Diffusion vs. EffusionDiffusion - The tendency of the molecules of a given substance to move from regions of higher concentration to regions of lower concentrationExamples: A scent spreading throughout a room
or people entering a theme park
Effusion - The process by which gas particles under pressure pass through a tiny holeExamples: Air slowly leaking out of a tire or
helium leaking out of a balloon
Effusion
Particles in regions of high concentrationspread out into regions of low concentration,
filling the space available to them.
NETNET MOVEMENT
To use Graham’s Law, both gases must be at same temperature.
diffusiondiffusion: particle movement from
high to low concentration
effusioneffusion: diffusion of gas particles
through an opening
For gases, rates of diffusion & effusion obey Graham’s law: more massive = slow; less massive = fast
Graham’s Law
KE = ½mv2
Speed of diffusion/effusionSpeed of diffusion/effusion– Kinetic energy is determined by the temperature of
the gas.
– At the same temp & KE, heavier molecules move more slowly.
• Larger m smaller v
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Derivation of Graham’s Law
• The average kinetic energy of gas molecules depends on the temperature:
KE 1
2mv 2
where m is the mass and v is the speed
• Consider two gases:
KE1 1
2m1v1
2
KE2 1
2m2v2
2
Graham’s Law
2
21
222
2112
21 v m
1v mv m
v m
1
1
22
2
21
mm
v
v
1
2
2
1
mm
vv
Consider two gases at same temp.Gas 1: KE1 = ½ m1 v1
2
Gas 2: KE2 = ½ m2 v22
Since temp. is same, then… KE1 = KE2
½ m1 v12 = ½ m2 v2
2
m1 v12 = m2 v2
2
Divide both sides by m1 v22…
Take square root of both sides to get Graham’s Law:
Graham’s Law
Graham’s LawGraham’s Law– Rate of diffusion of a gas is inversely related to the
square root of its molar mass.
– The equation shows the ratio of Gas A’s speed to Gas B’s speed.
A
B
B
A
m
m
v
v
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Graham’s Law• The rate of diffusion/effusion is
proportional to the mass of the molecules
– The rate is inversely proportional to the square root of the molar mass of the gas
v 1
m
250 g
80 g
Large molecules move slower than small molecules
Step 1) Write given information
GAS 1 = helium
M1 = 4.0 g
v1 = x
GAS 2 = chlorine
M2 = 71.0 g
v2 = x
He Cl2
Step 2) Equation
1
2
2
1
mm
vv
Step 3) Substitute into equation and solve
v1
v2
=71.0 g
4.0 g
4.21
1
Find the relative rate of diffusion of helium and chlorine gas
He diffuses 4.21 times faster than Cl2
Cl35.453
17
He4.0026
2
If fluorine gas diffuses at a rate of 363 m/s at a certain temperature, what is the rate of diffusion of neon gas at the same temperature?
Step 1) Write given information
GAS 1 = fluorine
M1 = 38.0 g
v1 = 363 m/s
GAS 2 = Neon
M2 = 20.18 g
v2 = x
F2 Ne
Step 2) Equation
1
2
2
1
mm
vv
Step 3) Substitute into equation and solve
363 m/s
v2
=20.18 g
38.0 g498 m/s
Rate of diffusion of Ne = 498 m/s
Ne20.1797
10
F18.9984
9
Find the molar mass of a gas that diffuses about 4.45 times faster than argon gas.
What gas is this?
Hydrogen gas: H2
Step 1) Write given information
GAS 1 = unknown
M1 = x g
v1 = 4.45
GAS 2 = Argon
M2 = 39.95 g
v2 = 1
? Ar
Step 2) Equation
1
2
2
1
mm
vv
Step 3) Substitute into equation and solve
4.45
1=
39.95 g
x g2.02 g/mol
H1.00794
1
Ar39.948
18
HCl and NH3 are released at same
time from opposite ends of 200 cm horizontal tube. Where do the two gases meet?
Ammonium hydroxide (NH4OH) is ammonia (NH3) dissolved in water (H2O)
NH3(g) + H2O(l) NH4OH(aq)
Stopper
1 cm diameter
Cotton plug
Cotton plug
Stopper
Clamps
70-cm glass tube
Graham’s Law of Diffusion
HCl NH3
100 cm 100 cm
Choice 1: Both gases move at the same speed and meet in the middle.
NH4Cl(s)
Diffusion
HCl NH3
81.1 cm 118.9 cm
NH4Cl(s)
Choice 2: Lighter gas moves faster; meet closer to heavier gas.
Calculation of Diffusion Rate
NH3V1 = XM1 = 17 amu
HCl V2 = XM2 = 36.5 amu
Substitute values into equation
V1 (NH3) moves 1.465x for each 1x move of V2 (HCl)
t = 81.1 s
1 2
2 1
v m
v m
1
2
36.5
17
v
v
DISTANCE = RATE x TIME
Velocities are relative; pick easy #s:
3200 cm HCl dist. NH dist.
cmHCl sv 1.000
3
cmNH sv 1.465
cm cms s200 cm = (1.000 t) (1.465 t)
So HCl dist. …1.000 cm/s (81.1 s) = 81.1 cm
Calculation of Diffusion Rate
NH3V1 = XM1 = 17 amu
HCl V2 = XM2 = 36.5 amu
Substitute values into equation
V1 moves 1.465x for each 1x move of V2
NH3 HCl
1.465 x + 1x = 2.465
200 cm / 2.465 = 81.1 cm for x
1 2
2 1
v m
v m
1
2
36.5
17
v
v
1
2
1.465v
xv
Determine the relative rate of diffusion for krypton and bromine.
1.381
Kr diffuses 1.381 times faster than Br2.
Kr
Br
Br
Kr
m
m
v
v2
2
A
B
B
A
m
m
v
v
g/mol83.80
g/mol159.80
Graham’s Law
The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “vA/vB”.
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Kr83.80
36
Br79.904
35
A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions?
A
B
B
A
m
m
v
v
2
2
2
2
H
O
O
H
m
m
v
v
g/mol 2.02
g/mol32.00
m/s 12.3
vH 2
Graham’s Law
3.980m/s 12.3
vH 2
m/s49.0 vH 2
Put the gas with the unknown
speed as “Gas A”.
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
O15.9994
8
H1.00794
1
An unknown gas diffuses 4.0 times faster than O2. Find its molar mass.
Am
g/mol32.00 16
A
B
B
A
m
m
v
v
A
O
O
A
m
m
v
v2
2
Am
g/mol32.00 4.0
16
g/mol32.00 mA
2
Am
g/mol32.00 4.0
g/mol2.0
Graham’s Law
The first gas is “Gas A” and the second gas is “Gas B”. The ratio “vA/vB” is 4.0.
Square both sides to get rid of the square
root sign.
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
O15.9994
8
H1.0
1H2 = 2 g/mol