Transcript Zero, First & Second Rate orders - Caldwell

Zero, First & Second
Rate orders
Chapter 14 Part III
Integrated Rate Law
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For any reaction:
Every reaction has a
rate law of this form.
We will develop the
integrated rate law for
n=1: first order
n=2: second order
n=0 : zero order
aA  Products
Rate= -∆[A]/∆t= k[A]m
First Order Rate Laws
2N2O5(soln) -> 4NO2(soln) + O2(g)
 Rate =(- ∆[N2O5]/∆t) = k[N2O5]
 As it is first order, this means that if
the [N2O5] doubles the production of
also NO2 & O2 doubles.
 By integrating the formula (calculus)
 ln [N2O5] = -kt + [N2O5]t=0
 This is still the integrated rate law
with [Reactant] as a function of time
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1st order integrated rate law:
Rate as function of time
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For any reaction aA-> products
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Rate =(- ∆[A]/∆t) = k[A]
 ln
[A] = -kt + [A]t=0
 The equation depends on time. If
[A] initial is known and k is known,
the [A] may be calculated at t.
ln [A] = -kt + [A]t=0
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The equation is in the form y = mx +b.
A plot of x (time) versus y (ln[A]) will yield a
straight line.
Therefore another way to determine rate
order, if the plot of ln[A] versus time gives a
straight line, then it is first order. If it is not
a straight line, it is not first order.
This rate law may also be expressed:
ln ([A]t=0/[A])=kt
2N2O5(g) -> 4NO2(g) + O2(g)
[N2O5]Mtime
0.10000
0
0.07070 50
0.05000 100
0.25000 200
0.01250 300
0.00625 400
The decomposition
(s)of this gas was
studied at a
constant volume to
collect these
results.
 Verify the rate
order and find the
rate constant for
rate =(
∆[N2O5]/∆t)
First Order
time (s)
0
50
100
200
300
400
First order
450
400
350
300
ln[N2O5]
ln[N2O5
-2.303
-2.649
-2.996
-3.689
-4.382
-5.075
-5
250
200
150
100
50
0
-4.5
-4
-3.5
-3
time (s)
-2.5
-2
-1.5
-1
Calculate k
Slope = ∆y/∆x
 = ∆(ln[N2O5])/∆t
 Using first and last points
 Slope = -5.075-(-2.303)/(400s-0s)
 = -2.772/400s
 k = -6.93 x 10-3 s-1
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Half life of a first order
reaction
 The time required to
[N2O5]Mtime (s)
reach one half the
0.10000
0
original concentration.
0.07070 50  Designated by t 1/2
 We can see the half
0.05000 100
life from the data is
100 sec.
0.25000 200
0.01250 300  Note is always takes
100 seconds.
0.00625 400
Second Order Rate
Laws
Butadiene forms its dimer
2C4H6 (g)  C8H12 (g)
aA-> products
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If the reaction is first order:
ln([A]0/[A])=kt
By definition, when t= t 1/2 then,
[A] = [A]0/2
ln([A]0/[A]0/2) = kt 1/2
ln(2)=kt 1/2
t 1/2= 0.693/k
Second Order Rate Laws
 Butadiene
forms its dimer
 2C4H6 (g) - > C8H12 (g)
2nd Order Data
[C4H6] (mol/L)
Time (sec)
0.01000
0
0.00625
1000
0.00476
1800
0.00370
2800
0.00313
3600
0.00270
4400
0.00241
5200
0.00208
6200
2nd Order Reaction
0.01200
Concentration (mol/L)
0.01000
0.00800
0.00600
0.00400
0.00200
0.00000
0
1000
2000
3000
4000
Time (sec)
5000
6000
7000
Time (sec) [C4H6] (mol/L)
ln[C4H6]
1/[C4H6]
0
0.01000
-4.60517
100.0
1000
0.00625
-5.07517
160.0
1800
0.00476
-5.34751
210.1
2800
0.00370
-5.59942
270.3
3600
0.00313
-5.76672
319.5
4400
0.00270
-5.9145
370.4
5200
0.00241
-6.02813
414.9
6200
0.00208
-6.17539
480.8
2nd order reaction using
integrated law for first order
2nd order reaction; First Order rate law
-3
0
1000
2000
3000
4000
-3.5
ln[C4H6]
-4
-4.5
-5
-5.5
-6
-6.5
Time (sec)
5000
6000
7000
2nd order reaction and 2nd order
integrated rate law
2nd order using 2nd order rate law
600
500
1/C4H6
400
300
200
100
0
0
1000
2000
3000
4000
Time(sec)
5000
6000
7000