General Chemistry Principles and Modern Applications Petrucci Harwood Herring 8 th Edition Chapter 6: Gases Philip Dutton University of Windsor, Canada N9B 3P4 Prentice-Hall 2002 Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 1 of 41
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 Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 2 of 41
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 Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 3 of 41
6-1 Properties of Gases: Gas Pressure Gas Pressure Force (N) P (Pa) = Area (m2 ) Liquid Pressure P = g h d Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 4 of 41
Barometric Pressure Standard Atmospheric Pressure 1.00 atm 760 mm Hg, 760 torr 101.325 kpa 1.01325 bar 1013.25 mbar Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 5 of 41
Manometers Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 6 of 41
6-2 Simple Gas Laws Boyle 1662 P % 1 V PV = constant Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 7 of 41
Example 5-6 Relating Gas Volume and Pressure Boyle s Law. P 1 V 1 = P 2 V 2 V 2 = P 1V 1 P 2 = 694 L V tank = 644 L Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 8 of 41
Charles s Law Charles 1787 Gay-Lussac 1802 V % T V = b T Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 9 of 41
STP Gas properties depend on conditions. Define standard conditions of temperature and pressure (STP). P = 1 atm = 760 mm Hg T = 0 C = 273.15 K Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 10 of 41
Avogadro s Law Gay-Lussac 1808 Small volumes of gases react in the ratio of small whole numbers. Avogadro 1811 Equal volumes of gases have equal numbers of molecules and Gas molecules may break up when they react. Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 11 of 41
Formation of Water Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 12 of 41
Avogadro s Law At an a fixed temperature and pressure: V % n or V = c n At STP 1 mol gas = 22.4 L gas Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 13 of 41
6-3 Combining the Gas Laws: The Ideal Gas Equation and the General Gas Equation Boyle s law V % 1/P Charles s law V % T Avogadro s law V % n V % nt P PV = nrt Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 14 of 41
The Gas Constant R = PV nt PV = nrt = 0.082057 L atm mol -1 K -1 = 8.3145 m 3 Pa mol -1 K -1 = 8.3145 J mol -1 K -1 Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 15 of 41
The General Gas Equation P R = = P 2 V 1 V 1 2 n 1 T 1 n 2 T 2 If we hold the amount and volume constant: P 1 T 1 = P 2 T 2 Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 16 of 41
6-4 Applications of the Ideal Gas Equation Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 17 of 41
Molar Mass Determination PV = nrt and n = m M PV = m M RT M = m RT PV Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 18 of 41
Example 6-10 Determining a Molar Mass with the Ideal Gas Equation. Polypropylene is an important commercial chemical. It is used in the synthesis of other organic chemicals and in plastics production. A glass vessel weighs 40.1305 g when clean, dry and evacuated; it weighs 138.2410 when filled with water at 25 C (ä=0.9970 g cm -3 ) and 40.2959 g when filled with propylene gas at 740.3 mm Hg and 24.0 C. What is the molar mass of polypropylene? Strategy: Determine V flask. Determine m gas. Use the Gas Equation. Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 19 of 41
Example 5-6 Determine V flask : V flask = m H 2O ) d H2O = (138.2410 g 40.1305 g) ) (0.9970 g cm-3 ) = 98.41 cm 3 = 0.09841 L Determine m gas : m gas = m filled - m empty = (40.2959 g 40.1305 g) = 0.1654 g Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 20 of 41
Example 5-6 Use the Gas Equation: PV = nrt PV = m M RT M = m RT PV M = (0.6145 g)(0.08206 L atm mol -1 K -1 )(297.2 K) (0.9741 atm)(0.09841 L) M = 42.08 g/mol Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 21 of 41
Gas Densities PV = nrt and d = m V, n = m M PV = m M RT m = d = V MP RT Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 22 of 41
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. Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 23 of 41
Example 6-10 Using the Ideal gas Equation in Reaction Stoichiometry Calculations. The decomposition of sodium azide, NaN 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 NaN 3 is decomposed. 2 NaN 3 (s) 2 Na(l) + 3 N 2 (g) Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 24 of 41
Example 6-10 Determine moles of N 2 : n N 2 = 70 g N 3 H 1 mol NaN 3 65.01 g N 3 /mol N 3 H 3 mol N 2 2 mol NaN 3 = 1.62 mol N 2 Determine volume of N 2 : nrt V = = P (1.62 mol)(0.08206 L atm mol -1 K -1 )(299 K) (735 mm Hg) 1.00 atm 760 mm Hg = 41.1 L Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 25 of 41
6-6 Mixtures of Gases Gas laws apply to mixtures of gases. Simplest approach is to use n total, but... Partial pressure Each component of a gas mixture exerts a pressure that it would exert if it were in the container alone. Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 26 of 41
Dalton s Law of Partial Pressure Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 27 of 41
Partial Pressure P tot = P a + P b + V a = n a RT/P tot and V tot = V a + V b + V a V tot n a RT/P = tot = n a n tot RT/P tot n tot Recall n a n tot = χ a P a P tot n a RT/V = tot = n a n tot RT/V tot n tot Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 28 of 41
Pneumatic Trough P tot = P bar = P gas + P H 2O Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 29 of 41
6-7 Kinetic Molecular Theory Particles are point masses in constant, random, straight line motion. Particles are separated by great distances. Collisions are rapid and elastic. No force between particles. Total energy remains constant. Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 30 of 41
Pressure Assessing Collision Forces Translational kinetic energy, Frequency of collisions, Impulse or momentum transfer, e = k v I = = 1 2 u mu N V mu 2 Pressure proportional to impulse times frequency P N V mu 2 Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 31 of 41
Pressure and Molecular Speed Three dimensional systems lead to: P = 1 3 N V 2 m u u m is the modal speed u av is the simple average u rms = 2 u Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 32 of 41
Pressure Assume one mole: PV = 1 3 N A m u 2 PV=RT so: 3RT = N A m u 2 N A m = M: 3RT = M u 2 Rearrange: u rms = 3RT M Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 33 of 41
Distribution of Molecular Speeds u rms = 3RT M Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 34 of 41
Determining Molecular Speed Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 35 of 41
Temperature Modify: PV = 1 3 N A m u 2 = 2 3 N A 1 ( 2 m u 2 ) PV=RT so: RT = 2 3 N A e k Solve for e k : e k = 3 2 R N A (T) Average kinetic energy is directly propotional to temperature! Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 36 of 41
6-8 Gas Properties Relating to the Kinetic-Molecular Theory Diffusion Net rate is proportional to molecular speed. Effusion A related phenomenon. Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 37 of 41
Graham s Law rateof rateof effusionof effusionof A B = (u (u rms rms ) ) A B = 3RT/MA 3RT/MB = M M B A Only for gases at low pressure (natural escape, not a jet). Tiny orifice (no collisions) Does not apply to diffusion. Ratio used can be: Rate of effusion (as above) Molecular speeds Effusion times Distances traveled by molecules Amounts of gas effused. Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 38 of 41
Example 5-10 Standardizing a Solution for Use in Redox Titrations. A piece of iron wire weighing 0.1568 g is converted to Fe 2+ (aq) and requires 26.42 ml of a KMnO 4 (aq) solution for its titration. What is the molarity of the KMnO 4 (aq)? 5 Fe 2+ (aq) + MnO 4- (aq) + 8 H + (aq) 4 H 2 O(l) + 5 Fe 3+ (aq) + Mn 2+ (aq) Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 39 of 41
Example 5-6 5 Fe 2+ (aq) + MnO 4- (aq) + 8 H + (aq) 4 H 2 O(l) + 5 Fe 3+ (aq) + Mn 2+ (aq) Determine KMnO 4 consumed in the reaction: n H O 2 = 0.1568gFe 1molFe 55.847gFe 1molFe 1molFe 2+ 4 2+ 1molMnO 5molFe 1molKMnO 1molMnO 4 4 = 5.615 10 4 molkmno 4 Determine the concentration: 4 5.615 10 mol KMnO4 [ KMnO 4] = = 0. 2140 M KMnO 0.2624 L 4 Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 40 of 41
Chapter 6 Questions 9, 13, 18, 31, 45, 49, 61, 63, 71, 82, 85, 97, 104. Prentice-Hall 2002 General Chemistry: Chapter 6 Slide 41 of 41