Final Study Guide
Final Study Guide chem 10061-001
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This 8 page Study Guide was uploaded by Matthew Goetz on Saturday May 7, 2016. The Study Guide belongs to chem 10061-001 at Kent State University taught by David bowers in Summer 2015. Since its upload, it has received 40 views. For similar materials see general chemistry 2 in Chemistry at Kent State University.
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Date Created: 05/07/16
Gen Chem 2 Final Exam Study Guide: Hydrocarbons are organic compounds composed of carbon and hydrogen. Know how to name hydrocarbons: Alkanes are hydrocarbons with only single bonds and a suffix of –ane. Root is due to the longest chain of carbons. Meth = 1, eth = 2, prop = 3, but = 4, pent = 5, hex = 6, hept = 7, oct = 8, non = 9. An example of an alkane which you may use to test your naming abilities is below. For this alkane, the longest chain is 6 carbons long, and it has 2, one carbon branches on the second and third carbons. Therefore, this is 2,3- methylhexane. Alkenes are when there are double bonds between carbons in the hydrocarbon. -These have a suffix of –ene but are otherwise named similarly. -A note must be made when naming these of whether they are cis or trans. This depends on whether the carbon chain is on the same side following the double bond, cis, or if it rotates, trans. Alkynes are when there are triple bonds between carbons in the hydrocarbons. -These have a suffix of –yne but are otherwise named similarly. Another important topic from chapter 15 is how to recognize functional groups. -These are specific combinations of atoms that react in set manners. -They also determine physical and chemical properties of compounds. Functional groups: -Alcohol: this is when there is an –OH bound to a carbon. Has an –ol suffix in the name. -Haloalkane: this is when there is a halogen bound to a carbon. Has a halo- prefix. -Amine: this is when a terminal N is bound to carbons and hydrogens. Has an –amine suffix. -Aldehyde: this is when an O is double bound to a carbon with an H as well. Has an –al suffix. -Ketone: this is when an O is double bound to a carbon without an H. Has an – one suffix. -Carboxylic acid: This is when there is a –COOH present. Has an –oic suffix. -Ester: This is when there is there is one O which is double bound to a C and another O in the carbon chain itself. Has an –oate suffix. -Amide: This is when there is an N in the carbon chain itself. Has an –amide suffix. -Nitrile: This is when there is a terminal N in the carbon chain. Has a –nitrile suffix. -Examples of several of these functional groups are seen below. Two important equations which are important to know are Q=mcΔt, and Q=mΔH. Types of Force: Bonding: Ionic, covalent, or metallic. Ion dipole: An ion is attracted to a dipole. H bonding: H bonds with an N, O, or F. Dipole Dipole bonding: Two dipoles are attracted to each other. Ion induced dipole: An ion creates a dipole in a nonpolar covalent molecule. Dipole induced dipole: A dipole creates a dipole in a nonpolar covalent molecule. Dispersion forces: Weak forces present in every compound. Properties of liquids: Surface tension: Energy required to increase the surface area of a liquid. Capillarity: Rising of liquid against gravity due to competition of cohesion and adhesion forces. Viscosity: Resistance of a fluid to flow. Higher temps reduce viscosity and small, spherical molecules have low viscosity. Solid Types: Atomic solids: Atoms held together only by dispersion forces. Bound to same atoms. Molecular solids: Held together by combinations of IF’s. Ionic solids: Arrays of cations and anions. Metallic solids: Organic crystalline structure with a sea of electrons. Network covalent solids: Atoms covalently bond in 3D lattice, like diamonds. Intermolecular forces determine solubility. Like Dissolves Like: Nonpolar dissolves nonpolar, polar dissolves polar, and polar dissolves ionic. Know how to use major equations from chapter 14 such as: Molality = mol of solute / kg of solvent Parts by mass = mass of solute / total mass Parts by volume = volume of solute / total volume Mole fraction = mol of solute / total moles Know colligative properties: Vapor pressure reduction, ΔP = Xsolvent (Psolvent) X is the mole fraction, and Psolvent is vapor pressure of pure solvent. Boiling point elevation, ΔTb = Kbm Kb is the molar boiling point and m is the molality. Freezing point depression, ΔTf = Kfm. Same are boiling point elevation. Osmotic pressure depression, π = mrt M is molarity, r is .0821, and t is the temperature in kelvin. For strong electrolytic solutions, you must multiply your answer to the colligative properties by I, which is the number of dissociated ions. Molarity x Seconds is the speed of chemical reactions. Chemical kinetics is the study of the speed of chemical reactions. In a chemical reaction you may control 4 factors to determine speed: Concentration of reactants which will allow more collisions. Physical states, for the reactants must mix to react. Temperature Presence of a catalyst. Rate calculations with coefficients must be performed in specific manners, as shown below: 2H2 + O2 2H2O. 1/2 (H2/t), (O2/t), ½(H2O/t). I am unsure whether this will be on the test but it is good to mention. Rate law expression: Rate law = k[A]^x [B]^y K is the rate constant. X and y are the orders of reaction with respect to the reactants. Products are not included in rate law. Reaction orders are determined experimentally by examining a chart of data. First order reactions are when the rate of A is directly proportional to the concentration. So, if A doubles, then the concentration doubles as well. Second order reactions are when the rate is the square of the concentration. So, if A doubles, then the concentration quadruples. Or, if A triples, then the concentration is multiplied by 9. Zero order reactions are when the Rate has no effect on the concentration. Determining the rate constant K is done by K = rate/ [A][B] Units differ depending on the orders. Zero order is mol/LxS First order is 1/S Second order is L/molxS Integrated rate laws show how time and rate are related, and are shown here: Collision theory: Reactants must collide to react. Increasing concentration causes more collisions. Effective collisions must have high enough activation energy. They must also collide at the correct orientation. Raising temperature adds energy to the colliding particles and increased frequency with which they collide, so it is very important to reactions. The Clausius Clapeyron equation may be used to calculate the activation energy if you know the rate constant at 2 different temperatures. Should be a given equation. Transition state theory: Instantaneous transition states form in reactions and disappear to form the products. Intermediates are high energy and unstable. Mechanism: A series of steps in an overall reaction. To combine each of these two ideas, here is an example: 2O3 (g) 3O2 (g) First step in this process is O3 O2 + O The second step is O + O3 2O2 (g). In this process, O(g) is the intermediate. In these reactions, either the first step or the second step is limiting, meaning it occurs more slowly. If step one is limiting, that is because it takes a lot of energy to become transition state 1, and once it has become transition state 1 then step 2 occurs very quickly. If step two is limiting then step 1 occurs very quickly because it requires little energy but transition state 1 gets stuck at an equilibrium until it gains enough energy to perform step 2. Catalysts lower the Ea and speed up the forward and reverse reactions at equilibrium. Does not produce more product though. Increase the rate but are not consumed and aren’t in the rate function. Equilibrium: The extent of a reaction. No net change is occurring at this point. Kc is the equilibrium constant and is equal to the concentration of products over the concentration of the reactants. If K is small then the reactants are favored. Reaction quotient (Q) shows the ratio of products to reactants at any point, not just equilibrium. If Q is greater than K then the reaction must proceed towards reactants. If Q is lesser than K then the reaction must proceed towards products. Pure solids and liquids are not considered when solving for K or Q. Kp is the pressure equilibrium constant and may be related to Kc with the equation Kp = Kc (RT)^Δn. Sometimes you may not have enough variables to solve for K or Q, which is when you must construct an ICE table. An example of a simple ICE table is shown below: Simplifying assumptions may be made in this process as well, which is when you assume that the x in the above example which is being subtracted from the .500 is insubstantial and you can act as if it were 0. Le Chatelier’s Principle: If a system at equilibrium is disturbed then the system will work to reattain equilibrium. -Changes in the system that we discussed were concentration, pressure/volume, and temperature. For concentration, if you add more product to the system then it will turn to product into an appropriate amount of reactant to reattain equilibrium. For pressure/volume, if you increase pressure or decrease volume then the side with fewer moles is favored. The opposite happens if you decrease pressure or increase volume. Effects of temperature: If heat is a reactant and I add heat to the system then products will be favored. If heat is a product and I add heat then reactants are favored. Arrhenius acid is an acid that donates H+ in solution. Bases create OH. Ka is the acid equilibrium constant and is equal to [H3O][A]/[HA]. Know weak and strong acids: Weak are: HF, organic acids, hydrated metal ions, acids where H is not bound to O or a halogen, and acids where the number of oxygens is equal to the number of OH bonds. Strong bases are bases with group one or two metals. Weak bases are NH3 and H2O. Amines are also weak bases. pH = log (H3O+) pOH = log (OH) pKa = log (Ka) Bronsted Lowry Acid is an acid that donates H+. Bronsted Lowry Base is a base that accepts H+ into a lone electron pair. Any amine, fluorine, and ammonia. Conjugate acid/base pairs sho uld also be known. Salts may affect pH depending on if they have the cation or anion of a weak acid or base. For example, if Zn(CH3COO)2 were placed into a solution, it would make the solution more basic because ZnOH is a strong base while C2H3O2H is a weak acid. Buffers are solutions that don’t react greatly to addition of a strong acid or base. Contain a weak acid/base and its conjugate pair which allow them to convert any acid or base which is added into one of the conjugate pairs. HA + H2O A + (H3O+) A + H2O HA + OH There are examples of how buffer systems work. Buffers simply follow Le Chatelier’s Principle! Know how to calculate pH changes when additions are made to buffer systems as we did in class. Know how to calculate how to make a buffer using the HendersonHasselbach equation which will be given on the exam. Spontaneous reactions occur continuously without continuous input of energy. If a reaction is spontaneous in one direction it is not spontaneous in the other. S is entropy, and if it is positive then the system is spontaneous. H is enthalpy, and if it is negative then the system is spontaneous. G is gibbs free energy, and if it is negative then the system is spontaneous. Calculating entropy changes: ΔS = ΣmSproducts – ΣnSreactants. ΔH may be calculated in the same manner as S. Then, ΔG may be calculated with the equation: ΔG = ΔH – TΔS. This is everything that I think will be important to know for the final exam, so good luck everybody!
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