1. Integrated rate laws
zero order
t1/2 =
[A]0 2k
rate = k

2. Integrated rate laws 1st order ln[A] = -kt + ln[A]0 rate = k[A] t1/2 =

3. 2nd order
1/[A] = kt + 1/[A]0
rate = k[A]2
t1/2 = 1
k [A]0
can also get negative or fractional orders

4. Reaction mechanism reaction =  Elementary steps Molecularity
rate law
unimolecular
A  product
k[A]
bimolecular
A+ B  product
k[A][B]
termolecular
A+B+C  product
k[A][B][C]
A+A+A  product
k[A]3
chemical reaction
= sum of elementary steps
rate law
and stoichiometry

5. Reaction mechanism increase rate of reaction
2H2O2 (aq)  2H2O(l) + O2(g)
Gorxn =
[ 2(-237.9)]
- [2( )]
= kJ
spontaneous reaction
experimental rate law:
rate = k[H2O2]
[I-]
I- =
catalyst
increase rate of reaction
not consumed in the overall reaction
reactant in early elementary step
product in later elementary step

6. Reaction mechanism 2H2O2 (aq)  2H2O(l) + O2(g) rate = k[H2O2] [I-]
step 1
H2O2
+ I-
 H2O
+ OI- k2
step 2
H2O2 +
OI- 
H2O + O2
+I-
2H2O2 (aq) 
2H2O(l)
+ O2(g)
step 1
rate = k1
[H2O2]
[I-]

7. Reaction mechanism step 1 H2O2 + I-  H2O + OI- step 2k1
step 1
H2O2 + I-  H2O + OI- k2
step 2
H2O2 + OI-  H2O + O2 +I-
what about step 2 ?
assume k2
>> k1
step 1
rate determining step
I- catalyst
consumed in early elementary step
regenerated in later elementary step
OI-
intermediate
formed in early step,
consumed in later step

8. Rate determining step
on Labor Day weekend
“Big Mac” bridge

9. Reaction Mechanism H2(g) + I2(g)  2HI(g) rate = k [H2] [I2]
reaction faster in light
free radical
unpaired electron

10. Reaction Mechanism H2(g) + I2(g)  2HI(g) rate = k [H2] [I2]  I2 2I.
step 1 
forward rate
= kf [I2]
reverse rate =
kr [I.]2
equilibrium
kf[I2] =
kr[I.]2
Keq=
[I.]2
= kf
[I2] kr

11. Reaction Mechanism H2(g) + I2(g)  2HI(g) rate = k [H2] [I2] I2 2I. 
step 1
step 2
H2 +
2I.
2HI H2
+ I2 
2HI
need H2 in the rate determining step
from step 2
rate =
k [H2] [I.]2
I. =
intermediate

12. Reaction Mechanism H2(g) + I2(g)  2HI(g) rate = k [H2] [I2] I2 2I. 
step 1
step 2 H2 + 2I. 2HI
rate = k [H2] [I.]2
Keq =
[I.]2
= kf
[I2] kr
rate =
k [H2]
Keq [I2]
[I.]2 =
Keq [I2]
rate = k’ [H2] [I2]

13. Reaction Mechanism 2NO + O2 2NO2 step 1 2NO  N2O2 step 2
overall reaction:
2NO + O2  2NO2
intermediates:
N2O2
[N2O2] = constant
d[N2O4]
= 0
rate = k2
[N2O2]
[O2] dt
steady state approximation

14. 2NO + O2 2NO2 step 1 2NO  N2O2 step 2 N2O2 + O2 2NO2
rate = k2
[N2O2]
[O2]
= k2[O2]
k1[NO]2
k2[O2]
+ k-1
k1[NO]2 =
k2 [N2O2][O2]
+ k-1[N2O2]
produce N2O2
consume N2O2
[N2O2] =
k1 [NO]2
k2[O2]
+ k-1

15. 2NO + O2 2NO2
step 1
2NO  N2O2
step 2
N2O2 + O2 2NO2
2NO + O2  2NO2
low [O2]
rate
= k2[O2]
k1[NO]2
rate =
k'[O2][NO]2
k2[O2]
+ k-1
high [O2]
rate =
k1[NO]2