1. Review Reaction mechanism Br2(l) + C5H12(l)  C5H11Br(l) + HBr(l)
step 1
Br2 
step 2
Br. +
C5H12 
HBr + C5H11.
C5H11. +
Br. 
C5H11Br
step 3
overall
Br2
+ C5H12 
C5H11Br
+ HBr

2. Br2(l) + C5H12(l)  C5H11Br(l) + HBr(l) 2 Br.  step 1 Br2
HBr + C5H11. 
step 3
C5H11. +
Br.
C5H11Br
assume step 2 is rate determining
(slow)
rate = k2
[Br.]
[C5H12]
Br.= intermediate
Keq =
[Br.]2
[Br2]
rate = k2
k’[Br2]1/2
[C5H12]
rate =
k’[Br2]1/2
[C5H12]
[Br.] =
Keq1/2
[Br2]1/2
11/2 order reaction

3. bimolecular elementary steps
increase [react] 
increase rate of reaction
increase T 
increase rate of reaction
increase number of collisions
increase force of collisions

6. minimum energy required for reaction:
T2 > T1
# molecules
Kinetic Energy
minimum energy required for reaction:
activation energy
= Ea

7. Arrhenius Equation k = z p e-Ea/RT T dependence of a rate constant, k
a) increases
b) decreases
with T k
a) decreases
b) increases
with Ea
Ea = activation energy (kJ/mol)
R = gas constant (8.314 x 10-3kJ/K mol)
T = temperature (K)
z = collision frequency
p = steric factor (<1)

8. k = A e-Ea/RT p = steric factor z = collision frequency
combine to give A
k = A
e-Ea/RT

9. Arrhenius Equation k = A e-Ea/RT ln k = - (Ea/R) (1/T) + ln A ln A
y = m x b
plot ln k
v.s. 1/T
slope = -Ea/R
intercept = ln A
ln (k2/ k1) =
(Ea /R)
(1/T1
- 1/T2)

10. A-B + C A...B...C A + B-C activated complex P.E. is at a maximum
transition state
A + B-C

11. Hrxn A-B + C A...B...C A + B-C Eaf Eab activated complex reactants
P.E.
Hrxn
products
Reaction coordinate
A-B + C
A...B...C
A + B-C

12. Hrxn exothermic Eab Eaf > endothermic Eaf Eab activated complex
reactants
P.E.
Hrxn
products
Reaction coordinate
exothermic
Eab Eaf
>
endothermic

13. exothermic Eab Eaf > large Ea = slow rate endothermic Eab Eaf <
activated complex
products
reactants
Eaf
Eab
P.E.
Reaction coordinate
exothermic
Eab Eaf
>
large Ea
= slow rate
endothermic
Eab Eaf
<

14. faster forward reaction ( kf ) lowers Ear 
- catalyst
+ catalyst
lowers Eaf 
faster forward reaction ( kf )
lowers Ear 
faster reverse reaction ( kr )
Keq unchanged
and H kf =
Keq kr

15. Br2(l) + C5H12(l)  C5H11Br(l) + HBr(l)
Hrxn
< 0 Ea
Reaction coordinate
P.E.
Br2 + C5H12
Hrxn
C5H11Br + HBr

16. Br2(l) + C5H12(l)  C5H11Br(l) + HBr(l) step 1 Br2 2 Br.hn
Br2(l) + C5H12(l)  C5H11Br(l) + HBr(l)
step 1
Br2
2 Br. h
step 2
Br. +
C5H12 
HBr + C5H11.
step 3
C5H11. +
Br. 
C5H11Br Ea Ea Ea
HBr + C5H11.
+ Br.
2Br.
+ C5H12
C5H12
+ Br2
P.E.
C5H11Br + HBr