© university of south carolina board of trustees trial[no] m [h 2 ] m initial rate, m/s...
TRANSCRIPT
© University of South Carolina Board of Trustees
Trial [NO]M
[H2]M
InitialRate, M/s
1 0.057 0.130 4.50x10-3
2 0.057 0.260 9.00x10-3
3 0.11 0.130 1.80x10-2
Student Problem
Write the rate law for the reaction given the following data
2NO + 2H2 N2 + 2H2O
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Chapt. 13
Sec. 13.3Properties of
Common Rate Laws
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Common Rate Laws
A products
a) First-orderRate = k [A]
b) Second-orderRate = k [A]2
c) Zero-orderRate = k [A]0 = k
© University of South Carolina Board of Trustees
Common Rate Laws
A products
a) First-orderRate = k [A]
b) Second-orderRate = k [A]2
c) Zero-orderRate = k [A]0 = k
© University of South Carolina Board of Trustees
1st-Order Rate Law
Differential Form
Rate = k [A](t)
© University of South Carolina Board of Trustees
1st-Order Rate Law
Differential Form
Rate = k [A](t)
Integral Forms
[A](t) = [A]0 e-kt
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[A] = [A]0 exp(-k t)
Time0 2 4 6 8 10
[A]
0
1
2
3
4
5
6Time [A]
0.00 5.001.00 3.352.00 2.253.00 1.514.00 1.015.00 0.686.00 0.457.00 0.308.00 0.209.00 0.14
10.00 0.09
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Concentration at a Later Time
C12H22O11 + H2O ® C6H12O6 + C6H12O6 sucrose glucose fructose
This reaction is 1st order with a rate constant of 6.2 x10-5 s-1. If the initial sucrose concentration is 0.40 M, what is the concentration after 2 hrs (7200 s)?
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Time to Reach a Concentration
C12H22O11 + H2O ® C6H12O6 + C6H12O6 sucrose glucose fructose
This reaction is 1st order with a rate constant of 6.2 x10-5 s-1. If the initial sucrose concentration is 0.40 M, at what time does the concentration fall to 0.30 M?
© University of South Carolina Board of Trustees
1st-Order Rate Law
Differential Form
Rate = k [A](t)
Integral Forms
[A](t) = [A]0 e-kt
ln [A](t) = ln [A]0 - k t y = b +mx
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Graphing a 1st -Order Reaction
Time0 2 4 6 8 10
[A]
0
1
2
3
4
5
6Time [A]
0.00 5.001.00 3.352.00 2.253.00 1.514.00 1.015.00 0.686.00 0.457.00 0.308.00 0.209.00 0.14
10.00 0.09
© University of South Carolina Board of Trustees
Graphing a 1st -Order Reaction
Time0 2 4 6 8 10
ln [A
]
-3
-2
-1
0
1
2
Time0 2 4 6 8 10
[A]
0
1
2
3
4
5
6Time [A] ln [A]
0.00 5.00 1.611.00 3.35 1.212.00 2.25 0.813.00 1.51 0.414.00 1.01 0.015.00 0.68 -0.396.00 0.45 -0.797.00 0.30 -1.198.00 0.20 -1.599.00 0.14 -1.99
10.00 0.09 -2.39
© University of South Carolina Board of Trustees
Graphing a 1st -Order Reaction
Time0 2 4 6 8 10
ln [A
]
-3
-2
-1
0
1
2
Time0 2 4 6 8 10
[A]
0
1
2
3
4
5
6Time [A] ln [A]
0.00 5.00 1.611.00 3.35 1.212.00 2.25 0.813.00 1.51 0.414.00 1.01 0.015.00 0.68 -0.396.00 0.45 -0.797.00 0.30 -1.198.00 0.20 -1.599.00 0.14 -1.99
10.00 0.09 -2.39
straight line this is a first-order reaction.
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[A] vs t Data Rate Law
Method of Initial Rates (Sec. 13.2)
Trial and Error with Common Laws (Sec. 13.3)
© University of South Carolina Board of Trustees
ln [A] = ln [A]0 - k t
Time0 2 4 6 8 10
ln [A
]
-3
-2
-1
0
1
2[A]0
Time0 2 4 6 8 10
[A]
0
1
2
3
4
5
6Time [A] ln [A]
0.00 5.00 1.611.00 3.35 1.212.00 2.25 0.813.00 1.51 0.414.00 1.01 0.015.00 0.68 -0.396.00 0.45 -0.797.00 0.30 -1.198.00 0.20 -1.599.00 0.14 -1.99
10.00 0.09 -2.39
intercept = ln [A]0
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ln [A] = ln [A]0 - k t
Time0 2 4 6 8 10
ln [A
]
-3
-2
-1
0
1
2
x
y
[A]0
Time0 2 4 6 8 10
[A]
0
1
2
3
4
5
6Time [A] ln [A]
0.00 5.00 1.611.00 3.35 1.212.00 2.25 0.813.00 1.51 0.414.00 1.01 0.015.00 0.68 -0.396.00 0.45 -0.797.00 0.30 -1.198.00 0.20 -1.599.00 0.14 -1.99
10.00 0.09 -2.39
( 1.2) 0.8
7.0 2.0
0.4 k
-1
yslope
x
s s
s
© University of South Carolina Board of Trustees
1st-Order Rate Law
Differential Form
Rate = k [A](t)
Integral Forms
[A](t) = [A]0 e-kt
ln [A](t) = ln [A]0 - k t
Half-life
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1st -Order Half-Life
Time0 2 4 6 8 10
[A]
0
1
2
3
4
5
6 Time [A] ln [A]
0.00 5.00 1.611.00 3.35 1.212.00 2.25 0.813.00 1.51 0.414.00 1.01 0.015.00 0.68 -0.396.00 0.45 -0.797.00 0.30 -1.198.00 0.20 -1.599.00 0.14 -1.99
10.00 0.09 -2.39
[A] [A]/2 t1/2
© University of South Carolina Board of Trustees
1st -Order Half-Life
Time0 2 4 6 8 10
[A]
0
1
2
3
4
5
6 Time [A] ln [A]
0.00 5.00 1.611.00 3.35 1.212.00 2.25 0.813.00 1.51 0.414.00 1.01 0.015.00 0.68 -0.396.00 0.45 -0.797.00 0.30 -1.198.00 0.20 -1.599.00 0.14 -1.99
10.00 0.09 -2.39
[A] [A]/2 t1/2
5 2.5 1.75 s
t1/2
© University of South Carolina Board of Trustees
1st -Order Half-Life
Time0 2 4 6 8 10
[A]
0
1
2
3
4
5
6 Time [A] ln [A]
0.00 5.00 1.611.00 3.35 1.212.00 2.25 0.813.00 1.51 0.414.00 1.01 0.015.00 0.68 -0.396.00 0.45 -0.797.00 0.30 -1.198.00 0.20 -1.599.00 0.14 -1.99
10.00 0.09 -2.39
[A] [A]/2 t1/2
5 2.5 1.75 s
3 1.5 1.72 s
t1/2
© University of South Carolina Board of Trustees
Half-life
Time0 2 4 6 8 10
[A]
0
1
2
3
4
5
6 Time [A] ln [A]
0.00 5.00 1.611.00 3.35 1.212.00 2.25 0.813.00 1.51 0.414.00 1.01 0.015.00 0.68 -0.396.00 0.45 -0.797.00 0.30 -1.198.00 0.20 -1.599.00 0.14 -1.99
10.00 0.09 -2.39
[A] [A]/2 t1/2
5 2.5 1.75 s
3 1.5 1.72 s
1 0.5 1.73 s
t1/2
© University of South Carolina Board of Trustees
1st-Order Rate Law
Differential Form
Rate = k [A](t)
Integral Forms
[A](t) = [A]0 e-kt
ln [A](t) = ln [A]0 - k t
Half-life
always the same