2 × 10–7
g/cm3) is less than the original concentration (5.0 × 10–7 g/cm3), as it should be.
In (b) the given concentration (3.0 × 10–7 g/cm)2 is greater than that remaining
after 1.00 yr, indicating that the time must be less than a year. Thus, t = 0.35 yr is
a reasonable answer.
ln(3.0 × 10–7) = –(1.45 yr–1)(t) + ln(5.0 × 10–7)
t = –[ln(3.0 × 10–7) – ln(5.0 × 10–7)]/1.45 yr–1
= –(–15.02 + 14.51)/1.45 yr–1 = 0.35 yr
© 2015 Pearson Education, Inc.
14.2 Reaction Rates (review) Average rates could represent the average rate of the
disappearance of a reactant (let’s say [A])
Change in [A] - ∆[A]
change in time ∆ t
Average rates could represent the average rate of the
appearance of a product (let’s say [B])
Change in [B] ∆[B]
change in time ∆ t=
=
© 2015 Pearson Education, Inc.
14.2 Change in Rate with Time Instantaneous Rate (or Initial Rate): Rate at a
particular instant during the reaction
Also the slope at any particular instant
Slope also ∆Y
∆ X
There is a better way to calculate the
slope, but it involves calculus, which
we will not do in this class.
=
Note: Reactions typically slow down over time!
© 2015 Pearson Education, Inc.
14.2 Reaction Rates and Stoichiometry
The rate of disappearance of a reactant is equaled to the
rate of appearance of a product because the
stoichiometric ratio is 1:1
What if it’s not 1:1
2 HI(g) H2(g) + I2(g)
Relative Reaction Rate Equation
-1 ∆[HI] ∆[H2] ∆[I2]
2 ∆ t ∆ t ∆ t= =
© 2015 Pearson Education, Inc.
14.2 Reaction Rates and Stoichiometry
2 HI(g) H2(g) + I2(g)
Relative Reaction Rate Equation
-1 ∆[HI] ∆[H2] ∆[I2]
2 ∆ t ∆ t ∆ t
Relative reaction rate equations let us calculate how
molecules’ relative concentrations change over time.
These equations do not let us calculate actual
concentrations: only proportional concentrations of
reactants relative to products.
= =
© 2015 Pearson Education, Inc.
14.3 Concentration and Rates Laws
Reaction Orders (General Rate Law Equation)
Rate k[A]m[B]n
The terms m and n are called reaction orders
and have to be determined experimentally.
Note that m and n are not always the same as
the coefficients in front of A and B in the
balanced chemical reaction.
=
© 2015 Pearson Education, Inc.
14.3 Concentration and Rates Laws
Reaction Orders (General Rate Law Equation)
Rate k[NH4+][NO2
-]
The exponents above are both 1. We say, then, that
the reaction is first order in NH4+ and first order in NO2
-.
These numbers are called the individual reaction
orders for each reactant. The overall reaction order is
obtained by adding these together (1+1=2). This
reaction, then, is second order overall.
=
© 2015 Pearson Education, Inc.
14.4 The Change of Concentration with Time
The general rate law does not let us
calculate how molecules’ concentrations
change over time. It only tells us how the
rate will be affected by altering [A] and [B].
So what if we want to be able to calculate
the actual concentrations of reactants and
products for a reaction over time? Can we
do that?
© 2015 Pearson Education, Inc.
14.4 The Change of Concentration with Time
Yes, we can.
The equations we use are called
integrated rate laws, and they
combine elements from our relative
reaction rate equations and our
general rate laws.
© 2015 Pearson Education, Inc.
14.4 The Change of Concentration with Time
These equations vary, depending on
whether the reaction is first order,
second order or third order and so
forth. We will only discuss first and
second order in this class.
© 2015 Pearson Education, Inc.
14.4 The Change of Concentration with Time
A simple reactions looks like A products, it can sometimes
be (not always) first order. For such a reaction
- ∆[A] K[A]∆ t
=
We can take this
equation and use
calculus…… calculus
…to transform it
into this equation ln [A] = − kt + ln [A]o
This is the
integrated rate
law for 1st order
reactions
© 2015 Pearson Education, Inc.
14.4 The Change of Concentration with Time
You should notice that the integrate rate law for first
order reactions…..
…..Has the formula y = mx + b: Y = mx + b
ln [A] = − kt + ln [A]o
So the graph of a first order reaction will show a
straight line if you plot ln [A] vs. time. It’s slope is − k
and it’s y-intercept is ln [A]o
© 2015 Pearson Education, Inc.
Integrate Rate Law for Second Order Reactions
In contrast, a simple reaction that looks like A
products, A + B products can sometimes be (but
isn’t always) second order. For such a reaction
- ∆[A] K[A]2
∆ t
Or if both [A] and [B] contribute to the reaction rate
then, Rate = k[A]m[B]n . For this class we will consider
rate = k[A]2
=
© 2015 Pearson Education, Inc.
14.4 The Change of Concentration with Time
For this type of reaction
- ∆[A] K[A]2
∆ t=
We can transform this
using calculus……calculus
…to obtain this
[A] = kt + [A]o
This is the
integrated rate
law for 2nd order
reactions
1 _ 1 _
© 2015 Pearson Education, Inc.
14.4 The Change of Concentration with Time
This equation also has the formula y = mx + b:
Y = mx + b
[A] = − kt + [A]o
So the graph of a second order reaction will show a
straight line if you plot 1/[A] vs. time. It’s slope is k and
it’s y-intercept is 1/[A]o If you need to distinguish
between 1st & 2nd graph both.
1 _ 1 _
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Half-life• Definition: The amount of time
it takes for one-half of a
reactant to be used up in a
chemical reaction.
• First Order Reaction:
ln [A] = − k t + ln [A]o
ln ([A]o/2) = − k t½ + ln [A]o
− ln ([A]o/2) + ln [A]o = k t½
ln ([A]o / [A]o/2) = k t½
ln 2 = k t½ or t½ = 0.693/k
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Second Order Reactions
• Some rates depend only on a reactant
to the second power.
• These are second order reactions.
• The rate law becomes:
Rate = k [A]2
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solving the Second Order
Reaction for A → Products
• rate = k [A]2
• rate = − Δ [A] / Δ t
• So, k [A]2 = − Δ [A] / Δ t
• Rearranging: Δ [A] / [A]2 = − k Δ t
• Using calculus: 1/[A] = 1/[A]o + k t
• Notice: The linear relationships for first
order and second order reactions differ!
Chemical
Kinetics
© 2015 Pearson Education, Inc.
An Example of a Second
Order Reaction:
Decomposition of NO2A plot following NO2
decomposition shows
that it must be second
order because it is
linear for 1/[NO2], not
linear for ln [NO2].
Equation:
NO2 → NO + ½ O2
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Half-Life and Second Order
Reactions
• Using the integrated rate law, we can
see how half-life is derived:
1/[A] = 1/[A]o + k t
1/([A]o/2) = 1/[A]o + k t½
2/[A]o −1/[A]o = k t½
t½ = 1 / (k [A]o)
• So, half-life is a concentration
dependent quantity for second order
reactions!
Chemical
Kinetics
© 2015 Pearson Education, Inc.
The reaction of C4H9Cl
with water is a first-order
reaction. (a) Use Figure
14.4 to estimate the half-
life for this reaction. (b)
Use the half-life from (a) to
calculate the rate
constant.
Sample Exercise 14.9 Determining the Half-Life of a First-Order
Reaction (pg. 10)
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solution
Analyze We are asked to estimate the
half-life of a reaction from a graph of
concentration versus time and then
to use the half-life to calculate the
rate constant for the reaction.
The reaction of C4H9Cl with water is a first-order reaction. (a) Use
Figure 14.4 to estimate the half-life for this reaction.
Sample Exercise 14.9 Determining the Half-Life of a First-Order Reaction
Plan
(a)To estimate a half-life, we can select a
concentration and then determine the time required
for the concentration to decrease to half of that
value.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Solve
(a) From the graph, we
see that the initial value of
[C4H9Cl] is 0.100 M. The
half-life for this first-order
reaction is the time
required for [C4H9Cl] to
decrease to 0.050 M, which
we can read off the graph.
This point occurs at
approximately 340 s.
Sample Exercise 14.9 Determining the Half-Life of a First-Order Reaction
Chemical
Kinetics
© 2015 Pearson Education, Inc.
(b) Use the half-life from (a) to calculate the rate constant.
Solve
(a) 340 s.
(b) Solving Equation 14.15 for k, we have
Check At the end of the second half-life, which should
occur at 680 s, the concentration should have decreased by
yet another factor of 2, to 0.025 M. Inspection of the graph
shows that this is indeed the case.
Sample Exercise 14.9 Determining the Half-Life of a First-Order Reaction
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Zero Order Reactions
• Occasionally, rate is
independent of the
concentration of the
reactant:
• Rate = k
• These are zero order
reactions.
• These reactions are
linear in
concentration.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Factors That Affect Reaction
Rate
1. Temperature
2. Frequency of collisions
3. Orientation of molecules
4. Energy needed for the reaction to take
place (activation energy)
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Factors That Affect Reaction Rate
1. Temperature? Most reactions speed up with
increased temperature.
Why? Let’s look at the rate law equation again:
Rate = k [A]m[B]nDoes changing the temperature change [A]? Does
changing the temperature change [B]? So why does
increasing the temperature increase a reactions' rate?
No
No
Because it increases the rate constant k
So how does it do that? The answer is explained
by using something called the collision model.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Factors That Affect Reaction Rate
Molecules have to collide to react. The greater
the number of collisions per second, the greater
the chance that molecules will get together in a
way that leads them to react.
Increasing a reaction’s temperature increases
individual molecules’ speeds. This increases the
total number of collisions they experience and,
hence, the overall likelihood that molecules will
get together in a way that leads them to react.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Factors That Affect Reaction Rate
But colliding by itself, isn't enough to
automatically cause a reaction to occur. For
example, in a mixture of H2 and I2, the molecules
undergo about 1010 collisions per second, but the
reaction still proceeds very slowly. In fact, only
one in every 1013 collisions causes a reaction.
Why?
The answer comes down to two additional factors:
orientation and activation energy.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Temperature and Rate
• Generally, as
temperature increases,
rate increases.
• The rate constant is
temperature dependent:
it increases as
temperature increases.
• Rate constant doubles
(approximately) with
every 10 ºC rise.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Frequency of Collisions• The collision model is based on the kinetic
molecular theory.
• Molecules must collide to react.
• If there are more collisions, more reactions
can occur.
• So, if there are more molecules, the
reaction rate is faster.
• Also, if the temperature is higher, molecules
move faster, causing more collisions and a
higher rate of reaction.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
The Collision Model
• In a chemical reaction, bonds are
broken and new bonds are formed.
• Molecules can only react if they collide
with each other.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Orientation of Molecules
• Molecules can often collide without
forming products.
• Aligning molecules properly can lead to
chemical reactions.
• Bonds must be broken and made and
atoms need to be in proper positions.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Energy Needed for a Reaction to
Take Place (Activation Energy)
• The minimum energy needed for a reaction to take
place is called activation energy.
• An energy barrier must be overcome for a reaction
to take place, much like the ball must be hit to
overcome the barrier in the figure below.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Transition State
(Activated Complex)
• Reactants gain energy as the reaction
proceeds until the particles reach the
maximum energy state.
• The organization of the atoms at this
highest energy state is called the
transition state (or activated complex).
• The energy needed to form this state is
called the activation energy.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Reaction Progress
• Plots are made to show
the energy possessed
by the particles as the
reaction proceeds.
• At the highest energy
state, the transition state
is formed.
• Reactions can be
endothermic or
exothermic after this.
transition
state
transition state – Very
unstable (transition
products are also
called intermediates)
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Reaction Progress• Suppose you could
measure the rates for
both the forward and
reverse reactions of the
process in the figure. In
which direction would
the rate be larger?
• Why? The energy barrier is lower in the forward
direction than in the reverse. Thus more
molecules will have energy sufficient to
cross the barrier in the forward direction.
The forward rate will be greater.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Reaction Progress
• Suppose we have two
reactions A B and
B C. You could isolate
B, and it is stable. Is B
the transition state for
the reaction A C?
No. If B can be isolated, it can‘t correspond
to the top of the energy barrier. There would
be transition states for each of the individual
reactions above.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Distribution of the Energy
of Molecules
• Gases have an average temperature, but
each individual molecule has its own energy.
• At higher energies, more molecules possess
the energy needed for the reaction to occur.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
The Relationship Between
Activation Energy & Temperature
• Like everything else in chemistry, there exists an
equation we can use to calculate activation
energy (Ea). It is called the Arrhenius Equation:
k = Ae−Ea/RT
• Where k is the rate constant, R is the ideal gas
constant (8.314 J/mol-K), and yes, you have to
use this value of R, instead of the others in your
book, for this particular equation), T is the reaction
temperature, Ea is its activation energy, and A is a
constant called a frequency factor.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
The Relationship Between Activation Energy & Temperature
• Reaction constant k changes with temperature.
By manipulating the Arrehenius equation in some
ways we can calculate what the new k value will
be if we changed temperature, by using the
following equation:
• Allows us to calculate the temperature if we have
rate constants at two different temperatures.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Try this problem:
A certain first-order reaction has a rate constant of
2.75x10-2 s-1 at 20 C. What is the k at 60 C if
Ea = 75.5 kj/mol?
Chemical
Kinetics
© 2015 Pearson Education, Inc.
The Relationship Between
Activation Energy & Temperature
• Activation energy can be determined graphically
by reorganizing the equation: ln k = −Ea/RT + ln A
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Law vs. Theory
• Kinetics gives what happens. We call
the description the rate law.
• Why do we observe that rate law? We
explain with a theory called a
mechanism.
• A mechanism is a series of stepwise
reactions that show how reactants
become products.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Reaction Mechanisms
• Reactions may occur all at once or
through several discrete steps.
• Each of these processes is known as an
elementary reaction or elementary
process.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Molecularity
The molecularity of an elementary reaction
tells how many molecules are involved in that
step of the mechanism.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Termolecular?
• Termolecular steps require three
molecules to simultaneously collide with
the proper orientation and the proper
energy.
• These are rare, if they indeed do occur.
• These must be slower than
unimolecular or bimolecular steps.
• Nearly all mechanisms use only
unimolecular or bimolecular reactions.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
What Limits the Rate?
• The overall reaction cannot occur faster than
the slowest reaction in the mechanism.
• We call that the rate-determining step.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
What is Required of a
Plausible Mechanism?
• The rate law must be able to be devised
from the rate-determining step.
• The stoichiometry must be obtained when
all steps are added up.
• Each step must balance, like any equation.
• All intermediates are made and used up.
• Any catalyst is used and regenerated.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
A Mechanism With a Slow
Initial Step
• Overall equation: NO2 + CO → NO + CO2
• Rate law: Rate = k [NO2]2
• If the first step is the rate-determining step,
the coefficients on the reactants side are the
same as the order in the rate law!
• So, the first step of the mechanism begins:
NO2 + NO2 →
Chemical
Kinetics
© 2015 Pearson Education, Inc.
A Mechanism With a Slow
Initial Step (continued)
• The easiest way to complete the first
step is to make a product:
NO2 + NO2 → NO + NO3
• We do not see NO3 in the stoichiometry,
so it is an intermediate, which needs to
be used in a faster next step.
NO3 + CO → NO2 + CO2
Chemical
Kinetics
© 2015 Pearson Education, Inc.
A Mechanism With a Slow
Initial Step (completed)
• Since the first step is the slowest step, it
gives the rate law.
• If you add up all of the individual steps
(2 of them), you get the stoichiometry.
• Each step balances.
• This is a plausible mechanism.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
A Mechanism With a
Fast Initial Step
• Equation for the reaction:
2 NO + Br2 ⇌ 2 NOBr
• The rate law for this reaction is found to be
Rate = k [NO]2 [Br2]
• Because termolecular processes are rare,
this rate law suggests a multistep
mechanism.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
A Mechanism With a Fast
Initial Step (continued)
• The rate law indicates that a quickly
established equilibrium is followed by a
slow step.
• Step 1: NO + Br2 ⇌ NOBr2
• Step 2: NOBr2 + NO → 2 NOBr
Chemical
Kinetics
© 2015 Pearson Education, Inc.
What is the Rate Law?
• The rate of the overall reaction depends
upon the rate of the slow step.
• The rate law for that step would be
Rate = k2[NOBr2] [NO]
• But how can we find [NOBr2]?
Chemical
Kinetics
© 2015 Pearson Education, Inc.
[NOBr2] (An Intermediate)?
• NOBr2 can react two ways:
– With NO to form NOBr.
– By decomposition to reform NO and Br2.
• The reactants and products of the first
step are in equilibrium with each other.
• For an equilibrium (as we will see in the
next chapter):
Ratef = Rater
Chemical
Kinetics
© 2015 Pearson Education, Inc.
The Rate Law (Finally!)
• Substituting for the forward and reverse rates:
k1 [NO] [Br2] = k−1 [NOBr2]
• Solve for [NOBr2], then substitute into the rate
law:
Rate = k2 (k1/k−1) [NO] [Br2] [NO]
• This gives the observed rate law!
Rate = k [NO]2 [Br2]
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Catalysts
• Catalysts increase the rate of a reaction by
decreasing the activation energy of the
reaction.
• Catalysts change the mechanism by which
the process occurs.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Types of Catalysts
1) Homogeneous catalysts
2) Heterogeneous catalysts
3) Enzymes
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Homogeneous Catalysts
• The reactants and catalyst are in the
same phase.
• Many times, reactants and catalyst are
dissolved in the same solvent, as seen
below.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Heterogeneous Catalysts• The catalyst is in a
different phase than
the reactants.
• Often, gases are
passed over a solid
catalyst.
• The adsorption of
the reactants is
often the rate-
determining step.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Enzymes
• Enzymes are
biological catalysts.
• They have a region
where the reactants
attach. That region
is called the active
site. The reactants
are referred to as
substrates.
Chemical
Kinetics
© 2015 Pearson Education, Inc.
Lock-and-Key Model
• In the enzyme–substrate model, the
substrate fits into the active site of an
enzyme, much like a key fits into a lock.
• They are specific.