Graham’s Law of Diffusion

Graham’s Law of Diffusion

HCl NH3

100 cm 100 cm

NH4Cl(s)

Choice 1: Both gases move at the same speed and meet in the middle.

Diffusion

HCl NH3

81.1 cm 118.9 cm

NH4Cl(s)

Choice 2: Lighter gas moves faster; meet closer to heavier gas.

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

DiffusionDiffusion– Spreading of gas molecules throughout

a container until evenly distributed.

EffusionEffusion– Passing of gas molecules through a tiny

opening in a container

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

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

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

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

H2.0

1

http://www.unit5.org/christjs/tempT27dFields-Jeff/GasLaw1.htm

Graham's Law

        Graham's Law

        Graham's Law

Diffusion

Gas Diffusion and Effusion

Graham's law governs effusion and diffusion of gas molecules.

Thomas Graham(1805 - 1869)

Rate of effusion is inversely proportional to its molar mass.

Rate of effusion is inversely proportional to its molar mass.

Aof massB of mass

B of Rate Aof Rate

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

Diffusion

Particles in regions of high concentrationspread out into regions of low concentration,

filling the space available to them.

Weather and Diffusion

Map showing tornado risk in the U.S.

HighestHigh

LOWAir Pressure

HIGHAir Pressure

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

Calculation of Diffusion Rate

V1 m2

V2 m1

= NH3V1 = XM1 = 17 amu

HCl V2 = XM2 = 36.5 amu

Substitute values into equation

V1 36.5V2 17

=

V1 V2

= 1.465

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