Some Important Industrial Gases

Transcription

1 Gaseous state

2 Table 5.1 Some Important Industrial Gases Name (Formula) Methane (CH 4 ) Ammonia (NH 3 ) Chlorine (Cl 2 ) Oxygen (O 2 ) Ethylene (C 2 H 4 ) Origin and Use natural deposits; domestic fuel from N 2 +H 2 ; fertilizers, explosives electrolysis of seawater; bleaching and disinfecting liquefied air; steelmaking high-temperature decomposition of natural gas; plastics Atmosphere-Biosphere Redox Interconnections

3 An Overview of the Physical States of Matter The Distinction of Gases from Liquids and Solids 1. Gas volume changes greatly with pressure. 2. Gas volume changes greatly with temperature. 3. Gases have relatively low viscosity. 4. Most gases have relatively low densities under normal conditions. 5. Gases are miscible.

4 Three states of the matter

5 Vacuum Effect of atmospheric pressure on objects at the Earth s surface.

6 A mercury barometer.

7 closed-end Two types of manometer open-end

8 Table 5.2 Common Units of Pressure Unit pascal(pa); kilopascal(kpa) atmosphere(atm) Atmospheric Pressure x10 5 Pa; kpa 1 atm* Scientific Field SI unit; physics, chemistry chemistry millimeters of mercury(hg) 760 mm Hg* chemistry, medicine, biology torr 760 torr* chemistry pounds per square inch (psi or lb/in 2 ) 14.7lb/in 2 engineering bar bar meteorology, chemistry, physics *This is an exact quantity; in calculations, we use as many significant figures as necessary.

9 Converting Units of Pressure PROBLEM: A geochemist heats a limestone (CaCO 3 ) sample and collects the CO 2 released in an evacuated flask attached to a closedend manometer. After the system comes to room temperature, Dh = mm Hg. Calculate the CO 2 pressure in torrs, atmospheres, and kilopascals. PLAN: Construct conversion factors to find the other units of pressure. SOLUTION: mmhg 1torr 1 mmhg torr 1 atm 760 torr = torr = atm atm kpa 1 atm = kpa

10 The relationship between the volume and pressure of a gas. Boyle s Law

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12 Figure 5.6 The relationship between the volume and temperature of a gas. Charles s Law

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15 Boyle s Law V a 1 P n and T are fixed V x P = constant V = constant / P Charles s Law V a T P and n are fixed V T = constant V = constant x T Amontons s Law P a T V and n are fixed P T = constant P = constant x T combined gas law V a T P V = constant x T P PV T = constant

16 An experiment to study the relationship between the volume and amount of a gas.

17 Standard molar volume.

18 The volume of 1 mol of an ideal gas compared with some familiar objects.

19 Equal volumes of different gases contain, in the same conditions of P and T, the same number of particles 100 ml H 2 double amount of moles of 50 ml of O 2

20 Avogadro s Law Gas Volume at certain T and P is straight proportional to the gas quantity, that is to its moles number ( n = g/mm) V n

21 THE IDEAL GAS LAW R = PV nt R is the universal gas constant PV = nrt 1atm x L 1mol x K = IDEAL GAS LAW atm*L mol*k 3 significant figures PV = nrt or V = nrt P fixed n and T fixed n and P fixed P and T Boyle s Law Charles s Law Avogadro s Law V = constant P V = constant X T V = constant X n

22 Applying the Volume-Pressure Relationship PROBLEM: Boyle s apprentice finds that the air trapped in a J tube occupies 24.8 cm 3 at 1.12 atm. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64 atm. Assuming constant temperature, what is the new volume of air (in L)? PLAN: V 1 in cm 3 1cm 3 =1mL V 1 in ml 10 3 ml=1l V 1 in L xp 1 /P 2 V 2 in L unit conversion gas law calculation V 2 = SOLUTION: P 1 V 1 P 2 P 1 = 1.12 atm V 1 = 24.8 cm cm 3 L P 1 V 1 P 2 V 2 = n 1 T 1 n 2 T 2 = L P and T are constant 10 3 ml 1.12 atm 2.46 atm P 2 = 2.64 atm V 2 = unknown = L P 1 V 1 = P 2 V 2 = L

23 Applying the Temperature-Pressure Relationship PROBLEM: A steel tank used for fuel delivery is fitted with a safety valve that opens when the internal pressure exceeds 1.00x10 3 torr. It is filled with methane at 23 0 C and atm and placed in boiling water at exactly C. Will the safety valve open? PLAN: P 1 (atm) T 1 and T 2 ( 0 C) 1atm=760torr P 1 (torr) x T 2 /T 1 P 2 (torr) K= 0 C T 1 and T 2 (K) atm SOLUTION: P 1 = 0.991atm T 1 = 23 0 C P 1 V 1 P 2 V 2 = n 1 T 1 n 2 T torr = 753 torr 1 atm P 2 = unknown T 2 = C P 1 P 2 = T 1 T 2 P 2 = P 1 T 2 T 1 = 753 torr 373K 296K = 949 torr

24 Applying the Volume-Amount Relationship PROBLEM: A scale model of a blimp rises when it is filled with helium to a volume of 55 dm 3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm 3. How many more grams of He must be added to make it rise? Assume constant T and P. PLAN: We are given initial n 1 and V 1 as well as the final V 2. We have to find n 2 and convert it from moles to grams. n 1 (mol) of He x V 2 /V 1 n 2 (mol) of He subtract n 1 mol to be added x M g to be added SOLUTION: V 1 n 1 = 1.10 mol n 2 = unknown V 1 = 26.2 dm 3 V 2 = 55.0 dm 3 V 2 V = n n 1 n 2 = n n 2 = 1.10 mol V 1 P and T are constant 55.0 dm 3 g He = 2.31 mol dm3 mol He P 1 V 1 P 2 V 2 = n 1 T 1 n 2 T 2 = 9.24 g He

25 PROBLEM: Solving for an Unknown Gas Variable at Fixed Conditions A steel tank has a volume of 438 L and is filled with kg of O 2. Calculate the pressure of O 2 at 21 0 C. PLAN: V, T and mass, which can be converted to moles (n), are given. We use the ideal gas law to find P. SOLUTION: V = 438 L T = 21 0 C (convert to K) n = kg (convert to mol) P = unknown 0.885kg 103 g kg nrt P = V mol O g O 2 = 27.7 mol O 2 atm*l 24.7 mol x x K mol*k 438 L 21 0 C = K = 1.53 atm

26 Using Gas Laws to Determine a Balanced Equation PROBLEM: The piston-cylinders below depict a gaseous reaction carried out at constant pressure. Before the reaction, the temperature is 150K; when it is complete, the temperature is 300K. New figures go here. Which of the following balanced equations describes the reaction? (1) A 2 + B 2 2AB (2) 2AB + B 2 2AB 2 (3) A + B 2 AB 2 (4) 2AB 2 A 2 + 2B 2 PLAN: SOLUTION: We know P, T, and V, initial and final, from the pictures. Note that the volume doesn t change even though the temperature is doubled. With a doubling of T then, the number of moles of gas must have been halved in order to maintain the volume. Looking at the relationships, the equation that shows a decrease in the number of moles of gas from 2 to 1 is equation (3).

27 The Density of a Gas density = m/v n = m/m PV = nrt PV = (m/m)rt m/v = M x P/ RT The density of a gas is directly proportional to its molar mass. The density of a gas is inversely proportional to the temperature.

28 PROBLEM: Calculating Gas Density To apply a green chemistry approach, a chemical engineer uses waste CO 2 from a manufacturing process, instead of chlorofluorocarbons, as a blowing agent in the production of polystyrene containers. Find the density (in g/l) of CO 2 and the number of molecules (a) at STP (0 0 C and 1 atm) and (b) at room conditions (20. 0 C and 1.00 atm). PLAN: Density is mass/unit volume; substitute for volume in the ideal gas equation. Since the identity of the gas is known, we can find the molar mass. Convert mass/l to molecules/l with Avogadro s number. M x P d = mass/volume PV = nrt V = nrt/p d = RT SOLUTION: g/mol x 1atm d = = 1.96 g/l (a) atm*l x K mol*k 1.96 g mol CO x10 23 molecules = 2.68x10 22 molecules CO 2 /L L g CO 2 mol

29 Calculating Gas Density continued (b) d = g/mol x 1 atm atm*l mol*k x 293K = 1.83 g/l 1.83g L mol CO x10 23 molecules 44.01g CO 2 mol = 2.50x10 22 molecules CO 2 /L

30 Determining the molar mass of an unknown volatile liquid. based on the method of J.B.A. Dumas ( )

31 Finding the Molar Mass of a Volatile Liquid PROBLEM: An organic chemist isolates a colorless liquid from a petroleum sample. She uses the Dumas method and obtains the following data: Volume of flask = 213 ml Mass of flask + gas = g T = C Mass of flask = g P = 754 torr Calculate the molar mass of the liquid. PLAN: Use unit conversions, mass of gas, and density-m relationship. SOLUTION: m = ( ) g = g atm*l m RT g x 373K M = mol*k VP L x atm = 84.4 g/mol

32 Mixtures of Gases Gases mix homogeneously in any proportions. Each gas in a mixture behaves as if it were the only gas present. Dalton s Law of Partial Pressures P total = P 1 + P 2 + P P 1 = c 1 x P total c 1 = n 1 n 1 + n 2 + n = n 1 n total

33 Applying Dalton s Law of Partial Pressures PROBLEM: In a study of O 2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mol% N 2, 17 mol% 16 O 2, and 4.0 mol% 18 O 2. (The isotope 18 O will be measured to determine the O 2 uptake.) The pressure of the mixture is 0.75atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18 O 2 in the mixture. PLAN: Find the c and P from P total and mol% 18 O O 2 18 O 2 mol% 18 O 2 divide by 100 SOLUTION: c 18 O 2 = 4.0 mol% 18 O c 18 O 2 P = c x P total = x 0.75 atm = atm 18 O 18 2 O 2 multiply by P total

34 The Molar Mass of a Gas n = mass M = PV RT M = m RT VP d = m V M = d RT P

35 Table 5.3 Vapor Pressure of Water (P H 2 O ) at Different T T( 0 C) P (torr) T( 0 C) P (torr)

36 Collecting a water-insoluble gaseous reaction product and determining its pressure.

37 PROBLEM: Calculating the Amount of Gas Collected Over Water Acetylene (C 2 H 2 ), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC 2 ) reaction with water: CaC 2 (s) + 2H 2 O(l) C 2 H 2 (g) + Ca(OH) 2 (aq) For a sample of acetylene that is collected over water, the total gas pressure (adjusted to barometric pressure) is 738torr and the volume is 523mL. At the temperature of the gas (23 0 C), the vapor pressure of water is 21torr. How many grams of acetylene are collected? PLAN: The difference in pressures will give us the P for the C 2 H 2. The ideal gas law will allow us to find n. Converting n to grams requires the molar mass, M. P P SOLUTION: total P C2 H = (738-21)torr = 717torr 2 C2 H 2 H 2 O n = PV atm 717torr = 0.943atm RT 760torr n g C2 H 2 C2 H 2 x M

38 Calculating the Amount of Gas Collected Over Water continued n C2 H 2 = = mol mol 26.04g C 2 H 2 mol C 2 H 2 = g C 2 H 2

39 Summary of the stoichiometric relationships among the amount (mol,n) of gaseous reactant or product and the gas variables pressure (P), volume (V), and temperature (T). P,V,T of gas A amount (mol) of gas A amount (mol) of gas B P,V,T of gas B ideal gas law molar ratio from balanced equation ideal gas law

40 Using Gas Variables to Find Amount of Reactants and Products PROBLEM: Dispersed copper in absorbent beds is used to react with oxygen impurities in the ethylene used for producing polyethylene. The beds are regenerated when hot H 2 reduces the metal oxide, forming the pure metal and H 2 O. On a laboratory scale, what volume of H 2 at 765 torr and C is needed to reduce 35.5 g of copper(ii) oxide? PLAN: Since this problem requires stoichiometry and the gas laws, we have to write a balanced equation, use the moles of Cu to calculate mols and then volume of H 2 gas. mass (g) of Cu SOLUTION: CuO(s) + H 2 (g) Cu(s) + H 2 O(g) divide by M mol Cu 1 mol H mol of Cu 35.5 g Cu 2 = mol H g Cu (da 1 mol Cu molar ratio vcorrggere!!!) mol H mol of H 2 x atm*l x 498K = 22.6 L 2 mol*k use known P and T to find V 1.01 atm L of H 2

41 Using the Ideal Gas Law in a Limiting-Reactant Problem PROBLEM: The alkali metals [Group 1A(1)] react with the halogens [Group 7A(17)] to form ionic metal halides. What mass of potassium chloride forms when 5.25 L of chlorine gas at atm and 293K reacts with 17.0 g of potassium? PLAN: After writing the balanced equation, we use the ideal gas law to find the number of moles of reactants, the limiting reactant and moles of product. SOLUTION: 2K(s) + Cl 2 (g) 2KCl(s) P = atm V = 5.25 L PV atm X 5.25 L T = 293K n = unknown = = mol Cl 2 RT atm*l x 293K mol*k 2 mol KCl 17.0g mol K mol Cl = mol K 2 = mol 1 mol Cl g K 2 KCl formed 2 mol KCl Cl 2 is the limiting reactant mol K = mol 2 mol K g KCl KCl formed mol KCl = 30.9 g KCl mol KCl

42 Diffusion of a gas

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44 Effusion of a gas It is the movement of gas through a tiny opening in a container into another containerwhere the pressure is very low.

45 Graham law Rate of effusion of gas1 (molar mass of gas 2) 1/ = Rate of effusion of gas2 (molar mass of gas 1) 1/2 By considering a constant path size for both gases: t2 (molar mass of gas2) 1/ = t1 (molar mass of gas1) 1/2 where t = effusion time

46 Kinetic molecular theory of gases It describes the behavior of matter at the molecular or atomic level as concerns the gaseous state. The molecular model allows us to interpret the macroscopic evidences by a statistical mathematical model.

47 Postulates of the Kinetic-Molecular Theory Postulate 1: Particle Volume Because the volume of an individual gas particle is so small compared to the volume of its container, the gas particles are considered to have mass, but no volume. Postulate 2: Particle Motion Gas particles are in constant, random, straight-line motion except when they collide with each other or with the container walls. Postulate 3: Particle Collisions Collisions are elastic therefore the total kinetic energy(e k ) of the particles is constant.

48 Distribution of molecular speeds at three temperatures.

49 A molecular description of Boyle s Law.

50 A molecular description of Dalton s law of partial pressures.

51 A molecular description of Charles s Law.

52 Avogadro s Law V a n E k = 1/2 mass x speed 2 E k = 1/2 mass x u 2 u 2 is the root-mean-square speed u rms = 3RT M R = 8.314Joule/mol*K Graham s Law of Effusion The rate of effusion of a gas is inversely related to the square root of its molar mass. rate of effusion a 1 M

53 A molecular description of Avogadro s Law.

54 Relationship between molar mass and molecular speed.

55 Applying Graham s Law of Effusion PROBLEM: Calculate the ratio of the effusion rates of helium and methane (CH 4 ). PLAN: The effusion rate is inversely proportional to the square root of the molar mass for each gas. Find the molar mass of both gases and find the inverse square root of their masses. SOLUTION: M of CH 4 = 16.04g/mol M of He = 4.003g/mol rate rate He CH 4 = = 2.002

56 Diffusion of a gas particle through a space filled with other particles. distribution of molecular speeds mean free path collision frequency

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59 Table 5.4 Molar Volume of Some Common Gases at STP (0 0 C and 1 atm) Gas Molar Volume (L/mol) Condensation Point ( 0 C) He H 2 Ne Ideal gas Ar N 2 O 2 CO Cl 2 NH

60 The behavior of several real gases with increasing external pressure.

61 The effect of intermolecular attractions on measured gas pressure.

62 The effect of molecular volume on measured gas volume.

63 Table 5.5 Van der Waals Constants for Some Common Gases Van der Waals equation for n moles of a real gas (P n2 a )(V nb) nrt 2 V Gas a atm*l 2 mol 2 b L mol He Ne Ar Kr Xe H 2 N 2 O 2 Cl 2 CO 2 CH 4 NH 3 H 2 O