CHEM1001 - Chemical Dynamics
Text: Petrucci, Harwood, Herring, Madura and Bissonnette
(10th edition)
Chemical Kinetics:
How fast are reactions?
Chemical Equilibrium:
In what direction do reactions proceed?
When do reactions stop? (equilibrium)
H2(g) + CO2(g) → H2O(g) + CO(g)
Thermodynamics and Electrochemistry:
What determines the position of equilibrium?
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Chemical Kinetics
Reading: Chapter 14 of Petrucci, Harwood, Herring,
Madura and Bissonnette (10th edition).
Examples: 14-1, 14-3 through 14-10.
Assigned Problems: Chapter 14, questions 7, 12, 14, 23a,
24, 41, 53, 70.
Suggested Problems: Chapter 14, questions 56, 57, 60, 62,
87.
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Aims of Chemical Kinetics
1) Describing chemical reaction rates:
 Definition of “rate of reaction”.
 How rates can be measured.
 How rates depend on concentrations.
 How rates depend on temperature.
2) Understanding chemical reaction rates:
 What sequence of steps is involved?
 Why do rates depend on concentrations?
 Why do rates depend on temperature?
 Think about reactions at the molecular level.
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Types of Reactions
Chemical reactions involve:
 Rearranging atoms in a molecule.
 Breaking up a molecule into two or more molecules.
 Transferring atoms from one molecule to another.
Reactions may take place in:
 One step (elementary reactions).
 More than one step. Each step is an elementary
reaction. The sequence of steps is the reaction
mechanism.
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Elementary Reactions
Elementary reactions are either:

unimolecular (a molecule dissociates or isomerizes).
Examples: N2O3 → NO + NO2
HCN → CNH

bimolecular (two molecules collide and react).
Example: NO + O3 → NO2 + O2
In an elementary reaction, all stoichiometric coefficients
are integers.
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Reaction Mechanisms
A mechanism for a reaction consists of all the elementary
reactions that take place.
Overall reaction: 2NO + O2 → 2NO2
Mechanism:
NO + O2 → NO3
step (1)
NO3 + NO → 2NO2
step (2)
There are two elementary reactions in this mechanism.
 The elementary reactions occur in a definite sequence.
 Adding the equations for the elementary reactions gives
the equation for the overall reaction.

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Reversible Elementary Reactions
Sometimes a mechanism includes reversible reactions.
Overall reaction: 2NO + O2 → 2NO2
Mechanism: NO + O2 → NO3
(1)
NO3 → NO + O2
(−1)
NO3 + NO → 2NO2
(2)


Step (−1) is the reverse of step (1).
Step (1) can be followed by either (−1) or (2).
o Adding equations for the sequence which leads to
products (1+2) gives the equation for the overall reaction.
o Step (−1) affects the rate, not the stoichiometry.
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Speed of Chemical Reactions
Section 14.1
An important aspect of chemistry is how fast (or slow)
reactions are.
The reaction of sugar with oxygen is very favourable.
 It is usually slow at room temperature.
 Enzymes can control the rate of reaction.
The reaction of gasoline with oxygen is very favourable.
 It is usually slow at room temperature.
 A spark can make the reaction occur very rapidly.
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Rates of Chemical Reactions
Expressed as the rate of change of concentrations.
 Concentrations could be reactants or products.
 Properly expressed as derivatives.
 May be approximated in terms of differences.
The symbol "Δ" indicates the change in a quantity:
Example: Sn2+ + 2Fe3+ → Sn4+ + 2Fe2+
Δ[Sn 4+ ] change in concentration of Sn 4+
rate 

Δt
change in time

Units are usually M s-1 = mol L-1 s-1.
Dimensions are always 'amount' / 'time'.
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Rate is a Slope
The rate is easy IF a plot
of concentration vs. time
gives a straight line:



Rate = Δ[Sn4+] / Δt =
slope
In this graph, the
slope is constant.
Usually things are not
so simple.
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Rates Change as Reactions Proceed
The plot of concentration vs. time is not usually a straight line.
Slope is not constant.
d[Sn4+]/dt = slope of the
tangent line.
d [ Sn 4 ] [ Sn 4 ]

dt
t
The derivative gives the
slope at a point.
Rate = slope = d[Sn4+]/dt at any point.
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Estimating Rates



The derivative (slope) of a curve is different at every point.
The derivative
can be estimated
from the slope of
a straight line
through two
points.
This provides an
estimate of the
rate.
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Stoichiometry is Important
Rates can be different for different species.
Example: Sn2+ + 2Fe3+ → Sn4+ + 2Fe2+
4+
4+
d[Sn ] Δ[Sn ]
Rate of forming [Sn ] 

dt
Δt
4+
2+
2+
d[Fe ] Δ[Fe ]
Rate of forming [Fe ] 

dt
Δt
2+
2 moles of Fe2+ are formed for each mole of [Sn4+]. So
d [ Fe 2 ]
d [ Sn 4 ]
2
dt
dt
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Definition of Reaction Rate
Reaction rates are defined in terms of rate of change of
concentration.
We can write the reaction rate as equal to the rate of producing
a product (or reactant) with a stoichiometric coefficient of
unity. Let’s use product for now.
Example:
Sn2+ + 2Fe3+ → Sn4+ + 2Fe2+
d[Sn 4+ ] Δ[Sn 4+ ]
Reaction rate =

dt
Δt
1 d[Fe 2+ ] 1 Δ[Fe 2+ ]
Reaction rate =

2 dt
2 Δt
The chemical equation must be properly balanced.
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Reaction rate (continued)
The reaction rate depends on how the reaction is written.
Example: Sn2+ + 2Fe3+ → Sn4+ + 2Fe2+
d[Sn 4+ ] 1 d[Fe 2+ ]
Reaction rate 

dt
2 dt
½Sn2+ + Fe3+ → ½Sn4+ + Fe2+
4+
2+
d[Sn ] d[Fe ]
Reaction rate = 2

dt
dt
The rates of producing Sn4+ or Fe2+ do not depend on how the
reaction is written.
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Reaction rate (continued)
Example:
N2 + 3H2 → 2NH3
d[N 2 ]
1 d[H 2 ]
1 d[NH 3 ]
Reaction rate  


dt
3 dt
2 dt
Reaction rates are always positive.

Rates of change of products are positive.

Rates of change of reactants are negative.

Divide by the stoichiometric coefficient.

Units are usually M s-1 = mol L-1 s-1. ('amount' / 'time')
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General Example of Reaction Rate
aA + bB → G + hH
1 d[A] 1 d[B]
Reaction rate==a dt
b dt
1 d[G]
1 d[H]
=+
=+
g dt
h dt
The rate of reaction may be expressed using either

Consumption of reactants or

Production of products.
It is not necessarily the same as these rates.
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Reaction Rates are not Constant
The reaction rate changes as the reaction proceeds.
Reaction rates depend on:
 Concentrations of reactants.
 Concentrations of products.
 Concentrations of other species (catalysts).
 Temperature.
 Solvent, phase, pressure.
 Other reactions.
There are ways to describe these dependencies (rate laws).
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Measuring Reaction Rates
Section 14.2
Concept is simple:
(1) Mix reactants together.
(2) Measure concentration at a particular time. Tricky!
Method depends on the species in the reaction.
(3) Repeat step (2) over and over.
(4) Make graph or table of concentration vs. time.
(5) Determine the slope. (several options)
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Example (Experiment)
H2O2(aq) → H2O(l) + ½ O2(g)
Could measure volume of O2(g) produced, or
could measure H2O2(aq) consumed at different times.
To do the latter:
(1) Withdraw aliquot from reaction solution.
(2) Add aliquot to permanganate solution:
2MnO4- + 5 H2O2 + 6 H+ → 2Mn2+ + 8H2O + O2(g)
(3) Measure loss of H+ (change of pH) electrochemically.
(4) Repeat steps (1) to (3) at intervals (say, every 400
seconds).
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Experimental Data (Tabulated)
Table 14.2
from Petrucci
et al.,
Each value for
the reaction
rate is an
average over
400 seconds.
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Average and Instantaneous Rates
The average rate, Δ[H2O2]/Δt, is determined by the slope
between two points, as in the previous table.
The instantaneous rate, d[H2O2]/dt, is determined by the
slope (tangent) at one point.
The instantaneous rate is the correct value.
To obtain the instantaneous rate, d[H2O2]/dt:
· Approximate by Δ[H2O2]/Δt (OK if Δ[H2O2] is small).
· Graph the data in a way that gives a straight line.
· Use integrated rate laws and calculus.
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Same Experimental Data (Plotted)
Figure 14.2 from
Petrucci et al.
Rate of reaction
decreases as
H2O2
is consumed.
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The Rate Law
Section 14.3
Reaction rates depend on concentrations of the reagents.
The relationship between reaction rate and concentrations
is called the rate law (or rate equation).
Two type of rate laws:
· Differential: Reaction rate as function of concentration.
· Integrated: Concentration as function of time.
Rate laws are especially simple for elementary reactions.
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Unimolecular Reactions
These are elementary reactions of a single molecule.
N2O3 → NO + NO2 (dissociation)
HCN → CNH
(isomerization)
Reactions require breaking chemical bonds (and perhaps
forming new ones).
An input of energy is normally needed:
· Collisions with other molecules.
· Absorbing radiation (light).
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Rate Law for Unimolecular Reactions
The rate of reaction is directly proportional to the concentration
of the reacting molecule.
The reaction must be written so the reacting molecule has a
stoichiometric coefficient of unity.
Correct: N2O4 → 2NO2
Incorrect: ½N2O4 → NO2
Rate law:
d[N 2 O 4 ]
Reaction rate = = k[N 2 O 4 ]
dt
k is a proportionality factor called the rate constant or rate
coefficient
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Unimolecular Rate Law (continued)
N2O4 → 2NO2
Rate law: reaction rate =
· Units of concentration are usually M, but not always.
· Units of time are seconds, or minutes, or hours, etc.
· The reaction rate has units of concentration time-1.
· The rate constant, k, has units of time-1. k is different for
each reaction and may depend on temperature, solvent, phase,
and pressure.
This is an example of a first-order reaction.
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Half-Life
Implications of the rate law for unimolecular reactions:
· In any interval of time, a molecule has a certain
probability of reacting.
· In any interval of time, the fraction of remaining
molecules that react is the same.
· The time that it takes for one half the remaining
molecules to react is constant.
Definition: The half-life is the time it takes for one half of a
reactant to be converted into product.
Later we will see how half-life is related to rate constant.
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Bimolecular Reactions
These are elementary reactions between two molecules.
NO + O3 → O2 + O2
HO2 + HO2 → H2O2 + O2
The reaction must be written so that all stoichiometric
coefficients are integers.
Correct:
2HO2 → H2O2 + O2
Incorrect:
HO2 → ½H2O2 + ½O2
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Rate Law for Bimolecular Reactions
The rate of reaction is directly proportional to the
concentrations of each of the reacting molecules.
NO + O3 → NO2 + O2
Rate law:
d[NO 2 ]
Reaction rate =
= k[NO][O3 ]
dt
2HO2 → H2O2 + O2
Rate law:
d[H 2 O 2 ]
Reaction rate =
= k[HO 2 ]2
dt
k is the rate constant or rate coefficient.
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Bimolecular Rate Law (continued)
NO + O3 → NO2 + O2
Rate law:
·
d[NO 2 ]
Reaction rate =
= k[NO][O3 ]
dt
Units of concentration are usually M, but not always.
· Units of time are seconds, or minutes, or hours, etc.
· The reaction rate has units of concentration time-1.
· The rate constant has units of concentration-1 time-1.
k is different for each reaction and may depend on temperature,
solvent, phase, and pressure.
This is an example of a second-order reaction.
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Rate Laws for Multi-Step Reactions
Rate laws for elementary reactions:
· Are very simple.
· Depend only on the concentrations of the reactants.
· Can be determined from the equation for the reaction.
Reactions that require more than one step are not so simple.
But there are similarities:
· Reaction rates still depend on the concentrations of the
reagents (but not just the reactants).
· The rate law gives the relationship between the reaction
rate and concentrations.
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Properties of Rate Laws
Typically (but not always), the rate law is of the form:
rate of reaction = k[A]m[B]n
· [A] and [B] represent reagent molarities.
· k, m, and n are constants for any particular reaction.
· There may be more or fewer than two reagents in the rate
law.
· The reagents are usually, but not always, reactants or
products.
· m and n are often small, positive integers. But they may
be zero, fractional, and/or negative.
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The Constants m, n and k
rate of reaction = k[A]m[B]n
Reaction order:
· The reaction is of order m for A, order n for B, etc.
· The overall reaction order is m + n.
Rate constant:
· k is a proportionality constant.
· k is called the rate constant.
· The faster the reaction, the larger the value for k.
· k depends on temperature, pressure, solvent, and phase.
· The units for k depend upon the reaction order.
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Examples of Reaction Order
reaction rate = k[N2O4]
· first order in N2O4
· first order overall
reaction rate = k[NO][O3]
· first order in NO
· first order in O3
· second order overall
reaction rate = k[HO2]2
· second order in HO2
· second order overall
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Units of the Rate Constant
Depends on overall reaction order
Reaction order
·
·
·
Rate constant units
0
concentration time-1
1
time-1
2
concentration-1 time-1
3
concentration-2 time-1
Units of concentration are usually M, but not always.
Units of time are seconds, or minutes, or hours, etc.
The rate always has units of concentration time-1.
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Reaction Order and Molecularity
The terms unimolecular (a single molecule dissociates or
isomerizes) and bimolecular (two molecules collide and
react) apply only to elementary reactions.
Reaction order applies to all reactions.
·
All unimolecular reactions are first-order.
·
Not all first-order reactions are unimolecular.
·
All bimolecular reactions are second-order.
·
Not all second-order reactions are bimolecular.
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Establishing the Rate Law
This consists of determining k, m, n, ...
The rate law is established by analyzing experimental data.
We will consider two methods:
·
Method of initial rates.
·
Method of integrated rate laws.
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Method of Initial Rates
Principles:
·
Initial concentrations of reagents are easily obtained.
· Initial slope of the concentration-time curve provides
the initial rate of reaction.
Procedure:
· Measure initial rates for different concentrations of
each reagent while keeping concentrations of other
reagents constant.
· Compare relative rates and initial concentrations to
find k, m, n, ...
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Example: Method of Initial Rates
Common to assume that initial rate depends only on reactant
concentrations.
Then: rate of reaction = k[HgCl2]m[C2O42-]n.
Experiment 2 has twice as much C2O42- as experiment 1.
Experiment 3 has half as much HgCl2 as experiment 2.
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Determination of m
rate of reaction = k[HgCl2]m[C2O42-]n
Use rates and initial concentrations for experiments 2 and 3.
From the rate law:
m
n
m
R 2 k(0.105) (0.3)  0.105 
m
=
=
 =(2.02)
m
n
R 3 k(0.052) (0.3)  0.052 
From the initial rates:
R 2 7.1×10-5
=
=2.03
-5
R 3 3.5×10
2.03 = 2.02m, therefore m = 1.
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Determination of n
rate of reaction = k[HgCl2]m[C2O42-]n
Use experiments 1 and 2:
from the rate law
from initial rates
R 2 k(0.105) (0.3)
7.1 10
n
=
2 
=3.94
m
n
-5
R1 k(0.105) (0.15)
1.8  10
m
n
-5
Since 2n ≈ 4, n = 2 (by inspection).
Or, take logs of both sides of the equation:
n log(2.0) = log(3.94)
n (0.30) = 0.595 therefore n = 1.99 ≈ 2
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Determination of k
The rate law is:
rate of reaction = k[HgCl2][C2O42-]2
Solve for the rate constant, k:
2- 2
k = (rate of reaction) / ([HgCl2][C2O4 ] )
Using data from experiment 2 (any one will do):
k = (7.1×10-5 M min-1) / (0.105 M) (0.30 M)2
k = 7.5×10-3 M-2 min-1
Therefore the rate law is:
rate of reaction = (7.5×10-3 M-2 min-1)[HgCl2][C2O42-]2
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Method of Integrated Rate Laws
Uses the full concentration-time curve from one experiment
(not just the initial rates).
The integrated rate equation relates concentration and time:
[A] = F(t)
The function, F(t), depends on the reaction order.
Here we will consider only certain simple cases:
· Assume that only one concentration is changing.
· Consider only zero, first, and second order reactions.
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Method of Integrated Rate Laws (continued)
Consider the reaction: aA → products
Assume that the rate law is of the form:
1 d[A]
Reaction rate = = k[A]n
a dt
Here n is the reaction order and k is the rate constant.
Integrating this expression for different values of n gives the
integrated rate laws.
Note: Petrucci et al. assume that the stoichiometric
coefficient, a=1. We use the more general approach.
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Integrated Rate Laws
Calculus helps
A rate law is a differential equation:
d[A]
n
= -ak[A]
dt
It can be integrated to give the concentration as a
function of time:
[A] = F(t)
In the lab, we work the other way: from [A] vs. time to the
differential equation.
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Zero-Order Reactions: n = 0 - Section 14.4
−d[A] / dt = ak[A]0 = ak = constant
−d[A]/dt = ak slope is a constant  straight line
Slope = −ak < 0.
[A]0 = concentration at t = 0
Equation: [A] = [A]0 − akt
This is the integrated rate law for
a zero order reaction.
Derivative: d[A]/dt = −ak
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Zero-Order Reactions: Example
Certain bacteria produce penicillinase, an enzyme that
destroys the antibiotic penicillin. The data below are
obtained when penicillin ('P') is added to a suspension of
these bacteria.
a) What is the reaction order?
b) Find the rate constant for
penicillin →products
c) Find [P] after 1200 seconds.
d) Find [P] after 4000 seconds.
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Zero-Order Reaction Example (continued)
a) The change in concentration is the same in each 500
second interval.  Reaction is zero order.
b) [P] = [P]0 − akt
[P]0 = 5.0×10−3 M, a = 1, use any pair of [P] and t.
Solve for k. k = 1.6×10−6 M s−1
c) Let t = 1200 s; use [P] = [P]0 − akt.  [P] = 3.1 mM
d) Let t = 4000 s; use [P] = [P]0 − akt.  [P] = −1.4 mM
This is impossible! The reaction must eventually stop
being zero order.
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First-Order Reactions: n = 1
−d[A] / dt = ak[A]
Integrated rate law: [A] = [A]0exp(−akt)
or
ln([A]/[A]0) = −akt
Differentiation yields: −d[A]/dt = ak[A]
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Half-Life
Definition: The half-life is the time it takes for one half of
a reactant to be converted into product.
Integrated rate law:
or
ln([A]/[A]0) = −akt
ln([A]0/[A]) = akt
By definition, [A] = ½[A]0 when t = t½ = half-life
so
ln([A]0/(½[A]0)) = kt½
ln(2) = kt½ or
t½ = ln(2)/k = 0.693/k
For a first-order reaction, the half-life is independent of
concentration. Not true for any other reaction order.
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Examples of First-Order
Reactions
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Example Calculation for First-Order Reaction
CH3-N=N-CH3(g) → N2(g) + CH3CH3(g)
The partial pressure of azomethane was measured over time
and gave the data on the right.
Determine:
a) the reaction order
b) the rate constant
c) the half-life
d) the time at which the partial pressure
is 71 torr
e) the partial pressure after one hour
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Gas Phase First-Order Reactions
At constant temperature, we can replace concentration
with partial pressure.
Integrated rate law: ln([A]0/[A]) = akt
Ideal gas law:
Therefore:
N P PN A
[A]= =
=
V kT RT
ln(PA,0/PA) = akt
For a first-order reaction:
· k has dimensions of time−1 (e.g. s−1)
· k does not depend on the units of concentration
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First Order Plot for Azomethane Decomposition
Plot of ln(PA) versus time is linear, so reaction is first-order.
ln[A] = −akt + ln[A]0  k = −slope = 2.55×10−3 s−1
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Example Calculations (continued)
a) Graph shows that the reaction is first order.
b) Solve ln([A]0/[A]) = akt for k.
[A]0 = 284 torr, a = 1, and [A] = 150 torr at t = 250 s
k = 2.55×10-3 s−1
c) t½ = ln(2)/k = 272 seconds
d) Solve ln([A]0/[A]) = akt for t.
[A]0 = 284 torr, a = 1, [A] = 71 torr, k = 2.55×10-3 s-1
t = 543 s (two half lives)
e) [A] = [A]0exp(−akt) = 0.029 torr at t = 3600 s.
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Second-Order Reactions: n = 2 - Section 14.6
−d[A] / dt = ak[A]2
Integrated rate law: 1/[A] = 1/[A]0 + akt
or
[A] = [A]0 / (1+[A]0akt)
2
Differentiation yields: −d[A]/dt = ak[A]
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Second-Order Reaction Example
2I(g) → I2(g)
This is a second order reaction with k = 7.0×109 M−1 s−1.
The initial concentration of I(g) is 0.086 M.
a) Find the concentration after 2.0 minutes.
[I] = [I]0 / (1+[I]0akt)
[I]0 = 0.086 M, a = 2, t = 120 s [I] = 5.9×10−13 M
b) Find the instantaneous reaction rate at 2.0 minutes.
rate = k[I]2 = (7.0×109 M−1 s−1)(5.9×10−13 M)2
rate = 2.5×10−15 M s−1
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Using Integrated Rate Laws
To use integrated rate law equations, make the appropriate plot.
· get reaction order from type of plot that gives straight line
· get rate constant from slope
· get initial concentration from intercept
Zero Order First Order Second Order
plot vs. time
[A]
ln[A]
1 / [A]
slope
−ak
−ak
ak
intercept
[A]0
ln[A]0
1 / [A]0
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Integrated Rate Laws for Multiple Reagents
We have assumed that the rate depends on a single reactant.
· What if there is more than one reactant?
· What if a product affects the rate?
This can get complicated. But there is a way to make it easy.
Consider the following rate law:
rate of reaction = k[A]m[B]n
Set up the experiment so that [B] >> [A].
· Δ[B] ≈ Δ[A] so Δ[B] << [B] and [B] ≈ [B]0 (constant)
· rate law becomes:
rate of reaction ≈ k[A]m[B]0n = k/[A]m
(k= k/[B]0n)
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Example
BrO3-(aq) + 5Br-(aq) + 6H+(aq) → 3Br2(l) + 3H2O(l)
Rate = k[BrO3-][Br-][H+]2
if [Br-]0=1.0 M, [H+]0=1.0 M, [BrO3-]0 = 1.0×10-3 M
-
-
-
then [Br ] >> [BrO3 ] so [Br ] ≈ constant
+
+
and [H ] >>[BrO3 ] so [H ] ≈ constant
Rate = k[BrO3-][Br-]0[H+]02 = k/[BrO3-]
where k/ = k [Br-]0[H+]02
A plot of ln[BrO3-] vs. t will be straight line with slope = k/
Doubling [Br-]0 doubles k/. Doubling [H+]0 quadruples k/.
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Describing Reaction Rates - Summary
Section 14.7
Reaction rates are expressed as rates of change of
concentrations. They are derivatives (slopes).
We must be careful about stoichiometry.
Rate laws give the dependence of the rate on concentration:
rate of reaction = k[A]m[B]n...
Initial rates are easy to analyze.
· reaction orders (rate laws)
· rate constants
· uses only part of available data
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Reaction Rate Summary (Continued)
Integrated rate laws describe concentration as a function of
time.
· Form depends on reaction order.
· Slope of appropriate graph gives rate constant.
· Simple if only one concentration is changing or if one
concentration is much less than the others.
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Radioactive Decay
Certain atomic nuclei are unstable and spontaneously
decompose. These reactions are first order.
238
U →
234
222
4
Th + He
218
Po + 4He
Rn →
3
H → 3He + e−
14
C→
14
N + e−
9
t½ = 4.5×10 years
t½ = 3.8 days
t½ = 12.3 years
t½ = 5730 years
The half-lives do not depend on temperature or
chemical environment. Some of these reactions can
be used as clocks to date materials.
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14
C Dating
14
C is produced in the atmosphere by cosmic rays
(neutrons):
14
n+ N →
14
C+p
A small portion of atmospheric CO2 contains 14C. Plants
convert CO2 into carbohydrates. Animals eat the plants.
A predictable fraction (about 1 part in 1012) of all the carbon in
living organisms is radioactive.
After the organism dies, the radioactivity decays. How
much is left tells you when the organism died.
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14
C Example
In pre-industrial air, the fraction of 14C was 1.33×10−12.
If a sample of charcoal from an archeological dig has
a 14C fraction of 8.5×10−13, when was the site occupied?
14
C →
14
−
N+e
First order: kt = ln([14C]0/[14C])
[14C]0 = 1.33×10−12
[14C] = 8.5×10−13
k = ln(2)/t½ = ln(2) / (5730 yr) = 1.21×10−4 yr−1
t = 3700 years ago
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