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Reactions that cause change of oxidation numbers are called redox reactions. Element loses electrons → its oxidation number increases → element is oxidized.

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Presentation on theme: "Reactions that cause change of oxidation numbers are called redox reactions. Element loses electrons → its oxidation number increases → element is oxidized."— Presentation transcript:

1 Reactions that cause change of oxidation numbers are called redox reactions. Element loses electrons → its oxidation number increases → element is oxidized → oxidation reaction Substance that contains the oxidized element is call the reducing agent. Substance that contains the reduced element is call the oxidizing agent. Element gains electrons → its oxidation number decreases → element is reduced → reduction reaction

2 1.Write down the oxidation number for each element above its symbol. 2. Which element in what substance is oxidized? Which element in what substance is reduced? 3. What is the reducing agent? What is the oxidizing agent? 2SO 2 + O 2 → 2SO 3..

3 Chapter 5 Gases

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5 Four variables used to describe a gas Amount — n — unit: mol Volume — V — unit: m 3, L, mL Temperature — T — unit: °C, K Pressure — P — unit:

6 1 atm = pressure exerted by the air at 0 °C at sea level 1 atm = 760 mmHg = 760 torr = 101325 Pa Pressure = Force / Area Pa = N / m 2 e.g. 125 torr = ? atm = ? Pa demo

7 Four variables used to describe a gas Amount — n — unit: mol Volume — V — unit: m 3, L, mL Temperature — T — unit: °C, K Pressure — P — unit:atm, torr, Pa

8 How is the volume of gas defined?

9 Volume of a gas equals the volume of the container in which it is held because gas expands to fill the container.

10 Cu

11 Orange juice

12 Relationships among the four variables Gas Laws P, V, T, n: study the relationship between two variables while keep the other two constant.

13 Boyle’s Law Relationship between P and V under constant n and T. P V T n

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15 Boyle’s Law Relationship between P and V under constant n and T. PV = k

16 (1)(2) P 1 V 1 = P 2 V 2 Boyle’s Law

17 Charles’s Law Relationship between V and T under constant n and P. P V T n

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19 Plots of V Versus T(°C) for Several Gases 2732 − 273 °C

20 Plots of V versus T as Before, Except Here the Kelvin Scale is Used for Temperature

21 Charles’s Law Relationship between V and T under constant n and P. V = bT T must be in K or

22 A sample of gas at 15 °C and 1 atm has a volume of 2.58 L. What volume will this gas occupy at 38 °C and 1 atm?

23 P V T n Relationship between V and n under constant P and T. Avogadro’s Law

24 Relationship between V and n under constant P and T. Avogadro’s Law V = an

25 V = bTV = k/P PV = nRT Ideal Gas Law

26 Units in ideal gas law PV = nRT P — atm, V — L, n — mol, T — K Option 1 R = 0.082 atm · L · mol −1 · K −1 P — Pa, V — m 3, n — mol, T — K Option 2 R = 8.314 J · mol −1 · K −1 Chem 1211

27 Calculate the volume occupied by 0.845 mol of nitrogen gas at a pressure of 1.37 atm and a temperature of 315 K. Example 5.5, page 192

28 A sample of diborane gas (B 2 H 6 ) has a pressure of 345 torr at a temperature of −15 °C and a volume of 3.48 L. How many B 2 H 6 molecules are in this sample? If conditions are changed so that the temperature is 36 °C and the pressure is 468 torr, what will be the volume?

29 What is the volume in liters of exactly 1 mol ideal gas at exactly 0 °C and 1 atm? 0 °C and 1 atm: standard temperature and pressure (STP) At STP, the volume of one mole ideal gas is 22.4 L

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31 For a gas sample, the density is 1.95 g/L at 1.50 atm and 27 °C. What is the molar mass of the gas?

32 Calculate the density of nitrogen gas at 125 °C and a pressure of 755 torr. Example 5.7, page 195 d = 0.853 g/L

33 What about a gas mixture?

34 Dalton’s Law of partial pressure The total pressure of a mixture of gases equals the sum of the pressures that each gas would exert if it were present alone. pressure that each gas would exert if it were present alone = partial pressure of that gas P t = P 1 + P 2 + P 3 + · · · = n t RT/V Each component and the whole gas mixture obey ideal gas law.

35 6.00 g O 2 and 9.00 g CH 4 are placed in a 15.0 L vessel at 0 °C. What is the partial pressure of each gas and what is the total pressure? P O2 = 0.280 atm P CH4 = 0.837 atm P total = 1.117 atm

36 How to describe the composition of a mixture?

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38 6.00 g O 2 and 9.00 g CH 4 are placed in a 15.0 L vessel at 0 °C. What is the partial pressure of each gas and what is the total pressure? What is the mole fraction of O 2 in the mixture? P O2 = 0.280 atm P CH4 = 0.837 atm P total = 1.117 atm A similar example on page 200. practice.

39 Air: 78 % N 2, 21 % O 2, 1 % other gases. What is the partial pressure of N 2 in torr?

40 The air on Pandora consists of N 2, O 2, CO 2, Xe, CH 4, and H 2 S. The mole fractions of CO 2 and H 2 S are 18 % and 1.0 %, respectively. If the partial pressure of CO 2 is 164 torr, what is partial pressure of H 2 S in Pa?

41 The partial pressure of CH 4 (g) is 0.175 atm and that of O 2 (g) is 0.250 atm in a mixture of the two gases. a)What is the mole fraction of each gas in the mixture? b)If the mixture occupies a volume of 10.5 L at 65 °C, calculate the total number of moles of gas in the mixture. c) Calculate the mass of each gas in the mixture.

42 How is ideal gas law related to chemical reactions?

43 PV = nRT Link between a gas and chemical reaction aA + bB  cC + dD

44 Methanol (CH 3 OH) can be synthesized by the following reaction: CO(g) + 2H 2 (g)  CH 3 OH(g) What volume (in liters) of hydrogen gas, at a temperature of 355 K and a pressure of 738 torr, is required to synthesize 35.7 g of methanol? Example 5.12, page 203

45 The metabolic oxidation of glucose, C 6 H 12 O 6, in our bodies produces CO 2, which is expelled from our lungs as a gas: C 6 H 12 O 6 (aq) + 6O 2 (g)  6CO 2 (g) + 6H 2 O(l) Calculate the volume of CO 2 produced at body temperature (37 °C) and 0.970 atm when 24.5 g of glucose is consumed in this reaction.

46 FOR PRACTICE 5.12, page 204 In the following reaction, 4.58 L of O 2 was formed at P = 745 torr and T = 308 K. How many grams of Ag 2 O must have decomposed? 2Ag 2 O(s)  4Ag(s) + O 2 (g)

47 FOR PRACTICE 5.13, page 205 How many liters of oxygen (at STP) are required to form 10.5 g of H 2 O? 2H 2 (g) + O 2 (g)  2H 2 O(g)

48 What is the theory behind ideal gas law?

49 Kinetic Molecular Theory The size of gas molecules is assumed to be zero Gas molecules do not attract or repel each other Gas molecules constantly move, collisions with the wall of a container cause pressure Average kinetic energy of gas molecules is directly proportional to absolute temperature

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51 Kinetic Molecular Theory The size of gas molecules is assumed to be zero Gas molecules do not attract or repel each other Gas molecules constantly move, collisions with the wall of a container cause pressure Average kinetic energy of gas molecules is directly proportional to absolute temperature Remember these four assumptions

52 Problems for Chapter 5 Work Exercises: 11, 15, 18, 20, 23, 29a-b, 35, 41, 43, 49, 65, 67, 69, 75, 77, & 79.


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