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

77

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

17

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.

20

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

21

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

22

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?

23

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