Chem. 116 Prof. T.L. Heise 1 CHE 116: General Chemistry uCHAPTER TWENTY ONE Copyright © Tyna L....

Chem. 116 Prof. T.L. Heise

1

CHE 116: General Chemistry

CHAPTER TWENTY ONE

Copyright © Tyna L. Heise 2002

All Rights Reserved


2

Nuclear Chemistry

Nuclear Reactions: changes in matter that occur in the nucleus of an atom

- spontaneous changes of nuclei, which emit radiation, are said to be radioactive


3

Radioactivity

Nucleus - made up of two subatomic particles

Chap. 21.1

PROTONNEUTRON

Both molecules are called nucleons


4

Radioactivity

All atoms of a given element have the same number of protons, known as atomic number

All atoms of a given element can have different numbers of neutrons, and therefore different mass numbers

- mass number is the number of nucleons in nucleus

- same atomic number, different mass number is an ISOTOPE

Chap. 21.1


5

Radioactivity

Different isotopes have different abundancies in nature.

Different nuclei also have different stabilities:

- nuclear properties of an atom depend on the number of protons and neutron - nuclei that are radioactive are called radionuclides - atoms containing these nuclei are called radioisotopes

Chap. 21.1


6

Radioactivity

The vast majority of nuclei found in nature are stable and remain intact indefinately

Radionuclides - unstable and spontaneously emit particles and electromagnetic energy.

- emission of radiation is one way an unstable nuclide can become a stable

nuclide with less energy - when a nuclide spontaneously decomposes, it is called radioactive decay

Chap. 21.1


7

Radioactivity

Alpha decay ()- emission of the nucleus of a helium atom : 4He2

238U92 ---> 234Th90 + 4He2

** all mass numbers and atomic numbers are similarly balanced in all nuclear equations

Chap. 21.1


8

Radioactivity

Sample exercise: What element undergoes alpha decay to form lead-208?

Chap. 21.1


9

Radioactivity


X ---> 208Pb82 + 4He2

Chap. 21.1


10

Radioactivity


X ---> 208Pb82 + 4He2

atomic numbers add up to 212

Chap. 21.1


11

Radioactivity


X ---> 208Pb82 + 4He2

atomic numbers add up to 212

mass numbers add up to 84

Chap. 21.1


12

Radioactivity


212X84 ---> 208Pb82 + 4He2

look up atomic number 84 to identify symbol

Chap. 21.1


13

Radioactivity


212Po84 ---> 208Pb82 + 4He2

Chap. 21.1


14

Radioactivity

Beta decay ()- emission of the nucleus of a high speed electron : 0e-1

131I53 ---> 131Xe54 + 0e-1

** beta emission is equivalent to the conversion of a neutron to a proton, thereby increasing the atomic number by 1

1n0 --> 1p1 + 0e-1

the electron only comes into existence during nuclear reaction, it was NOT there all along

Chap. 21.1


15

Radioactivity

Gamma radiation ()- emission of the nucleus of a high energy photons : 00

** not shown when writing nuclear equations

Chap. 21.1


16

Radioactivity

nope

Chap. 21.1


17

Radioactivity

Positron emission - emission of the nucleus of a high speed positive electron : 0e+1

11C6 ---> 11B5 + 0e+1

** positron emission is equivalent to the conversion of a proton to a neutron, thereby decreasing the atomic number by 1

1p1 --> 1n0 + 0e+1

the positron only comes into existence during nuclear reaction, it was NOT there all along

.

Chap. 21.1


18

Radioactivity

Electron capture - capture by the nucleus of a high speed electron : 0e-1

81Rb37 + 0e-1 --> 81Kr36

** electron capture is equivalent to the conversion of a proton to a neutron, thereby decreasing the atomic number by 1

1p1 + 0e-1 --> 1n0

.

Chap. 21.1


19

Radioactivity

along.

Chap. 21.1


20

Radioactivity

Write a balanced nuclear equation for the reaction in which oxygen-15 undergoes positron emission.

Chap. 21.1


21

Radioactivity


15O8 --> 0e+1 + X

Chap. 21.1


22

Radioactivity


15O8 --> 0e+1 + 15X7

Chap. 21.1


23

Radioactivity


15O8 --> 0e+1 + 15N7

Chap. 21.1


24

Patterns of Nuclear stability

The stability of a particular nucleus depends on a variety of factors, and no single rule allows us to predict whether a particular nucleus is radioactive and how it might decay, however empirical observations can be made

- neutron to proton ratio is most important

Chap. 21.2


25


neutron to proton ratio

- the more protons packed into the nucleus, the more neutrons needed to bind the nucleus together

stable nuclei with low atomic numbers have approximately equal numbers of neutrons and protons

Chap. 21.2


26


neutron to proton ratio

- the more protons packed into the nucleus, the more neutrons needed to bind the nucleus together

nuclei with higher atomic numbers, the number of neutrons exceeds the number of protons because the number of neutrons necessary to create a stable nucleus increases more rapidly than the number of protons

Chap. 21.2


27


The belt of stability ends at 83

- above the belt can lower their ratio by emitting a

beta - below the belt can increase their ratio by either positron emission or electron capture - nuclei with atomic numbers above

84 tend to undergo alpha emission

Chap. 21.2


28


Sample exercise: Predict the mode of decay of

(a) plutonium-239

Chap. 21.2


29



(a) plutonium-239

atomic number of 94, alpha emission

Chap. 21.2


30



(a) indium-120

Chap. 21.2


31



(a) indium-120

atomic number of 49, neutrons are 71, above the belt of stability; beta emission

Chap. 21.2


32


Keep in mind that the previous slides describe guidelines to follow, and not all nuclei abide by the guidelines given.

Certain nuclei can not gain stability by a single emission. Elements like this have a series of emissions called a disintegration series.

Chap. 21.2


33


Uranium-238 is an excellent example of

a nuclei which has a disintegration series

Chap. 21.2


34


Two other observations have proven useful in the determination of stable nuclei

Nuclei with 2, 8, 20, 28, 50, or 82 protons OR 2, 8, 20, 28, 50 or 82 neutrons are generally more stable. These numbers have been called the magic numbers

Nuclei with even numbers of both protons and neutrons are generally more stable than those with odd numbers of nucleons

Chap. 21.2


35


Sample exercise: Which of the following nuclei would you expect to exhibit a special stability:

118Sn50, 210At85, 208Pb82

Chap. 21.2


36


Sample exercise: Which of the following nuclei would you expect to exhibit a special stability:

118Sn50208Pb82

Chap. 21.2


37

Nuclear Transmutations

Another way a nucleus can change identity is to be struck by a neutron or by another nucleus. Nuclear reactions that have been induced this way are called Nuclear (Artificial) Transmutations

Nuclear Transmutations are listed in the following order: target nucleus + bombarding particle --> ejected particle + product nucleus

14N7 + 4He2 --> 1H1 + 17O8

14N7 (, p) 17O8

Chap. 21.3


38


Charged particles must be moving very fast in order to overcome the electrostatic repulsion between them and the target nucleus.

- the higher the nuclear charge on either the projectile or the target, the faster the

particle must be going

- Strong magnetic and electric fields are used to accelerate the particles.

Chap. 21.3


39


Particle Accelerators

Chap. 21.3


40


Particle Accelerators

Chap. 21.3


41


Most synthetic isotopes in quantity in medicine and scientific research are made using neutrons as projectiles

- neutrons are neutral so there is no nuclear repulsion to overcome

- no need to be accelerated

Chap. 21.3


42

Rates of Radioactive Decay

Different nuclei undergo radioactive decay at different rates.

Radioactive decay is a first order kinetic process

- characteristic half life

- independent of initial concentration

- unaffected by external forces such as temperature, pressure, or state of chemical combination

- radioactive atoms cannot be rendered harmless by a chemical reaction or by any other practical treatment

Chap. 21.4


43


Sample exercise: Carbon-11, used in medical imaging, has a half life of 20.4 min. The carbon-11 nuclides are formed and then incorporated into a desired compound. The resulting sample is injected into the patient, and the image is obtained. The entire process takes five half lives. What percentage of original carbon remains at this time?

Chap. 21.4


44



100 50

Chap. 21.4


45



100 50 25

Chap. 21.4


46



100 50 25 12.5

Chap. 21.4


47



100 50 25 12.5 6.25

Chap. 21.4


48



100 50 25 12.5 6.25 3.125

Chap. 21.4


49


Due to the constancy of half lives, they can be used as a molecular clock to determine the ages of different objects

Chap. 21.4


50


Shroud of Turin - face

Chap. 21.4


51


Shroud of Turin - hands

Chap. 21.4


52


Calculation based on Half-livesRate = kNthe first order rate constant is called a decay

constantThe rate at which a sample decays is called its

activity, units are disintegrations/secln(Nt/No) = -kt

k = 0.693/t1/2

Chap. 21.4


53


Sample exercise: A wooden object from an archeological site is subjected to radiocarbon dating. The activity of the sample due to carbon-14 is measured to be 11.6 disintegration per second. The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second. The half-life of 14C is 5,715 yr. What is the age of the archeological sample?

Chap. 21.4


54


Sample exercise: A wooden object from an archeological site is subjected to radiocarbon dating. The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second. The half-life of 14C is 5,715 yr. What is the age of the archeological sample?

k = 0.693/t1/2

Chap. 21.4


55



k = 0.693/t1/2

k = 0.693/5,715 yr

Chap. 21.4


56



k = 0.693/t1/2

k = 0.693/5,715 yr

k = 1.21 x 10-4 yr-1

Chap. 21.4


57



k = 0.693/t1/2 t = (-1/k)ln(Nt/No)

k = 0.693/5,715 yr

k = 1.21 x 10-4 yr-1

Chap. 21.4


58



k = 0.693/t1/2 t = (-1/k)ln(Nt/No)

k = 0.693/5,715 yr t = (-1/1.21x10-4)ln(11.6/15.2)

k = 1.21 x 10-4 yr-1 t = (-8264)(-0.2702)

Chap. 21.4


59



k = 0.693/t1/2 t = (-1/k)ln(Nt/No)

k = 0.693/5,715 yr t = (-1/1.21x10-4)ln(11.6/15.2)

k = 1.21 x 10-4 yr-1 t = (-8264)(-0.2702)

t = 2233 yr

Chap. 21.4


60


Sample exercise: A sample to be used for medical imaging is labeled with 18F, which has a half-life of 110 minutes. What percentage of the original activity in the sample remains after 300 minutes?

Chap. 21.4


61



k = 0.693/t1/2

Chap. 21.4


62



k = 0.693/t1/2

k = 0.693/110 min.

Chap. 21.4


63



k = 0.693/t1/2

k = 0.693/110 min.

k = 0.0063 min.-1

Chap. 21.4


64



k = 0.693/t1/2 ln(Nt/No) = -kt

k = 0.693/110min

k = 0.0063 min-1

Chap. 21.4


65



k = 0.693/t1/2 ln(Nt/No) = -kt

k = 0.693/ 110 min ln(x/100g) = -0.0063(300)

k = 0.0063 min-1

Chap. 21.4


66



k = 0.693/t1/2 ln(Nt/No) = -kt

k = 0.693/ 110 min ln(x/100g) = -0.0063(300)

k = 0.0063 min-1 x/100 g = e-1.89

Chap. 21.4


67



k = 0.693/t1/2 ln(Nt/No) = -kt

k = 0.693/110 min ln(x/100g) = -0.0063(300)

k = 0.0063 min-1 x/100 g = e-1.89

x/100 g = 0.151

Chap. 21.4


68



k = 0.693/t1/2 ln(Nt/No) = -kt

k = 0.693/ 110 min ln(x/100g) = -0.0063(300)

k = 0.0063 min-1 x/100 g = e-1.89

x/100 g = 0.151

x = 15.1 g or 15.1%

Chap. 21.4


69

Detection of Radiation

A variety of methods have been designed to detect emissions from radioactive substances.Photographic film and plates, the greater the

exposure, the darker the area exposedGeiger counters, uses the conduction of electricity by

ions and electrons produced by radioactive substancesPhosphors glow when as electrons excited by

radiation fall back down to ground stateScintillation counter detects tiny flashes of light from

phosphors

Chap. 21.5


70


Geiger counters

Chap. 21.5


71


Radiotracers: a radioactive element that can be traced so easily they are used to follow the pathway a chemical reaction takes

- ability to do this comes from the fact that all isotopes of an element have essentially

identical chemical properties

- the chemicals pathway is revealed by the radioactivity of the radioisotope

Chap. 21.5


72

Energy Changes

The energies involved in nuclear reactions must be considered using Einstein’s famous equation

E = mc2

This equation states that the mass and energy of an object are proportional, if a system loses mass, it loses energy and vice versa.

The proportionality constant c2 is so large, even small changes in mass cause large changes in energy

Chap. 21.6


73

Energy Changes

The mass changes and the associated energy changes in nuclear reactions are much greater than those in chemical reactions.

- the mass change in the decay of 1 mole of Uranium-238 is 50,000 times

greater than that for the combustion of one mole of methane.

238U92 --> 234Th90 + 4He2

Chap. 21.6


74

Energy Changes

238U92 --> 234Th90 + 4He2

mass of nuclei: 238.0003 233.9942 + 4.0015

(amu)

238.0003 = 237.9957

Chap. 21.6


75

Energy Changes

238U92 --> 234Th90 + 4He2

mass of nuclei: 238.0003 233.9942 + 4.0015

(amu)

238.0003 = 237.9957

0.0046 amu are LOST, so proportional energy is LOST

**Lost energy is exothermic

Chap. 21.6


76

Energy Changes

238U92 --> 234Th90 + 4He2

mass of nuclei: 238.0003 233.9942 + 4.0015

(amu)

238.0003 = 237.9957

0.0046 amu

If 1 mole of U-238 is considered, amu turns into grams

Chap. 21.6


77

Energy Changes

238U92 --> 234Th90 + 4He2

mass of nuclei: 238.0003 233.9942 + 4.0015

(g)

238.0003 = 237.9957

0.0046 g

E = mc2

E = 0.0000046 kg(3.00x108m/s)2

E = 4.14x1011 kg m2/s2

E = 4.14x1011 J

Chap. 21.6


78

Energy Changes

Sample exercise: Positron emission form 11C,11C6 --> 11B5 + 0e1

occurs with release of 2.87x1011 J per mole of 11C. What is the mass change per mole of 11C in this nuclear reaction?

Chap. 21.6


79

Energy Changes



E = mc2

Chap. 21.6


80

Energy Changes



E = mc2

2.87x1011 J = m(3.00x108m/s)2

Chap. 21.6


81

Energy Changes



E = mc2

2.87x1011 J = m(3.00x108m/s)2

2.87x1011 J = m

(3.00x108m/s)2

Chap. 21.6


82

Energy Changes



E = mc2

2.87x1011 J = m(3.00x108m/s)2

2.87x1011 J = m

(3.00x108m/s)2

3.18x 10-6 kg = m

Chap. 21.6


83

Energy Changes



E = mc2

2.87x1011 J = m(3.00x108m/s)2

2.87x1011 J = m

(3.00x108m/s)2

3.19x 10-6 kg = m

0.00319 g = m

Chap. 21.6


84

Energy Changes

Scientists discovered in the 1930’s that the masses of nuclei are always less than the masses of the individual nucleons of which they are composed.The mass difference between a nucleus and its

constituent nucleons is called the mass defectThe origin of the mass defect is readily understood if

we consider that energy is used to break into the nucleons

The larger the binding energy, the more stable the nucleus

Chap. 21.6


85

Energy Changes

nuclei of intermediate mass numbers are more tightly bound than those with smaller or larger mass numbers

- a larger atom will break up into two intermediates

- 2 or more smaller atoms will fuse into an intermediate

Chap. 21.6


86

Nuclear Fission

Chap. 21.7


87

Nuclear Fission

2.4 neutrons produced by every fission of uranium-235.

Number of fissions and energy released quickly escalates exponentially is unchecked

In order for a fission chain reaction to occur a minimum mass of material must be present

(critical mass) - with minimum present only one neutron is effective in producing another fission

Chap. 21.7


88

Nuclear Fission

2.4

Chap. 21.7


89

Nuclear Fission

To trigger the fission reaction, two subcritical masses are slammed together using chemical explosives.

The two combined masses are supercritical which rapidly leads to an uncontrolled nuclear explosion

Chap. 21.7


90

Nuclear Fission

Nuclear Reactors: Uranium is enriched to about 3% U-235 and

then used to form UO2 pellets that are encased in zirconium or stainless steel tubes

Rods composed of materials such as cadmium or boron control the fission process by absorbing neutrons

Moderators slow down neutrons so they can be captured more readily by the fuel

Chap. 21.7


91

Nuclear Fission

Nuclear Reactors: A cooling liquid is circulated through the core

to carry off heat generated by the nuclear fission.

Cooling liquid and moderator could be one and the same substance

Steam is used to drive a turbine connected to an electrical generator, however steam must be condensed so additional cooling liquid is required, generally acquired from lake or river

Chap. 21.7


92

Nuclear Fission

Nuclear Reactors: Reactor is surrounded by a concrete shell to shield

personnel and nearby residents from radiationReactor must be stopped periodically so that the

fuel can be replaced or reprocessedSpent fuel rods are being kept in storage at

reactor sites20 half-lives are required for their radioactivity to

reach levels acceptable for biological exposure (600 years)

Chap. 21.7


93

Nuclear Fission

Chap. 21.7


94

Nuclear Fusion

Fusionappealing as an energy source because of

availability of light isotopes and because fusion products are generally not radioactive

not presently used to generate energy because high energies are needed to overcome the repulsion between nuclei

reaction requires temps of about 40,000,000 Kthese temps have only been achieved using a

hydrogen bomb

Chap. 21.8


95

Nuclear Fusion

Fusionalso a problem with confining the reaction - no

known structural material can withstand such temps

possibilities? Tokamak

Lasers

Chap. 21.8


96

Nuclear Fusion

Tokamak

Chap. 21.8


97

Biological Effects

We are continually bombarded with radiation!

When matter absorbs radiation, the energy of radiation can cause either excitation or ionization of the matter

- ionizing radiation is more harmful

When living tissue is irradiated, most of the energy is absorbed by the 70% water by mass of living tissue

Chap. 21.9


98

Biological Effects

Ionizing radiationelectrons are removed from water forming

highly reactive H2O+ ions

H2O+ + H2O --> H3O+ + OH

the unstable and highly reactive OH molecule is an example of a free radical due to the unpaired electron, •OH

in tissue, free radicals attack a host of surrounding biomolecules to produce more free radicals

Chap. 21.9


99

Biological Effects

Damage depends onactivity and energy of the radiationlength of exposurewhether source is inside or outside the body

Tissue that shows most damagereproduce at rapid ratesbone marrowblood forming tissuelymph nodes

Chap. 21.9


100

Biological Effects

Extended Exposure to Low Dosescancerdamage to growth regulation mechanism in cell,

inducing cells to reproduce in an uncontrolled manner

Chap. 21.9


101

Biological Effects

Units used to measure radiationbecquerel (Bq) = 1 nuclear disintegration per

secondcurie (Ci) = 3.7 x 1010 disintegrations per

secondgray (Gy) = 1 J absorbed per kilogram of tissuerad (radiation absorbed dose) = 1 x 10-2 J per

kilogram of tissueto correct for differences in strengths of varying

radiation, a multiplication factor is used

Chap. 21.9


102

Biological Effects

RadonRn-222 is a product of nuclear disintegration of

U-238being a noble gas, radon is extremely unreactive

and easily escapes the groundradon has a short half life and emits alpha

particles222Rn86 --> 218Po84 + 4He2

polonium is also an alpha emitter

Chap. 21.9

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Chem. 116 Prof. T.L. Heise 1 CHE 116: General Chemistry uCHAPTER TWENTY ONE Copyright © Tyna L....

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