•Thus far we have derived rate laws and rate constants on the basis of empirical observations of macroscopic properties. These facts do not rely upon, or require, any knowledge or theory of molecular structure. •To understand chemical kinetics, we need to consider what is happening at the molecular level. The temperature dependence of chemical reactions provides a useful clue. Theoretical Models for Chemical Kinetics •Collision Theory - highly energetic molecular collisions produce chemical reactions. •Transition State Theory - examines changes in molecular structure during a reaction. It postulates an “activated complex” in transitory equilibrium with reactants. TS theory provides a powerful way to think about chemical reaction kinetics.
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Theoretical Models for Chemical Kinetics · 2011. 3. 28. · 4) intermediates produced in an elementary reaction do not appear in the net chemical reaction or the rate law. Intermediates
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•Thus far we have derived rate laws and rate constants on the basis of empirical
observations of macroscopic properties. These facts do not rely upon, or require,
any knowledge or theory of molecular structure.•To understand chemical kinetics, we need to consider what is happening at the
molecular level. The temperature dependence of chemical reactions provides a
useful clue.
Theoretical Models for Chemical Kinetics
•Collision Theory - highly energetic molecular
collisions produce chemical reactions.
•Transition State Theory - examines changes in
molecular structure during a reaction. It postulates an
“activated complex” in transitory equilibrium with
reactants. TS theory provides a powerful way to
think about chemical reaction kinetics.
Arrhenius Equation
• In 1889 Svante Arrhenius observed that the rate constant for
chemical reactions fit the following equation.
a-E /RTk = Ae
•A is the pre-exponential factor and carries the units of k
•Ea is the activation energy in kJ/mole
•R =8.314 Jmol-1K-1
•T =temp in Kelvin
•Given k vs T we can obtain A and Ea OR given A & Ea we can
compute k at any temperature.
Diffusion and Activation
The Pre-Exponential factor A is related to collision frequency and
orientation.
Molecules have to get together in order to react. In solution or in the gas phase the
diffusion controlled rate constant is typically about 1011 M-1s-1 rate = k[A][B]
For two gases (P = 1 atm) rateD = 1011[1/22.4)2] = 106 M/s
For two solutes at 1 M rateD = 1011 [1]2 M/s
millimolar rate = 1011 [.001]2 = 105 M/s
Eact is related to the energetics of collisions.
Collisions must have enough energy to weaken or break bonds.
average kinetic energy = 1/2mv2 = 3/2 kBT (T = Kelvin)
Typically only a very small fraction of collisions are energetic enough for reaction
(1 in 1010 for gases at 1 atm would give rate = 10-4 M/s ).
Boltzmann population
1/2mv2 = 3/2 kT : where v2 is avg squared
speed
-The minimum energy required for a
reaction to occur is indicated by the
arrow in the figure. The fraction of
molecules possessing this energy will be
greater at T2 than at T1
- Collisions continually re-establish the
distribution at constant T.
- at 298 K RT = NkT= 2.5kJ/mol
while Ea is typically 50-100 kJ/mol
•Kinetic Molecular Theory (6.7) tells us that the
average kinetic energy of molecules increases with
temperature AND is distributed as shown. * Don’t confuse Boltzmann’s constant
k = 1.38 x 10-23 J/K per molecule with
the rate constant. R is just a molar
version of Boltzmann’s. k = R/N
• The rate of reaction is also limited by the orientation
of molecules at the time of collision
Example: N2O(g) + NO(g) → N2(g) + NO2(g)
k = Zpe-Ea/RT where Z = collision freq. & p = fraction of
collisions with suitable orientation.
Transition State Theory (Eyring 1935)
A + B AB*
• The activated complex (AB*) is a transitory molecule. The reaction rate is
proportional to the concentration of activated complexes.
• TS theory lets us think about reaction rates much like equilibria. The higher
the energy (less stable) of the activated complex, the slower the reaction.
• The TS lies at a maximum in energy and does not have a finite lifetime. A
reactive intermediate lies at a local minimum and has a finite lifetime.
Example: N2O(g) + NO(g) → N2(g) + NO2(g)
Transition State Theory
•The pass (transition state) is the maximum altitude along the
minimum altitude path from Calgary to Vancouver. Trains do
not go over mountain peaks and neither do chemical reactions.
•The red line is the
reaction progress. It
is not time and it is
not fraction reacted.
It traces the route
taken by a molecule
to get from reactants
to products. One
molecule makes the
trip in ~ 10-13 sec, the
duration of a sticky
collision.
Transition State TheoryThis N3O2 activated
complex lies at the
maximum energy
along this path. The
molecule gains
energy as we
break one bond
and releases it as
we form the new
bond.
Thermodynamics deals only with reactants and products.
Kinetics concerns only reactants and the transition state.
Applying TS Theory
Potential Energy Surfaces
- A topographical map displays contours of equal altitude allowing a 2 dimensional picture of altitude changes.
- Reaction progress can be represented on a multidimensional potential energy surface. Advanced theoretical methods can compute this surface for simple reactions such as : H2 + Br � HBr + H
Multiple pathways, intermediates and TS’s can be identified on this surface.
. Reaction Profile diagrams are a useful way of displaying and explaining the
energetics of chemical reactions. Try the following
• Show the effect of catalase on the reaction : H2O2 � H2O + ½ O2
• Sketch the profile for the uphill reaction : H2O � H2 + ½ O2
• Does the reverse rxn take a different path or give a different TS than the
forward rxn?
Potential Energy Surface for H2 + Br � HBr + H contours are energy in kcal/mole 1 = reactants 2 = TS 3 = products
PE vs reaction path for H2 + Br � HBr + H
Ea = +18 kcal/mole ∆Ho = +15 kcal/mole
Applying the Arrhenius Equation
a-E /RTk = Ae
1. Graphical : A plot of lnk vs 1/T is linear:
ln(k) = ln A - Ea/RT
2. k is given at two temps :
ln (k1) = lnA - Ea/RT1
ln (k2) = lnA -Ea/RT2
note : ln(x/y) = ln x - ln y
3. Ea and k1 given; find k2 at T2.
a2
1 1 2
Ek 1 1ln = -
k R T T
•HINT: Use 1000/T for
kJ/mole answer. Make sure
the higher T has the larger k.
Example 1. N2O5 → N2O4 + ½ O2
Plot of ln k versus 1/T : slope of -Ea/R and
intercept = ln(A) (see Lab # 1)
Ea = - slope X R in J/mol
Ea = 12000 X 8.314/1000
= 106 kJ/mol
Units K X J/mol/K = J/mol
J/mol / 1000 J/kJ= kJ/mol
remember T decreases as 1/T
increases and rates almost
always increase with T.
*For log(k) : Ea = -slope x 2.303 R
Example 2. Cricket chirping roughly doubles for every 10 °C
increase in temperature. What Ea does this correspond to?
a2
1 1 2
Ek 1 1ln = -
k R T T
ln 2 = Ea/R [ 1000/300 -1000/310)]
Ea = .693 X 8.314 / (3.333-3.225)
Ea = 53.4 KJ/mol. HINT#1: use 1000/T to convert to kJ/mole and
avoid math blunders.
HINT#2: You need to pick a reasonable
temperature to complete the problem.
HINT #3: A ratio can often remove a variable-
in this case A is not required.
Reaction Mechanisms
• The detailed step-by-step pathway by which a
reaction occurs is called the reaction mechanism
• a plausible reaction mechanism must be consistent
with the
1) stoichiometry of the overall reaction
2) experimentally determined rate law
• The steps in a mechanism are called elementary
reactions.
Elementary Reactions
1) The number of reactant molecules involved in an elementary
rxn is called the molecularity. Examples:
unimolecular H2 → 2H rate = k [H2]
bimolecular H + H → H2 rate = k[H]2
CH3I + OH- → CH3OH + I- rate = k[CH3I][OH-]
2) Note that for elementary reactions the molecularity is the same
as the kinetic order in the rate law.
3) Mechanisms typically consist of several elementary steps and
the kinetic order is often not related to the overall reaction
stoichiometry.
4) intermediates produced in an elementary reaction do not appear in the net chemical reaction or the rate law. Intermediates are produced by one elementary reaction and consumed by another. d[I]/dt = 0 is the steady state approximation.
5) The rate of the overall reaction is largely determined by theslowest step- the rate-determining step = RDS.
TIP: When deriving the rate law for a given mechanism:1. First find the RDS and write the rate expression for it.
2. Then replace any species which is not a primary reactant by using equilibrium relations or by using the steady state approx.for intermediates. The rate law should only include the “stuff” you measured out- not something produced after mixing.
3. Ignore entirely, fast reactions occurring after the RDS.
4. The kinetic orders tell you how many of each reactant are involved before and including the RDS.