Chem 1101 Where are we going? examinable A/Prof Sébastien Perrier Room: 351 Phone: 9351-3366 Email: s.perrier@chem.usyd.edu.au Prof Scott Kable Room: 311 Phone: 9351-2756 Email: s.kable@chem.usyd.edu.au A/Prof Adam Bridgeman Room: 222 Phone: 9351-2731 5 main topics Gases Properties, molecular picture Thermochemistry Fuels, environmental effects of fuel combustion Equilibrium Gas phase, acid-base, solubility Electrochemistry Galvanic and concentration cells 3 interleaving applications Nitrogen chemistry e.g. smog, explosives Sources of chemicals e.g. ammonia, polymers Batteries, Electrolysis Intermolecular Forces e.g. corrosion, fuel cells Polymers, biomolecules Email: a.bridgeman@chem.usyd.edu.auslide 20-1 Slide 20-2 for interest Lecture resources Gases Access from WebCT or Blackman, Bottle, Schmid, Mocerino & Wille, Chapter 6 http://www.kcpc.usyd.edu.au/chem1101.html Login: chem1101 Password: carbon12 Slide 20-3 Blackman Figure 16.8 Slide 20-4 Boyle s Law Charles Law The volume of a gas is inversely proportional to its pressure The volume of a gas is directly proportional to its Kelvin temperature Figure 6.5 Blackman Slide 20-5 Figure 6.6 Blackman Slide 20-6 1
Avogadro s Law Lorenzo Romano Amedeo Carlo Avogadro di Quaregna e di Cerreto Equal volumes of gas at the same pressure and temperature contain equal numbers of molecules Empirical Gas Laws Boyle s Law: the volume of a gas is inversely proportional to its pressure V=k 1 /P Charles Law: the volume of a gas is directly proportional to its Kelvin temperature V=k 2 T Slide 20-7 Avogadro s Law: equal volumes of gas at the same pressure and temperature contain equal numbers of molecules V=k 3 n Slide 20-8 Ideal Gas Equation Ideal Gas Equation pressure nt V = k P P V = n R T absolute temperature volume number moles gas constant Universal gas constant R = 8.314 J mol 1 K 1 (R has different values with different units see later) Slide 20-9 Slide 20-10 Common Units of Pressure R in Different Units unit pascal (Pa); kilopascal (kpa) ; hectopascal (hpa) atmosphere mm mercury = torr pounds/sq inch (psi) bar atmospheric pressure 1.01325 10 5 Pa; 101.325 kpa; 1013.25 hpa 1 atm 760 torr 14.7 lb in 2 1.01325 bar SI unit only used by primitives Slide 20-11 R = 8.314 J mol 1 K 1 Energy = J = kg m 2 s 2 (eg E = ½ mv 2 ) Pressure = Pa = force/area = kg m -1 s 2 (force = kg m s 2 ) Pa m 3 has units kg m 2 s 2, same as J. = 8.314 Pa m 3 mol 1 K 1 1 kpa = 10 3 Pa, 1 L = 10 3 m 3, 1 kpa L = 1 Pa m 3 i.e. we can use P in kpa, V in L (= dm 3 ) = 8.314 kpa L mol 1 K 1 Slide 20-12 2
R in Different Units cont d 1 atmosphere = 101.3 kpa R = 8.314/101.3 atm L mol 1 K 1 = 0.08206 atm L mol 1 K 1 Example: Car Air Bags Often made with sodium azide: 6NaN 3 (s) + Fe 2 O 3 (s) 3Na 2 O(s) + 2Fe(s) + 9N 2 (g) Example: How many grams of NaN 3 are required to produce 75 L of nitrogen at 25 C and 1.31 atm? P V = n R T n = P V / (R T) example n = 1.31 atm x 75 L / (0.082 x 298 K) = 4.0 mol N2 Stoichiometry => 6/9 x 4 = 2.67 mol NaN3 Mass NaN3 = 2.67 x 65.0 = 174 g Slide 20-13 Slide 20-14 Gas Densities Gases have low densities because the molecules are far apart. For example: N 2 (g) at 20 C, ρ = 1.25 g/l N 2 (l) at 196 C, ρ = 808 g/l N 2 (s) at 210 C, ρ = 1030 g/l The density of the liquid is 100-1000 times greater than the gas state. n = P V / (R T) The density of the solid is usually 10% greater than the liquid. Slide 20-15 Example Find the density of oxygen gas at 25 C and 0.85 atm Density (ρ) = m/v = M n / V M = Molar mass So ρ = M P / RT (from PV = nrt) 32.0 0.85 Density of O2 = = 1.1 g/l 0.082 298 (check units balance!) g mol 1 L atm mol 1 K 1 atm 273+25 K Slide 20-16 The Molecular Picture (or Kinetic Theory of Gases) Boyle s Law The Molecular Picture or Kinetic Theory of Gases Figure 6.11 Blackman Slide 20-17 Slide 20-18 3
Molecular Picture of Boyle s Law Charles Law V increases Figure 6.12 Blackman P decreases Slide 20-19 Slide 20-20 Molecule Picture of Charles Law Temperature Dependence T increases P increases Slide 20-21 Blackman 6.9 Slide 20-22 Speeds of Different Molecules Kinetic energy E k = 1/2 mv 2 T = constant at fixed T Pressure P E k = 1/2 mv 2 Τ = same for all gases Gas Behaviour at STP (STP = Standard Temperature and Pressure) Standard Molar Volume He N 2 O 2 Blackman Figure 6.8 Slide 20-23 Slide 20-24 4
Dalton s Law of Partial Pressure So far we have considered pure gases. What happens when we combine them? Dalton s Law of Partial Pressures Consider 2 vessels with different gases Depress the piston and force gas A into the other cylinder. Dalton s Law: The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone." Ptot = Pa + Pb +... where a, b, are different gases Slide 20-25 Assume equal cylinder volumes What is the total pressure, P, and the partial pressures of A and B? Slide 20-26 Dalton s Law of Partial Pressures The number of moles of each gas in the cylinder is trivial n tot = n A + n B = 0.3 + 0.6 = 0.9 mol The mole fraction of A is Χ A = 0.3 / 0.9 = 0.33 The mole fraction of B is Χ B = 0.6 / 0.9 = 0.67 Note the relationship between mole fraction and partial pressure: p A = Χ A P tot Slide 20-27 Implications of PV = nrt Boyle s Law: PV = constant for constant temperature Charles Law: V/T = constant for constant pressure Lowest temperature: -273 ºC (ie V = 0 at this temperature) Gay-Lussac: (combining volumes) Avogadro s hypothesis: (equal volumes of gas have the same number of molecules) Avogadro s constant: N A = # molecules/mol = 6.02 10 23 mol 1 Dalton s law of partial pressures (see later) Diverse observations from single unifying theory: one of the goals of science! Slide 20-28 Example questions CONCEPTS Classical behaviour of gases at different pressure, volume, temperatures and moles Boyle s Law, Charles Law, Avagadro Law CALCULATIONS PV=nRT Gas densities Slide 20-29 Slide 20-30 5
PV = nrt from Kinetic Theory Beyond examinable: Pressure arises from the force experienced on vessel wall by gas molecules bouncing off it L Square of side L momentum of gas molecule = mass velocity = mv change in momentum from bouncing off wall: p = 2mv Slide 20-31 PV = nrt from Kinetic Theory no. collisions/second with wall = ½ N v L 2 half molecules moving towards wall dmv changein momentum force= = dt timeincrement 2 = 2mv 1 2 Nv L force 2 pressure= = mv N area no. molecules / unit volume time increment t = 1/(no. collisions/second) no. moles N = no. molecules/unit volume = n N A / V area of wall Avogadro constant Slide 20-32 PV = nrt from Kinetic Theory P V = n N A m v 2 define temperature T average energy = ½ mv 2 by T = N A m v 2 / R (defines absolute temperature scale in terms of molecular energy) P V = n RT Learn equation, not derivation Slide 20-33 6