### Lecture 18-Kinetics 2

```Kinetics pt 2: Temperature Dependence of
Rate Constants
Lecture 18 (Ch 18)
HW: Ch 18: 1, 3, 15, 41
Temperature and Rate
 The rates of most chemical reactions increase with temperature.
 How is this temperature dependence reflected in the rate expression?
 Rates increase with temperature because rate constants increase with
temperature. An example is the 1st order reaction:
CH3NC ---> CH3CN
= [ ]
Variation in k with
temperature.
Collision Model
 The collision model makes sense of this. This model is based on kinetic
theory
 We’ve seen that the thermal energy of a molecule is converted to kinetic
energy in order to facilitate motion, and we’ve seen that the velocity of a
molecule increases with T.
 The central idea of the collision model is that molecules must collide to
react. The more collisions per second, the faster the reaction goes.
 This model also rationalizes the concentration dependence. The more
molecules present, the more collisions you have.

Activation Energy
 Of course, there is more to a chemical reaction than just collisions of molecules.
 Molecules must have some minimum energy in order to overcome the energy
barrier of reaction
 Upon colliding, the kinetic energy of molecules is used to stretch, bend, and
break bonds in order to cause a reaction.
 But if the molecules don’t have enough kinetic energy, they simply bounce off
one another.

E too low
 This minimum energy that molecules must have is called the activation
energy,(Ea)

Activation Energy Needed to Reach Transition State
Transition state
• The molecule passes through a
transition state, a high-energy
intermediate in which the CN
bond gets rotated 90o.
• This is an unstable configuration
and requires an activation
energy to be reached.
Enthalpy
• Lets go back to the reaction
CH3NC ---> CH3CN
ΔH
nitrogen
carbon
• Once this transition state is reached, the process is energetically downhill. The
transition state is always the highest energy point in the reaction pathway. As
drawn, this reaction is exothermic because the energy of the products is less
than the energy of the reactants.
Arrhenius Equation
 The fraction of molecules that have an energy equal to or greater than the
activation energy is given by the expression:
=

−

 Arrhenius noted that reaction-rate data depended on three aspects
 (1) the fraction of molecules possessing an energy Ea or greater
 (2) the number of collisions per second
 (3) the fraction of molecules oriented in the right way for a reaction to
occur
 These factors are incorporated into the Arrhenius Equation
=

−

 This equation relates k to temperature. A is the frequency factor, and is
related to aspects (2) and (3) listed above.
Rearranging the Arrhenius Equation
 Taking the natural log of both sides, the Arrhenius equation becomes:
= −

+

• As you see, we once again have an equation in y=mx+b form.
y = ln k

= −

1
=

Plotting ln k vs 1/T yields a linear
plot with slope –Ea/R and a yintercept of ln A
Showing Changes in k Graphically
= −

+

 From the data below, determine the activation energy of the 1st order reaction
189.7
2.52 x 10-5
198.9
5.25 x 10-5
230.3
6.30 x 10-4
251.2
3.16 x 10-3
Plot ln k vs 1/T to get the slope. We
know R, so we can calculate Ea. T must
be in Kelvin.
0
0.00185 0.0019 0.00195 0.002 0.00205 0.0021 0.00215 0.0022
-2
-4
ln k
Temperature, oC K (s-1)
-6
y = -19038x + 30.517
-8
1/T (oK-1)
ln K
0.002161
0.002119
0.001987
-10.5887
-9.8547
-7.36979
0.001908
-5.75718
-10
-12
1/T

=−

(8.314
)

= .

= −19038 = −
Calculating Changes in k Mathematically
 The Arrhenius equation can be rearranged once again to compare k
values at different temperatures
2
2 − 1
ln =
1

1 2
ln() =
Example
 The reaction above has an activation energy of 43.5 kJ/mol.
The
reaction occurs at 298 K with a rate constant of 110 s-1. What will the
rate constant be if we increase the temperature to 308 K?
2
2 − 1
ln =
1

1 2
2
43500   −1 308 − 298
ln
=
110
8.314   −1  −1
91784
ln

2
= .57
110
2
110
=  .57
2
= 1.77
110
2 = 194.7  −1
Reaction Mechanisms
 In a previous lecture, I alluded to the following reaction:
+   → () +  ()
 In looking at this equation, and applying your understanding of collision
theory, you would assume that this reaction proceeds when
 NO2 and CO collide
 The activation energy is exceeded
 The atoms rearrange to form NO and CO2
• Thus, we would expect the rate of the reaction to depend on both [NO2]
and [CO]
Reaction Mechanisms
+   → () +  ()
 However, according to experimental data, the reaction is 2nd order with
respect to NO2 and zero order with respect to CO
= [2 ]2 []0 = [2 ]2
• This means that the rate of the reaction does not depend on [CO] at all.
How can this be? What does this tell us about the reaction mechanism?
• This particular reaction must proceed through multiple steps, and the
rate-determining step must only depend on NO2
Reaction Mechanisms
+   → () +  ()
• For this particular reaction, there are two steps:
1 2  + 2  → 3  +
()
2 3  +   → 2  + 2
( )
3 2  +   →   + 2
• This step-by-step description of the molecular pathway is the reaction
mechanism. From reaction (1), we can physically understand why the
reaction is 2nd order with respect to [NO2].
• The 1st step is substantially slower than the 2nd. Therefore, step (1) acts
as the “reaction bottleneck”, and the overall reaction speed can only be
as fast as its slowest step.
• Thus, step (1) is the rate determining step.
 Each step in a reaction mechanism is called an elementary reaction.
 An elementary reaction is any reaction that proceeds in a single step.
Mechanisms consist of sums of elementary reactions.
 Going back to rate laws, each elementary reaction must have a
corresponding rate constant
intermediate (transition step)
k1
1 2  + 2  −−→ 3  +
()
k2
2 3  +   −−→
2  + 2  ( )
 *** For elementary reactions only, you can assume that the rate law
depends directly on the number of species present. Thus, the rate law of
an elementary reaction follows the stoichiometry.
1 = 1 2 2 = 1 [2 ]2
rate of overall reaction
Reaction Mechanisms Involving a Fast Equilibrium Step
 Let’s consider reaction mechanisms with a fast equilibrium step.
 To date, we have discussed reactions occurring in only one direction.
However, in many cases, the forward and back reactions are both
important. For example:
2
k1
k-1
2 2 (g)
• When a reaction is in equilibrium, like the one shown above, the
forward and back rates are equal, so there is no change in the
amount of reactant or product once equilibrium is established.
Reaction Mechanisms Involving a Fast Equilibrium Step
 For many reaction mechanisms, a rapid equilibrium step is involved. Lets
look at the following reaction:
2  + 2  → 22 ()
 This reaction goes through two elementary steps:
1 2
k1
k-1
2 2
k2
2 2 2  + 2  −−→
22
( )
()
 Overall reaction rate would be written as:
= 2 2 2 [2 ]
 [N2O2] is an intermediate. Intermediates are never included in the overall
rate law because they are too short-lived, so their concentrations can’t
easily be measured.
Mathematical Substitution For [Intermediate]
1 2
k1
2 2
k-1
2 2 2  + 2
k2
22
( )
()
= 2 2 2 [2 ]
• Since we have an equilibrium in step (1), the forward and back rates are equal
1 []2 = −1 [2 2 ]
• Solving for [N2O2], we get:
1
[]2 = [2 2 ]
−1
• Substitute this term into our rate law expression:

1
= 2
[]2 [2 ]
−1
may be written in the book as simply K
Examples
 Write the rate law for the following reaction given the elementary reaction
steps of the mechanism:
22
1
2
2  + 2 ()
2  + 2
3
k2
k1
+ 3
+ 2
()
()
• Write the rate law for the following reaction given the elementary
reaction steps of the mechanism. Substitute any intermediates.
2  + 2
2   + 2
k1
1
2
2
2 2  + 2
k-1
2 2
k2
( )
2   + 2 ()
()
```