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

Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

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