1. Factors Affecting Rate
Collision model: in order to react, reactant molecules must collide
Chemical reactions speed up when temperature increases, but not as much as one would think
Only a small fraction of collisions actually react……..

2. Activation Energy
Threshold energy that must be overcome in order for chemicals to react
According to the collision model, energy comes from kinetic energy stored in reactants’ bonds. Kinetic energy becomes potential energy during intermediates.
Collisions between reactant molecules must have enough energy to allow products to form

4. # of collisions with AE = (total # collisions)e-Ea/RT
Reaction Rate
Temperature will exponentially increase the number of collisions with enough activation energy to react
# of collisions with AE = (total # collisions)e-Ea/RT
Ea = activation energy
R = constant (8.3145)
T = kelvin
e-Ea/RT = fraction of collisions with energy Ea or greater at temperature, T.

5. Still Not Enough
Even though the reactants collide with enough energy, they still often will not react…
Molecular orientations can play a roll in reactions

6. Successful Collisions:
Needs enough energy to meet or exceed the activation energy
Orientations must allow for new bonds to form
NEW RATE CONST. (Arrhenius) EQUATION:
K = Ae-Ea/RT
A = frequency factor
e-Ea/RT = fraction of collisions with enough energy

7. Natural Log ln(k) = -(Ea/R) (1/T) + ln(A) Y = m x + B
Plotting ln(k) versus 1/T will give a line

8. 2N2O5(g) -> 4NO2(g) + O2(g)
K(s-1)
T(°C)
2.0 X 10-5 20
7.3 X 10-5 30
2.7 X 10-4 40
9.1 X 10-4 50
2.9 X 10-3 60
Example
2N2O5(g) -> 4NO2(g) + O2(g)
Calculate the value for Ea for this reaction.
Answer: Answer: 1.0 X 105 J/mol

9. Rearranging the Equation
ln(k2/k1) = (Ea/R)[(1/T1)-(1/T2)]

10. CH4(g) + 2S2(g) -> CS2(g) + 2H2S(g)
Example
CH4(g) + 2S2(g) -> CS2(g) + 2H2S(g)
At 550°C the rate constant for this reaction is 1.1 L/mol S, and at 625°C the rate constant is 6.4 mol/L s. Using these values calculate Ea for this reaction.
Answer: 1.4 X 105 J/mol

11. Example #2 H2(g) + I2(g) -> 2HI(g)
If the activation energy for the reaction is 167 kJ/mol and the rate constant at 302°C is 2.45 X 10-4 L/mol, what is the rate constant for the reaction at 205°C?
Answer: 2.06 X 10-7 L/mol s

12. Example #3
A second order reaction has rate constants of 8.9 X 10-3 L/mol and 7.1 X 10-2 L/mol at 3°C and 35°C respectively. Calculate the value of the activation energy for the reaction.
Answer: 46 kJ/mol

13. Reaction Rate: Catalysis
Sometimes, raising the temperature is not feasible (temp. regulation or $)
Catalysts: substances that speed up reactions without being used up
New pathway formed with a LOWER ACTIVATION ENERGY

14. Types of Catalyst
Homogeneous Catalysts: Present in the same phase as the reacting molecules
Heterogeneous Catalysts: In a different phase, usually a solid

15. Heterogeneous Catalysis
Usually involves gases being absorbed on a solid’s surface
Adsorption: on surface
Absorption: penetrates surface
Example: hydrogenation - conversion of CC double bonds to CC single bonds through the addition of hydrogen

16. Homogeneous Catalysis
Same phase (usually gases or liquids)
Example: ozone gases, freons, etc.