The Rate of Chemical Reactions
1. Rate Laws
a.
For generic reaction: aA + bB
b.
c.
Rate = k[A]x[B]y [Units of Rate always = M/s = mol/L • s]
Details
1. Reactants decrease (-); Products increase (+)
2. For simple forward only reactions: ONLY REACTANTS IN RATE LAW
3. Exponents (x and y) in the rate law must be determined experimentally
N2O4 -------> 2 NO2
Rate = k[N2O4]1 (first order)
3 NO -------> N2O + NO2
Rate = k[NO]2 (second order)
4. Properties of the rate constant “k”
Specific to each reaction and changes with Temperature
Units of “k” depend on what order the reaction is
mol
 k[A]
Ls
mol
1 -1
 mol
 k

k

s

Ls
L
s


First Order : Rate 
cC + dD
Second Order : Rate 
mol
 k[A]2
Ls
2
mol
L
 mol
 k

k

 L  mol-1  s -1

Ls
mol s
 L 
2. Method of Initial Rates
a.
b.
Method for determining the “order” of the reactants (exponents in rate law)
First order reactants = [A]1
Double the reactant concentration -------> Doubles the rate of reaction
[A]exp 2
[A]exp 1
2
Rateexp 2
Rateexp 1
2
(Conc.Ratio)   Rate Ratio
c.
First Order
Second order reactants = [A]2
Double the reactant concentration -------> quadruples the rate of reaction
[A]exp 2
Rateexp 2
2
4
[A]exp 1
Rateexp 1
(Conc.Ratio)   Rate Ratio
d.
21  2
2 2   4
Second Order
Zero order reactants = [A]0 (changing concentration has no effect on rate)
[A]exp 2
Rateexp 2
2
1
[A]exp 1
Rateexp 1
(Conc.Ratio)   Rate Ratio
20   1
ZeroOrder
3. Temperature, Reaction Rates, and the Arrhenius Equation
k = rate constant
A = frequency factor (combines z and p)
Ea = activation energy
T = temperature in Kelvins
R = gas constant = 8.3145 J/K.mol
a. Taking the natural log of each side gives us another form of the equation that
gives a linear plot. lnk vs. 1/T gives straight line with slope = -Ea/R and
intercept = ln(A)
Ea
ln(k) 
R
1
   ln(A)
T
b. Example:
2 N2O5
4 NO2 + O2 Ea?
T(oC)
T(K)
1/T(K)
k (s-1)
ln(k)
20
293
3.41x10-3
2.0x10-5
-10.82
30
303
3.30x10-3
7.3x10-5
-9.53
40
313
3.19x10-3
2.7x10-4
-8.22
50
323
3.10x10-3
9.1x10-4
-7.00
60
333
3.00x10-3
2.9x10-3
-5.84
Ea
 E a  (slope)(R) -(-1.2x 104 K)(8.3145J/mol K)
R
E a  99,774 J/mol  100 kJ/mol
slope  
4. Reminder about the Dilution Equation
Example: 5 ml of 0.015 M KIO3 is added to 2.5 ml HSO3- and 7.5 ml H2O
M1V1  M 2 V2
M2 
M1V1 (0.015M)(5ml)

 0.005M KIO3
V2
(15 ml)
5. Today’s Reactions: KIO3 + NaHSO3
rate = k[IO3-]n[HSO3-]m
IO3- + 3HSO3- -------> I- + 3SO42- + 3H+
IO3- + 8I- + 6H+ -------> 3I3- + 3H2O
I3- + HSO3- + H2O -------> 3I- + SO4- + 3H+
fast
fast
slow
Starch + I3- -------> Colored Complex
a)
The “slow” reaction occurs until all of the HSO3- is gone, then Blue color
forms
b) We will determine the rate of the “slow reaction” by timing how long it takes
for the Blue Complex to appear.
c)
Rate = ([HSO3-]initial)/(time till Blue) = [0.0125 M]/73s = 1.72 x 10-4 mol/L•s
6. Reminder:
a) Experiments 1-5 at different concentrations give us Rate Law and k
b) Experiments 6-9 at different temperatures give us Ea