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
a) 1x10-3Ms-1
b) 3x10-3Ms-1
c) 0.33x10-3Ms-1
what is
-[B] t

2. Integrated rate laws differential rate laws are differential equations
t = t
rate = k  
= -d[A]/dt
zero order reaction
t = 0
t = t
rate = k[A]
= -d[A]/dt
first order reaction
t = 0
t = t
rate = k[A]2
= -d[A]/dt
second order reaction
t = 0
differential rate eqn
integrated rate eqn

3. Integrated rate laws differential rate laws are differential equations
rate = k  
= -d[A]/dt  
zero order reaction
[A] =
-kt
+ [A]0
y = mx + b
slope = -k
intercept = [A]0
differential rate eqn
integrated rate eqn

4. CH3CH2OH + NAD+  CH3CHO + NADH + H+
t (min) [ethanol] M
0.051
zero order in ethanol
[CH2CH2OH] = -kt + [CH3CH2OH]0
rate = k
k = -
= 4.7 x 10-4
60 - 0
rate = 4.7 x 10-4
M min-1
t1/2 =
[CH3CH2OH]0
= 70 min
2 k

5. Integrated rate laws differential rate laws are differential equations
rate = k[A]  
= -d[A]/dt  
first order reaction
ln [A] =
-kt
+ ln [A]0
y = mx + b
slope = -k
intercept = ln [A]0
differential rate eqn
integrated rate eqn

6. cis-[Pt(NH3)2Cl2] + H2O  [Pt(NH3)2Cl(H2O)]Cl-
t (min) [cis-platin] M
first order in cis-platin
ln [cis-platin] = -kt +ln [cis-platin]0
rate = k [cis-platin]
k = -
(-6.32)-(-5.12)
= 1.5x10-3
rate = 1.5 x 10-3
min-1
ln 2
t1/2 =
= 462 min k

7. cis-[Pt(NH3)2Cl2] + H2O  [Pt(NH3)2Cl(H2O)]Cl-
t (h) [cis-platin] M
first order in cis-platin
ln [cis-platin] = -kt +ln [cis-platin]0
rate = k [cis-platin]
k = -
(-6.32)-(-5.12)
= 1.5x10-3
rate = 1.5 x 10-3
min-1
radioactive decay 1st order
ln 2
t1/2 =
= 462 min k
14C dating
t1/2 = 5730 years

8. Integrated rate laws differential rate laws are differential equations
rate = k[A]2  
= -d[A]/dt  
second order reaction
1/[A] = kt
+ 1/[A]0
y = mx + b
slope = k
intercept = 1/[A]0
differential rate eqn
integrated rate eqn

9. t (s) [X] (M)
0.392
0.294
0.261
1/.235 – 1/.393
= .213
= k
10 - 2
1/.235 =
(.213)(10) + 1/[X]0 1
[X]0 = 0.471
t1/2 =
k[X]0
second order in X
1/[X] = kt + 1[X]0
rate = [X]2
0.213
M-1s-1

10. Integrated rate laws 1 = kt + 1 second order reactions rate = k [A]2
[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

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

12. X + 2Y  2Z second order in X t (s) [X] (M) [Y] (M) 0 0.470 2.0 0.448
0.448
0.427
0.409
0.392
second order in X
t (s) [X] (M) [Y] (M)
0.427
0.391
0.361
0.335

13. X + 2Y  2Z rate = k [X]2[Y] rate = k’ [X]2 k = k’ k = .028 M-2 [Y] s
k’ =
= 0.057
10 - 0
second order in X
[Y] = 2.0 M
first order in Y
.106
= (4.0)y
y =1
.057
(2.0)y
k’’ =
= .106
[Y] = 4.0 M
10 - 0