Post on 05-Jan-2016
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Chapter 13 Integrated Rate Laws
I. Integrated Rate LawsA. Preview: So far, we have looked at rate as a function of concentration
Next, we look at concentration as a function of time
aA products
B. The Integrated First Order Rate Law
1. Integration is a calculus operation
2. Integration of the first order rate law:
3. Gives the Integrated first-order rate law:
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).
nk[A]t
]A[- rate
k[A]t
]A[-
0ln[A]-ktln[A]
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
b) A plot of ln[A] vs t gives a straight line with
slope = -k and intercept = ln[A]0
c) If plot is a straight line, reaction is1st order
d) If not a straight line, not 1st order
5) Example: 2 N2O5 4 NO2 + O2 Find k
0ln[A]-ktln[A]
bmxy
[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
6) Half-life of a first order reaction
a) The integrated first order rate law can be rewritten as:
b) The half-life of a reaction = time it takes for [A] = [A]0/2
kt]A[
[A]ln 0
c) [N2O5] 0.1000.050 = 100 s
[N2O5] 0.0500.025 = 100 s
[N2O5] 0.0250.0125 = 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:
g) Example: A first order reaction has t1/2 = 20 min. k? How much time until reaction reaches 75%completion?
k
0.693 t
kt693.0kt ln2
kt/2[A]
[A]lnkt
]A[
[A]ln
1/2
1/21/2
1/20
00
C. The Integrated Second Order Rate Law
1. For aA product that is second order in [A]:
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
2k[A]t
]A[rate
0[A]
1kt
[A]
1
4) Half-life of Second Order Reactions
a) Now, the half-life depends on the concentration
b) Every half life is twice as long as the previous one
5) Example: 2 C4H6 C8H12 Order? k? t1/2?
01/21/2
0
1/2000
1/20
0
0
k[A]
1tkt
]A[
1
kt]A[
1
]A[
2
[A]
1kt
/2[A]
1
2
[A][A]
[A]
1kt
[A]
1
[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
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
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:
kk[A]t
[A]rate 0
2k
[A]t
2
[A]kt
[A] kt- 2
[A]
[A] kt - [A]
01/2
01/2
01/20
0
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) Rate Law:
b) Integration would be difficult for this complicated rate law
c) 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:
23
3 ]][H][Brk[BrOΔt
]Δ[BrOrate
200
32
003
][H]k[Brk'
][BrOk'][H]][Brk[BrOrate
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’
h) Then we can find k from k’
i) Once we have k we can calculate anything else we need to know
][BrOk'rate 3
0ln[A]t-k'ln[A]
ln[BrO3-]
200
200
][H][Br
k'k
][H]k[Brk'