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CH 102 | Detailed Study Guide | Exam 1

by: Isabella Ryerson

CH 102 | Detailed Study Guide | Exam 1 CH 102

Marketplace > University of Alabama - Tuscaloosa > Chemistry > CH 102 > CH 102 Detailed Study Guide Exam 1
Isabella Ryerson

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These are detailed notes for the material that will be on Exam 1 on September 19, 2016.
General Chemistry II
Paul Rupar
Study Guide
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This 11 page Study Guide was uploaded by Isabella Ryerson on Sunday September 18, 2016. The Study Guide belongs to CH 102 at University of Alabama - Tuscaloosa taught by Paul Rupar in Fall 2016. Since its upload, it has received 97 views. For similar materials see General Chemistry II in Chemistry at University of Alabama - Tuscaloosa.

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Date Created: 09/18/16
CH 102 Exam 1 Study Guide Lectures 1, 2, & 3: Phases of Matter and Intermolecular Forces  Describe the differences between the phases of matter (solids, liquids, and gasses). o Solids: high density, fixed shape, fixed volume, intermolecular forces strong.  A solid is a solid because the molecules are being held together and the thermal motion/energy of the molecules cannot overcome intermolecular forces o Gasses: low density, shape not fixed, volume not fixed, intermolecular forces weak.  A gas is a gas because the molecules are not being held together and thermal motion/energy of the molecules does overcome the intermolecular forces o Liquids: density moderately high, shape not fixed, fixed volume, intermolecular forces moderately strong.  A liquid’s intermolecular strength is in between that of a gas and a solid  Changing between states heat heat or reduce pressure SOLI LIQUI GA D cool D cool or increase S pressure  What causes intermolecular attraction? o All attractions between molecules are due to Coulombic attractions o Caulombic attractions are just charges o If particles are close, then the attraction is strong o If particles are far away, then the attraction is weak o Larger opposite charges = more attraction  List the types of intermolecular forces in order of weak to strong and describe them at the molecular level o Dispersion Forces (London Forces):  Attractive forces that arise as a result of temporary dipoles induced in atoms or molecules  all molecules are subject to dispersion forces  the instantaneous dipole of one molecule can induce a dipole in neighboring atoms (look at my “Lecture 1” notes to see an illustration of this)  it is the attraction of these dipoles that cause dispersion forces  as molar mass increases, so does the boiling point  this is because the magnitude of the dispersion force will increase with more electrons that can move around easily  larger electron cloud = larger dispersion forces  generally, higher molecular weights = larger dispersion forces  Polarizability: how easy the electron distribution in the atom or molecule can be distorted (or “pushed around”)  Increases with: greater number of electrons, more diffuse the electron could o Dipole-Dipole Forces:  Attractive forces between polar molecules  Many molecules have permanent dipoles  Molecules with permanent dipoles have a stronger attraction to each other o Hydrogen Bonding:  A special dipole-dipole interaction between the hydrogen atom in a polar N-H, O-H, or F-H bond and an electronegative O ,N, or F atom o Ion-Dipole Forces:  Attractive forces between an ion and a polar molecule o Ion-Ion Forces:  Attractive forces between ions Lecture 3: Surface Tension, Viscosity, and Capillary Action  Define and explain molecular factors that influence tension, viscosity, and capillary action o Surface Tension:  the tendency of liquids to minimize surface area  measured in mJ/ m^2  how much surface area is takes to increase the surface area of a liquid  think about how you see bugs running on top of a pond  generally, the higher the surface tension, the stronger the intermolecular forces o Viscosity:  the resistance of a liquid to a flow  measured in units of centipoise (cP)=g*cm^-1*s^-1)  think of the thickness of honey and how it takes a long time to pour  Decreases with increasing temperature o Capillary action:  combination of intermolecular attraction of molecules in a liquid and the attraction of a liquid to a surface  think of what happens when you touch a paper towel to water  this also causes water to travel up narrow tubes  the narrower the tube, the farther it travels Lecture 4: Vapor Pressure, Boiling Point, Critical Point, Sublimation, Fusion, and the Heating Curve of Water  Describe the process of vaporization and condensation at the molecular level o Vaporization: molecules leave the liquid state and become a gas o Condensation: molecules leave the gas state and become a liquid  Vaporization is an endothermic process o What does endothermic mean?  A system that absorbs energy from the surroundings  Condensation is an exothermic process o What does exothermic mean?  A system that releases energy into the surroundings  Water is held together by hydrogen bonds o A certain amount of energy is required to overcome the hydrogen bonds o At any temperature, some molecules have more kinetic energy than others o Those with enough energy can escape and form a vapor  Heat of Vaporization (enthalpy of vaporization) o Amount of energy needed to vaporize 1 mol of a liquid to a gas o Always exothermic o Know how to calculate this  This is a dimensional analysis problem  You will be given the H vap  Your answer will be in kJ  You can see example problems in my “Lecture 4” notes  Apply the concept of dynamic equilibrium to the vapor pressure in a closed system o Dynamic Equilibrium: the rate of evaporation is equal to the rate of condensation o So in a closed system in dynamic equilibrium, the number of water molecules that are vaporizing is equal to the number of molecules condensing o Think about pulling a piston up: volume increases, pressure falls, more gas vaporizes, and then the pressure is restored o Then think about pushing a piston down: volume is decreased, pressure rises, more gas condenses, and then pressure is restored o There is an illustration in this in our powerpoint and in my “Lecture 4” notes o This piston example perfectly explains Le Chatelier’s Principle:  When a system in a dynamic equilibrium is disturbed, the system responds so as to minimize the disturbance and return to a state of equilibrium o Temperature Dependence  As you increase temperature, the vapor pressure increases until it reaches its boiling point.  Use the Clausius-Clapeyron equation (both forms) o The Clausius-Claperion Equation relates vapor pressure to vaporization o This is in the form of y=mx+b o You can plot and solve for this o ???????????? ????????????= (−∆???? ????????????/????????) + ln ???? o ???????????? ????????????= (−∆???? ????????????????) (1/ ????) + ln ???? o You can see a worked out example in my “Lecture 4” notes  Describe the critical point o A point on a phase diagram at which both the liquid and gas phases of a substance have the same density, and are therefore indistinguishable  What is sublimation? o Direct transition of a solid to a gas o Think of dry ice  What is fusion? o Fancy word for melting o Heat of fusion (∆???? fus)  The energy required to melt one molecule of a substance  The Heating Curve of Water o You can see a good diagram of this in my “Lecture 4” notes o Or you can google it and look at those diagrams (those may exclude some of the info Dr. Rupar provided in class and expects us to know) Lecture 5: Phase Diagrams  Calculate the quantify the heat needed for a substance to undergo phase transitions o You can see a worked out example in my “Lecture 5” notes  What is a phase diagram? o a diagram representing the limits of stability of the various phases in a chemical system at equilibrium, with respect to variables such as composition and temperature o You can see a good diagram of this in my Lecture 5 notes taken on September 1, 2016 o Or you can google it and look at those diagrams. Just make sure the picture you study includes all of the “key parts” listed below  Describe key parts of a phase diagram o y-axis: pressure o x-axis: temperature o Gas: low pressure, high temperature o Liquid: increase pressure o Solid: increase pressure, lower temperature o Critical Point: the point where the distinction between liquids and gasses cease to exist o Fusion Curve: liquid/solid barrier, also known as freezing/melting o Vaporization Curve: liquid/vapor barrier, also known as evaporation/condensation o Sublimation Curve: solid/vapor barrier o Triple Point: the temperature and pressure at which the three phases (gas, liquid, and solid) of that substance coexist in thermodynamic equilibrium  Use a phase diagram to predict phase changes in a substance  What makes the phase diagram of water special? o Water expands when it freezes  What is regelation? o Melting ice by pressure Lecture 6: Crystalline Solids  Qualitatively describe x-ray crystallography o X-ray particles go under constructive (dot) and destructive (not dot) interference o Constructive interference occurs when Bragg’s Law is satisfied  What is Bragg’s Law? o there is a definite relationship between the angle at which a beam of x-rays must fall on the parallel planes of atoms in a crystal in order that there be strong reflection  Use Bragg’s law to calculate spacing at the atomic level o n=2dsin o Look in my “Lecture 6” notes for a worked out example  What is a crystal lattice? o the symmetrical three-dimensional arrangement of atoms inside a crystal o a crystal lattice is made from unit cells  What is a unit cell? o A particular arrangement of atoms that repeat over and over and over again to form a crystal lattice o Tiles on a bathroom floor are the unit cells in a 2D space  Describe the three cubic unit cells (way more notes on this in “Lecture 6” notes and in “Recitation 2” notes) o Simple Cubic Unit Cell:  atoms at each corner  coordination number of 6, packing efficiency of 52%, and contains 1 atom per unit cell o Body-Centered Unit Cell:  Atoms at each corner and one in the center  Coordination number of 8, packing efficiency of 68%, and contains 2 atoms per unit cell o Face-Centered Unit Cell:  Atoms at each corner and one on each face  Coordination number of 12, packing efficiency of 74%, and contains 4 atoms per unit cell  What is a coordination number? o The number of atoms with which each atom is in contact with  What is packing efficiency? o The % of volume of the unit cell occupied by the atoms o The higher the coordination number, the higher the packing efficiency Lecture 7: Classifying Crystalline Solids | Solutions and Solubility  Describe the three crystalline solids o Molecular Solids:  Made from molecules  Geld together by dipole-dipole bonds, hydrogen bonds, and dispersion forces  They have a low melting point  CO 2 and H 2O are good examples o Ionic Solids:  Made up of cations and anions  They have a high melting point  Interactions are much stronger  NaCl is a good example o Atomic Solids:  Non-Bonding Atomic Solids:  Held together by dispersion forces  Very low melting points  He and Ne are good examples  Metallic Atomic Solids:  Held together by metallic bonds  Melting points have a huge range: from less than room temperature to 2000C  Network Covalent Atomic Solids:  Held together by covalent bonds  Tightly woven network of atoms  High melting points  Typically very hard  Diamond (carbon) and SiO 2 (quartz) are good examples  List the components of a solution o Solvent: the majority component of a solution o Solute: the minority component of a solution  Qualitatively describe entropy o The pervasive tendency for energy to spread out o Related to an increase in disorder o And important role in solution creation o Entropy drives a huge number of processes o Entropy is always increasing  Describe the energetics of the dissolution process  Using the concept of polarity, identify whether a compound will dissolve in a solvent o Solvents are broadly categorized as being polar or non-polar o Like dissolves like  Polar solvents dissolve in polar compounds  Non-polar solvents dissolve in non-polar compounds  Polar solvents do not dissolve in non-polar compounds  Non-polar solvents do not dissolve in polar compounds Lecture 8: Ionic and Metallic Atomic Solids  In Lecture 8, the three crystalline solids were described in even more detail. In order to prevent redundancy, I’m going to leave this out of my notes. You can look at my “Lecture 8” notes if you want to read about the crystalline solids in more detail. However, I will provide an outline and describe the new information given in the lecture: o Molecular Solids o Ionic Solids o Atomic Solids  Non-Bonding Atomic Solids  Metallic Atomic Solids  Network Covalent Atomic Solids  Graphite o Sp2 hybridized carbon atom that forms fused six-membered rings arranged in sheets o The sheets can slide past each other which makes graphite a good lubricant (used on automobiles) o Trigonal planar geometry for each C atom o The sheets are held together by dispersion forces o Graphite’s melting point is around 3800C o It’s hard to melt because you have to disrupt covalent bonds between each individual sheet o Forms hexagonal crystals o Graphite is an electrical conductor parallel to the sheets (only one direction)  Diamond o Also made up of carbon atoms but sp3 hybridized o Tetrahedral o Every carbon is bonded to 4 other carbon atoms o A diamond crystal is basically just one carbon molecule- that’s why it’s so strong o Diamond’s melting point is around 3800C o Very hard and rigid o Diamond is an electrical insulator- it doesn’t conduct electricity o But it is the best know thermal conductor  Buckminsterfullerene: Buckyball o Forms 60 carbon atoms in the shape of a soccer ball o C 36-C100 but C60 is the most common form  The whole family of structures are called “Fullerenes” o If stacked, the balls are held together by dispersion forces o Individual balls are held together by covalent bonds o Because the stacked balls are held together by dispersion forces, they can be dissolved  Nanotubes o These are like rolled up graphite layers o Very strong  Quartz o SiO 2 o Similar in structure to diamond o 3D array of each Si atom covalently bonded to 4 oxygen atoms o Tetrahedral o Melting point is around 1600C o Very hard  Silicates o SiO 2units but sometimes Al in place of Si  The ones with aluminum are called “aluminosilicates”  Aluminosilicates make up 90% of Earth’s crust o Glass is SiO 2  Amorphous = not crystallized  Melts at much lower temperature  Know how to classify crystalline solids  Describe the Cesium Chloride, rock salt, zinc blend, and fluorite structures o Cesium Chloride (CsCl)  Body-Centered Cubic  Coordination number = 8  Cs+ in the center  Cl- on the edges o Rock Salt (NaCl)  Face-Centered Cubic  Coordination number = 6  4 Cl- anions  4 Na+ cations o Zinc Blend  Face-Centered Cubic  Coordination number = 4  4 S anions  4 Zn cations o Fluorite  Face-Centered Cubic  4 Ca cations  8 F anions  Coordination number = 8 Lecture 9: Solutions and Solubility Continued | Solution Concentrations  Energetics of solution formation o H solutio= Hs olut+ H solven+ H mix o  H mixdetermines the whether the solution is endothermic or exothermic o Typically, H solven+ H mix are combined and called H hydration  Exothermic: energy cost of breaking solute-solute and solvent-solvent interactions is less than energy released upon mixing  Endothermic: energy cost of breaking solute-solute and solvent- solvent interactions is greater than energy released upon mixing  Heats of Solution o Because lattice energy is always exothermic, (therefore H solution is always endothermic) the sign and the size of H solutiotells us something about H hydration o H hydratio< H latti, then dissolution will be endothermic o H hydratio> H latti, then dissolution will be exothermic  Solution Equilibrium o Imagine you have a glass of water and you pour salt (NaCl) in. At a certain point, the salt will stop dissolving. This is called saturation. o Saturation: max amount of solute in a solvent o Solution Equilibrium: rate of dissolution and rate of recrystallization (un-dissolving) is equal o Supersaturated Solution: solutions that contain more than equilibrium saturated solution (very rare)  solution with more dissolved solute than the solvent would normally dissolve in its current conditions  The effect of temperature on the solubility of solids and gasses o The solubility of solvents increases with temperature o For gasses this is opposite: low temperature = higher solubility  This is why little bubbles form as you heat water (below 100C)  Gas Solubility: Henry’s Law o The higher the gas pressure, the more soluble a gas is in a liquid o S gas = K Hx Pgas o S gas: solubility in mol/L o K H: Henry’s Law gas sonstant o P gas: pressure of the gas  Perform calculations with Henry’s Law o Look in my “Lecture 9” notes for a worked out example  Concentrations of solutions: o mol/L o Molality=(mols solute)/(mass of solvent in kg) o % mass=(mass solute)/(mass solvent) o ppm=(mass solute)/(mass solvent) x 10^6 o ppb=(mass solute)/(mass solvent) x 10^9  Calculate the concentration of a solution using molarity, molality, ppm and ppb o Look in my “Lecture 9” notes for worked out examples


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