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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.
Calculating rates
• When we travel at a speed of 70 mph, we will cover a distance of 70 miles in one hour.
• The rate or speed can change while you are driving a car.
• The rate of a chemical reaction changes over time as well.
Rate of reaction• Chemists measure the change in concentration per unit time.
• The reaction rate is used to describe how reactions evolve over time.
• We can measure the rate at which the reaction proceeds by measuring the decrease in the concentration of reactant A over time, or by measuring the increase in concentration of product C over time.
5
Chemical Kinetics
Thermodynamics – does a reaction take place?
Kinetics – how fast does a reaction proceed?
Reaction rate is the change in the concentration of a
reactant or a product with time (M/s).
A B
rate = -D[A]
Dt
rate = D[B]
Dt
D[A] = change in concentration of A over
time period Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is negative.
6
A B
rate = -D[A]
Dt
rate = D[B]
Dt
Solved problem
RelationshipsMultiply the rate by the mole ratio to solve for each unknown rate
SolveRate of N2 appearance:
Answer2AgNO3(aq) + MgBr2(aq) → 2AgBr(s) + Mg(NO3)2(aq)
According to the reaction below, how quickly will N2 appear if the rate of disappearance of oxygen is 0.050 mol/s? 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(g)
Factors affecting rate
• The following factors affect the reaction rate.
1. The concentration of the reacting chemicals
2. The temperature at which the reaction is carried out
3. The surface area of the reacting particles
4. The addition of a catalyst
• A catalyst is a substance that facilitates a chemical reaction without being used up itself.
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.
Reactant Concentrations
• Increasing reactant concentration generally increases reaction rate.
• Since there are more molecules, more collisions occur.
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.
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.
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
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.
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.
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!
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.
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)
19
Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
time
393 nm
light
Detector
D[Br2] a D Absorption
red-brown
t1< t2 < t3
20
Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
average rate = -D[Br2]
Dt= -
[Br2]final – [Br2]initial
tfinal - tinitial
slope of
tangentslope of
tangentslope of
tangent
instantaneous rate = rate for specific instance in time
21
rate a [Br2]
rate = k [Br2]
k = rate
[Br2]= rate constant
= 3.50 x 10-3 s-1
22
2H2O2 (aq) 2H2O (l) + O2 (g)
PV = nRT
P = RT = [O2]RTnV
[O2] = PRT
1
rate = D[O2]
Dt RT
1 DP
Dt=
measure DP over time
23
24
Reaction Rates and Stoichiometry
2A B
Two moles of A disappear for each mole of B that is formed.
rate = D[B]
Dtrate = -
D[A]
Dt
1
2
aA + bB cC + dD
rate = -D[A]
Dt
1
a= -
D[B]
Dt
1
b=
D[C]
Dt
1
c=
D[D]
Dt
1
d
13.1
Write the rate expressions for the following reactions in terms of
the disappearance of the reactants and the appearance of the
products:
13.1
Strategy To express the rate of the reaction in terms of the
change in concentration of a reactant or product with time, we
need to use the proper sign (minus or plus) and the reciprocal
of the stoichiometric coefficient.
Solution
(a) Because each of the stoichiometric coefficients equals 1,
(b) Here the coefficients are 4, 5, 4, and 6, so
13.2
Consider the reaction
Suppose that, at a particular moment during the reaction,
molecular oxygen is reacting at the rate of 0.024 M/s.
(a) At what rate is N2O5 being formed?
(b) At what rate is NO2 reacting?
Strategy To calculate the rate of formation of N2O5 and
disappearance of NO2, we need to express the rate of the
reaction in terms of the stoichiometric coefficients as in
Example 13.1:
We are given
where the minus sign shows that the concentration of O2 is
decreasing with time.
13.2
13.2
Solution
(a) From the preceding rate expression we have
Therefore
13.2
(b) Here we have
so
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.
32
The Rate Law
The rate law expresses the relationship of the rate of a reaction
to the rate constant and the concentrations of the reactants
raised to some powers.
aA + bB cC + dD
Rate = k [A]x[B]y
Reaction is xth order in A
Reaction is yth order in B
Reaction is (x +y)th order overall
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
−]
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.
Rate laws
• The speed of reactions can be classified into categories or orders.
• The reaction order is an exponent in a rate law. Rate laws have the following general form:
• Typically chemical reactions are either zero, first or second order.
13.3
The reaction of nitric oxide with hydrogen at 1280°C is
From the following data collected at this temperature, determine
(a) the rate law
(b) the rate constant
(c) the rate of the reaction when [NO] = 12.0 × 10−3 M and
[H2] = 6.0 × 10 −3 M
13.3
Strategy We are given a set of concentration and reaction rate
data and asked to determine the rate law and the rate constant.
We assume that the rate law takes the form
rate = k[NO]x [H2]y
How do we use the data to determine x and y?
Once the orders of the reactants are known, we can calculate k
from any set of rate and concentrations.
Finally, the rate law enables us to calculate the rate at any
concentrations of NO and H2.
13.3
Solution (a) Experiments 1 and 2 show that when we double the concentration of NO at constant concentration of H2, the rate quadruples. Taking the ratio of the rates from these two experiments
Therefore,
or x = 2, that is, the reaction is second order in NO.
13.3
Experiments 2 and 3 indicate that doubling [H2] at constant
[NO] doubles the rate. Here we write the ratio as
Therefore,
or y = 1, that is, the reaction is first order in H2. Hence the rate
law is given by
which shows that it is a (2 + 1) or third-order reaction overall.
13.3
(b) The rate constant k can be calculated using the data from
any one of the experiments. Rearranging the rate law, we
get
The data from experiment 2 give us
13.3
(c) Using the known rate constant and concentrations of NO
and H2, we write
Comment Note that the reaction is first order in H2, whereas
the stoichiometric coefficient for H2 in the balanced equation
is 2. The order of a reactant is not related to the stoichiometric
coefficient of the reactant in the overall balanced equation.
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]
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.
An Example: Conversion of Methyl Isonitrile to Acetonitrile
• The equation for the reaction:
CH3NC → CH3CN
• It is first order.
Rate = k [CH3NC]
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.
46
First-Order Reactions
A product rate = -D[A]
Dtrate = k [A]
k = rate
[A]= 1/s or s-1M/s
M=
D[A]
Dt= k [A]-
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
[A] = [A]0e−kt
ln[A] = ln[A]0 - kt
13.4
The conversion of cyclopropane to propene in the gas phase is a first-order reaction with a rate constant of 6.7 × 10−4 s−1 at 500°C.
(a) If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 min?
(b) How long (in minutes) will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M?
(c) How long (in minutes) will it take to convert 74 percent of the starting material?
13.4
Strategy The relationship between the concentrations of a reactant at different times in a first-order reaction is given by Equation (13.3) or (13.4).
In (a) we are given [A]0 = 0.25 M and asked for [A]t after 8.8 min.
In (b) we are asked to calculate the time it takes for cyclopropane to decrease in concentration from 0.25 M to 0.15 M.
No concentration values are given for (c). However, if initially we have 100 percent of the compound and 74 percent has reacted, then what is left must be (100% − 74%), or 26%. Thus, the ratio of the percentages will be equal to the ratio of the actual concentrations; that is, [A]t/[A]0 = 26%/100%, or 0.26/1.00.
13.4
Solution (a) In applying Equation (13.4), we note that because k is given in units of s−1, we must first convert 8.8 min to seconds:
We write
Hence,
Note that in the ln [A]0 term, [A]0 is expressed as a dimensionless quantity (0.25) because we cannot take the logarithm of units.
13.4
(b) Using Equation (13.3),
(c) From Equation (13.3),
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
52
First-Order Reactions
The half-life, t½, is the time required for the concentration of a
reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2
ln[A]0
[A]0/2
k=t½
ln 2
k=
0.693
k=
53
A product
First-order reaction
# of
half-lives [A] = [A]0/n
1
2
3
4
2
4
8
16
13.6
The decomposition of ethane (C2H6) to methyl radicals is a first-
order reaction with a rate constant of 5.36 × 10−4 s−1 at 700°C:
Calculate the half-life of the reaction in minutes.
13.6
Strategy To calculate the half-life of a first-order reaction, we
use Equation (13.6). A conversion is needed to express the
half-life in minutes.
Solution For a first-order reaction, we only need the rate
constant to calculate the half-life of the reaction. From
Equation (13.6)
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
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!
An Example of a Second Order Reaction: Decomposition of NO2
A plot following NO2
decomposition shows that it must be second order because it is linear for 1/[NO2], notlinear for ln [NO2].
Equation:
NO2 → NO + ½ O2
59
Second-Order Reactions
A product rate = -D[A]
Dtrate = k [A]2
k = rate
[A]2= 1/M•s
M/sM2=
D[A]
Dt= k [A]2-
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
1
[A]=
1
[A]0+ kt
t½ = t when [A] = [A]0/2
t½ =1
k[A]0
13.7
Iodine atoms combine to form molecular iodine in the gas
phase
This reaction follows second-order kinetics and has the high
rate constant 7.0 × 109/M · s at 23°C.
(a) If the initial concentration of I was 0.086 M, calculate the
concentration after 2.0 min.
(b) Calculate the half-life of the reaction if the initial
concentration of I is 0.60 M and if it is 0.42 M.
13.7
Strategy
(a) The relationship between the concentrations of a reactant at
different times is given by the integrated rate law. Because
this is a second-order reaction, we use Equation (13.7).
(b) We are asked to calculate the half-life. The half-life for a
second-order reaction is given by Equation (13.8).
Solution (a) To calculate the concentration of a species at a
later time of a second−order reaction, we need the initial
concentration and the rate constant. Applying Equation (13.7)
13.7
where [A]t is the concentration at t = 2.0 min. Solving the
equation, we get
This is such a low concentration that it is virtually undetectable.
The very large rate constant for the reaction means that nearly
all the I atoms combine after only 2.0 min of reaction time.
(b) We need Equation (13.8) for this part.
For [I]0 = 0.60 M
13.7
For [I]0 = 0.42 M
Check These results confirm that the half-life of a second-
order reaction, unlike that of a first-order reaction, is not a
constant but depends on the initial concentration of the
reactant(s).
Does it make sense that a larger initial concentration should
have a shorter half-life?
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!
Zero Order Reactions• Occasionally, rate is
independent of the concentration of the reactant:
• Rate = k
• These are zero orderreactions.
• These reactions are linear in concentration.
66
Zero-Order Reactions
A product rate = -D[A]
Dtrate = k [A]0 = k
k = rate
[A]0= M/s
D[A]
Dt= k-
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t = 0
t½ = t when [A] = [A]0/2
t½ =[A]02k
[A] = [A]0 - kt
Rate orders
• Zero Order Rate= k[A]0
– When a reactant is zero order, its concentration has no effect on the reaction rate.
• First Order Rate= k[A]1
– When a reactant is first order, its concentration has a direct effect on reaction rate.
• Second Order Rate= k[A]2
– When a reactant is second order, its concentration has an exponential effect on reaction rate.
68
Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order Rate Law
Concentration-Time
Equation Half-Life
0
1
2
rate = k
rate = k [A]
rate = k [A]2
ln[A] = ln[A]0 - kt
1
[A]=
1
[A]0+ kt
[A] = [A]0 - kt
t½ln 2
k=
t½ =[A]02k
t½ =1
k[A]0
Solved Problem (Part 1)
Given
RelationshipsRate = k[CO]m[NO2]n
SolveStep 1: Determine the order with respect to CO and NO2
Compare Trials 1 & 2. The concentration of CO does not change but the concentration of NO2 doubles. The reaction rate also doubles, so the reaction is 1st order for NO2.
Compare Trials 1 & 3. the concentration of NO2 does not change but the concentration CO doubles. The reaction rate also doubles, the reaction is also 1st order for CO.
Therefore, Rate = k[CO]1[NO2]1
What is the rate for Trial 4?
CO + NO2 → CO2 + NO
Trial # [CO] [NO2] Rate mol/L⋅hr
1 1.0 x 10-3 7.2 x 10-3 3.4 x 10-8
2 1.0 x 10-3 3.6 x 10-3 1.7 x 10-8
3 2.0 x 10-3 7.2 x 10-3 6.8 x 10-8
4 3.0 x 10-3 1.4 x 10-2 ?
Solved problem (Part 2)
Given
SolveStep 2: Determine the Rate Constant and UnitsUsing Trial 3, substitute values into the rate law.Rate = k[CO]1[NO2]1
6.8 x 10-8 mol/L⋅hr = k [2.0 x 10-3]1[7.2 x 10-3]1
What is the rate for Trial 4?CO + NO2 → CO2 + NO
Trial # [CO] [NO2] Rate mol/L⋅hr
1 1.0 x 10-3 7.2 x 10-3 3.4 x 10-8
2 1.0 x 10-3 3.6 x 10-3 1.7 x 10-8
3 2.0 x 10-3 7.2 x 10-3 6.8 x 10-8
4 3.0 x 10-3 1.4 x 10-2 ?
Solved Problem (Part 3)
Relationships Rate = k[CO]m[NO2]n
Solve Step 3: Determine the rate of Trial 4Put trial four values into the rate law.
Answer Trial four will occur at 2.0 x 10-7 mol/L⋅hr.
What is the rate for Trial 4?
Using the rate law
• Trails 1 and 2: The concentration of NO was doubled and the concentration of O2 was kept constant.
– the rate increased by a factor of 4 from 2.5 x 10-5 M/s to 1.0 x 10-4 M/s. This is an exponential increase. The reaction is second order with respect to NO.
• Trails 1 and 3: The concentration of O2 was doubled and the concentration of NO was kept constant.
– the rate increased by a factor of 2 from 5.0x10-5 M/s to 1.0x10-4 M/s. This is a direct increase. The reaction must be first order with respect to O2.
• To determine the rate constant, we will substitute the data from any one of the trials into the rate expression.
Zero order
• The orders of reactions can also be determined graphically.
• If the plot of Concentration vs Time is the most linear the reaction is zero order.
First order
• If the plot of ln(Concentration) vs Time is the most linear the reaction is first order.
Second order
• If the plot of 1/(Concentration) vs Time is the most linear the reaction is a second order reaction.
Solved problem (Part 1)
Given
Relationships The shape of the graph helps you determine reaction order. • If the plot of Concentration vs Time is the most linear
the reaction is zero order.• If the plot of ln(Concentration) vs Time is the most linear
the reaction is first order.• If the plot of 1/(Concentration) vs Time is the most linear
the reaction is second order.
The following data was collected for NO2. What is the order with respect to NO2? Graph the following data and determine order of the reaction.
Time (s) [NO2] ln[NO2] 1/[NO2]
0 1.00 x 10−2 −4.605 100
60 6.83 x 10−3 −4.986 146
120 5.18 x 10−3 −5.263 193
180 4.18 x 10−3 −5.477 239
240 3.50 x 10−3 −5.655 286
300 3.01 x 10−3 −5.806 332
360 2.64 x 10−3 −5.937 379
Solved problem (Part 2)Solve Graph each expression of concentration in the table:
• If the plot of Concentration vs Time is the most linear the reaction is zero order.
• If the plot of ln(Concentration) vs Time is the most linear the reaction is first order.
• If the plot of 1/(Concentration) vs Time is the most linear the reaction is second order.
Answer The reaction is second order with respect to NO2, because the plot of 1/[NO2] is the most linear.
Overall order
The overall order of a simple reaction is calculated by adding the reactant exponents in the rate law.
2NO + O2 → 2NO2
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)
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.
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.
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.
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.
The collision theory• Reactants must collide with each other in order for a
reaction to occur.
• A successful collision is one that causes products to be made. It must satisfy two conditions.
1. The collisions must be hard enough to break the chemical bonds of the reactants.
2. The particles must be aligned so that once the bonds are broken they will form products.
The reaction energy profile
• The minimum amount of energy required for molecules to react is called the activation energy, Ea.
• At the highest point of the reaction profile the activated complex, Ac, forms.
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.
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.
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.
89
Exothermic Reaction Endothermic Reaction
The activation energy (Ea ) is the minimum amount of
energy required to initiate a chemical reaction.
A + B AB C + D++
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.
The Relationship Between Activation Energy & Temperature
• Arrhenius noted relationship between activation energy and temperature: k = Ae−Ea/RT
• Activation energy can be determined graphically by reorganizing the equation: ln k = −Ea/RT + ln A
92
Temperature Dependence of the Rate Constant
Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature
A is the frequency factor
ln k = -Ea
R
1
T+ lnA
(Arrhenius equation)
)/( RTEaeAk
Alternate format:
13.8
The rate constants for the decomposition of acetaldehyde
were measured at five different temperatures. The data are
shown in the table. Plot ln k versus 1/T, and determine the
activation energy (in kJ/mol) for the reaction. Note that the
reaction is “3/2” order in CH3CHO, so k has the units of
1/M½ ·s.
13.8
Strategy Consider the Arrhenius equation written as a linear
equation
A plot of ln k versus 1/T (y versus x) will produce a straight line
with a slope equal to −Ea/R. Thus, the activation energy can be
determined from the slope of the plot.
13.8
Solution First we convert the data to the following table
A plot of these data yields the graph in Figure 13.18. The slope
of the line is calculated from two pairs of coordinates:
13.8
From the linear form of Equation (13.13)
Check It is important to note that although the rate constant
itself has the units 1/M½ · s, the quantity ln k has no units (we
cannot take the logarithm of a unit).
97
Alternate Form of the Arrhenius Equation
At two temperatures, T1 and T2
or
13.9
The rate constant of a first-order reaction is 3.46 × 10−2 s−1 at
298 K.
What is the rate constant at 350 K if the activation energy for
the reaction is 50.2 kJ/mol?
13.9
Strategy A modified form of the Arrhenius equation relates two
rate constants at two different temperatures [see Equation
(13.14)]. Make sure the units of R and Ea are consistent.
Solution The data are
Substituting in Equation (13.14),
13.9
We convert Ea to units of J/mol to match the units of R. Solving
the equation gives
Check The rate constant is expected to be greater at a higher
temperature. Therefore, the answer is reasonable.
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.
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.
103
Importance of Molecular Orientation
effective collision
ineffective collision
104
Reaction Mechanisms
The overall progress of a chemical reaction can be represented
at the molecular level by a series of simple elementary steps
or elementary reactions.
The sequence of elementary steps that leads to product
formation is the reaction mechanism.
2NO (g) + O2 (g) 2NO2 (g)
N2O2 is detected during the reaction!
Elementary step: NO + NO N2O2
Elementary step: N2O2 + O2 2NO2
Overall reaction: 2NO + O2 2NO2
+
Reaction mechanisms• Chemical reactions do not always go straight from reactants
to products.
• In fact, a balanced equation should be thought of as an abridged version of a reaction.
• The actual pathway is often several steps long.
• Reactions occur in a series of elementary steps.
• The series of steps is referred to as a reaction mechanism.
Two-step reaction mechanisms
• The sum of the elementary steps must yield the overall balanced equation.
• A reaction intermediate is a temporary product formed in an elementary step that is consumed in a later elementary step such as N2O2(g) in the example above.
• Reactions occur in several steps because it is unlikely that 3 reactant molecules will collide with the desired energy and position.
107
2NO (g) + O2 (g) 2NO2 (g)
Mechanism:
108
Elementary step: NO + NO N2O2
Elementary step: N2O2 + O2 2NO2
Overall reaction: 2NO + O2 2NO2
+
Intermediates are species that appear in a reaction
mechanism but not in the overall balanced equation.
An intermediate is always formed in an early elementary step
and consumed in a later elementary step.
The molecularity of a reaction is the number of molecules
reacting in an elementary step.
• Unimolecular reaction – elementary step with 1 molecule
• Bimolecular reaction – elementary step with 2 molecules
• Termolecular reaction – elementary step with 3 molecules
Molecularity
The molecularity of an elementary reaction tells how many molecules are involved in that step of the mechanism.
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 onlyunimolecular or bimolecular reactions.
111
Sequence of Steps in Studying a Reaction Mechanism
13.10
The gas-phase decomposition of nitrous oxide (N2O) is believed to occur via two elementary steps:
Experimentally the rate law is found to be rate = k[N2O].
(a) Write the equation for the overall reaction.
(b) Identify the intermediate.
(c) What can you say about the relative rates of steps 1 and 2?
13.10
Strategy
(a) Because the overall reaction can be broken down into
elementary steps, knowing the elementary steps would
enable us to write the overall reaction.
(b) What are the characteristics of an intermediate? Does it
appear in the overall reaction?
(c) What determines which elementary step is rate
determining? How does a knowledge of the rate-
determining step help us write the rate law of a reaction?
13.10
Solution
(a) Adding the equations for steps 1 and 2 gives the overall
reaction
(b) Because the O atom is produced in the first elementary step
and it does not appear in the overall balanced equation, it is
an intermediate.
(c) If we assume that step 1 is the rate-determining step, then
the rate of the overall reaction is given by
and k = k1.
13.10
Check There are two criteria that must be met for a proposed
reaction mechanism to be plausible.
(1) The individual steps (elementary steps) must sum to the
corrected overall reaction.
(2) The rate-determining step (the slow step) must have the
same rate law as the experimentally determined rate law.
Reaction pathways
• A unimolecular elementary step refers to a process with a single reactant dissociating or breaking apart. (Step 3)
• A bimolecular elementary step involves the collision of just two molecules. (Step 1 and 2)
What Limits the Rate?• The overall reaction cannot occur faster than the
slowest reaction in the mechanism.
• We call that the rate-determining step.
The rate determining step
• The rate determining step is the elementary process that takes the longest, and acts as a bottleneck in the overall reaction mechanism.
• Even though the later steps may be fast, they cannot begin until the rate determining step is complete.
A biological pathway
• Throughout the process of fermentation the sugar molecule is modified by several enzyme-facilitated chemical reactions
• The breakdown of the glucose molecule via fermentation requires 12 enzymes!
• First step:C6H12O6 → 2CH3COCO2H
• Second step:2CH3COCO2H → 2C2H5OH + 2CO2
• Overall reaction:C6H12O6 → 2C2H5OH + 2CO2 + energy
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.
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 →
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
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.
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.
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
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]?
[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
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]
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.
Types of Catalysts
1) Homogeneous catalysts
2) Heterogeneous catalysts
3) Enzymes
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.
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.
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.
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.
Reaction profiles
• In the graph above, the reaction A + B → C + D is shown catalyzed and non-catalyzed.
• The uncatalyzed reaction energy is depicted in orange and the catalyzed version is in green.
• With the lower activation energy, less energy input is needed to complete the reaction and more molecules have enough energy to react.
• The reaction now occurs at a faster rate overall.
Inhibitors
• An inhibitor is a substance that slows a reaction down by increasing the activation energy.
Enzymes
• Enzymes are biological catalysts, responsible for speeding up the chemical reactions inside our bodies.
• The natural enzyme, bromelain, is contained in certain fruits. Pineapple contains bromelain which helps to digest or “break down” proteins.
Catalysts and the environment
• The Cl⋅ free radical is a catalyst because it is present in the same form at the start of the reaction and at the end of the reaction.
Catalysts and the World Around Us
• Approximately 90% of all commercially produced chemical products use catalysts at some point in their process.
• Environmental chemists also design ways to speed up reactions that decrease the amount of a high-level pollutant.
