File - Olson Chemistry
SECTION 11.4 Diffusion and Effusion Teacher Notes and Answers SECTION 4 Diffusion and Effusion 1.The nitrogen gas will diffuse more quickly. 2.lower; lower 3.nitrogen Practice A.162 g/mol Review 1.Effusion is the process by which gas molecules under pressure pass through a tiny opening. Diffusion is the mixing of two gases by random molecular motion. 2.Graham’s law states that the rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses. 3.16 g/mol 4.256:1 5.BrF, NO2 , HCl, H2 O, He 6.128 g/mol; HI Gases 1 SECTION 11.4 Diffusion and Effusion The kinetic-molecular theory of matter, as applied to gases, states that a gas is a large group of fast moving particles that are far apart and travel in all directions. Two processes that can be explained by examining a gas at the particle level are diffusion and effusion. Key Terms Graham’s law of effusion In diffusion, the particles from two different gases mix together. In the diagram below, perfume molecules diffuse into the air when a bottle of perfume is opened. At the same time, the nitrogen gas and oxygen gas in the air diffuse into the perfume bottle. Gas molecule from perfume Oxygen molecule from the air Air molecules diffuse into a bottle of perfume when it is opened. At the same time, perfume molecules diffuse into the air outside the bottle. Nitrogen molecule from the air The rate of diffusion depends on the speed of the molecules. If the particles in a gas are moving faster than the particles in another gas, then the first gas will diffuse more quickly. Since the two gases are usually at the same temperature, then the average kinetic energy of the particles in two mixed gases is the same. Therefore, the gas whose particles have a smaller mass will have a greater speed. For the perfume example, the perfume molecules have a greater mass than the particles in the air, so they diffuse into the air more slowly than the air particles diffuse into the bottle. 2 CHAPTER 11 Critical Thinking 1. Apply A container filled with oxygen gas is connected to a container filled with nitrogen gas. Which gas will diffuse more quickly to fill both containers? The rates of effusion and diffusion for gases depend on the velocities of their molecules. In the mid-1800s, the Scottish chemist Thomas Graham studied the effusion and diffusion of gases. Recall that effusion occurs when the random motions of a gas result in some of the molecules passing through a small opening. One of Graham’s discoveries was a relationship between the rate of effusion of gases and their molar mass. The mass and velocity of a gas molecule determines its kinetic energy, through the expression mv2 /2. If two different gases are at the same temperature, then the following relationship is true. __ A2 = __ 1 MB v vB2 1 MA 2 2 where v A and vBare the average velocities of the particles in are the molar masses of the the two gases and MA and MB two gases. The rate of effusion of a gas is proportional to the average velocity of its particles. Therefore, the equation above can be used to show that the following relationship is true. READING CHECK 2. A gas that has a greater mass than another gas will have a ____ √ B M rate of effusion of A __________________ _____ = ____ rate of effusion of B √ A M This equation, Graham’s law of effusion, states that the rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses. So, a gas that effuses twice as fast as a second gas has one-fourth of the mass of the second gas. rate of diffusion and a rate of effusion than the other gas. 3. Which gas, nitrogen or oxygen, will effuse out of the tire in the photograph faster? Molecules of oxygen and nitrogen from inside the bicycle tire effuse out through a small nail hole. Oxygen molecule, O2 Nitrogen molecule, N2 Gases 3 SAMPLE PROBLEM Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure. SOLUTION 1 ANALYZE Determine the information that is given and unknown. 2 Given: identities of two gases, H 2 and O Unknown: relative rates of effusion 2 PLAN Write the equation that can be used to find the unknown. The molar masses of the two gases can be determined from the periodic table. Graham’s law can then be used to write the relative rates of effusion. ____ B √ M of A = _____ __________________ ____ rate of effusion rate of effusion of B 3 SOLVE √ M A Calculate the unknown using the given information. ____ ____ √ √ M rate of effusion of H 2 ______ 32.00 g/mol O 2 ___________________ = ____________ __________ = = 3.98 rate of effusion of O 2 √ M H 2 √ 2.02 g/mol ___________ Hydrogen effuses 3.98 times faster than oxygen. 4 CHECK YOUR WORK Check the answer to see if it is reasonable. Oxygen is about 16 times more massive than hydrogen. The square root of 16 is 4, and 3.98 is very close to 4. PRACTICE A.A sample of hydrogen effuses through a porous container about 9 times faster than an unknown gas. Estimate the molar mass of the unknown gas. rate of effusion of H2 ? What value is given for ____________________________ rate of effusion of unknown gas rite an equation for M W B and then substitute the known values. 4 CHAPTER 11 SECTION 11.4 REVIEW VOCABULARY 1. Compare diffusion with effusion. REVIEW 2. State Graham’s law of effusion. 3. Estimate the molar mass of a gas that effuses at 1.6 times the effusion rate of carbon dioxide. 4. Determine the molecular mass ratio of two gases whose rates of effusion have a ratio of 16:1. 5. List the following gases in order of increasing average molecular velocity at 25°C: H2 O, He, HCl, BrF, and NO2 . Critical Thinking 6. ANALYZING INFORMATION An unknown gas effuses at one-half the speed of oxygen. What is the molar mass of the unknown? The gas is known to be either HBr or HI. Which gas is it? Gases 5 Math Tutor Algebraic Rearrangements of Gas Laws When using the gas laws, you do not need to memorize all of the equations, because they can be derived from the combined gas law. In each of Boyle’s, Charles’s, and Gay-Lussac’s laws, one of the quantities—T, P, or V—does not change. By eliminating that factor from the equation, you obtain the equation for one particular gas law. The conditions stated in the problem should make clear which factors change and which are held constant. This information will tell you which law’s equation you need to use. Gas law Held constant Cancellation Result Combined gas law none P V1 _____ P V2 _____ = 2 1 P V2 P V1 _____ _____ 1 = 2 Boyle’s law temperature P V1 _____ P V2 _____ = 2 1 P1 V1= P2 V2 Charles’s law pressure P V1 _____ P V2 _____ 1 = 2 V V ___ 1 = ___ 2 Gay-Lussac’s law volume P V1 _____ P V2 _____ = 2 1 P P ___ 1 = ___ 2 T1 T1 T1 T1 T2 T2 T2 T2 T1 T1 Problem-Solving TIPS • When you solve problems in chemistry, it is usually a bad idea to just start entering numbers into a calculator. • Try to manipulate an equation in variable form to solve for the unknown variable before you start computation. This way, you can substitute all of the values at the very end and save having to keep track of intermediate results. SAMPLE A cylinder of nitrogen gas has a volume of 35.00 L at a pressure of 11.50 atm. What pressure will the nitrogen have if the contents of the cylinder are allowed to flow into a sealed reaction chamber whose volume is 140.0 L, and if the temperature remains constant? Start with the combined gas law, and cancel the temperature. P V1 _____ P V2 _____ = 2 → P1V1= P2 V2 1 T1 T2 The unknown value is P 2, so solve the above equation for P 2 . P 1 V P2 = _____ 1 2 V Substitute the given values into the equation to find the answer. (11.50 atm)(35.00 L) P2 = __________________ = 2.875 atm 140.0 L Practice Problems: Chapter Review practice problems 11 and 12 6 CHAPTER 11 T2 T1 T2 T2
© Copyright 2024