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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
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:
2
2
22
2
2
2
2
Cl
COCOCl
Cl
CO
CO
Cl
1
2
2
1
m
m v v
m
m
v
v
mm
vv
m/s 320 g 71g 44
m/s 410 v2Cl
On average, carbon dioxide travels at 410 m/s at 25oC.
Find the average speed of chlorine at 25oC.
**Hint: Put whatever you’re looking for in the numerator.
2
2
2
2
F
unk
2
unk
F
F
unk
unk
F
1
2
2
1
m
m
v
v
m
m
v
v
mm
vv
Kr amu 82.9 394582
amu 38 v
v m m
2
unk
FFunk
2
2
At a certain temperature fluorine gas travels at 582 m/s
and a noble gas travels at 394 m/s. What is the noble gasnoble gas?
unkCH v 1.58 v4
444
4
CH
unk2
CH
unk
unk
unk
CH
unk
unk
CH
m
m (1.58)
m
m
v
v 1.58
m
m
v
v
Ar amu 39.9 amu) (16 (1.58) m (1.58) m 2CH
2unk 4
CH4 moves 1.58 times faster than which noble
gas?Governing relation:
So HCl dist. = 1.000 m/s (0.487 s) = 0.487 m
HClNHNH
HCl
HCl
NH v 1.465 v 1.465 17
36.5
m
m
v
v3
3
3
m/s 1.000 vHCl m/s 1.465 v
3NH
s 0.487 t t 1.465 t 1.000 dist. NH dist. HCl 1.20 3
HCl NH3
1.20 m
DISTANCE = RATE x TIME
HCl and NH3 are released at same time from opposite ends
of 1.20 m horizontal tube. Where do gases meet?
Velocities are relative; pick easy #s:
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:
“mouse in the house”
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
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
Where should the NH3 and the HCl meet in the tube if it is approximately 70 cm long?
41.6 cm from NH3
28.4 cm from HCl
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 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
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