111
Kinetics Part III:Integrated Rate Laws
Dr. C. Yau
Spring 2013
Jespersen Chap. 14 Sec 4
222
Integrated Rate Laws
How are “integrated rate laws” different from “rate laws?”
A rate law gives the speed of reaction at a given point in time, and how it is affected by the concentration of the reactants.
An integrated rate law gives the change in concentration from time zero up to a given point in time.
33
Integrated Rate Law for First-Order Reactions
We already know the rate law for a 1st order reaction is
Rate = k[A]
Using calculus, the integrated rate law is
Where [A]o = molarity of A at time zero
[A]t = molarity of A at time t
and k = rate constant for the reaction
o
t
[A]ln = kt
[A]
44
Integrated Rate Law of First-Order Reactions
o
t
kto
t
-ktt
o
-ktt o
[A]anti ln ln = anti ln (kt)
[A]
[A] = e
[A]
Finding the reciprocal of both sides of the equation gives us
[A] = e
[A]
and finally [A] = [A] e
It is often more useful to rewrite the equation by finding the anti natural log of both sides of the equation:
o
t
[A]ln = kt
[A]
5
Quick Reminder of Treatment of Sig. Fig. in Log
Log 3.4 x 10-1 = -0.468521083
= -0.47
2 sig.fig. 2 decimal places
antilog -3.213 = 10-3.213= 6.123503x10-4
= 6.12x10-4
3 decimal places 3 sig. fig.
6
Treatment of Sig. Fig. in Ln
Ln 3.4 x 10-1 = -1.078809
= - 1.08
antiln -4.273 = e-4.273 = 1.393990x10-2
= 1.39x10-2
2 sig. fig.
3 decimal places
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Use of the Integrated Rate of 1st Order Rxns
Example 14.7 p. 655Dinitrogen pentoxide is not very stable. In the gas phase or dissolved in a nonaqueous solvent, like CCl4, it decomposes by a 1st order rxn into dinitrogen tetroxide and molecular oxygen.
2 N2O5 2 N2O4 + O2
The rate law is Rate = k[N2O5]
At 45oC, the rate constant for the rxn is 6.22 x 10-4s-1. If the initial concentration of N2O5 at 45oC is 0.500 M, what will its concentration be after exactly one hour?
Ans. 0.053 M
Do Practice Exercises 15 & 16 p.656
88
The rate constant can be determined graphically. Again we manipulate the rate law to allow us to obtain a linear graph: Variables are [A]t and t
Determination of the Rate Constantfor the Integrated Rate Law
0
t
0 t
t 0
t 0
[A]ln = kt
[A]
ln[A] - ln [A] = kt
- ln [A] = kt ln[A]
ln [A] = kt + ln [A]
How does this allow us to determine the rate constant, k, graphically?First determine which are variable & which are constants.
9
Determination of the Rate Constantfor the Integrated Rate Law
t oln [A]= +
=
l
- k
m
n [A
+
] t
xy b
What is the slope equal to?slope = - kSo, k = - slopeNote: (-) does not mean k is negative but it has the opposite sign of the slope.
Slope is calculated to be - 6.0x10-4 s-1.So, k = 6.0x10-4 s-1
Rate = k[N2O5] Units are correct!
10
Half-life (t½ ) of a Reactant"Half-life" of a reactant is the time it takes for ½
of the reactant to disappear.It is NOT half the time it takes for all of the
reactant to disappear.
For example, t½ of the radioisotope I-131 is 8.0 days. Starting with 20 g of I-131, after 8.0 days, there would be 10 g left.
After a total of 16.0 days (two half-lives), there would be……
5 g left
11
• Half-life is a measure of the speed of reaction: The shorter the half-life, the faster is the reaction.
• By definition, after one half-life,[A]t = ½ [A]o
For a 1st order rxn,
Note that half-life isnot affected byconcentration.
How do you know that?There is no concentrationterm in the equation.
o1/2
o
1/2
1/2
1
2
[A]ln = k t
[A]
ln 2 = k t
ln 2 t =
k
12
Application of Half-lifeFig.14.7 p.658
I-131 has a half-life of 8 days.
After 4 half-lives (32 days), it is down to 1/16 of its original concentration.
If we start with 20.0 g of I-131, after 40 days, how much of I-131 is still there?
(Half-life of I-131 = 8 days.)
What percent of I-131 remains after 24 days?
What fraction of I-131 remains after 48 days?
13
14
Example 14.8 A patient is given a certain amt of I-131 as part of a diagnostic procedure for a thyroid disorder. What fraction of the initial I-131 would be present in a patient after 25 days if none of it were eliminated through natural body processes?
t1/2 = 8.02 days What to do if # days is not an exact multiple of the half-life?
We solve this problem using the integrated rate law for 1st order rxn?
o
t
[A]ln = kt
[A]
15Do Practice Exercises 15, 16, 17 p. 659
Half-life = 8.02 days, fraction after 25 days?
After 25 days, 1/9 of the original initial I-131 would be present.
16
The half-life of I-132 is 2.295h. What percentage remains after 24 hours? (Previous problem was for I-131.)
1/2
1/2
-1
ln 2t =
kln 2 0.693
k = = = 0.302 ht 2.295 h
-1o
t
7.25 3o
t
3
t
[A]ln = kt = 0.302 h 24 h 7.25
[A]
[A] = e = 1.4x10
[A]
100 = 1.4x10
[A]
t3
100 and = [A]
1.4 10x
Ans. 0.071 % I-132 remains.
% is based on 100
3
100 = 0.071
1.4 10x
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C-14 dating: Determination of the age of organic substances.
• When object is still living, it is ingesting C with a constant ratio of
C-14/C-12 = 1.2x10-12 (given) = ro
• Once the object dies, the amount of C becomes constant and no more C-14 is incorporated. As a result the C-14/C-12 ratio begins to decrease due to the decay of C-14 (to give rt = ratio at time t).
• By measuring the C-14/C-12 ratio (rt) of the object, we can estimate how long it has been dead.
C-14 has a half-life of 5730 years and
radioactive decay is a first-order process.
This means…
ando
t
[A]ln = kt
[A]1/2
ln 2 k =
tWhat is the rate constant, knowing that the half-life is 5730 yrs?
o
t
-12
o
t
1.2x1
where r is C-14/C-12 ratio at time of death =
and r is C-14/C-12 ratio at time of analysis ("now")
0
(measured)
rln = ktr
REMEMBER THIS!
= 1.21x10-4 yr -1
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Example 14.9 p.660
A sample of an ancient wooden object was found to have ratio of 14C to 12C equal to 3.3x10-13 as compared to a contemporary biological sample, which has a ratio of 1.2x10-12. What is the age of the object?
Ans. 1.07x104 years
Do Pract Exer 20, 21 p.661
o
t
rln = ktr
Calculate k from t1/2 for C-14 = 1.21x10-4 yr-1
Find t.
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Integrated Rate Law of 2nd Order Reactions
• They are of several types:
Rate=k[A]2,
Rate=k[A]1[B]1 and
Rate=k[A]2[B]0, etc…• 2nd order means the powers must add up to 2.• The integrated equation is of the form
t 0
1 1t
[A] [A]k
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Example 14.10 p.661
Nitrosyl chloride, NOCl, decomposes slowly to NO and Cl2.
2NOCl 2NO + Cl2
The rate law shows that the rate is second order in NOCl. Rate = k[NOCl]2
The rate constant k equals 0.020 L mol-1 s-1 at a certain temp. If the initial conc of NOCl in a closed reaction vessel is 0.050 M, what will the concentration be after 35 minutes?Ans. 0.016 M Do Pract Exer 22 & 23 p.662
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Integrated Rate Law of 2nd Order Reactions
How can we determine rate constant graphically? Which are the variables?
t 0
1
1
[A]t
[A]k
Rearrange
y = mx + b
t 0
1 1t
[A] [A]k
What do we plot on the x-axis? y-axis? What does the slope represent?
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Half-life of 2nd Order Reactions
Note that unlike 1st order reactions, half-life of 2nd order rxns depend on the initial conc of the reactant.
Ex 14.11 p.664 2HI(g) H2(g) + I2 (g)
has the rate law, Rate = k[HI]2 with k = 0.079 L mol-1s-1 at 508oC. What is the half-life of this rxn at this temp when the initial HI concentration is 0.10 M?
1/ 2o
1
k [A]t
Ans. 1.3x102s
Do Prac Ex 24 & 25 p.664
t 0
1 1t
[A] [A]k Derived from
Summary of Integrated Rate Laws
o2/1 [A]k
1 t
1 1t
[A] [A]ok
1 1t
[A] [A]t o y mx b
k
o
t
[A]ln = kt
[A]
t oln [A] = - kt + ln [A]
y = mx + b
1/2
ln 2 t =
k
1st Order Rxn 2nd Order Rxn