1. Guest demonstration ! Dr. Simpson does the thermite reaction…..

2. SPECTACULAR REDOX REACTION BEHAVIOR 2Al o (s) + Fe 2 O 3 (s)  2Fe 0 (liq)+Al 2 O 3 (s) 3)let’s see the reaction in the `flesh’ (why we don’t do this one in class…) 0 +3-2 0 +3 -2 Which element oxidized ? Which element reduced ? Al Fe 1)assign oxidation # 2) Who does what? Thermite reaction https://www.youtube.com/watch?v=qqvQwfH_wGQ

3. One other unique and highly useful feature of redox reactions … The oxidation and reduction can be separated in space and you can still run the reaction as a `cell’ or ‘battery’

4. Example: Cu-Mg cell Cu 2+ + Mg o  Cu o +Mg 2+ The complete redox reaction Reduction half reaction: Cu 2+ +2e -  Cu o cathode Oxidation half reaction: Mg o  Mg 2+ +2e- Gains 2e - Loses 2e- anode Cu 2+ + Mg o  Cu o +Mg 2+

5. The experimental set-up of (Mg|Mg 2+ ||Cu 2+ |Cu) cell anode cathode LOAD Electron flow out of anode OxidationReduction Mg 0  Mg 2+ +2e - Cu 2+ +2e -  Cu o NaCl bridge Na +   Cl -

6. Cu +2 | Cu Mg|Mg 2+ The Potato clock relies on O 2 to cause Cu to oxidize to CuO which dissolves to form Cu +2 (aq). Any `juicy’ vegetable or fruit can serve as electrolyte source e - out e - in OXRED ~1.5 V

7. Ford Focus all electric car High surface area lithium ion batteries

8. Li x C 6  xLi + + 6C + xe- Cathode (right side=red) 0 0 Power cycle – what happens as you drive +1 Anode (left side =ox) Li o embedded in `intercalation’ compounds on both sides Li 1-x CoO 2 + xe - + xLi +  LiCoO 2 Key Chemistry trick: +1 LOAD

9. Our Chemistry Road trip so far … Moles, molecules and reactions Molecules and Moles acting alone

10. Chapter 5:Gases (pp. 189-230) an exploration of `aggregate’ behavior Next up: the Chemistry of the Crowd

11. Pressure (P) concepts & measures in “Physics-speak “(see p 192 ) P = Force Area Physics speak Physics units P = newtons m 2 = kg*m/s 2 m*m = kg m s 2

12. Pressure (P) concepts & measures in “Physics-speak “(see p 191 ) 1 kg = 1 Pascal (Pa) m s 2 

13. Some lowbrow chem insights into Pressure Getting hammered Bed of nails

14. Pressure expresses how much a (constant) impressed force is spread out… MORE AREA, LOWER PRESSURE LESS AREA, HIGHER PRESSURE What the bed of nails demonstration illustrates….

15. outer space Earth ~ 20 mile tall column of air in Earth’s atmosphere creates pressure we all feel Pressure =1 atm at sea level Pressure in `lowbrow’ Chem speak: barometers Edge of Earth’s atmosphere

16. Pressure in `lowbrow’ Chem speak (continued): barometers ( see fig. 5.3, 5.4 of text) Evacuated space P in mm Hg=760 External atmospheric pressure, P Hg A Torricelli barometer

17. Atmosphere demonstrated the Italian Way ( BIG time Torricelli Barometers) Height of Hg columns is independent of cross- sectional areas of tubes Lesson from 17 th century picture Vat of mercury (Hg)

18. 1 atmosphere =760 mm Hg 1 atmosphere=10336 mm H 2 O =33.9 ft H 2 O = 20 miles of air = 15 pounds/in 2 (psi) = 760 torr Pressure units in CHEMISTRY LAND =29.92 inches Hg Other common measures of an atmosphere

19. 1 atmosphere =760 mm Hg = 760 torr =29.92 inches Hg = 15 pounds/in 2 (psi) 28 inches Hg= atm = torr = psi 0.936 711 14.0 P conversions

20. Physics speak 1 atm = 101.3 kPa Confusing Compromise 1 bar =100 kPa= 0.986 atm Pressure units in PHYSICS COUNTRY WE ARE GOING TO STICK WITH ATMOSPHERES (atm) IN THIS COURSE

21. Empiric gas law derivations: the common sense way T down  ?? P, T, V, n n P V down V = bT T up  ?? V up CHARLES’ LAW (p. 195) V 1 /T 1 = V 2 /T 2 V 1 = T 1 V 2 T 2 => T, V vary V T What variables do we follow when describing a gas ???

22. Empiric gas law derivations: the common sense way (cont.) n T V down ??  P up  P down V up ?? BOYLE’S LAW () P = k/V P 1 V 1 =P 2 V 2 P 1 = 1/V 1 P 2 1/V 2 => P, V vary V P

23. Empiric gas law derivations: the common sense way (cont.) n V T up  ?? P up T down  ?? P down P = aT GAY-LUSSAC’S LAW (not in text) P 1 /T 1 = P 2 /T 2 T 1 = P 1 T 2 P 2 => T, P vary P T T in K

24. How Boyle, Gay-Lussac and Charles Laws are reflected in the Combined Gas Law (when n is constant) P 1 V 1 = P 2 V 2 T 1 T 2 constant n,P P 1 V 1 = P 2 V 2 T 1 T 2 Charles’ Law (P 1 =P 2 ) constant n, T Boyle’s Law (T 1 =T 2 ) constant n, V Gay-Lussac’s Law (V 1 =V 2 ) Combined Gas Law constant n P 1 V 1 = P 2 V 2 T 1 T 2 P 1 V 1 = P 2 V 2 T 1 T 2 ConditionsName of Gas Law Gas Law Equation

25. An ideal gas at constant V and P=2 atm is heated from 300 to 600 K. What is the final P ? A.1 atm B.2 atm C.3 atm D.4 atm

26. 1.A sample of oxygen gas is expanded from 20 to 50 liters at constant temperature. The final pressure is 4 atm. What was the initial pressure ? P 1 =10 atm COMBINED GAS LAW PROBLEMS…BOARD WORK

27. A sample of ideal gas arrives at 300K when expanded from 3 to 9 L at constant P. What was the original temperature ? A.900 K B.100 K C.600 K D.150 K

28. 2. A child’s balloon originally occupies 5 liters at sea level (P=1 atm) and room temperature (300 K). It is released and is allowed to rise to an altitude where the pressure is 0.25 atm and the temperature is 150 K. What is the balloon’s new volume ? 10 L COMBINED GAS LAW PROBLEMS…BOARD WORK (CONT.)

29. An ideal gas in a fixed volume and an initial pressure of 10 atm and initial T of 177 C has a final pressure of 3.33 atm. What is the final T(K) (K=C+273) A.150 K B.59 K C.450 K D.531 K

30. 3.The volume of a piston at fixed pressure changes as it is cooled from 500 o C to 250 o C. If the final volume is 6.76 L, what is the initial volume ? V 1 =10 L COMBINED GAS LAW PROBLEMS…BOARD WORK (CONT.)

31. 4. Autoclaves are essentially pressure cookers. At 1 atm, steam has a temperature of 100 o C. Would you expect the pressure to double if the autoclave to attains a steam temperature of 200 o C ? a) NO…must convert C  K…ratio is not 200/100 What pressure do you actually expect to reach at 200 o C? 473.15 =1.73 atm 373.15 COMBINED GAS LAW PROBLEMS…BOARD WORK (CONT.)

32. When n varies…. Blowing up a balloonWhat varies ? n, V (1st time n changes) => P and T are constant n up ? => V up n down ? => V down n V Avogadro’s Law (pp. 289-90) V=an

33. Trickier problem 2.00 L H 2 at 0.625 atm 1.00 L N 2 at 0.200 atm Stopcock closed Stopcock open Final P = ??? 0.483 atm