1. Agenda 3/10/2014 Slip Quiz 5 AdayPart 1 Gas Law calculations- any questions? Avogadro’s Principle – notes Ch 14 Study Guide Ch 14 Assessment Question: Applying Scientific Methods The Haber Process Homework Slip quiz 5 AdayPart 2

2. 5A Day Part 1 1.Which of the following processes involves the release of energy when a material undergoes the changes? A evaporation and melting B evaporation and freezing C condensation and freezing D condensation and evaporation 2. The Combined Gas Law is shown by A.KE = ½ mv 2 B PV = constant C PV = nRTD P 1 V 1 = P 2 V 2 T 1 T 2

3. 5 Aday- Part 1 1.Condensation and freezing involve the release of energy when a material undergoes the changes. 2. The Combined Gas Law is shown by … P 1 V 1 = P 2 V 2 T 1 T 2 Gas Liquid release energy Solid

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

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

6. 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

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 If V gas1 = V gas2 then n gas1 = n gas2

8. 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)

9. 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:

10. 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

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 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

12. Test type questions: 1) Standard temperature and pressure (STP) are defined as Possible answers?

13. Test type questions: 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

14. Test type questions: 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

15. Test type questions: 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

16. Test type questions: 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?

17. Test type questions: 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

18. Test type questions: 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

19. Test type questions: 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?

20. 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

21. 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

22. 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.

23. Test type questions: 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

24. Test type questions: 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

25. Test type questions: 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

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

27. CH 4(g) + 2 O 2(g)  CO 2(g) + 2 H 2 O (g) 1 volume2 volumes 1volume2 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.

28. 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) + 7O 2(g)  4 CO 2(g) + 6 H 2 O (g) 2 volumes :7 volumes Or 1 vol : 7/2 volumes 1L : 7/2 L 4L : 7/2 x 4

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) + 7 O 2(g)  4 CO 2(g) + 6 H 2 O (g) 2 volumes : 7 volumes Or 1 vol : 7/2 volumes 1.00L : 7/2 L 4.00L : 7/2 x 4 L Answer: 4.00 L of ethane will require 14.00 L of oxygen for complete combustion.

30. 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 107.92 g x 1 mol = 4 mol 26.98g Which is what we need according to the equation (yeah!)

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 107.92g Al needs 3 moles oxgyen to react completely At STP Molar volme = 22.4 L/mol 3 mol x 22.4 L = 67.2 L 1mol

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 Answer: 67.2 L of oxygen at STP will be needed for complete reaction with 107.92 g of aluminum.

33. Ch 14 Assessment Sheet Applying Scientific Methods 1.Write a balanced chemical equation for the chemical reaction that occurs in the Haber process. Show the state of each reactant and product. 3 H 2(g) + N 2(g)  2 NH 3(g) 3x2 = 6 H2N 2N2x3=6 H (balanced)

34. Ch 14 Assessment Sheet Applying Scientific Methods 2. Heated nitrogen and hydrogen gases are reduced in volume in the compressor. What effect do these changes in temperature and volume have on the pressure of the gas. As the molecules are heated they will move faster (higher average kinetic energy) and as they are compressed into a smaller volume they are closer together and will collide with each other and the walls of the container more often. These 2 effects will both serve to increase the pressure of the gas.

35. Ch 14 Assessment Sheet Applying Scientific Methods 3. The return pump moves any unreacted hydrogen and nitrogen gases from the cooler back into the compressor. What effect will removing gases from the cooler have on the pressure of the gases in the cooler? Explain. The pressure of the gases in the cooler will decrease as the unreacted gases are removed since there will be fewer gas molecules left in the cooler and pressure is directly proportional to the number of gas particles (ideal gas equation PV=nRT, where P is pressure, V is volume, n is number of moles of gas, R is the gas constant, and T the temperature in Kelvins).

36. Ch 14 Assessment Sheet Applying Scientific Methods 4. Ammonia is a real gas. What will happen to it if the pressure continues to be increased and the temperature continues to be decreased. At high pressures and low temperatures real gases deviate more and more from ideal behavior as the intermolecular forces start to influence the behavior of the real gas under these conditions. Ammonia has hydrogen atoms bonded to a small and highly electronegative atom (nitrogen) which also has a lone pair of electrons available. This means that one ammonia molecule can form a hydrogen bond with another molecule of ammonia. The intermolecular attractions between ammonia molecules will result in it liquefying at high pressures and low temperatures.

37. Ch 14 Assessment Sheet Applying Scientific Methods 5. What volume of ammonia gas will be produced from 2400 L of hydrogen gas at the same temperature and pressure? From the balanced equation we know 3 volumes of H 2 produce 2 volumes of NH 3 1 vol. : 2/3 vol. 2400 L: 2/3 (2400 l) = 1 600 L NH 3(g) produced from 2 400 L of H 2(g)

38. Ch 14 Assessment Sheet Applying Scientific Methods 6. The Haber process occurs at a pressure of approximately 800.0 atm and a temperature of 400.0°C. Assume a gas sample occupies 25.0 L at these conditions. What volume will the sample occupy at STP? P 1 = 800.00 atmP 2 = 1atm V 1 = 25.0LV 2 = ? T 1 = 400.00 + 273 T 2 = 273K = 673K Using combined gas law P 1 V 1 = P 2 V 2 T 1 T 2 Solving for V 2 V 2 = T 2 P 1 V 1 = 800atm 25L 273K T 1 P 2 673K 1atm V 2 = 8 110 L (3 sig fig)

39. Ch 14 Assessment Sheet Applying Scientific Methods 7. If 126 L of ammonia is produced at 800.0 atm and a temperature of 400.0°C in the Haber process, what mass of hydrogen was used in the reaction. Thinking ahead to needing to convert volume to mass, we want to know the volume of gas at STP so we can compare it to the molar volume (22.4 l/mol). So first let’s find the volume that 126L of ammonia would occupy at STP P 1 = 800 atmP 2 = 1atm V 1 = 126 LV 2 = ? T 1 = 400.00 + 273 = 673KT 2 = 273K Use combined gas lawP 1 V 1 = P 2 V 2 T 1 T 2

40. Ch 14 Assessment Sheet Applying Scientific Methods 7. If 126 L of ammonia is produced at 800.0 atm and a temperature of 400.0°C in the Haber process, what mass of hydrogen was used in the reaction. Solving for V 2 V 2 = T 2 P 1 V 1 = 800 atm 126 L 273K T 1 P 2 673K 1atm V 2 = 40 900L (3 sig fig) 40 900L x 1mol = 1 830 mol 22.4L 126 L of ammonia 800 atm and 400°C is 1 830 mol at STP From equation 2 mol NH 3 : 3 mol H 2

41. Ch 14 Assessment Sheet Applying Scientific Methods 7. If 126 L of ammonia is produced at 800.0 atm and a temperature of 400.0°C in the Haber process, what mass of hydrogen was used in the reaction. 126 L of ammonia 800 atm and 400°C is 1 830 mol at STP From equation 2 mol NH 3 : 3 mol H 2 1 mol: 3/2 mol Now convert mol of H 2 to mass using atomic mass of H = 1amu 1830 mol: 3/2 mol x 1830 x 2g 1mol = 5490g The answer will depend on how many steps you combine before rounding your answer to 3 sig figs, and whether you use 1amu or 1.01 for atomic mass of H.

42. 5 Adays 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