1. Review Differential Rate Laws ... rate (M s-1) = k [A]a [B]b
A + 3B  2C
rate =
-[A] t
= [B] t
1/3
Assume that A is easily detected
initial rate =
1x10-3 M s-1

2. [A] [A] Exp. 1 [B]i [A]i initial rate (M) (M) (M s-1) 1.0 1.0
t (ms)
Concentration (M)
[A]
t (ms)
Concentration (M)
[A]
[B]i
[A]i
initial rate
(M)
(M)
(M s-1)
1.0
1.0
1.0 x 10-3
Exp. 2
[A]i [B]I initial rate
(M) (M) (M s-1)
2.0
1.0
2.0 x 10-3
Exp. 3
[A]i [B]I initial rate
(M) (M) (M s-1)
1.0
2.0
1.0 x 10-3

3. rate = k [A] 1st order reaction Exp. 1 rate = k [A]a [B]b
[A]i [B]I initial rate
(M) (M) (M s-1)
a = 0
a = 1
a = 2
rate 2 =
rate 1
2 x 10-3 =
1 x 10-3
[2.0]a
[1.0]a
1.0 x 10-3
Exp. 2
rate 3 =
rate 1
1 x 10-3 =
1 x 10-3
[2.0]b
[1.0]b
b = 0
b = 1
b = 2
[A]i [B]I initial rate
(M) (M) (M s-1)
2.0 x 10-3
rate = k
[A]
1st order reaction
Exp. 3
[A]i [B]I initial rate
(M) (M) (M s-1)
1x10-3 (M s-1) = k
[1.0 M]
k = 1 x 10-3 s-1
1.0 x 10-3

4. rate = k [A] [B] 2nd order reaction Exp [A]i [B]i initial rate
(M) (M) (M s-1)
Concentration (M)
t (ms)
1.0 x 10-3
2.0 x 10-3
2.0 x 10-3
4.0 x 10-3
rate 2 =
rate 1
2.0 x 10-3 =
1.0 x 10-3
[2.0]a
[1.0]a
a = 1
1st order in [A]
rate 3 =
rate 1
2.0 x 10-3 =
1.0 x 10-3
[2.0]b
[1.0]b
b = 1
1st order in [B]
rate = k [A] [B]
2nd order reaction

5. Experiment [NO]0 (M) [H2]0 (M) initial rate (M/s)
2NO + H2  N2O + H2O
Experiment [NO]0 (M) [H2]0 (M) initial rate (M/s)
x x X10-5
x x x10-4
x x x10-5
rate = k [NO]
rate = k [NO]2
rate = k [NO]2 [H2]
rate = k [NO][H2]
rate = k [NO][H2]2

6. Integrated rate laws differential rate laws are differential equations
t = t
rate = k[A]  
= -d[A]/dt
t = 0
rate = k[A]2
= -d[A]/dt
differential rate eqn
integrated rate eqn
rate = k[A] ln
[A]t =
- kt
[A]0

7. Integrated rate laws ln [A]t = - kt [A]0 ln[A]t = ln[A]t - ln[A]0 -kt
y = mx + b
y = ln[A]t
x = t
m = -k
b = ln[A]0
plot
ln[A]t
v.s. t
linear

8. rate = k[A] ln[A]t = - kt + ln[A]0 y = ln[A]t x = t m = -k b = ln[A]0
Concentration (M)
t (ms)
[A]
rate = k[A]
ln [A]
ln[A]t = - kt + ln[A]0
ln [A]
y = ln[A]t
x = t
m = -k
b = ln[A]0

9. 1st order reactions 2N2O5(g)  4NO2(g) + O2(g) rate = (M s-1)
ln [A]t = - kt
[A]0
ln[A]t = - kt + ln[A]0
2N2O5(g) 
4NO2(g)
+ O2(g)
rate =
(M s-1)
k [N2O ]
(M)
k = 5.1 x 10-4
s-1
What is [N2O5] after 3.2 min if
[N2O5] =
0.25M ln
[N2O5]
= -
(5.1x10-4 s-1)
(192 s)
3.2
[0.25]
[N2O5]3.2
= 0.23 M

10. 1st order reactions How long will it take for [N2O5] to go from
ln [A]t = - kt
[A]0
ln[A]t = - kt + ln[A]0
How long will it take for [N2O5] to go from
0.25 M to M ? ln
[0.125] =
- (5.1x10-4 s-1 ) t
t =
23 min
[0.25]
The half-life (t1/2) is the time required for
[reactant] to decrease to 1/2 [reactant]i
t1/2 =
ln 2 k

11. 1st order reactions t1/2 = ln 2 k 700 ms = ln 2 k k  1 s-1
Radioactive decay 1st order
14C dating
t1/2 = 5730 years

12. Integrated rate laws differential rate eqn integrated rate eqn
________________________
________________________
rate = k[A]
ln [A]t = - kt
[A]0
1st order 1
[A]t = kt + 1
[A]0
rate = k[A]2
2nd order
y = mx + b
y = 1
[A]t
x = t
m = k
A plot of v.s. is linear
1/[A]t t
b = 1
[A]0

13. [A]
rate = k[A]2
t1/2 =
k[A]0
1 = kt + 1
[A]t [A]0
1/[A]

14. Integrated rate laws second order reactions rate = k [A]2 1 = kt + 1
[A]t [A]0
many second order reactions
rate = k
[A]
[B]
A + B  C
A and B consumed
stoichiometrically
[A]0 = [B]0
if not,
no analytical solution

15. Pseudo 1st order reactions
high order reactions
difficult to analyze
put in large excess of all but one reagent
rate = k [A]a [B]b
[A] [B]0
1.0 x 10-3 M
1.0 M
[B] 
constant
-0.5 x 10-3 mol
-0.5x 10-3 mol
rate =
k’ [A]a
0.5 x 10-3 M
0.999 M
k = k’
k[A]a[B]b
= k’[A]
[B]b