Chapter 13 Integrated Rate Laws
I.
Integrated Rate Laws
A.
Preview: So far, we have looked at rate as a function of concentration
Next, we look at concentration as a function of time
aA
B.
products
[A]
rate   k[A]n
t
The Integrated First Order Rate Law
1.
2.
3.
Integration is a calculus operation
Integration of the first order rate law:
[A]
 k[A]
t
Gives the Integrated first-order rate law: ln[A]  -kt  ln[A] 0
a) Integration and Differentiation are connected. If we know the
integrated rate law, we could differentiate it to get the rate law
b) If we know the initial concentration [A]0 and k, we can now
calculate [A] at any time (t).
4)
How is the integrated rate law of any more help?
a) The integrated first order rate law is in the form of a straight line
ln[A]  -kt  ln[A] 0
y  mx  b
b)
c)
d)
5)
A plot of ln[A] vs t gives a straight line with
slope = -k and intercept = ln[A]0
If plot is a straight line, reaction is1st order
If not a straight line, not 1st order
Example: 2 N2O5
[N2O5]
ln[N2O5]
Time (s)
0.1000
-2.303
0
0.0707
-2.649
50
0.0500
-2.996
100
0.0250
-3.689
200
0.0125
-4.382
300
0.00625
-5.075
400
4 NO2 + O2 Find k
6)
Half-life of a first order reaction
 [A]0
a) The integrated first order rate law can be rewritten as: ln 
 [A]
b)
The half-life of a reaction = time it takes for [A] = [A]0/2

  kt

c)
[N2O5] 0.1000.050 = 100 s
[N2O5] 0.0500.025 = 100 s
[N2O5] 0.0250.0125 = 100 s
d)
The half life does not depend on concentration for a first order
reaction
e)
t1/2 = 100 s from the graph. Is there another way to calculate it?
f)
The new form of the integrated rate law helps:
 [A]0 
 [A]0 
  kt1/2
  kt  ln 
ln 
 [ A] 
 [A]0 /2 
 ln2  kt1/2  0.693  kt1/2
 t1/2 
g)
0.693
k
Example: A first order reaction has t1/2 = 20 min. k? How much time
until reaction reaches 75%completion?
C.
The Integrated Second Order Rate Law
1. For aA
product that is second order in [A]:
[A]
rate  
 k[A]2
t
1
1
 kt 
[A]
[A]0
2.
The integrated form is:
3.
This is also in the form of a straight line. A plot of 1/[A] vs t gives a
straight line with slope = k, and intercept = 1/[A]0
4)
Half-life of Second Order Reactions
[A]0
1
1
 kt 
[A] 
[A]
[A]0
2
1
1
2
1

 kt1/2 


 kt1/2
[A]0 /2
[A]0
[A]0 [A]0
1
1

 kt1/2  t1/2 
[A]0
k[A]0
a)
b)
5)
Now, the half-life depends on the concentration
Every half life is twice as long as the previous one
Example: 2 C4H6
[C4H6]
ln [C4H6]
1/ [C4H6]
time
0.01000
-4.605
100
0
0.00625
-5.075
160
1000
0.00476
-5.348
210
1800
0.00370
-5.599
270
2800
0.00313
-5.767
320
3600
0.00270
-5.915
370
4400
0.00241
-6.028
415
5200
0.00208
-6.175
481
6200
C8H12 Order? k? t1/2?
D.
The Integrated Zero-Order Rate Law
1. Some reactions don’t depend on concentration at all
a) These are often reactions that occur on a surface
b) Once the surface is full, more reactant doesn’t help
c) Example: 2 N2O
2 N2 + O2
rate  
2.
The rate law:
3.
Integrated Zero Order Rate Law: [A] = -kt + [A]0
4.
A plot of [A] vs. t gives a straight line with slope = -k and intercept = [A]0
5.
Half-life of a zero order reaction:
[A]  - kt  [A]0
[A]0
 - kt1/2  [A]0
2
[A]0
[A]0
 kt1/2 
 t1/2 
2
2k

[A]
 k[A]0  k
t
E.
Pseudo-First-Order Rate Laws
1) So far, we have only had one reactant aA
products
2) How do we develop the rate law and the integrated rate law if we have
more than one reactant?
3) Example: BrO3- + 5 Br- + 6 H+
3 Br2 + 3 H2O

a)
Δ[BrO 3 ]

 k[BrO 3 ][Br  ][H  ]2
Rate Law: rate  
Δt
b)
c)
Integration would be difficult for this complicated rate law
We do the reaction with Pseudo-First-Order Conditions:
i. [Br-] = very large ~ constant = [Br-]0
ii. [H+] = very large ~ constant = [H+]0
iii. [BrO3-] = very small
d)
Rate law can be expressed more simply:


rate  k[BrO 3 ][Br  ]0 [H  ]0  k'[BrO 3 ]
k'  k[Br  ]0 [H  ]0
2
2

rate  k'[BrO 3 ]
e)
Rate law is now just like the first order case:
f)
Integrated Pseudo-First-Order rate law:
g)
This can be graphed the same way as first order law to get k’
ln[A]  -k' t  ln[A] 0
ln[BrO3-]
h)
Then we can find k from k’
k'  k[Br  ]0 [H  ]0
k
i)
2
k'
2
[Br  ]0 [H  ]0
Once we have k we can calculate anything else we need to know