File - Olson Chemistry

Transcription

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