Chemistry part 18, Julia Burdge,2e (2009) ppt

422 CHAPTER 11 Gases Re member that the symbol (X m ea ns" is proportional to ." 110 100 -+- - , 90 ~ 80 ~ 70 ;'] 60 50 P (mmHg) 760 855 950 1045 1140 1235 1330 1425 1520 2280 V ( mL ) 100 89 78 72 66 59 55 54 50 33 Show n in Shown in Shown in Figure Figure Figure 11.7(a) 11.7(b) 11.7(c) by a column of mercury. The apparatus functions as an open-end manometer. When the mercury levels on both sides are equal [Figure 11.7(a)], the pressure of the confined gas is equal to atmo- spheric pressure. When more mercury is added through the open end, the pressure of the confined ga s is increased by an amount proportional to the height of the added mercury and the volume of the ga s decreases. If, for example, as shown in Figure 11.7(b), we double the pressure on the confined gas by adding enough mercury to make the difference in mercury levels on the left and right 760 mm (the height of a mercury column that exerts a pressure equal to 1 atm), the volume of the gas is reduced by half. If we triple the original pressure on the confined gas by adding more mercury, the volume of the ga s is reduced to one-third of its original volume [Figure 11.7(c)]. Table 11.3 gives a set of data typical of Boyle's experiments. Figure 11.8(a) and (b) shows some of the volume data plotted as a function of pressure and as a function of the inverse of pressure, re spectively. Th ese data illustrate Boyle's law, which states that the pressure of a fixed amount of gas at a constant temperature is inversely proportional to the volume of the gas. This inverse relationship between pressure and volume can be expressed mathematically as follows: , - . . or 1 V cx: - P Equation l1.1(a) 1 V = k] P (at constant temperature) where k] is a proportionality constant. We can rearrange Equation 11.1(a) to get Equation 11.1 (b) PV = k] (at constant temperature) According to this form of Boyle's law, the product of the pressure and the volume of a given sample of ga s (at constant temperature) is a constant. Although the individual values of pressure and volume can vary greatly for a given sample of gas, the product of P and V is always equal to the same constant as long as the temperature is held constant and the amount of gas does not change. Therefore, for a given sample of gas under two different sets of conditions at constant temperature, we can write • ;3 110 100 90 5 80 8 '" 70 - ;'] 60 50 • • • 40 -+ r , ~~ r-~~ ~ ~~-,~~ r- ~~ ~ 40 -+-==~ ~ ~~ ·· - · ~ · ==~·=···=·····=·= · ····=···~·····=····=· . = ~ ~ = - ~ ~ 700 800 900 1000 1100 1200 1300 1400 1500 1600 Pressure (mmHg ) (a) 0.0006 0.0007 0.0008 0.0009 0.0010 0.0011 0.0012 0.0013 0.0014 1 Pressure (mmHg) (b) Figure 11.8 Plots of volume (a) as a function of pre ss ure and (b) as a function of lIpressure. or PI VI = P 2 V 2 (at constant temperature) Equation 11.2 where VI is the volume at pressure PI and V? is the volume at pressure P? Sample Problem 11.2 illustrates the use of Boyle's law. If a skin diver takes a breath at the surface, filling his lungs with 5.82 L of air, what volume will the air in hi s lungs occupy when he dives to a depth where the pressure is 1.92 atm? (Assume constant temperature and that the pressure at the surface is exactly 1 atm.) Strategy Use Equation 11.2 to solve for V 2 . Setup P I = 1.00 atm, VI = 5.82 L, and P2 = 1.92 atm. Solution v? = PI X VI = 1.00 atm X 5.82 L = 3.03 L - P 2 1.92 atm SECTION 11.2 The Gas Laws 423 Think About It At higher pr essure, the volume should be smaller. Therefore, the answer I makes sense. Practice Problem A Calculate the volume of a sample of gas at 5.75 atm if it occupies 5.14 L at I 2.49 atm. (Assume constant temperature.) Practice Problem B At what pressure would a sample of gas occupy 7.86 L if it occupies 3.44 L at 4.11 atm? (Assume constant temperature.) ~ Charles's and Gay-Lussac's Law: The Temperature-Volume Relationship If you took a helium-filled Mylar balloon outdoors on a cold day, the balloon would shrink some- what when it came into contact with the cold air. This would occur because the volume of a sample of gas depends on the temperature. A more dramatic illustration is shown in Figure 11.9, where liquid nitrogen is being poured over a balloon. The large drop in temperature of the air in the bal- loon (the boiling liquid nitrogen ha s a temperature of -196 °C) results in a significant decrease in its volume, causing the balloon to shrink. The first to study the relationship between gas volume and temperature were French scien- tists Jacques Charles 3 and Joseph Gay-Lussac. 4 Their studies showed that, at constant pressure, the (a) (b) 3. Jacques Alexandre Cesar Charles (1746- 1823 ). French physicis t. Charles was a gifted l ect urer, an inventor of scientific apparatus, and the first person to use hydrogen to inflate balloons. 4. Joseph Louis Gay-Lussac (1778 -1850 ). French chemist and physicist. Like Charles, Gay-Lussac was a balloon enthusi- as t. Once he ascended to an altitude of 20,000 ft to co ll ect air samples for analysis. : Mu lti me d ia Gas Laws - Charles's law. Figure 11.9 (a) Helium balloon. (b) Lowering the temperature with liquid nitrogen causes a volume decrease. 424 CHAPTER 11 Gases 4000 - 4000 3500 - 3500 ~ ~ '" '" - 3000 - .~ '" - 3000 .~ ~ - ;::l ;::l ~ 2500 - ;>-, 2500 ~ - 2000- .~ • 0 ~ - 2000 .~ 0 '" ~ 1500 - " ~ 1500 " E 8 ;::l 1000 - - ;::l 1000 - ~ 500 - ~ 500 0 1 1 . 1 . 1 I I . I 0 -150 -100 -50 0 50 100 150 200 -273.15 -200 -100 0 100 200 Temperature (0C) Temperature CC) (a) (b) ~ Figure 11 .10 (a) Plot of the volume of a sample of gas as a function of temperature. (b) Plot of the vo lume of a sample of gas as a function of temperature at three different pressures. Remember that a kelvin and a degree Celsius have the same magnitude. Thus, while we add 273.15 to the temperature in °C to get the temperature in K, a change in temperature in Celsius is equal to the change in temperature in K. A temperature of 20°C is the same as 293.15 K. A change in temperature of 20°C, however, is the same as a change in temperature of 20 K. Don't forget that volume is proportional to absolute temperature. The volume of a sample of gas at constant pressure doubles if the temperature increases from 100 K to 200 K-but not if the temperature i ncreases from 100°C to 200°C I vo lume of a gas sample increases when heated and decreases when cooled. Figure 11.1O(a) shows a plot of data typical of Charles's and Gay-Lussac's experiments. Note that with pressure held constant, the vo lum e of a sample of gas plotted as a function of temperature yields a straight line. These experiments were carried out at several different pressures [Figure 11.1 O(b)], each yielding a different straight line. Interestingly, if the lines are extrapolated to zero volume, they all meet at the x axis at the temperature -2 73.15 d c. The implication is that a gas sample occupies zero vol- ume at -273.15 °C. This is not observed in practice, however, because all gases condense to form liquids or solids before -273.15 °C is reached. In 1848 Lord KelvinS realized the significance of the extrapolated lines all meeting at -273. 15 °C. He identified -273.15 °C as absolute zero, theoretically the lowest attainable tempera- ture. Then he set up an absolute temperature scale, now called the Kelvin temperature scale, with absolute zero as the low est point [ ~. Section 1.3] . On the Kelvin sca le, 1 kelvin (K) is equal in magnitude to 1 degree Celsius. The difference is simp ly an offset of 273.15. We obtain the abso- lute temperature by adding 273.15 to the temperature expressed in Celsius, although we often use . . . . . . . . . . , '" . . simply 273 instead of 273.15. Several important points on the two scales match up as follows: Absolute zero Freezing point of water Boiling point of water Kelvin Scale (K) OK 273.15 K 373.15 K Celsius Scale (0C) -273.1SoC O°C 100°C The dependence of the vo lume of a sample of gas on temperature is given by Equation 11.3(a) v = k2T (at constant pressure) . . . . . . . . . . . . . . . . . . . . or Equation 11.3(b) V T = k2 (at constant pressure) where k2 is the proportionality constant. Equations 11.3(a) and (b) are expressions of Charles's and Gay-Lussac's law, often refelTed to simply as Charles's law, which states that the volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas. 5. William Thomson, Lord Kelvin (1824-1907). Scottish mathematician and physicist. Kelvin did important work in many branches of physics. Just as we did with the pressure-volume relationship at constant temperature, we can com- pare two sets of volume-temperature conditions for a given sample of gas at constant pressure. From Equation 11.3 we can write or (at constant pressure) Equation 11.4 where VI is the volume of the gas at T] and V 2 is the volume of the gas at T 2 • Sample Problem 11.3 shows how to use Charles's law. ~ Sample ~ Problemli.3 •• I I I I I A sample of argon gas that originally occupied 14 .6 L at 25.0°C was heated to 50.0°C at constant pressure. What is its new vo lume? Strategy Use Equation 11.4 to solve for V 2 . Remember that temperatures mu st be expressed in kelvin. Setup TI = 298.15 K, VI = 14 .6L, and T2 = 323.15 K. Solution 14.6 L X 323.15 K = 15. 8 L 298.15 K Practice Problem A A sample of gas originally occupies 29 .1 L at O.O °C. What is its new volume when it is heated to 15.0°C? (Assume constant pressure.) Practice Problem B At what temperature ( in 0 c) will a sam pl e of gas occupy 82.3 L if it occupies 50.0 L at 75.0°C? (Assume constant pressure.) I~ " Avogadro's law: The Amount-Volume Relationship In 1811, the Italian scientist Amedeo Avogadro proposed that equal vo lumes of different gases contain the sa me number of particles (molecules or atoms) at the same temperature and pressure. This hypothesis gave rise to Avogadro's law, which states that the volume of a sample of gas is directly proportional to the number of moles in the sample at constant temperature and pressure: Vo:n or V = k3n (at constant temperature and pressure) Equation 11.5 Avogadro's law makes it possible to predict the volumes of gaseous reactants and products. Con- sider the reaction of H2 and N2 to form NH 3: The balanced equation reveals the ratio of combination of reactants in terms of moles [ ~~ Section 3.4]. However, because the volume of a gas (at a given temperature and pressure) is directly proportional to the number of moles, the balanced equation also reveals the ratio of com- bination in terms of volume. Thus, if we were to combine three volumes (liters, milliliters, etc.) of hydrogen gas with one volume of nitrogen gas, assuming they react completely according to the balanced equation, we would expect two volumes of ammonia gas to be produced (Figure 11.11). The ratio of combination of H2 and N2 (and production of NH 3)' whether expressed in moles or units of volume, is 3: 1 :2. SECTION 11.2 The Gas Laws 425 Think About It When temperature increases at constant pressure, the volume of a gas sample increases. 426 CHAPTER 11 G ases , Think About It Remember that the coefficients in balanced chemi ca l equations indicate ratios in molecules or moles. Under conditions of constant temperature and pressure, the volume of a gas' is proportional to the number of moles. Therefore, the coefficients in balanced equations containing only gases also indicate ratios in liters, provided the reactions occur at constant temperature and pressure. It is important to re co gn ize that coe ff ici ents indi c at e ra ti os in liters o nl y in ba la nc ed equations in which ali of the r eactants and products are gases. We cannot apply the s ame approach to react io ns in whi ch there ar e s ol id, l iquid , or aq u eous species. 3 molecules 3 moles 3 volumes + + + + + 1 molecule 2 molecules 1 mole 2 moles 1 volume 2 volumes Figure 11 . 11 Illustration of Avogadro's law. The volume of a sample of gas is directly proportional to the number of mole s. Sample Problem 11.4 s hows how to apply Avogadro's law. Samp'le If we combine 3.0 L of NO and 1.5 L of O 2 , and they react according to the balanced equation 2NO(g) + 0 2( g) • 2N0 2 (g), what volume of N0 2 will be produced? (Assume that the reactants and product are all at the same temperature and pressure.) Strategy Apply Avogadro's law to determine the volu me of a gaseous product. Setup Because volume is proportional to the number of moles, the balanced equation determines in what volume ratio the reactants combine and the ratio of product volume to reactant volu me . The amounts of reactants given are stoichiometric amounts [ ~~ Section 3.6]. Solution According to the balanced equation, the volume of N0 2 formed will be equal to the volume of NO that reacts. Therefore, 3.0 L of N0 2 will foml. Practice Problem A What volume (in liters) of water vapor will be produced when 34 L of H2 and 17 L of O 2 react according to the equation 2H 2 (g) + 0 2(g) • 2H 2 0(g)? Practice Problem B What volumes (in liters) of carbon monoxide and oxygen gas must react according to the equation 2CO(g) + O zC g) , 2C0 2 (g) to form 3.16 L of carbon dioxide? Checkpoint 11.2 The Gas Laws 11.2.1 11.2.2 Given PI = 1.50 atm, VI = 37.3 mL, and P 2 = 1.18 atm, calculate V 2 . Assume that nand T are constant. a) 0.0211 mL b) 0.0341 mL c) 29.3 mL d) 12.7 mL e) 47.4 mL Given TI = 21.5°C, VI = 50.0 mL, and T2 = 316°C, calculate V 2 . Assume that nand P are constant. a) 100 mL b) 73.5 mL c) 25.0 mL d) 3.40 mL e) 26.5 mL 11 .2.3 11 .2.4 At what temperature will a gas sample occupy 100.0 L if it originally occupies 76.1 L at 89.5°C? Assume constant P. a) 276°C b) 118°C c) 203°C d) 68.1 °C e) 99.6°C What volume ofNH 3 will be produced when 180 mL ofH 2 reacts with 60.0 mL of N2 according to the following equation: Assume constant T and P for reactants and products. a) 120 mL b) 60 mL c) 180 mL d) 240 mL e) 220 mL SECTION 11.3 The Ideal Gas Equation 427 The Ideal Gas Equation Recall that the state of a sample of gas is described completely using the fo ur variables T, P, V, and n. Each of the gas laws introduced in Section 11.2 relates one variable of a sample of gas to another while the other two variables are held constant. In experiments with gases, however, there are usually changes in more than ju st two of the variables. Therefore, it is useful for us to combine the equations representing the gas laws into a single equation that will enable us to account fo r changes in any or all of the four variables. Deriving the Ideal Gas Equation from the Empirical Gas Laws Summarizing the gas law equations from Section 11.2: Bo yl e's law: 1 V ex - P Charles's law: V ex T Avogadro's law: Vex n We can combine these equations into the following general equation that describes the physical behavior of all gases: or Vex nT P V= R nT P where R is the proportionality constant. This equation can be rearranged to give: PV = nRT Equation 11.6 Equation 11.6 is the mo st commonly used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T. An ideal gas is a hypothetic al sample of gas whose pr essure-volume-temperature behavior is predicted accurately by the ideal gas equation. Although the behavior of real gases generally differs slightly from that predicted by Equation 11.6, in most of the cases we will encounter, the differences are usually small enough for us to use the ideal gas equation to make reasonably good predic ti ons about the behavior of gases. . . . The proportionality constant, R, in Equation 11.6 is called the gas constant. Its value and units depend on the units in which P and V are expressed. ( The variables nand T are always expressed in mol and K, respec ti vel y.) Recall from Section 11.1 that pressure is commonly expressed in atmospheres, rnmHg (torr ), pascal s, or bar. Volume is typically expressed in liters or milliliter s, but can also be expressed in other unit s, such as m 3 . Table 11.4 lists several different expressions of the gas constant, R. Numerical Value 0.08206 62.36 0.08314 8.314 8.314 1.987 . . . . . . . Unit L . atmIK . mol L . torr/K . mol L . barlK . mol m 3 • PaiK . mol 11 K· mol callK · mol Note that the product of volume and pressure gives units of energy (i.e., joules and calories ). . ~ Mult imedia Gas Laws - ideal gas law (interactive). We will discuss the conditions that result in deviation from idea l behavior in Section 11.7. Wh en you us e R in a calculation, use the version that facilitates proper cancellation of units. • 428 CHAPTER 11 Gases In thermochemistry we often used 25°( as the" standard " temperature although temperature is not actually part of the definition of the standard state [ ~~ Section 5. 6] . The standard temperature for gases is defined specifically as O°C. In this problem, because they are specified rather than measured, O°C, 1 mole, and 1 atm are exact numbers and do not affect the number of significant figures in the result [ ~~ Section 1.5]. Think About It With the pressure held constant, we should expect the volume to increase with increased temperature. Room temperature is higher than the standard temperature for gases (O °C), so the molar volume at room temperature (25 °C) should be higher than the molar volume at O° C-and it i s. n X At = m, where m is mass in grams. Another way to arrive at Equation 11.7 is to substitute ml J iIt for n in the ideal gas equati on and rearrange to solve for mlV ( denSity): m m PAt PV = -RT and - = d = '""== At V RT • Keep in mind that all these expressions of R are equal to one another, ju st as I yard is equal to 3 ft. They are simply expressed in different units. One of the simplest uses of the ideal gas equation is the calculation of one of the variables when the other three are already known. For example, we can calculate the volume of 1 mole of an ideal gas at ooe and 1 atm, conditions known as standard temperature and pressure (STP). In this case, n, T, and P are given. R is a constant, leaving Vas the only unknown. We can reanange Equation 11.6 to solve for V, V= nRT P enter the information that is given, and calculate V. Remember that in calculations using the ideal gas equation, temperature must always be expressed in kelvins. (1 mol)(O.08206L . atrnJK . mol)(273.15K) V = = 22.41L 1 atm Thu s, the volume occupied by 1 mole of an ideal gas at STP is 22.41 L, a volume slightly less than 6 gal. Sample Problem 11.5 shows how to calculate the molar volume of a gas at a temperature other than O°e. Calculate the vo lume of a mole of ideal gas at room temperature (25°C) and 1 atm. Strategy Convert the temperature in °C to temperature in kelvins, and use the ideal gas equation to sol ve for the unknown vo lume. Setup The data given are n = 1 mol, T = 298.15 K, and P = 1.00 atm. Because the pressure is expressed in atmosphere s, we use R = 0.08206 L . atmlK . mol in order to solve for volume in liters. Solution (1 mol)(0.08206 L . atmlK . mol)(298.1S K) V = = 24.5 L 1 atm Practice Problem A What is the vo lume of 5.12 moles of an ideal gas at 32°C and 1.00 atm? Practice Problem B At what temperature (in 0 c) would 1 mole of ideal gas occupy 50.0 L (P = 1.00 atm)? Applications of the Ideal Gas Equation Using some simple algebraic manipulation, we can solve for variables other than those that appear explicitly in the ideal gas equation. For example, if we know the molar mass of a gas (g/mol), we can determine its density at a given temperature and pressure. Recall from Section 11.1 that the density of a gas is generally expressed in units of gIL. We can reanange the ideal gas equation to solve for mol!L; n P - - V RT If we then multiply both sides by the molar ma ss, .AIt, we get . . . . . . . . . . . . . . ' " . . • • • .AIt X !!:. = P x.AIt V RT where.Alt X nlV gives gIL or density, d. Therefore, • Equation 11.7 • d=P.AIt RT SECTION 11.3 The Ideal Gas Equation 429 92.0 13 JiI. • 92.029 " = = (a) (b) Conversely, if we know the density of a gas, we can determine its molar ma ss: .M = dRT P Equation 11.S In a typical experiment, in which the molar mass of a gas is determined, a flask of known volume is evacuated and weighed [Figure 11.12(a)]. It is then filled (to a known pressure) with the gas of unknown molar mass and reweighed [Figure 11.12(b)]. The difference in mass is the mass of the gas sample. Dividing by the known volume of the flask gives the density of the gas, and the molar mass can then be determined using Equation ll.S. Similarly, the molar mass of a volatile liquid can be determined by placing a small volume of it in the bottom of a flask, the mass and volume of which are known. The flask is then immersed in a hot-water bath, causing the volatile liquid to completely evaporate and its vapor to fill the flask. Because the flask is open, some of the excess vapor escapes. When no more vapor escapes, . . . . . . . . the flask is capped and removed from the water bath. The flask is then weighed to determine the mass of the vapor. (At this point, some or all of the vapor has condensed but the mass remains the same.) The density of the vapor is determined by dividing the mass of the vapor by the volume of the flask. Equation II.S is then used to calculate the molar mass of the volatile liquid. Sample Problems 11.6 and 11.7 illustrate the use of Equations 11.7 and 11 .S. Sample Problem 11.6 Carbon dioxide is effective in fire extinguishers partly because its density is greater than that of air, so CO 2 can smother the flames by depriving them of oxygen. (Air has a density of approximately 1.2 gIL at room temperature and 1 atm.) Calculate the density of CO 2 at room temperature (2S°C) and 1.0 atm. Strategy Use Equation 11.7 to solve for density. Because the pressure is expressed in atm, we should use R = 0.OS206 L . atmIK . mol. Remember to express temperature in kelvins. Setup The molar mass of CO 2 is 44.01 g/mol. Solution 44.01g (1 atm) d=P.M= mol =1. SglL RT (0 .OS206 L . atm)(29S.1SK) K'mol Practice Problem A Calculate the density of helium in a helium balloon at 2S.0°C. (Assume that the pressure inside the balloon is 1.10 atm.) Practice Problem B Calculate the density of air at O°C and 1 atm. (Assu me that air is SO percent N2 and 20 percent O 2 ,) ~ ~ • Figure 11 .12 (a) Evacuated flask. (b) Flask filled with gas. The mass of the gas is the difference between the two masses. The density of the gas is determined by dividing mass by volume . Because the flask is open to the atmosphere whi le the volatile liquid vaporizes, we can use atmospheric pressure as P. Also, because the flask is capped at the water bath temperature, we can use the water bath temperature as T. Think About It The calculated densi ty of CO 2 is greater than that of air under the same conditions (as expected). Although it may seem tedious, it is a good idea to write units for each and every entry in a problem such as this. Unit cancellation is very useful for detecting errors in your reasoning or your solution setup. 430 CHAPTER 11 Gases Think About It Because more than one compound can ha ve a particular molar mass, this method is not definitive for identification. However, in this circumstance, further testing of the proprietary formula certainly would be warranted. Samp'le A company has just patented a new synthetic alcohol for alcoholic beverages. The new product is said to have all the pleasant properties associated with ethanol but none of the undesirable effects such as hangover, impairment of motor skills, and risk of addiction. The chemical formula is proprietary. You analyze a sa mple of the new product by placing a small volume of it in a round- bottomed flask with a volume of 511.0 rnL and an evacuated mass of 131.918 g. You submerge the flask in a water bath at 100.0°C and allow the volatile liquid to vaporize. You then cap the flask and remove it from the water bath. You weigh it and determine the mass of the vapor in the flask to be 0.768 g. What is the molar mass of the volatile liquid, and what doe s it mean with regard to the new product? (Ass ume the pressure in the laboratory is 1 atm.) Strategy Use the measured ma ss of the vapor and the given volume of the flask to determine the density of the vapor at 1 atm and 100.0°C, and then use Equation 11.8 to determine molar ma ss. Setup P = 1 atm, V = 0.5110 L, R = 0.8206 L . atm/mol . K, and T = 373. 15 K. Solution 0.768 g d = = 1.5029 aIL 0.5110 L b ( 1.5029 g )( 0.08206 ]0' · atnf)(373.15){) Y v/ . mol oM = ~ = 46.02 glmol 1 Jlli1f The result is a molar ma ss suspiciously close to that of ethanol! Practice Problem A Determine the molar mass of a gas with a density of 1.905 gIL at 80.0°C and 1.00 atm. Practice Problem B Determine the molar mass of a gas with a density of 5.14 gIL at 73.0°C and 1.00 atm. Checkpoint 11.3 The Ideal Gas Equation 11.3.1 Calculate the volume occupied by 8.75 11.3.3 Determine the density of a gas with moles of an ideal gas at STP. oM = 146.07 glmol at 1.00 atm and a) 196 L 100.0°C. b) 268 L a) 6.85 X 10- 3 giL c) 0.718 L b) 4.77 gIL d) 18.0 L c) 146 gIL e) 2.56 L d) 30.6 gIL e) 17 .8 giL 11.3.2 Calculate the pressure exerted by 10.2 moles of an ideal gas in a 7.5-L 11.3.4 Determine the molar mass of a gas vessel at 150°C. with d = 1.963 gIL at 1.00 atm and a) 17 atm 100.0° C. b) 31 atm a) 0.0166 g/mol c) 0.72 atm b) 60.1 g/mol d) 1.3 atm c) 16.1 glmol e) 47 atm d) 6.09 X 10 3 glmol e) 1.63 X 10 3 gl mol Reactions with Gaseous Reactants and Products In Chapter 3 we used balanced chemical equations to calculate amounts of reactants and/or prod- ucts in chemical reactions expressing those amounts in mass (usually grams). However, in the SECTION 11.4 Reactions with Gaseous Reactants and Products 431 case of reactants and products that are gases, it is more practical to measure and express amounts in volume (liters or milliliters). This makes the ideal gas equation useful in the stoichiometric analysis of chemical reactions that involve gases. Calculating the Required Volume of a Gaseous Reactant According to Avogadro's law, the volume of a ga s at a given temperature and pressure is propor- tional to the number of moles. Moreover, balanced chemical equations give the ratio of combina- tion of gaseous reactants in both moles and volume (Figure 11.11). Therefore, if we know the volume of one reactant in a gaseous reaction, we can determine the required amount of another reactant (at the same temperature and pressure). For example, consider the reaction of carbon monoxide and oxygen to yield carbon dioxide: 2CO(g) + 02(g) +. 2C0 2 (g) The ratio of combination of CO and O 2 is 2: 1, whether we are talking about moles or units of volume. Thus, if we want to determine the stoichiometric amount [ ~~ Section 3.6] of O 2 required to combine with a particular volume of CO, we simply use the conversion factor provided by the balanced equation, which can be expressed as any of the following: 1 mol O 2 2 mol CO or 1 L0 2 2LCO or 1 mL0 2 2mLCO Let's say we want to determine what volume of O 2 is required to react completely with 65.8 mL of CO at STP. We could use the ideal gas equation to convert the volume of CO to moles, use the stoichiometric conversion factor to convert to moles O 2 , and then use the ideal gas equation again to convert moles O 2 to volume. But this method involves several unnecessary steps. We get the same result simply by using the conversion factor expressed in milliliters: 65.8.rnL-eO X = 1 mL O 2 = 32.9 mL 0 2~ 2 In cases where only one of the reactants is a gas, we do need to use the ideal ga s equation in our analysis. Recall, for example, the reaction of sodium metal and chlorine gas used to illustrate the Born-Haber cycle [I ~~ Section 8.2 ]: 2Na(s) + ClzCg) +. 2NaCl(s) Given moles (or more commonly the mass) of Na, and information regarding temperature and pressure, we can determine the volume of C1 2 required to react completely: , grams Na X 1 mol Na = 22.99 g Na moles Na X 1 mol Cl2 = 2 mol Na moles C1 2 X RT P liters C12 Sample Problem 11.8 shows how to use the ideal gas equation in a stoichiometric analysis. . . . Sample Problem 11.8 Sodium peroxide (Na202) is used to remove carbon dioxide from (a nd add oxygen to) the air supply in spacecrafts. It works by reacting with CO 2 in the air to produce sodium carbonate (Na2C03) and O 2 , What volume (in liters) of CO 2 (at STP ) will react with a kilogram of Na202? Strategy Convert the given mass of Na202 to moles, use the balanced equation to determine the stoichiometric amount of CO 2 , and then use the ideal gas equation to convert moles of CO 2 to liters. Setup The molar mass of Na202 is 77.98 glmol (1 kg = 1000 g). (Treat the specified mass of Na202 as an exact number.) (Continued) [...]... sum of the two partial pressures: Ptotal = Po2 + PH20 By subtracting the partial pressure of water from the total pressure, which is equal to atmospheric pressure, we can determine the partial pressure of oxygen and thereby determine how many moles are produced by the reaction We get the partial pressure of water, which depends on temperature, from a table of values Table 11.5 lists the partial pressure... 11.11 shows how to apply Dalton's law of partial pressures ~ ~ ~ Sample Problem 11.11 A 1.00-L vessel contains 0.215 mole of N2 gas and 0~0118 mole of H2 gas at 25.5 °C Determine the partial pressure of each component and the total pressure in the vessel Strategy Use the ideal gas equation to find the partial pressure of each component of the mixture, and sum the two partial pressures to find the total... sum of the partial pressures exerted by each component of the mixture: Thus, the total pressure exerted by a mixture of 1.00 mol N2 and 1.00 mol O2 in a 5.00-L vessel at oo is e Ptotal = PN 2 + Po2 = 4.48 atm + 4.48 atm = Figure 11.13 illustrates Dalton's law of partial pressures Figure 11.13 Schematic illustration of Dalton's law of partial pressures Total pressure is equal to the sum of partial pressures... the partial pressure and number of moles of each gas in a 15.75-L vessel at 30.0°C containing a mixture of xenon and neon gases only The total pressure in the vessel is 6.50 atm, and the mole fraction of xenon is 0.761 Using Partial Pressures to Solve Problems The volume of gas produced by a chemical reaction can be measured using an apparatus like the one shown in Figure 11.14 Dalton's law of partial... exerted by each gas is known as the partial pressure (PJ of the gas We use subscripts to denote partial pressures: P N2 P02 = (1.00 mol)(0.08206 L atmlK mol)(273 15 K) 5.00 L = 4.48 atm = (1.00 mol)(0.08206 L atmlK mol)(273 15 K) 5.00 L = 4.48 atm and we can solve the ideal gas equation for each component of any gas mixture: nRT p = _Ie:-I V Dalton's law of partial pressures states that the total... of partial pressures to determine the amount of gas produced in a chemical reaction and collected over water " "Sample Problem 11.13 Calcium metal reacts with water to produce hydrogen gas [~ Section 7.7] : Determine the mass of H2 produced at 25°C and 0.967 atm when 525 mL of the gas is collected over water as shown in Figure 11.14 • Strategy Use Dalton's law of partial pressures to detemrine the partial... fraction of O2 is necessary for the partial pressure of O 2 to be 2.8 atm when the total pressure is 4.6 atm? - SECTION 11.6 Checkpoint 11.5 11.5.1 The Kinetic Molecular Theory of Gases 439 Gas Mixtures What is the partial pressure of He in a 5.00-L vessel at 25°C that contains 0.0410 mole of He, 0.121 mole of Ne, and 0.0922 mole of Ar? 11 5.3 a) 1.24 atm What is the partial pressure of oxygen in a gas... atm 1 L Practice Problem A Determine the partial pressures and the total pressure in a 2.50-L vessel containing the following mixture of gases at 15.8°C: 0.0194 mol He, 0.0411 mol H 2, and 0.169 mol Ne Practice Problem B Determine the number of moles of each gas present in a mixture of CH4 and C2H 6 in a 2.00-L vessel at 25.0°C and 1.50 atm, given that the partial pressure of CH4 is 0.39 atm L - ... KCl0 3 at 35.0°C and 1.08 atm Bringing Chemistry to Life Hyperbaric Oxygen Therapy One unfortunate group of patients was undergoing treatment when the power to the chamber was sh ut off accidentally All the patients died At the time, their deaths were attributed to influenza, but they almost certainly died as the result of the unintended rapid decompression In 1 918, during the Spanish flu epidemic that... compressed air to 2.8 atm and the patient breathes pure O2 , Determine the partial pressure of O2 in each treatment protocol and compare the results Strategy Because the gas the patient breathes is inside the hyperbaric chamber, its total pressure is Catalina Hyperbaric Chamber the same as the chamber pressure To obtain the partial pressure of O2 , multiply the mole fraction of O2 in the breathing . 428 CHAPTER 11 Gases In thermochemistry we often used 25°( as the" standard " temperature although temperature is not actually part of the definition of the standard. Figure 11.13 Schematic illustration of Dalton's law of partial pressures. Total pressure is equal to the sum of partial pressures. • Gas Mixtures So far our discussion of. of gases, the pressure exerted by each gas is known as the partial pressure (PJ of the gas. We use subscripts to denote partial pressures: (1.00 mol)(0.08206 L . atmlK . mol)(273.15