1. Agenda 4/2/2015 Slip Quiz Part 1 Avogadro’s Law and Calculations Check Gas Law Calculations Continued Test Preparation Packet Homework Slip quiz Part 2

2. Slip Quiz Part 1 State the ideal gas law and the assumptions of the Kinetic Molecular- theory for ideal gases that underpin it.

3. Slip Quiz Part 1 State the ideal gas law and the assumptions of the Kinetic Molecular- theory for ideal gases that underpin it. The Ideal gas law is PV = nRT where P is pressure, V is volume, n is amount of gas (or number of moles of gas), R is the ideal gas constant (determined experimentally) and P is pressure.

4. The assumptions of the kinetic molecular theory that underpin the ideal gas law are that for an ideal gas: The volume of each particle is negligible compared the distances between the particles There are negligible forces of attraction between particles The kinetic energy of the particles is ½ mv 2 and is proportional to the temperature Collisions between particles are perfectly elastic

5. Avogadro’s Principle Chapter 14 Section 14.2 Starts at Page 430 And Gas Stoichiometry 14.4 (p 440)

6. Avogadro’s Principle Equal volumes of gases at the same temperature and pressure contain equal numbers of particles.

7. Avogadro’s Principle Equal volumes of gases at the same temperature and pressure contain equal numbers of particles. Ideal Gas Equation PV = nRT V = nRT P

8. Avogadro’s Principle Equal volumes of gases at the same temperature and pressure contain equal numbers of particles. Ideal Gas Equation PV = nRT V = nRT P If V gas1 = V gas2 then n gas1 = n gas2

9. Remember one mole of anything contains 6.02 x 10 23 particles Molar volume for a gas is the volume that one mole occupies at 0.00°C and 1.00 atm pressure. –These conditions of temperature and pressure are known as standard temperature and pressure (STP)

10. One mole of any gas occupies a volume of 22.4 L at STP Very useful!!!! Example: 1 mole oxygen occupies 22.4 L at STP and contains 6.02 x 10 23 molecules of O 2 Avogadro showed experimentally that:

11. 1 mole oxygen occupies 22.4 L at STP and contains 6.02 x 10 23 molecules of O 2 1 mole Carbon dioxide occupies…….L at STP and contains………………………molecules CO 2 2 moles oxygen - ….. at STP and contains ……………….. molecules O 2 3 moles Carbon dioxide - …….L at STP and contains………………………molecules CO 2

12. 1 mole oxygen occupies 22.4 L at STP and contains 6.02 x 10 23 molecules of O 2 1 mole Carbon dioxide occupies 22.4 L at STP and contains 6.02 x 10 23 molecules CO 2 2 moles oxygen - 44.8L at STP and contains 1.20 x 10 24 molecules O 2 3 moles Carbon dioxide - 67.2 L at STP and contains 1.80 x 10 24 molecules CO 2

13. Assessment: 1) Standard temperature and pressure (STP) are defined as Possible answers?

14. Assessment: 1) Standard temperature and pressure (STP) are defined as Possible answers? 0°C and 1 atm pressure 273 K and 1 atm pressure 0°C and 760 mm Hg pressure 273 K and 760 mm Hg pressure 0°C and 101.325 kPa pressure

15. Assessment: 2) Under which of the following sets of conditions will a 1 mole sample of {any gas} occupy a volume of 22.4 liters? Possible answers? This is the definition of molar volume – that 1 mole of any gas occupies 22.4 L at STP – appears on your reference sheet under Constants Volume of Ideal Gas at STP 22.4 L mol -1

16. Assessment: 2) Under which of the following sets of conditions will a 1 mole sample of {any gas} occupy a volume of 22.4 liters? Possible answers? STP 0°C and 1 atm pressure (typically this) 273 K and 1 atm pressure 0°C and 760 mm Hg pressure 273 K and 760 mm Hg pressure 0°C and 101.325 kPa pressure

17. Assessment: 3) Under which of the following sets of conditions will a 0.5 mole sample of {any gas} occupy a volume of 11.2 liters? Possible answers? Look carefully If 0.5 mole occupies 11.2 liters at this T and P What would 1 mole of this gas occupy?

18. Assessment: 3) Under which of the following sets of conditions will a 0.5 mole sample of {any gas} occupy a volume of 11.2 liters? Possible answers? Look carefully If 0.5 mole occupies 11.2 liters at this T and P What would 1 mole of this gas occupy? twice as much = 22.4 liters = molar volume

19. Assessment: 3) Under which of the following sets of conditions will a 0.5 mole sample of {any gas} occupy a volume of 11.2 liters? Possible answers? Any answer that shows Standard temperature and pressure (STP) 0°C and 1 atm pressure Or equivalent in other common units

20. Assessment: 4) A sample of carbon dioxide gas occupies a volume of 20 L at standard temperature and pressure at (STP). What will be the volume of a sample of argon gas that has the same number of moles and pressure but twice the absolute temperature?

21. Using PV = nRT For the problem we have same number of moles of CO 2 and Argon = n, P doesn’t change PV = nRT or P = T nR V If T is doubled V must double to keep the ratio the same

22. OR Combined gas law – is for a fixed amount of gas – a fixed number of moles of gas P 1 V 1 = P 2 V 2 = constant T 1 T 2 P.V = (Constant).T

23. OR 4) A sample of carbon dioxide gas occupies a volume of 20 L at standard temperature and pressure at (STP). What will be the volume of a sample of argon gas that has the same number of moles and pressure but twice the absolute temperature? Use Avogadro’s principle – If equal volumes of gas contain equal numbers of particles under same conditions of T and P, then Equal numbers of moles of gas will occupy the same volume at the same T and P.

24. Assessment: 4) A sample of carbon dioxide gas occupies a volume of 20 L at standard temperature and pressure at (STP). What will be the volume of a sample of argon gas that has the same number of moles and pressure but twice the absolute temperature? CO 2 V = 20 L at STP (0°C and 1 atm) Ar (same number of moles as the CO 2 ) so would also be same volume at STP V 1 = 20 L at STP

25. Assessment: 4) …What will be the volume of a sample of argon gas that has the same number of moles and pressure but twice the absolute temperature? Ar V 1 = 20 L at STPV 2 = P 1 = 1atmP 2 = 1 atm T 1 = 273KT 2 = 2 x 273K = 546 K

26. Assessment: 4) cont. Ar V 1 = 20 L at STPV 2 = P 1 = 1atmP 2 = 1 atm T 1 = 273KT 2 = 2 x 273K = 546 K 1atm. 20L = 1atm.V 2 273 K546 K V 2 = 1atm. 20L. 546 K= 40L 1atm. 273 K

27. Gas Stoichiometry CH 4(g) + 2 O 2(g)  CO 2(g) + 2 H 2 O (g) 1 mole2 moles 1 volume2 volumes1volume2 volumes

28. CH 4(g) + 2 O 2(g)  CO 2(g) + 2 H 2 O (g) 1 volume2 volumes1volume2 volumes Ex. Qu.: What volume of oxygen gas is needed for the complete combustion of 4.00 L of methane gas? (Assume constant temperature and pressure.) Look at relevant ratios in equation 1 : 2 or 1 volume of CH 4 will need 2 volumes O 2 1L:2L so4L:2L x 4 = 8L Answer: 4 L of methane will require 8.00 L of oxygen for complete combustion.

29. Ex. Qu. 2: What volume of oxygen gas is needed for the complete combustion of 4.00 L of ethane gas? (Assume constant temperature and pressure.) 2 C 2 H 6(g) + 5 O 2(g)  4 CO 2(g) + 6 H 2 O (g) 2 volumes : 5 volumes 5L Oxygen = xL oxygen 2L ethane 4.00L ethane 5L O 2 4.00LC 2 H 6 = xL oxygen 2LC 2 H 6

30. Ex. Qu. 2: What volume of oxygen gas is needed for the complete combustion of 4.00 L of ethane gas? (Assume constant temperature and pressure.) 2 C 2 H 6(g) + 5 O 2(g)  4 CO 2(g) + 6 H 2 O (g) 5L Oxygen = xL oxygen 2L ethane 4.00L ethane 5 L O 2 4.00LC 2 H 6 = xL oxygen 2 LC 2 H 6 = 10 L Oxygen

31. Ex. Qu. 3: What volume of oxygen gas at STP is needed for the complete reaction of 107.92g of aluminum in the reaction below: 4 Al (s) + 3 O 2(g)  2 Al 2 O 3(s) 4 moles: 3 moles Look up atomic mass of aluminum 26.98 amu – remember 1 mole is atomic mass in g So here, 4 moles of Al would be 4 x (26.98g) = 107.92g Which is what we are provided with (yeah!)

32. Ex. Qu. 3: What volume of oxygen gas at STP is needed for the complete reaction of 107.92g of aluminum in the reaction below: 4 Al (s) + 3 O 2(g)  2 Al 2 O 3(s) 4 moles: 3 moles 107.92g Al needs 3 moles Oxgyen to react completely At STP 1 mole of any gas = 22.4 L 3 moles = 3 x 22.4 L = 67.2 L

33. Ex. Qu. 3: What volume of oxygen gas at STP is needed for the complete reaction of 107.92g of aluminum in the reaction below: 4 Al (s) + 3 O 2(g)  2 Al 2 O 3(s) 4 moles: 3 moles Answer: 67.2 L of oxygen at STP will be needed for complete reaction with 107.92 g of aluminum. Better to use this approach – always correct: 107.92gAl x 1mol Al x 3 mol O 2 x 22. 4L = 67.2L O 2 26.98g 4mol Al 1molO 2

34. Assessement Section 13.4 Phase changes 1.Why are fountains often found in enclosed courtyards of houses in hot and dry countries. (Hint: evaporation requires energy.) 2.Rank the boiling points of water for each of the following locations increasing order. Daytona Beach, Florida (altitude:sea level); Death Valley, California (altitude 86m below sea level); Mount Shasta, California (altitude: 669 m). Explain your reasoning.

35. 1. Why are fountains often found in enclosed courtyards of houses in hot and dry countries? All write Fountains are often found in enclosed courtyards of houses in hot and dry countries because as the water from the fountain falls through the air it absorbs energy from the hot air which cools the air. The liquid water uses the energy it absorbs from the air to change phase and become water vapor molecules via evaporation. (This process also makes the air less dry.)

36. 2.Mount Shasta (669m) lowest b.pt (less than 1 atm, less than 100°C) Daytona Beach (sea level, 1 atm) normal boiling point for water, 100°C. Death Valley (86 m below sea level, greater than 1atom) boiling point higher than 100°C. “The temperature at which the vapor pressure of a liquid equals the external or atmospheric pressure is called the boiling point.” (p. 406) At high altitude (like Mount Shasta) water will boil at a temperature below 100°C because

37. …its molecules will escape from the liquid phase in sufficient numbers to exert a vapor pressure equal to the external pressure at a lower temperature since the atmospheric pressure at high altitude will be less than 1 atm. At Daytona Beach (sea level) the atmospheric pressure will be 1 atm and so the water will boil at its normal value, 100°C. However, in Death Valley (86 m below sea level) the external pressure will be great than 1 atm and so the boiling point of the water will be higher than 100°C as more…

38. …energy will be required to cause enough molecules to escape from the liquid phase in sufficient numbers to exert a vapor pressure equal to the external pressure.

39. Structured Response Preparation 1.a. The arrow represents the phase transition from solid (graphite) to liquid which is melting.

40. Structured Response cont. 1b. What is the phase of carbon at a temperature of 1000 K and a pressure of 10 8 Pa? Phase is solid graphite.

41. Structured Response cont. 1c. How many phases of carbon are represented in the phase diagram and what are they? 4 phases of carbon Diamond(solid) Graphite (solid) Liquid Vapor

42. More than one possible correct answer. Graphite at a pressure of 10 9 Pa, heat to 2000K and then decrease pressure below 10 7 Pa. Continue to heat graphite past 5000K and then increase pressure of the vapor to the 10 7 Pa required.

43. Structured Response Preparation 2. Using Dalton’s law of partial pressures Pressure total = Pressure CO2 + Pressure H2O Or PressureCO 2 = Pressure total –PressureH 2 O =100.00kPa –3.15kPa =

44. 2. b. Collecting carbon dioxide using Displacement of Water http://www.docbrown.info/page13/ChemicalTests/GasPrepa ration.htm Why does the collection vessel have to be inverted?

45. Structured response 3. Differentiate among dispersion forces, dipole-dipole forces, and hydrogen bonds. Dispersion forces are momentary attractions between molecules that occur as a result of temporary shifts in the electron clouds in the molecules which cause temporary dipoles in molecules. All molecules are subject to dispersion forces.

46. Structured Response 3. cont. Dipole-dipole forces only occur between polar molecules where the molecules have oppositely charged ends as a result of unequal sharing of electrons in covalent bonds as a result of differences in electronegativity of the atoms involved. Dipole-dipole forces can exist for longer than dispersion forces.

47. Cont. Hydrogen bonds are special dipole-dipole forces that can only be formed when a hydrogen atom is covalently bonded to a small, highly electronegative atom with at least one unbonded pair of electrons on the electronegative atom (O, F or N). Hydrogen bonds result in intermolecular attractions that last for the longest amount of time and have greatest effects on physical properties of materials.

48. Hydrogen bonding in water Accounts for relatively high melting point, boiling point, properties as a solvent

49. Homework Review, study, practice – Ch 13 and 14 test preparation. Use notes, use your assessment packet and study guide, use multiple choice questions, use online test questions, flash cards etc. Actively, interactively, thoughtfully study. Just looking over your book or notes is of little use.

50. Slip Quiz Part 2 1.The temperature at which all molecular motion stops is A 273 KB 273°C C 0KD freezing pt. And this temperature is called ___________. (2 words) 2. 1 mole of any gas occupies…at STP. A.6.03 x 10 23 L B 22.4 L C PV = nRTC 1 L