Atomic and Molecular Structure, Equilibria, Acids, and Bases
Atomic and Molecular Structure, Equilibria, Acids, and Bases CHEM 14A
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Date Created: 09/04/15
Weekly Lecture Schedule CHEM 14B Instructor Dr Laurence Lavelle SUMMER 2000 WEEKS 5 amp 6 Functional Groups Alkanes and Cycloalkanes Read Brown amp Foote Ch 13 Do Problems 41 42 43 Read Brown amp Foote Ch 21 to 27A Omit 248 Do Problems 21 22 23 25 26 28 29 33 35 36 40 43 Optional problems do not need to be handed in Review of molecular structure and bonding from Chem 14A Ch 1 Brown amp Foote Lewis structures formal charges resonance VSEPR model hybrid orbitals sp sp2 sp3 and their geometry39s molecular orbitals electronegativity dipole moments polarity After going through the readings amp problems and attending the lectures amp discussion groups you should be able to Identify and name functional groups 0 Identify primary secondary tertiary and quaternary carbons Identify and draw alkanes and cycloalkanes 0 Give IUPAC and common names for alkanes and cycloalkanes 0 Identify constitutional isomers o Distinguish and name alkanes alkenes alkynes o Recognize and name a simple haloalkane alcohol ether aldehyde ketone carboxylic acid given a molecular structure 0 Draw and interpret two and threedimensional representations of molecules and describe the structural information each reveals 0 Use ChemDraw and Chem3D to build manipulate and energy minimize molecular structures 0 Draw Newman structures and visualize molecules from Newman structures 0 Understand and give examples or identify the different terms used to describe molecular conformations dihedral angles staggered eclipsed gauche anti 0 Understand graphs of potential energy vs dihedral angle and why the potential energy depends on dihedral angles WEEKS 5 amp 6 CONTINUED Understand the terms torsional strain and steric strain and be able to identify strained conformations Calculate the ratio of two conformations given their relative energy AGO RT ln Keq Describe the conformations of cyclopropane cyclobutane cyclopentane cyclohexane and their relative stabilities Relate the degree of strain to the size of a ring Identify axial and equatorial substituents Visualize and draw the chair and boat conformations of cyclohexane and identify signi cant nonbonded interactions Identify and draw cistrans isomers in substituted cyclohexanes and determine their relative stabilities ChemOffice Worksheet 1 is due the end of week 7 Discussion Week 5 Lectures from 26 Apr to 3 May Ch 10 and 11 Ch 10 7 Homologous Recombination Homologous recombination 7 swapping genetic material between like DNA ie crossingover Just for fun check out pg 281 and this site to see the empirical evidence for crossing over httpusersrcncomjkimballmaultranetBiologyPagesHHarlequinhtml H olliday junction 7 shown on ppg 261 263 the Holliday model simply explains what we ve been drawing all quarter The X between single strands of DNA is formed by nicks that close with another homolog H eteroduplex 7 This is DNA that contains complementary strands from two different DNA molecules with similar sequences Double Stranded Breaks DSBs pg 265 When both strands break a more complicated model is required 1 A restriction enzyme clips dsDNA leaving a blunt end no overhang 2 Exonucleases chew up the DNA to leave a 3 overhand sticky end 3 Cut DNA anneals to homologous chromosome one Holliday junction is formed 4 New DNA is synthesized 5 9 3 on the cut DNA using the homolog as a template Note in a DSB new DNA is integrated into the old chromosome and no swapping occurs 5 Two Holliday junctions are formed the original DNA strand attaches the new DNA That s it for Ch 10 Dr Simons said he would emphasize selfleaming for this chapter As always read the book thoroughly Ch 11 7 SiteSpeci c Recombination and Transposition Recombination sites 7 places in a genome where crossingover is frequent Crossingover at these loci is called sitespeci c recombination Types of exchange at recombination sites are Conservative 7 both sites are recombination sites DNA is readily moved in and out of these sites Reversal of DNA directionality may also occur Insertions deletions and inversions are all conservative Transpositional 7 one site of insertion is a nonspeci c sequence ie not a recombination site EX Bacteriophage enter the lysogenic cycle in this manner Direct vs Inverted indirect insertion Direct 7 polarity of DNA sequence is maintained Inverted 7 polarity of DNA sequence is reversed Ex The polarity of one of these sequences in a promoter will change expression of the gene B 9A protein A will be expressed After conservations inversion of promoter sequence genome will look like this B A protein B will be expressed Direct multiple replication in circular chromosomes pg 307 An origin of replication Ori exists in prokaryotic circular genomes Replication of both strands starts here In order for these sites to create a new circular DNA two crossovers must occur 1 the loopingback Dr Simons drew on the board 2 a simple Holliday junction at a recombination site to split DNA into two genomes How do bacteria acquire antibiotic resistance Mutation anal recombination Wantanabe expts Shigella E coli 7 normally sensitive to killed by tetracycline tetR 7 tetracyclineresistant tetS 7 tetracyclinesensitive nice guy bacteria 1 108 tetS cells plated on media with tetracycline 7 all die 2 108 tetS cells and l tetR cell on media with tetracycline 7 all live In addition cells from expt 2 are resistant to streptomycin spectromycin and high Hg How did this happen Recombination This is how bacteria are able to spread a good gene to other bacteria when it s needed Bacteria Sex 7 this stu wasn t explicitly covered but it s fun Hey big guy nice plasmid might score points with some toothsome cytologist Bacteria pass genetic material by a plasmid a selfreplicating circular piece of DNA found in some bacterial cells F39 7 describes bacteria without a plasmid the women F 7 describes bacteria with one or more plasmids the guys Hfr 7 describes bacteria with a plasmid incorporated into the bacterium s genome the studs 1 The plasmid codes for pili which reach out to nd bacteria babes 2 Once connected the male s plasmids cross over into the female bacteria An Hfr bacterium will also incorporate some of his own genome in the conjugal visit The pili break and the plasmid is incorporated into the new bacterium making it an F and passing all genes in the plasmid eg tetracycline resistance 3 V Wantanabe s second experiment really looked like this 108 F39 cells 1 F cell on media with tetracycline are left to have the ultimate frat party for 12 hrs At the end of 12 hours the one F cell has given its plasmid to all 4x108 cells present Every cell living on the plate after 12 hrs has the plasmid with the tetracycline resistance gene Mechanisms oftransposition 7 ppg 315321 DNA transposition is s nonreciprocal incorporation of DNA into a new source A lysogenic Virus relies on its ability to incorporate itself into the host s genome and excise itself as needed We can imagine the mechanism of action for the Virus on pg 315 to look like this 1 The Virus breaks free of the hosts DNA 2 The target DNA is broken open with a restriction enzyme 3 The Viral DNA attaches to the sticky ends leaVing unpaired bases 4 DNA polymerases ll the gaps and ligase seals them Ch 14 7 Translation Going ahead of lecture Open Reading Frames ORFs 3 possible for ssRNA 6 possible for dsDNA For the following sequence of DNA several mutations are observed GATTACAGATTACA Which of the following mutations would be most detrimental to an offspring with the following DNA Hunk in terms of ORFs and get eXcited about LS4 Normal sequence GATTACAGATTACA GATTA GATTACA Deletion of 2 base pairs GAGTACAGATTACA Mutation of one base pair GATTACAG ACA Deletion of 3 base pairs 4 types of RNA listed chronologically are needed to translate the message from DNA to protein 0 hnRNA 7 heteronuclear refers to the primary transcript in the nucleus before modifications 1 mRNA 7 messenger transfers DNA into RNA 2 tRNA 7 transfer or translating transfers RNA into amino acids 3 rRNA 7 ribosomal transfers amino acids into proteins Protein is written in a degenerate triplet code Degenerate here means redundant Why 20 amino acids and a stop signal are required for translation but a triplet code means that 64 amino acids are possible Base pairs per amino acid Possible amino acids 2 l6 3 64 4 256 leat would be an advantage ofba Ving 6 base pair per amino acid One disadvantage 7 leat would be an advantage ofba Ving four base pairs per amino acid One disadvantage 7 Our triplet code allows for 63 amino acids and a stop signal Wli y do we onlyba ve 20 Although there are 64 possible codes present in DNA signi cantly less than 64 tRNAs exist This should help answer the above question Why The wobble position Codons are present in mRNA and complimented by an anticodon in tRNA Codon def 7 3 bases on mRNA that code for a speci c amino acid Anticodon def 7 3 bases on tRNA that code for the amino acid on tRNA Wobble position def 7 5 end of an anticodon 3 end of a codon a place in which unique base pairing can exist and a new more versatile base Inosine can be found A new nucleotide code is needed to transcribe this reglon Wobble position For example UUU and UUC both code for phe What does this mean It means that the we will need to synthesize signi cantly fewer than 64 tRNA molecules to code for the all 20 amino acids At the wobble position in tRNA a unique base exists to make translation more ef cient Inosine deaminated Guanine The unique code for tRNA at the wobble position is this Codon Anticodon U or C G G C U A A or G U A U or C I Ex Alanine in your reader Codons Anticodons GCU CGI GCC CGI GCA CGI GCG CGC This means that we only need two tRNA molecules to code for all possible alanine codons What s the minimum number of tRNA molecules needed to code for every amino acid You know it has to be greater than twenty because alanine and others need at least two but less than 64 because of this redundancy 3 sites exist in the rRNA A P and E A 7 Amino acyl charged tRNA binding site charged or amino acyl tRNA is bound to the mRNA Via a codonanticodon interaction P 7 Eeptide bond formation site the next peptide bond bt amino acids occurs here E 7 Exit site the last aa that has just left the rRNA complex it s codonanticodon are no longer bound CHAPTER 10 ACIDS AND BASES 102 a H30 b H20 c C6H5NH3 d HS e P043 0c1o4 104 a H200 CNaq 9HCN1 0Haq acidl base2 acid2 basel b H200 NH2NH2aq 9 NH2NH3aq 0Haq acidl base2 acid2 basel c H200 C032 aq 9 HC03 aq 0Haq acidl base2 acid2 basel d H200 HP042 aq 0 H2P04 aq 0Haq acidl base2 acid2 basel e H200 NH2CONH2 aq 9NH2CONH3 aq 0Haq acidl base2 acid2 basel 106 a Brznsted acid NH4 Brznsted base H503 b Conjugate base to NH4 NH3 Conjugate acid to HSO3 H2503 108 a as an acid H2P04aq H200 9 H30aq HP042 aq Conjugate acidbase pairs H2P04aq HP042 aq H30aq H200 as a base H2P04aq H200 9 0Haq H3P04aq Conjugate acidbase pairs H3P04aq H2P04aq H200 0Haq c as an acid H2C40aq H200 9 H30aq C2042aq Conjugate acidbase pairs HC204aq C2042 aq H30aq H200 as a base HC204aq H200 0Haq H2C204aq Conjugate acidbase pairs H2C204aq HC204aq H20l OHaq 1010 a acidic b basic c acidic d amphoteric 10141n each case use KW H30OH10x1014 H30 10x10 14 012 83 x 10 13 b 16x1010 c 43x1012 1016 a H30OH39x108 molL Kw H30OH 15 x1015 pKW 1482 b pHlog 39x108741 pOH 1018 KNH20 nominal concentration of KN39HZ KNH20 05gKNH2250 L1molKN39H25513gKN39H2 0036 molL Because KN39HZ is a soluble salt KNH20 K NH20 Where NH20 is the nominal concentration of NH2 thus K and N39HZ 0036 molL N39HZ reacts with water N39HZaq H20l7N39H3aq OHaq Because N39HZ is a strong base this reaction goes essentially to completion therefore N39HZ0 OH0036molL H30 KwOH 28 x 1013 molL 1020 a H30 antilog5 1x105 molL b H30 antilog23 5x103 molL c H30 antilog74 4x108 molL d H30 antilog105 3x1011 molL Acidity increases as pH decreases The order is thus milk ofmagnesia lt blood lt urine lt lemon juice 1022 a pH log003561448 pOH 1255 b le log 00725112pOH14001121286 c BaOHBa2 20H OH2x346x103molL692x103molL pOHlog692x1032 160 pH140021601184 d 109x103g5611gmol 194x10 4 molKOH OH194X104molL00100 L 194 x102 molL pOH log194x1021712 pH140017121229 e OH 100mL 2500 mL 500molL00200molL pOH log020017039 pH14000699123039 f H3050mL 250 mL 935 x104 molL 70 x105 molL pH log70x10541539 pOH1400415985 1026 a Kb CH32NH2 OHCH32NH Ka H30CH32NH CH32NH2 b Kb C14H10N2H OH C14H10N2 Ka H30 C14H10N2 C14H10N2H c Kb C6H5NH3 OH C6H5NH2 Ka H30 C6H5NH2 C6H5NH3 1028 Decreasing pKa will correspond to increasing acid strength because pKa log Ka The pKa values given in parentheses determine the following ordering CH32NH 1400419971ltN2H51400577823 lt HCOOH 375 lt HF 345 Remember that the pKa for the conjugate acid of a weak base will be given by pKa pr 14 1030 Decreasing pr will correspond to increasing base strength because pr logKb The pr values given in parentheses determine the following ordering N2H4577lt BrOl400869531ltCN1400931469 lt C2H52N299 Remember that the pr for the conjugate base of a weak acid will be given by pKa pr 14 1032 Any base whose conjugate acid lies below water in Table 103 will be a strong base that is the conjugate acid of the base will be a weaker acid than water and so water will preferentially protonate the base Based upon this information we obtain the following analysis a 02 strong39 b Br weak39 c HSO4 weak39 d HCO3 weak39 e CH3NH2 weak39 f H strong39 g CH3 strong 1034 In oxoacids with the same number of oxygen atoms attached to the central atom the greater the electronegativity of the central atom the more the electrons of the 07H bond are withdrawn making the bond more polar This allows the hydrogen of the OH group to be more readily donated as a proton to H20 due to the stronger hydrogen bonds that it forms with the oxygen of water Therefore HClO is the stronger acid with the lower pKa 1036 a H3PO4 is stronger39 it has the more electronegative central atom b HBrO3 is stronger39 there are more 0 atoms attached to the central atom in HBrO3 making the H70 bond in HBrO3 more polar than in HBrO c We would predict H3PO4 to be a stronger acid due to more oxygens on the central atom d H2Te is the stronger acid because the HiTe bond is weaker than the HiSe bond e HCl is the stronger acid Within a period the acidities of the binary acids are controlled by the bond polarity rather than the bond strength and HCl has the greater bond polarity due to the greater electronegativity of Cl relative to S f HClO is stronger because Cl has a greater electronegativity than 1 1038 a Methylamine is CH3NH2 ammonia is NH3 Methylamine can be thought of as being formed from NH3 by replacing one H atom with CH3 Because CH3 is less electron withdrawing than H CH3NH2 is a weaker acid and therefore a stronger base b Hydroxylamine is HONH2 hydrazine is H2N7NH2 The former can be thought of as being derived from NH3 by replacement of one H atom with OH39 the latter by replacement of one H atom with NH2 Because the hydroxyl group is more electron withdrawing than the amino group NH2 hydroxylamine is the stronger acid and therefore a weaker base 1040 The solution of 010 M H2SO4 would have the higher pH would be the weaker acid because the conjugate base HSO4 is less electronegative than the conjugate base of hydrobromic acid namely Br 1042 The smaller the value of pr the stronger the base39 hence aniline is the stronger base 4Chloroaniline is the stronger acid due to the presence of the electronwithdrawing Cl atom making it the weaker base39 and again we see that aniline is the stronger base 1044 The higher the pKa of an acid the stronger the corresponding conjugate base39 therefore the order is 3hydroxyaniline lt aniline lt 2hydroxyaniline lt 4hydroxyaniline No simple pattern exists but the position of the 70H group does affect the basicity 1046 a Ka 84 x104 x212x 1212 Here we have assumed that x is small enough to neglect it relative to 012 This is a borderline case We will also solve this exercise without making this approximation and compare the results below X H30 0010 molL pH 200 POH 1400200 1200 Without the approximation the quadratic equation that must be solved is 84 x 104 x212 7x x 0096 H30 X H30 00096 molL pH 202 b Ka 84 x104 x212 x103x The negative root can be eliminated X 11x103 H30 pH 296 pOH1104 c 84 x104 x212 x105 x The negative root can be eliminated pH 492 pOH 908 1052 a KaHCN49x1010 H30 antilog535x106molL Let x nominal concentration of HCN then Concentration HCN nominal x equilibrium x 5x10 6 H20 nom inal equilibrium H30 nominal 0 equilibrium 5x106 CN nominal 0 equilibrium 5x106 x 005 molL b Kbpyridine18x109 p0H5 2 OH 6x106molL Let x nominal concentration of C5H5N then C5H5N H20 C5H6N OH Concentration C5H5N nominal x equilibrium x 6x10 6 Concentration C5H6N equilibrium 6x106 Concentration 0H equilibrium 6x106 x 002m olL 1054 veronal H20 H30 veronalate ion The equilibrium concentrations are veronal 002000014x 00200020molL H30veronalateion00014x002028x105molL pHlog28x105455 Ka 39 x 108 1056 cacodylic acid H20 H30 cacodylate ion The equilibrium concentrations are cacodylic acid 0011000077x00110001100000080109molL H30cacodylate ion00077x0011085x105molL pH log85x105 407 Ka 66x107 1058 a CH3NH2 CH3NH3 The proton occupies the lon pair of the nitrogen in the conjugated base b First determine the concentration of methylam ine following dilution Moles of methylamine are found by 50 mL085gmL 425 g solution 425035149 g methylamine 149 g3106 gmol 0479 mol methylamine diluted to 1000 L the resulting solution is 0479 M in methylamine Concentration molL CH3NH2 H20 CH3NH3 0H initial 0479 70 0 equilibrium 0479 x i x x Ka 36 x 104 x2479x X 13x102 OH pOH19 pH140191 21 1060 a pHgt 7basic b 23pH7neutralCa2 is not an acid and N03 is not a base c pHlt 7acidic d pHgt7basic e pHlt7acidic f pHlt7acidic 1062 a Concentration CH3NH3 H20l 0 CH3NH2 H30 initial 025 i 0 0 change x 7 x x equilibrium 025x i x x See Table 102 for Kb of CH3N39H2 the conjugate base of CH3NH3 Ka KWKb 28 x1011 x225x 1225 X 26 x106 molL H30 pH 558 b Concentration 03 H20l 0 3HSO OH initial 013 i 0 0 changex ix x equilibrium 013x 7X x See Table 109 for Ka ofKa HSO3 Ka2H2SO3 Kb KwKa283 x108 x213xlt213 X 10 x104 molL OH pOH 400 pH 1000 c Concentration FeH2063 aq H20l H30aq FeH205OHaq initial 0071 7 0 0 change 7 x x x equilibrium 0071x i x x Ka 35 x10 3 x2071x x 0014 0018 The negative root can be discarded H30 0014 pH 85 1064 a Initial concentration of C6H5NH3 is 78g194133 gmol 350 L 24 M Given this initial concentration and the Ka for this acid found in Table 107 the percent deprotonation is found as follows Concentration C6H5NH3 H20l 0 C6H5NH22 H30 initial 024 i 0 0 change xi x x equilibrium 024x i x x Ka 23 x10 5 x224xlt224 X 23 x103 M H30 Percent dissociation 10 1066 a HSO3can act as both an acid and a base Both actions need to be considered simultaneously HSO3 H20 H30 SO32 HSO3 H20 OH H2503 Summing HSO3 HSO3 H2SO3 SO32 The simplest approach is to recognize that there are two conjugate acid base pairs and to use the HendersonHasselbalch equation twice once for each pair PH pKa1 log HSO3 H2SO3 pH Pka2 log SO32 HSO3 Summing 2 PH pKa1 Pka log HSO3 H2SO3 SO32 HSO3 5032 H2SO3 2 pH PKa1 pKa2 Therefore pH 2 PKaH pKa2 436 Note that it is not necessary to know the concentration of because it cancels out of the equation At extremely low concentrations of H503 however the approximations upon which the use of this equation are based are no longer valid b Neither silver ion nor nitrate ion is acidic or basic in aqueous solution Therefore the pH is that of neutral water 700 1068 a 0250molLKCN00350L01000L 00875molL CN0 Concentration H20l CNaq HCNaq OHaq initial 7 00875 0 0 change i x x x equilibrium 7 00875x x x Kb KwKa 20 x105 x20875xlt20875 Fundamental Aspects of Chemical Equilibrium 1 Chemical equilibrium is dynamic forward and reverse reactions are always occurring but at the same rate 2 No macroscopic evidence of change ie there is no net change in the amounts of reactants and products at equilibrium 3 Equilibrium reached by spontaneous processes which minimize the energy of a particular chemical system 4 A given equilibrium condition is the same regardless of the direction of approach Chemical equilibria are numerically represented by an equilibrium constant K The equilibrium constant is temperature dependent constant at constant temperature regardless of equilibrium amounts of reactants and products Important When a system approaches equilibrium the 39s or P39s change in such a way as to conform to the value of K at that temperature General Types of Equilibrium Problems calculating K given initial partial pressures or concentrations calculating equilibrium partial pressures or concentrations given K and initial conditions calculating new equilibrium partial pressures or concentrations following a shift from an initial equilibrium condition Example Calculating Kp A sample of gaseous PCls was introduced into an evacuated ask so that the pressure of pure PC15 would be 050 atm at 523 K However PC15 decomposes to gaseous PC13 and C12 and the actual pressure in the ask was found to be 084 atm Calculate Kp for the decomposition reaction PC15g PC13g C12g at 523 K The key to this problem is to determine how much PC15 is consumed in order to reach equilibrium ie we need to determine the degree of dissociation If we let X the amount of PC15 consumed then from the balanced chemical equation we see that X amount of PC13 formed and the amount of C12 formed The variable x is the degree of dissociation 0fPCl5 Also the problem states that at equilibrium Ptotal Thus PPC15 PPC13 PC12 084 atm Manipulating Equilibrium Expressions For a given equilibrium system the K for the reverse reaction equals the inverse of the K for the forward reaction If a balanced chemical equation is multiplied by a factor then the equilibrium constant is raised to a power equal to that factor If chemical equations are added together or subtracted from each other the equilibrium constants are multiplied or divided Equilibrium Expression Stoichiometric Coef cients Other Than quot1quot Suppose gaseous H2 and F2 are introduced into an evacuated bulb such that their initial partial pressures are both 100 atm A reaction then occurs in which HF gas is produced The reaction proceeds until equilibrium is established and the chemical equation is H2g F2g 2HFg Set up the expression for the equilibrium constant Kp Note the initial partial pressure of HF is O Example Calculating Equilibrium Conditions Large K The equilibrium constant at 350 K for the reaction Br2g 12g 211mg has a value of 322 Bromine at an initial partial pressure of 00500 atm is mixed with iodine at an initial partial pressure of 00400 atm and held at 350 K until equilibrium is reached Calculate the equilibrium partial pressure of each of the gases Let X amt Br2 reacted amt 12 reacted Therefore 2X amount of IBr formed Example Calculating Equilibrium Conditions Small K At 25 0C the equilibrium constant for the reaction 2N02g 2N0g 02g is 59 X 1013 Suppose a container is lled with nitrogen dioxide at an initial partial pressure of 089 atm Calculate the partial pressures of all three gases after equilibrium is reached at this temperature Let X degree of dissociation of N02 Thus 2X total amount of N02 consumed 2X amount of NO formed X amount of 02 formed K and Extent of Reaction or Equilibrium Position Consider aA bB cC dD K C Dd ATTBb What does the magnitude of K tell us If K gtgt 1 then C Dd gt A3 Bb therefore products are favored and equilibrium is said to quotlie to the rightquot eg H2g F2g 2HFg K115x102 The reaction mixture is predominantly HF product If K ltlt 1 then C Dd lt A3 Bb therefore reactants are favored and equilibrium is said to quotlie to the leftquot eg 2 NOClg 2 NOg C12g K 16 x 10395 The reaction mixture is predominantly NOCl reactant Q and the Direction of Change follows the law of mass action except pressures and concentrations are nonequilibrium values Q is not a constant It changes with time as a reaction system approaches equilibrium When 0 K the reaction system has reached equilibrium the comparison of Q to K tells you whether the reaction system will proceed towards products or reactants to reach equilibrium 1 When Q lt K too few products compared to the equilibrium condition Thus the net direction of change will be towards products to reach equilibrium 2 When Q gt K too much products compared to the equilibrium condition Thus the net direction of change will be towards reactants to reach equilibrium Example An evacuated vessel is charged with 010 atm of PCl5g 20 atm of PC13g and 60 atm of Clzg at 300 0C A reaction occurs an equilibrium is eventually attained according to PC15g PC13g C12g Kp 115 Which direction does the reaction proceed to reach equilibrium What are the equilibrium partial pressures Chem l4A2 The FactorLabel Method Fall 2000 The Factor Label method also known as Dimensional Analysisis a very powerful very intuitive problem solving method for Chemistry it is primarily used for unit conversions and for stoichiometry problems How it works You are given a problem with a starting point in units of quotXquot you need to get the answer which is in units of quotYquot In order to do this you must find one or more conversion factors that will get you to your final answer39s units A conversion factor could be quot12 inches 1 footquot or the mole relationships in a balanced chemical equation Often to get to the desired unit several conversion factors are used in sequence When the final conversion factor with the correct units is used the numerical answer is found by doing the math i V WN Example How many inches are in 32 kilometers Given the following conversions 1 kilometer 062137 miles 1 mile 5280 feet 1 foot 12 inches First draw an empty grid with what you are given in the top left and the unit of what you want after the equals sign Leave yourself plenty of room since you usually don39t know how many steps you39ll need 32 kilometers 2 inches Rules of the grid everything on top is multiplied everything on bottom is divided the vertical lines are used to separate terms and all relevant rules pertaining to significant digits apply here as well 500 500 300 We have no conversion factor between kilometers and inches but we have one for kilometers to miles so 32 kilometersI 062137 miles I inches so 500 x 5003oo would look like this 2 833 I l kilometer Note that the units of kilometer cancel and we are left with the unit of miles 32kirometei s1 062137 miles I lkilometer I which leads us to the next conversion and the next 32 kilometersI 062137 miles I 5280 feetI12 inches mches 2 inches I l kilometer I 1 mile I 1 foot all units except inches are canceled out now do the math 32 kllometersI 062137 mlles I 5280 feetI12 mches 13 X 106 inches I lkilometer I lmile I lfoot same as above just with the units canceled for clarity 32kilometersl 062137miles I5280feetI12 inches 6 I 13X10 mches lkilometer lmile I 1 Using the Conversion factors found in your text Convert 125 x 10392 metric tons to pounds 125 x 10392 metric tonsI 1000 kg 1 1b 276 lb I1 metrrc ton I045359 kg CHAPTER 9 CHEMICAL EQUILIBRIA 92 a True b False Changing the rate of a reaction will not affect the value of the equilibrium constant it merely changes how fast one gets to equilibrium c True d False The STANDARD reaction free energy is not 0 at equilibrium The reaction free energy which depends upon the concentrations of the products and reactants is 0 at equilibrium 96 a K PN022PNO2 P02 b K PSbCl3 PCl3 PCl2PSbC15 c K PN2H4PN2 PH22 98 All values should be the same because the same amounts of the substances are present at equilibrium It doesn t matter whether we begin with reactants or with products the equilibrium composition will be the same if the same amounts of materials are used If different amounts had been used only e would be the same in the two containers A more detailed analysis follows H2g Br2g 2 HBrg The equilibrium constant expression for this system is K HBr2H2Br2 1 In terms of the change in concentration X of H2 and Br2 that has come about at equilibrium we may write K 2X205X05X 2 in the rst container and in terms of the change in concentration y of HBr that has come about at equilibrium in the second container we may write K lOy2y2y2 Because KC is a constant and because the relative amounts of starting materials are in the ratio of their stoichiometric factors we must have K 2x205 x2 lOy2y22 2x05 x lOyy2 Cross multiplying Ky 005X010y Ky 0005 005 y010x xy y 0102X 3 This may also be seen by solving quadratic equations for Eqs 1 and 2 for X and y to obtain any value of KC a Br2005X in the rst container Br2 y 2 in the second container Because y2005X satis es the conditions of Eq 3 the concentrations and hence the amounts of Br2 are the same in the two cases b H2005X y2 as above hence the concentrations of H2 are the same in both systems c Because all concentrations are the same in both cases this ratio will also be the same d For the same reason as in part c this ratio is the same in both cases e This ratio is the equilibrium constant so it must be the same for both systems 1 Because all the concentrations and amounts are the same in both cases and because the volumes and temperatures are the same the total pressure must be the same P nH2 nBr2 nHBr RT V 910 K PH2S PNH3 Forconditionl K 0307 X030700942 Forcondition2K 0364X 0258 00939 Forcondition3K 0539X 0174 00938 912 a Q C1 2ClO3 ClO3 b Q CO2 c Q HCH3COOCH3COOH 928 H2g C12g 2 9 HClg KC 51 X 10 8 51 X 10 8 HCl2H2Cl2 51 X 10 8 145 X 10 32H2245 X 103 H2 145 X 10 3251 X 10 8 245 X 103 H2 17 X 1012 molL 930 K PSbCl3 PC12 Pst15 35 X 104 502 X 103 PC12 0072 PC12 35 x 10 4 0072 502 x 10 3
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