1/7/2013. Chapter 11. Gases. Chemistry: Atoms First Julia Burdge & Jason Overby. Gases Properties of Gases. Gases

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1 1/7/013 Chemistry: Atoms First Julia Burdge & Jason Overby 11 Gases Chapter 11 Gases Kent L. McCorkle Cosumnes River College Sacramento, CA Copyright (c) The McGraw-Hill Companies, Inc. ermission required for reproduction or display roperties of Gases 11. of Gases Molecular Speed Diffusion and Effusion 11.3 Gas ressure Definition and Units of ressure Calculation of ressure Measurement of ressure 11.4 Boyle s Law: The ressure-olume Relationship Charles s and Gay-Lussac s Law: The Temperature-olume Relationship Avogadro s Law: The Amount-olume Relationship and Kinetic Molecular Theory The Combined Gas Law: The ressure-temperature-amount-olume Relationship 11.5 The Ideal Gas Equation Applications of the Ideal Gas Equation 11 Gases 11.1 roperties of Gases 11.6 Real Gases Factors That Cause Deviation from Ideal Behavior The van der Waals Equation van der Waals Constants 11.7 Gas Mixtures Dalton s Law of artial ressures Mole Fractions 11.6 Reactions with Gaseous Reactants and roducts Calculating the Required olume of a Gaseous Reactant Determining the Amount of Reactant Consumed Using Change in ressure Using artial ressures to Solve roblems Gases differ from solids and liquids in the following ways: 1) A sample of gas assumes both the shape and volume of the container. ) Gases are compressible. 3) The densities of gases are much smaller than those of liquids and solids and are highly variable depending on temperature and pressure. 4) Gases form homogeneous mixtures (solutions) with one another in any proportion. 11. The kinetic molecular theory explains how the molecular nature of gases gives rise to their macroscopic properties. The basic assumptions of the kinetic molecular theory are as follows: 1) A gas is composed of particles that are separated by large distances. The volume occupied by individual molecules is negligible. ) Gas molecules are constantly in random motion, moving in straight paths, colliding with perfectly elastic collisions. 3) Gas molecules do not exert attractive or repulsive forces on one another. 4) The average kinetic energy of a gas molecules in a sample is proportional to the absolute temperature: E k T Gases are compressible because molecules in the gas phase are separated by large distances (assumption 1). ressure is the result of the collisions of gas molecules with the walls of their container (assumption ). Decreasing volume increases the frequency of collisions. ressure increases as collision frequency increases. 1

2 1/7/013 Heating a sample of gas increases its average kinetic energy (assumption 4). Gas molecules must move faster. Faster molecules collide more frequently and at a greater speed. ressure increases as collision frequency increases. The total kinetic energy of a mole of gas is equal to: The average kinetic energy of one molecule is: For one mole of gas: 3 RT 1 m is the mass mu u is the mean square speed 1 3 NA mu RT rearrange and Take the square root (m x N A = M) u rms 3RT M The root-mean-square (rms) speed (u rms ) is the speed of a molecule with the average kinetic energy in a gas sample. u rms is directly proportional to temperature The root-mean-square (rms) speed (u rms ) is the speed of a molecule with the average kinetic energy in a gas sample. u rms is inversely proportional to the square root of M. u rms 3RT M u rms 3RT M When two gases are at the same temperature, it is possible to compare the the u rms values of the different gases. u u rms rms () 1 M () M1 Worked Example 11.1 Determine how much faster a helium atom moves, on average, than a carbon dioxide molecule at the same temperature. urms () 1 M Strategy Use equation u and the molar masses of He and CO to rms () M1 determine the ratio of their root-mean-square speeds. When solving a problem such as this, it is generally best to label the lighter of the two molecules as molecule Think 1 and About the heavier It Remember as molecule that. the This relationship ensures that between the result molar will be greater mass than and 1, which molecular is relatively speed is easy reciprocal. to interpret. A CO molecule has approximately 10 times the mass of an He atom. Therefore, we Solution should The expect molar an masses He atom, of He on and average, CO are to be moving and 44.0 approximately g/mol, respectively. 10 times (~3. times) as fast as a CO molecule g u rms (He) 1 mol = u rms (CO ) = g 1 mol On average, He atoms move times as fast as CO molecules at the same temperature.

3 1/7/013 Diffusion is the mixing of gases as the result of random motion and frequent collisions. Effusion is the escape of gas molecules from a container to a region of vacuum. Graham s law states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass. Rate 1 M 11.3 Gas ressure Determine the molar mass and identity of a diatomic gas that moves 4.67 times as fast as CO. Step 1: Use the equation below to determine the molar mass of the unknown gas: u u rms rms () 1 M () M1 u rms (unknown gas) = 4.67 x u rms (CO ) urms ( unknown) 44.0 g/mol 4.67 urms ( CO) M unknown ressure is defined as the force applied per unit area: force pressure = area The SI unit of force is the newton (N), where 1 N = 1 kg m/s The SI unit of pressure is the pascal (a), defined as 1 newton per square meter. 1 a = 1 N/m M =.018 g/mol The gas must be H. Gas ressure Gas ressure A barometer is an instrument that is used to measure atmospheric pressure. Standard atmospheric pressure (1 atm) was originally defined as the pressure that would support a column of mercury exactly 760 mm high. h = height in m hdg d = is the density in kg/m 3 g = is the gravitational constant ( m/s ) 3

4 1/7/013 Gas ressure A barometer is an instrument that is used to measure atmospheric pressure. 101,35 a 1 atm* = = 760 mmhg* = 760 torr* = bar = 14.7 psi * Represents an exact number Gas ressure A manometer is a device used to measure pressures other than atmospheric pressure. Worked Example 11. Worked Example 11. (cont.) Calculate the pressure exerted by a column of mercury 70.0 cm high. Express the pressure in pascals, in atmospheres, and in bars. The density of mercury is g/cm 3. Strategy Use = hdg to calculate pressure. Remember that the height must be expressed in meters and density must be expressed in kg/m 3. Solution 1 m h = 70.0 cm = m 100 cm d = g cm 3 g = m/s 1 kg cm 1000 g = m 4 kg/m 3 Solution pressure = m 4 kg m = kg/m s m 3 s a 1 atm a = 0.91 atm 101,35 a bar 0.91 atm = bar 1 atm Think About It Make sure your units cancel properly in this type of problem. Common errors include forgetting to express height in meters and density in kg/m 3. You can avoid these errors by becoming familiar with the value of atmospheric pressure in the various units. A column of mercury slightly less than 760 mm is equivalent to slightly less than 101,35 a, slightly less than 1 atm, and slightly less than 1 bar Boyle s law states that the pressure of a fixed amount of gas at a constant temperature is inversely proportional to the volume of the gas = at constant temperature (a) (b) (c) (mmhg) (ml) Calculate the volume of a sample of gas at 5.75 atm if it occupies 5.14 L at.49 atm. (Assume constant temperature.) Step 1: Use the relationship below to solve for : 1 1 = atm 5.14 L.6 L 5.75 atm 4

5 1/7/013 Worked Example 11.3 If a skin diver takes a breath at the surface, filling his lungs with 5.8 L of air, what volume will the air in his lungs occupy when he dives to a depth where the pressure of 1.9 atm? (Assume constant temperature and that the pressure at the surface is exactly 1 atm.) Charles s and Gay-Lussac s law, (or simply Charles s Law) states that the volume of a gas maintained at constant pressure is directly proportional to the absolute temperature of the gas. Strategy Use 1 1 = to solve for. Solution 1 = 1.00 atm, 1 = 5.8 L, and = 1.9 atm. = atm 5.8 L = = 3.03 L 1.9 atm Think About It At higher pressure, the volume should be smaller. Therefore, the answer makes sense. Higher temperature Lower temperature Charles s and Gay-Lussac s law, (or simply Charles s Law) states that the volume of a gas maintained at constant pressure is directly proportional to the absolute temperature of the gas. T T T 1 1 at constant pressure Worked Example 11.4 A sample of argon gas that originally occupied 14.6 L at 5 C was heated to 50.0 C at constant pressure. What is its new volume? Strategy Use 1 /T 1 = /T to solve for. Remember that temperatures must be expressed in kelvin. Solution T 1 = K, 1 = 14.6 L, and T = K. = 1 T 14.6 L K = = 15.8 L T K Think About It When temperature increases at constant pressure, the volume of a gas sample increases. Avogadro s law 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. n n n 1 1 at constant temperature and pressure Worked Example 11.5 If we combine 3.0 L of NO and 1.5 L of O, and they react according to the balanced equation NO(g) + O (g) NO (g), what volume of NO will be produced? (Assume that the reactants and products are all at the same temperature and pressure.) Strategy Apply Avogadro s law to determine the volume of a gaseous product. Solution 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 volume. The amounts of reactants given are stoichiometric amounts. According to the balanced equation, the volume of NO formed will be equal to the volume of NO that reacts. Therefore, 3.0 L of NO will form. Think About It When temperature increases at constant pressure, the volume of a gas sample increases. 5

6 1/7/013 A sample of gas originally occupies 9.1 L at 0.0 C. What is its new volume when it is heated to 15.0 C? (Assume constant pressure.) Step 1:Use the relationship below to solve for : (Remember that temperatures must be expressed in kelvin. T T T 9.1 L K 30.7 L T K 1 What volume in liters of water vapor will be produced when 34 L of H and 17 L of O react according to the equation below: H (g) + O (g) H O(g) Assume constant pressure and temperature. Step 1: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 volume. The amounts of reactants given are stoichiometeric amounts. 34 L of H O will form. Cooling at constant volume: pressure decreases Heating at constant volume: pressure increases Cooling at constant pressure: volume decreases Heating at constant pressure: volume increases The presence of additional molecules causes an increase in pressure. The combined gas law can be used to solve problems where any or all of the variables changes. n T n T

7 1/7/013 The volume of a bubble starting at the bottom of a lake at 4.55 C increases by a factor of 10 as it rises to the surface where the temperature is C and the air pressure is atm. Assume the density of the lake water is 1.00 g/ml. Determine the depth of the lake. Step 1:Use the combined gas law to find the pressure at the bottom of the lake; assume constant moles of gas. 1 1 n T n T 1 1 T atm K atm T K The volume of a bubble starting at the bottom of a lake at 4.55 C increases by a factor of 10 as it rises to the surface where the temperature is C and the air pressure is atm. Assume the density of the lake water is 1.00 g/ml. Determine the depth of the lake. Step :Use the equation = hdg to determine the depth of the lake. The pressure represents the difference in pressure from the surface to the bottom of the lake. ressure exerted by the lake = bottom of lake air = 9.19 atm atm = 8.5 atm The volume of a bubble starting at the bottom of a lake at 4.55 C increases by a factor of 10 as it rises to the surface where the temperature is C and the air pressure is atm. Assume the density of the lake water is 1.00 g/ml. Determine the depth of the lake. Step (Cont): Convert pressure to pascals and density to kg/m ,35 a 8.5 atm = 833,398 a 1 atm 1 g 1 kg 100 cm 1000 kg = 3 3 cm 1000 g 1 m m = hdg 833,398 a = h(1000 kg/m 3 )(9.81 m/s ) h = 85.0 m 3 Worked Example 11.6 If a child releases a 6.5-L helium balloon in the parking lot of an amusement park where the temperature is 8.50 C and the air pressure is 757. mmhg, what will the volume of the balloon be when it has risen to an altitude where the temperature is C and the air pressure is mmhg? Strategy In this case, because there is a fixed amount of gas, we use 1 1 /T 1 = /T. The only value we don t know is. Temperatures must be expressed in kelvins. We can use any units of pressure, as long as we are consistent. Solution T 1 = K, T = K. = 1 T mmhg K 6.5 L = = 10. L T mmhg K Think About It Note that the solution is essentially multiplying the original volume by the ratio of 1 and, and by the ratio of T to T 1. The effect of decreasing external pressure is to increase the balloon volume. The effect of decreasing temperature is to decrease the volume. In this case, the effect of decreasing pressure predominates and the balloon volume increases significantly The Ideal Gas Equation The Ideal Gas Equation The gas laws can be combined into a general equation that describes the physical behavior of all gases. 1 T n Boyle s law Charles s law Avogadro s law The ideal gas equation (below) describes the relationship among the four variables,, n, and T. = nrt nt R nt rearrangement = nrt An ideal gas is a hypothetical sample of gas whose pressurevolume-temperature behavior is predicted accurately by the ideal gas equation. R is the proportionality constant, called the gas constant. 7

8 1/7/013 The Ideal Gas Equation The gas constant (R) is the proportionality constant and its value and units depend on the units in which and are expressed. = nrt The Ideal Gas Equation Standard Temperature and ressure (ST) are a special set of conditions where: ressure is 1 atm Temperature is 0 C (73.15 K) The volume occupied by one mole of an ideal gas is then.41 L: = nrt (1 mol)( Latm/K mol)(73.15 K) =.41 L 1 atm Worked Example 11.7 Calculate the volume of a mole of ideal gas at room temperature (5 C) and 1 atm. Strategy Convert the temperature in C to kelvins, and use the ideal gas equation to solve for the unknown volume. Solution The data given are n = 1 mol, T = K, and = 1.00 atm. Because the pressure is expressed in atmospheres, we use R = L atm/k mol in order to solve for volume in liters. nrt (1 mol)( L atm/k mol)(98.15 K) = = = 4.5 L 1 atm 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 (0 C), so the molar volume at room temperature (5 C) should be higher than the molar volume at 0 C and it is. The Ideal Gas Equation Using algebraic manipulation, it is possible to solve for variables other than those that appear explicitly in the ideal gas equation. = nrt n RT n M RT M M d is the density (in g/l) d RT M is the molar mass (in g/mol) The Ideal Gas Equation What pressure would be required for helium at 5 C to have the same density as carbon dioxide at 5 C and 1.00 atm? Step 1:Use the equation below to calculate the density of CO at 5 C and 1 atm. The Ideal Gas Equation What pressure would be required for helium at 5 C to have the same density as carbon dioxide at 5 C and 1 atm? Step :Use the density of CO found in step 1 to calculate the pressure for He at 5 C. d M RT 1.00 atm44.01 g/mol d = = g/l Latm/mol K98.15 K g/mol g/l = Latm/mol K K = 11.0 atm 8

9 1/7/013 Worked Example Real Gases Carbon dioxide is effective in fire extinguishers partly because its density is greater than that of air, so CO can smother the flames by depriving them of oxygen. (Air has a density of approximately 1. g/l at room temperature and 1 atm.) Calculate the density of CO at room temperature (5 C) and 1.0 atm. Strategy Use d = M/RT to solve for density. Because the pressure is expressed in atm, we should use R = L atm/k mol. Remember the express temperature in kelvins. Solution The molar mass of CO is g/mol. M d = = RT (1 atm)(44.01 g/mol) L atm/k mol)(98.15 K) = 4.5 L Think About It The calculated density of CO 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. The van der Waals equation is useful for gases that do not behave ideally. Experimentally measured pressure an nb nrt corrected pressure term Container volume corrected volume term Real Gases The van der Waals equation is useful for gases that do not behave ideally. an nb nrt Real Gases Calculate the pressure exerted by 0.35 mole of oxygen gas in a volume of 6.50 L at 3 C using (a) the ideal gas equation and(b) the van der Waals equation. Step 1:Use the ideal gas equation to calculate the pressure of O. = nrt nrt 0.35 mol Latm/mol K K 1.3 atm 6.50 L Real Gases Calculate the pressure exerted by 0.35 mole of oxygen gas in a volume of 6.50 L at 3 C using (a) the ideal gas equation and(b) the van der Waals equation. Step :Use table 11.6 to find the values of a and b for O. Real Gases Calculate the pressure exerted by 0.35 mole of oxygen gas in a volume of 6.50 L at 3 C using (a) the ideal gas equation and (b) the van der Waals equation. Solution Step 3:Use the van der Waals equation to calculate. an nb nrt nrt an 0.35 mol Latm/mol K K 1.36atmL /mol ( 0.35 mol) nb 6.50 L 0.35 mol L/mol 6.50 L = 1.3 atm 9

10 1/7/013 Worked Example Worked Example (cont.) A sample of 3.50 moles of NH 3 gas occupies 5.0 L at 47 C. Calculate the pressure of the gas (in atm) using (a) the ideal gas equation and (b) the van der Waals equation. Strategy (a) Use the ideal gas equation, = nrt. (b) Use ( + an / )( nb) = nrt and a and b values for NH 3 from Table 11.5 Solution T = K, a = 4.17 atm L/mol, and b = L/mol. nrt (a) = = (3.50 mol)( L atm/k mol)(30.15 K) 5.0 L (b) Evaluating the correction terms in the van der Waals equation, we get an (4.17 atm L/mol = )(3.50 mol) (5.0 L) = 1.89 atm nb = (3.50 mol) L mol = L = 17.7 atm Solution Finally, substituting these results into the van der Waals equation, we get ( atm)(5.0 L L) = (3.50 mol)( L atm/k mol)(30.15 K) = 16. atm Think About It As is often the case, the pressure exerted by the real gas is lower than predicted by the ideal gas equation. Worked Example Gas Mixtures Consider a gas sample consisting of molecules with radius r. (a) Determine the excluded volume defined by two molecules and (b) calculate the excluded volume per mole (b) for the gas. Compare the excluded volume per mole with the volume actually occupied by a mole of the molecules. Strategy (a) Use the formula for volume of a sphere to calculate the volume of a sphere of radius r. This is excluded volume defined by two molecules. (b) Multiply the excluded volume per molecule by Avogardo s number to Think About It The excluded volume per mole is four times the determine excluded volume per mole. volume actually occupied by a mole of molecules. 4 Solution The equation for volume of a sphere is 3 πr 3 ; N A = (a) Excluded volume defined by two molecules = π(r) = 8( 3 πr 3 ). 4 8( πr (b) Excluded volume per mole = ) N A molecules 4N = A ( 3 πr 3 ) molecules 1 mole 1 mole The volume actually occupied by a mole of molecules with a radius r is 4 4N A ( 3 πr 3 ). When two or more gases are placed in a container, each gas behaves as though it occupies the container alone mole of N in a 5.00 L container at 0 C exerts a pressure of 4.48 atm. (1mol)( Latm/K mol)(73.15 K) N 4.48 atm 5.00 L Addition of 1.00 mole of O in the same container exerts an additional 4.48 atm of pressure. (1mol)( Latm/K mol)(73.15 K) O 4.48 atm 5.00 L The total pressure of the mixture is the sum of the partial pressures ( i ): total = N + O = 4.48 atm atm = 8.96 atm Gas Mixtures Dalton s law of partial pressure states that the total pressure exerted by a gas mixture is the sum of the partial pressures exerted by each component of the mixture: total = i Gas Mixtures Determine the partial pressures and the total pressure in a.50-l vessel containing the following mixture of gases at 15.8 C: mol He, mol H, and mol Ne. Step 1:Since each gas behaves independently, calculate the partial pressure of each using the ideal gas equation: mol He Latm/mol K88.95 K He atm.50 L mol H Latm/mol K88.95 K H atm.50 L mol Ne Latm/mol K88.95 K Ne 1.60 atm.50 L 10

11 1/7/013 Gas Mixtures Determine the partial pressures and the total pressure in a.50-l vessel containing the following mixture of gases at 15.8 C: mol He, mol H, and mol Ne. Step :Use the equation below to calculate total pressure. Gas Mixtures Each component of a gas mixture exerts a pressure independent of the other components. The total pressure is the sum of the partial pressures. total = i total = atm atm atm =.17 atm Worked Example 11.1 A 1.00-L vessel contains 0.15 mole of N gas and mole of H gas at 5.5 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 Think of About the mixture, It The and total sum pressure the two in partial the vessel pressures can also to find be the total pressure. determined by summing the number of moles of mixture components ( = 0.7 mol) and solving the ideal gas Solution equation T = for K total : (0.15 mol)( L atm/k mol)(98.65 K) N = (0.7 mol)( L atm/k mol)(98.65 K) = 5.7 atm total = 1.00 L = 5.56 atm 1.00 L ( mol)( L atm/k mol)(98.65 K) H = = 0.89 atm 1.00 L total = N + H = 5.7 atm atm = 5.56 atm Gas Mixtures The relative amounts of the components in a gas mixture can be specified using mole fractions. ni i = n total There are three things to remember about mole fractions: 1) The mole fraction of a mixture component is always less than 1. ) The sum of mole fractions for all components of a mixture is always 1. 3) Mole fractions are dimensionless. Χ i is the mole fraction. n i is the moles of a certain component n total is the total number of moles. Worked Example Reactions with Gaseous Reactants and roducts In 1999, the FDA approved the use of nitric oxide (NO) to treat and prevent lung disease, which occurs commonly in premature infants. The nitric oxide used in this therapy is supplied to hospitals in the form of a N /NO mixture. Calculate the mole fraction of NO in a L gas cylinder at room temperature (5 C) that contains 6.0 mol N and in which the total pressure is atm. Strategy Think Use About the ideal It To gas check equation your to work, calculate determine the total χ N number by subtracting of moles in the cylinder. χ NO Subtract from 1. moles Using of each N mole from the fraction total and to determine the total pressure, moles of NO. Divide moles NO calculate by total the moles partial to pressure get mole of fraction. each component using χ i = i / total and verify that they sum to the total pressure. Solution The temperature is K. (14.75 atm)(10.00 L) total moles = = = 6.09 mol RT ( L atm/k mol)(98.15 K) mol NO = total moles N = = mol NO n χ NO = NO mol NO = = n total 6.09 mol What mass (in grams) of Na O is necessary to consume 1.00 L of CO at ST? Na O (s) + CO (g) Na CO 3 (s) + O (g) Step 1:Convert 1.00 L of CO at ST to moles using the ideal gas equation. = nrt (1 atm)(1.00 L) = n( Latm/mol K)(73.15 K) n = moles CO 11

12 1/7/013 Reactions with Gaseous Reactants and roducts Worked Example What mass (in grams) of Na O is necessary to consume 1.00 L of CO at ST? Na O (s) + CO (g) Na CO 3 (s) + O (g) Step :Determine the stoichiometric amount of Na O. mol Na O g moles CO = 3.48 g mol CO mol Sodium peroxide (Na O ) is used to remove carbon dioxide from (and add oxygen to) the air supply in spacecrafts. It works by reacting with CO in the air to produce sodium carbonate (Na CO 3 ) and O. Na O (s) + CO (g) Na CO 3 (s) + O (g) What volume (in liters) of CO (at ST) will react with a kilogram of Na O? Strategy Convert the given mass of Na O to moles, use the balanced equation to determine the stoichiometric amount of CO, and then use the ideal gas equation to convert moles of CO to liters. Worked Example (cont.) Solution The molar mass of Na O is g/mol (1 kg = 1000 g). (Treat the specified mass of NaO as an exact number.) 1 mol Na 1000 g Na O O = 1.8 mol Na O g Na O mol CO 1.8 mol Na O = 1.8 mol Na mol Na O O CO = (1.8 mol CO )( L atm/k mol)(73.15 K) 1 atm = 87.4 L CO Think About It The answer seems like an enormous volume of CO. If you check the cancellation of units carefully in ideal gas equation problems, however, with practice you will develop a sense of whether such a calculated volume is reasonable. Reactions with Gaseous Reactants and roducts Although there are no empirical gas laws that focus on the relationship between n and, it is possible to rearrange the ideal gas equation to find the relationship. = nrt rearrangement n RT The change in pressure in a reaction vessel can be used to determine how many moles of gaseous reactant are consumed: n RT Worked Example Another air-purification method for enclosed spaces involves the use of scrubbers containing aqueous lithium hydroxide, which react with carbon dioxide to produce lithium carbonate and water: LiOH(aq) + CO (g) Li CO 3 (s) + H O(l) Consider the air supply in a submarine with a total volume of L. The pressure of atm, and the temperature 5 C. If the pressure in the submarine Think drops About to It Careful atm as cancellation the result of of carbon units is dioxide essential. being Note consumed that by this amount of CO an aqueous lithium hydroxide corresponds to 16 moles or 3.9 kg of LiOH. scrubber, how many moles of CO are consumed. (It s a good idea to verify this yourself. Strategy Use Δn = Δ (/RT) to determine Δn, the number of moles CO consumed. Solution Δ = atm atm = atm, = L, and T = K. Δn CO = atm 5 L ( L atm/k mol) (98.15 K) = 81 moles CO consumed Reactions with Gaseous Reactants and roducts The volume of gas produced by a chemical reaction can be measured using an apparatus like the one shown below. When gas is collected over water in this manner, the total pressure is the sum of two partial pressures: total = collected gas + H O 1

13 1/7/013 Reactions with Gaseous Reactants and roducts Gas Mixtures The vapor pressure of water is known at various temperatures. Calculate the mass of O produced by the decomposition of KClO 3 when 81 ml of O is collected over water at 30.0 C and atm. Step 1:Use Table 11.5 to determine the vapor pressure of water at 30.0 C. Reactions with Gaseous Reactants and roducts Reactions with Gaseous Reactants and roducts Calculate the mass of O produced by the decomposition of KClO 3 when 81 ml of O is collected over water at 30.0 C and atm. Step :Convert the vapor pressure of water at 30.0 C to atm and then use Dalton s law to calculate the partial pressure of O C total = O + H O O = total H O 1 atm = 31.8 torr = atm 760 torr O = atm atm = atm Calculate the mass of O produced by the decomposition of KClO 3 when 81 ml of O is collected over water at 30.0 C and atm. Step 3:Convert to moles of O using the ideal gas equation and then find mass. = nrt atm0.81 L n moles O RT Latm/mol K K 3.00 g moles O 1.03 g O mole Worked Example Chapter Summary: Key oints Calcium metal reacts with water to produce hydrogen gas: Ca(s) + H O(l) Ca(OH) (aq) + H (g) Determine the mass of H produced at 5 Cand atm when 55 ml of the gas is collected over water as shown in Figure Strategy Use Dalton s law of partial pressures to determine the partial pressure of H, use Think the About ideal gas It equation Check unit to determine cancellation moles carefully, of H, and remember then use the molar mass of that H the to convert densities to of mass. gases (ay are careful relatively attention low. The to units. mass of Atmospheric pressure approximately is given in atmospheres, half a liter of whereas hydrogen the at vapor near pressure room of temperature water is tabulated in torr.) and 1 atm should be a very small number. Solution H = total H O = atm atm = atm moles of H = ( atm)(0.55 L) ( L atm/k mol)(98.15 K) = mol moles of H = ( mol)(.016 g/mol) = g H Characteristics of Gases Gas ressure: Definition and Units Calculation of ressure Measurement of ressure Boyle s Law: The ressure-olume Relationship Charles s and Gay-Lussac s Law: The Temperature-olume Relationship Avogadro s Law: The Amount-olume Relationship Deriving the Ideal Gas Equation from the Empirical Gas Laws Applications of the Ideal Gas Equation Calculating the Required olume of a Gaseous Reactant Determining the Amount of Reactant Consumed Using Change in ressure redicting the olume of a Gaseous roduct Dalton s Law of artial ressures Mole Fractions Using artial ressures to Solve roblems Application of the Gas Laws Molecular Speed Diffusion and Effusion Factors That Cause Deviation from Ideal Behavior The van der Waals Equation 13

14 1/7/013 Group Quiz #4 What mass of solid sodium azide (NaN 3 ) is needed to generate 75. L of nitrogen gas (and solid sodium) at o C and 1.00 atm? (Hint: Start with a balanced chemical equation!) 14