 Molecules of a cold sample of liquid have lower kinetic energy than those in a warmer sample  If a particle near the surface has enough kinetic energy, it can overcome the attractive forces in a liquid and escape into the gaseous state  Known as a phase change

 Viscosity:  The friction or resistance to motion that exists between the molecules of a liquid when they move past one another  The stronger the attraction between the molecules in a liquid, the greater the resistance to flow  Liquids with large intermolecular forces tend to be highly viscous

 Surface Tension:  The resistance of a liquid to an increase in its surface area  Which liquids will have high surface tensions and why?  Because of decreased volume and increased molecular interaction, liquids expand and contract only very slightly with temperature change  Boiling Point:  The point at which the liquid’s vapor pressure is equal to the atmospheric pressure  Rapidly converting from liquid to the vapor phase within the liquid as well as at the surface

 The attraction of the surface of a liquid to the surface of a solid  Liquids will rise very high in a narrow tube if a strong attraction exists between the liquid molecules and the molecules that make up the tubing  Pulls liquid up against force of gravity  Concave meniscus  Polar liquids exhibit capillary action  The spontaneous rising of a liquid in a narrow tube, due to:  Cohesive forces – the intermolecular forces among the molecules of the liquid  Adhesive forces – the forces between the liquid and its container  Which of these are stronger for water?  Adhesive

Evaporation (vaporization) – a process by which the molecules of a liquid can escape the liquid’s surface and form a gas Endothermic process Heat of vaporization (enthalpy of vaporization) – energy required to vaporize one mole of a liquid at a pressure of 1 atm Symbol: Δ h vap

 Condensation – process by which vapor molecules re-form a liquid

 Eventually, enough vapor molecules are present so that the rate of condensation equals the rate of evaporation  The system is said to be at equilibrium  The pressure of the vapor present at equilibrium is called vapor pressure

 What will happen if the temperature is increased?  The number of liquid molecules will be reduced  The number of gaseous molecules will be increased  The rates of evaporation and condensation will become equal again  This illustrates what is known as ; Le Ch â telier’s Principle

 Reversible reactions – conversion of reactants to products and vice versa occur simultaneously  Change in conditions is imposed on a system at equilibrium, the equilibrium will shift in the direction that tends to reduce that change in conditions.  CHANGES IN CONCENTRATION:  Substance is added  reaction consumes added substance  Substance is removed  reaction shifts to produce more 2NO 2 (g)  N 2 O 4 (g)  Which direction will the above reaction shift if we add NO 2 ?  Which direction will the above reaction shift if we remove N 2 O 4 ?

 CHANGES IN PRESSURE:  Increase : shift in direction that produces fewer molecules (moles) of gas.  Decrease : shifts in direction that produces more molecules of gas.  In the reaction below, if we increase the pressure which direction will the reaction shift? NH4Cl (s)  NH3 (g) + HCl (g)  CHANGES IN TEMPERATURE:  EXOTHERMIC: Reaction gives off heat (product)  ENDOTHERMIC: Reaction absorbs heat (reactant)  Consider heat as a component of the reaction. H 2 (g) + I 2 (g)  2 HI (g) + Heat  If we raise the temperature (add heat) which way will the reaction shift?  If we want the reaction to go to the right do we add or remove heat?

DDuring WWII, Allied forces blocked the Germans from acquiring sodium nitrates used for explosives from mines in Chile in hopes of shortening the war. FFritz Haber used Le Chatelier’s Principle to come up with a new process of making ammonia. NN 2 (g) + 3H 2 (g) ―› 2NH 3 (g) + Heat

 Concentration  Concentration: Remove the products (NH 3 )  Pressure  Pressure: Increased the pressure  Temperature  Temperature: Kind of complicated; reducing heat meant less pressure, but increasing heat would shift toward reactants therefore kept temperature moderate (500C)

 Can be classified into very broad categories: 1. Crystalline solids – highly regular arrangement of components 2. Amorphous solids – have considerable disorder in their structure 3. Polycrystalline solid – an aggregate of a large number of small crystals in which the structure is regular but the crystals are arranged in random fashion

Crystalline Solid Amorphous Solid Polycrystalline Solid

 Lattice structure – a 3D system of points designating the positions of the components  Unit cell – the smallest portion of a crystal lattice that is repeated throughout the crystal

 Atomic solid containing strong directional covalent bonds  Allotropes – forms of the same element that differ in crystalline structure Differ in properties because of differences in structure Example: Diamond is one allotrope of carbon in which each carbon is covalently bonded to four other carbon atoms in a tetrahedral direction. Graphite is another allotrope of carbon, covalently bonded to form hexagonal sheets What is a buckyball?

 Melting point – the temperature at which atomic or molecular vibrations of a solid become so great that the particles break free from their fixed positions and start to slide past each other in a liquid state  Heating curve – a plot of temperature versus time for a substance where energy is added at a constant rate  Sublimation – when a solid goes directly to a gaseous state without passing through the liquid phase

 Heat of fusion( ΔH fus ) – the amount of energy required at the melting point temperature to cause the change of phase to occur  Heat of vaporization (ΔH vap ) – the amount of heat needed to vaporize 1 gram of a liquid at constant temperature and pressure

 Way to represent the phases of a substance as a function of temperature and pressure  Triple point – the point at which all three states of a substance are present  Critical temperature – the temperature above which the vapor cannot be liquefied no matter what pressure is applied  Critical pressure – pressure required to produce liquefication at the critical temperature  Together, the critical temperature and critical pressure define the critical point

o2%20phase%20diagram.GIF

 Both solids and liquids are condensed states of matter  Relatively weak forces which occur between molecules  Both dipole-dipole and London dispersion forces are known as Van der Waals forces *It is important to recognize that when a substance such as water changes from solid to liquid to gas, the molecules remain intact. The changes in state are due to changes in the forces among the molecules rather than within the molecules*

Dipole-dipole Forces The attractive force resulting when polar molecules line up so that the positive and negative ends are close to each other Try to maximize the (+)----(-) interactions In the gas phase, these forces are unimportant Weaker than ionic or covalent bonds

London Dispersion Forces Forces which exist among all covalent molecules but is the only force for nonpolar molecules. Weak attractive forces between molecules resulting from the small, instantaneous dipoles that occur because of the varying positions of the electron during their motion about nuclei.

Hydrogen Bonding Unusually strong dipole- dipole attractions that occur among polar molecules in which hydrogen is bonded to a highly electronegative atom such as O-H, N-H, F-H

 Nonpolar tetrahedral hydrides show a steady increase in boiling point  Polar tetrahedral hydrides, the lightest member has an unexpectedly high boiling point  This is due to hydrogen bonding that exist among the smallest molecule with the most polar X—H bond.

 Bonds are WAY stronger than forces  Ionic>Ionic/Dipole>H>Dipole/Dipole> Dipole/Induced> Induced/Induced  The stronger the intermolecular forces the higher the melting and boiling points  Solids have highest intermolecular forces followed by liquids and gases.

1 st Law of Thermodynamics (AKA Law of conservation of Energy): Energy cannot be created or destroyed. It remains constant in the universe.  E = q + w  E = change in system’s internal energy q = heat w = work  Heat: energy that flows into or out of a system because of difference in temperature between the system and its surrounding

 Thermodynamics: Science of the relationships between heat and other forms of energy  Thermochemistry: Study of heat absorbed or given off by chemical reactions.  Energy: Ability to do work (measured in Joules)  Work = Force (N) x Distance (m)  Types of energy:  Kinetic =1/2 mv 2  Potential = mgh

 Heat of Reaction (q)  Before any reaction the system and its surrounding are at the same temperature.  When the reaction starts, the temperature changes.  The value of q needed to return the system to the given temperature at the completion of the reaction is known as the heat of reaction.

 Exothermic Reaction (-q): Heat is given off. Products contain less energy than reactants.  Endothermic Reaction(+q): Heat is absorbed. Reactants have less energy than products.

 Measured in joules and calories  Heat capacity – the amount of heat needed to raise the temperature of an object exactly 1 ˚ C q = C ∆ t  Depends on mass and chemical make-up  Specific heat (s) – the amount of heat it takes to raise the temperature of 1g of the substance 1 ˚ C q = s x m x ∆t

 Can be measured using a calorimeter  The heat released by the system is equal to the heat absorbed by the surrounding and vice versa

 Enthalpy (H): Extensive (meaning it depends on the amount of substance) property of a substance that can be used to obtain the heat absorbed or evolved in a chemical reaction.  All chemical reactions absorb or give off heat.  This change in energy is known as the change in enthalpy (heat content) of a system. ΔH = H (products) – H (reactants)

 The entropy (S) of the universe increases for any spontaneous process ΔS universe = ΔS system + ΔS surroundings  Entropy: A measure of the degree of disorder  Reactions are driven by the need for a greater degree of disorder and the drive towards the lowest heat content  Reactions with negative Δ H’s are exothermic and those with positive Δ S’s are proceeding to greater disorder

 When a gas is formed from a solid CaCO 3 (s)  CaO(s) + CO 2 (g)  When a gas is evolved from a solution Zn(s) + 2H+  H 2 (g) + Zn 2+ (aq)  When the number of moles of gaseous product exceeds the moles of gaseous reactant 2C 2 H 6 (g) + 7O 2  4CO 2 (g) + 6H 2 O(g)  When crystals dissolve in water NaCl(s)  Na+(aq) + Cl-(aq)

 Δ H for an endothermic reaction is positive  Δ H for an exothermic reaction is negative  Changes in enthalpy are independent of the path taken to change a system from the initial to final state  Heat absorbed or given off varies with the temperature  Standard enthalpies of formation are given at 25°C and 1 atm pressure ( Δ H 0 )

 Standard enthalpy of formation – the change in enthalpy that accompanies the formation of 1 mole of a compound from its elements with all substances in their standard states at 25°C  These values are known

 Example:  How much heat is liberated when 10.0 grams of CH 4 (g) reacts with excess O 2 (g)? CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2 H 2 O(l): ∆H = kJ Convert grams CH 4 → moles of CH 4 → kilojoules of heat

 Free energy – energy available to do work  The Gibb’s Free Energy Equation:  ΔG = ΔH –TΔS  The sign of Δ G can be used to predict the spontaneity of a reaction

ΔHΔH ΔSΔS ΔGΔGWill it happen Comment Exothermic (-) +Always negative YesNo exceptions Exothermic (-) -At lower temperatures ProbablyAt low temperature Endothermic (+) +At higher temperatures ProbablyAt high temperatures Endothermic (+) -NeverNoNo exceptions

 If a series of reactions are added together, the enthalpy for the total reaction is the sum of the enthalpy changes for the individual steps What is the ∆H f of the following reaction? 2C(graphite) + O 2 (g)→ 2CO(g) 2C(graphite) + 2 O 2 (g)→ 2 CO 2 (g) 2 CO 2 (g)→ 2CO (g) + O 2 (g)

2C(graphite) + 2 O 2 (g)→ 2 CO 2 (g) 2(0) + 2(0) 2( kJ) (-787.0kJ) – 0 = kJ 2 CO 2 (g)→ 2CO (g) + O 2 (g) 2 (-393.5kJ) 2 (-110.5) + 0 (-221.0kJ) – (-787.0kJ) = kJ ∆H f = kJ kJ = kJ