Intermolecular Forces I

Intermolecular Forces I
Phases of Matter Liquids, Solids (Crystals) & Solutions Colligative Properties Dr. Ron Rusay

Intermolecular Forces: Phases of Matter & Colligative Properties
Changes of State Phase transitions Phase Diagrams Liquid State Pure substances and colligative properties of solutions, which depend upon the ratio of the number of solute particles to the number of solvent molecules in a solution. They are independent of the nature of the solute particles.

Intermolecular Forces: Phases of Matter
Solid State Classification of Solids by Type of Attraction between Units Crystalline solids; crystal lattices and unit cells Structures of some crystalline solids Determining the Crystal Structure by X-ray Diffraction

Phase Transitions Melting: change of a solid to a liquid.
H2O(s)  H2O(l) H2O(l)  H2O(s) H2O(l)  H2O(g) H2O(s)  H2O(g) H2O(g)  H2O(l) H2O(g)  H2O(s) Melting: change of a solid to a liquid. Freezing: change a liquid to a solid. Vaporization: change of a solid or liquid to a gas. Change of solid to vapor often called Sublimation. Condensation: change of a gas to a liquid or solid. Change of a gas to a solid often called Deposition.

Phases of Matter / Intermolecular Forces

Phase Changes Surface molecules are only attracted inwards towards the bulk molecules. Sublimation: solid  gas. Hsub > 0 (endothermic). Vaporization: liquid  gas. Hvap > 0 (endothermic). Melting / fusion: solid  liquid. Hfus > 0 (endothermic). Deposition: gas  solid. Hdep < 0 (exothermic). Condensation: gas  liquid. Hdep < 0 (exothermic). Freezing: liquid  solid. Hdep < 0 (exothermic). All phase changes are possible under the right conditions. The sequence heat solid  melt  heat liquid  boil  heat gas is endothermic. cool gas  condense  cool liquid  freeze  cool solid is exothermic.

QUESTION

ANSWER B) crystallization
As a substance crystallizes, generally from a liquid, but a gaseous starting point is possible, the molecules, atoms, or ions lose kinetic energy to allow for bonding. This kinetic energy is emitted as heat. STUDENT CD: Understanding Concepts: Changes of State

Bonds vs. Intermolecular Forces
( kJ/mol) (Ionic bond 700-4,000 kJ/mol) 16 kJ/mol 431 kJ/mol

Intermolecular Forces
Ion-Dipole Forces ( kJ/mol) Interaction between an ion and a dipole (e.g. NaOH and water = 44 kJ/mol) Strongest of all intermolecular forces.

Dipole-Dipole Forces (permanent dipoles) Dipole-Dipole Forces Dipole-dipole forces exist between neutral polar molecules. Polar molecules need to be close together. Weaker than ion-dipole forces. There is a mix of attractive and repulsive dipole-dipole forces as the molecules tumble. If two molecules have about the same mass and size, then dipole-dipole forces increase with increasing polarity. 5-25 kJ/mol

Dipole-Dipole Forces

London or Dispersion Forces An instantaneous dipole can induce another dipole in an adjacent molecule (or atom). The forces between instantaneous dipoles are called London or Dispersion forces ( kJ/mol). London Dispersion Forces Weakest of all intermolecular forces. It is possible for two adjacent neutral molecules to affect each other. The nucleus of one molecule (or atom) attracts the electrons of the adjacent molecule (or atom); a dipole forms. London dispersion forces increase as molecular weight increases. London dispersion forces exist between all molecules. London dispersion forces depend on the shape of the molecule.

London Dispersion Forces Which has the higher attractive force?

London Dispersion Forces

Gecko: toe, setae, spatulae 6000x Magnification
Full et. al., Nature (2000) 5,000 setae / mm2 600x frictional force; 10-7 Newtons per seta Geim, Nature Materials (2003) Glue-free Adhesive 100 x 10 6 hairs/cm2 GECKOS tiny tropical lizards are able to run up walls and along ceilings extremely fast, yet they can stick to a sheet of polished glass with only one foot. The secret of their success lies in the rows of tiny hairs on the bottom of their feet. Thousands of these hairs, called setae, are arrayed like the bristles of a toothbrush across a gecko’s toes. Microscopy reveals that the tip of each seta is divided into hundreds of tiny“spatulae”, each pointing in a different direction and tipped with a cone-shaped structure. This shape suggests a suction mechanism, but suction relies on air pressure—and gecko feet are known to stick to walls even in a vacuum. Robert Full of the University of California, Berkeley, Nature 2000. Using a tiny micro-electro-mechanical force sensor, they conducted various experiments to measure the stickiness of a single seta. The maximum adhesive force that could be exerted by a single seta had already been estimated, by measuring the total force exerted by a foot and dividing by the number of setae (around 5,000 per square millimetre). But to their surprise, researchers found that a single seta can actually exert ten times as much force as this. Setae are, in other words, even stickier than expected—giving thegecko a surprisingly large safety margin.adhesive force is about 600 times greater than the simple frictional force between lizard skin and the surface. And a seta will stick to a surface most firmly if it is first pushed into the surface and then pulled along it by a few millionths of a metre. These findings suggest that setae operate at a molecular level, and exploit intra-molecular forces, called van der Waals forces, for their stickiness. They detach by curling up the tips of their toes before moving, forming a sort of reverse fist. This allows them to peel their feet off the surface gently at a critical angle without damage, much like peeling a sticky label off a jar withot tearing it. The researchers found that setae reliably detach from the surface at an angle of about 30°.

Boiling Points & Hydrogen Bonding Hydrogen Bonding
Special case of dipole-dipole forces. By experiments: boiling points of compounds with H-F, H-O, and H-N bonds are abnormally high. H bonded to an electronegative element : F, O, and N. H-bonds are strong. Eg. Ice Floating Solids are usually more closely packed than liquids; Therefore, solids are usually more dense than liquids. However, Ice is an exception due to H-bonding. Ice (solid) is less dense than water (liquid). The H-O bond length is 1.0 Å. The O…H hydrogen bond length is 1.8 Å. Ice has waters arranged in open, regular hexagons.

Hydrogen Bonding Hydrogen bonds, a unique dipole-dipole (10-40 kJ/mol).

Which pure substances will not form hydrogen bonds?
QUESTION Which pure substances will not form hydrogen bonds? I) CH3CH2OH II) CH3OCH3 III) H3C−NH−CH3 IV) CH3F A) I and II B) I and III C) II and III D) II and IV

Which pure substances will not form hydrogen bonds?
ANSWER Which pure substances will not form hydrogen bonds? I) CH3CH2OH II) CH3OCH3 III) H3C−NH−CH3 IV) CH3F A) I and II B) I and III C) II and III D) II and IV Correct Answer: D Question Number: 6

Hydrogen Bonding

DNA: Size, Shape & Self Assembly
Views & Algorithms 10.85 Å 10.85 Å 15-23 Title: Base Pairs Caption: Representation of the base pairings of thymine and adenine, and cytosine and guanine, in line as well as ball-and-stick models. Notes: Base pairs occur through hydrogen bonding.

Protein Shape: Forces, Bonds, Self Assembly,
Folding 15-16 Title: Tertiary Interactions Caption: The four distinct interactions that stabilize tertiary protein structures. Notes: Note that the disulfide bridge is a covalent bond. Ion-dipole (Dissolving) 40-600kJ/mol 10-40kJ/mol kJ/mol kJ/mol 700-4,000kJ/mol

QUESTION Predict which liquid will have the strongest intermolecular forces of attraction (neglect the small differences in molar masses). A) CH3COCH2CH2CH3 (molar mass = 86 g/mol) B) CH3CH2CH2CH2CH2OH (molar mass = 88 g/mol) C) CH3CH2CH2CH2CH2CH3 (molar mass = 86 g/mol) D) HOH2C−CH=CH−CH2OH (molar mass = 88 g/mol)

ANSWER Predict which liquid will have the strongest intermolecular forces of attraction (neglect the small differences in molar masses). A) CH3COCH2CH2CH3 (molar mass = 86 g/mol) B) CH3CH2CH2CH2CH2OH (molar mass = 88 g/mol) C) CH3CH2CH2CH2CH2CH3 (molar mass = 86 g/mol) D) HOH2C−CH=CH−CH2OH (molar mass = 88 g/mol) Correct Answer: D Question Number: 9

Vapor Pressure on the Molecular Level
Some of the molecules on the surface of a liquid have enough energy to escape the attraction of the bulk liquid. After some time the gas and liquid will be at equilibrium and pressure is constant, which is referred to as the vapor pressure at temperature T. If equilibrium is never established then the liquid evaporates. Substances with high vapor pressures evaporate rapidly in an open system. Increasing the temperature, increases the average kinetic energy, and the liquid evaporates faster. Would water have a higher or lower vapor the same temperature? (bp H2O > CH3CH2OH; bp = oC when Vapor Pressure = Atmospheric Pressure)

Vapor Pressure Explaining Vapor Pressure on a Molecular Level

Volatility, Vapor Pressure, and Temperature

Vapor Pressure Solid Liquid Gas

QUESTION

ANSWER E) The vapor pressure of a liquid.
Molecules at the surface of a liquid will be held tighter by stronger intermolecular forces, making it more difficult for them to escape into the vapor phase.

Temperature & Vapor Pressure
The boiling point (b.p.) of a pure liquid is the temperature at which the vapor pressure above the liquid equals the external pressure. Could water 0oC?

Temperature Dependence of Vapor Pressures
The vapor pressure above the liquid varies exponentially with changes in the temperature. The Clausius-Clapeyron equation shows how the vapor pressure and temperature are related. (R = J K−1 mol−1)

Clausius – Clapeyron Equation
A straight line plot results when ln P vs. 1/T is plotted and has a slope of Hvap/R. Clausius – Clapeyron equation is true for any two pairs of points.

QUESTION

ANSWER D) diethyl ether, CH CH O
Diethyl ether boils first as the temperature increases, leading one to believe that its vapor pressure at any temperature is also the highest of the given group of compounds. – – 3 2 2 3 HMCLASS PREP: Figure 10.42

Heating Curve

Energy (Heat) and Phase Changes
Heat of vaporization: heat needed for the vaporization of a liquid. H2O(l) H2O(g) DH = 40.7 kJ/mol Heat of fusion: heat needed for the melting of a solid. H2O(s) H2O(l) DH = 6.01 kJ/mol Temperature does not change during the change from one phase to another. 50.0 g of H2O(s) and 50.0 g of H2O(l) were mixed together at 0°C. Determine the heat required to heat this mixture to 100.0°C and evaporate half of the water.

Intermolecular Forces II
Phases of Matter / Phases Diagrams Solids (Crystals) & Solutions Colligative Properties Dr. Ron Rusay

Phase Diagrams Graph of pressure-temperature relationship; describes when 1,2,3 or more phases are present and/or in equilibrium with each other. Lines indicate equilibrium state between two phases. Triple point- Temp. and press. where all three phases co-exist in equilibrium. Critical temp.- Temp. where substance must always be gas, no matter what pressure. Critical pressure- vapor pressure at critical temp. Critical point- system is at its critical pressure and temp.

Phase Diagrams Phase diagram: plot of pressure vs. temperature summarizing all equilibria between phases. Given a temperature and pressure, phase diagrams tell us which phase will exist. Features of a phase diagram: Any temperature and pressure combination not on a curve represents a single phase. Triple point: temperature and pressure at which all three phases are in equilibrium. Critical temperature: the minimum temperature for liquefaction of a gas using pressure. Critical pressure: pressure required for liquefaction. Critical point: critical temperature and pressure for liquefaction of the gas. Melting point curve: as pressure increases, the solid phase is favored if the solid is more dense than the liquid. Normal melting point: melting point at 1 atm.

The Phase Diagrams of H2O and CO2
Water: The melting point curve slopes to the left because ice is less dense than water. Triple point occurs at C and 4.58 mmHg. Normal melting (freezing) point is 0C. Normal boiling point is 100C. Critical point is 374C and 218 atm. Carbon Dioxide: Triple point occurs at -56.4C and 5.11 atm. Normal sublimation point is -78.5C. (At 1 atm CO2 sublimes it does not melt.) Critical point occurs at 31.1C and 73 atm.

Oscillatory Vapor-Liquid-Solid growth of sapphire nanowires (α-Al2O3)
660°C, Pressure = 10–6 Pa In vapor-liquid-solid (VLS) growth of nanowires, the liquid phase acts as a transporter to bring material from the gas phase to the growing solid. By heating a single crystal of sapphire in a high-resolution transmission microscope, Oh et al. (p. 489) monitored the growth of sapphire ({alpha}-Al2O3) nanowires out of an aluminum droplet. The liquid aluminum brings oxygen to the growing wire surface, in alternating growth and dissolution reactions at the edge of the wire. The oscillation created an optimum face at the self-catalytic site for atomic stacking and regenerated the junction between the VLS phases, allowing growth of the nanowire. Fig. 3 Oscillatory mass transport in the VLS growth of sapphire nanowires. A series of HRTEM images outlines a period of oscillatory growth and dissolution reactions. (A to F) Images were captured from a real-time movie (movie S1) at the elapsed times shown. Local crystal growth of a rim at the triple-junction region by mass diffusion from the LV surface to the -Al2O3 facet is shown in (A) to (C). During this process, the facet shrinks. Shortly after (C) is reached, the LS (0001) interface comes in a direct contact with the LV interface. Dissolution of the previously grown crystalline rim on the facet is shown in (D) to (F). Dissolution continues to supply oxygen until a complete (0006) layer is added on the interface. The end of dissolution is shown in (F). Arrows in each figure indicate local crystal growth (white), dissolution (gray), and oxygen diffusion (yellow). (G) Difference image obtained by subtracting the video image (C) from (E). The locations of LV and LS interfaces before and after dissolution are delineated by white and yellow lines, respectively. (H and I) Schematic illustration of the triple-junction configuration during local crystal growth on the facet (H) and at the end of growth (I). Note that the dihedral angles are almost the same for growth and dissolution. The surface (and interface) tension forces (not to scale) and the measured values of dihedral angles are indicated. S. H. Oh et al., Science 330, (2010) Published by AAAS

Critical Temperature and Pressure
Phase Changes Critical Temperature and Pressure

Solutions Archimedes’ Mirrors Bonus: Explain how a 50:50 molar solution of molten potassium & sodium nitrates could be used to store solar energy. Freezing Point: 238°C Melting Point: 221°C Heat of Fusion: 161 kJ/kg

Solutions Bonus: Explain how a 50:50 molar solution of molten potassium & sodium nitrates could be used to store solar energy. Freezing Point: 238°C Melting Point: 221°C Heat of Fusion: 161 kJ/kg

Solana Bonus: http://www.youtube.com/watch/?v=G1hdoWk17wU $ 2 billion
The heat from a solution of molten potassium & sodium nitrates is used to store the energy, which turns a turbine in 6 hrs of darkness/ day. $ 2 billion 3 years 2,200,000 m2 2,700 mirrors 280 MegaWatts 70,000 homes - 475,000 MT CO2 Equivalent

Solutions

QUESTION

ANSWER B) decreases over time.
The concentration of the solution increases as the water evapora tes. The higher the concentration of salt, the lower the vapor pressure.

Factors Affecting Solubility
Concentration Gradients: Diffusion Movement of molecules from an area of high concentration to an area of lower concentration. Factors that affect the rate of diffusion: size of molecules, size of pores in membrane, temperature, pressure, and concentration. Osmosis Osmosis is the movement of solvent across a semi-permeable membrane Initially the concentration of solute is very high, but over time, the solvent moves across the semi-permeable membrane and dilutes the particles.

Osmosis – A Special kind of Diffusion
Diffusion of water across a selectively permeable membrane (a barrier that allows some substances to pass but not others). The cell membrane is such a barrier. Small molecules pass through – eg. water Large molecules can’t pass through – eg. proteins and complex carbohydrates

Over time molecules move across the membrane until the concentration of solute is equal on both sides. This type of solution is called ISOTONIC.

Factors Affecting Solubility
Pressure Effects The higher the pressure, the more molecules of gas are close to the solvent and the greater the chance of a gas molecule striking the surface and entering the solution. Therefore, the higher the pressure, the greater the solubility. The lower the pressure, the fewer molecules of gas are close to the solvent and the lower the solubility. Carbonated beverages are bottled with a partial pressure of CO2 > 1 atm. As the bottle is opened, the partial pressure of CO2 decreases and the solubility of CO2 decreases. Therefore, bubbles of CO2 escape from solution. If Sg is the solubility of a gas, k is a constant, and Pg is the partial pressure of a gas, then Henry’s Law gives:

QUESTION A minimum of 1.3  10–4 M O2 must be maintained in freshwater supplies to sustain aquatic life. In the mountains of Montana, the partial pressure of O2 may drop to 0.15 atm. What is the water concentration of O2 there? Henry’s constant for O2 = 1.3  10–3 mol/L-atm. At the lower elevations at the base of those mountains, would more or less O2 be dissolved in water? M = 2.0  10–4; more dissolved M = 8.7  10–4; more dissolved M = 2.0  10–4; less dissolved M = 8.7  10–4; less dissolved

ANSWER A) provides the correct M and the correct change in concentration. Henry’s Law relates pressure of a gas over a solution to the concentration of the gas in the solution: C = k  P. At lower altitudes, the partial pressure of O2 would be higher, thus more O2 would dissolve. The huge fishing population of Montana is very appreciative of Henry’s Law.

Phase Diagrams for Pure Water (Red) and for an Aqueous Solution Containing a Nonvolatile Solute (Blues)

Sugar Dissolved in Water to Make Candy Causes the Boiling Point to be Elevated

Spreading Salt on a Highway
The Addition of Antifreeze Lowers the Freezing Point of Water in a Car's Radiator

Colligative Properties Lowering Vapor Pressure
Colligative properties depend on quantity and type of solute/solvent molecules. (E.g. freezing point depression and melting point elevation.) Lowering Vapor Pressure Non-volatile solvents reduce the ability of the surface solvent molecules to escape the liquid. Therefore, vapor pressure is lowered. The amount of vapor pressure lowering depends on the amount of solute.

Colligative Properties Lowering Vapor Pressure
Raoult’s Law: PA is the vapor pressure with solute, PA is the vapor pressure without solvent, and A is the mole fraction of A, then Recall Dalton’s Law:

QUESTION

ANSWER C) 0.800 Remember to convert grams to moles before
attempting to find the mole fraction.

Concentration molality and Molarity
Molality relates to colligative properties. Converting between molarity (M) and molality (m) requires density. Therefore Molarity and molality are most often not equal

QUESTION

ANSWER A) 5.47 m Using the density, the mass of the solution is
found. Don’t forget that molality has units of kg of solvent and the mass of the solute must be subtracted from the calculated mass of solution.

Molal Boiling-Point Elevation Constants (Kb) and Freezing-Point Depression Constants (Kf) for Several Solvents

QUESTION Household bleach is an aqueous solution of sodium hypochlorite. If 5.25 g of NaOCl (molar mass = 74.5 g/mol) were placed in g of water, what would you calculate as the molality? The density of the solution is slightly greater than water. Would the molarity of the solution be greater, less or the same as the molality? m; M would be greater 0.705 m; M would be the same 0.744 m; M would be greater 0.744 m; M would be less

ANSWER D) provides the correct answer to both parts of the question. The molality involves moles/kg of water, so the given mass of solute can be converted to moles of solute and then divided by kg to obtain molality. The molarity of the solution will be greater than the molality because the density of the solution shows that one milliliter of solution has a mass greater than one. So one liter of solution will contain more mass of solute than would be found mixed with one kilogram of water.

QUESTION Suppose you want to keep the water in your car cooling system from freezing during a cold Alaska winter night. If you added 5.00 kg of ethylene glycol (C2H4(OH)2 MM = 62.0 g/mol) to 5.50 kg of water, what would be the freezing temperature of the coolant/water mixture in your automobile? k f.p. H2O = –1.86°C kg/ mol –0.0367°C –7.90°C –14.7°C –27.3°C HMClassPresent Freezing Point Depression A,B and C

ANSWER D) oF provides the correct, although very cold, answer for that Alaska night. The molality of ethylene glycol must be calculated from the mass (in grams) and molar mass. This is then multiplied by the freezing point depression constant for the solvent to obtain the drop in freezing point. Since water normally freezes at zero degrees Celsius, the change is the actual new freezing point.

Atomic Solid / Ionic Solid / Molecular Solid

Structure of Solids Structure of Solids Types of solids:
Crystalline – a well defined arrangement of atoms; this arrangement is often seen on a macroscopic level. Ionic solids – ionic bonds hold the solids in a regular three dimensional arrangement. Eg. “Galena”, lead sulfide ore Molecular solid – solids like ice that are held together by intermolecular forces. Covalent network – a solid consists of atoms held together in large networks or chains by covalent networks. Eg. Diamond & graphite Metallic – similar to covalent network except with metals. Provides high conductivity. Amorphous – atoms are randomly arranged. No order exists in the solid. Eg. Glass, gels, thin films

Closest Packing Arrangement of Uniform Spheres
Arrays of atoms act as if they are spheres. Two or more layers produce a 3-D structure.

Close Packed: The Red Sphere has 12 Nearest Neighbors

Closest Packed Structure

Unit Cells Crystals are made up regular arrays – the smallest repeating array of atoms is called the unit cell. There are 14 different unit cells that are observed which vary in terms of the angles between atoms some are 90°, but others are not.

Unit Cells Length of sides a, b, and c as well as angles a, b, g vary to give most of the unit cells.

Cubic Hexoctahedral cF8, space group Fm3m, No. 225
QUESTION Cubic Hexoctahedral cF8, space group Fm3m, No. 225 Consider an interior atom in the simple cubic crystal lattice. What is the maximum number of unit cells that share this atom in the three-dimensional crystal lattice? A) 2 B) 4 C) 6 D) 8

ANSWER Consider an interior atom in the simple cubic crystal lattice. What is the maximum number of unit cells that share this atom in the three-dimensional crystal lattice? A) 2 B) 4 C) 6 D) 8 Correct Answer: D Question Number: 11

Unit Cells Simple-cubic Body-centered cubic Face-centered cubic

QUESTION The number of atoms per unit cell in the body-centered cubic lattice is A) 1 B) 2 C) 3 D) 4

ANSWER The number of atoms per unit cell in the body-centered cubic lattice is A) 1 B) 2 C) 3 D) 4

Unit Cells Face Centered Cubic structure has a-b-c-a-b-c stacking. It takes three layers to establish the repeating pattern and has 4 atoms per unit cell and the coordination number is 12.

The Net Number of Spheres in a Face-Centered Cubic Unit Cell

Unit Cells Simple-cubic shared atoms are located only at each of the corners. 1 atom per unit cell. Body-centered cubic 1 atom in center and the corner atoms give a net of 2 atoms per unit cell. Face-centered cubic corner atoms plus half-atoms in each face give 4 atoms per unit cell.

Cubic Hexoctahedral cF8,
Unit Cells Can you hear anything? Galena, lead sulfide Cubic Hexoctahedral cF8, space group Fm3m, No. 225

Connecting: Science, Technology, Engineering and Mathematics (STEM)
Crystals for the Classroom Bridging the realms of the macro and atomic/nano scale Connecting: Science, Technology, Engineering and Mathematics (STEM)

Crystals for the Classroom Bridging the realms of the macro and atomic/nano scale The story of NIF ( The National Ignition Facility)

Crystals for the Classroom Bridging the realms of the macro and atomic/nano scale Time lapsed KDP Growth

Crystals for the Classroom Bridging the realms of the macro and atomic/nano scale Simulation: Fusion

X-ray Crystallography
Rosalind Franklin’s Photo 51 46 Å 12 base sequence (1953)

Intermolecular Forces I

Similar presentations

Presentation on theme: "Intermolecular Forces I"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Intermolecular Forces I

Similar presentations

Presentation on theme: "Intermolecular Forces I"— Presentation transcript:

Similar presentations

About project

Feedback