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Principles of Chemistry

by: Kendrick O'Keefe

Principles of Chemistry CHE 115

Marketplace > Davidson College > Chemistry > CHE 115 > Principles of Chemistry
Kendrick O'Keefe
GPA 3.97

William Nutt

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William Nutt
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This 21 page Class Notes was uploaded by Kendrick O'Keefe on Sunday October 11, 2015. The Class Notes belongs to CHE 115 at Davidson College taught by William Nutt in Fall. Since its upload, it has received 24 views. For similar materials see /class/221261/che-115-davidson-college in Chemistry at Davidson College.


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Date Created: 10/11/15
cm us mmin IV Chemical Equilibrium A Law of Chemical Equilibrium Concentration expressed as rnolarity mole mitts Mulanty an A s amp vulume m tiers L M Consider the hypothetical reaction Ag e 83 Cg Dag nmnnlmd D i t plotted as afun ion oftime Plot of the Concentrations of A B and C or D Velsus Time l E E um E g 008 e S u we claim a n4 am e i a on t t t t t o m w so so ms 170 Timelseel Note A er eleven seconds the concentrations do not change with tirne tiittittt products no longer change with tinne chernical equilibrium isadynannic state ltoccurswhen t AiB h t B 7 CD Alg 89 f 69 Dg indicates snhilihnhm CHE 115 chemical Equilibrium At equilibrium there is a speci c relationship between the concentrations of the reactants and pro ducts Consider three experiments for the reaction AgBg C t D in which the initial concentrations of A and B vary from experiment to experiment Table of Equilibrium Concentrations Exp Aleq Blul CLq Dleq 1 300 200 100 100 2 960 100 400 400 3 0500 300 0500 0500 The following relationship ts the data in each experiment C A D eq 9 1 0167 constant K B 0 9 1 eq For the general reaction mAg nBg quot pCg qDg Kc is de ned CPeq Drieq Aimeq B The magnitude of the constant Kc depends on the nature of the reaction and the temperature The units of Kc depend on the speci c reaction Example A 3 B 2 C male 2 K 7 29 units W712 4 male Example A B C D Em us Term gammamin mlrtm 7 T 5 ma 11 l 5 When Klt 1 then the equilibrium is said to lie to the le When Kgt 1 then the equilibrium is said to lie to the rlght B Equilibrium Calculations Example 1 Initially 02660 mole ochlg and 03340 mole of cl2 are placed in a 200 L 300 c container at Ifthe moles ofPCls are found experimentally to be 00660 at equiliorium calculate Kc for the reaction 1ch c12 Pcl5 at 300 C Plot of the Concentrations of PCIJ 22 and FCI5 versus Time 0 20 01B Clzlm 0 334 molyz 00 L i 0 IE 7 014 s f 7 7 cnruneumousarpcl 012 a Pc alu 39 D 265 W D L 0 10 ZUEL Pol 7 x 2 iquot 005 s 006 s Pct 00060 mole2 00 L 004 s Mmuunngc 002 7 ant chrslt a 000 i i i i 0 20 40 so 00 100 120 Timesec wherewn m l Important observations from the above plot CHE 115 ChamcalEquilibnum Initial moles ofa substance change in moles ofsubstance equilibrium moles ofsubstance Example initial moles of PC15 change in moles of PC15 equilibrium moles of PC15 00660 moles of PC15 00 moles of PC15 00660 moles of PC15 The changes in number of moles of reactants and products are related by male ratios These observations can be utilized in an Equilibrium Table to solve equilibrium problems Equilibrium Table Change in moles of PCI5 equilibrim moles PCI5 initial PCI3 CI2 PCI5 1 moles of PCI5 initial 02660 03340 00 1 mole change 00660 1 00660 x 00660 mole equil 02000 quot 02680 xx 00660 mole 4 l I r quot Negative sign i39 X indicates moles of CI2 are l Change in moles of CI2 consumed change in moles PCI51 mole CI2 1 mole PCI5 Iquot J Equilibrium moles of CI2 initial moles CI2 change in moles CI2 Substituting the equilibrium moles into the equilibrium expression 00660mole PClsleq 7 i 0268mole mole 0 200mole 20 C 1303 Clzeq 2 m Example 2 Initially 300 moles of PC13 and 300 moles of C12 are placed in a 500 L container at 300 C Calculate the equilibrium concentration of PC15 PC15eq Kc is 246 Lmole Unknown PC15eq Let x moles of PC15 at equilibrium Knowns initial moles of PC13 300 mole initial moles of C12 300 mole volume of container CHEHS Chamcal Equilibrium 500L KC 246 Lmole Concepts chemical equilibrium molarity Substitute known values into equilibrium table PCl3 ClZ PCl5 initial 300 300 0 mole change mole equil x mole Use the concepts presented above to fill in the table PCl3 ClZ PCl5 initial 300 300 0 mole change x x x mole equil 300 x 300 x x mole Substituting equilibrium moles into the equilibrium expression a K 7 Pclslq 7 m 7 46 PCl3lzq C12q 1300 loor 39 mole 50011 50011 246x2 1976x 1 2211 7 0 For the quadratic equation ax bx c 0 the solutions are x 7 biVb2 4ac 2a Substituting x 7 bigb2 4ac 7 197Ei197E2 424622IZ 2a 2246 x 7 135 667 Although both roots 135 and 667 are mathematically correct only one root satis es the physical constraints of the problem and that root is 135 If root 667 is used then the moles of PC13 and C12 at equilibrium would be 367 300 667 and the equilibrium concentrations of PC13 and ClZ would be 0734 M A negative or zero equilibrium concentrationfor reactants orproducts is physically impossible CHEHS chamai Equilibrium x 7 135 male 7 males afPCls at equilibrium 135male PC 7 x 7 5 500L 500L 300 x 300 135 165male mg Q g 500L 500L 500L 165male Cl 7 2 500L The degree ofdissociation a ofB de2A change in moles of B initial moles ofB and the percent dissociation is 0L100 Example 3 If 300 moles ofB are placed in a 100 L ask at 25 C calculate the degree of dissociation 0L ofB Kc for B at 25 C is 0100 moleL Unknown change in moles ofB Let x change in moles ofB Knowns initial moles ofB 300 mole volume of ask 100 L KC 0100 moleL Concepts chemical equilibrium degree of dissociation Substitute known values into the equilibrium table B 2 A initial 300 0 mole change x mole equil mole The reaction must go to the right in order to produce A Moles of B will be consumed and thus the change in moles ofB will be x Use the concepts presented above to fill in the table B 2 A initial 300 0 mole change x 2 x mole equil 300 x 2 x mole Substitute equilibrium moles into the equilibrium expression CHE 115 chemical Equlllbnum 2x 2 KC 7 A22g 10011 7 0100 B 300 L 2g 1 100 400x2 1 0100x 0300 7 0 x 7 0261 0287 x 0261 mole change in moles ofB 7 change in males ofB 7 x 7 0261m0le 7 8 70102 initial males ofB 300m0le 300m0le I C Factors Affecting Equilibrium Concentrations at Constant Temperature Le Chatelier s principle when a system is in equilibrium a change in any one of the factors upon which the equilibrium depends will cause the equilibrium to shift in such a way as to diminish the effect of the change To determine if the reaction aAg bBg cCg dDg will go to the le or to the right to reestablish the equilibrium Q is de ned as Clcm Dldm Aain Brim If Q lt KC then the reaction will go to the right to reestablish the equilibrium Moles of reactants A and B will be consumed and moles of products C and D will be produced If Q gt KC then the reaction will go to the left to reestablish the equilibrium Moles of products C and D will be consumed and moles of reactants A and B will be produced 1 Addition or Removal of Moles of Reactants or Products Example 4 If 100 mole of PC15 is added to the equilibrium in Example 2 what will be the NEW equilibrium concentration of PC15 Unknown new equilibrium molarity of PC15 after the addition of 100 mole of PC15 Knowns equilibrium moles of PC13 C12 and PCl5 165 165 and 135 mole before the addition of 100 mole of PC15 see Example 2 KC 246 Lmole39 volume of the container 500 L chamai Equilibrium CHE 115 Concepts chemical equilibrium Q molarity Step 1 Evaluate Q anal determine whether the reaction will go to the left or right to reestablish the equilibrium PCl3 ClZ PCl5 initial 165 165 135 100 mole The moles in the table are the equilibrium values from Example 2 and the 100 mole of PCl5 added 235moiz Q 7 PCZSLH 7 5001 7 431 PC13m cl2m 1165m012 165m012 39 male 5001 5001 Q 7 431 L gt 246L KC male male Since Q gt KC the reaction will go to the le to reestablish the equilibrium Moles of PC15 will be consumed and moles of PC13 and C12 will be produced Step 2 Solve the equilibrium problem Let x change in moles of PC15 Since the reaction goes to the left to reestablish the equilibrium moles of PC15 Will be consumed and the entry for PCl5 in the row labeled change of the equilibrium table will be x PCl3 ClZ PCl5 initial 165 165 235 mole change x x x mole equil l65x l65x 235 x mole 235 gtC 7 PClszq 7 1 5001 7 2 i c PCZSLQ CHM 1165 w 39 155 a male 5001 5001 0492x2 262 1 1016 7 0 x 0361 mole change in moles ofPCl5 CHE 115 chemical Equilibrium PCZSLQ 7 235m0le x 7 235m0le 036m0le 7 199m0le 500L 500L 500L 2 Change in Volume of the Container Example 5 If the volume of the container in Example 2 is decreased from 500 L to 2 00 L will the moles of PC15 at the NEW equilibrium be greater or less than the moles ofPCl5 in Example 2 Unknown new equilibrium moles of PC15 after the volume of the container in Example 2 is reduced to 200 L Knowns equilibrium moles of PC13 C121 and PCl5 in Example 2 new volume 200 L KC 246 Lmole Concept Q molarity Step 1 PCl3 C12 1 PCl5 initial 165 165 135 mole 135m012 7 PClslm 7 1 200 L 7 0992 L pg m Cum 1165m012 165m012 male 2001 2001 Q 7 0992 L lt 246 7 K5 male male Since Q lt KC the reaction will go to the right to reestablish the equilibrium Moles of PC13 and C12 will be consumed and moles of PC15 will be produced The moles of PC15 at the NEW equilibrium will be greater than the moles of PC15 in Example 2 D Dependency of K on Temperature Plot of lnKp versus lT for the reaction N204 g 1 2 N02 g Em us mmin Plot of mm versus 1IT him 2 72 siuue AHR WK 4 mm g a i i 011031 0 on o 0053 00934 0 0035 ii 1N2 T K 1 mi Equation for a straight line is y Slupe zmemept lil c where C is a constant Consider K2 at T2 and K at T for a speci c reaction then Chamcal Equilibrium cm mag l C 4MP lli quot C we 111091 i Rearranging mag 111091 When AH gt 0 heat is absorbed and the reaction is said to be endothermic When AH lt 0 heat is evolved and the reaction is said to be exothermic Example Ipr 009823 at 20 C and 04554 at 40 C calculate AH for the reaction N204 g 1 2 N02 g AH T2 T139 In K 7 l l 1 In K Q2 R TZTI Q1 Rearranging KPH TZTl T2 T 04554 313K293K 7 58510 KP K 7 8314i1nl K 009823 20 AH 7 Rlnl 1 1 E Entropy and Free Energy CHEHS Chamcal Equilibrium The thermodynamic function entropy S is a measure of the disorder or randomness of a system Entropy is a state function The second law of thermodynamics states that an increase in entropy of the universe is associated with every spontaneous process The change in entropy of the universe is the sum of the changes in the entropy of the system and surroundings AS AS AS univ 3y surr 1 For a process to proceed spontaneously AS must be a positive number If AS umv is a negative 0 number then the process will proceed spontaneously in the reverse direction and when AS then the system is at equilibrium univ The change in the entropy of the surroundings ASsm is dependent on the heat that is transferred between the system and the surroundings For a constant pressure and constant temperature process AS M 2 surr T T T As an example consider the reaction N2 g 3 H2 g 2 NH3 g AHquot 924kJ ASquot 1985 JK The change in the entropy of the surroundings with the formation of two moles of NH3 is 92I4kJ 1000J AS51471 AH 1 k T 3 10 i T 298 K K and the change in the entropy of the universe is AS AS AS l9851 3101 1111 14an YyS 31477 K K K The entropy of the system AS 1985 J K decreases primarily because there are fewer product molecules than reactant molecules The reduction in number of molecules results in a lowering of the positional disorder Although AS for the reaction is a negative number AS niv gt 0 due to the large increase in the entropy of the surroundings and the reaction proceeds spontaneously to the right For a constant pressure and constant temperature process eq 2 can be substituted into eq 1 ASMV ASJ A3 3 W T T Multiply eq 3 by T and rearrange CHEHS Chamcal Equlllbnum TASumV AH TAS 4 Gt bbs free energy G is de ned as G H TS For a constant pressure and constant temperature process AG Comparing eqs 4 and 5 AH 7 TAS 5 AG 2 TAS univ A constant pressure and constant temperature process will be spontaneous when AS lt 0 m gt 0 or AG Example Calculate AG at 25 C for the following reaction H2 g 39 2 H g AH 436 kl AS 985 JK AG 7 AH TAS 7 436k 298K9851W 7 407k K 1000J F Reaction Free Energy For the generic reaction aAg bBg quot cCg dDg the reaction free energy AG is de ned as Partial Pressure of Gas C Standard ReaCtion Absolute Partial Pressure of Gas D Free Energy Temp P C P d AG AG RT n Db P a P f A B 39 Ideal Gas T 223 Free constant I Partial Pressure of Gas B P artial Pressure of Gas A A plot of G for the above reaction versus the extent of reaction yields Em us mmle Floi of 5 versus Degree of Reaction G slaps AS u u n o 5 10 A mm nZmuitsmk anlesm e momenta 39 quot h 393 namemu 4 elzmnlema dmaiewH Degree 0 Readan Thedegreeuf 39 H 39 39 39 39 39 le wright AG wnentne a and the reaction will is 10 only 39 39 a L L l L t L degree ofreaetion is 0 only the reaetants A and B are present AG 3 L L L L 1 c present 40gt U reaetion is 016 n10 a and he macnun 15 at eqmlzbrmm At equilibrium AG 0 and Fe 039 a n 1 AG 0 AG RT1n 6 KP is de ned as CHE 115 chemical Equilibrium PC Fade P PA my 7 Substituting eq 7 into eq 6 and rearranging AG0 2 RTlnKp 0r KP exp 7AG RT Relationship between Kc and Kp For the generic reaction aAg bBg quot cCg dDg PC 13 P PA 05 Assuming ideal gas behavior A P 7AltRT ART PB BRT PC CRT PD DRT Substituting K Pcgt2qltPngtdeq39 cimzqqmmdeq39 creqDdeq39RTcda b P Piveqwbeq lltARTgt2qltBRTgt2qlt 911me and KP KCR T where Ancdab G Standard Reaction Free Energy and Standard Absolute Entropy 1 AGquot Em us nominan VII Spectrophotometry A Absorption Spectrum of light absorbed by a substance quot L I CL i n t a in th ofthe incident light Example Absorption spectrurn ofan aqueous so1uton of CoHzog Absorption Spectrum Aquaous Solution 0 captionquot Absorbing species Cum201 52 Absuvbance B BeerLambert Law Beer s Law 1 A sbc The arnount of light absorbed by a solution is an exponential function ofthe concentration ofthe absorbing species 6 Ix man Gull CHE 115 Spectrophotomeh39y where PD is the radiant power energyareasec of the incident light of known wavelength P is the radiant power of the light a er it has been transmitted through the solution c is the molarity moles of solute liters of solution of the absorbing species and b is the path length Beer s Law P lo s bc g P 0 where s is a constant called the molar absorptivity The magnitude of 5 depends on the wavelength of the incident monochromatic light and the nature of the absorbing species Transmittance T T 1 Percent Transmittance T T TXIOO AbsorbanceA A logT 200 log T A sbc or 200 elogT ebc Note When A and b are held constant then s is constant and A or logT varies linearly with c The plot of A or logT versus c at constant k yields a calibration curve When c and b are held constant then s and hence A varies with k The plot of A versus 9 at constant c and b gives an absorption spectrum 2 Calibration Curve Plot of A or lo gT versus c at a speci c h Example Calibration Curves for Aqueous Solutions of MnO439 cm us icnl le Plot of logT versus Molarlty oi MnOquot 2 wle tD39 where MnO 10 15 rneansthatMno rblng species humanism Plot ui Ahsorbance versus Moarity of mmquot has ins ht 39 no t t t a i R 2 a MHDJ m selo M er s Law If 39L 39 curveto uc L 39 quot 39 L the plot oflogT or A versus c gives a straight line then Beer s Law is said to hold for the so Use the calibration curve to assist in the determination ofthe unknown molarity ofthe absorbing species in other solu 39 absorbin speciesinasolut39 Ion ion measure 0 determine the unknown Conc rat the an en ion ofan ce Am ofthe solution and read the concentration cm from the calibration curve see example above 3 Deviations for Beer s Law The absorbance at a constant 1 does not vary linearly with rnolarity a Chemical Apparent Deviation ii 3944 dissociative or associative reaction or in areaction with the solvent Example Consider the absorbing species D that dissociates into B and c in aqueous solutions D W 39 8 ac C W Em us where Althou initial molarity ofD eq 1 When solvent is added to asolution ofDt decreases but mg does not decrease linearly with mm due to the eq A ebD shim 7V 0 K a W 2 gh D obeys Beer s Law the absorbance A does n 3 humminan 1 D n uilibrium b polychromatic Light Beer s Law is wlid only when monochromatic light is used The incident light in 39 39 iman 39 39 quot quotof measurements are made in the 0 nm or greater For this reason absorbance region ofthe absorption spectrum where A changes little with a change in x um usually W en two or speci c 7 is the sum h at amaxlm c Mixtures ofAbsorbing Species asolutionL L L cuuu cuu Mixture mix and v n Jim at x ot vary linearly with the CHE 115 Spectrophotomeh39y At 7 410 nm A410 A410X A410Y 84101ch 84101bci At 7 495 nm A495 A495X A495Y 84951ch 84951bci Two equations are required to determine the two unknowns Cx and CV The molar absorptivities of X at the selected wavelengths 410 and 495 nm are determined from absorbance measurements of a solution with a known amount of X solution of X The molar absorptivities of Y are obtained from absorbance measurements of a solution with a known amount of Y solution on D Analytical Procedure To determine the concentration of an absorbing species in a sample solution 39 Obtain an absorption spectrum of the absorbing species to be studied 39 Select the wavelength 7 at which the absorbance measurements for the calibration curve will be made Usually 7 is selected at a maximum in the absorption spectrum Reasons A Greatest change in absorbance with change in molarity of the absorbing species A Least change in absorbance with change in 7 39 Determine if other species in the sample solution absorb at the chosen wavelength 39 Prepare a calibration curve A set of standard solutions of the absorbing species are prepared and absorbances of these solutions are measured at the chosen wavelength The absorbances are plotted as a lnction of the concentrations The concentrations of the standard solutions are chosen so that the T of these solutions lie in the range of 16 to 60 transmittance 39 And measure the absorbance of the sample solution at the chosen wavelength The concentration of the sample solution may be obtained from the calibration curve E Solution Calculations Solute is the substance in a solution in the smaller amount Solvent is the substance in a solution in the larger amount CHE 115 Spectropholomeh39y Solute Solvent gt Solution Example Aqueous solution of silver nitrate Silver nitrate AgNO3 is the solute and water is the solvent Concentration expressed in molarity Molarity of Solute Molarity solute oxalate W iters 0 so ution units M M L To determine the number of moles in a given volume of solution moles of solute Molarityliters of solution Example How many liters of a 0500 M AgNO3 solution stock solution are needed to prepare 100 mL ofa 0200 M AgNO3 solution new solution Step 1 Calculate the moles of solute needed for the new solution mole solute 7 Molarity liters new new mole AgNOS 7 Molarity AgNOQnEW litersnzw mole AgNOS 7 0200AI0100L 7 20010 2mole AgNO3 Step 2 Calculate the volume of the stock solution that gives the moles of solute needed for the new solution mole solute Molaritymk mole AgNOS Molarity AgNO 3 Stock litersxmk litersmk 20010 2 mole AgNOS 0500 Mstock liters 40010 2L stock Step 3 Add solvent to obtain the desired volume of the new solution 400 mL ofa 0500MAgNO3 solution 1 600 mL of H20 a 100 mL ofa 0200M AgNO3 solution


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