1. The Collision Model The reaction rate depends on: collision frequency
a probability or orientation factor
activation energy (Ea)
The reaction rate increases as the number of collisions between reacting species increase.
Concentration
temperature

2. The Collision Model
Collisions must occur in a particular orientation for reactions to occur.
For the reaction: Cl . + H - Br H - Cl + Br .
Desired rxn cannot occur.
Cl . Br H Br H
Cl .
Desired rxn cannot occur.
Desired rxn can occur.
Cl . Br H

3. The Collision Model
Collisions must occur with a specific minimum amount of energy in order for a reaction to take place.
Activation energy (Ea)
the minimum energy the reactants must have for a reaction to occur
the energy difference between the reactants and the transition state

4. The Collision Model Transition state:
a particular arrangement of atoms of the reacting species in which bonds are partially broken and partially formed
the state of highest energy between reactants and products
a relative maximum on the reaction-energy diagram.

5. Reaction Energy Diagrams
a plot of potential energy changes that occur as reactants are converted to products

6. Reaction Energy Diagrams
Given a reaction energy diagram for a chemical reaction, you should be able to identify the reactants, products, transition state, activation energy, the heat of reaction, and whether the reaction is endothermic or exothermic.

7. Reaction Energy Diagrams
Example: For each reaction energy diagram below, mark the location of the reactants, products and transition state. Identify the magnitude of Ea and DHrxn. Is each reaction endothermic or exothermic?

8. Arrhenius Equation Reaction rate increases with temperature because:
molecules have more kinetic energy
more collisions occur
greater number of collisions occur with enough energy to “get over the hill”
i.e. with energy greater than or equal to Ea

9. Arrhenius Equation
The Arrhenius Equation relates the value of the rate constant to Ea and the temperature:
k = Ae
where k = rate constant
Ea = activation energy
R = gas constant (8.314 J/mol.K)
T = temperature in Kelvin
A = frequency factor (a constant)
A is related to the frequency of collisions and the probability that the collisions are oriented favorably for reaction.
-Ea/RT

10. Arrhenius Equation
The Arrhenius Equation can be converted to another useful (experimentally useful, that is) form by taking the natural log of both sides:
ln k = - Ea + ln A RT
A graph of ln k vs. 1/T (in K) gives a straight line with a slope of - Ea/R.
Ea = - slope x R

11. Arrhenius Equation
The activation energy for a reaction can also be found in a non-graphical way if the rate constant at two or more temperatures is known:
ln k1 = Ea
k2 R T2 T1

12. Arrhenius Equation
Example: Calculate the activation energy for the rearrangement of methyl isonitrile to acetonitrile using the following data.
You can solve this in one of several ways:
Graph ln k vs. 1/T and determine the slope;
Find ln k and 1/T and determine the slope using two well spaced points using D(ln k)/D(1/T);
Use the non-graphical method

13. Arrhenius Equation Personally, I prefer the non-graphical method:
ln k1 = Ea
k2 R T2 T1
Use two points that are well separated and convert T to Kelvin (K = oC )

14. Arrhenius Equation
Plug the values of k and T into the equation. Be careful to put k1 and T1 in the appropriate order:
ln k1 = Ea
k2 R T2 T1
ln 2.52 x = Ea __1__
3.16 x (8.314 J/mol K) K K
Solve for Ea.

15. Arrhenius Equation ln 2.52 x 10-5 = Ea 1 - __1__
3.16 x (8.314 J/mol K) K K
Solve for Ea.
= ( x 10-5 mol/J) Ea
Ea = 1.59 x 105 J/mol

16. Arrhenius Equation Alternately, you can solve it graphically.
First, pick two well-separated points and calculate 1/T (in Kelvin) and ln k.

17. Arrhenius Equation Find the slope of the line: Dy = D (ln k)
Dx D (1/T)
Slope = ( ) = x 104 K
1.907 x x 10-3

18. Arrhenius Equation Since slope = - Ea/R - Ea = - 1.910 x 104 K R
Ea = - ( x 104K) x J/mol.K
Ea = x 105 J/mol

19. Arrhenius Equation
Once you find the value for Ea, you can use it to find the frequency factor (A) for the reaction.
k = A e
A = k e
-Ea/RT
-Ea/RT

20. Arrhenius Equation
To find A for the previous example, pick one of the data points given and plug in the values of k, Ea, and T.
A = x 10-5 s-1 e
A = x 1013 s-1
x 105J/ (8.314 J/molK x 462.9K)

21. Arrhenius Equation
Once you have the value for A and Ea for the reaction, you can calculate the value of the rate constant at any temperature using the Arrhenius equation:
For example, the value of k at 25oC (298 K) is:
k = (2.096 x 1013 s-1) e
k = 2.82 x s-1
- (1.59 x 105 J/mol)/(8.314 J/mol K x 298 K)