3: Find 3 concepts from within the article and relate them to 3 concepts within CHEM 210 we have discussed in class and cite 3 textbook references using the chapter and page number. 2: Find 2 concepts from within the article that you want to know more about (i.e. muddy points, have questions about, did not quite understand). 1: Write an exam question with the answer about 1 concept discussed from within the article. The exam question must be well thought out and appropriate to the subject matter.
Chapter 11. Intermolecular Forces and
the Liquid State
11.1 Kinetic Molecular Theory, States of Matter, and Phase Changes
11.2 Vapor Pressure
11.3 Other Properties of Liquids
11.4 The Nature of Intermolecular Forces
11.5 Intermolecular Forces and the Properties of Liquids
To this point, we have described the structures of individual atoms and individual
molecules in great detail. For atoms, we have focused on electronic structure, and
for molecules, we have studied the arrangement of atoms with respect to one
another. We are now ready to move to the next level of complexity in chemical
systems: how collections of molecules interact and how those interactions control
the physical properties of matter.
Intermolecular Forces and the
Liquid State
11.1 Kinetic Molecular Theory, States of
Matter, and Phase Changes
2
11.1
KMT for Solids and Liquids
• KMT also applies to liquids and solids
– Molecules in liquids and solids are in constant
random motion
• Liquids and solids are called condensed phases
– Particles are packed in close proximity
– Density is a measure of the differences between the
phases
– Except for water, the density increases when going
from solid to liquid to gas
3
11.1
Intermolecular Forces (IMFs)
• Forces between particles that hold one molecule
near another molecule
– Relative strengths of IMFs closely mirror that of the
density ranking
– Greatest in solids and the weakest in gases
4
Table 11.1.1 – Properties of Solids, Liquids, and Gases
Physical
state
IMFs
between
particles
Compressibility
Shape and
volume
Ability
to flow
Gas
Generally weak
High
Takes on shape
and volume
of container
High
Liquid
Generally
intermediate
Very low
Takes on shape of
container;
volume limited by
surface area
Moderate
Solid
Generally
strong
Almost none
Maintains own
shape and
volume
Almost
none
5
11.1
Intermolecular Forces
Influence chemistry, directly related to chemical
and physical properties:
– Melting point
– Boiling point
– Energy to convert a solid to liquid
– Energy to convert liquid to vapor
– Solubility of gases, liquids, and solids in various
solvents
– Structures of biologically important molecules, such
as DNA and proteins
– Much weaker than covalent bonds
6
Intermolecular Forces and the
Liquid State
11.4 The Nature of Intermolecular
Forces
7
11.4
Intermolecular Forces
Attractive forces that hold particles together in the
condensed phases are called Intermolecular Forces (IMFs).
Magnitude of intermolecular forces is what determines if the
particles in the substance are in a gas, liquid, or solid phase.
van der Waals forces -dipole-dipole interactions, including
hydrogen bonding and dispersion forces
8
11.4
Intermolecular Forces
Dipole-dipole interactions are attractive forces that act
between polar molecules.
As the dipole forces increase, intermolecular forces increase.
As the intermolecular forces increase, boiling points increase.
9
11.4
Intermolecular Forces
Hydrogen bonding is a special type of dipole-dipole
interaction between H-F, H-O, and H-N ONLY .
H bonded to a small, highly electronegative atom, such as N,
O, or F .
10
11.4
Hydrogen Bonding in H2O
• H-bonding is especially
strong in water O—H bond
is very polar
• 2 e- lone pairs on the O
atom attract H-atoms from
other H2O molecules
• Accounts for many of
water’s unique properties.
11.4
Hydrogen Bonding in H2O
• Ice has open latticelike structure.
• Density Ice < liquid • solid water floats on water. Snow flake: www.snowcrystals.com 11.4 Intermolecular Forces • Electrons in a molecule have some freedom to move about • molecule may have a nonuniform distribution of electron density, resulting in temporary dipole— instantaneous dipole or induced dipole An instantaneous dipole in one molecule can induce dipoles in neighboring molecules. 14 11.4 Intermolecular Forces London dispersion forces or simply dispersion forces result attractive forces As dispersion forces increase, intermolecular forces increase. As the intermolecular forces increase, boiling points increase. 15 11.4 Polar Covalent Bonds: Electronegativity Polarizability: Relative charge distribution, the electron cloud of an atom or molecule, is distorted from its normal shape by an external electric field, which may be caused by the presence of a nearby ion or dipole. 11.4 Intermolecular Forces Ion-dipole interactions are Coulombic attractions between ions (either positive or negative) and polar molecules. Magnitude of ion-dipole interactions depends on: Ion’s the charge and the ion size Polar molecule’s dipole moment and size Na+ Cl ? 17 11.4 Summary of IMF in descending Order Type of interaction Factors Responsible for Interaction Example Ion-Ion Charge of the ion, 400-4000 kJ/mol K+ and Cl- Ion–dipole Ion charge, magnitude of dipole, 40-600 kJ/mol K+ in H2O Hydrogen bonding Very polar X-H bond to atom Y (X,Y = F,N,O), 10-40 kJ/mol H2O–H2O Dipole–dipole Dipole moment, 5-25 kJ/mol CH2Cl2–CH2Cl2 Dipole-induced dipole Dipole moment of polar molecule & O2–H2O polarizability of nonpolar molecule, 2-10 kJ/mol Induced dipole– Polarizability, 0.05-40 kJ/mol induced dipole (LDF) Br2–Br2 18 11.4 Concept Map - IMFs Interacting particles No Are polar molecules involved? No London forces (induced dipoles) No Are ions involved? Are Hydrogen atoms bonded to O, N or F involved? Dipole– dipole forces Hydrogen bonding Yes Are polar molecules involved? No Yes Yes Ion–dipole forces Ionic bonding 19 What kind(s) of intermolecular forces exist in 11.1 (a) CCl4(l) (b) CH3COOH(l) Draw Lewis dot structures and apply VSEPR theory to determine whether each molecule is polar or nonpolar. (a) CCl4 is nonpolar, so the only intermolecular forces are dispersion forces. (b) CH3COOH is polar and contains an O-H bond, so it exhibits dipole-dipole interactions (including hydrogen bonding) 20 and dispersion forces. 11.1 • • • • • Properties of Liquids Particles are in constant motion Particles are in close contact Liquids are almost incompressible Liquids do not fill the container Intermolecular forces are relevant molecules are inter-dependent 2 11.1 Phase Changes • Energy needed to break the intermolecular bonds in liquids and solids – Energy is released when intermolecular “bonds” are formed • N is blue and H is white What kind of change is represented? Deposition g?s Is energy absorbed or released when the change occurs? Released 2 11.1 Phase Changes Phase Change (State Change): A change in physical form but NOT the chemical identity of a substance. NO bonds are broken! Break IMF’s 23 11.1 Enthalpies of Physical and Chemical Change Enthalpy of Fusion (?Hfusion opposite ?Hfreezing): The amount of heat necessary to melt a substance without changing its temperature, s ? l or l ? s ?Hfusion = - ?Hfreezing +6.02 kJ/mol -6.02 kJ/mol Enthalpy of Vaporization (?Hvap opposite ?Hcondensation): The amount of heat required to vaporize a substance without changing its temperature, l ? g or g ?l ?Hvaporization = - ?Hcondensation +40.7 kJ/mol -40.7 kJ/mol Enthalpy of Sublimation (?Hsublimation opposite ?Hdeposition): The amount of heat required to convert a substance from a solid to a gas without going through a liquid phase, s ? g or g ?s ?Hsublimation = - ?Hdeposition Chapter 8/24 11.5 Table 12.5.1 - Selected Heats of Fusion and Vaporization at the Temperature of the Normal Phase Transition ?Hfusion (kJ/mol) Melting point (°C) ?Hvap (kJ/mol) Boiling point (°C) Methane 0.94 –182.5 8.2 –161.6 Ethane 2.86 –182.8 14.7 –88.6 Propane 3.53 –187.6 19.0 –42.1 Methanol 3.16 –97.0 35.3 64.7 Ethanol 5.02 –114.3 38.6 78.4 1-Propanol 5.20 –127 41.4 97.2 Water 6.01 0.0 40.7 100.0 Na 2.60 97.82 97.42 881.4 NaBr 26.11 755 160.7 1390 25 Phase Diagram summarizes the conditions (temperature and pressure) at which a substance exists as a solid, liquid, or gas. 11.1 Phase Diagrams Normal Boiling Point and Melting Point : Occurs at 1 atm. Critical Point: A combination of temperature and pressure beyond which a gas cannot be liquefied. • Critical Temperature: The temperature beyond which a gas cannot be liquefied regardless of the pressure. • Critical Pressure: The pressure beyond which a liquid cannot be vaporized regardless of the temperature. Supercritical Fluid: A state of matter beyond the critical point that is neither liquid nor gas. Triple Point: A point at which three phases coexist in equilibrium. 11.1 28 Intermolecular Forces and the Liquid State 11.2 Vapor Pressure 29 11.2 Dynamic Equilibrium and Vapor Pressure • Vapor Pressure - Pressure of the liquid molecules in the vapor state • For a liquid to vaporize, sufficient energy must be added (endothermic) to overcome the break IMFs • Dynamic equilibrium: State in which rate of forward is equal to the rate of reverse • Equilibrium vapor pressure: Pressure exerted by a vapor over a liquid in a closed container at a given temperature What is the chemical composition of the vapor? 30 11.2 Vapor Pressure and IMFs • Liquid with the stronger IMFs has lower vapor pressure at a given T Inverse relationship S=C=S 31 Table 11.2.1 - Vapor Pressure of Some Common Liquids 32 Boiling Point 11.2 • Temperature at which Pvap = Patm – Normal boiling point: Patm = 760 mmHg (1 atm) • As strength of IMF increases, the boiling point increases • At reduced external pressure (higher altitude) Boiling occurs at a lower temperature 500 0 -40 Normal bp = 78.5°C Salt Lake City (4400 ft) P=650 mmHg -20 0 20 40 Temperature, °C 60 80 Normal bp Sea level 1 atm = 760 mmHg bp = 34.6°C Vapor pressure, mmHg 1000 Water boils at 100°C at sea level 100 Water boils at 95°C in Salt Lake City 120 11.2 Boiling Point and Baking Baking at Higher elevations (lower air pressure) is a SCIENCE! Liquids evaporate faster, amounts of flour, sugar and liquids are changed to prevent batter that is too moist, dry or gummy Gases expand more - doughs and baked goods rise faster easier for gas bubbles to rise. Leavening agents (baking soda and baking powder) are decreased At elevations over 3500 feet, the oven temperature for batters and doughs should be 25oF higher than at sea level, WHY? …leavening and evaporation are faster, use a higher temperature to set the structure before overexpansion and dry out 34 11.2 IMF Summary • IMFs affect- Enthalpy of vaporization, Vapor pressure, Boiling point of liquids • For a series of liquids, as IMF strength increases: – Energy needed to vaporize the liquids increases (?Hvap increases) – Liquid vapor pressure decreases – Liquid boiling point increases Molecules in the liquid state ?Hvap Volatility Equilibrium vapor pressure Boiling point Strong IMFs More endothermic Low Low High Weak IMFs Less endothermic High High Low 35 11.2 Clausius–Clapeyron Equation • Mathematical relationship between vapor pressure (P), temperature (T), and strength of IMFs (related to ?Hvap) - ?H vap ? 1 ? In P = +C ? ? R ?T ? y = m x + b R - Ideal gas constant (R = 8.314×10–3 kJ/K·mol) 36 11.2 Clausius-Clapeyron Equation Measured values to obtain the relationship between VP & T: Plot ln P versus 1/T slope = – ?Hvap R NOTE: T must be in Kelvin! R = 8.3145 J/mol-K Ethanol 11.2 Clausius-Clapeyron Equation What is the ?Hvap for diethyl ether, CH3CH2OCH2CH3 if P1 = 57.0 mm Hg at T1= -22.8oC and P2 = 534 mm Hg at T2= 25.0oC ? ln P2 P1 534 Ln 57.0 = -?Hvap R 1 – T2 1 T1 -?Hvap 1 = – -1 -1 8.315 J K mol 298.15 ?Hvap = 29048.8 J/mol = 29.0 kJ/mol 1 250.35 Intermolecular Forces and the Liquid State 11.3 Other Properties of Liquids 39 11.3 Properties of Liquids Surface Tension, measure of the elastic force, energy required to stretch or increase the surface of a liquid by a unit area (1 cm2). Measure of force required to "break" the surface of a liquid, “skin” A liquid with strong intermolecular forces has a high surface tension. Molecules on the surface are not pulled upward but are pulled downward and sideways, causes tightening of surface molecules http://www.npr.org/blogs/krulwich/2013/04/21/17794960 5/a-wet-towel-in-space-is-not-like-a-wet-towel-on-earth 40 11.3 Properties of Liquids Viscosity, is a measure of a fluid’s resistance to flow, units of N · s/m2, Higher the viscosity, the more slowly a liquid flows. Liquids that have strong IMFs have higher viscosities than those that have weaker IMFs. Higher Temperature = Lower viscosity • Ek > Ebarrier required to move past another
molecule.
Longer molecule = Higher viscosity
• Flexible, longer molecules become tangled
and hinder flow
41
11.3
Properties of Liquids
Capillary action – liquid is pulled up into a narrow glass cylinder
A competition between:
• Adhesion: attractions between unlike molecules.
• Cohesion: attractions between like molecules.
Shape of the meniscus illustrates the relative strength of the
adhesive and cohesive forces
Adhesion > Cohesion
Concave meniscus is formed
H2O adheres
to the wall
Adhesion < Cohesion Convex meniscus is formed Intermolecular Forces and the Liquid State 11.5 Intermolecular Forces and the Properties of Liquids 43 11.5 Table 11.5.1 - Properties of Some Common Nonpolar Species Compound Molar mass (g/mol) He 4.0 0.08 –268.9 Ne 20.3 1.7 –246.1 N2 28.0 5.6 –195.8 O2 32.0 6.8 –183.0 Ar 39.9 6.4 –185.9 Cl2 70.9 20.4 –34.0 Br2 159.8 30.0 58.8 ?Hvap (kJ/mol) Boiling point (°C) 44 11.5 Enthalpy of Vaporization • Weaker the IMF the lower the vapor pressure • Consider a series of atoms and molecules that have only dispersion forces – As the molar mass increases, enthalpy of vaporization increases – Larger the molecule, greater the number of electrons, and greater the polarizability – Greater the surface area available for contact, greater the dispersion forces 45 11.5 Quantitative Comparison of IMFs Nonpolar N2 SiH4 GeH4 Br2 Polar Molar mass Boiling point Molar mass Boiling point (g/mol) (°C) (g/mol) (°C) 28 –196 CO 28 –192 32 –112 PH3 34 –88 77 –90 AsH3 78 –62 160 59 ICl 162 97 • If the molar masses are similar: – Molecule with the greater IMF will have the higher boiling point • In each pair below, the molecule that is polar has the higher boiling point 46 11.5 Summary of IMF in descending Order Type of interaction Factors Responsible for Interaction Example Ion-Ion Charge of the ion, 400-4000 kJ/mol K+ and Cl- Ion–dipole Ion charge, magnitude of dipole, 40-600 kJ/mol K+ in H2O Hydrogen bonding Very polar X-H bond to atom Y (X,Y = F,N,O), 10-40 kJ/mol H2O–H2O Dipole–dipole Dipole moment, 5-25 kJ/mol CH2Cl2–CH2Cl2 Dipole-induced dipole Dipole moment of polar molecule & O2–H2O polarizability of nonpolar molecule, 2-10 kJ/mol Induced dipole– Polarizability, 0.05-40 kJ/mol induced dipole (LDF) Br2–Br2 47 11.5 Concept Map - IMFs Interacting particles No Are polar molecules involved? No London forces (induced dipoles) No Are ions involved? Are Hydrogen atoms bonded to O, N or F involved? Dipole– dipole forces Hydrogen bonding Yes Are polar molecules involved? No Yes Yes Ion–dipole forces Ionic bonding 48 Nitrogen in Your Car Tires 3: Find 3 concepts from within the article and relate them to 3 concepts within CHEM 210 we have discussed in class and cite 3 textbook references using the chapter and page number. The three-concept discussed in the article are; how volume get affected as the temperature increase, the basic behind why nitrogen does not mix with any gas at any temperature, and finally the relation between water and temperature and how the reaction taking place in tire leads to corrosion. Nitrogen is synthetically a non-combustible, non-lethal dormant gas. An idle gas does not meld with some other gas at any temperature. The fundamental thought of Nitrogen gas is that it is cooler than compacted air and along these lines is valuable in any driving conditions. The higher temperature in tires additionally tends to blast after extended periods of running the idea driving this is the higher the temperature the higher the volume, and along these lines an abundant volume of gas will result in a burst that implies temperature is straightforwardly corresponding to the volume at the consistent pressure (Kibar, et al, p542-554). Nitrogen decreases the odds of tire blasted by 90% implying that temperature is steady, so the volume won't increment. The atomic structure of nitrogen varies from that of air so that it escapes through the tire's inward liner or cylinder at a slower rate than ordinary compacted air. Accordingly, the outcome is a drastically slower rate of weight misfortune in a tire loaded up with nitrogen. For instance, it may take as long as a half year to lose 0.14kpa with nitrogen, contrasted with only one month with compacted air. As tires heat up, their swelling pressure expands, which at that point lessens the span of the tire's impression, the tire at that point loses hold in light of this little impression. So the cooler they run the better the tires will grasp the street and this can be accomplished by means of the utilization of nitrogen gas (Falgout, Zachary, et al. p 14-21.). The actualities are clear and there is proof to propose that nitrogen tire expansion is desirable over air swelling. The best three purposes behind victories are poor condition, overburdening and under - expansion. Poor condition originates from drivers not being mindful and running their tires excessively far. Over-burdening originates from a similar flippancy and the law not being authorized as it ought to be on the two checks. We can anyway take care of underexpansion. The appropriate response may be to fill tires with nitrogen tire gas. Water vapor assimilates and holds heat, and when it changes from a fluid to vapor, the water grows in volume (Elbaba, Ibrahim F., and Paul T. Williams, p528-536). So tires swelled with wet air will, in general, run more smoking and the weight vacillates more. Nitrogen has next to no water vapor. And if there is trace of water then it will lead to corrosion (Falgout, Zachary, et al. p 14-21.). 2: Find 2 concepts from within the article that you want to know more about (i.e. muddy points, have questions about, did not quite understand). Despite being stated that nitrogen will help in conserving of fuel, what I would like to learn more and understand better is how will use of nitrogen in tires help in conserving the fuel and what is the relation between nitrogen use in the tires and conserving of fuel. Also why is that tires that are filled with oxygen tend to heat up easily thus increasing their temperature while tires filled with nitrogen tend not to heat easily and thus their temperature remain constant. 1: Write an exam question with the answer about 1 concept discussed from within the article. The exam question must be well thought out and appropriate to the subject matter. 2.0mol of an ideal gas is contained in a 3.0L tire at a temperature of 25?C. The gas applies a weight of 16atm on the tire. In the event that weight is kept consistent, what is the last volume of the gas if the temperature of the tire increments to 200?C? Solution Since pressure is kept consistent, the main variable that is controlled is temperature. This implies we can utilize Charles' law so as to look at volume and temperature. Since volume and temperature are on inverse sides of the perfect gas law, they are legitimately relative to each other. As one variable build, the other will increment also. Charles's law is written as follows: V1T1=V2T2 To use this law, we must first convert the temperatures to Kelvin. 25oC+273=298K 200oC+273=473K Use these temperatures and the initial volume to solve for the final volume. 3L298K=V2473K V2=4.8L Work cited Falgout, Zachary, et al. "Gas/fuel jet interfaces under high pressures and temperatures." Fuel 168 (2016): 14-21. Elbaba, Ibrahim F., and Paul T. Williams. "High yield hydrogen from the pyrolysis–catalytic gasification of waste tyres with a nickel/dolomite catalyst." Fuel 106 (2013): 528-536. Kibar, Zeynep Bak, Fatma Yaman, and Alipa?a Ayas. "Assessing prospective chemistry teachers' understanding of gases through qualitative and quantitative analyses of their concept maps." Chemistry Education Research and Practice14.4 (2013): 542-554.