Transcript 6 Intermolecular Forces of Attraction and Phases

STATES OF
MATTER
1
Intermolecular Forces
of Attraction
Kinetic Molecular Theory
• All matter is composed
of atoms that are in
constant motion
Kinetic Theory Facts
• All phases of matter express the degree that
they reflect the kinetic theory through their
kinetic energy
• kinetic energy is measured by temperature
• phase changes involve changes in
temperature due to the existence threshold
temperature of each phase (i.e. ice naturally
is found at cold not hot temperatures)
Solids
Liquids
Gases
Definite
size
Definite size
No Definite
size
Definite
shape
No Definite
shape
No Definite
shape
Low
Kinetic
Energy
More Kinetic High Kinetic
Energy than
Energy
Solids, but less
than Gases
• While gases have a great deal of random
motion, solids and liquids exist at lower
temperatures, thus allowing other forces of
attraction to act upon them
• these forces are the van der Waals forces
Definitions
• Bonds are intramolecular forces of
attraction
• Forces of attraction between molecules are
called intermolecular forces of attraction
• intermolecular forces of attraction are
commonly called van der Waals forces
The Condensed Phases
• Solids and Liquids
• Physical properties of the condensed phases
reflect the degree of intermolecular forces
(i.e. boiling point)
Interacting substances
Are ions present?
NO
van der Waals
Are polar molecues involved?
Yes
Is H bonding to N, O, or F?
no
yes
dipole-dipole forces Hydrogen bonding
no
London dispersion forces
(induced dipoles)
Yes
Are polar molecules and ions present?
yes
Ion-dipole forces
no
Ionic bonding
Dipole-dipole forces
• Exist between neutral polar molecules
• work best the closer the molecules are to
each other
• the greater the polarity of the molecules, the
greater the force of attraction
H bonding
• Special case of dipole-dipole interaction
specifically between H of one polar
molecule with N, O or F and an unshared
electron pair of another nearby small
electronegative ion (usually N, O, or F on
another molecule)
• VERY STRONG
London dispersion forces
• Induced dipoles
• not really dipoles on the AVERAGE, but
instantaneously dipole conditions can exist
thus allowing for pseudopolar regions to
occur
• No matter how strong the van der Waal
force of attraction is, it is still not stronger
than attractions involving ions
Ion-dipole forces
• Attraction between ions and the partial
charge on the end of a polar molecule
• ex. NaCl in water solution
INTRAMOLECULAR: FORCES WITHIN A MOLECULE
FORCES
INTERMOLECULAR: FORCES BETWEEN MOLECULES
O = O STRONG INTRAMOLECULMR FORCES
VERY WEMK INTERMOLECULAR FORCES
MUCH WEAKER THAN EITHER
IONIC OR COVALENT BONDS
ELECTROSTATIC
DICTATE WHETHER MOLEULAR
SUBTANCE IS GAS, LIQUID OR SOLID
AT ROOM CONDITIONS
GAS: NEGLIGIBLE
LIQUID: WEAK TO MODERATE
SOLID: MODERATE TO STRONG
15
HELP DEFINE STANDARD STATE OF
A SUBSTANCE
INTERMOLECULAR
FORCES:
HELP DEFINE COMMON PHYSICAL
PROPERTIESOF A SUBSTANCE
ARE MUCH
WEAKER
THAN
BONDS
BOILING/MELTING
SURFACE
CAPILLARY
VISCOSITY
TENSION
ACTION
POINTS
(INTRA MOLECULAR FORCES)
IONIC BONDS:
COVALENT BONDS:
INTERMOLCULAR FORCES:
~300 - 1000’S kJ/MOL
~150 - ~800 kJ/MOL
~1 - 40 kJ/MOL
STRENGTH DIMINISHES WITH INCREASING DISTANCE
16
d+
IONIC COMPOUNDS:
MOLECULAR
COMPOUNDS:
d+
d-
dMOLECULAR
DIPOLE
“INDUCED” TEMPORARY DIPOLE
DISPERSION
FORCES
d-
d+ d-
d+ d-
d+
PRESENT IN ALL MOLECULAR SUBSTANCES!
INCREASES WITH MOLECULAR MASS
F2 -188 oC
Cl2 58.8
I2 184
17
DIPOLE-DIPOLE FORCES
NATURALPERMANENT
ASSYMETRICDIPOLE
CHAGE DISTRIBUTION
POLAR
MOLECULE
OCCURS IF CENTERS OF CHARGES DO NOT COINCIDE
O
C
O
2dO
Cl
Cl C
Cl
F
H
H
d+
d+
Cl
THE MORE POLAR THE MOLECULE,
THE STRONGER THEINTERMOLECULAR FORCE
CH3 - CH2 - CH3 44 = MASS = 41 CH3CN
0.1==BOILING
DIPOLE POINT
MOMENT
3.9
231
(K) == 355
18
SPECIAL CASE OF DIPOLE FORCES
HYDROGEN
BOND
H2O
H2S
H2Se
H2Te
18
34
81
130
H ATOM ATTACHED TO F, O, N
LARGE ELECTRONEGATIVITY DIFFERENCE
SMALL SIZE ALLOWS H TO GET CLOSE
~ 100
- 73 oC
- 60.7
-41.5
-2
THIS PROPERTY AFFECTS
LIFE AND MANY OTHER
PROPERTIES
19
DIFFUSION:
VISCOSITY:
MOLECULES MOVING THRU MOLECULES
RESISTANCE TO FLOW
SURFACE
TENSION:
20
GAS (VAPOR)
SUBLIMATION
SYSTEM ENERGY
LIQUID
MELTING
(FUSION)
DEPOSITION
VAPORIZATION
SYSTEM ENERGY
CONDENSATION
FREEZING
SOLID
21
ENTHALPY
TEMPERATURE
DHf = HEAT REQUIRED TO MELT
SUBSTANCEAT ITS MELTING POINT
HEAT REQUIRED TO BREAK DOWN
VAPOR
> 0 J/g OR J/mol
INTERMOLECULAR FORCES
LIQUID +
VAPOR
LIQUID
LIQUID +
SOLID
SOLID
MELTING POINT
FREEZING POINT
TIME
22
GASES
LIQUIDS
SOLIDS
ASSUMES
SHAPE &
VOLUME OF
CONTAINER
DEFINITE
DEFINITE SHAPE
VOLUME;
AND VOLUME
ASSUMES SHAPE
OF CONTAINER
COMPRESSIBLE
VIRTUALLY
VIRTUALLY
INCOMPRESSIBLE INCOMPRESSIBLE
DIFFUSES
RAPIDLY IN
ANOTHER GAS
DIFFUSES
SLOWLY IN
ANOTHER
LIQUID
DIFFUSES VERY
SLOWLY IN
ANOTHER SOLID
FLOWS READILY FLOWS READILY DOES NOT FLOW
23
F/A =
PRESSURE
Patm
Dh
14.7 lb/in2 = 1 ATM = 760 mm
torrHg
GAS
P = Patm - Dh
bar
24
WHAT IS THE PRESSURE OF A GAS (ATM) IF IT SUPPORTS
A COLUMN OF MERCURY TO A HEIGHT OF 535 mm?
535 mm Hg x
1 atm
760 mm Hg
= 0.704 atm
IF THE ATMOSPHERIC PRESSURE DROPS TO 0.645 atm,
HOW HIGH IS THE COLUMN OF MERCURY SUPPORTED?
0.645 atm x 760 mm Hg = 490 OR 4.9 x 102 mm Hg
1 atm
IF THE BAROMETRIC PRESSURE IS 755 mm Hg AND A GAS
CREATES A Dh OF 275 mm IN A MANOMETER, WHAT IS
THE PRESSURE OF THE GAS?
P = Patm - Dh
480 torr x
= 755 - 275 = 480 mm Hg = 480 torr
1 atm
760 torr
= 0.632 atm
25
P x V = k (n, T)
VOLUME
OR P a 1/V
BOYLE’S
LAW
PRESSURE
26
VOLUME
CONSTANT PRESSURE!!!
CHARLES’ LAW
V = k(n, P) x T
OR V a T
TEMPERATURE
27
ABSOLUTE OR KELVIN SCALE
n=4
VOLUME
K = oC + 273
n=2
-273.15 oC
n = 1.0
n = 0.5
0 oC
USE IN ALL CALCULATIONS!!!
TEMPERATURE
28
AVOGADRO: EQUAL VOLUMES OF GASES AT SAME
T & P CONTAIN EQUAL NUMBER OF MOLECULES!
V = k(n, T) x n
P x V = k (n, T)
V = k(n, P) x T
IDEAL GAS
LAW
PV
PV == nRT
nkT
. .
.
R = 0.0821 L atm mol-1 K-1
29
THE PRESSURE IN AN AEROSOL CAN AT 25 oC IS 1.5 atm.
A FIRE CAN REACH 1200 oC. WHAT IS THE PRESSURE OF
THE CAN AT THAT TEMPERATURE?
INIT
FINAL
P V
1.5 Vi
X Vi
=
n
n
n
R
T
0.0821 25 + 273
0.0821 1200 + 273
Pf / Tf = R=
Pi / TPi i / Ti
Pf =
STP
1.57.4
atm
atmx 1473 K
298 K
WHAT IS THE V OF 1.0 MOL GAS AT 1.0 atm AND 0 oC?
. .
.
1nRT
mol x 0.0821 L atm mol-1 K-1 x 273 K
V=
P
1.0 atm
= 22.4 L
30
IDEAL VS. REAL GAS
MOLECULES FAR APART
NO COLLISIONS
NO INTERACTIONS
OCCUPY NO SPACE &
HAVE NO VOLUME
FAR APART
SOME COLLISIONS
LOW INTERACTIONS
MATTER; MUST OCCUPY
SPACE & HAVE VOLUME
PV/RT = 1
PV/RT > 1
SINCE GASES ARE REAL:
CAN NEVER ACHIEVE ABSOLUTE 0
APPROACH IDEAL GAS AT HIGH VOLUMES
LOW P, HIGH T
31
KINETIC MOLECULAR THEORY
1. THE VOLUME OF GAS MOLECULES IS NEGLIBLE
COMPARED TO THE VOLUME OF THE CONTAINER
2. PARTICLES UNDERGO CONSTANT RANDOM MOTION
AND DO NOT INTERACT WITH ONE ANOTHER
3. AVERAGE KINETIC ENERGY OF THE PARTICLES IS
PROPORTIONAL TO ABSOLUTE TEMPERATURE
TEMPERATURE IS THE MEASURE OF THE AVG.
KINETIC ENERGY OF THE PARTICLES IN THE SYSTEM
E ~ RT
.
.
R = 8.314 J mol-1 K-1
32
LIQUID VAPOR EQUILIBRIA
DYNAMIC EQUILIBRIUM!
EVAPORATION
VAPOR PRESSURE
ALL NON-GASEOUS
MATERIALS EXERT A
VAPOR PRESSURE.
FOR SOLIDS: VERY LOW
ASSUMED TO BE 0
33
VAPOR
PRESSURE
(mm Hg)
DHVAP
VOLATILE NON-VOLATILE
760
AMOUNT OF HEAT
REQUIRED TO
VAPORIZE SOME
AMOUNT OF LIQUID
T, oC
NORMAL BOILING POINT
VS BOILING POINT34
BOILING POINT
LIQUID
1.0
MELTING POINT
SOLID
P,
a
t
m
VAPOR (GAS)
TRIPLE POINT
T, oC
35
CAPILLARY ACTION
INTERMOLECULAR
ADHESIVE VS COHESIVE FORCES
ADHESIVE >
< COHESIVE
36
IONIC
B
O
I
L
I
N
G
P
O
I
N
T
X
DISPERSION: INCREASING MASS
DIPOLE-DIPOLE: INCREASING POLARITY
HYDROGEN BOND: INCREASING “NUMBER”
X
X
X
DISPERSION, DIPOLE
& H-BONDING
DISPERSION + DIPOLE
OR ANY OF THE PHYSICAL
PROPERTIES
DISPERSION ONLY
INTERMOLECULAR FORCES
37
WHAT FORCES ARE PRESENT IN:
A) NaBr IONIC
B) NF3 DISPERSION + DIPOLE-DIPOLE
C) CH2OHCH2CH2OH DISPERSION + DIPOLE + H-BOND
D) Ar
DISPERSION
WHICH OF THE FOLLOWING HAVE THE HIGHER BP?
A)
C6H14
C10H22
C2H6
B)
C6H14
C6H13OH
C6H12(OH)2
C)
CCl4
CCl3F
CF4
D)
HCl
HF
F2
38
UNIT CELLS
CRYSTAL LATTICE:
A A A A
A A A A A
A A A A
A A A A A
A A A A
A A A A A
THE REPEATING PATTERN IN A
THREE DIMENSIONAL ARRAY
NOTE: THIS ATOM
IS SHARED BY MORE
THAN ONE UNIT CELL
NOT 5 ATOMS PER CELL
IS 1 + 1/4(4) = 2 FULL ATOMS
CONSIDER FACES, EDGES,
CORNERS AND THOSE
TOTALLY WITHIN CELL
39
CRYSTALLINE SOLIDS
WELL DEFINED POSTIONS
FOR EMCH ATOM
ORDERED REPETITION OF
PATTERN
LONG RANGE ORDER!
AMORPHOUS SOLIDS
ILL DEFINED POSTIONS
ORDER EXTENDS OVER SHORT RANGE
LOCAL ORDER!
40
UNIT CELL STOICHIOMETRY
ATOM LOCATED ENTIRELY WITHIN CELL CONTRIBUTES
1 FULL ATOM TO CELL STOICHIOMETRY
6 FACES
2 CELLS
12 EDGES 4 CELLS
EIGHT CORNERS
8 CELLS
FACE ATOM CONTRIBUTES 1/2 x 6 = 3 ATOMS TO UNIT CELL
EDGE ATOM CONTRIBUTES 1/4 x 12 = 3 ATOMS TO UNIT CELL
CORNER ATOM CONTRIBUTES 1/8 x 8 = 1 ATOM TO UNIT CELL
WHAT ARE THESE UNIT CELLS?
41
b
c
g
SIMPLE CUBIC (SC)
a=b=c
a
a = b = c = 90o
SC = 1 ATOM/UNIT CELL
b
a
BODY CENTERED CUBIC (bcc)
CONTAINED WITHIN CELL
BCC = 2 ATOMS/UNIT CELL
FACE CENTERED CUBIC (fcc)
FCC = 4 ATOMS/UNIT CELL
42
COORDINATION NUMBER (CN) OR GEOMETRY
4 PARTICLES CONNECTED TO CENTRAL ATOM
CN = 4 = TETRAHEDRON
CN = 8
CN = 6
CN = 12
52%
68%
74%
PACKING
% BY
OF HOW
UNITCLOSE
CELL OCCUPIED
BY
DENSITY EFFICIENCY:
IS DETERMINED
(EFFICIENCY)
ATOMS,
OR MOLECULES
PARTICLES
AREIONS
PACKED
INTO A UNIT CELL
DENSITY IS A MEASURE OF HOW CONCENTRATED IS THE
MASS OF A PURE SUBSTANCE ....OR HOW TIGHTLY PACKED
d=
ATOMS/UNIT
CELL x MASS ATOM
MASS OF SUBSTANCE
SIDE3
VOLUME OF SUBSTANCE
43
FOUR TYPES OF CRYSTALLINE SOLIDS:
METALLIC:
METALLIC
IONIC
COVALENT
MOLECULAR
COVALENT TYPE BOND IN METALS
DELOCALIZED “SEA” OF ELECTRONS
MOLECULAR
ORBITALS
22ATOMIC
ORBITALS
4
8
16
ENERGY
BAND
6 x 1023
44
ENERGY
EMPTY
ORBITALS
FERMI
LEVEL
BAND
GAP
FILLED
ORBITALS
METAL
NONCONDUCTOR
SEMICONDUCTOR
45
IONIC
CRYSTALS
COVALENT
SOLIDS
MOLECULAR
SOLIDS
EXTENDED SOLIDS
HIGH MELTING & BOILING POINTS
IONIC COMPOUNDS
LARGE NETWORKS
HIGH MELTING, HARD SOLIDS
DIAMOND, MOST SEMICONDUCTORS, SiO2
INDIVIDUAL MOLECULES IN A LATTICE
LOW MELTING, SOFT
ICE, SUGAR, IODINE, SOLID HYDROGEN
46
GASES
Importance of Gases
• Airbags fill with N2 gas
in an accident.
• Gas is generated by the
decomposition of sodium
azide, NaN3.
• 2 NaN3 ---> 2 Na + 3
N2
THREE STATES OF
MATTER
General Properties of
Gases
• There is a lot of “free” space
in a gas.
• Gases can be expanded
infinitely.
• Gases fill containers
uniformly and completely.
• Gases diffuse and mix rapidly.
Properties of Gases
Gas properties can be modeled using
math. Model depends on—
• V = volume of the gas (L)
• T = temperature (K)
– ALL temperatures in the entire
chapter MUST be in Kelvin!!!
No Exceptions!
• n = amount (moles)
• P = pressure
(atmospheres)
Pressure
Pressure of air is
measured with a
BAROMETER
(developed by
Torricelli in 1643)
Hg rises in tube until force of Hg
(down) balances the force of
atmosphere (pushing up). (Just like
a straw in a soft drink)
P of Hg pushing down related to
• Hg density
• column height
Pressure
Column height measures
Pressure of atmosphere
• 1 standard atmosphere
(atm) *
= 760 mm Hg (or torr) *
= 14.7 pounds/in2 (psi)
= 101.3 kPa (SI unit is
PASCAL)
= about 34 feet of water!
Pressure Conversions
A. What is 475 mm Hg expressed in atm?
475 mm Hg x
1 atm
760 mm Hg
= 0.625 atm
B. The pressure of a tire is measured as 29.4 psi.
What is this pressure in mm Hg?
760 mm Hg
29.4 psi x
= 1.52 x 103 mm Hg
14.7 psi
Pressure Conversions
A. What is 2 atm expressed in torr?
B. The pressure of a tire is measured as 32.0 psi.
What is this pressure in kPa?
Boyle’s Law
P α 1/V
This means Pressure and
Volume are
INVERSELY
PROPORTIONAL if
moles and temperature
are constant (do not
change). For example, P
goes up as V goes down.
P1V1 = P2 V2
Robert Boyle
(1627-1691).
Son of Earl of
Cork, Ireland.
Boyle’s Law and Kinetic
Molecular Theory
P proportional to 1/V
Boyle’s Law
A bicycle pump is a
good example of
Boyle’s law.
As the volume of
the air trapped in
the pump is
reduced, its
pressure goes up,
and air is forced
into the tire.
Charles’s
Law
If n and P are constant,
then V α T
V and T are directly
proportional.
V1
V2
=
T1
T2
• If one temperature
goes up, the volume
goes up!
Jacques Charles (17461823). Isolated boron and
studied gases. Balloonist.
Charles’s original balloon
Modern long-distance balloon
Charles’s Law
Gay-Lussac’s Law
If n and V are constant,
then P α T
P and T are directly
proportional.
P1
P2
=
T1
T2
• If one temperature
goes up, the pressure
goes up!
Joseph Louis GayLussac (1778-1850)
Gas Pressure, Temperature, and
Kinetic Molecular Theory
P proportional to T
Combined Gas Law
• The good news is that you don’t have to
remember all three gas laws! Since they
are all related to each other, we can
combine them into a single equation.
BE SURE YOU KNOW THIS
EQUATION!
P1 V1
P2 V2
=
T1
T2
No, it’s not related to R2D2
Combined Gas Law
If you should only need one of the other gas laws,
you can cover up the item that is constant and you
will get that gas law!
P1 V1 =
T1
P2 V2
T2
Boyle’s Law
Charles’ Law
Gay-Lussac’s
Law
Combined Gas Law Problem
A sample of helium gas has a volume of 0.180 L, a
pressure of 0.800 atm and a temperature of 29°C.
What is the new temperature(°C) of the gas at a
volume of 90.0 mL and a pressure of 3.20 atm?
Set up Data Table
P1 = 0.800 atm
V1 = 180 mL
T1 = 302 K
P2 = 3.20 atm
V2= 90 mL
T2 = ??
Calculation
P1 = 0.800 atm
P2 = 3.20 atm
P1 V1
V1 = 180 mL
V2= 90 mL
T1 = 302 K
T2 = ??
P2 V2
=
T1
P1 V1 T2 = P2 V2 T1
T2
T2 = P2 V2 T1
P1 V1
T2 = 3.20 atm x 90.0 mL x 302 K
0.800 atm x 180.0 mL
= 604 K
Learning Check
A gas has a volume of 675 mL at 35°C and
0.850 atm pressure. What is the temperature in
°C when the gas has a volume of 0.315 L and
a pressure of 802 mm Hg?
One More Practice Problem
A balloon has a volume of 785 mL
on a fall day when the temperature
is 21°C. In the winter, the gas
cools to 0°C. What is the new
volume of the balloon?
And now, we pause for this
commercial message from
STP
OK, so it’s really not THIS kind
of STP…
STP in chemistry stands for
Standard Temperature and
Pressure
Standard Pressure = 1
atm (or an equivalent)
Standard Temperature
= 0° C (273 K)
STP allows us to compare
amounts of gases between
different pressures and
temperatures
Try This One
A sample of neon gas used in a neon sign has a volume
of 15 L at STP. What is the volume (L) of the neon gas
at 2.0 atm and –25°C?
Avogadro’s Hypothesis
Equal volumes of gases at the same T and
P have the same number of molecules.
V = n (RT/P) = kn
V and n are directly related.
twice as many
molecules
Avogadro’s Hypothesis and
Kinetic Molecular Theory
The gases in this
experiment are all
measured at the
same T and V.
P proportional to n
IDEAL GAS LAW
PV=nRT
Brings together gas
properties.
Can be derived from
experiment and theory.
BE SURE YOU KNOW
THIS EQUATION!
Using PV = nRT
P = Pressure
V = Volume
T = Temperature
N = number of moles
L • atm
Mol • K
R is a constant, called the Ideal Gas Constant
Instead of learning a different value for R for
all the possible unit combinations, we can
just memorize one value and convert the
units to match R.
Using PV = nRT
How much N2 is required to fill a small room with a
volume of 960 cubic feet (27,000 L) to 745 mm
Hg at 25 oC?
Solution
1. Get all data into proper units
V = 27,000 L
T = 25 oC + 273 = 298 K
P = 745 mm Hg (1 atm/760 mm Hg)
= 0.98 atm
And we always know R, 0.0821 L atm / mol
K
77
How much N2 is required to fill a small room with a volume of 960 cubic
feet (27,000 L) to P = 745 mm Hg at 25 oC?
Solution
2. Now plug in those values and solve for the
unknown.
PV = nRT
RT
RT
n = 1.1 x 103 mol (or about 30 kg of gas)
Learning Check
Dinitrogen monoxide (N2O), laughing gas, is
used by dentists as an anesthetic. If 2.86 mol of
gas occupies a 20.0 L tank at 23°C, what is the
pressure (mm Hg) in the tank in the dentist
office?
Learning Check
A 5.0 L cylinder contains oxygen gas at
20.0°C and 735 mm Hg. How many
grams of oxygen are in the cylinder?
Deviations from
Ideal Gas Law
• Real molecules have volume.
The ideal gas consumes the entire
amount of available volume. It
does not account for the volume of
the molecules themselves.
• There are intermolecular
forces.
An ideal gas assumes there are no
attractions between molecules.
Attractions slow down the
molecules and reduce the amount
of collisions.
– Otherwise a gas could
not condense to
become a liquid.
Gases in the Air
The % of gases in air
Partial pressure (STP)
78.08% N2
593.4 mm Hg
20.95% O2
159.2 mm Hg
0.94% Ar
7.1 mm Hg
0.03% CO2
0.2 mm Hg
PAIR = PN + PO + PAr + PCO = 760 mm Hg
2
2
2
Dalton’s Law of Partial
Pressures
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
0.32 atm
0.16 atm
What is the total pressure in the flask?
Ptotal in gas mixture = PA + PB + ...
Therefore,
Ptotal = PH2O + PO2 = 0.48 atm
Dalton’s Law: total P is sum of
PARTIAL pressures.
Dalton’s Law
John Dalton
1766-1844
Health Note
When a scuba diver is several
hundred feet under water, the
high pressures cause N2 from the
tank air to dissolve in the blood.
If the diver rises too fast, the
dissolved N2 will form bubbles
in the blood, a dangerous and
painful condition called "the
bends". Helium, which is inert,
less dense, and does not dissolve
in the blood, is mixed with O2 in
scuba tanks used for deep
descents.
Collecting a gas “over
water”
• Gases, since they mix with other gases readily, must be
collected in an environment where mixing can not occur. The
easiest way to do this is under water because water displaces the
air. So when a gas is collected “over water”, that means the
container is filled with water and the gas is bubbled through the
water into the container. Thus, the pressure inside the container
is from the gas AND the water vapor. This is where Dalton’s
Law of Partial Pressures becomes useful.
Table of Vapor Pressures for Water
Solve This!
A student collects
some hydrogen
gas over water at
20 degrees C and
768 torr. What is
the pressure of
the H2 gas?
768 torr – 17.5 torr = 750.5 torr
GAS DENSITY
22.4 L of ANY gas
AT STP = 1 mole
High
density
Low
density
Gases and Stoichiometry
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
Decompose 1.1 g of H2O2 in a flask with a volume of
2.50 L. What is the volume of O2 at STP?
Bombardier beetle
uses decomposition of
hydrogen peroxide to
defend itself.
Gases and Stoichiometry
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the
volume of O2 at STP?
Solution
1.1 g H2O2
1 mol H2O2 1 mol O2
22.4 L O2
34 g H2O2 2 mol H2O2 1 mol O2
= 0.36 L O2 at STP
Gas Stoichiometry: Practice!
A. What is the volume at STP of 4.00 g of CH4?
B. How many grams of He are present in 8.0 L of gas at STP?
What if it’s NOT at STP?
• 1. Do the problem like it was at STP. (V1)
• 2. Convert from STP (V1, P1, T1) to the
stated conditions (P2, T2)
Try this one!
How many L of O2 are needed to react 28.0 g NH3 at
24°C and 0.950 atm?
4 NH3(g) + 5 O2(g)
4 NO(g) + 6 H2O(g)
GAS DIFFUSION AND
EFFUSION
• diffusion is the gradual
mixing of molecules of
different gases.
• effusion is the
movement of molecules
through a small hole
into an empty container.
GAS DIFFUSION AND
EFFUSION
Graham’s law
governs effusion
and diffusion of gas
molecules.
Rate for A
Rate for B
M of B
M of A
Rate of effusion is
inversely proportional to
its molar mass.
Thomas Graham, 1805-1869. Professor
in Glasgow and London.
GAS DIFFUSION AND
EFFUSION
Molecules effuse thru holes in a
rubber balloon, for example, at
a rate (= moles/time) that is
• proportional to T
• inversely proportional to M.
Therefore, He effuses more
rapidly than O2 at same T.
He
Gas Diffusion
relation of mass to rate of diffusion
• HCl and NH3
diffuse from
opposite ends of
tube.
• Gases meet to form
NH4Cl
• HCl heavier than
NH3
• Therefore, NH4Cl
forms closer to HCl
end of tube.