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Chemistry SM-1131 Week 1 Lesson 1 Chapter 9 Dr. Jesse Reich Assistant Professor of Chemistry Massachusetts Maritime Academy Fall 2008

Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

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Chemistry SM-1131 Week 1 Lesson 1 Chapter 9. Dr. Jesse Reich Assistant Professor of Chemistry Massachusetts Maritime Academy Fall 2008. Class Today. Poem Test Review Finish the Timeline Lecture Electron Configuration One more exam the week of Dec 8 No class Friday. Quote. - PowerPoint PPT Presentation

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Page 1: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Chemistry SM-1131Week 1 Lesson 1

Chapter 9Dr. Jesse Reich

Assistant Professor of ChemistryMassachusetts Maritime Academy

Fall 2008

Page 2: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Class Today• Poem• Test Review• Finish the Timeline Lecture• Electron Configuration• One more exam the week of Dec 8• No class Friday

Page 3: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Quote

• “Go Big or Go Home”

Page 4: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Test

• Test was “bimodal”• A good chunk of you guys got ridiculous As• A good chunk of you guys haven’t learned a

thing• There wasn’t too much in between.

Page 5: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Test Review

• Q1: Aluminum sulfate and sodium phosphate double displacement rxn.

Al2(SO4)3 + Na3PO4 AlPO4 + Na2SO4

Al2(SO4)3 + 2Na3PO4 2AlPO4 + 3Na2SO4

Al2(SO4)3(aq) + 2Na3PO4(aq) 2AlPO4(s) + 3Na2SO4(aq)

2Al + 3 SO4 + 6 Na + 2 PO4 2AlPO4 +6 Na + 3SO4

2 Al + 2 PO4 2AlPO4

Page 6: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Test Review

• Q2: 3 FeCl2 + 2Al 2 AlCl3 + 3 Fe

a. Single Displacementb. 20.2g FeCl2 x 1 mole FeCl2 2 AlCl3 132g AlCl3

127g FeCl2 3 FeCl2 1 mole AlCl3

= 14.1g AlCl3

c. 300g Fe x 1 mole Fe x 2 mol Al x 27g Al 56g Fe 3 mol Fe 1 mol Al= 96.4g Al

Page 7: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Test Review

• 2d: Percent Yield = 100 X actual / theoretical• Percent = 50%• Actual = X• Theoretical = 30050% = x/ 300.5 x 300 = x150 = x150g of Fe

Page 8: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Test Review

• Q3: C2H6 + O2 H2O + CO2

2C2H6 + 7 O2 4 CO2 + 6 H2O

26g C2H6 x 1 mol C2H6 x 7 mol O2 x 32g O2

30g C2H6 2 mol C2H6 1 mol O2

= 97g O2… therefore 97g O2 would be required, but I only have 26g so O2 is the limiting reagent.

Page 9: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Test Review

• Q3: C2H6 + O2 H2O + CO2

2C2H6 + 7 O2 4 CO2 + 6 H2O

26g C2H6 x 1 mol C2H6 x 4 mol CO2

30g C2H6 2 mol C2H6

= 1.73 mol CO2

26g O2 x 1 mol O2 x 4 mol CO2

32g O2 7 mol O2

= 0.46 mol CO2 This must be limiting!

Page 10: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Test Review

• Q4:• A: Decomp• B: Combination• C: Combustion• D: Single Displacement• E: Double Displacement

Page 11: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Test Review

• Q5: CH4 + 2O2 CO2 + 2H2O

6 moles O2 x 1 mol CO2

2 mol O2

= 3 mol CO2

23.5 mol H2O x 1 mol CH4 x 16 CH4

2 mol H2O 1 mol CH4

= 188g CH4

Page 12: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Test Review

• Q5: CH4 + 2O2 CO2 + 2H2O

• 20g CH4 x 1 mol CH4 x 2 mol O2

16 g CH4 1 mol CH4

= 2.5 mol O2

Page 13: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Test Review

• Q6: Li2SO4 + CaCl2 CaSO4 + 2 LiCl

Li2SO4(aq) + CaCl2(aq) CaSO4(s) + 2 LiCl (aq)

2Li + SO4 + Ca + 2Cl CaSO4 + 2 Li + 2 Cl

Ca + SO4 CaSO4

Page 14: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Timeline

Page 15: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

A Nagging Concern

Rutherford’s Model

Positive Protons and Oppositely Charged ElectronsWhat should happen?Collapse!

Page 16: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Niels Bohr

Oct. 7, 1885, Denmark – Nov. 18, 1962

http://education.jlab.org/qa/atom_model_03.gif

Page 17: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

This is the quantized idea!

• He’s basically saying it can’t just be anywhere. It has to be in a specific space!

• Quantum means specific amount of energy, so not all energies are allowed, only certain ones.

• In this case it means that the electron has to stay a certain distance from the nucleus at all times… it’s not like standard physics.

Page 18: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

A problem in physics: Spectrum of Hydrogen

Page 19: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

A colorful spectrum of hydrogen

Page 20: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Bohr’s Idea

What if the path of the electrons was fixed…

Page 21: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

What energies would exist? What would that mean we would see?

Page 22: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

A story of what would happen

www.usm.maine.edu/

Page 23: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

A colorful spectrum of hydrogen

Page 24: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

This is what he thought it would look like!

http://pittsford.monroe.edu/pittsfordmiddle/rountree/rounweb_2_02/scottimage4.jpg

Page 25: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

This model describes small atoms

• But………• When they looked at bigger atoms then

hydrogen the model didn’t predict the light patterns accurately. Something was wrong with the model.

Page 26: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Erwin Schrodinger and Werner Heisenberg

1877- 1961 1901 - 1976

Page 27: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Let’s talk about light, Baby!

Page 28: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

What if it’s a particle?

Page 29: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

1 Photon leads to 1 electron

Page 30: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Let’s talk about light, Baby!

Page 31: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

But light moves in waves

Page 32: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Double Slit

http://homepage.univie.ac.at/Franz.Embacher/KinderUni2005/waves.gif

Page 33: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Here’s the experiment

Page 34: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9
Page 35: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Light or Wave?

• It’s both. It’s called the “wave particle duality.”

• Light sometimes acts like a particle and sometimes acts like a wave.

• It moves like a wave!• Sometimes it hits like a particle.

Page 36: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Heisenberg Uncertainty

• This uncertainty leads to some strange effects. For example, in a Quantum Mechanical world, I cannot predict where a particle will be with 100 % certainty. I can only speak in terms of probabilities. For example, I can only say that an atom will be at some location with a 99 % probability, and that there will be a 1 % probability it will be somewhere else

Page 37: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Probability Density

An abstract function, called wave function or probability amplitude (formula sign Y), describes the states of a particle or a physical system. Y is dependent on position and time. The wave function itself does not have any explicit meaning, but its square (more precisely the square of its absolute value) describes the probability of measuring the various possible positions of a particle. In addition, it provides information about the probability distribution of all other physical quantities (for example impulse and energy). Y satisfies the Schrödinger Equation, whose solution describes the behavior over time of a physical system. Schrödinger’s original formulation for particles with a given total energy E is as follows:

Page 38: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

A compilation of individual electrons

In 1927 Heisenberg formulated an idea, which agreed with tests, that no experiment can measure the position and momentum of a quantum particle simultaneously. Scientists call this the "Heisenberg uncertainty principle." This implies that as one measures the certainty of the position of a particle, the uncertainty in the momentum gets correspondingly larger. Or, with an accurate momentum measurement, the knowledge about the particle's position gets correspondingly less.

The visual concept of the atom now appeared as an electron "cloud" which surrounds a nucleus. The cloud consists of a probability distribution map which determines the most probable location of an electron. For example, if one could take a snap-shot of the location of the electron at different times and then superimpose all of the shots into one photo, then it might look something like the view at the top.

Page 39: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

This is what the probability density looks like for the first orbits described

Page 40: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

James Chadwick

Oct. 20, 1891, England – July 24, 1974

www.particlephysics.ac.uk

Page 41: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Chadwick’s Setup• In 1930, the German physicists Walther Bothe and Herbert Becker noticed something odd.

When they shot alpha rays at beryllium (atomic number 4) the beryllium emitted a neutral radiation that could penetrate 200 millimeters of lead. In contrast, it takes less than one millimeter of lead to stop a proton. Bothe and Becker assumed the neutral radiation was high-energy gamma rays.

• Marie Curie's daughter, Irene Joliot-Curie, and Irene's husband, Frederic, put a block of paraffin wax in front of the beryllium rays. They observed high-speed protons coming from the paraffin. They knew that gamma rays could eject electrons from metals. They thought the same thing was happening to the protons in the paraffin.

• Chadwick said the radiation could not be gamma rays. To eject protons at such a high velocity, the rays must have an energy of 50 million electron volts. The alpha particles colliding with beryllium nuclei could produce only 14 million electron volts.

• Chadwick had another explanation for the beryllium rays. He thought they were neutrons. He set up an experiment to test his hypothesis.

Page 42: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

Chadwick’s Experiment

Chadwick collided the radiation emerging from the (Po-Be) source not only with proton (paraffin), but also with helium and nitrogen. Comparing the results of these experiments with each other, Chadwick concluded that this mysterious radiation from the (Po-Be) source cannot be interpreted by assuming it to be a gamma ray. He finally concluded that all were able to be understood without any contradiction by assuming that the mysterious radiation is electrically neutral particles with almost the same mass as a proton. This is the confirmation of the existence of the "neutral proton" predicted by Rutherford. Chadwick named this particle "neutron" (1932).

www2.kutl.kyushu-u.ac.jp

Page 43: Chemistry SM-1131 Week 1 Lesson 1 Chapter 9

In his own words

• The properties of the penetrating radiation emitted from beryllium (and boron) when bombarded by the a-particles of polonium, have been examined. It is concluded that the radiation consists, not of quanta as hitherto supposed, but of neutrons, particles of mass 1, and charge 0. Evidence is given to show that the mass of the neutron is probably between 1·005 and 1·008.