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Exam 1 Study Guide

by: Erik Roseberry

Exam 1 Study Guide CHEM 0970

Marketplace > Engineering and Tech > CHEM 0970 > Exam 1 Study Guide
Erik Roseberry
GPA 3.0
Chemistry 2
Dr. Maleckar

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To help you study for that big first exam coming up soon! Has notes from all material, and practice problems to make sure you know your material for the exam.
Chemistry 2
Dr. Maleckar
Study Guide
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This 7 page Study Guide was uploaded by Erik Roseberry on Saturday January 31, 2015. The Study Guide belongs to CHEM 0970 at a university taught by Dr. Maleckar in Fall. Since its upload, it has received 324 views.


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Date Created: 01/31/15
Erik Roseberry Dr Maleckar Chemistry 2 Study Guide Exam 1 1 Calculate the percent by mass of the solute in each of the following aqueous solutions a 575 g of NaBr in 679 g of solution b 246 g of KCI in 114 g of water c 48 g of toluene in 39 g of benzene 2 What is the osmotic pressure in atm of a 157 M aqueous solution of urea NH22CO at 270 C 3 Calculate the amount of water in grams that must be added to a 500 g of urea NH22CO in the preparation of a 162 percent by mass solution b 262 g of MgCl2 in the preparation of a 15 percent by mass solution 4 Write the reaction rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of products a H2g 12g DZHKg b 5Braq BrO 3 aq 6Haq n 3Br2aq 3H20l 5 For the reaction NOg 0 g I NO g 0 g the frequency factor A is 87 x 1012 s 1 and the activation energy is 63 kJmol What is the rate constant for the reaction at 75 C 6 What are the units of the rate constant for a thirdorder reaction 7 The equilibrium constant K P for the reaction 280 g II 280 g 0 g is 18 x 10 5 at 350 C What is K for this reaction 8 One mole of N2 and three moles of H2 are placed in a ask at 375 C Calculate the total pressure of the system at equilibrium if the mole fraction of NH3 is 021 The KP for the reaction is 431 x 104 ghapter Notes Chapter 13 Solutions Solution homogenous mixture of two or more substances Solute the smaller component of a solution Solvent greater component of solution NaClsolid gt Naaqueous solutionCl39aq salt dissolving a physical process 0 NaaqCl39aq gt NaCls salt reforming crystal lattice In order to properly dissolve solute in solvent 1 Solute must separate by overcoming Intermolecular forces 2 Solvent molecules must separate to open spaces for solute lceberg Effect 3 Energy is released upon formation of solutesolvent particles Solubility Unsaturated able to hold more solute in solution Saturated at max capacity for solute in solution Supersaturated more solute than able to dissolve Some solute remains not dissolved Non polar molecules do not dissolve in water Solids become more soluble at high temperatures Gases become less soluble at high temperatures Henry39s Law gas dissolved in solvent is proportional to pressure of gas above surface S9 solubility of gas in solution phase P9 pressure of gas above Molality m Moles of SoluteKilograms of Solvent Mole Fraction XA Mole SoluteTotal mole Colligative Properties Vapor pressure goes down as you add more solute Boiling point goes up as you add more solute Melting point goes down as you add more solute Evaporation lessens as you add more solute Raoult39s Law PA partial pressure solvent XAmole fraction of solvent in solution PAquot vapor pressure of pure solvent Boiling Point Elevation after adding solute vapor pressure lowers more energy required to boil it K constant dependent on solute i van t Hoff factor m molality Freezing Point Depression impurities in solution make it harder to freeze Van39t Hoff factor number of particles which a solute dissociates into Osmosis ow of solvent into a solution through semipermeable membrane lsotonic two solutions have same salt concentration on both sides of membrane Hypotonic uid in cell has higher salt concentration than other side Hypertonic uid outside cell has higher salt concentration than cell Osmotic Pressure pressure by ow of water through semi permeable membrane quot iMRT l39l osmotic pressure i Van t Hoff Factor MMolarity R gas constant T temp Kelvin Chapter 14 Chemical Kinetics Study of Reaction Rates Fast Reaction Combustion Slow Reaction Iron Rusting Cement Setting Food Cooking Reactant Rate amount of product formedtime Measured in MolaritySeconds As concentration rate goes up reaction rate also increases 0 As temperature goes up reaction rate increases Gases react quicker than Liguids which are faster than Solids Catalysts increase rate of reaction by lowering energy needed to start a reaction activation energy energy needed to start a reaction Gets regenerated in process of reaction Rate Law aAbB l with C as a potential catalyst dDeE Will form rate law kAmBquotCp k rate constant varies with temp m n and 0 determined experimentally Necessary to have data on reaction in order to determine rate law Impossible to determine from aA I bBcC alone The following are examples C3H6 I CH2CHCH3 given Rate Law kC3H61 rst order because power of 1 2NO2H2 l N22H20 given Rate Law kNO2H2 second order NO rst order H2 third order overall CH3COCH32 H catalyst CH3COCH2H Rate Law kCH3COCH3H second order reaction rate not dependent on l2 because its not in rate law Will only increase reaction rate if increased catalyst or rst reactant Integrated Rate Law acquired through complex not important integration Or for half life reactions Collision Theory For every successful reaction activation energy must be met through collision of two or more particles Transition State Theory There must be a state of unstable transition in between the reaction of one molecule into another where all elements involved are separated and held together by weak bonds Potential Energy Diagrams 1 Endothermic starts with low energy and must peak over high activation energy to react Ends with higher energy than beginning 2 Exothermic generally lower activation energy this reaction loses energy through the reaction and ends in more stable low energy position Arrhenius Equation relates k and temperature Basic Ea Activation Energy Joules A Frequency factor dif cult to obtain often given R Gas Constant T Temperature Kelvin Rearrangedr or Mechanism details about speci c reaction including steps in between AA l CE often slower reaction EB AD often fast AB l CD nal Strikethrough means they cancelled each other out Molecularity number of molecules on the reactant side of any elementary reaction Elementary Reaction individual reactions in mechanism Unimolecular 1 reactant molecule 0 Rate Equation kA Bimolecular 2 reactant molecules colliding at same time 0 Rate Equation kAB Termolecular 3 reactants very rare 0 Rate Equation kABC In order to graph a reaction order sometimes it is best to use linear A v Time or ogarithmiclnA v Time or inverse 1A v Time all depends which graph will give you the most linear relationship for the speci c equation Chapter 15 Chemical Equilibrium Concept of Equilibrium chemical equilibrium is the point where both reactants exist together The rate at which reactants go to products is equal it the rate in which products turn into reactants and reactions continue to happen so it is dynamic it is a reversible process this chapter focuses on chemical changes Equilibrium Constant ll my notation for equalibrium If N204 N02 then kfN204krN022 50 kfkr N0212N204D kfkr Kc Kc is the equilibrium constant while the entire problem is the equilibrium expression can never determine reaction rate from expression alone KC CCDdAaBb is also another way to express the equilibrium constant products on top reactants on bottom if x is solution use molarity for x if x is gas use partial pressure for x is x is solid or pure liquid do not use for Keq it adds no part to equann Equilibrium Expressions if KC is huge the concentration of products will be greater than of the reactants lf KC is small the concentration of reactants will be greater than of the products K is used for gas partial pressure equilibrium Kp KC RT 239 1 Rgas constant TTemperatureK n difference in stoichiometric coefficientsprodreact Solving Problems with Equilibrium Constant Qc reaction quotient To calculate initial productsinitial reactants If KC Qc then process is at equilibrium If QC gt KC then process has more products and will move to left towards reactants to reach equilibrium 0 If QC lt KC then process has more reactants and will move to right towards products to reach equilibrium Equations to solve for missing variable At ktAO for zero order problems ttime lnAtAokt rst order problems krate constant 1At kt 1AO second order problems A concentration How Chemical Equilibrium is Affected Le Chatker39s Principle when chemical system at equilibrium gets disturbance it will shift to left or right to minimize disturbance Disturbances o Increasedecrease pressure Increasedecrease temperature Increasedecrease concentration If pressure increases changes the side that has fewer moles If temperature increases chill change towards reactants for exothermic towards products for endothermic reaction


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