Chemical Kinetics
Chapter 14
AP Chemistry
Chemical Kinetics
• Kinetics – the area of
chemistry concerned
with the rate (or speed)
of a reaction.
• Kinetics vs.
Thermodynamics
• Applications: Medicine,
Chemical Engineering
Reaction Rate Factors
• Physical state of reactants
– Surface area
• Concentration
– Rate increases with concentration increase
• Temperature
– Rate doubles every 10oC increase
• Catalyst
– Increase the reaction rate w/o being used up
Reaction Rates
• Speed is the change in a particular quantity
with respect to a change in time.
• In chemistry, we define the reaction rate
– The change in concentration of the reactants or
products over time
– Units are usually M/sec
– Rate =
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
• A graph of
concentration vs. time
is often plotted.
• Slope of the tangent
line at any point along
the curve is the
instantaneous rate.
• Rate decreases with
time.
– Reactants decrease with
time
Average Rate
• Because the rate of reaction changes with
time, it is useful to consider an average rate.
• Rate =
• The average rate for a reaction is usually take
as the early stages of a reaction.
Measuring Rates
• To determine the progress of a reaction, we
can measure two quantities:
1. Disappearance of the reactant
2. Formation of products
•
•
Reaction rate is a positive value.
Reaction rate is the same, no matter the method of
measurement.
aA + bB  cC + dD
• The rate of reaction is given by the following
equalities:
– A: Rate =
– B: Rate =
– C: Rate =
– D: Rate =
– Rate =
H2O2(g)  H2(g) + O2 (g)
• Write the rate in terms of each species.
SO2(g) + O2(g)  SO3(g)
• Write the rate in terms of each species.
H2(g) + O2(g)  H2O(g)
• Hydrogen is burning at the rate of 0.85 M/sec.
Rate of oxygen consumption? Rate of water
vapor formation?
How?
• How is it possible to measure the
concentration of reactants or products?
• There are a variety of methods.
• One of the more common methods is
spectroscopy.
• Measures the ability to absorb/transmit light
and converts the data to a concentration.
Spectrophotometer (Spec-20)
Beer-Lambert Law
• There is a linear relationship between the
concentration of a sample and its absorbance.
– A = -logT
– Beer’s Law: A = εbC
– Standards to find slope
– Convert T to A to C
NH3(g)  N2(g) + H2(g)
• Nitrogen is forming at the rate of 0.264 M/sec.
Rate of ammonia consumption? Rate of
hydrogen formation?
Rate Law
• General Equation: aA + bB  cC + dD
• Rate =
•
•
•
•
m is the order of A
n is the order of B
(m+n) is the overall reaction order
k is the rate constant
– specific for a rxn, Temperature dependent
Rate = k[A]m[B]n
• The rate law must be experimentally
determined.
• m and n are NOT the stoichiometric coefficient
• Unit of rate constant k
– M-p s-1 or 1/(Mp s )
– p = (m+n) – 1
• Rate depends on reactant conc…k does not depend
on reactant conc.
Rate = k[A][B]
• What happens to the rate if we…
a) Double conc of A (everything else the same)?
b) Double conc of B (everything else the same)?
c) Triple conc of A and double conc of B?
d) Order of A? B? Overall?
Rate = k[A]2[B]
• What happens to the rate if we…
a) Double conc of A (everything else the same)?
b) Double conc of B (everything else the same)?
c) Triple conc of A and double conc of B?
d) Order of A? B? Overall?
Rate = k[A]0[B]3
• What happens to the rate if we…
a) Double conc of A (everything else the same)?
b) Double conc of B (everything else the same)?
c) Triple conc of A and double conc of B?
d) Order of A? B? Overall?
Rate = k[A]m[B]n
• Two ways to determine the rate law
1. Initial rate method
– Can have many reactants
2. Graphical Method
– Can only have one reactant
• Solving a rate Law:
– Need to determine the orders of the reactants
– Need to determine the rate constant k
• Determine
the rate
law:
2 NO(g) + 2 H2(g)  N3(g) + 2H2O(g)
Experiment
[NO]
[H2]
Initial Rate (M/s)
1
0.10
0.10
1.23 x 10-3
2
0.10
0.20
2.46 x 10-3
1
0.20
0.10
4.92 x 10-3
• Determine
the rate
law:
a A(g) + b Bg)  c C(g) + d D(g)
Experiment
[A]
[B]
Initial Rate (M/s)
1
0.40
0.30
1.00 x 10-4
2
0.80
0.30
4.00 x 10-4
1
0.80
0.60
1.60 x 10-3
• Determine
the rate
law:
S2O82-(aq) + 3 I1-(aq)  2 SO42-(aq) + I31-(aq)
Experiment
[NO]
[H2]
Initial Rate (M/s)
1
0.080
0.034
2.2 x 10-4
2
0.080
0.017
1.1 x 10-4
1
0.160
0.017
2.2 x 10-4
Change in Conc. with time
• So far, we have considered rate based on the
change in concentration and rate constants.
• Using calculus, we can convert these same
equations to more useful forms.
• This is the graphing method to determining
the order of a reaction.
• Specific for only one reactant: [A]
Rate Laws
• Differential Rate Law:
∆𝐴
𝑅𝑎𝑡𝑒 = −
= 𝑘[𝐴]
∆𝑡
• Expressed how rate depends on concentration.
• Integrated Rate Law
ln[𝐴]𝑡 = −𝑘𝑡 + ln[𝐴]0
• Integrated form of the differential. Has specific variables.
Zero Order Reaction
• Rate only depends on the rate constant…not
on the concentration of A
• Differential:
• Integrated:
First Order Reaction
• Rate only depends on the rate constant and
on the concentration of A
• Differential:
• Integrated:
First Order
Plot of [A] vs. t
Plot of ln[A] vs. t
Second Order Reaction
• Rate only depends on the rate constant and
on the square concentration of A
• Differential:
• Integrated:
Second Order
Plot of ln[A] vs. t
Plot of 1/[A] vs. t
Usefulness of the Integrated Rate Laws
[A]t = -kt + [A]0
• We can know the concentration at any time
point for a given reaction.
• We can determine the order of a reaction.
• We can determine the half life of a reaction.
Determining the Order
[A]t = -kt + [A]0
y = mx + b
• This tells us that a plot of concentration of A
vs time will yield a straight line.
• Because this is the zero order rate equation, a
plot of [A] vs. t will yield a straight line.
Determining the Order
• First Order:
ln[A]t = -kt + ln[A]0
– A plot of ln[A] vs. t will give a straight line.
• Second Order:
– A plot of 1/[A] vs. t will give a striaght line.
Half life, t1/2
• The time required for the concentration of a
reactant to reach one-half its value:
– [A]t1/2 = ½[A]0
• This is a convenient way to describe the rate
of a reaction.
• A fast reaction will have a short half life.
Derivation: t1/2 of First Order
ln[A]t = -kt + ln[A]0
Summary
Order
Rate Law
Integrated Rate Law
Half Life
Zero
Rate = k
[A]t = -kt + [A]0
t1/2 =
First
Rate = k[A]
ln[A]t = -kt + ln[A]0
t1/2 =
Second
2
Rate = k[A]
1
1
= 𝑘𝑡 +
[𝐴]𝑡
[𝐴]0
t1/2 =
[𝐴]0
2𝑘
0.693
𝑘
1
𝑘[𝐴]0
** Note: The half life of a first order says that it does NOT depend
the concentration of the reactant A. So, the concentration
decreases by ½ each regular time interval, t1/2.
A first order reaction has k = 6.7 x 10-4 s-1.
• How long will it take for the conc to go from
0.25M to 0.15M?
• If the initial conc is 0.25M, what is the conc
after 8.8 min?
A first order reaction has [A] = 2.00M initially.
After 126 min, [A] = 0.0250M.
• What is the rate constant k?
• What is the half life?
A second order reaction has k = 7.0 x 10-9 M-1s-1.
• If the initial conc is 0.086M, what is the conc
after 2.0 min?
A first order reaction has t1/2 = 35.0 sec.
• What is the rate constant k?
• How long would it take for 95%
decomposition of the reactant?
A first order reaction has a half life of 19.8 min.
What is the reaction rate when [A] = 0.750M?
Temperature
Collision Model
• Based on Kinetic Molecular Theory
• Molecules must collide to react
• Greater the collisions, greater the rate
• As concentration increases, rate increases
• As temperature increases, rate increases
Orientation
• Most collisions do not lead to reactions
• Molecular orientation of collision is important
Still not enough
• Usually, a collision in the correct orientation is
still not enough to cause a reaction.
• Kinetic energy of a collision must cause bonds
to break.
• For a reaction to occur, there must be enough
kinetic energy to be greater than some energy.
• Activation Energy, Ea, is the minimum energy
required to initiate a reaction.
Activation Energy
Transition State
• Transition state is also called the activated
complex
• High energy intermediate state
• The activation energy represents the higher
energy state of the transition state
A
A
A + B
A
B
‡
A
B
B
B + A
B
Arrhenius Equation
• This equation combines all factors contributing to
the reaction rate.
k = A e−Ea/RT
•
•
•
•
•
k = rate constant
Ea = activation energy
R = Gas constant
T = Temperature in Kelvin
A = frequency factor “constant”
– probability of correctly oriented collisions
Other versions
• Original Equation:
• Linear Equation:
k = A e−Ea/RT
y = m * x + b
• If we know the Ea and k2 at T2, we can
calculate k1 at T1:
Catalysis
• Catalyst – speed up a reaction without being
consumed
• Beneficial or harmful, ex. Body
• Homogenous catalyst – same phase as
reactant
– Phase transfer catalyst
• Heterogeneous catalyst – different phase than
reactant
– Saturation of alkenes and alkynes
Catalyst
• Lowers the activation energy
• Potential energy diagram is 3-D
Enzymes
•
•
•
•
•
•
Biological Catalyst
Reactants called “substrate”
S + E  SE  P + E
Active site – binding location
SE = enzyme-substrate complex
Lock and Key Model
vs.
Induced Fit Model
Reaction Mechanism
• Balanced rxn tells us the species before a rxn
starts and after the rxn ends.
– Does NOT show how the reaction occurs.
• Reaction Mechanism – the steps a reaction
progresses.
• The steps can be of varying speeds
Elementary Reactions
• Reactions occur because of collisions
– Must be correctly oriented
– Must have enough energy to reach TS


A+BC+D
This is a single collision reaction
Elementary reactions – rxn occurring in a single
step
Molecularity
• Number of molecules that participate in a
single reaction.
• Unimolecular, bimolecular, termolecular
• Probability decreases with molecularity
AP
2A  P
A+BP
A + 2B  P
Multiple Steps
• Some reactions occur in multiple steps
• Multistep Mechanism – a reaction consisting
of a series of elementary reactions
– Must add to give the overall reaction.
– Items that are produced within a mechanism and
consumed within a mechanism are intermediates.
– Intermediates are not R or P.
NO2(g) + CO(g)  NO(g) + CO2(g)
Rate Laws
• Earlier we stated that rate laws can not be
determined from a chemical equation.
• Why?
– There is a possibility for multistep mechanism
– We must consider the speed of each step
• However, we can derive the rate law from a
the mechanism of a reaction.
It’s Elementary, my dear chemists!
• If a rxn is an elementary rxn, the rate law is
based off the molecularity (coefficients)
• AP
Rate =
• 2AP
Rate =
• A+BP
Rate =
• 3AP
Rate =
• A+2BP
Rate =
• A + B + C  P Rate =
Rate-Determining Step (RDS)
•
•
•
•
•
Let’s go shopping.
Mechanisms can have a slow step.
RDS = slowest step – governs rate law
If first step is slow, intermediates short lived
If later step is slow, there is a buildup if
intermediates.
• Each step has its own rate constant
Slow Initial Step
NO2(g) + CO(g)  NO(g) + CO2(g)
Experimentally Determined Rate Law: Rate =
k[NO2]2[CO]0
Fast Initial Step
• The slow step rate law still governs the rate
law as before.
• However, because the slow step is the second
step, there are intermediates in the rate law.
• Intermediates are short lived and can not be
measured easily – not good in rate law
Fast Initial Step
• Make assumptions: there is a dynamic
equilibrium established in the fast step
– The intermediate is more likely to decompose
(reverse of fast step is also fast = k-1) than be
consumed in step 2 (k2)
– The forward rxn rate in step one (k1) equals the
rate of the reverse reaction:
Rate forward = Rate Reverse
2 NO(g) + Br2(g0  2 NOBr(g)
• Experimental Rate Law: Rate = k[NO]2[Br2]