1. Honors Chemistry, Chapter 11 Page 1 Chapter 11 – Molecular Composition of Gases

2. Honors Chemistry, Chapter 11 Page 2 Gay-Lussac’s Law of Combining Volumes of Gases hydrogen gas + oxygen gas  water vapor 2 volumes + 1 volume  2 volumes hydrogen + chlorine  hydrogen chloride gas 1 volume + 1 volume  2 volumes Gay-Lussac’s law of combining volumes of gases states that at constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers.

3. Honors Chemistry, Chapter 11 Page 3 Avogadro’s Law Avogadro’s law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. V = kn Where V is volume n is the number of moles k is a constant

4. Honors Chemistry, Chapter 11 Page 4 Avogadro’s Law Assume hydrogen, oxygen and chlorine are diatomic molecules. H 2 (g) + Cl 2 (g)  2HCl(g) 1 vol. + 1 vol.  2 vol. and 2H 2 (g) + O 2 (g)  2H 2 O(g) 2 vol. + 1 vol.  2 vol.

5. Honors Chemistry, Chapter 11 Page 5 Molar Volume The volume occupied by one mole of a gas at STP is known as the standard molar volume of a gas and has a volume of 22.41410 L (22.4 L for our purposes). (STP – Standard Temperature and Pressure: 0 o C. and 1 atm.)

6. Honors Chemistry, Chapter 11 Page 6 Sample Problem 11-1 What is the volume of 0.0680 mole of oxygen gas at STP? moles of O 2  volume of O 2 in liters 0.068 mol O 2 x 22.4 L/mol = 1.52 L O 2

7. Honors Chemistry, Chapter 11 Page 7 Sample Problem 11-2 What is the mass in grams of 98.0 mL of SO 2 at STP? Vol. of SO 2 in liters  mol of SO 2  mass of SO 2 mol mass of SO 2 =32.00+32.07= 64.07 g/mol 98.0 mL x 1L/1000 mL x 1 mol SO 2 / 22.4 L x 64.07 g SO 2 / mol SO 2 = 0.280 g SO 2

8. Honors Chemistry, Chapter 11 Page 8 Thought Problem Consider a container with mass of 1 kg and an internal volume of 22.4 L. : If the container is filled with air, how much water would the container displace if placed in a container of water?

9. Honors Chemistry, Chapter 11 Page 9 Thought Problem How much would the container weigh when filled with air? How much would the container weigh when evacuated? How much would the container weigh when filled with nitrogen? oxygen? hydrogen?

10. Honors Chemistry, Chapter 11 Page 10 Chapter 11, Section 1 Review 1.State the law of combining volumes. 2.State Avogadro’s law and explain its significance. 3.Define standard molar volume of a gas, and use it to calculate gas masses and volumes. 4.Use standard molar volume to calculate the molar mass of a gas.

11. Honors Chemistry, Chapter 11 Page 11 Idea Gas Law Boyle’s Law: V  1/P Charles’s Law: V  T Avogadro’s Law: V  n or V  nT/P orV = nRT/P or PV = nRT Where R is the Ideal Gas Constant

12. Honors Chemistry, Chapter 11 Page 12 Ideal Gas Constant R = _PV_ = _(1 atm)(22.4140 L)_ nT (1 mol)(273.15 K) = 0.082057 L atm/(mol K)

13. Honors Chemistry, Chapter 11 Page 13 Numerical Values of the Ideal Gas Constant Numerical Value of RUnits 0.082(L atm)/(mol K) 62.4[L (mm of Hg)]/(mol K) 8.314(L kPa) / (mol K) 8.314( J ) / (mol K) Note: 1L atm = 101.325 J 1 J = 1 Pa m 3

14. Honors Chemistry, Chapter 11 Page 14 Sample Problem 11-3 What is the pressure in atmospheres exerted by 0.500 mol of nitrogen gas in a 10.0 L container at 298 K? V = 10.0 L n = 0.500 mol of N 2 T = 298 K P = nRT/V = 0.500 mol N 2 x 0.0821 L atm/ (mol K) x 298 K / 10.0 L = 1.22 atm

15. Honors Chemistry, Chapter 11 Page 15 Sample Problem 11-4 What is the volume, in liters, of 0.250 mol of oxygen at 20 o C. and 0.974 atm pressure? P = 0.974 atm n = 0.250 mol of Oxygen T = 20 o C. or 273.2 + 20 = 293.2 K V = nRT/P = 0.250 x 0.0821 (L atm)/(mol K) x 293.2 K / 0.974 atm = 6.17 L O 2

16. Honors Chemistry, Chapter 11 Page 16 Molar Mass PV = nRT = m RT / M (n = m / M) Where m is mass and M is molar mass Solve for M M = m RT/(PV)

17. Honors Chemistry, Chapter 11 Page 17 Gas Density D = m/V m = DV PV = m RT/ M PV = DV RT/M D = M P / (RT)

18. Honors Chemistry, Chapter 11 Page 18 Problem 11-6 At 28 o C. and 0.974 atm, 1.00 L of a gas has a mass of 5.16 g. What is the molar mass of this gas? What is the gas? P = 0.974 atm V = 1.00 L T = 28 o C. + 273 = 301 K m = 5.16 g. M = m RT/PV = 5.16 g x [0.0821 L atm / (mol K)] x 301 K / ( 0.974 atm x 1.00 L) = 131 g/mol

19. Honors Chemistry, Chapter 11 Page 19 Example Density Problem The density of a gas was found to be 2.0 g/L at 1.50 atm and 27 o C. What is the molar mass of the gas? What is the gas? D = 2.0 g/L T = 27 o C. + 273 = 300 K P = 1.50 atm D = M P / (RT) M = DRT / P = 2.0 g/L x [0.0821 L atm / (mol K)] x 300 K / 1.50 atm = 33 g/mol

20. Honors Chemistry, Chapter 11 Page 20 Chapter 11, Section 2 Review 1.State the ideal gas law. 2.Derive the gas constant and discuss the units. 3.Using the ideal gas law, calculate pressure, volume, temperature or amount of gas when the other three quantities are known.

21. Honors Chemistry, Chapter 11 Page 21 Chapter 11, Section 2 Review 4.Using the ideal gas law, calculate the molar mass or density of a gas. 5.Reduce the ideal gas law to Boyle’s law, Charles’s law, and Avogadro’s law. Describe the conditions under which each applies.

22. Honors Chemistry, Chapter 11 Page 22 Stoichiometry of Gases Coefficients indicate molecule, mole, and volume ratios in gas reactions. For Example: 2CO(g) + O 2 (g)  2CO 2 (g) 2 molecules 1 molecule 2 molecules 2 mol 1 mol 2 mol 2 volumes 1 volume 2 volumes

23. Honors Chemistry, Chapter 11 Page 23 Volume Ratios Volume Ratios from the CO + O 2 Reaction 2 vol CO / 1 vol O 2 or 1 vol O 2 / 2 vol CO 2 vol CO / 2 vol CO 2 or 2 vol CO 2 / 2 vol CO 1 vol O 2 / 2 vol CO 2 or 2 vol CO 2 / 1 vol O 2

24. Honors Chemistry, Chapter 11 Page 24 Sample Problem 11-7 What is the volume, in liters, of oxygen required for the complete combustion of 0.350 L of propane? What will the volume of CO2 be? (Assume constant T and P.) C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) 0.350L C 3 H 8 x 5 vol O 2 / 1 vol C 3 H 8 = 1.75 L O 2 0.350L C 3 H 8 x 3 vol CO 2 / 1vol C 3 H 8 =1.05 L CO 2

25. Honors Chemistry, Chapter 11 Page 25 Volume-Mass and Mass-Volume Calculations We can use the ideal gas law to calculate problems like: gas vol A  moles A  moles B  mass B or mass A  moles A  moles B  gas vol B

26. Honors Chemistry, Chapter 11 Page 26 Sample Problem 11-8 How many grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide at STP? CaCO 3 (s)  CaO(s) + CO 2 (g) n = PV/RT = (1atm)(5.00 L CO 2 ) / [ 0.0821 L atm /(mol K)]/(273 K) = 0.223 mol CO 2 or n = 5.00L CO 2 / (22.4 L/mol)=0.233 mol CO 2

27. Honors Chemistry, Chapter 11 Page 27 Sample Problem 11-8 continued Molecular Weight of CaCO 3 ? 100.09 g CaCO 3 /mol 0.223 mol CO 2 x 1 mol CaCO 3 / (1 mol CO 2 ) x 100.09 g CaCO 3 /1 mol CaCO 3 = 22.4 g CaCO 3

28. Honors Chemistry, Chapter 11 Page 28 Sample Problem 11-9 How many liters of hydrogen at at 35 o C. and 0.980 atm are needed to completely react with 875 grams of tungsten oxide? WO 3 (s) + 3H 2 (g)  W(s) + 3H 2 O(l) Molar Mass of WO 3 ? 231.84 g/mol 875 g WO 3 x 1 mol WO 3 / (231.84 g WO 3 ) x 3 mol H 2 / (1 mol WO 3 ) = 11.3 mol H 2

29. Honors Chemistry, Chapter 11 Page 29 Sample Problem 11-9 Continued V = nRT/P = (11.3 mol H 2 ) x [0.0821 L atm / (mol K)] x 308 K / (0.980 atm) = 292 L H 2

30. Honors Chemistry, Chapter 11 Page 30 Chapter 11, Section 3 Review 1.Explain how Gay-Lussac’s law and Avogadro’s law apply to volumes of gases in chemical reactions. 2.Use a chemical equation to specify volume ratios for gaseous reactants or products, or both. 3.Use volume ratios and the gas laws to calculate volumes, masses, or molar amounts of gaseous reactants or products.

31. Honors Chemistry, Chapter 11 Page 31 Diffusion and Effusion Diffusion is the process where two gases gradually mix spontaneously due to the constant motion of the gas molecules. Effusion is the process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container.

32. Honors Chemistry, Chapter 11 Page 32 Average Kinetic Energy For two gases at the same temperature: Avg kinetic energy = ½ M A v A 2 = ½ M B v B 2 M A v A 2 = M B v B 2 v A 2 / v B 2 = M B / M A v A M B --------- = ------------ v B M A

33. Honors Chemistry, Chapter 11 Page 33 Graham’s Law of Effusion Graham’s law of effusion states that the rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses. Rate of effusion of A M B Density B ---------------------------- = ------- = ------------- Rate of effusion of B M A Density A

34. Honors Chemistry, Chapter 11 Page 34 Sample Problem 11-10 Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure. Rate of effusion of H 2 M O2 32.00 g/mol ---------------------------- = ------- = ------------- Rate of effusion of O 2 M H2 2.02 g/mol = 3.98

35. Honors Chemistry, Chapter 11 Page 35 Chapter 11, Section 4 Review 1.State Graham’s law of effusion. 2.Determine the relative rates of effusion of two gases of know molar masses. 3.State the relationship between the molecular velocities of two gases and their molar masses.