Chem 122 first exam study guide
Chem 122 first exam study guide Chem 122
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This page Study Guide was uploaded by Alisha Notetaker on Monday February 8, 2016. The Study Guide belongs to Chem 122 at University of North Dakota taught by Dr. Harmon Abrahamson in Spring 2016. Since its upload, it has received 43 views. For similar materials see General Chemistry 2 in Chemistry at University of North Dakota.
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Date Created: 02/08/16
Ch 10 Intermolecular forces attractive forces between molecules determined by differences in electronegativity types 1 iondipole between an ion and the partial charge on the end of a polar molecule 2 dipoledipole when polar molecules are close together 3 hydroqen bondinq hydrogen must be directly bonded to a small atom of high electronegativity NOF have higher boiling points more energy is needed to break the intermolecular forces 4 London dispersion forces very weak attraction formed by the uneven electron cloud in a molecule 9increase with polarizability 9increase with both size and molecular weight Hydrogen bond gt Iondipole gt Dipoledipole gt London dispersion forces Electronegativity atom s ability to attract electrons to itself when it s bonded to another atom higher electronegativityhigher ability to attract electrons trend on periodic table increases going across to the right and decreases going down Bond polarity measure of how unevenly the electrons in a bond are shared and is indicated by the difference between the electronegativity of each atom involved arger differencepoarcontains dipole 9ionic dipole exists when two equal but opposite charges are separated smaer differencemore covalent bond character Covalent bond electrons are shared Ionic bond electrons are transferred Electronegativity differences can determine type of bond Nonpolar covalent lt 05 Polar covalent 05 20 Ionic gt 20 Nonpolar when the bond dipoles are equal in magnitude and opposite in direction then there is no molecular dipole Polar when bond dipoles are present but are not equal then there is a molecular dipole Electric fields can orient molecules based upon their molecular dipoles and partial charges 9 partial charge will align toward negative end 9 partial charge will align toward positive end Nonspontaneous reaction does not occur Spontaneous reaction does occur irreversibe a process in which the energy content of the system decreases tends to occur spontaneously Eguations Gibbs free energy Exothermic temp increases Endothermic temp decreases AH is the enthalpy T is the temp in Kelvin AS is the change in entropy pontaneous decrease in H AH Increase in S AS nonspontaneous increase in H AH Decrease in S AS Ch 11 Solution when one substance disperses uniformly throughout another 9what forms is a homogenous mixture solvent and solute Solvent substance present Solute substance dissolved in solvent Solvent Solute 9 Solution the energy gained AHsom needs to be greater than the energy required in order for the intermolecular forces to be broken in the solvent and solute if AHmlx gt AHsolv AHsol then AH50m 9 favorable if AHmlx lt AHsolv AHsol then AH50m 9 unfavorable Equilibrium the forward reaction rate is equal to the reverse reaction rate Saturated a solution in equilibrium with undissolved solute Solubility amount of solute in a saturated solution ugersaturated excess solid present saturated equilibrium unsaturated no solid present 9the stronger the interactions are between solute and solvent molecules the greater the solubility 9LKE DISSOLVE LIKE 9attractions that can lead to hbonding will increase the solubility of a solute 9the presence of a carboxyl group will increase the strength of hbonding 9poar OH groups readily form hbonds in water 9as length of carbon chain increases the solubility decreases Not energetically favorable breaking a strong interaction hbonding to form a weak interaction London dispersion Miscible when two solvents mix in all proportions 9to be miscible in water there must be strong enough interactions to overcome the nonpolar portions of the molecule Endothermic energy entering a system and have a positive enthalpy Exothermic energy released from the system and have a negative enthalpy Hydration when water molecules cluster around solute particles Eguations Henry s Law 3959 is the so u i ity in molL k is Henry s constant in molLatm P g is the pressure Example How many grams of C02 dissolve in 250mL of bottled water under 1900mm Hg at 25 C k3110392 molLatm 59 kPg 2 mol latm mol 31 10 m 1900 007757 77510 2 M 0250L w 08525 coz 1 mol C02 Expressing Solution Concentration mass of concentration 1 by mass 100 total mass of solution 2 by volume 3 Molarity M quot10135 of solute Liters of solution dependent on temp moles of solute 4 MOIallty m kg of solvent independent of temp moles of component 5 Mole fraction X total moles of all components wieght or volume of solute 6 6 Parts per million ppm weight or volume of solution lg solute 1mg solute 1 ppm 1069 Solution 1 L solution 7 Parts per billion ppb 1ppb M o I 109solution Example 86 by mass of nitric acid What is the Molarity and Molality Density1425 gmol gmol gmL 86g HN03 1 kg Mass 1009 solution 986g HN03 14g H20 14g H20 10009 00140kg 86g HN03 quot01 1365 mol HN03 100g soln 1quot 1L 2 007042 L 63029 14259 1000mL Molarity L65 194 M Molality 1365 007042 00140 Colligative Properties 1 Vapor pressure lowering Raoult s Law XA M moles A moles B PA is the vapor pressure of solution XA is the mole factor of solvent P A is the vapor pressure of pure solvent as concentration increases vapor pressure decreases if contains volatile solutes Ptotal 2 PA PB 2 XAP A XBP B Example 1 0340 mol nonvolatile dissolved in 370 mol H20 P A of water is 238 370 mol 2 0916 404 mol Mole fraction Vapor pressure 0916 238 Example 2 180 mol Cyclohexane P A 976 torr and 200 mol Acetone P B 2295 torr 18 mol 38 mol Ptml 0474 976 0526 2295 6698 torr 20 mol 38 mol XA 0526 0474 XB 2 Boiling point elevation k is the moa boiling point constant m is the molality Example 50 by mass C2H602 kb0512 Cm 1 k C2H602 1000 1 mol g 9 161 molkg 1 kg H20 1 kg 6208 g ATb 0512 161 824 C boiling point 100 824 10824 C 3 Freezing point elevation k is the moa freezing point constant m is the molality Example 50 by mass C2H602 kf186 Cm 1 kg C2H602 10009 1 mol 1 kg H20 1 kg 6208g 16391 molkg ATf 186 161 2995 C freezing point 0 2995 2995C moleso articles insolution 9Van t Huff Factor I f p moles of solute dissolved used in both boiling and freezing point if the solute is ionic Example How many grams of KNOB have to be added to 275mL H20 in order for it to freeze 2 mol solute particles 1 mol solute dissolved 145 ATf 0 145 145 C m 1862 39 molkg 1 kg 275mL H20 275g 0275 kg H20 10009 390 molkg xmoz 21390725 mOl 10725 mo w 108g KNOB 0275 kg 1mol E 4 OsmOSIS 7139 v RT 9 7139 is the pressure required to prevent osmosis M is the molarity R is 008206 Latm kmo T is the temp Ch 12 Reaction rates the change in concentration of a reactant or product per unit of time measured in Msec A 23 9 C 2D reactant concentration will decrease roduct concentration will increase Rate Instantaneous rate rate at a particular time Obtained from the slope at a point Initial rate rate at t0 Rate law expresses how the rate depends on the concentration A B 9 C D x and y are the order of reaction x and y can only be determined by experimental data not the stoichiometric coefficient orders 0 0 order means changing the concentration will not change the rate 0 1 order means that if concentration doubles the rate doubles f concentration triples then the rate triples And so on o 2 order means that if the concentration doubles then the rate quadruples If the concentration triples then the rate increases by a factor of 9 And so on x2 units of k 0 1st order 3391 o 2ncl order M39ls391 o 3rel order M39zs391 Example 1 2 N205 9 4 N02 02 N205 Initial rate 09 M 5410 4 Msec RatekN 205quot 045 M 2710 4 Msec x 4 4 Zl ii lxi L 213134 9 2quot 2 9 xan In 2 9 X1 X1 is a first order reaction ratekN2051 541O394 k09 9 k 60 10394 3391 Rate law 60 103943391N2051J Exp 52 082 139 Initial rate 1 015M 021M 114 Ms 2 x 045M 021M 341 Ms Ratek5208 I 1y 3 045M 012M 195 Ms Example 2 52082 31 gt 2504 1339 N Exp 2 045M 021M y 341MS x x Exp 1 015M 021M 114 Ms 3 3 9 Xn3 In 3 9 X 1 Ex 2 03945Mx 3921My 3quot M5 9 1753 175 9 yIn175 In175 9 x1 Exp3 045M 012M 195 Ms x 114 k52082 I y 9 k01510211 114 MS 9 k015021362s 1 Rate law 362 Ms3208211391J Halflife 01 kt 9 0693 ktlZ Example certain 1st order has halflife of 40 min What is rate constant How much time is required for the reaction to be 75 complete 0693 kt12 9 0693 k 40 9 k 00173 min 1 01 001731 At t0 100 975 complete 9 At 25 9 025 Ini00173t 9 t1114 9 025 00173 Inteqrated rate laws relates concentration to reaction time For 1St order rearranged in y mx b form Order Integrated rate law Graph slope 0 A kt A0 A v t k 1 Ln A kt In A0 Ln A v t k 2 1 1 i k kt A V39 t lnlA A10 1 Example 1 0th order A 700 10392 M after 125 sec and A 100 10392 M after 300 sec What is k What was rise the initial concentration K m run Rise 100 10 2 700 10 2 006 M Run 300 125 175 sec k m 03906M 343 10394 MsecJ 175 A A0e kt 9 700 10 2M34310 4125 A0 9 A00113 MJ Example 2 1St order A 800 10392 M after 500 sec and A 750 10393 M after 650 sec What is k Make a data table rise 4893 2501 2392 run 65 50 15 sec A Ln A t 2 392 800 10 2 2501 50 39 1 k m 15 0159 s 750 10 3 4893 65 Example 3 2ncl order A 0860M after 285 sec and A 510 10392M after 750 sec What is k Make a data table 1 1 Rise 510102 0860 1845 M 1 A i t l Run 750 285 465 sec 0860 1 285 0860 1845 510 10 2 1 750 7lt H 2 1 1 465 39610 M 5 J 510 102 The Collision Model molecules must collide to react and the number of collisions is proportional to concentration and temp increased concentration increased molecules increased probability of collision increased temp increased velocity increased frequency of collisions o For a reaction to occur the collision energy must be greater than the activation energy 0 Collision E gt EA o If temp increases there is a much larger fraction of molecules that could react Ea levur temp high temp Fraction ef meleculee Kinetic Energyr The Orientation Factor the probability of a particular high energy collision resulting in a reaction depends on the molecular orientation during the reaction Example ZBrNO 9 Br2 2N0 1 MD ferm will net all w the reectien D n Br e n Hr e H WED3 Br N will ferm 2 Br and the hi will reek eff 1 tr ferm encl hreelt eff Transition state theory focuses on hypothesized intermediate species called an activated complex which forms during an energetic collision Exists very briefly and dissociates either back to reactants or to products intermediate Reacts ntr petential energyr Preclucts extent ef reactien Examgle 1 If EA 171 kJmol for forward reaction and AH 13 kJmol What is EA for the reverse reaction Exoth ermine ea ction Ea in mm Ea for forward reaction E3 1m V 113 13a ikamoil Ea for revereee i reaction REHEIEI H1151 potential en erg a 43 Mfume Pmdmct extent of reaction For exothermic reaction the AH R P EA for the forward reaction intermediate R EA for the reverse reaction EA of forward AH Examgle 2 If the EA for A 9 B is 150 kJmol And EA for B 9 A is 100 kJmol Then what is AH Endothermic Reaction pate ntia energy y 1 Il k JJ39mUI A reaction program For endothermic reaction AH B A EA for the forward reaction intermediate A EA for the reverses reaction intermediate B Arrhenius Equation slo e F I A is the requency actor R is the gas constant 8314 Jmol If we have two data points then the equation will be Example Calculate rate constant at 25 C for hydrolysis of sucrose given that it s equal to 1 103M1S1 at 37 C R 8314 JmoK EA 108 kJmol 9 108000 Jmol 25 C 308 K 27 C 310 K 110 3 1 10 3 108000 1 1 1 108000 i In 9 3 k1 3 8314 308 310 k1 8314 308 310 lo 361299020910 5 9 1103 30272 R1 R1 110 3 k1 131 9 k1763104M131J Reaction Mechanisms many times a reaction doesn t occur in one step And it may not show us the other steps involved those hidden steps are known as the elementary steps together the steps make up a mechanism the slowest step is known as the rate determininq step rds Example N02 co 9 N0 coz rate law kN022 Takes 2 steps 1 NozNoi9uo coz 2 mmoacozme z N02 CO 9 NO C02 the parts crossed out are known as the intermediates based on rate law rds Molecularity how molecules collide in a reaction biomolecular collision of 2 molecules unimolecular involves only one molecule in the reactants termolecular very rare Collision of 3 molecules Order and rate law overal order of elementary steps depends on molecularity rate law can be determined by the number of each type of molecule in the reactants of the elementary steps Example 1 3x 9 E F 2 4 2M 9 F N 3X 2M 9 2F N intermediate E rate law for step 1 kX3 rate law for step 2 kE1M2 Steady state approximation solve for concentration of intermediate by assuming that an equilibrium is established in the fast step which precedes the slowest step Example 1 Clz lt gt 267 fast 2 9H CHCl3 lt gt HCl 9amp3 slow 3 6F lt gt CCl fast Cl2 CHCl3 lt gt HCl CCl4 Predicted rate law kCl CHCl3 but Cl is an intermediate Clz lt gt ZCl Rate of Clz decomposition k1 C12 put them equal to each other because it s in equilibrium Rate of Clz formation k1Cl2 solve for CI k1 C12 k1Cl2 9 Cl2lc 1 C12 9 take square root of both sides 1 Cl C12 9 CI k Clz12 a plug into rate law 1 Rate law 1 k Cl212CHCl3 J Catalyst speeds up a reaction by reducing the activation energy for both forward and reverse reactions encore original Ea original reaction Ea after catalysis l catalvaecl reaction energyF reaction prooeaa
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