Chemical Kinetics

CHEM 102 T. Hughbanks

Integrated Rate Laws

 From initial concentrations & rate law, we can predict all concentrations at any time t.

 Mathematically, this is an initial value problem involving a (usually) simple differential equation.

Simplest Case: First Order

A → 2P

By integrating, we can get an equation relating concentration and time:

rate = −

d[A]dt

= k [A]

d[A ′ ] [A ′ ] [A]0

[A]t∫ = −k d ′ t 0

t

∫ln [A]t

[A]0 = − k t

First Order Reactions

 From this, see that a plot of ln[A] vs. t will be a line with a slope of -k .

ln[A]t[A]0

= −k t

[A]t[A]0

= e−k t so [A]t = [A]0 e−k t

Half-Lives of radioisotopes

12.3 y 7.1 × 108 y

5.73 × 103 y 4.5 × 109 y

2.4 s 30.17

1.26 × 109 y 8.05 d

28.1 y 1.60 × 103 y

5.26 y

13H

614C

615C

1940 K

92235U

92238 U

55137Cs

53131I

88226 Ra 38

90Sr

2760Co

Hydrogen peroxide decomposes into water and oxygen in a first-order process.

H2O2(aq) → H2O(l) + ½ O2(g)

At 20.0 °C, the ½-life for the reaction is 3.92 × 104 seconds. If the initial concentration of hydrogen peroxide is 0.52 M, what is the concentration after 7.00 days (6.048 × 105 s)?

Example

Second Order, one reactant

  If second order kinetics apply, a plot of 1/[A] vs. t will be a line with slope k .

rate = −

d[A]dt

= k [A]2

d[A][A]2

= − k dt ⇒ d[A ′][A ′] 2[A]0

[A]t∫ = −k d ′t0

t∫

This leads to:

1

[A]t

−1

[A]0= k t or

1[A]t

= k t + 1

[A]0

Second Order, one reactant

rate = k [A]2

d[A][A]2

= − k dt

This leads to:

1

[A]t

−1

[A]0= k t or

1[A]t

= k t + 1

[A]0

  If second order kinetics apply, a plot of 1/[A] vs. t will be a line with slope k .

1st vs. 2nd Order Kinetics

Both cases shown with k = 0.693 (Which is "not terribly meaningful since units differ!) "

00.10.20.30.40.50.60.70.80.9

1

0 1 2 3 4 5 6 7 8 9 10Time (s)

[A] r

emai

ning

1st order2nd order

1st Order Test Plot: ln[A] vs. t

-7

-6

-5

-4

-3

-2

-1

01 2 3 4 5 6 7 8 9 10

time (s)

ln [A

]

1st order2nd order

2nd Order Test Plot: 1/[A] vs. t

0

200

400

600

800

1000

1200

1 2 3 4 5 6 7 8 9 10time (s)

1/[A

]

1st order

2nd Order Test Plot: 1/[A] vs. t

0

1

2

3

4

5

6

7

8

1 2 3 4 5 6 7 8 9 10time (s)

1/[A

]

2nd order

A Real Example ... 2 N2O5(g) → 4 NO2(g) + O2(g)

 Experiment at T = 338 K gives: time (s) [N2O5] (M)

0 0.100100. 0.0620300. 0.0221600. 0.0053900. 0.0013

 Find rate law?

Zero Order? y = -0.0001x + 0.0765

R2 = 0.7868

-0.02

0

0.02

0.04

0.06

0.08

0.1

0 100 200 300 400 500 600 700 800 900

time

[N2O

5]

[N2O5] vs time

First Order?

y = -0.0048x - 2.3187R2 = 0.9997

- 7

- 6

- 5

- 4

- 3

- 2

- 1

00 100 200 300 400 500 600 700 800 900

time

ln[N

2O5]

Second Order?

y = 0.7839x - 92.022R2 = 0.8059

10

110

210

310

410

510

610

710

810

0 100 200 300 400 500 600 700 800 900

time

1/[N

2O5]

Example...

 Graphs show us that the reaction is first order, so:

rate = k [N2O5]  We can also find k from slope of graph:

ln [N2O5] = ln [N2O5]0 – k t  So slope is equal to –k . Fit gives us:

k = 0.0048 s–1

Example …. (half life)  At what time will the [N2O5] be equal to

one-half of its original value?  Use integrated rate law, solved for t:

t = 1

k ln

[N2O5]0

[N2O5] The question asks, what is t when

[N2O5] = (1/2)[N2O5]0?

Elements that decay via radioactive processes do so according to 1st order kinetics:

Element: Half-life:

92238 U → 92

238Th + 24α 4.5×109 years

614C→ 7

14 N + −10β 5730 years

53131I → 54

131Xe + −10β 8.05 days

notes: −10β is an electron, 2

4α is a 24 He nucleus.

Tritium decays to helium by beta (β) decay:

The half-life of this process is 12.3 years

Starting with 1.50 mg of 3H, what quantity remains after 49.2 years?

Recall that that the rate constant for a 1st order process is given by:

   k = ln(2)

t1/2

= 0.693t1/2

[A]t[A]0

= e−k t [A]t = [A]0 × e−k t

[3H]t = 1.50 mg ×

= 0.094 mg

Notice that 49.2 years is 4 half-lives…

After 1 half life: = 0.75 mg remains

After 2 half life's: 1.50 mg

4= 0.38 mg remains

After 3 half life's: 1.50 mg

8= 0.19 mg remains

After 4 half life's: 150 mg

16= 0.094 mg remains

Another Example … (half life)   14C measurements on the linen wrappings from

the Dead Sea Scrolls suggest that the scrolls contain about 80.3% of the 14C expected in living tissue. How old are the scrolls if the half-life for the decay of 14C is 5.73 × 103 years?

Note: Radiocarbon dating relies on the uptake of 14C into living tissue and subsequent decay after the death of the organism. 14C is produced at a steady rate by cosmic ray absorption:

01n + 7

14 N → 614C + 1

1 p (decay: 614C→ 7

14 N + −10e− + ν)