General Chemistry Principles and Modern Applications Petrucci Harwood

General Chemistry Principles and Modern Applications Petrucci • Harwood • Herring 8 th Edition Chapter 6: Gases Philip Dutton University of Windsor, Canada N 9 B 3 P 4 Prentice-Hall © 2002 (modified 2003 by Dr. Paul Root and 2005 by Dr. David Tramontozzi)

Contents 6 -1 Properties of Gases: Gas Pressure 6 -2 The Simple Gas Laws 6 -3 Combining the Gas Laws: The Ideal Gas Equation and The General Gas Equation 6 -4 Applications of the Ideal Gas Equation 6 -5 Gases in Chemical Reactions 6 -6 Mixtures of Gases

Contents 6 -6 Mixtures of Gases 6 -7 Kinetic—Molecular Theory of Gases 6 -8 Gas Properties Relating to the Kinetic—Molecular Theory 6 -9 Non-ideal (real) Gases Focus on The Chemistry of Air-Bag Systems

6 -5 Gases in Chemical Reactions • Stoichiometric factors relate gas quantities to quantities of other reactants or products. • Ideal gas equation used to relate the amount of a gas to volume, temperature and pressure. • Law of combining volumes can be developed using the gas law (volumes of gases involved in reactions are in the ratio of small whole numbers).

Example 6 -12 Using the Ideal gas Equation in Reaction Stoichiometry Calculations. The decomposition of sodium azide, Na. N 3, at high temperatures produces N 2(g). Together with the necessary devices to initiate the reaction and trap the sodium metal formed, this reaction is used in air-bag safety systems. What volume of N 2(g), measured at 735 mm Hg and 26°C, is produced when 70. 0 g Na. N 3 is decomposed. Δ 2 Na. N 3(s) → 2 Na(l) + 3 N 2(g)

Example 6 -12 Given: P = 735 mm Hg T = 26˚C m. Na. N 3 = 70 g Na. N 3 R = 0. 08206 L atm mol-1 K-1 Determine volume of N 2(g): PV = n. RT Need: Mols of N 2(g) P in atm T in Kelvin

Example 6 -12 Determine moles of Na. N 3: n. Na. N 3 = 70 g Na. N 3 65. 01 g mol-1 = 1. 076 mol Relate to mols of N 2: 3 mol N 2 = 1. 62 mol N 2 1. 076 mol Na. N 3 × 2 mol Na. N 3 Convert P into atm. : 1. 0 atm = 0. 967 atm 735 mm Hg × 760 mm Hg Convert T into Kelvin 26 ˚C + 273. 15 K = 299. 15 K

Example 6 -12 Determine volume of N 2: n. RT (1. 62 mol)(0. 08206 L atm mol-1 K-1)(299. 15 K) V= = P 0. 967 atm = 41. 1 L

6 -6 Mixtures of Gases • Gas laws apply to mixtures of gases. • Simplest approach is to use ntotal, but. . • Partial pressure – Each component of a gas mixture exerts a pressure that it would exert if it were in the container alone.

Dalton’s Law of Partial Pressure

Partial Pressure Ptot = Pa + Pb +… Va = na. RT/Ptot and na na. RT/Ptot Va = = ntot. RT/Ptot ntot Vtot na na. RT/Vtot Pa = = ntot. RT/Vtot ntot Ptot Vtot = Va + Vb+… Recall na = a ntot Mole fraction

Example 6 -15 Calculating partial pressures in a gaseous mixture. What are the partial pressures of H 2 and He in the gaseous mixture described in the following diagram? Method # 1 PH 2 = n. H 2 (RT / V) (0. 5 mol)(0. 0821 L atm mol-1 K-1)(293. 15 K) 5. 0 L = 2. 4 atm PHe = n. He (RT / V) (1. 25 mol)(0. 0821 L atm mol-1 K-1)(293. 15 K) 5. 0 L = 6. 0 atm

Example 6 -15 Calculating partial pressures in a gaseous mixture. What are the partial pressures of H 2 and He in the gaseous mixture described in the following diagram? Method # 2 n. H 2 (Ptotal) ntotal 0. 5 mol (8. 4 atm) 1. 75 mol PH 2 = PHe = = 2. 4 atm n. He (Ptotal) ntotal 1. 25 mol (8. 4 atm) 1. 75 mol = 6. 0 atm

Pneumatic Trough Ptot = Pbar = Pgas + PH 2 O Pgas = Pbar - PH 2 O

Example 6 -16 Collecting gas over a liquid (water). In the following reaction, 81. 2 m. L of O 2(g) is collected over water at 23 ˚C and barometric pressure 751 mm Hg. What must have been the mass of Ag 2 O(s) decomposed? (the vapour pressure of water at 23 ˚C is 21. 1 mm Hg) 2 Ag 2 O(s) 4 Ag(s) + O 2(g) Strategy: 1. Calculate moles of O 2(g) 2. Relate moles of O 2(g) to moles of Ag 2 O 3. Convert moles of Ag 2 O to mass

Example 6 -16 1. Calculating Moles of O 2(g) PO 2 = Pbar – PH 2 O = 751 mm Hg – 21. 1 mm Hg = 729. 9 mm Hg PO 2 = 729. 9 mm Hg (1 atm / 760 mm Hg) = 0. 960 atm V = 81. 2 m. L = 0. 0812 L R = 0. 08206 L atm mol-1 K-1 T = 23 ˚C + 273. 15 = 296. 15 K n= (0. 960 atm)(0. 0812 L) (0. 08206 L atm mol-1 K-1)(296. 15 K) 0. 00321 mol O 2 gas

Example 6 -16 2. Relate to moles of Ag 2 O n. Ag 2 O = 0. 00321 mol O 2 (2 mol Ag 2 O / 1 mol O 2) = 0. 00642 mol Ag 2 O 3. Convert moles Ag 2 O to mass m = 0. 00642 mol (231. 7 g/mol) = 1. 49 g Ag 2 O

6 -7 Kinetic Molecular Theory of Gases • Gas particles are point masses in constant, random, straight line motion. • Gas particles are separated by great distances. • Collisions are rapid and elastic. • No force between particles. • Total energy remains constant.

Pressure – Assessing Collision Forces • Translational kinetic energy • Frequency of collisions • Impulse or momentum transfer • Pressure proportional to impulse times frequency m = mass u = speed N = # molecules V = Volume

Pressure and Molecular Speed • Three dimensional systems lead to: um is the modal speed uav is the simple average urms

Pressure Assume one mole (and multiply by V): 1 2 = PV NAm u 3 PV=RT so: 2 = 3 RT N A m u NAm = M: 2 = 3 RT M u Rearrange: u rms = 3 RT M

Distribution of Molecular Speeds Use R = 8. 3145 J mol-1 K-1 and MW in kg mol-1

Determining Molecular Speed