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Chem 116 Lecture Note Outlines
Set 06: Integrated Rate Laws
Integrated Rate Laws
♦ An integrated rate law includes time as a variable
Differential Form of rate law
First-order rate equation:
− ∆[A ]
− d[A ]
= k[A ]
rate =
= k[A ] as ∆t goes to zero, rate =
∆t
dt
A
ln  t
 A0
Second-order rate equation
∆[A ]
− d[A ]
2
2
as ∆t goes to zero, rate =
= k[A ]
rate = k[A ] = −
∆t
dt
1
1
−
= kt
At A0
Zero-order rate equation
∆[A ]
− d[A ]
as ∆t goes to zero, rate =
rate = k = −
= −k
∆t
dt
Integrated form

 = − kt

A t − A 0 = − kt
Graphical determination of the reaction order for the decomposition of A; A → B + C
♦
♦
♦
1.
2.
3.
♦
♦
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Page 1 of 9
Chem 116 Lecture Note Outlines
Set 06: Integrated Rate Laws
d[A ]
dt
plot of ln [A] vs time yields straight line
First-order reaction, rate = k[A ] = −
Second-order reaction
plot of 1A vs time yields straight line
[ ]
Zero-order reaction
plot of [A] vs. time gives a straight line
Spring 2015 (Ratcliff)
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Page 2 of 9
Chem 116 Lecture Note Outlines
Set 06: Integrated Rate Laws
Graphical determination of the reaction order for the decomposition of A; A → B + C
Time (min)
[A]
ln [A]
1
[ A]
0
10
20
30
40
50
60
0.0165
0.0124
0.0093
0.0071
0.0053
0.0039
0.0029
−4.10
−4.39
−4.68
−4.95
−5.24
−5.55
−5.84
60.6
80.6
1.1×102
1.4×102
1.9×102
2.6×102
3.4×102
♦
Concept Check
The following graph shows the kinetics curves for the reaction of nitrogen with hydrogen to form ammonia.
Which curve is hydrogen?
N2(g) + 3 H2(g) → 2 NH3(g).
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Page 3 of 9
Chem 116 Lecture Note Outlines
Set 06: Integrated Rate Laws
At 1000oC, cyclobutane (C4H8) decomposes to two molecules of ethylene (C2H4).
The rate constant of the reaction is 87 s−1.
C4H8 → 2 C2H4
If the initial C4H8 concentration is 2.00 M, what is the concentration after 0.010 s?
0.83 mole/L
Reaction Half-life
♦
For a first-order reaction, the half life does not depend on the starting concentration
♦
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Page 4 of 9
Chem 116 Lecture Note Outlines
Set 06: Integrated Rate Laws
Cyclopropane is the smallest cyclic hydrocarbon. Because its 60° bond angles allow poor orbital
overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 1000°C
via a first-order reaction. The rate constant is 9.2 s−1
What is the half-life of the reaction?
0.075 s
How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value?
0.15s
For a second-order reaction, half life is inversely proportional to the initial concentration
t1 =
2
1
k [A ]0
For a zero-order reaction, half life is directly proportional to the initial concentration
t1 =
2
[A ]0
2k
An Overview of Zero-Order, First-Order, and Simple Second-Order Reactions
Rate Law
Zero order
0
rate = k [ A] = k
Integrated rate law
[A] t − [A] 0 = −kt
Units for k
mole
M
or
L ⋅s
s
[A]0
t1 =
2
2k
[A]t = −kt + [A]0
ln[ A]t = − kt + ln[ A]0
[A]t vs t
ln[A]t vs t
Half life
Integrated rate law in
straight line form
Plot for straight line
First Order
rate = k [ A]
ln
[A]0
[A]t
1
s
= kt
or s -1 s
t1 =
2
ln 2 0.693
=
k
k
slope
-k
-k
y-intercept
[A]0
ln[A]0
Spring 2015 (Ratcliff)
Second order
2
rate = k [A]
1
1
−
= kt
[A] t [A]0
L
1
or
mole ⋅ s
M ⋅s
1
t1 =
2
k [A ]0
1
1
= kt +
[A]t
[A]0
1
vs t
[A]0
k
1
[A]0
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Page 5 of 9
Chem 116 Lecture Note Outlines
Set 06: Integrated Rate Laws
Concept Check
Which of the following plots indicates that the reaction is zero order?
a.
b.
c.
d.
The decomposition of hydrogen peroxide in dilute sodium hydroxide solution is described by the
equation:
2H2O2(aq) → 2H2O(l) + O2(g)
The reaction is first order in H2O2, the specific rate constant for the consumption of H2O2 at 20 °C is
1.8×10−5s−1 and the initial concentration is 0.30 M.
What is the half-life (in hours) of the reaction at 20 °C?
11 hrs
What is the molarity of H2O2 after four half-lives?
0.019 M
How many hours will it take for the concentration to drop to 25% of its original value?
21 hrs
What will be the concentration of H2O2 after 555 minutes?
0.16 M
What is the initial rate of the reaction?
5.4×10−6 M/s
Spring 2015 (Ratcliff)
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Page 6 of 9
Chem 116 Lecture Note Outlines
Set 06: Integrated Rate Laws
At elevated temperatures, nitrogen dioxide decomposes to nitric oxide and molecular oxygen:
2NO2(g) → 2NO(g) + O2(g)
Concentration-time data for the consumption of NO2 at 300 °C are as follows
Time (s)
0
50
100
150
200
300
400
500
[NO2]
0.00800 0.00658 0.00559 0.00485 0.00429 0.00348 0.00293 0.00253
From the prepared plots below determine the order of the reaction?
second order in NO2, rate = k[NO2]2
How can you determine the rate constant?
0.54 M−1·s−1
What is the concentration of NO2 at t = 20.0 min?
1.3×10−3 M
What is the half-life of the reaction when [NO2]0 = 6.00×10−3 M?
3.1×102 s
−3
What is the half life when [NO2]0 = 3.00×10 M?
6.2×102 s
How much time will it take for 25 % of NO2 to react when [NO2]0 = 8.00×10−3 M?
77 s
Spring 2015 (Ratcliff)
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Page 7 of 9
Chem 116 Lecture Note Outlines
Set 06: Integrated Rate Laws
A reaction is first order in the concentration of reactant i.e. has the rate law Rate = k[A]. This reaction
has a half-life of 102 seconds
What is the numerical value of k?
k = 6.80 × 10–3 s–1
How much time (in minutes) will it take for 84.0% of the reactant to react?
4.49 mins
If 0.50 M of reactant is present initially, how much reactant would remain after 5 minutes?
0.065 M
After three half-lives what percent of reactant will remain?
13 %
The specific rate constant for a certain first order reaction is 0.0447 hr−1. What is the half-life of the
reaction?
15.5 hours
The active ingredient in an over the counter pain killer analgesic decomposes with a rate constant,
k= 9.05×10−4 day−1. How many days does it take for 15 % of the ingredient to decompose?
180. days
Spring 2015 (Ratcliff)
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Page 8 of 9
Chem 116 Lecture Note Outlines
Set 06: Integrated Rate Laws
A certain reaction is known to be second order in A with k = 4.36×10−2 M−1hr−1. What is the numerical
value of the half-life when the initial concentration of A is 0.250 M?
91.7 hours
A certain reaction is known to be second order in A with k = 0.014 M−1s−1. What will be the
concentration of A after 3.0 hours if the initial concentration is 0.025 M
0.0052 mole/ L
The decomposition of SOCl2 is first-order in SOCl2. If the half-life for the reaction is 4.1 hr, how long
would it take for the concentration of SOCl2 to drop from 0.36 M to 0.045 M?
12 hours
Spring 2015 (Ratcliff)
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Page 9 of 9