A little more Thermodynamics, Redox, Electrochemistry, and Radioactivity

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A little more Thermodynamics, Redox, Electrochemistry, and Radioactivity. --Whew. Entropy -a measure of disorder in a system. In general, in terms of entropy: (s)

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A little more Thermodynamics, Redox, Electrochemistry, and

Radioactivity

--Whew.

Entropy -a measure of disorder in a system

• In general, in terms of entropy:

(s)<(l)<(aq)<<(g)

• Entropy (S) measured in (J/mol k)

• Def: entropy of a pure substance, perfect crystal, at absolute 0 = 0 J/ mol k

Is S (+) or (-) ?

• 2NaHCO3 (s) Na2CO3 (s) + H2O (g)

• CaCO3 (s) CaO (s) +CO2 (g)

• H2 (g) +Cl2 (g) 2HCl (g)

• N2 (g) + 3H2 (g) 2NH3 (g)

Just check the states

Changes in entropy

A: A system can lose entropy, becoming more ordered.

B: As it does, the rest of the Universe becomes less ordered

• B is always greater than A

Entropy always increases

Gibbs Free Energy

• For any reaction:

G=H-TSor

Go=Ho-TSo

(“o”=standard conditions)

and G for a reaction = -G for the reverse reaction

Standard conditions (thermo)

25oC (298k)

Solutes at 1.0M

Gasses at 1.0 atm.

Go and Spontaneity

Spontaneity

• Go<0: the system will proceed forward from standard conditions. This is called spontaneous.

• G=0. concentrations are stable. A system in equilibrium has no gain or loss of enthalpy or entropy .

• Go>0: Rxn is not spontaneous, the reverse rxn is.

What is So , Ho , and Go?

• 2NaHCO3 (s) Na2CO3 (s) + H2O (g)

• CaCO3 (s) CaO (s) +CO2 (g)

• H2 (g) + Cl2 (g) 2HCl (g)

• N2 (g) + 3H2 (g) 2NH3 (g)

At what T does the spontaneity change?

• 2NaHCO3 (s) Na2CO3 (s) + H2O (g)

• CaCO3 (s) CaO (s) +CO2 (g)

• H2 (g) + Cl2 (g) 2HCl (g)

• N2 (g) + 3H2 (g) 2NH3 (g)

If Ho is & So is then Go…

- + - spontaneous at any T

+ - +Not spontaneous at any T

+ + - spontaneous at high T

(or) +Not spontaneous at low T

- - +Not spontaneous at high T

(or) - spontaneous at low T

In (other) words…

Situation 1

• If an exothermic reaction leads to an increase in entropy, then free energy is released, and the reaction is spontaneous at any temperature

In (other) words…

Situation 1

• If an exothermic reaction leads to an increase in entropy, then free energy is released, and the reaction is spontaneous at any temperature

• (Can you state the other three situations?)

In (other) words…

Situation 2

• If an endothermic reaction leads to an decrease in entropy, then free energy is absorbed, and the reaction is nonspontaneous at any temperature

In (other) words…

Situation 3

• If an endothermic reaction leads to an increase in entropy, then free energy is released at high temperatures, and the reaction is spontaneous only when it is hot enough

In (other) words…

Situation 4

• If an exothermic reaction leads to an decrease in entropy, then free energy is released only at low temperatures, and the reaction is spontaneous only when it is cool enough (It may be very slow at that temperature)

So, how do you make products?

N2 (g) + 3H2 (g) 2NH3 (g)

So, how do you make products?

• Non-spontaneous does not mean that reactants won’t make products. (It just won’t make a whole lot before they start decomposing just as fast.)

• Use Le Chatelier’s principle. Remove the products (including heat), and the system will keep making more.

Why must a non-spontaneous reaction make some product?

Why must a non-spontaneous reaction make some product?

• Entropy.

Why must a non-spontaneous reaction make some product?

• Entropy.

• Pure reactants have less entropy than a mixture of reactants and products

Come, let us reason together…

• Spontaneous=proceeding forward (more products) from standard conditions

• Standard conditions=conc. of 1 M (or 1 atm.)

Q (std) =[products] / [reactants]

= (1)x/(1)y=1

• and K=Q at eq.

therefore…

…if a reaction is spontaneous…

…K>1

G<0 and K>1for spontaneous reactions

G>0 and K<1 for non-spontaneous reactions

Gibbs free energy and equilibrium

• If a system is spontaneous:– It loses free energy if it proceeds forward from

standard conditions—so it will– and K>1

• In any case:

Go = -RT ln K

What is K (at 298k)?

• 2 NaHCO3 (s) Na2CO3 (s) + H2O (g)

• CaCO3 (s) CaO (s) +CO2 (g)

• H2 (g) +Cl2 (g) 2 HCl (g)

• N2 (g) + 3H2 (g) 2 NH3 (g)

Does it go forward from where it is?

If Q<K, yes! If you know Go, use the relationship:

G = Go + RT ln Q

If G<0, it proceeds forwards

If G=0, it is in equilibrium (Q=K)

If G>0, it proceeds in reverse

Three Laws of Thermodynamics1st Law: Energy is neither created nor

destroyed

2nd Law: Entropy (system and surroundings) always increases

3rd Law: The entropy of a pure substance, perfect crystal, at absolute 0 = 0 kJ/ mol k

Redox—Review and Ch. 19

• A reduction is a gain of electrons, an oxidation is a loss of electrons

• A reduction is always conjoined with an oxidation (e- ’ s are conserved, charges must balance)

• Remember: “OILRIG” or “LEO says GER”

OILRIG

Oxidation is loss of electrons

Reduction is gain of electrons

LEO says GER

Loss of electrons is oxidation

Gain of electrons is reduction

Does a redox reaction occur?• Look for an oxidizing agent and a reducing

agent.

• If there is one of each, then ask, “Can this oxidizing agent oxidize this reducing agent”

• Answer by comparing reduction potentials.

Don’t memorize a rule. Compare the values to a reaction you know will occur

Does a redox reaction occur?

If you combine…

• Na+ and Fe+3?

• Cl- and Ag?

• Cu and K+ ?

• Pb+2 and I- ?

• Fe+2 and Mg?

Does a redox reaction occur?

If you combine…

• Na+ and Fe+3?—No. There is no reducer.

• Cl- and Ag?—No. There is no oxidizer.

• Cu and K+ ?—No. This oxidizer can’t do it

• Pb+2 and I- ?—No, but it will precipitate.

• Fe+2 and Mg?—Yes.

Fe+2 + Mg Fe and Mg+2

Redox—half reactions

• Balance the atoms

• Rectify the electrons

• Add H2O and H+ to balance oxygen and hydrogen

• Check that charges are balanced

• (Add OH- if the reaction is specified as in a basic solution)

Try it.

Sodium thiosulfate and nitric acid yield…

Hydrogen peroxide and iron (II) sulfate

Potassium dichromate and potassium iodide

Potassium permanganate and ethanol

Try it.

• S2O3-2 + NO3

-

• H2O2 + Fe+2

• Cr2O7-2 + I-

• MnO4- + C2H5OH

Try it.

• S2O3-2 + NO3

- SO4-2 + NO

• H2O2 + Fe+2 H2O + Fe+3

• Cr2O7-2 + I- Cr+3 + I2

• MnO4- + C2H5OH Mn+2 + CO2 + H2O

Try it.

3x(S2O3-2 + 5H2O 2 SO4

-2 +10H + + 8 e-)

8x(NO3- +4H + + 3 e- NO + 2 H2O)

3S2O3-2 + 15H2O 6 SO4

-2 +30H + + 24 e-)

8NO3- +32H + + 24 e- 8NO + 16H2O)

3S2O3-2+8NO3

-+2H+6SO4-2 +8NO+H2O

Reduction potentials

• Standard reduction potentials are measured as compared to:

2H++2e-H2 (0.00V – by definition)

• Half reactions that accomplish this have (-) reduction potentials (Eo<0)

• Half reactions that force the reverse have (+) reduction potentials (Eo>0)

Reduction potentials

• Specifically: Magnesium reduces H+

• While bromine oxidizes hydrogen gas

Write each reaction.

What is the sign on Eo?

Reduction potentials

• Specifically: Magnesium reduces H+

• While bromine oxidizes hydrogen gas

2H+ + Mg Mg+2 + H2 which implies that Mg+2 + 2 e- Mg

has Eo<0

H2 + Br2 2H+ +2Br - which implies that; Br2 + 2 e- 2Br –

has Eo>0

Electrochemical Cells

• Half reactions are separated, and electrons are connected in a circuit.

• A salt bridge is needed to allow charges to migrate to offset the motion of electrons

• An electrode (anode or cathode) carries electrons to or from a half reaction

Cathode means reduction

Electrochemical Cells

Pb+2

Pb

Cd+2

Cd

The lead / cadmium battery

What is happening?

Electrochemical Cells

Pb+2

Pb

Cd+2

Cd

Pb+2 + Cd Cd+2 +Pb

What is happening?

Pb+2

Pb

Cd+2

Cd

Pb+2 Pb Cd Cd+2

Cathode =

reduction

Anode =

oxidation

Salt bridge

electrons

Cathode gains mass

Anode loses mass

Electrochemical Cells

Cu+2

Cu

Zn+2

Zn

The copper/zinc battery

What is happening?

Shorthand notation

• The Danielle Cell, using copper and zinc,

Zn|Zn+2||Cu+2|Cu

…makes 1.1 V

Zn|Zn+2||Cu+2|Cu

(or, in general)

product reactant

Anode of of cathode

oxidation reduction

If non-metals are used…

Pt|H2|H2O||O2|H2O|Pt

• The (non-reactive) metal electrode is noted outside the bars

Standard cell potentials

• Eo=Ered-Eox

Be able to:

• Sketch a cell (include salt bridge and circuit)

• Label anode and cathode• Write the half reactions, complete reaction• Calculate Eo, show direction of electron

flow• Describe the oxidation and reduction—

with mass changes, observations.• Read and write the shorthand notation

Practice

• Write the shorthand notation and find the standard cell potential for a chromium/chlorine cell

Practice

• 2Cr + 3Cl22Cr+3 + 6Cl-

• Cr|Cr+3||Cl2|Cl-|Pt

• E=Ered-Eox=1.36V-(-.74V)=2.10 V

The Nernst equation

Get real. When are all concentrations 1.00 M?

Ecell=Eo-(RT/nF) ln Q

(n=the number of electrons transferred in balanced reaction, F=96500C/mol=Faraday’s constant)

?Emf—Electromotive force, the voltage of the battery

• What is the emf of a Cu/Ag cell if [Cu+2]=.4M and [Ag+]=.1M?

?Emf

• What is the emf of a Cu/Ag cell if [Cu+2]=.4M and [Ag+]=.1M?

Cu + 2Ag+ Cu+2 +2Ag

n=2 and Q=[Cu+2]/[Ag+]2=40

?Emf

• What is the emf of a Cu/Ag cell if [Cu+2]=.4M and [Ag+]=.1M?

Cu + 2Ag+ Cu+2 +2Ag

n=2 and Q=[Cu+2]/[Ag+]2=40

You did remember to

square it, didn’t you?

?Emf

• What is the emf of a Cu/Ag cell if [Cu+2]=.4M and [Ag+]=.1M?

Cu + 2Ag+ Cu+2 +2Ag

n=2 and Q=[Cu+2]/[Ag+]2=40

• Ecell =Eo-(RT/nF) ln Q

?Emf

• What is the emf of a Cu/Ag cell if [Cu+2]=.4M and [Ag+]=.1M?

Cu + 2Ag+ Cu+2 +2Ag

n=2 and Q=[Cu+2]/[Ag+]2=40

• Ecell =Eo-(RT/nF) ln Q

=.46V-(8.31x298/2/96500)ln40

=.41V

?Emf

• What is the emf of a Cu/Ag cell if [Cu+2]=.4M and [Ag+]=.1M?

Cu + 2Ag+ Cu+2 +2Ag

n=2 and Q=[Cu+2]/[Ag+]2=40

• Ecell =Eo-(RT/nF) ln Q

=.46V-(8.31x298/2/96500)ln40

=.41V It’s running down. It will get worse as the

battery is used

?Emf

• What is the emf of a Cu/Ag cell if [Cu+2]=.02M and [Ag+]=.4M?

?Emf

• What is the emf of a Cu/Ag cell if [Cu+2]=.02M and [Ag+]=.4M?

Cu + 2Ag+ Cu+2 +2Ag

n=2 and Q=[Cu+2]/[Ag+]2=.125

• Ecell =Eo-(RT/nF) ln Q

=.46V-(8.31x298/2/96500)ln.125

=.49V

The Nernst equation

• You may find these equivalencies helpful

• Ecell =Eo-(RT/nF) ln Q

=Eo-2.303RT/nF log Q

=Eo-.0592/n log Q

Equilibrium and cell potential

Let us reason together.

• Q=K at equilibrium and

• Ecell=Eo-(RT/nF) ln Q and

• your batteries go dead (Ecell=0) eventually.

and these ideas imply that…

Equilibrium and cell potential

Let us reason together.

• Q=K at equilibrium and

• Ecell=Eo-(RT/nF) ln Q and

• your batteries go dead (Ecell=0) eventually.

and these ideas imply that…

Eo=(RT/nF) ln K

Gibbs free energy and cell potential

• As a battery operates, it gives off free energy in electrical work. There are two relationships of note:

Go = -RT ln K and Eo=(RT/nF) ln K

Therefore…

Gibbs free energy and cell potential

• As a battery operates, it gives off free energy in electrical work. There are two relationships of note:

Go = -RT ln K and Eo=(RT/nF) ln K

Therefore…

Go = -nFEo

K

EoGoGo = -nFEo

Go = -RT ln K Eo=(RT/nF) ln K

Use…

Use…

Use…

Quote: 4/10/09

“I have spent two years getting you to the point of using the Nernst equation, and now you tell me you don’t know how to subtract?”

--Tnichols, to an AP chemistry student

Lithium battery: Nicad battery: Dry cell:

Danielle cell: Alkaline cell:

Lead acid battery:

Mercury battery: Fuel cell:

You should know:

Lithium battery: Li+TiSLi++TiS-

Nicad battery: Cd+NiOCdO+NiDry cell:

Zn+2NH4++2MnO2Zn+2+2NH3+H2O+Mn2O3

Danielle cell: Zn+CuSO4ZnSO4+CuAlkaline cell:

Zn+2MnO2+H2OZn+2+Mn2O3+2OH-

Lead acid battery:

Pb+PbO2+H2SO42PbSO4+2H2OMercury battery: Zn+HgOZnO+Hg

Fuel cell: H2+O2H2O or CxHy+O2CO2+H2O

You should know:

Counting electrons—Faraday’s constant

• The unit of electrical charge, 1 coulomb, is a small fraction of a mole.

• We use Faraday’s constant

1F=96,500 C/mole

as the conversion.

Also:

• Potential energy (V): 1 Volt = 1J/C, and

• Current (I): 1 Amp = 1C/s

Electrolytic Cells

• Applying an external voltage will allow a non-spontaneous reaction to occur.

Electrolytic Cells

• Applying an external voltage will allow a non-spontaneous reaction to occur.

2H2O2H2+O2 is not spontaneous (Right?)

If you apply a voltage to water (with some electrolyte added to carry a charge), it will decompose (or electrolyse)

Electrolytic Cells

• Applying an external voltage will allow a non-spontaneous reaction to occur.

2H2O2H2+O2 is not spontaneous (Right?)

If you apply a voltage to water (with some electrolyte added to carry a charge), it will decompose (or electrolyse)

Anode reaction: 2H2O4H+ + O2 + 4e-

Cathode reaction: 4 H+ + 4 e-2 H2

PS

• On a battery, the anode is marked (-) –it produces electrons.

• On an electrolytic cell, the cathode is marked (-) –it receives electrons

Electrolytic Cells

• Stoichiometric relationships:

Current x time#electronsmass product

Electrolytic Cells

• Stoichiometric relationships:

Current x time#electronsmass product

If a copper (II) sulfate solution is electroplated at 15 amps for 12 minutes,

Electrolytic Cells

• Stoichiometric relationships:

Current x time#electronsmass product

If a copper (II) sulfate solution is electroplated at 15 amps for 12 minutes,

15 C/s x (12 x 60)s = 10,800 C

Electrolytic Cells

• Stoichiometric relationships:

Current x time#electronsmass product

If a copper (II) sulfate solution is electroplated at 15 amps for 12 minutes,

15 C/s x (12 x 60)s = 10,800 C

10,800 C x 1 mole / 96,500 C = .112 mole e-

Electrolytic Cells

• Stoichiometric relationships:Current x time#electronsmass productIf a copper (II) sulfate solution is

electroplated at 15 amps for 12 minutes,15 C/s x (12 x 60)s = 10,800 C10,800 C x 1 mole / 96,500 C = .112 mole e-.112 mole e- x 1 Cu+2/ 2 e- = .0560 mole Cu.

Electrolytic Cells

• Stoichiometric relationships:Current x time#electronsmass productIf a copper (II) sulfate solution is

electroplated at 15 amps for 12 minutes,15 C/s x (12 x 60)s = 10,800 C10,800 C x 1 mole / 96,500 C = .112 mole e-.112 mole e- x 1 Cu+2/ 2 e- = .0560 mole Cu.0560 mole Cu x 63.55 g/ mole= 3.6 g Cu

Electrolytic Cells

• Voltage concerns:

The easiest ox. & red. reaction will occur

Electrolytic Cells

• Voltage concerns:

The easiest ox. & red. reaction will occur

If you run an electrolytic cell with platinum electrodes, in a sodium nitrate solution:

Cathode reaction:

Na+ + e-Na or 4 H+ + 4 e-2 H2

?

Electrolytic Cells

• Voltage concerns:

The easiest ox. & red. reaction will occur

If you run an electrolytic cell with platinum electrodes, in a sodium nitrate solution:

Cathode reaction:

Na+ + e-Na or 4 H+ + 4 e-2 H2

Choose this one!

Electrolytic Cells

• Voltage concerns:The easiest ox. & red. reaction will occur

If you run an electrolytic cell with platinum electrodes, in a sodium nitrate solution:

Anode reaction:

Pt e- + Pt+2 or 2H2O4H+ + O2 + 4e-?

Electrolytic Cells

• Voltage concerns:The easiest ox. & red. reaction will occur

If you run an electrolytic cell with platinum electrodes, in a sodium nitrate solution:

Anode reaction:

Pt e- + Pt+2 or 2H2O4H+ + O2 + 4e-?

Choose this one!

Officially:

• You should be able to run an electrolytic cell with the same voltage that the battery using the reverse reaction would provide

• But.

• You have to add a small over-voltage to overcome surface effects.

Nuclear Chemistry

• --breaks the rules that one atom cannot be converted to another.

Chemistry is the dance of the electrons—nuclear reactions change the nuclei of atoms

• --charge and mass are still conserved.

Nuclide Notation

• A nuclide is a nucleus or atom of a specific isotope of an element

K39

19

• This is potassium-39. It has 19 protons (atomic number = 19), making it potassium, and 20 neutrons, making a mass number of 39

How many p, n, e- in each?What is the mass number atomic

number and EC?

Cl-36

17

Sr+290

38

I-131

53

Th228

90

H3

1

Fe+359

26

Natural decays

• —

• —

.

Natural decays

• —the loss of particle from a nuclide

• —emission of an electron ( particle) from the nucleus by the conversion of a n p + e-

.

Natural decays

• —the loss of particle from a nuclide--The particle is composed of 2p and 2n,

= 4He nucleus--decreases the mass by 4 and the atomic number by 2

• —emission of an electron ( particle) from the nucleus by the conversion of a n p + e-

--the electron is the particle--increases the atomic number by 1, does

not affect mass

Write the reaction

• Argon-39 undergoes a decay

• Thorium-228 undergoes an decay

• An decay forms lead-204

• A decay forms nitrogen-14

• A natural decay forms Sc-45 from Ca-45

• A natural decay forms Ac-227 from Pa-231

Notice what they do

• A decay lowers the n/p ratio in small nuclei, or when the ratio is too large.

• An decay lowers the total size, and raises the n/p in large nuclides, or when the ratio is too small.

What is “too large” or “too small”?

What is “too large” or “too small”?

decay

decay

Write a reaction for each transition marked on the graph

What is “too large” or “too small”?

What is “too large” or “too small”?

n:p = 1:1

n:p = 2:1

What is “too large” or “too small”?

n:p = 1:1

n:p = 2:1

Natural radioactive decay of U-238

The search for super-heavy stable nuclei continues

• Researchers report the first creation of the long-lived nucleus hassium-270, a "doubly magic" combination of 108 protons and 162 neutrons. Its long lifetime of 22 seconds supports the theory of an "island of stability" for the heaviest elements.

(J Dvorak et al. 2006 Phys. Rev. Lett. 97, 242501)

Nuclear reactions

• Many nuclear reactions involve colliding nuclei or smaller particles at some significant fraction of the speed of light,

• --find the missing particle by balancing mass and charge.

Particles might include…

• p

• n

• e- (AKA )

• d

• • (OK, it’s not a particle, but it’s often

written in)

Nuclear reactions

Condensed notationNuclide1(gets hit by…,produces…and)Nuclide2

• For example: 238U(n,2n)237U

means238U+1n21n+237U

(or so they say)

Condensed notation

• Write the reaction

• 7Li(1p,4He)4He

• Write the condensed notation

• 14N +41p +17O

In comparison

• Physical changes:

joules/mole range

• Chemical changes:

kilojoules/mole range

• Nuclear changes:

megajoules/mole range

(1 kiloton=4 TJ)

For example:

• If a nuclear reaction releases 45 MJ/mol, what is the wavelength of the photons produced?

Here is a problem

• The mass of a He-4 atom =4.00260 amu

Do you see the problem?

• 1p + 1e- has a mass of 1.007825 amu

• 1n has a mass 1.008665 amu

• The mass of a He-4 atom =4.00260 amu

Binding energy

• Careful measurements lead to the conclusion that matter and energy can be inter-converted.

• The mass lost in a nuclear reaction is converted to energy.

Nuclei have a mass defect representing the binding energy

in the nucleus

What is the binding energy?

• The mass of a He-4 atom =4.00260 amu

• 2 x p+e= 2.01565 amu

• 2 x n= 2.01733 amu

4.03298 amu

What is the binding energy?

• The mass of a He-4 atom =4.00260 amu

• 2 x p+e= 2.01565 amu

• 2 x n= 2.01733 amu

4.03298 amu

- 4.00260 amu0.03038 amu is missing!

0.03038 amu! Where did it go?

This is the binding energy. To convert this mass to energy, use E=mc2

.03038 amu x1g/6.022 x 1023amu= 5.045 x 10-26g

= 5.045 x 10-

29kg

E=mc2=(5.045x10-29kg)(3.00x108m/s)2=4.54x10-12J

(or 1.14x10-12J/nucleon since there are 4 particles in the nucleus)

What is the binding energy? (in J/nucleon)

1) The mass of I-127 atom=126.9004 amu

2) The mass of Bi-209 atom=208.9804 amu

3) The mass of a 0-16 atom =15.995 amu

What is the binding energy? (in J/nucleon)

1) The mass of I-127 atom=126.9004 amu

(1.36 x 10-12J/nucleon)

1) The mass of Bi-209 atom=208.9804 amu

(1.26 x 10-12J/nucleon)

1) The mass of a 0-16 atom =15.995 amu

(1.28 x 10-12J/nucleon)

What is the mass defect?

• Iron-56 is the most stable nuclide, having 1.40 x 10-12 J/ nucleon binding energy. What is the mass defect of the atom?

What is the mass defect?

• Iron-56 is the most stable nuclide, having 1.40 x 10-12 J/ nucleon binding energy. What is the mass defect of the atom?

(.52 amu)

Fission vs Fusion• Fission=breaking up large nuclei—

--natural radioactive decay of large atoms

--used for nuclear power

• Fusion=combining small nuclei

--occurs naturally in stars

--prospects for nuclear energy—no radioactive byproducts

Both are transmutations—one nuclide is converted into another

Rates of Radioactive decay

• Natural radioactive decays are first order reactions, so use the first order reaction rate laws

Rate= k[A] and

ln [A]t – ln [A]o= - k t

Consider the relationships

• Half life

• Original amount

• Final amount

• Time elapsed

• Rate constant

• Rate of decay

Try it.

• If you start with 1.38 mg of U-234 and k=2.84 x 10-6/yr for its decay

--how much is left after 20,000 years?

--how long will it take to reach .010 mg?

--what is the initial rate of decay?

--what is the final rate of decay?

--what is the half-life?

Try it.

• Br-82 has a half life of 35.3 hours. If you start with a 6.5 mg sample of Br-82

--how much is left after 2 days?

--what is the initial rate of decay?

--what is the rate constant?

--how long will it take to reach 1.5 mg?

Try it.

• A .350 mg sample of K-42 decays to only .066 mg in 29.7 hours.

--what is the rate constant?

--how much was left after 20.0 hours?

--what was the initial rate of decay?

--what is the half life?

--how long will it take to reach .010 mg?

Amount of radioactivity

• 1 rad=.00001 J/g (often converted to rems in medicine)

Officially:

• 1 Curie= 3.70 x 10 10 decays/s

…where decays/s = Rate x Avogadro’s #

Try it.

• A .350 mg sample of K-42 decays to only .066 mg in 29.7 hours.

--what is the rate constant?

--how much was left after 20.0 hours?

--what was the initial rate of decay?

--what is the half life?

--how long will it take to reach .010 mg?How radioactive is the sample?

Radionuclides of noteU-238 U-235 S-35

Pu-239 I-131 P-32

C-14 K-40 O-18

Am-241 Tc-99 Sr-90

Rn-222 Ra-226 Po-210

Th-th-that’s all, folks.

Atomic theory

• All matter is composed of atoms.

--atoms of one element are identical

--atoms of different elements are different

--reactions form different combinations of atoms, not different atoms

• Atoms are composed of protons, neutrons, and electrons.

(Are all of the little kids in bed?—Now we can tell you the real story…)

Modern Theory

In this Universe, you will find:

Fermions and Bosons (force mediating

particles)

Modern Theory

Bosons (force mediating particles) include:

g, (electromagnetic force)Zo, (weak nuclear force)Gluons (Strong nuclear force) Graviton (gravity) W±, (weak nuclear force) Higgs (mediates mass)

Modern Theory

Fermions are the fundamental

particles, including:

andQuarks Leptons

Modern Theory

Types of quarks include:

(and their antiparticles)

bottom (aka “beauty”)

down

top (aka “truth”)

strange

charm

up

Modern Theory

Leptons include:

Electrons, e-,Muons, ,Tauons, ,and three types of neutrinose, ,

(and their antiparticles)

Modern Theory

Combinations of quarks make hadrons, either:

Mesons (2 quarks each)

Including: +, K-, +, B0, c

Baryons (3 quarks each)

including p, n, p-, L, -

Modern TheoryIn this Universe, you will find

fermions which include the fundamental particles

and

Bosons (force mediating particles) g, (electromagnetic forceZo, (weak nuclear force)Gluons (Str. nuclear force) Graviton (gravity) W±,(weak nuclear force) Higgs (mediates mass)

Quarks u,d,c,s,t,b (and anti-particles) Quarks make up the hadrons, either

Leptons e-, , , e, , (and antiparticles)

Mesons (2 quarks each) +, K-, +, B0, c

baryons (3 quarks each) including p, n, p-,L,-

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