Chapter 21 Physical Properties Of Gases

Worked solutions to textbook questions 1

Chapter 21 Physical properties of gases

Q1. Use the kinetic molecular theory to explain the following observed properties of gases: a Gases occupy all the available space in a container. b Gases can be easily compressed compared with their corresponding liquid forms. c A given volume of a gaseous substance weighs less than the same volume of the substance in the liquid state. d Gases will mix together readily. e The total pressure of a mixture of gases is equal to the sum of the pressures exerted by each of the gases in the mixture. A1. a Molecules of gases are in constant, rapid, random motion and the forces between molecules are negligible. They continue to move outwards until stopped by the walls of the container, filling all the space available. b Most of the volume occupied by a gas is space, so compression can be achieved by reducing the space between the particles. c The molecules in a gas are spread much further apart than those of a liquid. A given mass of gas would occupy a much greater volume than the same mass of the liquid phase. Therefore, the density of the gas is less. d Gases mix easily together because of the large amount of space between the molecules. e The pressure exerted by a gas depends on the number of collisions of gas particles and the wall of the container. The pressure is independent of the type of gas involved. The total pressure exerted by a mixture of gases will depend on the total number of collisions each gas has with the container. Q2. Use the ideas of the kinetic molecular theory of gases to explain the following observations: a Tyre manufacturers recommend a maximum pressure for tyres. b The pressure in a car’s tyres will increase if a long distance is travelled on a hot day. c You can smell dinner cooking as you enter the house. d A balloon will burst if you blow it up too much. A2. a Tyres have a recommended maximum pressure to give a comfortable ride as well as good traction on the road. If the pressure in a tyre is too high, the gas inside cannot be compressed as easily and passengers will be more aware of bumps on the road. b During a long journey on a hot day, the air in a tyre warms up. This means the air molecules have increased kinetic energy, and collisions with the walls of the tyres will increase, and so the pressure will increase.

Heinemann Chemistry 1 (4th edition)  Reed International Books Australia Pty Ltd Worked solutions to textbook questions 2 c Particles from the cooking food escape the pot and move randomly through the house. If the food has an odour, and if there are enough particles in the air, you will detect the odour as you enter the house. d As air is pumped into a balloon, air molecules collide with the rubber of the balloon, forcing it to expand. If too much air is pumped in, the balloon reaches a stage where it cannot stretch any further. If the number of collisions by molecules per given surface area is increased still further, the rubber will break. Q3. Horse-drawn coaches had ‘tyres’ made from a solid strip of rubber attached to a wooden wheel. Why would these give a much bumpier ride than the modern pneumatic (air-filled) tyre? A3. When an air-filled tyre passes over a bump in the road, the air molecules inside the tyre move and so absorb some of the shock. A solid rubber tyre can distort and absorb some shock, but is not as effective as an air-filled tyre. Q4. The graph in Figure 21.3 (page 358) shows the distribution of molecular kinetic energies for oxygen at 20°C, 30°C and 40°C. Predict the shape of the graph of the distribution of molecular kinetic energies of the same sample of oxygen gas at –20°C. A4. The peak of the graph would be higher and closer to the Y-axis. Q5. Convert each of the following pressures to the units specified: a 1400 mmHg to atm, Pa and bars b 80 000 Pa to atm, mmHg and bars c 4.24 atm to mmHg and Pa d 120 kPa to mmHg, atm and bars e 140 kPa to Pa f 92 000 Pa to kPa A5. These conversions should be applied as needed: 1.00 atm = 760 mmHg = 101.3 kPa = 1.013 × 105 Pa = 1.013 bar a 760 mmHg = 1.00 atm 1400 So 1400 mmHg = 1.00 × atm = 1.84 atm 760 760 mmHg = 1.013 × 105 Pa 1400 So 1400 mmHg = 1.013 × 105 × Pa = 1.87 × 105 Pa 760 760 mmHg = 1.013 bar 1400 So 1400 mmHg = 1.013 × bar = 1.87 bar 760 b 1.013 × 105 Pa = 1.00 atm 80000 So 80 000 Pa = 1.00 × atm = 0.790 atm 1.013105

Heinemann Chemistry 1 (4th edition)  Reed International Books Australia Pty Ltd Worked solutions to textbook questions 3

1.013 × 105 Pa = 760 mm Hg 80000 So 80 000 Pa = 760 × mm Hg = 600 mmHg 1.013105 1.013 × 105 Pa = 1.013 bar 80000 So 80 000 Pa = 1.013 × bar = 0.800 bar 1.013105 c 1.00 atm = 760 mmHg So 4.24 atm = 760 × 4.24 mm Hg = 3220 mm Hg 1.00 atm = 1.013 × 105 Pa So 4.24 atm = 1.013 × 105 × 4.24 Pa = 4.30 × 105 Pa d 101.3 kPa = 760 mmHg 760 So 120 kPa = × 120 kPa = 900 mm Hg 101.3 101.3 kPa = 1.00 atm 1.00 So 120 kPa = × 120 atm = 1.18 atm 101.3 101.3 kPa = 1.013 bar 1.013 So 120 kPa = × 120 atm = 1.20 bars 101.3 e 140 kPa = 140 × 1000 Pa = 1.40 × 105 Pa 92000 f 92 000 Pa = kPa = 92 kPa 1000 Q6. Convert the following volumes to the unit specified: a 2 L to mL b 4.5 dm3 to mL c 2250 mL to L d l20 mL to L e 5.6 mL to L f 3.7 dm3 to m3 g 285 mL to m3 h 4.70 × 10–3 m3 to dm3 and cm3 A6. These conversions should be applied as needed: 1 mL = 1 cm3, 1 L = 1 dm3, 1 L = 1000 mL = 1000 cm3, 1 m3 = 103 dm3 = 106 cm3 a 2 L = 2 × 1000 mL = 2 × 103 mL b 4.5 dm3 = 4.5 L = 4.5 × 1000 mL = 4.5 × 103 mL 2250 c 2250 mL = L = 2.25 L 1000 120 d 120 mL = L = 0.12 L 1000 56 e 56 mL = L = 5.6 × 10–2 L 1000 3.7 f 3.7 dm3 = m3 = 3.7 × 10–3 m3 1000

285 g 285 mL = 285 cm3 = m3 = 2.85 × 10–4 m3 106 h 4.70 × 10–3 m3 = 4.70 × 10–3 × 1000 dm3 = 4.70 dm3 = 4.70 × 1000 cm3 = 4.70 × 103 cm3 Q7. In the kinetic molecular theory, pressure is described as the force per unit area of surface. Explain what happens to the pressure in each of the following situations: a The temperature of a filled aerosol can is increased. b A gas in a syringe is compressed. A7. a As temperature increases, the average kinetic energy of gas molecules in the can will increase. This will lead to an increase in the frequency and force of collisions of gas molecules with the inside walls of the aerosol cans. This will cause an increase in pressure. b As the syringe is compressed, the inside surface area of the syringe will decrease. The number of collisions of molecules per unit area per second with the inside walls of the syringe will increase. This will cause a pressure increase. Q8. Copy and complete the following table, which refers to a given mass of gas kept at a constant temperature. Initial conditions Final conditions Pressure Volume Pressure Volume a 2 atm 50 L 4 atm b 732 mmHg 20 L 500 mmHg c 800 mmHg 350 cm3 120 mmHg d 105 kPa 650 cm3 800 cm3 A8.

Remember to use Boyle’s law: P1V1 = P2V2. Each of the pressure and volume units needs to be the same. These conversions need to be applied as needed: 1.00 atm = 760 mmHg = 101.325 kPa = 101 325 Pa = 1.01 bar 1 mL = 1 cm3, 1 L = 1 dm3, 1 L = 1000 mL = 1000 cm3 2 50 a V2 = = 25 L 4 732 20 b V2 = = 29 L 500 800 350 c V = = 2.3 × 103 cm3 or 2.3 L 2 120 105 650 d P2 = = 85.3 kPa 800

Q9. A sample of air in a syringe has a volume of 120 mL when the pressure is 100 000 Pa at room temperature. If the temperature of the gas remains constant, what volume would the air occupy if the pressure was changed to: a 200 000 Pa? b 45 000 Pa? c 64 kPa? A9.

Remember to use Boyle’s law: P1V1 = P2V2. Each of the pressure and volume units needs to be the same. 100 000120 a V = = 60 mL 2 200 000 100 000120 b V = = 267 mL 2 45 000 This conversion needs to be applied: 101.325 × 103 kPa = 101 325 Pa. 100 000120 c V = = 188 mL 2 641000 Q10. The pressure on a 9.0 mL sample of carbon dioxide is increased from 110 kPa to 280 kPa. Calculate the new volume occupied by the gas, assuming the temperature does not change. A10.

Remember to use Boyle’s law: P1V1 = P2V2. Each of the pressure and volume units needs to be the same. 110 9 V = = 3.5 mL 2 280 Q11. A sample of nitrogen occupies 100 mL at 40°C and a pressure of 80 kPa. What pressure would be needed to reduce the volume to 25 mL at 40°C? A11.

Remember to use Boyle’s law: P1V1 = P2V2. Each of the pressure and volume units needs to be the same. 80100 P = = 320 Pa 2 25 Q12. Convert the following Celsius temperatures to absolute temperatures: a 100°C b 175°C c –145°C A12. Remember to convert temperatures by using: T = t + 273, where T is the absolute temperature (in kelvin), and t is the temperature on the Celsius scale. It is also

Heinemann Chemistry 1 (4th edition)  Reed International Books Australia Pty Ltd Worked solutions to textbook questions 6 conventional not to use the degree symbol when writing the absolute temperature. For example, 25C would be written as 298 K. a T =100 + 273 = 373 K b T = 175 + 273 = 448 K c T = –145 + 273 = 128 K Q13. A syringe contains 90 mL of gas measured at 1 atm pressure at 25°C. a What volume would the gas occupy at: i 1 atm and 120°C? ii 1 atm and –90°C? b To what temperature would the gas need to be changed for the volume to be: i 30 mL? ii 120 mL? (The pressure remains at 1 atm.) A13. V V Remember to use Charles’ Law: 1 = 2 . The temperature must be in kelvin, the T1 T2 volume units must be the same, and the pressure is constant. 90 (120  273) a i V2 = = 120 mL 25  273 90 (90  273) ii V2 = = 55 mL 25  273

V2T1 b Charles’ law needs to rearranged to give: T2 = . V1 30 (25  273) i T = = 99 K or –174C 2 90 120 (25  273) ii T = = 397 K or 124C 2 90 Q14. A cylinder, volume 20 000 L, contains methane. A second cylinder, with volume 500 L, contains 40 mol of methane. Both gas samples are at the same temperature and pressure. Calculate: a the amount of methane in the first cylinder b the mass of methane in the first cylinder A14. V V This law can be used: 1 = 2 . The temperature and pressure must remain constant. n1 n2 20 000 40 a n(methane) = = 1600 mol 500

m b Since n = and the molar mass of methane is 16 g mol–1. M m(methane) = 1600 × 16 g = 25.6 kg Q15. A balloon contains 0.35 mol of helium and has a volume of 5.3 L at a certain temperature and pressure. A further 0.12 mol of helium is added, keeping the temperature and pressure constant. Calculate the new volume of the balloon. A15. V V This law can be used: 1 = 2 . The temperature and pressure must remain constant. n1 n2

V1n2 The law must be rearranged to give: V2 = n1 5.3 (0.35  0.12) V = = 7.1 L 2 0.35 Q16. Calculate the volume of the following gases at SLC. a 1.4 mol of chlorine (Cl2) –3 b 1.0 × 10 mol of hydrogen (H2) c 1.4 g of nitrogen (N2) A16. Remember that, under standard laboratory conditions (SLC), 1 mol of any gas has a V volume of 24.5 L. Use the formula: n = where n is the amount in mol, V is the V m , –1 volume in L and Vm is the molar volume in L mol . To calculate V the formula is rearranged to V = n × Vm. a V(Cl2) = 24.5 × 1.4 = 34 L –3 b V(H2) = 24.5 × 1.0 × 10 L = 2.5 × 10–2 L = 25 mL m c Since n = and the molar mass of nitrogen is 28.0 g mol–1. M 24.51.4 V(N ) = 2 28.0 = 1.2 L Q17. Calculate the mass of the following gas samples. All volumes are measured at SLC. a 2.8 L of neon (Ne) b 50 L of oxygen (O2) c 140 mL of carbon dioxide (CO2) A17. Remember, that under standard laboratory conditions (SLC), 1 mol of any gas has a volume of 24.5 L. These questions involve three steps:

Step 1: Calculate the amount (in mol) of gas at SLC. Step 2: Find the molar mass (M) of the gas. m Step 3: Use n = to find the mass by reorganising as m = nM. M 2.8 a n(Ne) = = 0.114 mol 24.5 m(Ne) = 0.114 × 20.1 = 2.3 g 50 b n(O ) = = 2.04 mol 2 24.5 m(O2) = 2.04 × 32.0 = 65 g 140103 c n(CO2) = = 0.0057 mol 24.5 m(CO2) = 0.0057 × 44.0 = 0.251 g Q18. A fixed amount of gas occupies a volume of 30 L at 20°C and a pressure of 100 kPa. What volume would it occupy at: a 50°C and a pressure of 250 kPa? b –45°C and a pressure of 70 000 Pa? c 80°C and a pressure of 1 atm? A18. PV P V The combined gas equation needs to be used: 1 1 = 2 2 . Temperature should be T1 T2 in kelvin, and each of the pressure and volume units needs to be the same.

P1V1T2 a V2 = T1P2 100 30 (50  273) = (20  273)  250 = 13.2 L

P1V1T2 b V2 = T1P2 100 30 (45  273) = (20  273)  70 = 33.3 L

P1V1T2 c V2 = T1P2 100 30 (80  273) = (20  273) 101 = 35.8 L

Q19. A balloon has a volume of 150 L at a pressure of 101 kPa and a temperature of 27°C. It rises to an altitude of 15 km, where the temperature is –30°C and the pressure is 12 kPa. What is the volume of the balloon at this altitude? A19. PV P V The combined gas equation needs to be used: 1 1 = 2 2 . Temperature should be T1 T2 in kelvin, and each of the pressure and volume units needs to be the same.

P1V1T2 V2 = T1P2 101150 (30  273) = (27  273) 12 = 1023 L Q20. A volume of 32.0 mL of hydrogen at 25°C and 75 kPa is compressed to 16 mL and heated to 50°C. What will be the new pressure exerted by the hydrogen? A20. PV P V The combined gas equation needs to be used: 1 1 = 2 2 . Temperature should be T1 T2 in kelvin, and each of the pressure and volume units needs to be the same.

P1V1T2 P2 = T1V2 75 32.0 (50  273) = (25  273) 16 = 160 kPa Q21. A 4.20 L sample of gas at 23°C and 0.25 atm is transferred to a 6.50 L vessel. To what temperature must the gas be heated so that its pressure increases to 0.60 atm? A21. PV P V The combined gas equation needs to be used: 1 1 = 2 2 . Temperature should be T1 T2 in kelvin, and each of the pressure and volume units needs to be the same.

P2V2T1 T2 = P1V1 0.60 6.5 (23  273) = 0.25 4.20 = 1099 K or 826C = 830C (to 2 significant figures)

Q22. 0.25 mol of nitrogen is placed in a flask of volume 5.0 L at a temperature of 5°C. What is the pressure in the flask? A22. The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. nRT P = V 0.258.31 (5  273) = 5.0 = 116 kPa Q23. Calculate the mass of helium in a balloon if the volume is 100 L at a pressure of 95 000 Pa and a temperature of 0°C. A23. The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. m The second step involves use of n = . M The molar mass of helium is 4.0 g mol–1. PV n(He) = RT 95100 = 8.31 (273  0) = 4.19 mol m(He) = 4.19 × 4.0 = 16.8 g Q24. What volume of gas, in litres, is occupied by: a 0.20 mol of hydrogen at 115 kPa and 40°C? b 12.5 mol of carbon dioxide at 5 atm and 150°C? c 8.50 g of hydrogen sulfide (H2S) at 100 kPa and 27°C? A24. The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. nRT a V = P 0.20 8.31 (40  273) = 115 = 4.5 L nRT b V = P

12.5 8.31 (150  273) = 5101.325 = 87 L m c Since n = and the molar mass of hydrogen sulfide is 34.1 g mol–1 M nRT 8.50 8.31 (27  273) V = = × = 6.23 L P 34.1 100 Q25. At a given temperature, a sample of nitrogen, of mass 11.3 g, exerts a pressure of 102 kPa in a gas cylinder of volume 10.0 L. Calculate the temperature of the gas. A25. The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. m Since n = and the molar mass of nitrogen is 28.0 mol–1, M PV T = nR 10210.028.0 = 11.38.31 = 304 K = 31.1C Q26. Which sample of gas will contain the greater amount (mol) of gas: 3.2 L of nitrogen at 25°C and a pressure of 1.2 bar or 2.5 L of helium at 23°C and a pressure of 1.2 atm? A26. The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. These conversions need to be applied as needed: 1.00 atm = 101.325 kPa = 1.01 bar. PV n(N ) = 2 RT 120 3.2 = 8.31 (25  273) = 0.155 mol PV n(He) = RT (1.2101)  2.5 = 8.31 (23  273) = 0.123 mol  there is a greater amount of nitrogen. Q27. Magnesium reacts with hydrochloric acid according to the equation:

Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)

Calculate the volume of hydrogen produced at SLC when the following masses of magnesium react with excess hydrochloric acid: a 2.4 g b 0.54 g c 0.0018 g d 1.6 g A27. The balanced equation shows that 1 mol of magnesium produces 1 mol of hydrogen gas.

Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) m Remember that n = . (The molar mass of magnesium is 24.3 g mol–1.) M Remember that under SLC, 1 mol of any gas has a volume of 24.5 L. V Use the formula n = rearranged as V = n × Vm. Vm 2.4 a n(Mg) = = 0.10 mol 24.3 1 n(H )/n(Mg) = 2 1

 V(H2) = 24.5 × 0.10 L = 2.45 L 0.54 b n(Mg) = = 0.022 mol 24.3

 V(H2) = 24.5 × 0.022 L = 0.54 L 0.0018 c n(Mg) = = 0.000 074 mol 24.3

 V(H2) = 24.5 × 0.000 074 L = 0.0018 L or 1.8 mL 1.6 d n(Mg) = = 0.0658 mol 24.3

 V(H2) = 24.5 × 0.0658 L = 1.61 L Q28.

Propane (C3H8) burns in oxygen according to the equation:

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) Calculate: i the volume of oxygen at SLC used ii the volume of carbon dioxide at SLC produced when the following masses of propane react completely with excess oxygen: a 22 g b 5.0 g c 0.145 g d 16.5 g e 3.4 kg A28. The balanced equation shows that 1 mol of propane reacts with 5 mol of oxygen, and produces 3 mol of carbon dioxide.

C3H8(g) + 5O2(g)  3CO2 (g) + 4H2O(g)

m Remember that n = . (The molar mass of propane is 44 g mol–1.) M Remember that under SLC, 1 mol of any gas has a volume of 24.5 L. V Use the formula n = rearranged as V = n × Vm. Vm 22 a i n(C H ) = = 0.50 mol 3 8 44 n(O ) 5 2  n(C3H8 ) 1

 V(O2) = 5 × 0.50 × 24.5 = 61 L n(CO ) 3 ii 2  n(C3H8 ) 1

 V(CO2) = 3 × 0.50 × 24.5 = 37 L 5.0 b i n(C H ) = = 0.114 mol 3 8 44

 V(O2) = 5 × 0.114 × 24.5 = 14 L

ii  V(CO2) = 3 × 0.114 × 24.5 = 8.4 L 0.145 c i n(C H ) = = 0.0033 mol 3 8 44

 V(O2) = 5 × 0.0033 × 24.5 = 0.404 L

ii  V(CO2) = 3 × 0.0033 × 24.5 = 0.242 L 16.5 d i n(C H ) = = 0.375 mol 3 8 44

 V(O2) = 5 × 0.375 × 24.5 = 45.9 L

ii  V(CO2) = 3 × 0.375 × 24.5 = 27.6 L 3400 e i n(C H ) = = 77.3 mol 3 8 44

 V(O2) = 5 × 77.3 × 24.5 = 9500 L

ii  V(CO2) = 3 ×77.3 × 24.5 = 5700 L Q29. Octane is one of the main constituents of petrol. It burns according to the equation:

2C8H18(g) + 25O2(g)  16CO2(g) + 18H2O(g) What mass of octane must have been used if 50.0 L of carbon dioxide, measured at 120°C and 1.10 atm, was produced? A29. The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. m The second step involves using n = . The molar mass of octane is 114 g mol–1. M The balanced equation shows that 2 mol of octane produces 16 mol of carbon dioxide. PV n(CO ) = 2 RT

(1.10101)  50.0 = 8.31 (120  273) = 1.70 mol n(C H ) 2 8 18  n(CO 2 ) 16 2 2  n(C H ) = × n(CO ) = × 1.70 mol = 0.213 mol 8 18 16 2 16  m(C8H18) = n(C8H18) × M = 0.213 × 114 = 24.2 g Q30.

What volume of NO2 is produced when 0.5 L of nitrogen(II) oxide reacts with excess oxygen? (All volumes are measured at 25°C and 100 kPa.) The equation for the reaction is:

2NO(g) + O2(g)  2NO2(g) A30.

The balanced equation shows that 2 mol of NO2 is produced from 2 mol of NO. Therefore, the volumes of gases produced will be equal to the volume of NO reacted (as long as the temperature and pressure are constant).  V(NO) = V(NO2) = 0.5 L Q31. Calculate the volume of oxygen needed to react completely with 150 mL of carbon monoxide according to the following equation. Assume all volumes are measured at the same temperature and pressure.

2CO(g) + O2(g)  2CO2(g) A31.

The balanced equation shows that 1 mol of O2 reacts with 2 mol of CO. Therefore, the volumes of O2 and CO used also will be in the ratio 1 : 2 (as long as the temperature and pressure are constant).  V(CO) = 150 mL  V(O2) = 75 mL Q32. Hydrogen gas reacts with chlorine gas according to the equation:

H2(g) + Cl2(g)  2HCl(g) 3.0 L of hydrogen and 7.0 L of chlorine are allowed to react as much as possible. What will be the volume of the gaseous mixture at the end of the reaction, assuming all volumes are measured at the same temperature and pressure? A32.

The balanced equation shows that 1 mol of Cl2 reacts with 1 mol of H2, producing 2 mol of HCl. All of the 3.0 L of hydrogen will react, using up 3.0 L of chlorine. So there will be 4.0 L of chlorine in excess. A total of 6.0 L of HCl will be produced.  final volume = 4.0 + 6.0 = 10.0 L

Chapter review

Q33. List the main features of the kinetic theory of gases. A33. Gases are composed of small particles (atoms or molecules). The volume of the particles themselves is very much smaller than the volume occupied by the gas. The particles move in a rapid, straight-line motion and collide with each other and with the walls of the container. The forces between particles are extremely weak. The collisions between particles are elastic. The average kinetic energy of the particles is directly proportional to the kelvin temperature of the gas. Q34. Use the kinetic theory of gases to explain why: a the pressure of a gas increases if its volume is reduced at constant temperature b the pressure of a gas decreases if its temperature is lowered at a constant volume c in a mixture of gases, the total pressure is the sum of the partial pressure of each gas d the pressure of a gas, held at constant volume and temperature, will increase if more gas is added to the container A34. a As volume is reduced, there is an increase in the frequency of molecular collisions per unit wall area. This is measured as an increase in pressure. b When the temperature of a gas is lowered, the average kinetic energy of the particles decreases. The rate of collisions between particles and the walls of the container decreases and particles collide with less force. As pressure is a measure of the force of molecular collisions per unit wall area of the container, pressure is found to decrease. c In a mixture of gases, the particles of each gas are moving and colliding with the walls of the container, independently of each other. Each gas therefore exerts a pressure. As the gases behave independently of each other, total pressure is simply the sum of the individual gas (or partial) pressures. d When more gas is added to a container, the total number of particles in the container increases. Provided that the volume of the container and the temperature have not changed, the collisions of these additional particles means that the total pressure in the container has increased. Q35. Referring to scuba diving and Boyle’s law, answer the following questions: a As a diver dives deeper into the sea, the pressure on the body increases. What happens to the pressure inside the diver’s body? b What happens to the pressure inside a diver’s lungs if he or she returns to the surface without breathing? c A diver ascending too rapidly suffers ‘the bends’. Describe the bends and how it is treated. d Why should a diver not spend much time below 10 m?

A35. a The pressure also increases inside the diver’s body. b The pressure inside the lungs is high. Under these conditions, expansion of the lungs can cause injury. c The ‘bends’ are caused by gases, mainly nitrogen, coming out of solution in a diver’s blood when he or she ascends from a dive too rapidly. The gas can form small bubbles in joints, muscle and other tissue, causing serious injury and great pain. To treat the bends, a diver is placed in a decompression chamber and the pressure is increased until nitrogen bubbles re-dissolve in the blood. Pressure is then slowly reduced to one atmosphere. d The longer a diver stays down on a dive, the more nitrogen will dissolve in the blood and the slower the ascent will need to be to avoid problems from the bends. Q36. a If a container of gas is opened and some of the gas escapes, what happens to the pressure of the remaining gas in the container? b Use the kinetic molecular theory to explain what happens to the gas pressure in part a. A36. a The pressure inside the container is reduced when some of the gas escapes. b There are fewer gas molecules to collide with each other and the walls of the container. Pressure is the force exerted by the molecules over a defined area, so this will decrease. Q37. Temperatures in space approach absolute zero (–273°C) and atoms are very widely spaced. How do these factors influence the clothing worn by astronauts who are working outside their spacecraft? What would happen to an astronaut who did not have this protective clothing? A37. Astronauts need to be protected from the very low temperatures and pressures. Without a pressurised space suit, all air would rapidly escape from their lungs and they would die from lack of oxygen. Body tissue would also freeze after a short time. Q38. ‘A jar of air is a solution of gases.’ Comment critically on this statement by comparing the behaviour of particles in this jar to those in an aqueous solution. A38. The gas particles can mix fully at any concentration and move in any direction to fill a container of any size. Particles in an aqueous solution are confined to a specific volume and have a limit to their solubility in each other. Q39. Explain why: a there does not seem to be an upper limit to temperature b we can define an absolute zero of temperature

A39. a There appears to be no upper limit to the temperature scale since particles can move to very high speeds. The speed of a particle is related to its kinetic energy, which is related to the temperature of the surrounds. b A plot of volume against temperature for a gas provides a straight-line graph, which intersects the temperature axis at a certain point. This point is called the absolute zero of temperature. Because this graph can be drawn, it is possible to define absolute zero. Q40. A sample of carbon dioxide occupies a volume of 750 mL at a pressure of 100.0 kPa. If the temperature of the gas remains constant, calculate the pressure when the volume of the sample is changed to: a 250 mL b 2.0 L c 10.0 L A40.

Remember to use Boyle’s law: P1V1 = P2V2. Each of the pressure and volume units needs to be the same. This conversion should be applied as needed: 1 L = 1000 mL.

P1V1 a P2 = V2 750 = 100.0 × 250 = 300 kPa

P1V1 b P2 = V2 750 = 100.0 × 2000 = 38 kPa

P1V1 c P2 = V2 750 = 100.0 × 10 000 = 7.5 kPa Q41. A 2.0 L vessel containing oxygen at a pressure of 30 kPa is connected via a closed tap to a 4.0 L vessel containing nitrogen at a pressure of 60 kPa. If the tap is opened and the gases mix, what will be the total gas pressure in the vessels (assuming the temperature remains constant)? A41. Both gases will mix completely. Oxygen and nitrogen will not react under normal conditions. Gas from the 2 L cylinder will spread into the new total volume of 6 L. In the 2 L cylinder: P1V1 = P2V2.

P1V1 2 30 P2 = = = 10 kPa V2 6 Gas from the 4 L cylinder will spread into the 6 L volume.

P1V1 4 60 In the 4 L cylinder: P2 = = = 40 kPa V2 6 The new total pressure = 10 + 40 = 50 kPa. Q42. A sample of 6.0 L of air at 0°C is warmed to 100°C. Calculate the new volume of the sample if the pressure remains constant. A42. V V Remember to use Charles’ Law: 1 = 2 . The temperature must be in kelvin, the T1 T2 volume units must each be the same and the pressure must be constant. 6.0(100  273) V = 2 (0  273) = 8.2 L Q43. A car tyre is inflated to a pressure of 200 kPa at 20°C. At the end of a journey on a hot sunny day the pressure has risen to 215 kPa. Calculate the temperature (°C) of air in the tyres, assuming that the volume of the tyre hasn’t changed. A43.

P1 P The combined gas equation needs to be used: = 2 . Temperature should be in T1 T2 kelvin, pressure units need to be the same and the volume is constant.

P2T1 T2 = P1 215 (20  273) = 200 = 315 K or 42C Q44. A cylinder of nitrogen, volume 15 L, contains gas at 15°C under a pressure of 4500 kPa. Calculate the volume of gas provided by the cylinder when the temperature is 25°C and the pressure is 100 kPa. A44. PV P V The combined gas equation needs to be used: 1 1 = 2 2 . Temperatures must be in T1 T2 kelvin. Both the volume and pressure units need to be the same.

P1V1T2 V2 = T1P2

450015 (25  273) = (15  273) 100 = 698 L Q45. A 1.00 m3 balloon of helium at 100°C has its temperature lowered to 50°C at constant pressure. What is the new volume of the balloon? A45. V V Remember to use Charles’ law: 1 = 2 . Temperature must be in kelvin and volume T1 T2 units must each be the same. The pressure must be constant and remember that 1 m3 = 1000 L. 1.00103  (50  273) V2 = (100  273) = 866 L Q46. Calculate the temperature (°C) needed to change the volume of 10 L of helium at 2.0 atm and 20°C to 5.0 L at 1.0 atm. A46. PV P V The combined gas equation needs to be used: 1 1 = 2 2 . Temperatures must be in T1 T2 kelvin. Both the volume and pressure units need to be the same.

P2V2T1 T2 = P1V1 1.0 5.0 (20  273) = 2.010 = 73 K or –200C Q47. A balloon of volume 9.4 L contains 1.0 g of hydrogen gas. Another balloon, at the same temperature and pressure, contains 2.0 g of oxygen gas. Calculate the volume of the second balloon. A47. V V This law can be used: 1 = 2 . The temperature and pressure must remain constant. n1 n2 m Remember that n = . M The molar mass of hydrogen is 2.0 g mol–1 and that of oxygen is 32 g mol–1. 1.0 n(H ) = = 0.50 2 2.0 2.0 n(O ) = = 0.063 2 32

V1n2 V2 = n1 9.4 0.063 = 0.50 = 1.2 L Q48. 3.0 × 1023 molecules of oxygen occupy 9.5 dm3 at a given temperature and pressure. How many molecules of carbon dioxide will occupy 50.0 dm3 at that same temperature and pressure? A48. V V This law can be used: 1 = 2 . The temperature and pressure must remain constant. n1 n2 The number of moles is directly proportional to the number of molecules, so the two are exactly interchangeable.

n1V2 Number of CO2 molecules = V1 3.01023  50.0 = 9.5 = 1.6 × 1024 molecules. Q49. Refer to Table 21.4 (page 389), which gives the molar volumes of real gases at SLC. a Suggest two ways in which real gases differ from an ideal gas. b Which of the gases listed in the table has a molar volume which deviates most from that of an ideal gas? Suggest a possible explanation for this deviation. A49. a Molecules of real gases have a small but finite volume; those of ideal gases are assumed to be point particles. There are some forces of attraction between molecules of real gases, whereas it is assumed there are none between those of ideal gases. b ammonia; hydrogen bonding exists between molecules Q50. a Hot-air balloons rise because hot air is less dense than cooler air. In terms of Charles’ law, explain why the density of a gas decreases as the temperature is raised. b A chemistry student was explaining to a friend that as air is heated, the added heat is converted to kinetic energy so that the kinetic energy of all the air molecules increases. Comment critically on this statement, identifying any inconsistencies and making appropriate corrections. A50. a As temperature rises, the volume of a gas increases. Therefore, the same mass of gas occupies a greater volume at a higher temperature. Because density = mass/volume, the higher the temperature, the larger the volume and the lower the density.

Heinemann Chemistry 1 (4th edition)  Reed International Books Australia Pty Ltd Worked solutions to textbook questions 21 b As the temperature of a gas increases, the average kinetic energy of molecules increases. However, there will still be some molecules at the higher temperature with low kinetic energy. Q51. How many molecules of oxygen are present in 4.0 L of gas at 30°C and 1.5 atm? A51. The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. 23 Avogadro’s number: NA = 6.02 × 10 . PV n = RT (1.5101.325)  4.0 = 8.31 (30  273) = 0.241 mol 23 Number of particles = n × NA = 0.241 × 6.02 × 10 molecules = 1.45 × 1023 molecules Q52. Calculate the volume, in litres, occupied by 10.0 g of carbon dioxide at 25°C and 101.3 kPa. A52. The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa, and volume in L. R = 8.31 J K–1 mol–1. m Remember that n = . The molar mass of carbon dioxide is 44 g mol–1. M nRT V = P 10.0 8.31 (25  273) = × 44 101.3 = 5.56 L Q53. What is the mass of oxygen present in a 10.0 L container of oxygen at a pressure of 105 kPa and at 20°C? A53. The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. m The second step involves using: n = . The molar mass of oxygen is 32 g mol–1. M PV n(O ) = 2 RT 10510.0 = 8.31 (20  273) = 0.431 mol

m(O2) = 0.431 × 32 = 13.8 g Q54. At what temperature will 0.2 g of helium exert a pressure of 80 kPa in a container with a fixed volume of 4.0 L? A54. The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. m The second step involves using n = . The molar mass of helium is 4.0 g mol–1. M 0.20 n(He) = = 0.050 mol 4.0 PV T = nR 80 4.0 = 0.050 8.31 = 770 K Q55. A sample of gas of mass 10.0 g occupies a volume of 5.4 dm3 at 300 K and 100 000 Pa. a Calculate the amount (mol) of gas in the sample. b Determine the molar mass of the gas. A55. The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. a Conversions are: P = 100 000 Pa = 100 kPa V = 5.4 dm3 = 5.4 L PV n = RT 100 5.4 = 8.31 300 = 0.217 mol m b The second step involves using n = . M m M = n 10.0 = 0.217 = 46.2 g mol–1 Q56. Which container holds more molecules of oxygen gas: container A of volume 40.0 L at 25°C at 770 mmHg or container B of volume 0.10 L at 45°C at 390 mmHg?

A56. The general gas equation needs to be used: PV = nRT Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. Container A: PV n(O ) = 2 RT (770 /(760100))  40.0 = 8.31 (25  273) = 1.65 mol Container B: PV n(O ) = 2 RT (390 /(760101))  0.10 = 8.31 (45  273) = 0.0020 mol  Container A has more oxygen. Q57. A room has a volume of 220 m3. a Calculate the amount, in moles, of air particles in the room at 23°C and at a pressure of 100 kPa. b Assume that 20% of the molecules in the air are oxygen molecules and the remaining molecules are nitrogen. Calculate the mass of air in the room. A57. The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. a Conversions are: 220 m3 = 220 × 103 dm3 = 220 × 103 L PV n(air particles) = RT 100 220103 = 8.31 (23  273) = 8940 mol m b The second step involves using n = . M The molar mass of oxygen is 32 g mol–1 and of nitrogen is 28 g mol–1. 20% of the room is oxygen. 20  n(O ) = × 8940 2 100 = 1788 mol

 m(O2) = 1788 × 32 = 57.2 kg 80% of the room is nitrogen. 80  n(N ) = × 8940 2 100 = 7152 mol

 m(N2) = 7152 × 28 = 200 kg  Total mass of gas = 57.2 + 200 = 257 kg

Q58. Use the molar volume of a gas at SLC to find: a the volume occupied by 8.0 g of oxygen at SLC b the mass of nitrogen(IV) oxide present in 10 L at SLC A58. m This involves using n = and V = n × V . You could combine these two formulas to M m m give V = × V . M m a The molar mass of oxygen is 32 g mol–1. 8.0 V(O ) = × 24.5 2 32 = 6.1 L –1 b The molar mass of NO2 is 46 g mol . V 10  n(NO2) = = mol Vm 24.5 = 0.408 mol

 m(NO2) = 0.408 × 46 = 18.7 g Q59. Calculate the number of molecules of gas in: a 5.0 L at SLC b 100 L at SLC c 10 mL at SLC d 10 mL at STP e 103 L at STP f 156 L at STP A59. V 23 Remember: NA = 6.02 × 10 . Number of particles = n × NA = × NA Vm a At SLC, molar volume is 24.5 L. 5.0  Number of molecules = × N 24.5 A = 1.2 × 1023 b At SLC, molar volume is 24.5 L. 100  Number of molecules = × N 24.5 A = 2.46 × 1024 c At SLC, molar volume is 24.5 L. 0.010  Number of molecules = × N 24.5 A = 2.4 × 1020 d At STP, molar volume is 22.4 L.

 Number of molecules = 22.4 × NA = 2.7 × 1020 e At STP, molar volume is 22.4 L. 1000

 Number of molecules = 22.4 × NA = 2.7 × 1025 f At STP, molar volume is 22.4 L. 156  Number of molecules = × N 22.4 A = 4.2 × 1024 Q60. a Calculate the mass of 1 mole of carbon dioxide. b What is the volume of 1 mole of carbon dioxide at SLC? c Given that density is defined as mass/volume, calculate the density of carbon dioxide at SLC. d Would you expect the density of carbon dioxide at STP to be less than, equal to or greater than its density at SLC? Justify your answer. A60.

–1 a The molar mass of CO2 = 44 g mol and m = n × M. m(CO2) = 1 × 44 = 44 g b The volume of any gas at SLC = 24.5 L. 44 c Density (CO ) = 2 24.5 = 1.8 g L–1 d Greater, because volume decreases as temperature decreases. Q61. Methane will burn in excess oxygen according to the equation:

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) This reaction produces 5 L of carbon dioxide at 200°C and 100 kPa. Assuming all volumes are measured at the same temperature and pressure, calculate: a the volume of methane used b the volume of oxygen used c the mass of water vapour produced A61. The balanced equation shows that 1 mol of methane reacts with 2 mol of oxygen to give 1 mol of carbon dioxide, and 2 mol of water. At constant temperature and pressure, mole and volume are directly proportional to each other. a  V(CH4) used = 5 L b  V(O2) used = 10 L

m c Use the general gas equation, PV = nRT, and n = . The molar mass of water is M 18 g mol–1. n( H O) 2 2  n(CO 2 ) 1

n(H2O) = 2 × n(CO2) PV 2100 5 n(H2O) = 2 × = = 0.254 mol RT 8.31 (200  273)  m(H2O) = 0.254 × 18 = 4.58 g Q62. a Write a balanced equation for the decomposition of calcium carbonate by heat. b How many mole of carbon dioxide can be produced by the decomposition of 20 g of calcium carbonate? c What volume of carbon dioxide would this represent at SLC? A62. a CaCO3(s)  CaO(s) + CO2(g) b The balanced equation shows that 1 mol of calcium carbonate produces 1 mol of carbon dioxide. m Remember: n = . The molar mass of calcium carbonate is 100 g mol–1. M 20 n(CaCO ) = 3 100 = 0.20 mol V c Because these are SLC conditions, use the formula n = rearranged to Vm

V = n × Vm. V = 0.20 × 24.5 = 4.9 L Q63. Propane undergoes complete combustion as follows:

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) All volumes are measured at 120°C and 102 kPa. When 80 mL of propane and 500 mL of oxygen are reacted: a one of the gases does not react completely. Which gas is it and how much of it is unreacted? b what volumes of carbon dioxide and water are produced in the reaction? c what change in the total volume of all the gases has occurred as a result of the reaction? A63. a The balanced equation shows that the mole ratio of propane to oxygen is 1 : 5. All temperatures and pressures are the same, therefore mole and volume are directly proportional to each other.  80 mL of propane would need 400 mL of oxygen to react completely, and 500 mL of oxygen would need 100 mL of propane to react completely.

 There is excess oxygen, which means that all of the propane will react, and that there will be 100 mL of oxygen in excess. b V(CO2) = 3 × 80 = 240 mL V(H2O) = 4 × 80 = 320 mL c The original volume of gases = 80 + 500 = 580 mL The final volume of gases = 100 + 240 + 320 = 660 mL.  There was an increase in volume of 80 mL. Q64. 0.240 g of aluminium is reacted completely with a solution of hydrochloric acid to form aluminium chloride and hydrogen gas. a Write a balanced equation for the reaction. b What volume of hydrogen is produced in the reaction at 101 kPa and 22°C? A64. a 2Al(s) + 6HCl(aq)  2AlCl3(aq) + 3H2(g) b The balanced equation shows that 2 mol of aluminium produces 3 mol of hydrogen. m Remember that n = . The molar mass of aluminium is 27 g mol–1. M 0.240 n(Al) = 27 = 0.0089 3 n(H )/n(Al) = 2 2 3 n(H ) = × n(Al) 2 2 3 0.0089 n(H ) = 2 2 = 0.013 mol Use the general gas equation, PV = nRT. nRT V = P 0.0138.31 (22  273) = 101 = 0.32 L Q65. There are many scientists investigating alternative fuels to replace fossil fuels. A group of Japanese chemists is investigating the following reaction as a source of methane:

CaCO3(s) + 4H2(g)  CH4(g) + Ca(OH)2(s) + H2O(g) At 400°C, 100 kPa and under suitable reaction conditions: a what volume of methane is produced if 100 L of hydrogen is reacted completely? b what mass of calcium carbonate is used in part a?

A65. The balanced equation is:

CaCO3(s) + 4H2(g)  CH4(g) + Ca(OH)2(s) + H2O(g) It shows the mole ratio is 1 : 4 : 1 : 1 : 1. a When temperature and pressure are constant, mole and volume are directly proportional to each other.  The volume of methane produced will be ¼ the volume of hydrogen used.  The volume of methane will be 25 L. b The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1mol–1. PV n(H ) = 2 RT 100100 = 8.31 (400  273) = 1.79 mol n(CaCO ) 1 The mole ratio shows: 3  n( H 2 ) 4 1.79  The number of mole CaCO = = 0.447 mol 3 4 –1 The molar mass of CaCO3 = 40 + 12 + 48 = 100 g mol m Since n = M m(CaCO3) = nM = 0.447 mol × 100 g mol–1 = 44.7 g Q66. Consider 1.0 mL of hydrogen gas at 25°C and 101 kPa. a Calculate the number of hydrogen molecules in this sample. b If the total number of collisions per second in this container is 1.5 × 1029, how many collisions does each molecule experience? c Describe what happens to the kinetic energy of a hydrogen molecule as a result of these collisions. d Describe an experiment you could perform to support the theory that kinetic energy is conserved in these collisions. e The kinetic molecular theory of gases ignores the existence of forces between molecules in an ideal gas. Briefly describe how your life would be different if the kinetic molecular theory held true for all gases. A66. a The general gas equation needs to be used: PV = nRT. Temperature should be in kelvin, pressure in kPa and volume in L. R = 8.31 J K–1 mol–1. PV n = RT 101 (1.010 3 ) = 8.31 (25  273) = 4.07 × 10–5 mol

23 Remember: Avogadro’s number, NA = 6.02 × 10 .

 Number of molecules = nNA = 4.07 × 10–5 × 6.02 × 1023 = 2.5 × 1019 1.51029 b Number of collisions for each molecule= 2.51019 = 6.1 × 109 c The average kinetic energy of the hydrogen molecules remains constant as long as temperature is constant. However, the actual kinetic energy of any one hydrogen molecule could vary significantly over time. d Place a sample of gas in a thermally insulated container. Use a sensitive temperature probe to monitor the temperature of the gas over time. If collisions are not elastic, the average kinetic energy of the molecules and, hence, the temperature of the gas, would decrease with time. e If there were no forces between particles in the gaseous state then there would be little tendency for particles to come together and form liquids. A substance, once in its gaseous state, would remain in that state. Q67. Consider two containers of equal size. One contains oxygen and the other carbon dioxide. Both containers are at 23°C and at a pressure of 1.0 atm. a Compare the average molecular kinetic energy of these two gases. b The molecules of which of the two gases have the greater average velocity? c Which of the two samples of gas contains more molecules? d Which of the two samples of gas contains the greater number of atoms? e Which of the two gases has the greater density? Give a reason for each of your answers. A67. a Equal. Average kinetic energy is proportional to the temperature of the gas, and at the same temperature is not dependent on the particular gas, so they will be equal. b Oxygen. As average kinetic energy = ½mv2, the lighter gas particles will have the –1 –1 greater average velocity. M(O2) = 32 g mol , M(CO2) = 44 g mol . So the oxygen molecules will have the greater average velocity. c Equal. With pressure, volume and temperature the same, n will be the same. d Carbon dioxide. Each CO2 molecule contains 3 atoms and each O2 molecule contains 2 atoms. As there are an equal amount of molecules of each gas, there are more atoms in the CO2 sample. e Carbon dioxide. Density = mass ÷ volume. The volume is the same for each gas, but the mass of CO2 is greater, so it has the greater density.

Unit 2 Area of Study 2

Multiple-choice questions

Q1. Which one of the following does not occur during the carbon cycle? A Oxygen is added to the atmosphere by the process of decomposition. B Oxygen is removed from the atmosphere by the process of combustion. C Carbon dioxide is added to the atmosphere by the process of respiration. D Carbon dioxide is removed from the atmosphere by the process of photosynthesis. A1. A. This response is incorrect because oxygen is removed from the air during the process of decomposition. Q2. In the laboratory nitrogen(II) oxide is usually prepared by: A the fractional distillation of air B combustion of nitrogen gas in oxygen C the reaction between copper and concentrated nitric acid D the reaction between calcium carbonate and dilute nitric acid A2. D. The equation for the reaction is:

3Cu(s) + 8HNO3(aq)  3Cu(NO3)2(aq) + 4H2O(l) + 2NO(g) Concentrated acid tends to form NO2(g). Q3. The main gases of the atmosphere, occurring in decreasing order of abundance, are: A argon, carbon dioxide, nitrogen, oxygen B nitrogen, oxygen, argon, carbon dioxide C carbon dioxide, argon, oxygen, nitrogen D nitrogen, oxygen, carbon dioxide, argon A3. B. See the following table, which shows the composition of clean, dry air near sea level.

Gas Content by amount (mol %) Total mass in atmosphere ( 1012 tonnes) Nitrogen 78.09 3900 Oxygen 20.94 1200 Argon 0.93 6.7 Carbon dioxide 0.035 2.5 Neon 0.0018 0.065 Helium 0.00052 0.004 Methane 0.00015 0.0042 Krypton 0.0001 0.017 Carbon monoxide 0.00001 0.0006 Ozone 0.000002 0.003 Nitrogen dioxide 0.0000001 0.000013 Others Very small amounts Q4. The first step in the formation of ozone in the upper levels of Earth’s atmosphere is: A oxygen atoms combining with oxygen molecules B oxygen molecules being converted to oxygen atoms C oxygen atoms being converted to oxygen molecules D oxygen atoms combining with each other to form ozone molecules A4. B. The equations for the reactions in which ozone is formed are: UV O2(g)  O(g) + O(g)

Then O(g) + O2(g) + M  O3(g) + *M, where *M is a molecule that takes up energy released in the second reaction. Q5. Which of the following is involved in the formation of photochemical smog? A neon gas B nitrogen gas C carbon monoxide D unburned hydrocarbons A5. D. Photochemical smog is formed from the interaction of sunlight with a mixture of oxides of nitrogen and unburned hydrocarbons. Q6. One test for a sample of carbon dioxide gas in a test-tube is that it will: A turn damp red litmus paper blue B cause a glowing splint to burst into flame C produce a ‘pop’ when a lighted splint is applied to it D form a white precipitate when bubbled through limewater A6. D. A, B and C are tests for, respectively, ammonia, oxygen and hydrogen.

Q7. Which of the following points of the kinetic molecular theory can be used to explain that the volume of a fixed mass of gas is inversely proportional to pressure at a constant temperature? A Collisions between particles are elastic. B Forces between particles are extremely weak. C The kinetic energy of the particles is proportional to the temperature of the gas. D The volume of gas particles is very small compared with the volume occupied by the gas. A7. D. Because in a gas the particles are widely separated, gases are compressible. As the volume of the gas decreases, the particles are forced closer together, so there are more collisions of particles per second with the vessel wall, and so pressure will increase. Q8. As a diver comes to the surface, the pressure inside his or her lungs changes from 200 to 100 kPa. Assuming that the volume of the gas in the lungs was initially 3.0 L, at the surface the volume of the same amount of gas, in L, would be: A 1.5 B 2.0 C 4.5 D 6.0 A8. D. Volume is inversely proportional to pressure under these conditions. If the pressure is halved, the volume will double. Q9. The pressure inside a gas cylinder is 800 kPa. The cylinder contains 4.00 mol nitrogen gas, 10.0 mol oxygen gas and 2.00 mol argon gas. What is the partial pressure exerted by the oxygen gas, in kPa? A 0.625 B 10.0 C 500 D 1.30  103 A9. C n(O ) Step 1: Find mole fraction of O by using 2 . 2 total number of moles of gases 10 Mole fraction of O2 = = 0.625 4 10  2 Step 2: Find partial pressure of O2.

P(O2) = mole fraction of O2  total pressure = 0.625  800 = 500 kPa

Q10. If 64.0 g of oxygen gas occupies a volume of 25.0 L when the temperature is 30.0C, then the pressure of the gas, in kPa, will be closest to: A 20.0 B 200 C 400 D 6.40  103 A10. B Step 1: Calculate n(O2). m n = M 64.0 = 32 = 2.00 mol Step 2: Convert given quantities to correct units. n = 2.00 mol R = 8.31 J K–1 mol–1 T = 303 K V = 25.0 L Step 3: Calculate P(O2). nRT P = V 2.00 8.31 303 = 25.0 = 200 kPa Q11. Under certain conditions, hydrogen sulfide will burn in oxygen according to the equation:

2H2S(g) + O2(g)  2S(s) + 2H2O(g) If all measurements were made at the same temperature and pressure, the volume of water vapour, in mL, produced if 100 mL hydrogen sulfide reacted with 25.0 mL oxygen would be: A 12.5 B 25.0 C 50.0 D 100 A11. C Step 1: Since all measurements are made at the same temperature and pressure, the equation gives us not only the mole ratio with which the gases react with each other, but also the volume ratio.

 2H2S(g) + O2(g)  2S(s) + 2H2O(g) Gas volume ratio: 2 : 1 : 2 (Note that the sulfur is a solid under these conditions.) Step 2: Calculate V(O2) needed to react with 100 mL H2S.

V (H2S) V(O2) = 2 100 = 2 = 50.0 mL But there is only 25.0 mL O2 available, so H2S is in excess. Therefore, to calculate the volume of water produced we must work from the volume of O2. Step 3: Calculate V(H2O), using ratio in equation.

V(H2O) = 2  V(O2) = 2  25.0 = 50.0 mL Q12. 4.0 L of hydrogen gas is collected in a syringe. Then the pressure and the temperature (in kelvin) of the gas inside the syringe are both doubled. When this happens the volume of the hydrogen gas, in L, in the syringe would be: A 1.0 B 4.0 C 6.0 D 16 A12. B. The volume of a gas is proportional to the temperature (in kelvin) but inversely proportional to the pressure. In this situation the two effects cancel, so the volume remains at 4.0 L. Q13. The gas in the Earth’s stratosphere that absorbs the most ultraviolet radiation is: A ozone B methane C carbon dioxide D chlorofluorocarbon A13. A. All four gases listed are significant greenhouse gases because they absorb infrared radiation given out from the Earth’s surface. The gas that absorbs most ultraviolet radiation in the stratosphere is ozone. Q14. In the process of nitrogen fixation: A animals obtain protein by eating plants B nitrates are converted to protein in plants C atmospheric nitrogen is converted into nitrates D protein breaks down into ammonium compounds A14. C. Nitrogen fixation is the process whereby nitrogen from the air is converted into a form (e.g. nitrates or ammonium compounds) that can be absorbed by plants.

–1 Container A is filled with 1 mol of liquid water (H2O) (molar mass 18 g mol ) at a temperature of 80°C and pressure of 1 atm. –1 Container B holds 1 mol of gaseous ethane (C2H6) (molar mass 30 g mol ) also at a temperature of 80°C and at a pressure of 1 atm. Compared with particles in container B, particles in container A will be: A further apart and moving faster B further apart and moving slower C closer together and moving faster D closer together and moving slower A15. C. A is a liquid, so particles will be closer together than in gas B. The temperature is 1 the same in both containers, so the kinetic energy ( mv2) will be the same for the 2 particles in each container. Since the mass of molecules of A is less than that of molecules of B, the average velocity of molecules of A must be greater than that for molecules of B. Q16. The diagram shows, for a given sample of a gas, the proportion of molecules with a particular kinetic energy at a temperature of 80°C.

Which of the following diagrams best represents the situation for the same sample of gas at 30°C?

A16. B. At the lower temperature, there are more molecules with a lower kinetic energy. Q17. This graph could represent the relationship, for a fixed mass of gas, between its: A pressure and volume at a constant temperature B pressure and temperature at a constant volume C volume and temperature in K at a constant pressure D volume and temperature in °C at a constant pressure

A17. A. Volume is inversely proportional to pressure. Q18. Methane burns in oxygen according to the equation:

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

If all measurements were made at 200°C and 1 atm pressure, what would be the total volume of gases after the reaction if 20 mL of methane burned in 20 mL of oxygen? A 20 mL B 30 mL C 40 mL D 50 mL A18. C. Since all measurements are made at the same temperature and pressure, the equation gives us not only the mole ratio with which the gases react with each other, but also the volume ratio. Oxygen is the limiting reagent. So after the reaction, there will be no oxygen remaining. 10 mL carbon dioxide is produced. 20 mL of water vapour is produced. 10 mL methane is unreacted. So total amount of gas after the reaction = 40 mL.

Short-answer questions

Q19. a The gases of the atmosphere are vital to life on Earth and yet, at times, they are taken for granted. For each of the gases listed, write a sentence outlining their importance to living things. i oxygen ii ozone iii water vapour iv carbon dioxide b Briefly explain the difference between the terms greenhouse effect and global warming. A19. a i Oxygen is required for the process of respiration in living things—plant and animal. ii Ozone in the stratosphere absorbs harmful ultraviolet radiation, which can have a damaging effect on living things. iii Water vapour in the atmosphere plays a vital part in our weather patterns. Water is essential to living things. iv Carbon dioxide is used by green plants to make their food in the process of photosynthesis. b The greenhouse effect is the warming of the Earth’s atmosphere by solar radiation being trapped in the Earth’s atmosphere. Life on Earth depends on the greenhouse effect to maintain the Earth at a moderate temperature. Global warming is a slow but steady increase in the average temperatures of the Earth’s atmosphere caused by an enhanced greenhouse effect. This, in turn, has been caused by an increase in the proportion of greenhouse gases, such as carbon dioxide, in the atmosphere. Q20. When petrol is burned in a car engine the major products of the combustion reaction are carbon dioxide and water.

a Taking octane (C8H18) as one of the constituents of petrol, write a balanced equation to represent its combustion to carbon dioxide and water. b Sometimes petrol does not burn completely—for example, in an engine that has not been tuned properly. This is because the oxygen supply is limited and is not sufficient to convert the octane completely to carbon dioxide and water. Under these conditions, one of the products of combustion will be carbon monoxide. Write a balanced equation to show the production of carbon monoxide when octane burns in a limited supply of oxygen. Assume that carbon monoxide is the only carbon-containing compound produced. c Nitrogen(II) oxide (NO) is formed when nitrogen gas (N2) and oxygen gas (O2) react with each other at the high temperatures that exist in car engines. Write a balanced equation for the reaction between N2 and O2 to form nitrogen(II) oxide. d One of the components of the brown haze in a city blanketed by smog is the gas nitrogen(IV) oxide (NO2). Nitrogen(IV) oxide is formed when nitrogen(II) oxide oxidises in air. Write an equation for the oxidation of nitrogen(II) oxide in air. e Which chemicals react in air to produce photochemical smog? A20. a 2C8H18(g) + 25O2(g)  16CO2(g) + 18H2O(l) b 2C8H18(g) + 17O2(g)  16CO(g) + 18H2O(l) Balanced equations that contain mixtures of CO and C or CO2 as products are also acceptable. c N2(g) + O2(g)  2NO(g) d 2NO(g) + O2(g)  2NO2(g) e Oxides of nitrogen and unburned hydrocarbons react in the presence of sunlight to form photochemical smog. Q21. Explain the following by using the kinetic molecular theory. a What is meant by the term diffusion? b Two strong-smelling gases, ammonia (NH3) and hydrogen sulfide (H2S), are released from containers at one end of a sealed room. Which gas would you expect to spread more quickly to the other end of the room? c A gas cylinder contains a mixture of two gases, methane and ammonia, under pressure. i Explain, in this situation, what is meant by the partial pressure of methane. ii In the cylinder there are 1.50 mol methane gas and 4.50 mol ammonia gas. The total pressure inside the cylinder is 500 kPa. What is the partial pressure of methane in the cylinder? A21. a Diffusion is the process by which a gas will spread throughout the space available to it, due to the random motion of the particles of the gas. b The molar mass of ammonia is 17 g mol–1 and of hydrogen sulfide is 34 g mol–1. In this situation, the temperature of the gases is the same, so the kinetic energy of 1 their molecules (= mv2) is the same. Since the mass of ammonia molecules is 2 less than that of hydrogen sulfide molecules, the velocity of ammonia molecules must be greater than that of hydrogen sulfide molecules. Therefore, it could be

expected that ammonia molecules would spread more quickly than hydrogen sulfide molecules. c i The individual particles of gases are completely separated from each other and are in a constant state of random motion. All particles will collide with the inside walls of the cylinder, but will do this independently of each other. The partial pressure of methane is that proportion of the total pressure that is caused by the particles of methane colliding with the cylinder walls. ii Step 1: Calculate the mole fraction of methane. n(CH ) Mole fraction = 4 total moles of gas 1.50 = 1.50  4.50 1.50 = 6.00 = 0.250 Step 2: Calculate partial pressure of methane. Partial pressure = mole fraction  total P = 0.250  500 = 125 kPa Q22. A particular rock is known to contain magnesium carbonate. There is no other metal carbonate present in the rock. A sample of the rock weighing 12.8 g is finely ground and then added to excess hydrochloric acid solution in a reaction vessel. The magnesium carbonate in the rock dissolves in the acid to produce 2.60 L of carbon dioxide gas at SLC. a Write an equation for the reaction between the magnesium carbonate in the rock and the hydrochloric acid. b Calculate the amount, in mol, of carbon dioxide produced in this reaction. c Calculate the amount, in mol, of magnesium carbonate in the rock. d Determine the percentage, by mass, of magnesium carbonate in the rock. e Describe a test for carbon dioxide. f Write an equation for a process that removes carbon dioxide from the atmosphere. A22. a MgCO3(s) + 2HCl(aq)  MgCl2(aq) + H2O(l) + CO2(g) b n(CO2) = V/24.5 = 2.60/24.5 = 0.106 mol c From the equation: n(MgCO3) = n(CO2) = 0.106 mol m(MgCO3) = n × M = 0.106 × 84.3 = 8.95 g d %MgCO3 = 8.95/12.8 × 100 = 69.9%

Heinemann Chemistry 1 (4th edition)  Reed International Books Australia Pty Ltd Worked solutions to textbook questions 40 e When carbon dioxide is bubbled through a solution of limewater, the solution will turn ‘milky’ due to the formation of a precipitate of calcium carbonate.

Ca(OH)2(aq) + CO2(g)  CaCO3(s) + H2O(l) f Photosynthesis: 6CO2(g) + 6H2O(l)  C6H12O6(aq) + 6O2(g) + – or dissolving in oceans, lakes, etc. CO2(g) + 2H2O(l)  H3O (aq) + HCO3 (aq) Q23. A gas storage plant holds natural gas in a storage vessel with a floating roof. The plant holds 4.01  106 L on a day when the air temperature is 24.0C and the air pressure is 759 mmHg. Calculate the volume of gas held on a day when the air temperature is 30.0°C and the air pressure is 764 mmHg. A23. Step 1: Convert given quantities to correct units. P1 = 759 mmHg 6 V1 = 4.01  10 L T1 = 297 K P2 = 764 mmHg T2 = 303 K Step 2: Calculate V2. P1V1T2 V2 = T1P2 759 4.01 106  303 = 297  764 = 4.06  106 L Q24. Carbon dioxide can be made in a number of different ways, including fermentation of glucose (C6H12O6), heating limestone (CaCO3) and adding hydrochloric acid (HCl) to limestone. a Write an equation for each of these three methods of preparing carbon dioxide. b Which of these three methods are used to produce carbon dioxide commercially? c Give three uses of carbon dioxide. A24. a Fermentation of glucose: C6H12O6(aq)  2C2H5OH(aq) + 2CO2(g)

Heating limestone: CaCO3(s)  CaO(s) + CO2(g) Adding limestone to hydrochloric acid:

CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g) b Fermentation of glucose and heating limestone. c Dry ice as a refrigerant, fire extinguishers, ‘fizzy’ drinks, super critical carbon dioxide as a replacement for halogenated organic solvent. Q25. Oxygen can be prepared in the laboratory by the catalytic decomposition of hydrogen peroxide. a Write a balanced equation for this reaction. b What volume of oxygen, measured at 27°C and 750 mmHg, can be obtained from the decomposition of 10.0 g of hydrogen peroxide?

Heinemann Chemistry 1 (4th edition)  Reed International Books Australia Pty Ltd Worked solutions to textbook questions 41 c Hydrogen peroxide acts as an acid in water. Write an equation to illustrate the acidic behaviour of hydrogen peroxide. A25. a 2H2O2(aq)  2H2O(l) + O2(g) b The balanced chemical equation shows the mole ratio is 2 : 2 : 1. m Since n = , and the molar mass of H O = 2 + 32 = 34 g mol–1, then M 2 2 10.0 n(H O ) = = 0.294 mol 2 2 34 From the equation, n(O ) 1 2  n( H 2O2 ) 2 0.294 n(O ) = = 0.147 mol 2 2 The volume of oxygen can be determined by using the general gas equation, PV = nRT. nRT  V = P 0.147  8.31 (27  273) = (750 /(760101)) = 3.68 L – + c H2O2(aq) + H2O(l)  HO2 (aq) + H3O (aq) Q26.

Sulfur trioxide (SO3) is a gas that reacts readily with water to produce sulfuric acid. a Write an equation for this reaction. b Sulfuric acid is a strong, diprotic acid. Explain the meaning of the words in italics and illustrate your answer with appropriate equations. c 1.68 L, measured at SLC, of sulfur trioxide was bubbled through 1.5 L of water. i What amount, in mol, of SO3 was bubbled through the water? ii Assuming that there is no increase in volume of solution as a result of bubbling the gas, calculate the concentration of the sulfuric acid produced. iii Calculate the pH of the sulfuric acid solution formed. What assumption did you make in reaching your answer? A26. a SO3(g) + H2O(l)  H2SO4(aq) b Sulfuric acid is described as a diprotic acid because it can donate two protons to a base such as water. The equations that show this are: + – H2SO4(l) + H2O(l)  H3O (aq) + HSO4 (aq) followed by: – + 2– HSO4 (aq) + H2O(l)  H3O (aq) + SO4 (aq) Sulfuric acid is a strong acid because the first of these reactions proceeds to – completion, that is, H2SO4 donates a proton readily (HSO4 is a much weaker acid).

Heinemann Chemistry 1 (4th edition)  Reed International Books Australia Pty Ltd Worked solutions to textbook questions 42 c The balanced chemical equation shows the mole ratio is 1 : 1 : 1. At SLC, 1 mol V of any gas has a volume of 24.5 L. Calculate the amount using n = . Vm Remember that 1 dm3 = 1 L. 1.68 i n(SO ) = = 0.0686 mol 3 24.5 n(H SO ) 1 ii The mole ratio shows: 2 4 = n(SO3 ) 1

 n(H2SO4) = n(SO3) n Calculate the concentration of H SO using c = . 2 4 V 0.0686  [H SO ] = M 2 4 1.5 = 0.0457 M + iii To determine pH, [H3O ] needs to be calculated. + If the sulfuric acid is completely ionised, the [H3O ] = 2 × [H2SO4]. +  pH = –log10[H3O ] = –log10[2 × 0.0457] = 1.04 The assumption made here is that the second stage of ionisation occurs completely. If you were to assume that only the first ionisation step occurs completely, and the second not at all, then a pH of 1.34 would be calculated. Q27. A compound of carbon and hydrogen only contains 82.8% by mass of carbon. a Determine the empirical formula of the compound. b At 100°C, the compound is a gas. At that temperature, 2.63 g of that compound occupies 1.34 L at a pressure of 105 kPa. i Calculate the number of moles of gas in the 2.63 g sample. ii Determine the molar mass of the compound. c Use your answers to parts a and b to determine the molecular formula of the gas. d This compound exists as two structural isomers. Draw the structural formulas of these isomers. A27. a If the compound is 82.8% carbon, then as it is a hydrocarbon, there is 17.2% hydrogen. The molar masses are 12 g mol–1 and 1 g mol–1, respectively. C H Mass 82.8 g 17.2 g Molar mass 12 g mol–1 1 g mol–1 m 82.8 17.2 Amount, using n = n = = 6.9 mol n = = 17.2 mol M 12 1 Divide all by smallest 6.9 17.2 = 1 = 2.5 amount 6.9 6.9 Round off to whole 2 5 numbers

 empirical formula is C2H5

Heinemann Chemistry 1 (4th edition)  Reed International Books Australia Pty Ltd Worked solutions to textbook questions 43 b i Use the general gas equation: PV = nRT. PV n = RT 1051.34 = 8.31 (100  273) = 0.0454 mol ii Find the molar mass of the molecular formula. m n = M m M = n 2.63 = 0.0454 = 57.9  58 g mol–1 –1 c The empirical formula, C2H5, has a molar mass of 24 + 5 = 29 g mol . The molecule must contain a whole number of C2H5 units. molar mass of molecule (molecular formula) Number of units in a molecule = molar mass of one unit (empirical formula) 58 = 29 = 2  Molecular formula has twice the number of atoms of each element as the empirical formula.

 The molecular formula is C4H10. d

Q28. The examples of gases studied in this Area of Study are all compounds found in Chapter 7. They are covalent molecular compounds. However, many other substances can exist in the gaseous state, provided the temperature is high enough. For example, sodium and sodium chloride have boiling temperatures of 892°C and 1465°C, respectively. a Why do sodium and sodium chloride have such high boiling temperatures? b In the gaseous state, the particles of these two substances will be moving randomly, as described by the kinetic molecular theory. What particles might be present in these two gases? A28. a Sodium consists of a lattice of sodium ions surrounded by a sea of electrons. There is strong metallic bonding in three dimensions throughout the lattice.

Sodium chloride consists of a three-dimensional lattice of sodium cations and chloride anions held together by strong ionic bonding. b Na(g) and NaCl(g) or Na+(g) and Cl–(g). Q29. Ethanol has been suggested as an alternative source of fuel for cars. When ethanol (C2H6O) undergoes complete combustion it forms carbon dioxide (CO2) and water (H2O). a Write a balanced equation that represents the complete combustion of ethanol. b What mass (g) of carbon dioxide would be formed if 1.000 kg of ethanol reacts? c What volume of carbon dioxide would be formed at STP? d What are the advantages and disadvantages of using ethanol as an alternative fuel for cars? e Suggest a structure for ethanol. A29. a C2H6O(l) + 3O2(g)  2CO2(g) + 3H2O(g)

m(C2H 5OH) b n(C2H5OH) = M (C2H5OH) 1000 = 46.0 = 21.7 mol From the equation: n(CO2) = 2 n(C2H5OH) = 1 n(CO2) = 2 × n(C2H5OH) = 43.5 mol m(CO2) = n(CO2) × M(CO2) = 43.5 × 44 = 1913 g = 1.91 kg (to 3 significant figures) c 1 mol of any gas occupies 22.4 L at STP. So V(CO2) at STP = 22.4 × 43.5 = 974 L d Advantages: clean burning fuel and renewable. Disadvantages: higher cost than petrol; not recommended by some car manufacturers in quantities above 10%. e

Q30. Hydrogen has also been suggested as an alternative fuel and can be generated by adding magnesium and hydrochloric acid. a Determine the relative atomic mass of magnesium given the following data.

Isotope Atomic mass Abundance % 24Mg 23.98504 78.99 25Mg 24.98584 10.00 26Mg 25.98259 11.01 b Write a balanced equation to represent the reaction of magnesium and hydrochloric acid. c Calculate the amount, in mol, present in 15.00 g of magnesium using the relative atomic mass determined in part a. d Calculate the volume of hydrogen formed at SLC if 15.00 g of magnesium reacts completely. A30. 78.9923.9850410.0024.9850411.0125.98259 a A (Mg) = r 100 = 24.3 b Mg(s) + 2HCl(aq)  MgCl2(s) + H2(g) 15.00 c n(Mg) = 24.3 = 0.617 d From equation: n(Mg) = n(H2) n(H2) = 0.617 1 mole of any gas occupies 24.5 L at SLC. So V(H2) produced = 0.617 × 24.5 = 15.1 L Q31. Electricity bills now contain a graph displaying the amount of greenhouse gas emissions produced in generating the electricity for a household. Consider the following graph. How many tonnes of greenhouse gases have been released in generating the electricity for this household in the past year?

A31. Tonnes of greenhouse gases emitted = 4.38 + 4.44 + 4.83 + 4.79 = 18.36 tonnes.

Heinemann Chemistry 1 (4th edition)  Reed International Books Australia Pty Ltd