Chapter 16
Kinetics
Dinosaurs generated enough heat to sustain its biochemical reactions at high rates.
Reaction rate = f(temperature)
What is chemical kinetics?
Deals with the speed (rate) of a chemical reaction and its reaction mechanism.
Describes the change in concentration as a function of time.
Qualitatively……
Quantitatively……
Section 16.1: Factors that influence reaction rate
Each specific reaction has its own characteristic reaction rate.
We can control 4 factors that affect the rate of a given reaction:
• Concentration of reactants
• Physical state of reactants
• Temperature of reaction
• Presence of a catalyst
Section 16.1: Factors that influence reaction rate
(1) Concentration – Molecules must collide in order to react.
The reaction rate changes throughout the course of a reaction.
Section 16.1: Factors that influence reaction rate
(2) Physical state of reactants – Molecules must collide in order to react.
Reactants in same physical state  random thermal motion brings them into contact
Reactants in different physical states  contact between reactants occurs only at the
interface between the phases
Example: reactants – orange + blue
The more finely divided a solid or liquid
reactant, the greater its surface area per
unit volume  More contact with other
reactants  Faster reaction.
solid + aqueous
(interface only)
aqueous + aqueous
Crushed coral versus whole coral.
Section 16.1: Factors that influence reaction rate
(3) Temperature of reaction – Molecules must collide with enough energy to react.
Two aspects to this:
(1) At higher temperatures, more collisions occur at a given time.
(2) At higher temperatures, the energy of collisions is higher (K.E. of
molecules is higher).
Example: refrigeration to preserve food
(4) Presence of a catalyst – A catalyst speeds up the reaction rate without being
consumed (chemically reacting) in the reaction.
Example: Biological catalysts
Section 16.2: Expressing reaction rate quantitatively
Rate – a change in some variable per unit of time
Analogy with speed
Reaction: A  B
reaction rate – the changes in concentrations of reactants or products per unit time
Reactant concentrations decrease, while product concentrations increase.
Reaction: A  B
Section 16.2: Expressing reaction rate quantitatively
In most reactions, not only the concentration changes, but the reaction rate also changes.
Therefore, we can define three reaction rates:
Average reaction rate – how fast the concentration changes over the entire time period
Instantaneous reaction rate – the reaction rate at an instant in time
Initial reaction rate – the instantaneous rate at the moment the reactants are mixed
Example: Reaction involved in the decrease of photochemical smog (ethylene + ozone)
Reaction rate (rate of decrease of reactants):
Section 16.2: Expressing reaction rate quantitatively
Average reaction rate – how fast the concentration changes over the entire time period
Instantaneous reaction rate – the reaction rate at an instant in time
Initial reaction rate – the
instantaneous rate the
moment the reactants are
mixed (slope of line that is
tangent to the curve at
t = 0)
Section 16.2: Expressing reaction rate quantitatively
Rates for reactants and products
General formula:
General equation:
(a, b, c, d  coefficients)
Section 16.2: Expressing reaction rate quantitatively
(a, b, c, d  coefficients)
General formula:
General equation:
(1) Express the rate in terms of changes in [H2], [O2], and [H2O] with time.
(2) When [O2] is decreasing at 0.23 mol/L sec, at what rate is [H2O] increasing?
2 N2O5 (g)  4 NO2 (g) + O2 (g)
(3) When [N2O5] is decreasing at 0.95 mol/L sec, at what rate is [NO2] increasing?
2 N2O5 (g)  2 NO2 (g) + N2O3 (g) + 3 O2 (g)
(4) When [O2] is increasing at 0.54 mol/L sec, at what rate is [N2O5] decreasing?
Section 16.3: Rate laws
Experimentally determined  not determined from reaction stoichiometry
General reaction: aA + bB + …  cC + dD + …
Rate law: rate = k[A]m[B]n…
where [A] and [B] are concentrations of reactants A and B
k is the rate law constant – specific for a given reaction at a given temperature
m and n are the reaction orders – defines how the rate is affected by reactant
concentration
Example: Specific reaction, specific temperature
Rate law constant
CH3COOCH2CH3 (ester) + H2O 
CH3COOH + CH3CH2OH
Reaction orders – m and n = rate change / concentration change
Examples:
If the rate doubles when [A] doubles  [A]1 and m = 1
If the rate quadruples when [B] doubles  [B]2 and n = 2
If the rate does not change when [A] doubles  [A]0 and m = 0
Section 16.3: Rate laws
All terms in the rate law (rate = k[A]m[B]n….) must be determined experimentally.
[A] and [B] measured and rate, rate constant (k), reaction orders (m, n)
are deduced from these measurements.
Measuring rates – many methods:
(1) Conductometric methods – used when a nonionic reactant forms ionic products
Example: (CH3)3C–Br (l) + H2O (l)  (CH3)3C–OH (l) + H+ (aq) + Br- (aq)
(2) Manometric methods – used when a reaction involves a change in the number of
moles of a gaseous reactant or product  reaction rate determined by the change
in pressure over time
Example: Zn (s) + 2 CH3COOH (aq)  Zn2+ (aq) + 2 CH3COO- (aq) + H2 (g)
Section 16.3: Rate laws
(3) Spectrometric methods – used when one of the reactants of products absorbs (or
emits) certain wavelengths of light
Transmittance
Example: NO (g, colorless) + 2 O3 (g, colorless)  O2 (g, colorless) + NO2 (g, brown)
Lightout
Transmittance =
Lightin
Lightin
Lightout
NO2
Light Source
Transparent Reaction Cell
Light Detector
Section 16.3: Rate laws
(4) Direct chemical methods – used for reactions that can be easily slowed or stopped
Example: Measure respiration rate by killing bacteria with HgCl2 to stop respiration.
Respiration: O2 + CH2O  CO2 + energy
BOD bottle
(Biological Oxygen Demand)
• Measure O2 concentration at t=0 and t=24 hrs.
• Change in O2 concentration =
O2 (t=0) – O2 (t=24 hrs)
• Respiration rate = change in O2 concentration
time (24 hours)
Section 16.3: Rate laws
Determining Reaction Order
First, some terminology………individual reaction order vs. overall reaction order
Example: 2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g)
Rate = k[NO]2[H2]
Individual: Reaction is second order with respect to NO and first order with respect to H2.
Overall: Reaction is third order overall (sum of individual reaction orders).
Rate = k[NO][O3]
Rate = k[(CH3)3CBr][H2O]0
Rate = k[CHCl3][Cl2]1/2
*A zero order reaction order means that the reaction does
not depend on the concentration of that reactant.
Section 16.3: Rate laws
Determining Reaction Order
We have been determining reaction orders from a known rate law (i.e. Rate = k[NO][O3])
When the rate law is not known, use data from a series of experiments
with different reactant concentrations to determine initial reaction rates.
Change one reactant concentration, while keeping the other constant.
Reaction is what order with respect to O2? With respect to NO? Overall?
Section 16.3: Rate laws
Determining Reaction Order
We have been determining reaction orders from a known rate law (i.e. Rate = k[NO][O3])
When the rate law is not known, use data from a series of experiments
with different reactant concentrations to determine initial reaction rates.
Change one reactant concentration, while keeping the other constant.
Reaction is what order with respect to O2?
Double O2, double reaction rate: 1st order w.r.t O2
*reaction order = rate change / concentration change*
Section 16.3: Rate laws
Determining Reaction Order
We have been determining reaction orders from a known rate law (i.e. Rate = k[NO][O3])
When the rate law is not known, use data from a series of experiments
with different reactant concentrations to determine initial reaction rates.
Change one reactant concentration, while keeping the other constant.
Reaction is what order with respect to NO?
Double NO, quadruple reaction rate: 2nd order w.r.t NO
*reaction order = rate change / concentration change*
Section 16.3: Rate laws
Determining the Rate Constant
Simply, solve for k.
Rate law: rate = k[O2][NO]2
What is k for this reaction?
Section 16.4: Integrated rate laws
So far, we have not considered the time factor in the rate law equations.
Rate = k[O2][NO]2
This equation says what the rate will be when [O2] is X and [NO] is Y, but does
not tell use how long it will take for X moles of [NO] to be used up (for example).
Integrated rate laws – consider the time factor and are derived from equations
we have already seen using calculus
Rate = k[A]
For reaction: A  B
Magic
Section 16.4: Integrated rate laws
Example: At 1000 ºC, cyclobutane (C4H8) decomposes in a first-order reaction, with
The very high rate constant of 87 s-1, to two molecules of ethylene (C2H4).
If the initial cyclobutane concentration is 2.00 M, what is the concentration after 0.010 s?
What fraction of cyclobutane has decomposed in this time?
At 25 ºC, hydrogen iodide breaks down very slowly to hydrogen and iodine: rate = k[HI]2
The rate constant at 25 ºC is 2.4 x 10-21 L/mol sec. If 0.0100 mol of HI(g) is placed in a
1.0 L container, how long will it take for the concentration of HI to reach 0.00900 mol/L?
Section 16.1 to 16.4: Solving Rate Problems – A Summary
Q: What is the average,
instantaneous, or initial
reaction rate?
Q: If a reactant/product is
increasing/decreasing at some
rate X, what is the rate of
increase/decrease of some
other reactant/product?
Use concentration versus
time data in either a Table
or a Graph to calculate each.
*Stoichiometry-based*
(a, b, c, d  coefficients)
If the rate law (i.e. rate =
k[A]2[B])
is known:
∆rate
∆concentration
If given a rate, can solve for k.
Reaction order (m, n) =
If the rate law is not known, but you
have [reactant] and initial rate data:
Use data to find reaction orders for
each reactant, then solve for k.
Questions related to rate laws
aA + bB + …  cC + dD + …
Rate law: Rate = k[A]m[B]n…
*Use experimental data*
*NOT stoichiometry*
Q:
• If the [A] doubles what will happen
to the reaction rate?
• To quadruple the reaction rate,
how would you need to change [A]?
• What is reaction order w.r.t. [A]?
• What is overall reaction order if
the rate law is ‘rate = k[A]2[B]’?
• What is k for this reaction?
No rate law, no initial
reaction rates. Have
[A] vs. time data.
Trial-and-error
graphical plotting
time
Q:
• What is the [A] after x time?
• How long will it take for [A] to
reach X mol/L?
Integrated rate laws
Section 16.4: Integrated rate laws (continued)
What if you have concentration and time data, but do NOT
have the rate law (rate = k[A]m[B]n…) or the initial rate data?
Trial-and-error graphical plotting
For zero-order reactions:
For first-order reactions:
For second-order reactions:
Section 16.4: Integrated rate laws (continued)
Trial-and-error graphical plotting
(Ocean Acidification!!!)
Kinetics of Coral Dissolution
Temp1
Temp2
2
1.8
Mass loss (g)
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
1
2
3
4
Time (Days)
5
6
7
Does coral dissolution
follow zero-order kinetics?
Kinetics of Coral Dissolution
Temp1
Temp2
2
1.8
1.6
1.2
1
0.8
0.6
0.4
0.2
0
0
1
2
3
4
Time (Days)
5
6
7
Kinetics of Coral Dissolution
Temp1
Does coral dissolution
follow first-order kinetics?
Temp2
Linear (Temp2)
Linear (Temp1)
2
y = -0.4x + 0.6931
R2 = 1
1
ln (Mass Loss)
Mass loss (g)
1.4
0
-1
-2
-3
0
1
2
3
4
y = -0.9x + 0.6931
R2 = 1
-4
-5
-6
Time (Days)
5
6
7
8
Kinetics of Coral Dissolution
Does coral dissolution
follow second-order kinetics?
Temp1
Temp2
300
1 / Mass Loss
250
200
150
100
50
0
0
1
2
3
4
5
6
7
Time (Days)
If coral dissolution followed first-order kinetics, what would that tell you
about the dependence of the overall reaction rate of CaCO3 dissolution?
(What does the rate depend on?)
8
Section 16.4: Integrated rate laws (continued)
Example #2: Trial-and-error graphical plotting
Section 16.4: Integrated rate laws (continued)
Half-life (t1/2)
It is the time required for a reactant concentration to reach half of its initial value.
Applies to first-order reactions only  independent of the starting concentration
(In other words, if independent of starting concentration  the t1/2 is independent
of the number of other particles present)
t1/2 = 0.693 / k
Section 16.4: Integrated rate laws (continued)
Radioactive decay – a common
application of half-life (t1/2)
Carbon-14
t1/2 = 5568 yrs
How old is the Ice Man? Found in
1991 in the Alps. (~3,330 yrs old)
Suggested Problems
16.39, 16.41, 16.43
[O2] (mol/L)
(1) What is the average reaction rate, the initial reaction rate, and the instantaneous
reaction rate for the reaction below given the data in the Table below?
Time
(min)
[O2]
(mol/L)
1.70
0
1.60
1.50
1
1.21
1.30
2
0.98
1.10
3
0.82
4
0.70
5
0.62
6
0.55
0.90
0.70
0.50
0
2
Time (Minutes)
4
6
(2) For the same reaction (above), how would the reaction rate change if you doubled the
H2 concentration? How would you need to change the O2 concentration if you wanted to
quadruple the reaction rate?
(3) If you know that the rate law for the reaction shown below is rate = k[C2H4]2[O3], then
if you doubled the [O3], how would the reaction rate change? How would you need to
Change the [C2H4] to quadruple the rxn rate?
Suggested Problems (Continued)
(4) At 50 ºC, H2Cl2 breaks down into to H2 and Cl2 gases: rate = k[H2Cl2]2 The rate
constant 1.8 x 10-2 L/mol sec. If 0.40 mol of H2Cl2 is placed in a 2.5 L container, how
long will it take for the concentration of H2Cl2 to reach 0.25 mol/L?
(5) At 50 ºC, C4H8 decomposes. The rate law is rate = k[C4H8]. If the concentration of
cyclobutane is 0.80 mol/L after 20 minutes, what the the rate constant (k) at this
temperature? The initial concentration of cyclobutane is 1.38 mol/L.
(6) At 700 ºC, H2S gas breaks down into to diatomic hydrogen and sulfur gases. The
rate = k[H2S]0 If the rate constant is 9.30 x 10-8 mol/L sec and the initial concentration
of H2S gas is 0.18 mol/L, what will be the concentration of this gas 1 minute after the
reaction is started?
Section 16.4: Half-life (continued)
Half-life
t1/2 = 0.693 / k
16.43 In a first-order decomposition reaction, 50.0% of a compound decomposes in
10.5 minutes. (a) What is the rate constant of the reaction? (b) How long does it take
for 75 % of the compound to decompose?
Suggested Problems
16.44, 16.45
Section 16.5: Effect of Temperature on Reaction Rate
An increase of 10 ºC (or K) causes a doubling or
tripling of the reaction rate.
Q10 – the factor by which the reaction rate is accelerated
by raising the temperature by 10 degrees (2 – 3)
Expressed by relationship to
the rate constant, k.
Rate Law
rate = k[A]m[B]n…
Half-life
t1/2 = 0.693 / k
Integrated Rate Laws
Q10 – the factor by which the reaction rate is accelerated
by raising the temperature by 10 degrees (2 – 3)
Metabolic rates – the rates at which organisms use energy and materials  The
fundamental rate of biology (it comes down to biochemistry)
Universal temperature dependence (UTD) of biological processes
“Despite a hundred
years of research,
ecology has little in
the way of universal
laws of gravity and
thermodynamics in
physics or the
Mendelian laws of
inheritance in biology.
Is ecology really devoid
of universal laws?”
2004, New Scientist
2001, Science, Volume 293
Back to Chemistry……
Section 16.5: Effect of Temperature on a Chemical Reaction Rate
Arrhenius equation (Svante Arrhenius)
where A is the frequency factor (Next  Section 16.6)
T is the absolute temperature (*Must be in units of Kelvins)
R is the Universal Gas Constant (8.31447 J/mol K)
Ea is the activation energy (the minimum energy the molecules must
have to react)
Using this equation to find Ea from experimental data…
First, the math
Section 16.5: Effect of Temperature on Reaction Rate
The decomposition of HBr has rate constants of 1.37 x 10-11 L/mol s at 350 K and
2.43 x 10-9 L/mol s at 450 K. Find Ea.
Practice problem for you: Find Ea for the reaction of an ester with water using the
following data from a kinetics experiment.
Section 16.6: Effect of Concentration on Reaction Rate
Collision Theory – Reactant particles (atoms, molecules, and ions) must collide with
each other in order to react. Therefore, the # of collisions per time puts an upper limit
on how fast a reaction can take place.
Explains several things:
(1) Why reactant concentrations are multiplied together in the rate law
A + B  products
Rate Law
rate = k[A]m[B]n…
Section 16.6: Effect of Concentration on Reaction Rate
(2) How temperature affects the rate
In most collisions, the molecules rebound without reacting.  Every reaction has an
energy threshold that colliding molecules must exceed in order to react.
This minimum energy threshold is called the activation energy (Ea). Only collisions
with E > Ea will react.
Temperature rise increases the number of collisions with enough energy to exceed Ea.
Section 16.6: Effect of Concentration on Reaction Rate
(3) the influence molecular structure has on rate
In addition to colliding, and colliding with enough energy, molecules must also collide
so that the reacting atoms make contact.
Arrhenius equation (Svante Arrhenius)
The effect of molecular orientation is contained in the term A (frequency factor)
A = pZ
Z  collision frequency
p  orientation probability
(effectively oriented collisions / all possible collisions)
Section 16.6: Effect of Concentration on Reaction Rate
Collision Theory – Reactant particles (atoms, molecules, and ions) must collide with
each other in order to react. Therefore, the # of collisions per time puts an upper limit
on how fast a reaction can take place.
Molecules must:
(1) Collide
(2) Collide with a certain minimum energy (Ea – activation energy)
(3) Collide with the right orientation (the A term in Arrhenius equation)
Does not explain:
Why the activation energy is crucial
How the activated molecules look
Transition state theory – focuses on the high-energy species that
form through an effective collision
Suggested Problems
16.58 – 16.46, 16.72, 16.74, 16.77, 16.78, 16.82