Transcript Graham’s Law of Diffusion
Graham’s Law
• The rate of diffusion/effusion is proportional to the mass of the molecules 250 g 80 g Large molecules move slower than small molecules
Graham’s Law –The rate is
inversely proportional
to the square root of the molar mass of the gas
v
1
m
Graham’s Law
Speed 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
KE = ½mv
2
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Graham’s Law
Consider two gases at same temp.
Gas 1: Gas 2: KE KE 1 2 = ½ m 1 = ½ m 2 v 1 2 v 2 2 Since temp. is same, then… ½ m 1 m 1 KE 1 v 1 2 v 1 2 = KE 2 = ½ m = m 2 2 v 2 2 v 2 2 Divide both sides by m 1 v 2 2 … 1 m 1 v 2 2 m 1 v 1 2 m 2 v 2 2 1 m 1 v 2 2 v 1 2 v 2 2 m 2 m 1 Take square root of both sides to get Graham’s Law: v 1 v 2 m 2 m 1
“mouse in the house”
Gas Diffusion and Effusion
Graham's law governs effusion and diffusion of gas molecules.
Rate of A Rate of B mass of B mass of A Rate of effusion is
inversely
proportional to its molar mass.
Thomas Graham
(1805 - 1869)
Graham’s Law
Graham’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 .
v A v B
m B m A
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He
4.0026
2 Find the relative rate of diffusion of helium and chlorine gas 17 Cl
35.453
Step 1) Write given information GAS 1 = helium M 1
v
1 = 4.0 g = x Step 2) Equation v 1 v 2 m 2 m 1 He Step 3) Substitute into equation and solve v v 1 2 GAS 2 = chlorine M 2
v
2 = 71.0 g = x = 71.0 g 4.0 g Cl 2 4.21
1
He diffuses 4.21 times faster than Cl 2
9 F
18.9984
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?
10 Ne
20.1797
Step 1) Write given information GAS 1 = fluorine F 2 M 1
v
1 = 38.0 g = 363 m/s GAS 2 = Neon M 2
v
2 = 20.18 g = x Ne Step 2) Equation Step 3) Substitute into equation and solve v 1 v 2 m 2 m 1 363 m/s v 2 = 20.18 g 38.0 g
Rate of diffusion of Ne = 498 m/s
498 m/s
Find the molar mass of a gas that diffuses 18 Ar
39.948
about 4.45 times faster than argon gas.
Step 1) Write given information GAS 1 = unknown ?
M 1
v
1 = x g = 4.45
GAS 2 = Argon M 2
v
2 = 39.95 g = 1 Ar Step 2) Equation Step 3) Substitute into equation and solve v 1 v 2 m 2 m 1 What gas is this?
4.45
1 = 39.95 g x g
Hydrogen gas: H 2
2.02 g/mol 1 H
1.00794
At a certain temperature fluorine gas travels at 582 m/s and a noble gas travels at 394 m/s. What is the noble gas?
v 1 v 2 m 2 m 1 v F 2 v unk m unk m F 2 v F 2 v unk 2 m unk m F 2 m unk m F 2 v F 2 v unk 2 38 amu 582 394 82.9
amu Kr
CH 4 moves 1.58 times faster than which noble gas?
Governing relation: v CH 4 1.58
v unk v CH 4 v unk m unk m CH 4 1.58
v v unk unk m unk m CH 4 (1.58) 2 m m unk CH 4 m unk (1.58) 2 m CH 4 (1.58) 2 (16 amu) 39.9
amu Ar
HCl and NH 3 are released at same time from opposite ends of 1.20 m horizontal tube. Where do gases meet?
HCl NH 3 1.20 m v NH 3 v HCl m HCl m NH 3 36.5
17 1.465
v NH 3 1.465
v HCl Velocities are relative; pick easy #s: v HCl 1.000
m/s v NH 3 1.465
m/s 1.20
HCl dist.
NH 3 dist.
1.000
t 1.465
t t 0.487
s DISTANCE = RATE x TIME So HCl dist. = 1.000 m/s (0.487 s) = 0.487 m
HCl
Graham’s Law of Diffusion
NH 4 Cl(s) NH 3 100 cm 100 cm
Choice 1: Both gases move at the same speed and meet in the middle.
HCl
Diffusion
NH 4 Cl(s) 81.1 cm 118.9 cm
Choice 2: Lighter gas moves faster; meet closer to heavier gas.
NH 3
Calculation of Diffusion Rate
v
1
v
2
m
2
m
1 NH 3 V 1 M 1 = X = 17 amu HCl V 2 M 2 = X = 36.5 amu Substitute values into equation
v
1
v
2 36.5
17
v
1
v
2 1.465
x
V 1 moves 1.465x for each 1x move of V 2 NH 3 HCl 1.465 x + 1x = 2.465
200 cm / 2.465 = 81.1 cm for x
Calculation of Diffusion Rate
V 1 V 2 = m 2 m 1 NH 3 V 1 M 1 = X = 17 amu Substitute values into equation V 1 V 2 = 36.5
17 V 1 V 2 = 1.465
HCl V 2 M 2 = X = 36.5 amu V 1 moves 1.465x for each 1x move of v 2 NH 3 HCl 1.465 x + 1x = 2.465
200 cm / 2.465 = 81.1 cm for x
Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.
Br
79.904
35
Graham’s Law
Kr
83.80
36 Determine the relative rate of diffusion for krypton and bromine.
The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “v A /v B ”.
v
A
v
B
m
B
m
A
v
Kr
v
Br 2
m m
Br 2 Kr
159.80 g/mol 83.80 g/mol
1.381
Kr diffuses 1.381 times faster than Br 2 .
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1 H
1.00794
Graham’s Law
8 O
15.9994
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?
v A v B
m B m A v H 2 v O 2
m O 2 m H 2 v H
2
12.3
m/s
32.00 g/mol 2.02
g/mol
Put the gas with the unknown speed as “Gas A”.
v
H
2
12.3
m/s
3.980
v H
2
49.0 m/s
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1.0
H 1
v A v B v A v O 2
H 2 = 2 g/mol
Graham’s Law
An unknown gas diffuses 4.0 times faster than O 2 . Find its molar mass.
8 O
15.9994
The first gas is “Gas A” and the second gas is “Gas B”. The ratio “v A /v B ” is 4.0.
2
m B m A
Square both sides to get rid of the square root sign.
m O 2 16
32.00 g/mol m A m A m A
32.00 g/mol
2.0 g/mol 16
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