Unit 5 notes
Popular in Chemistry 152
Popular in Chemistry
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92514 Unit2 How do we determine structure The central goal of this unit is to help you develop ways of thinking that can be used to predict the atomic and molecular structure of substances Using spectroscopy to derive structural information M1 Analyzing LightMatter Interactions Deducing atom connectivity I 39239 Lookmg for Patterns based on atomic structure M3 Predicting Geometry Predicting the three dimensional geometry of molecules Analyzing the distribution of electrons in molecules M4 Inferring Charge Distribution Chemical Thinking 39 39 Unit2 How do we determine structure Module 2 Central goal To use experimental LOOkmg for spectroscopic data Patterns together with information about atomic composition and electron configurations to deduce atom connectivity in a given molecule Chemical Thinking 39 39 The Challenge Molecular structure is the fundamental clue to determine the identity and properties of substances O H Ll The structure of a molecule HCVCQ C OH Aspirin Is determined by both n39c39Cc O atomic composition and H AIL H CQHso4 atom connectivity O How can we determine what is bonded to what in the molecules of a given substance How can we predict how many bonds may exist between the different atoms m E x E 39 Tc 2 E a 0 Chemical Thinking 39 39 A Useful Model The idea that molecules are comprised of atoms connected to each other in some way and arranged in particular geometries in space has been very useful to explain and predict the properties of chemical substances DNA What holds the atoms together 92514 Chemical Thinking 39 39 A Bonding Model Atoms are assumed to be bonded or connected to each other in molecular compounds How do we model it 1 Pairs of atoms in molecules are attracted to each other through forces that we call covalent bonds e39ectmn nucleus The formation of these bonds is induced by electrostatic attraction between electrons and protons in the different atoms attraction bond length on E x E 39 Tc 2 E a 0 Let s Think III III Go to httpllwwwchemarizonaedultpplchemthinklsim gt Analyze the formation of a covalent bond between two hydrogen atoms using the Covalent Bond simulation What happens when you move the atoms closer to each other What happens if you try to have the nuclei touch each other How easy is to break the bond A pair of e perbond CLICK AND DRAG ONE OF THE ATOMS 92514 I Let s Think D gt Use the simulation Bond Energy to analyze the changes in total internal energy ET when the two atoms are brought closer to each other How would you explain the changes in energy that you observe How would you use the information from the graph to estimate the bond length and the bond strength for the hydrogen molecule H2 ET kJmol Chemical Thinking 39 CLICK AND DRAG THE RIGHTHAND ATOM III Quantum Model Electrons do not behave as classical particles We cannot know their position and velocity simultaneously Electron Densities Measure of probabilities Heisenberg Uncertainty Principle Bond formation is favored by 0 Increased electron density between atoms 0 O 1 Electron delocalization CLlCK AND DRAG THE RlGHT HAND ATOM Chemical Thinking 39 Dynamic Nature Chemical bonds are not static entities The atomic nuclei and electrons involved are constantly moving The atoms in a bond vibrate around their equilibrium positions with a frequency that depends on the strength of the force between the atoms and their masses m E x E 39 Ti 2 E a c 0 92514 Analyzing Vibrations Molecules experience a wide variety of vibrational motions characteristic of their component atoms and the bonds between them Symmetrical Antisymmetrical Scissoring Rocking Stretching Stretching Energy Vibrational states are quantized and hv 9R transitions between them can be induced by providing the right AE lAE Chemical Thinking Infrared Spectroscopy l The absorptlon of IR radiation at 4 specific frequencies can then be I used to detect the presence of 39 39 IR If n spec c b0 ds Spectrophotometer 3 K c V 1165 cmquot V g 5 2705 cmquot CH2 wag E E 1435 1 IS 250 sym stretch CHcm 1250 1 IE E scisszor CHchock a o E x Y 2 asym stretch 1759 m391 GE Formaldehyde 552 311 4000 39 303900 39 2000 I 1000 39 500 0 Wavenumber 1m lncreasin Ener 9 9y i Waven um ber 0 0 CCQN 121 cm1 C N Cic 4000 N H O H 3200 2800 2300 2100 1800 1500 Fingerprint 1 l I 1 2 3801 1 114601380 l cm 1 4000 3000 CO 2000 nuon 1000 75l AK h W a R39X if II 391quot g 7 II p O39H cH fr 1 l E l l fr CC H1 1quot u 7 1 If 1 l UI l 1 f LR c v l 1 1391 0 1M cH l l E 7 J2 lb Ll I I g 3 O O O 2 O O O 1 O O O 0 Wavenumber cm l 92514 RMS The combination of IR Spectroscopy and Mass Spectrometry is a powerful tool of analysis 02 CH 04 stretch 0H stretch Transmittance 3000 2000 1000 Wavenumber cm L The elemental analysis of a poison m 31 that causes blindness QwLet 5 think reveals the formula CH4O What can we say about the molecular structure of this compound 1390 2390 35 39 40 N O I n O I a O A Relative Abundance N C 3 T 01 39 Chemical Thinking 39 39 a D 3 i Let s Think 43 IRIMS breath analysis can be used to explore fat metabolism for people on a diet on O o O I b O I 58 Relative Abundance N P 15 D II I i II I I i I l l l 20 30 40 50 SD mq H O The metabolism of fat produces a volatile compound with the formula C3H60 P a 9 m i P o i CH stretch CO Transmittance What is the molecular structure of this 9 3000 2000 1000 compound Wavenumber cm ti 17 N bend stretch Chemical Thinking 39 39 Bonding Capacity halogens have 1 bond lnferences about 5 carbon always has 4 atom connectivity are I II I facilitated by f39 4 quot N has 3 analyzing the bonding patterns at different oc ClP Cl a4 haw tapaway they aFe feltem nelele O I types of atoms Cl I F gasses Let sthink C39 c39 o F non metals covalent bonds Analyze the structure of these 0 molecules and identify bonding u I I patterns for different atoms How is C l339l bonding capacity related to the position of the atom in the periodic table m E x E 39 Ti 2 E CD 0 92514 fixed bonding capacity or number of bonds they can torm Molecular compounds result from the combination of nonmetallic elements 395 by column A 3 9 1 Valence Valence 4 3 2 1 0 6789 CNOF 15 16 17 P 8 CI 35 Se Br 53 Nonmetallic elements 35 At Valence is a periodic property elements from the same group behave similarly Chemical Thinking 39 39 Let 3 Think decant filter distill then run IR and mass spec Let s apply what we have learned to solve a prototypical identification problem A substance found in most salad dressings may be responsible for the development of stomach ulcers The substance is normally found dissolved in water and is most likely a V liquid in its pure form Propose a method to isolate separate the substance from salad dressing Once isolated what would you do Chemical Thinking 39 39 L t Th k m molecular formulaC2H402 e s in 45 43 m D I Elemental analysis reveals that the empirical formula of the compound 1 60 a a N O L 01 Relative Abundance D is I l i i J 10 20 30 40 50 60 7O mq What is the 2 molecular 08 392 structure of this m 9 E 06 C H 5 compound 3 stretch I 04 C 8 a 02 039 co co stretch stretch stretch g 3000 2000 10100 O Wavenumber kmquot 92514 Looking into Atoms In order to better infer and predict the molecular structure of a compound we need to understand why different atoms form different numbers of bonds Covalent bonding is Data on the result of interactions between Atomquot Rad39us 5 electrons and and 35 pmtons in the Ionization Ener 939 E bonded atoms Thus gy 39 5 exploring the internal 73 structure of single E atoms may provide 2 important clues are very revealing 0 Atomic Radius DerIOdIC table The size of atoms is not well defined given deCreaSG that we cannot know the position of In rease electrons with precision Atomic radius is derived v as It 9098 top to bOttOm Inc indirectly by measuring distance between bonded do 39 2 V atoms in molecules 3 V c May XRay Crystallography E source X V I 1723 6 Crystal Structural 9 D Information E 15 g 0 Diffraction Pattern Atomic Radius 250 Rb GrouplA GroupZA Group3A GroquA Group7A E 200 K m 2 us 116 m a Na gm Li Li Be 3 o F t 128 I 96 34 ea 57 39g 10 Ar Kr E i E 50 Ne He a o o 5 1o 15 20 25 so 35 4 N3 ME Al 5 CI AtomicNumber 155 r 141 139 121 102 K Ca Ga Se Br F 176 122 120 120 i Periodic Behavior R increases 1 pm 1 x 1012 m 1 x 103 nm 92514 Ionization Energy gt Insight into atomic structure can also be gained by decrease gt Increase analyzing the energy required to remove a first electron from a gaseous atom banal H 0 Op ICI 336 MgMge39 First Ionization Potential 2500 39 He 2 Ne b e g o g E 1500 Ar Kr x g E S moo PerIodIc BehaVIor 2 l E g 500 lt LI 8 39 a K Rb quot o GE 0 5 1O 15 20 25 30 35 40 C AtomicNumber 0 IE Increases 2 Let s Thmk K Rb closeLtheLad1usthehaLdeL Lstopull apart E Na 39 bigger the radius weaker the attraction What do the 100 K periodic trends for A 39 a atomic radius 50 Ne and b first 0 He a ionization energy 5 1 immijiumb 3 35 4 suggest about the 2m 395 electron structure A quote Ne b E of different atoms 3 2 x 2 1500 Ar KI39 2 What atomic model I would allow us to WW 3 explain these m U N quotg trends 0 a K Rb g 0 g 10 1 5 20 2395 3390 35 40 o AlomicNumber 2500 Shequot MOdel 2000 He Ne Shell gt of electrons g g 1500 n 1 2 g 1000 H gt 0 IE 50 Li Na n2 gt 8 0 0 1390 15 2390 2395 3390 3395 40 n 3 gt 8 Atomic Number I n4 gt 1 8 an IE 39 39 9 9 I x E E H He Li Ne l o 73 4 g L E E 6 o o 92514 Looking Deeper Xray h39v Additional clues about KE e electron structure can be derived using 1 a 39 39 photoelectron spectroscopy V 39 Ionization Energy hv KE The n th Ionization Energy is the energy needed to remove the nth electron from an atom This technique can be used to determine the number of electrons in each shell Chemical Thinking III Photoelectron Spectroscopy W I P e39e quot quot e quota x ionization energy further left is larger and i222 harder to remove bw peaKs is difference in H F x E 5 C E boron 2 2 1 NOT 2 3 2 2 2 GE What are these results telling us about electron carbon 5 5 5 Let s think configurations in the different atoms sodium 2 2 6 1 Modified Shell Model The trends in the results from photoelectron spectroscopy suggest that electrons arrange in subshells within a shell Shell Subshell of e39 n 1 s 2 e Electrons in an atom occupy different CD energy levels In the 35 n 2 2 S 2 9 ground state of an E 2 p 6 e atom the lower energy I levels are occupied g n 3 3 s 2 e39 rst E 3 p 6 e 1s 9 2s 2p 9 3s CD 5 3 d 10 e39 Chemical Thinking 39 Quantum Mechanical Model The existence of atomic subshells is predicted by the modern quantum mechanical model of an atom In fact this model reveals an even more complex distribution of electrons within atoms E P S O O n3 8 3p 0 quot2 25 Zpl va n1 15 Subsubshells 1 s 3 up 5 lid Orbitals with a maximum occupancy of 2 electrons Chemical Thinking 39 Occupancy The relative energy of different orbitals determines their filling sequence Electrons always occupy the lowest energy levels available 15 25 2p 35 3p 45 3d 4p 55 4d 5p 65 gt 5d 6p 75 5f gt6d gt 7p m E x E c 39 Tc 2 E a c 0 Electron Configurations H 9 1s1 He 9 1s2 Li 9 1s22s1 He 2s1 Be 91 s22s2 He 2s2 B 9 1s22s22p1 He 2s22p1 Photoelectron Spectra Spectral Mode How are electrons distributed in N Ne Na P Ar and Sc What periodic trends do you observe Let sthink 92514 N gt He 25quot2pquot3 Ne gt He 23A2 2pquot6 Na gt Ne 3sA1 P gt Ne 3sA2 3pquot5 Ar gt Ne 3sA2 3pquot6 WW 10 Chemical Thinking 39 39 Recognizing Trends 51 szd1d2 d3 d4 d5 d6 d7 d3 d9d1o I313932 3933 I34 3935 p6 5bl0Ck Elements E d block elements transition metals p block elements 92514 Chemical Thinking 39 39 Let s Think 393 I Photoelectron Spectra 39 Spectral Wl39lte the Mode electron configuration of Al and sketch is expected photoelectron spectrum AI gt Ne 3squot2 3pquot1 Ne gt 13A2 23A2 2pquot6 m E x E 39 Tc 2 E a 0 Understanding Bonding The bonding capacity or valence of different atoms can be explained based on their electron configurations lConsidercarbon C He 2s22p2 I We can divide its electrons into two main groups CORE electrons In lower energy levels spend Core e more time close to the nucleus not very accessible VALENCE electrons In high energy levels farther from the nucleus exposed to interaction with other atoms Valence e39 11 92514 Delocalization feHn te fill euennesi sheH blp wih bonding As two atoms get close to each other eIectronS delocalization of electrons favors the formation of chemical bonds The more valence electrons are delocalized through bond formation the lower the energy of the molecule that may form Electron delocalization is constrained by the number of available unfilled valence orbitals that the delocalized electrons can occupy H1s1 O 2322p4 Electron Density Law High Chemical Thinking 39 39 covalent bond sharin of2e Let 3 Think HCH Based on the electron configurations of these atoms H C and C CICCI one also above nd below predict the structural formula of the molecules that will form when combining the following pairs of atoms H and H H and C C and Cl Chemical Thinking 39 39 valency how many bonds form Valence Electrons valence e add up electrons not In noble gas Valency I 4 3 2 1 valenceeH3 4 5 6 7 1 11 B C Valency can be explained if we assume that Sl The number of covalent bonds that each atom forms is determined by the number of valence electrons that the atom can have in its valence shell m E x E 39 Tc 2 E CD 0 12 Chemical Thinking 39 A Powerful Rule OCTET RULE When atoms combine OFTEN the most stable structures are those in which each atom has a full valence shell Given that for most nonmetallic atoms a full valence shell corresponds to eight electrons this statement is known as the octet rule HAPPY ATOMS NEEDY ATOMS GREEDY ATOMS 92514 Chemical Thinking 39 Maximum Occupancy The limit in electron occupancy of atomic and molecular orbitals has a large impact on the molecular structure of chemical compounds S Experimental data suggests that this phenomenon is due to an intrinsic property of electrons that constraints the number of these particles that can simultaneously occupy the exact same region of space m E x E c 39 Tc 2 E CD 0 Exclusion Principle Wolfang Pauli proposed that no two electrons with exactly the same properties could occupy the same orbital simultaneously Pauli exclusion principle The only way this could it Let s Think happen was if each electron had a different How would the bonding spin Given that electrons capacity of oxygen can only exist in two atoms change if an possible spin states no electron had 3 spin more than two electrons states would we have can simultaneously be in an octet rule the same orbital 13 Assess wh Chemical Thinking 39 Let39s apply at you know 92514 3amp0 Let39s apply Breath analysis of mouth air using GCIMS reveals the presence of three compounds assumed to be responsible for halitosis bad breath A comparison search in a MS database allows to identify 3 1 IN J V 1 room air mouth air incub saliva Retention Time 39 the substances as 1 H28 2 CH4S 3 CZHGS Chemical Thinking 39 Propose a molecular structure for the first two compounds Justify your decisions based on electron configurations of the atoms involved 3amp0 Let39s apply 1quot 08 Based on your previous analysis and the additional information provided propose a Lb Transmittance P a 02 ch I I 000 2000 1000 Relative Abundance molecular structure for the third Wavenumber cnr39 compound 100 CZHGS 30 47 62 m E x E 39 Tc 2 E a 0 14 92514 a 5 Describe one important application of what you have learned in this module Chemical Thinking 39 Looking for Patterns Summary The formation of covalent bonds between atoms is the result of electrostatic interactions between electrons and protons in the bonding atoms Molecules experience characteristic vibrational l w motions The absorption of EM radiation at specific frequencies can then be used to detect the presence of specific 3500 20 who bonds W 1 TRANSMlTTANCE o a Chemical Thinking 39 Looking for Patterns Summary Models of atomic structure can be used to make predictions about electron configurations In turn electron configurations allow us to make predictions about bonding behavior When two atoms combine their valence electrons are reorganized The most stable structures tend to be those in g which each atom has eight E valence electrons full valence shell octet rule l O v l O I on 00 00 l O 39 I O O m E x E c 39 Tc 2 E a c 0 15 c E x E 39 Tc 2 E a O 92514 For next class Investigate how we can use the octet rule to derive the molecular structure of simple compounds How can we use the molecular structure to predict the molecular geometry of a molecule 16 92514 Unit2 How do we determine structure The central goal of this unit is to help you develop ways of thinking that can be used to predict the atomic and molecular structure of substances Using spectroscopy to derive structural information M1 Analyzing LightMatter Interactions Deducing atom connectivity based on atomic structure M2 Looking for Patterns Predicting the three dimensional geometry of molecules IM3 Predicting Geometry Analyzing the distribution of electrons in molecules M4 Inferring Charge Distribution Chemical Thinking 39 Unit2 How do we determine structure Module 3 Central goal P d t To deduce the Lewis a re 39c mg structure of molecules and E Geometry predict their three 5 dimensional geometry 5 k based on the analysis of 73 k H 3 the number and type of E t k 1 valence electron pairs 2 a surrounding each atom 0 The Challenge The properties of a substance are determined by the structure of its molecules Molecular structure depends on Q VL K 39KQL A Atomic Composition k k 35393quot Atom Connectivity L 5 9 8 4 Molecular geometry How can we predict molecular geometry given information about atomic composition and atom connectivity m E x E 39 Tc 2 E a 0 Chemical Thinking 39 Electron Distribution When atoms of nonmetallic elements bond their valence electrons are reorganized The number of covalent bonds that are formed are determined by the most stable electron configurations typically full valence shell The task of predicting the molecular geometry of the UserI T00 molecule formed can be LeWiS E39eCtron39dOt simplified by following a Structures systematic procedure to derive the molecular 39 structure of any given chemical compound 92514 Chemical Thinking 39 Electron Dot Diagrams We first need to learn how to represent single atoms Orbitals 1 The atomic nucleus and core H C F electrons of an atom are represented by using the l l 1 chemical symbol of the atom 2 Each valence electron is F represented using a dot placed H around the chemical symbol 3 Valence electrons will singly l occupy empty orbitals when available and then pair up H m E x E c F 76 2 E CD 0 Electron Pairs Electron dot diagrams for atoms explicitly show the number of unfilled energy states available for bonding Thus these representations help us make predictions about the bonds that an atom will form with other atoms E if F F gt quot quot 00 I F F LeWIs quot 1 Structure Bonding F2 Electron Pair lone Chemical Thinking 39 Letps Think As you know the chemical elements oxygen and nitrogen also exists as homonuclear composed of the same type of atoms diatomic molecules v How would you explain the particular Lewis structure of the O2 molecule Build the Lewis structure of the N2 molecule based on the number of unfilled valence energy states in the nitrogen atoms 6 92514 Lewis Structures There are some simple rules that facilitate the creation of Lewis structures A Choose the central atom never H it forms only one bond The central atom tends to be the one with the highest bonding capacity valency O is central in this case m E x E 39 Tc 2 E a 0 a E g 2 Count valence electrons H 1 and O 6 C 2 Total 2 x 1 6 8 valence electrons 2 This electrons will organize in 4 pairs GE spin pairing to minimize energy 6 Lewis Structures 3 Use as many pairs as needed to form single bonds between the central atom and the surrounding atoms 0 Each bond line represents a pair of electrons 4 Use the remaining pairs to satisfy the full valence shell rule in each atom as needed Start with terminal or outside atoms but not if H place any leftover electrons 8 on the central atom 396 e 939 Chemical Thinking 39 Lewis Structures Let s consider another case Carbon dioxide C02 1What is the central atom C 2 How many valence e39 How many pairs 4 2 x 6 16 valence e39 9 8 e39 pairs 3What is the backbone 0 6 0 4 How do we distribute the e39 pairs left O C 5 How do we satisfy the octet rule for all atoms A O 0 Form double u bonds 399c9 gt 96 92514 Chemical Thinking 39 Let s Think A variety of substances contribute to indoor air pollution Among the most common we find Build the Lewis structures of the following greenhouse gases CH4 CO NH3 CHZO STRATEGY 1 What is the central atom 2 How many valence e39 How many pairs 3 What is the backbone 4 How do we distribute the e39 pairs left 5 How do we satisfy the octet rule for all atoms m E x E 39 Tc 2 E a 0 More Central Atoms For molecules with more than one central atom we can apply the same procedure atom by atom However the task is simplified in molecules made of C H O N H I I Cl C E Bonding Patterns N N NE I 6 0 Chemical Thinking 39 Let s Think Chemical scientists often use semistructural formulas that convey partial information about the distribution and connectivity of atoms in molecules Ethanol CH3CH20H Acetic Acid CH3COOH Build the Lewis structures of these molecules Chemical Thinking 39 Interesting Cases For some molecules the derivation of their actual Lewis structure is not so straightforward Consider for example the ozone molecule 03 which plays a central role in our atmosphere Experimental data indicates that both bonds in the 03 molecule have the same length but the value is intermediate between those of single and double bonds Bond Length pm 0 0 148 How do we 03 1278 explain it 0 121 m E x E c 39 Tc 2 E CD 0 Let s Think Build the Lewis structure of 03 This molecule illustrates a structural feature that we need to take into account when deciding how to distribute electrons among atoms in a molecule What is it 92514 single bond always longest but 03 falls between 92514 2 lewis structures for one molecule means that you M I I H b 39ds ecu 3quot y r39 write both with a double headed arrow between The structure is a hybrid of Resonance Structures Intermediate 02020 3 e39 pairs I 2 bonds between single and double Resonance structures are drawn when a single Lewis structure cannot represent the actual electron distribution in a molecule Chemical Thinking Resonance In molecules that exhibit resonance the electrons are delocalized over the entire system This delocalization tends to stabilize the molecule reduces its potential energy Benzene C6H6 H H H l I H H C K C H Resonance C v C C C Iv ll lt gt II H H Hybrid HCCCH H CCH H l l Chemical Thinking Let 3 Think We often Which of these pollutants exhibits resonance stabilization How many resonance structures do they have m E x E 39 Tc 2 E a O Molecular lons and Radicals Many important molecular species in our surroundings and inside our own body are either charged particles ions or have unpaired valence electrons radicals Lewis Structure H3O are build using the same valence e H30 3x16 18 strategy but paying close H H attention to the I gt I number of HQH HQH electrons and their distribution Chemical Thinking 39 92514 positive charge means that its more positive so subtract 1 less electrons more positive atomic GleCIl Ol lS neg charge means more negative more electrons add g Let s Think Build the Lewis structure of the Carbonate Ion c032 Identify potential resonance structure Chemical Thinking 39 2 Let s Think Radical species also called free radicals have unpaired valence electrons and can be neutral or charged particles In general they are easily recognizable because they have an odd rather than an even number of valence electrons Build the Lewis structure of radicals No2 Identify potential resonance structure a E x E 39 Tc 2 E a c 0 very reactive 92514 Electron Repulsion Once the Lewis structure of a molecule is derived its geometry can be predicted applying a simple principle Regions of high electron density around any single atom will be located as far as possible due to electron repulsions Valence Shell Electron Pair Repulsion VSEPR Theory Minimizing repulsions allows us to find the most stable shape lower energy Chemical Thinking 39 9 7i Let 3 Think pontrnl ntnm le39I39F39l39I39PQ Consider the following Lewis structures for has tWO eleCtron domalnS these molecules in our atmos here 39 639 CF20I2 has 4 How many regions of 6 o u l 3 high electron density c9 39E I E electron domains QI 3 395 do you identify 0 4 5 around each central ll atom C 90 9 A E 4 I How will these 8 regions be located in r E space due to electron b39 L g repulsions lone pairs count too so do bonds lone pair molecular has diff name than electron pair geometry Bent in 03 case less than 120 Molecular Geometry e39 Example e39 pair Molecular regions geometry geometry 2 u Linear 9c9 180 00 Linear 3 Trigonal Planar I C 120 HH c 118 J Trigonal planar 3 Trigonal u Planar 9 9 lt120 1168 Bent or Angular m E x E 39 Ti 2 E a 0 92514 Molecular Geometry one lone pair trioonal ovramid less than 109 e Example Hair Molecular 2 lone pairs bent or angular regions quot geometry geometry less than 109 4 CF Tetrahedral CI I o EE 109 ka QF Tetrahedral 39 4 Tetrahedral 2 ll lt1090 H a o E N HltI Trigonal Pyi1a01il TE 4 Tetrahedral 7 lbl lt 109 7 H e H 1552 g BentorAngular 0 Leti 3 Think nitrogen can be bonded to other nitrogens so Apply VSEPR theory to derive the molecular central atom geometry of the following atmospheric molecules 02 03 CH4 N20 Estimate the bond angles in these molecules STEP 1 STEP 2 39l STEP 3 NH gt H N H gt N gt gt Follow 3 HPH H H 2 J the H H sequence Molecular Lewis Electronepuir Molecular geometry formula structure geometry triangular pyramid tetrahedral Chemical Thinking 39 Larger Molecules The same ideas can be applied to deduce the molecular geometry of larger molecules The task is simplified by recognizing the following patterns for some of the most common central atoms 39 Carbon C Nitrogen N Oxygen 0 39 109 2 C Tetrahedral o E W Vquotquot Triangular Pyramid 000 quot108 5 quot120 V C Trigonal Planar N105 39 on Bent or Angular t6 C z beaminar w 39E K44 quot180 U near quot118 0 C 0 92514 oxygen has four groups but 2 lone pairs so bent less Larger Molecules than 109 lul I Consider the molecule of ethanol CZHSO 1090 Bent f The molecule has three A main centers 1 v Tetrahedral j 105 The overall geometry is determined by the geometry around each of these centers Chemical Thinking Let 3 Think 3 centers 2 tetrahedral 109 4 bent lessthan409 Consider the molecule of acetone C3H60 How many centers i 0 are in this molecule I II What is the geometry I I around each of these quot centers What bond angle characterizes each center Chemical Thinking Let 3 Think Consider this representation of the molecule of nicotine In this case carbon and hydrogen atoms are not explicitly represented line structure Identify the location of the missing hydrogens and predict the value of the selected bond angles m E x E 39 Tc 2 E a O 10 92514 Let39s apply Assess what you know Chemical Thinking 39 Functionality A central idea in chemistry is that the chemical properties of many molecules are determined by the presence of distinctive arrangements of atoms that tend to behave as a single chemical entity during a reaction This distinctive iv 5 t arrangements of atoms are called functional groups E Q g and their properties are hi determined by their atomic 3332 composition connectivity and geometry R o Hydroxyl group Chemical Thinking 39 39 g Le sapp39y39 Functional Groups Determine the geometry around the atomic centers of the following functional groups Chemical Functional Lewis Chemical Functional Lewis Class Group Structure Class Group Structure ll39 ao 39 Ketone Carbonyl chR2 Ether Ether R1 9 R2 53 E Ald h d Ald h d I A Amine 5 e y e e y e mine R39andRican E beHatoms I 8 Hc c Carboxylic 39 0 Carboxvl Aromatic Phenyl RC CH GE Acid RIC g H gtcclt H H 0 11 2 Let39s app Predict Phenylalanine is an essential aminoacid needed by our body to biochemically synthesize a wide variety of proteins What functional lll I i 3 9r ups are H c c c c c o H present in this I 39439 molecule c H H H H 5 92514 Estimate the value of the marked bond angles and make an sketch of the geometry of this molecule Chemical Thinking a Summarize in once sentence the basic principle that determines molecular geometry Chemical Thinking Predicting Geometry Summary The octet rule can be used to deduce the distribution of valence electrons among the different atoms in a molecule Lone e39 pairs The distribution of electrons is represented through the LeWIS structure of the molecule Malequot 6 e G m E x E 39 Tc 2 E a O 12 Chemical Thinking 39 Predicting Geometry Summary Once the Lewis structure of a molecule is derived its geometry can be predicted applying a simple principle Regions of high electron density around any single atom will be located as far as possible due to electron repulsions VSEPR Theory We can deduce the entire molecular geometry of a complex molecule by analyzing the electron pair distribution around each of its atoms 92514 m E x E 39 Tc 2 E a O For next class Investigate how molecular composition and geometry affect the distribution of electrons within a molecule What is the difference between a polar and a nonpolar molecule 13 92514 Unit2 How do we determine structure The central goal of this unit is to help you develop ways of thinking that can be used to predict the atomic and molecular structure of substances Using spectroscopy to derive structural information M1 Analyzing LightMatter Interactions Deducing atom connectivity based on atomic structure M2 Looking for Patterns M3 Predicting Geometry Predicting the three dimensional geometry of molecules Analyzing the distribution of I M4 Inferring Charge Distribution electrons in molecules I Chemical Thinking 39 Unit2 How do we determine structure Module 4 Central 0a 9 Inferring To infer electron h r i ri ion in Charge Distribution 23333qu 5120 on both differences of electronegativity between bonded atoms and molecular geometry Chemical Thinking 39 The Challenge The properties of chemical compounds are determined by how valence electrons distribute among different atoms This charge distribution can be inferred from molecular structure Composition 9 Connectivity 9 Geometry 9 Charge Distribution High electron density How can we identify regions of high or low electron density in a molecule How can we use charge distribution to predict how a molecule will interact with light or other substance m E x E 39 Tc 2 E a 0 Low electron density 92514 Let s think How would you explain this contrasting behaviors ll t 439 quot392O C6H14 Q Water Hexane Uneven Distributions Some atoms exert stronger forces of attraction on electrons than others These types of atoms tend to have unfilled valence energy states that are lower in energy than equivalent states in other atoms hi Photoelectron Spectra E quotS S anp EC ra A V 55 ms np H M 113113931 1 O EN 23o44131189 6 Chemical Thinking Electronegativity Electronegativity x is a measure of the decrease in the energy of a system if a bonding electron was to be localized in the vicinity of a given type of atom M H Trend N 0 i Cl wig 21 quot5 Ti V Cr Mn Fe Co Ni Cu Z 63 Q As 2 13 I 5 L6 I6 Ls 18 l9 19 L9 L6 5 Ls Nb Mo 1 Ru Rh Pd Ag Cd In Sn Sb T 239 16 l s I9 22 u 7 2 quot9 L7 L7 18 19 7 1 25 Os lr Pl Au At R Tl Pb BI P0 c 2 2 12 72 2quot L8 19 L9 20 23 u Np Nu L4 m IJ x measure of the probability of finding the electrons in a bond localized in the vicinity of that atom 63 E x E c 39 Tc 2 E a c 0 92514 Partial Charges Partial charges on the atoms in a bond depend on xAXB Ax Drlve 8A 83 9w XA Xa ZXAV Resistance Let s think c C c Calculate partial LE charges in each atom 39 7 of a water molecule 2 xH 21 05 mm s 35 C 0 Bond Polarity Differences in electronegativities between atoms in a bond lead to an unequal distribution of charge Bond polarity Dipole moment 8 1 8 gt The exustence of 6 and 6 partial charges creates a dipole moment in the bond The dipole moment iii of the bond is represented by an arrow going from the to the partial charge The larger the difference of electronegativities Ax and the bond length the larger pi and the more polar the bond Chemical Thinking 39 5 Electrone quot gatIVIty 2 Let s Think H 21 Arrange the following c N o F molecules in orderof 25 30 35 40 increasing bond polarity 8 CI 25 30 t K m k E cur co N CH HO 0 so x 2 2 4 1 1 E C 39 Ti 9 E CD 0 Chemical Thinking 39 Molecular Polarity Interactions of a molecule with light and with other molecules depends on bond polarity AND overall molecular polarity For example the CO2 molecule m 5 6 5 Vector Sum 4 i II gt m 7 H 0 The net dipole moment of the C02 molecule is zero We say that the molecule is non polar although it has polar bonds 92514 Chemical Thinking 39 Molecular Polarity Bond dipole moments add like vectors Non D H 0 Polar so3 H20 Polar 5 Vector Sum r p 185 D 5 8 i m E x E 39 Tc 2 E a c 0 Molecular Polarity Bond dipole moments add like vectors CH4 Symmetry In many cases molecular polarity can be inferred by analyzing the symmetry of the charge distribution For a molecule to be non polar the charge distribution has to be fully symmetrical Carbon F Hexafluoro Cll Tetrachloride F F benzene C CCI4 C6F6 c1 Cl F F F Acetylene CZH2 Dimethyl ether C2H60 9 H H 39 I I H CEC H H H HCOGltH 92514 Let 3 Think Consider the following molecules CCI4 CHZO N20 NH3 CHCI3 and N02 Are they polar or non polar Draw the net dipole moment k A You may go to httpwwwcbcarizonaed utpp W J And click on Dipole Moment to use interactive tools to w draw bond dipole moments Chemical Thinking 39 39 Let 3 Think I Are these molecules polar or non polar Csza Ethane CsHsa Propane a C6H14 Hexane m E x E 39 Tc 2 E a 0 92514 Let39s apply Assess what you know Chemical Thinking 39 A Natural Phenomenon Earth s atmosphere normally traps 84 of the solar energy reemitted by the planet s surface This phenomenon is called greenhouse effect and helps to keep an average temperature of 15 C or 59 F Without this effect Earth would 60 F cooler The temperature in Tucson right now would be around 10 F m E x E 39 Tc 2 E a O Greenhouse Effect quotK d 92514 30 avv39w IR Absorption Substances considered greenhouse gases absorb and reemit IR radiation For a molecule to absorb IR radiation there should be a change in its net dipole moment when it vibrates Molecules with N2 02 non polar bonds do not absorb IR radiation Non polar molecules like C02 absorb only for those vibrations that generate a net dipole moment Which vibrations are not IR aetive Chemical Thinking Let39s apply Greenhouse Gases Many pollutants in our atmosphere can act as greenhouse gases amp y 50 cle srs g N20 amp 39 E cup Analyze the polarity of these molecules and discuss whether they can absorb IR radiation Chemical Thinking Relative Contributions The image shows the IR spectra of the main greenhouse gases in our planet Each spectrum is scaled considering the substance s concentration and global warming potential GWP The image also shows Earth s emission spectrum black line Identify substance can be expected to absorb outgoing IR radiation emitted by our planet Estimate the relative o 560 who ISIJU IDIJO 1560 soth 3500 contribution to the greenhouse effect Wavenumber cm m 3 I HO 5 amount of Earth s IR 5 N10 radiation absorbed by g H E each of the different g C F I gases and arrange 39 01 I 3 TS them in order of decreasing E a 0 92514 a 5 Come up with one important application of the ideas discussed in this entire unit Chemical Thinking 39 Inferring Charge Distribution Summary In general bonding electrons are not evenly distributed between the two atoms in a bond Electronegativity is a measure of the ability of an atom in a bond to attract electrons to itself Differences in electronegativities between atoms in a bond lead to an unequal distribution of charge bond polarity The larger the difference of electronegativities the more polar the bond Chemical Thinking 39 Inferring Charge Distribution Summary Molecular polarity refers to the charge distribution in the entire molecule It can be inferred by analyzing the symmetry of the charge distribution For a molecule to absorb IR radiation there should be a change in its net dipole moment when it vibrates Molecules with non polar bonds do not absorb IR radiation m E x E 39 Tc 2 E a 0 92514 Are You Ready Chemical Thinking Unit 2 How do we determine structure The central goal of this unit was to help you develop ways of thinking that can be used to predict the atomic and molecular structure of substances Can you use experimental data and chemical models to establish the molecular structure of an unknown substance Chemical Thinking The Challenge The determination of a substance 3 identity based on the analysis of its molecular structure is a fundamental analytical tool in forensic science Consider this case described in the SFC in May 13 2001 In the thick of evening traffic earlier this year a minivan broke from the gridlock on Interstate 580 leveled a call box and continued out of control onto the streets of Livermore The van struck several parked cars and pedestrians scattered as it shot through a shopping plaza and lurched onto a concrete island Police found the 30yearold driver drooling on himself his hands clenched to the steering wheel They assumed he was drunk but the quotbeveragequot police found in the van wasn39t alcohol m E x E 39 Tc 2 E a 0 92514 Let39s think The analysis of the beverage revealed the presence of a liquid substance with the following elemental composition and mass spectrum MASS SPECTRUM 100 of this compound 5331C 80 1119H g i c E 3551 0 g MX 90 1 gmol x 40 E g r lE 20 6 Determine the i M 2 empiricaland 0000 20 40 6 0 e o 150 GE molecular formulas m z J Let39s think The IR spectrum of the compound is shown below m if i d mi Mm i g 087 V If CC ti 3 irV i 067 V cm i 1i 04 CH H E 7 39I g 0H 03927 iM i i 4000 3000 2000 1000 Wavenumber cml Draw the Lewis structures of at least 3 possible compounds given the information that you have Chemical Thinking 39 39 Let39s think These are some of the possible structures What physical properties could we use as differentiating characteristic m E x E 39 Tc 2 E a 0 10 Let39s think Draw the line structure of the compound Provide the electron pair and molecular geometry around each C and O atom Chemical Thinking 39 39 92514 Let39s think Assign partial charges to the different bonds in this molecule and discuss whether you would expect the substance to be polar Chemical Thinking 39 39 Let39s think Because of its structure BD has similar alcohollike euphoriant effects in our body However it also has narcoleptic effects BB is transformed into GHB inside cells GHB is also famous because of its narcoleptic effects rape drug The more polar a substance the more difficult for it to go through the cell membrane Which of these substances would you then expect it to be a better narcoleptic c E x E 39 Tc 2 E a O 11 92514 Let39s think The presence of a bluish line along the gums of the intoxicated man led the forensic investigators to also consider metal poisoning Using the spectra shown on the next page determine the most likely identity of the metal in this beverage m E x E 39 Tc 2 E a O Let39s think Mg Pb 9 Cd nanometer 420 440 460 450 500 520 540 560 580 600 620 640 660 660 nanometer Chemical Thinking 39 Let39s think X and its compounds are extremely toxic and bioaccumulate in the organism Write the electron configuration of X and discuss what type of substance this is m E x E 39 Tc 2 E a 0 12 Chemical Thinking Unit3 How do we predict properties The central goal of this unit is to help you develop ways of thinking that can be used to predict the physical properties of chemical compounds based on their submicroscopic structure Soft and rigid SQ LightEmittin9 Plastics A Diodes do we Soaps and Detergents i care How do you think structure determines properties in each of these cases 102014 Chemical Thinking Unit3 How do we predict properties FOURMAIN MODULES M1 Analyzing Molecular Structure Premcung pmpemes based on molecular structure i 39 1 on spatial conformations M2 Considering Conformations Pred39Ct39quot9 pmpemes based Predicting properties based M3 Characterlzmg Ionic Networks I on ion charge and sin Predicting properties based on M4 Exploring Electronic Structure electroncon gurations m E x E 39 Tc 2 E a 0 The Path Application U1U2 Molecular Particulate 3 How do we predict properties StrUCture39 Electronic Properties 2 How do we determine Atomic structure IO 4 V g olecular 1 Partlculate 1 How do we distinguish substances 102014 Integ ration To illustrate the power of chemical ideas and models in predicting the physical properties of chemical substances we will focus our attention on materials analysis and design How can we use structural information to predict properties How can we use properties to derive structural information c E x E C 39 Ti 2 E a C O Unit3 How do we predict properties Module 1 Central goal Analyzing To explain and predict the phySIcal properties MOIeCUIar Struoture of molecular compounds based on the nature and strength of the intermolecular interactions among their molecules Chemical Thinking The Challenge One of advantages of chemical thinking is that it allows us to explain and predict the physical and chemical properties of molecular substances based on their molecular structure How can use what we know to make such predictions How can we take advantage of this way of thinking to design materials with the desired properties Chemical Thinking 102014 The Power of Classification Predictions about the properties of substances are greatly facilitated by identifying the types of elements present in the system I Metals I Metalloids 7quot 3quot I Nonmetals 2A N as SB 48 SB 73 l 83 13 Lanthanide Series ctinide Series m U Chemical Thinking 39 Moecuar Compounds El ionic bonds transferring of an electron instead of shanng For example if the substance contains nonmetallic elements it is likely to be molecular 247 2 47 Le Alom Electronegalivity low high l f P03 ruma neg Ax ixA m lt 20 and aim xAxB2 gt 20 are such that nonmetallic elements combine forming individual molecules with polar covalent bonds Chemical Thinking 39 Emergence In order to make predictions about the physical properties of a molecular substance we need to recognize that these properties emerge from the interactions of many molecules Atomic Intermolecular 8 Q5 09 b r c It st O quot9 ompOSl Ion Forces 00 039 it 9 H20 owe etcoccm 6 l a 329 is l l E Macroscopic Properties I State of matter density 8 boiling and melting points quot vapor pressure heat q Molecular Charge capacity solubility etc 6 Structure Distribution 102014 Common Patterns T 0C In general the stronger M P the intermolecular HC39 forces IMFs the more 100 energy is required to separate the molecules or displace them from one region to another up List factors that 63 may affect the I 100 type and strength of the IMFs Lets s between particles Think Chemical Thinking Stronger IMF Intermolecular Forces Interactions between different molecules arise because molecules contain charged particles electrons and protons that attract or repel each other with forces that vary depending on the intermolecular distance I m 3 s E I d E c G lt0 Tu 2 g 4 as o o C 0 Polarizability Molecular Polarity Bond Polarity I Polarizability Electrons are always affected by the presence of other charges which change their distribution within a molecule or parts of the molecule I o 3 Use a 1 Paint Charge 3 Use Another Molecule 8 Apply Electric Field Len Wall Charm The presence of other charges polarizes the molecule and creates an induced dipole moment which may be new or strengthen the one that already exists a E x E 39 Tc 2 E CD 0 Chemical Thinking 39 39 Dispersion Forces When two atoms or molecules get close together the interactions between electrons and protons induce instantaneous dipole moments that cause the particles to attract each other Dispersion Forces DisperSIon forces are the major contribution to the IMFs between the atoms and molecules of most substances Induceddipole ltgt Induceddipole Chemical Thinking 39 39 Let s Think Consider the following sets of different types of atoms and molecules He HF CFQ CH4 Ne HCI CCI4 CI Hlo Ar HBr CBr4 CsH18 00 Which of the atoms or molecules in each set would be the leastmost polarizable Why 00 How would the strength of the dispersion forces between molecules vary within and across these sets of particles m E x E 39 Tc 2 E a 0 Let s Think In which orientation would you expect dispersion interactions to be stronger between octane C8H18 molecules k L k L J k w k k J 102014 muner around dispersion force polarizability induced dipole induced dipole Ml39 shown by dotted lrne if permanent dipole moment so polar dipoledipole meteeleetFGHS meteBelaHi able StFGHg 39 er dispersion forces stronger IMF higher bp dispersion dipole dipole high bp 102014 DipoleDipole Forces The strength of the IMFs depends on three main factors Molecular Polarity Bond Polarity Polarizability In the case of polar molecules additional contributions to the IMFs arise from the presence of permanent dipole moments in the system c 35 These types of IMFs 5 5 called dipoledipole forces 5 39 n I are saId to be dIrectIonal 5 3 because their strength depends on quotg the relative orientation of the 2 C interacting particles 5 DipoleDipole Forces Dipoledipole forces Temperature K A 7 are an additional g c H attractive 79 5 4 1o 39 contribution besides a m dIsperSIon forces to e 3 11 291 D D c the IMFs between a C o E particles of polar 1 c H o 2 molecular gt compounds 250 270 290 310 330 76 2 E CD 0 Let s Think How would you explain the differences in boiling points Dimethyl ether C3HGO Acetone CsHeO Tb 23 C Tb 565 0c m E x E 39 Tc 2 E a 0 Q em Chemical Thinking 39 39 Let s Think How would you explain the trends in boiling points within and across groups of substances based on your understanding of dispersion and dipoledipole interactions Alkane T K Ether 231 273 309 Chemical Thinking 39 39 Hydrogen Bonding Molecular Polarity Bond Polarity Polarizability Bond polarity is particular relevant in molecules in which H is bonded to N O or F The particular dipoledipole interaction that occurs between a hydrogen H atom bonded to a N 0 or F atom and another N 0 or F atom is called a hydrogen bond not a real bond though Highly directional Interactions m E x E 39 Tc 2 E a 0 Letps Think Analyze the structure of the selected molecules and indicate what parts of the molecule can form Hbonds with a similar molecule or Only 3 molecules can be dragged with H20 102014 HO HCCOH HN HOH 102014 Let 3 Think less polar lower bp 100 yellow is all non polar but as you go down the column on periodic table increase BP cuz inc Lines connect molecules containing atoms from the same periodic group E 0 Howdowe HUIng 9 eleenglls Se 398 pgleulzaleillty g explain these 39 g trends in g fE b quot39 39 ts up the graph higher BP you have dipole dipole and 400 E exceptmm disperSIon more p0Iar t g period h20 hf nh3 has hydrogen bonding capabilities so have all three IMF Interaction Strength break IMFs when phase change occurs The energy needed to overcome IMFs interactions is smue than what 5 mm equed to break a cmtalentbonds ebatdidbreakmmich is why I b d Vaent quot chemical change doesnt occur With a phase change Covalent bond 2001000 lemol IMFseo0540kdlmol anything wuth e can have disperSIon forces 39 More a The relative common E strength of E different types Dispersion 00540 lemol l of IMFs varies DipoleDipole 540 lemol 73 depending on HBonding 1040 lemol 39g the type of g molecule Less 0 common Mixed Interactions When different types of substances are in contact with each other in a mixture additional types of interactions may arise Intermolecular Forces in Mixtures i I o f o quot m a I t 39 c U E IonDipole DipoleDipole Hbonding DipoleDipole 4600 kJmol 1040 kJmol 525 ldmol C l Tc 1 I I r E a 39 v w a n IonInduced Dipole DipoleInduced Dipole Induced DipoleInduced Dipole 0 e15 kJmol 210 kJmol 00540 kJmol Mixtures Whether different substances will mix or not with each other depends on how mixing affects a The potential energy of the system b The of configurations that particles can adopt 102014 most favorable in a pee diagram 39 E 1 HzOCH30H System 5 3 J I r I c PEC DIAGRAM 4 4 n 6 V p C d I v Q a 39 GE Unmlxed Mixed 39 of Con urations 0 8 6 y 2amp2 Let s Think 0 Analyze each diagram and make a prediction about whether each pair of substances will mix or not when they get in contact with each other 1 Predict the effect of increasing or decreasing the temperature on mutual solubility E E I as lo culfuon r HIOCH HICHpH E t HOCH39CHJCH CH CH2CH Unmixed Mixed Unmixed Mixed Mixed Unmixed 8 of Con gurations a of Con gurations of Configurations Chemical Thinking 39 39 Your Turn By now you should have developed the To test your ability to apply chemical thinking we have designed a set of challenges based on two very important set of substances Compounds of Oxygen Compounds of Carbon m E x E 39 Tc 2 E a 0 knowledge and skills that are needed to explain and predict the properties of molecular compounds and even to design substances with desired properties Are you ready low PE high Conf so mixed state is most favorable can have dispersion hydrogen bonding dipole dipole if h o E I t I 1st can form H bonds mixed favorable at higher temp at same temp unmixed is favorable euz lew Rl at high Genrmixed is faverable 3rd unmixed state favorablesame PE high Conf unmixedoil and water polar and non polar unlike 102014 Let 3 Think oxvoen has more e so disoersion F is greater polarizability higher bp substantially As we have seen oxygen and its compounds play antral m39eintheatmosphm f quot 39aquot t39 higher cuz p0Iar and oupore oupore IMI Sub TbK HOW do you N2 774 explain these N2 02 is linear so non polar results What 39 02 90392 do you predict 5 03 161 forO3 and why NC5 C 313134231233 355 182 Jbtmgaboutgeomelmjumbemjijnd dipole ustify your deci5ion N20 dp0e N02 is bent as well as a lot of e Let 3 Think Water H20 helps regulate the temperature in the troposphere This is due to its high heat capacity The specific heat SUbStance He jlfapgfity g 0 capacity is a measure of the amount of H20quot 439184 energy required to H20V 20 change the CH3CH20HI 244 temperature of 10 g Ethanol of substance by 1 C c6H6 Benzene 167 How would you explain these differences What would be the environmental impact of water having a smaller heat capacity Chemical Thinking 39 39 Let 3 Think molecule would be non polar but still 2 very polar bonds Q Imagine for a second that the water molecule was linear and not bent stillean have hyelregen leenelmleenel pelarity issue not What would be the impact of this change on molecular polarity issue solubility differences a the phySIcal properties of water b Earthfsclimate and if linear molecular packing would be more d life in our planet the ground up density of ice would be higher than liquid water phase transition temps may increase m E x E 39 Ti 2 E a 0 10 102014 Amazing Carbon Carbon is perhaps the most multifacetic element The electron configuration of C 9 He 252p2 the carbon39atom allovys for many bonding possIbIIItIes For example as an element Carbon exists in several allotropic forms Diamond Chemical Thinking Covalent Networks Macromolecular Nanotubes Fullerenes exhibit a wide range of properties Solar Photo panels Flat detectors screens Carbon Chemistry Carbon atoms can bond to themselves and other types of atoms to form chains and rings of different lengths and shapes That opens the possibility of creating millions of different compounds The physical properties of these J compounds are mainly o determined by J Size JJ Shape 5r Functionality 2 Substituents C m E x E 39 Tc 2 E a 0 11 Chemical Thinking Let s Think 102014 Linear Alkanes and Alcohols 200 180 O are commonly 3 160 used as g solvents in a 100 many labs and E 80 industries g 8 How would you 2 20 explain the 0 trends in their boiling points of Carbons What boiling points I would you predict for J C5H11Br 9 J these compounds CGHnBr C7H1sBr Chemical Thinking Let s Think Liquid alkanes are commonly used as lubricants both in the automotive and the cosmetic industries In general lubricants are J expected to have high viscosities and low volatilities Which of these isomers of decane do you think is the best lubricant m E x E 39 Tc 2 E a 0 Branching pg BP MP in Lower IMFs lead mL C C cP to lower nPentane MeIting points 0645 36 130 0267 Boiling Points quot Viscosities 3950Pquottaquot H 0639 28 160 0245 and hIgher J Volatilities Neopentane BUT J 0613 10 18 0150 J J J 12 102014 Let39s apply Assess what you know Chemical Thinking Fats Ester Functional H quotquot Group Fats belong to a group of HC iWR compounds called quotquot triglycerides which have AH3i fk yquot39 quot quotg39 the following general H C 3 TX structure 139 quot 739 The hydrocarbon chains can be 1 VI saturated t c cl c c Saturated no double bonds ll 397 39II between carbons or unsaturated at least one double bond x I l39 I I I I l Unsaturated l lt cc tl Er Ill 139 39 Chemical Thinking a0 Let39sapp39w Predictions H l39 1 I I 39 139 Around each l 39 gggcg5 unsaturation the l 139 I 39v I I molecule can have one of two m configurations Cis Trans Apply your knowledge to predict the effect that a the length of the hydrocarbon chains b the degree of unsaturation how many double bonds in a chain c the types of configurations cis versus trans will have on fats density and melting point m E x E 39 Tc 2 E a O 13 Chemical Thinking 39 39 a i In your groups come up with one idea from this module that you understood well and one idea that is still confusing 102014 Chemical Thinking 39 39 Analyzing Molecular Structure Summary The physical properties of a molecular compound emerge from the intermolecular interactions among the millions of molecules present in a macroscopic sample of the material Intermolecular forces arise Hydrogenk 39 because atoms molecules and W 39 W ions have charged particles that 3 are unevenly distributed and constantly moving if m E x E 39 Tc 2 E a O Analyzing Molecular Structure Summary The stronger the intermolecular forces the more energy is required to separate the molecules or displace them from one region to another Thus the physical properties of many molecular compounds are determined by the polarity and polarizability of their molecules and the possibility of H bo n d I n g l l r HEO n gt k gt gt Methano I Acetone WEOHI CaHso CcHia Ociane Octane DipoleDipole DipoleInd Dipole Ind DipoleInd Dipole 14 102014 Analyzing Molecular Structure Summary The element carbon reacts with other nonmetallic elements to form millions of different molecular compounds Carbon atoms can bond to themselves and other types of atoms to form chains and rings of different lengths and shapes The physical properties of these carbonbased compounds are mainly determined by molecular size shape and functionality Chemical Thinking 39 39 For next class Investigate what a polymer is and identify different types of natural and synthetic polymers In which ways are the properties of very large molecules different from those of smaller molecules m E x E 39 Tc 2 E a 0 15 11314 Unit 3 How do we predict properties The central goal of this unit is to help you develop ways of thinking that can be used to predict the physical properties of chemical compounds based on their submicroscopic structure on molecular structure M1 Analyzing Molecular Structure Predmung pmpemes based IM2 Considering Different Scales Predicting properties based on analysis at various scales M3 Characterizmg Ionic Networks on ion charge and sin Predicting properties based Chemical Thinking Predicting properties based on electronconfigurations M4 Exploring Electronic Structure Unit3 How do we predict properties MOdule 239 Central goal Considering To explain and predict the physical properties of leferent scales macromolecular compounds based on nature and strength of the interactions between different parts of a molecule or between different molecules The Challenge Large molecules macromolecules have properties that depend on their functionality and their structure ate different scales The interactions between inter and within different parts intra of macromolecules determine macroscopic properties How can we explain and predict the properties of macromolecular system a E x E 39 Tc 2 E a 0 11314 Macromolecules The most common types of macromolecules in our world are polymers They may be natural or synthetic Polyethylene Polymers Polymers natural or synthetic are composed of repeating structural units monomers connected by covalent bonds Monomer Polymer Polymer Chain Structural Formula Molecular Structure H H H H cc 9 H H H Hn Ethylene Polyethylene H Cl H Cl 1 00 c c l H H H H Vinyl Chloride Polyvinyl chloride PVC F F i F cc cc I I F F F F Tetra uoroethylene Te on Chemical Thinking Different Scales The prediction of the physical properties of polymers is facilitated by analyzing their composition and structure at different scales Chain Scale length shape Monomeric scale Functionality m E x E 39 Tc 2 E a c 0 Chemical Thinking Let s Think The properties of polymers are determined by composition and structural features at a the monomeric scale b the chain scale rt How would you expect the composition and geometry of a monomer to affect the properties of a polymer rt How would you expect the length and shape of polymeric chains to affect such properties Chemical Thinking Monomeric Scale The presence of specific functional groups in the polymeric chains affects how molecules of the material interact with each other and with the molecules of other substances PVC it t 39 K k 1 k L er K L L Nylon Teflon OH Oc N o Polyethylene Cl 2 t v e q 5quot 9 Cellulose F Ia 11 I K I r 1 Risa 1314439 a a 32 i c E x E 39 Tc 2 E a O Let s Think Consider three solid surfaces made with the types of polymeric molecules shown in the image Which of them are likely to be wetted by water H20 will stick to them and which ones are likely to be dissolved by water H20 will be able to separate the chains What would happen if we use acetone or hexane ff 99 FFquot 11314 Different regions and different polar regions can interact differently with substances WWW weaker plasticex C3H6O acetone polar C6H14 hexane If IMF between gt IMF bw diff molecules cant dissolve WW lt w l u s will dissolve styrofoam is non polar 11314 The Power of Many The composition and structure of the monomers in a polymer have a strong impact on its physical properties because the same features are repeated multiple times along each polymeric chain Nylon 66 39 6 Let s Think there are Hbonds bw the oxygen and hydrogen and nitrogen Kevlar is a high strength polymeric material with many applications from producing bicycle tires to body armor huge amount of surface area a lot of dispersion force in planar stacked up I I I II II rt How would you explain the high strength of this material rt Why is this material stronger than nylon 66 Chemical Thinking Chain Scale The length of the macromolecules and the presence of branching chains affect the shape that polymer chains adopt and how these chains interact with themselves and with other chains Chain length and branching are used in polymer synthesis to control material properties such as melting point density flexibility elasticity viscosity and thermo plasticity m E x E 39 Tc 2 E a O Q em m E x E 39 Tc 2 E CD 0 Let s Think Consider three different types of polymeric materials made up of molecules with different lengths and degrees of ramification branching A a B Q lt1 39 f LDPE V If V 1 W C 0 Predict how the following properties of the three types of polymeric materials will differ Melting point density flexibility elasticity and viscosity at In which case will the value of each of these properties be higher or lower Chemical Thinking Crosslinking One common way to alter the properties of a polymeric materials is by adding other substances that alter their properties additives E For example it is Covalent onding common to add molecules that can act as a binder between polymer chains which tends to increase strength and rigidity CrossLinking Clo sslinked polylsoprene Vulcanized Rubber Chemical Thinking Let s Think Consider v lt Polyvinyl alcohol I V 1 r w y I t lt H V y i V x l gt y k rquotLquot B H H H v n 9 o l I x y A H c c quot g l l Predict how the properties could change when mixing with a Boric Acid 11314 A has the highest melting point biggest surface area so more dispersion forces C doesnt have enough electrons ampis nrost elastic because nlost electronsbranching reduces disperison forces less amount of surface are A is most rigid hydrogen bond can only be made if there is already a hydrogen bonded to it Hydrogen has to be bonded nll n nrl LVIVVVI I Build the structure CH2F2 can39t hydrogen bonds are super possible so it turns to a 11314 Crosslinking NonCovalent Crosslinking Chemical Thinking Let39s apply Assess what you know Chemical Thinking Proteins Proteins are natural polymers made by the combination of smaller molecules called amino acids monomers Amino Acid H O H H T o I II I I 0 H2NC C N C C I lzN C C 1 OH 1 OH H CH3 f R Amine 39 Peptide bond Carboxyl Side Chain Every protein has a unique sequence of amino acids or primary structure m E x E c 39 Tc 2 E a c 0 11314 Primary Structure Backbone o o i o o i o i o l1 N c c N c c N c c C C N C C N C C O r I I El c c on o oCo j o cwr Side chain with residues Backbone and side chain interactions determine the 3D structure that proteins adopt Chemical Thinking 2amp0 Let39s apply The image represents two different sections of 4 e a polypeptide that are approaching each other 39 Analyze the geometry 1 r L and polarity of the a backbone Assign a 39 partial charges 6 and a E to the different 0 atoms What types of 770 IMFs can exist within 39 the backbone itself or with another chain Chemical Thinking Hydrogen Bonding The stronger interactions are hydrogen bonding in the same or in different chains A 0 HYDROGEN BONDING Oi R Nquot Although hydrogen bonds are HN 0 not real bonds and are weaker Rio quotquot H R than covalent bonds there are as NH39mO27 so many of them within a single LE 034 R protein that they determine the 39 HN F0 3D structure of these g Rfo quotquot H R macromolecules E NHOI27 a 0 Chemical Thinking Protein Folding When protein molecules are synthesized inside a cell they exist as unfolded polypeptides Intermolecular interactions with water molecules and intramolecular interactions between amino acid residues lead the macromolecules to fold into the conformation that is most stable in that environment 11314 50 Let39s apply Chemical Thinking Consider this folded protein What types of intermolecular forces are present in this system Arrange these interactions from strongest to weakest Why do you think the protein is folded in this particular way m E x E 39 Tc 2 E a c 0 Similar with Similar Although there are attractive interactions between all types of residues potential energy is minimized when parts with interactions of similar strength are close together Induced dipole ltgt Induced dipole The residues of similar types tend to segregate from each other lonltgtDipole DipoleltgtDipole Hbond benzene ring and CH2 form dispersion forcenonpolar weakest W W m t so y rogen on mg IS capa esrongest a0 Let39s app39V Solvent Effects Proteins do not exist isolated in our body They are surrounded by water or by lipids in cell membranes o HZN 0C co Ho CH CH2 Hzc 2 I l l HZC CH CH2 Q 5 g Discuss how you would expect this protein to fold if you immerse it in a water b lipid oil Chemical Thinking Justify your reasoning g0 Let s Apply Let s explore the validity of your predictions Go to httpwwwchemarizonaedutpplchemthinklMWfoldingjnlp Folding Explore and explain the effect of changing the solvent and the type of amino acids on protein folding The simulation allows you to change the nature of the solvent and of each amino acid in the chain Charged Polar Hydrophilic NonPolar Hydrophobic Chemical Thinking a i Identify one topic of your interest in which the ideas discussed in this module are relevant m E x E 39 Tc 2 E a O 11314 the non polar bottom will fold amongst water top that is capable of H bonding will be on outside sticking to H20 bondfng top39wiHbeinside rclerand the non polar Will n Wih h H i ll c E x E c 39 Tc 2 E a c O 11314 For next class Investigate what types of compounds are formed when metals react with nonmetals In which ways fundamental ways are these types of compounds different from molecular compounds 10 111014 Unit 3 How do we predict properties The central goal of this unit is to help you develop ways of thinking that can be used to predict the physical properties of chemical compounds based on their submicroscopic structure Predicting properties based M1 Analyzmg Molecular Structure on molecular structure x M 39 a IM3 Characterizing Ionic Networks I Pred39Ct39quot9 pmpemes based I M2 Considering Conformations Pred39Ct39quot9 pmpemes based on spatial conformations on ion charge and size Predicting properties based on electronconfigurations Chemical Thinking M4 Exploring Electronic Structure Unit3 How do we predict properties MOdUIe 3 Central goal Characterizing To explain and predict the physical properties of ionic compounds based on the charge and size of the ions present in the system Ionic Networks Chemical Thinking The Challenge Metals can chemically react with nonmetals to form new chemical compounds In most cases these compounds are not comprised of molecules but of ions arranged in solid crystalline networks How can we make predictions about the composition of these ionic compounds How can we predict what properties they may have m E x E 39 Tc 2 E a O 111014 Important Differences most stable form octet Nonmetals Metals F u m w IJ N M o I x c In general atoms of nonmetals have higher ionizations potentials and electronegativities x than those of metals Thus nonmetals have a higher affinity for electrons than metals Chemical Thinking From Polar to lonic The larger the difference in electronegativity Ax between bonding atoms the more polar the bond Nature of the Chemical Bond Non polar Covalent Polar Covalent Ionic Ionic Character Ax Left Atom Electrone tiv39 247 a quoty low high A difference 4 g le rlghq RIghtAtam Eleclmnegatmty p05 neutral neg bw m9 Click on each atom Use the sliders to to select an element set electronegativities For Ax gt 2 the bonding electrons spend most of their time close to the nonmetal atom Chemical Thinking lonic Compounds The properties of the compounds that results from the reaction of metals with nonmetals are better described assuming that independent ions are formed F2 HBr HF Increasing ionic character Increasing covalent cy mr Ionic bonds Nonpolar covalent l 0 05 0 15 20 25 30 Electronegativity difference Metal ion cation I I Nonmetal 9ion anion m E x E 39 Tc 2 E a 0 111014 Let 3 Think Electrostatic interactions between the anions and cations lead to the formation of a characteristic submicroscopic structure Go to httpllwwwcbcarizonaedultpplchemthinklsimU3lionichtml 0 Select an initial number of atoms of each type and analyze what happens as the atoms react and ions are formed 0 Analyze what happens as you increase the temperature Chemical Thinking Let 3 Think metals will form negative charges an as they collide they interact by alternating charges to decreasepoterrtrai errergy Number of metal atoms Ion charge I 39E Gravity Chemical Thinking Ionic Networks and ions interact with all other ions in the system and they do not form molecules They arrange into a crystalline network Na NaCl For these compounds the chemical formula only indicates the ratio of to ions in the lattice 11 m E x E 39 Tc 2 E a 0 111014 halogens1 would be most stable ion because it is closest to full octet Let 3 Think The charge that atoms of metals and nonmetals acquire when they react with each other is determined by their electron configurations El p1 p2 p3 p4 p5 I06 What charges would the atoms in the highlighted families will acquire when they react with each other Justify your reasoning Chemical Thinking Ion Charges or lose if on metallic side Based on the octet rule we can predict that the ions formed will have full shell 99mquot con gurations Lomuts1ghomzaiLoneneLgJes 39M 8A 1A 2A 3A 4A SA SA I SB 43 SE 6E 78 l SB IE 23 A A A A A A Chemical Thinking Transition Metals The prediction of ion charges for transition metals is more difficult because electrons in the dsubshell may also be transferred to the nonmetal atoms So they may form a variety of ions Ion charge Atomic Number Common 0 Less Common m E x E 39 Tc 2 E a c 0 Naming Examples Cul Cull Nickell I Chemical Thinking Charge Neutrality No matter what ions are formed the ionic network that is formed has no net charge These implies that cations and anions combine in ratios that ensure charge neutrality Consider this reaction Nas Cl2g Na Na Cl 6 Cl39 Stable Ions Sodium chloride Thus the ionic product is NaCl What happens in this case Als 02g Al 9 AI3 O 0239 Stable Ions Thus the ionic product is AIZO3 Aluminum oxide 111014 Ole Chemical Thinking Let s Think Predict the ratios in which the ions resulting from these reactions will combine Ks F2g Cas 029 Nas S8 s 9 Srs N2g Write the chemical formulas for the resulting ionic compounds KF 20 NaZS Sr3N2 m E x E 39 Tc 2 E a 0 Predictions Reaction Ions Chemical mixture Formula KS F29 CaS 029 Nas S8 5 Srs 29 In general given Am and Bquot the chemical A B formula that ensures neutrality will be h m 111014 Let s Think 2 Ionic compounds do not conduct electricity as solids but they do it when melted Solid ionic compounds shatter when struck How would you explain these properties based on their submicroscopic structure Chemical Thinking A Powerful Model In ionic compounds electrons are localized on ions and cannot flow In molten state the ions can move freely and establish an electric current Chemical Thinking Charge and Size In general many properties of ionic compounds are determined by the charge and the size of the ions in the network I39 lt Interactions between qlqz ions are determined by F 2 Coulomb s law r Thus it is of central importance to be able to predict the charge and size of the ions formed whenever a metal reacts with a nonmetal m E x E 39 Tc 2 E a c 0 111014 Let 3 Think lose an electrondecrease size of atom gain an electron increase size During the formation of Ionic compounds atoms of metals lose electrons while those of nonmetals pUbiitveiy bi ldl geu39 U39ebl Babe gain them Ame Am negatively charges increased size 3 e39 9 Bquot39 going down periodic table increases a How would you expect the 5 size of each type of atom in ii 15 to change when becoming II 4 s t IIIIIH E ions IIIIIIIIIIIIIIIIII 7 How would you expect the EEEEEEEEEEEIIIIIII periodic trend for ion 7 2 sizes to be 0 lar er char er more ener it takes to se arate P ropertles g g gy p Coulomb sLaw Them hlgher meltmg pglnt The properties of an ionic lowest compound are determined by H Sal I 399 939 Ial 99 99 3959 SEQ IaFgeSi S39l39zequot 5 glqz the electrostatic forces among F 2 39 39 its ions and between these r meltmg pomt ions any surrounding particle 0 atom ion or molecule smaller snze higher melting ponnt cuz smaller Based on Coulomb 3 Law one may expect these and An UIOLClI IUU forces to be stronger 1 the larger the charge of the ions larger q1 qz 1 the smaller the size of the ions smaller r Chemical Thinking hi h to low Phase Transitions NaF NaCI NaBr g Ionic compound tend to Interacflons Energy KJmOI C30 have high melting and mm 400 39 4000 boiling points because C Valent 15 39 quot00 electrostatic forces are Metamc 75 39 100 pretty strong Hbonding 10 40 I DipoleDipole 5 25 i Let s Think Dispersion 005 40 Arrange each of these sets of ionic compounds in order of increasing melting point 0 NaCl NaBr NaF 0 CaO NaCl MgO m E x E 39 Tc 2 E a 0 111014 because interactions between each other are stronger than that it can have With water Water Solubility Some ionic compounds are very soluble in water while others are almost completely insoluble Ionic compounds that are insoluble in water tend to be comprised of ions 2 with strong 3 interactions among 395 each other 39 large charge g small size as 6 NaCls H20l gt Naaq Cl39aq Electrolytes Solutions of soluble 5 9 S ionic compounds 9 g 0 J0 conduct electricity J5 J This implies that O 5 99 ions can freely move 5quot O 53 3 across the system 3 a Q 8 3 Solutions of slightly 039 e 9 E soluble ionic 39 fit 39 I 0 I compounds barely e c e To conduct electricity 9 3 NJ 0 439 g V Only a small fraction of I T 20 the ions ls ln solution x f r PE vs Configurations Again For an ionic compound to be soluble in water at least one of the following conditions needs to be met v The net interactions between ions and water molecules should be stronger than the net interactions between the ions themselves leads to lower potential energy v The number of configurations that the ions and the water molecules can adopt when mixed is larger than the total number of configurations that they can have if they do not mix m E x E 39 Tc 2 E CD 0 111014 Let s Think 1 C if change temp inc more soluable temp change is minimal temp independent Consider the following potential energyconfigurations diagrams for the dissolution of three different ionic compounds NaCls CaCl2s and ZnSs 5 NEWS Soluble E CICls Soluble E ZnSs Insoluble F 2 low PE but if inc temp less soluable g 5quotan 392l39aq ZnSs I 070 E 33 of Con gurations S of Con gurations V a of Con gurations L E C 2 9 Use these diagrams to explain the solubility in C and IOW at U nmixed SO insoluable is cgt water of each of these substances GE 0 Discuss how changes in temperature may affect favorable C 0 the solubility of each these compounds Let 3 Think transition metals are insoluable tst column and 7th column is soluablelow charge Consider the following examples of soluble and insoluble ionic compounds Soluble Insoluble NaCl MgBr2 MgO ZnS K28 Cal2 Al203 FeS NaF AICI3 CrS Ni203 Use this data to derive a simple rule to make predictions about the solubility of ionic compounds based on ion charges Justify your ideas using Coulomb s Law Chemical Thinking Solubility Trends if abs value of charge multiplies to 4 or less soluable otherwise insoluable The following rule is useful to make predictions about the solubility of ionic compounds in water Given Coulomb s law we may also expect that the larger the ions the more m soluble the compound will be weaker r2 interactions among ions in the lattice Although this rule works well in many cases there are always exceptions For example AgCl PbCl2 Mng CaF2 are insoluble m E x E I To 2 E a O 111014 Polyatomic Ions There are ionic compounds in which the ions in the lattice are not necessarily atomic one single atom but molecular several atoms covalently bonded quot I mg H 0 OH39 NO339 C03239 Hydroxide ion Nitrate ion Carbonate ion E I 39 o E1 u u H I 0amp8 0330 H J 8 o 39 5 5 Sulfate ion Phosphate ion Ammonium ion Co m pa riso ns No matter the type of ion we can still make predictions about the structure and properties of these ionic compounds based on similar ideas Ammonium 45quot Chloride 2f Let s Think NH4CI Make predictions about the solubility of these ionic compounds Barium Sulfate Magnesium Nitrate Calcium Phosphate Chemical Thinking Ammonium Nitrate Calcium Carbonate Applications Insoluble Soluble Ca3PO42 NH4N 3 oaco3 M9N032 I rt t Structural fmpi an components in y 39 em 39zers39 living things a C 3 Insoluble E BaSO4 39 Tv Facilitates 0 39g Xray a analysis 0 10 111014 Let39s apply Assess what you know Chemical Thinking g0 Let39s apply Some Melting Points All common ionic compounds have K NH4 melting points above Cl39 776 C 338 C 100 CIand thus are N03 334 0C 170 0C solId a room temperature 0C 0C c 3 Based on what you have learned C E 1 Hypothesize on the structural characteristics of l c IonIc compounds that could be liquid at room 3 temperature GE 1 Discuss potential applications of these types of 5 fluids Let39s apply Soaps are also ionic compounds in which the anions are pretty large k J hydrocarbon chains Explain 25 Q 1 g 39 0 Why are soaps a lt3 quotlt3 a soluble in water 399 i L if 3 39 quotI quot5 J 0 How would you a i quot Q quot If expect soap ions to ag ix I 6 arrange In solution quot2 39 3 U 39 I 39igi 39g 0 Why do soaps help i k g in removing grease quot 0 11 Chemical Thinking a i Come up with a question about the content of this module that you desperately want to be answered 111014 Chemical Thinking Characterizing Ionic Networks Summary Metals can chemically react with nonmetals to form new chemical compounds In most cases these compounds are not comprised of molecules but of ions arranged in solid crystalline networks No matter what ions are formed the ionic network that is formed has no net charge These implies that cations and anions combine in ratios that ensure charge neutrality 3 Cquot m E x E 39 Tc 2 E a 0 Characterizing Ionic Networks Summary The charge that atoms acquire when they react with each other is determined by their electron configurations The ions that are formed tend to be those with a full shell electron configuration Many properties of ionic coulomb 3 Law compounds are determined by the charge and the size of r F M the ions in the network These 4 r2 forces are stronger the larger the charge of the ions and the smaller their size 12 c E x E 39 Tc 2 E a O 111014 For next class When metals are combine they form alloys Investigate the basic characteristics of these materials Why is that metals conduct electricity while ionic and molecular compounds do not 13 111414 Beyond the Physical So far our focus has been on h understanding the submicroscopic 5 structure of chemical substances and its relationship with their macroscopic physical properties if However an essential question in chemistry is how structure determines chemical reactivity How when why do substances interact to produce new substances Chemical Thinking 39 Synthesis Unit 4 How do we characterize chemical processes The central goal of this unit is to help you understand and apply ways of thinking that can be used to model chemical change in a system Produce new What do you synthetic 3 We need to know materials abOUt Control the E chemical production of I Why do change to air pollutants 5 we 2 complete Select E care these tasks the best 5 fuel Essential Questions Stoichiometry Thermochemistry How much substance How much energy do do I need or do I get I need or do I get Kinetics Thermodynamics To what extent does it happen How fast does it occur Synthesis How do we explain predict and control it on E x E i 39 Tc 2 E a i 0 111414 Unit 4 How do we characterize chemical processes THREE MAIN MODULES Identifying factors that affect chemical processes M1Modeling Chemical Reactions 5 b 3393 M2 Understanding Proportions Determining the amount of substance formed or consumed ul M3 Tracking Energy Predicting the amount of energy absorbed or released Chemical Thinking 39 39 39 39 lnteg ration To illustrate the power of chemical ideas in modeling chemical change we will focus our attention on two important types of systems malaoicluul m C E E C l g Processes that Processes that E generate pOIIUtantS generate energy 0 Unit 4 How do we characterize chemical processes Module 1 Modeling Chemical Reactions Central goal To explore factors that determine the extent and rate of chemical reactions m E x E 39 Tc 2 E a 0 Chemical Thinking 39 Chemical Change The formation of one or more new chemical substances with a distinct composition or a change in the net amounts of some chemical species in a closed system are indications that a chemical process has occurred Chemical Thinking 39 Let s Think 393 Observe carefully each of these chemical reactions rt Identify differences and similarities among them What factors characterize chemical processes m E x E 39 Tc 2 E a O Signature Features Analysis of a wide variety of chemical reactions has revealed basic common features of chemical reactions 1 Chemical reactions may involve one or more initial substances called the reactants 2 As a result of the process new chemical substances with different chemical composition are formed called the products 3 Reactants combine to form products in specific definite proportions For example H2g always combines with 02g in a 2 to 1 21 ratio when forming H20l 111414 conservation of matterno destroying or creating same number of atoms 2H2 02 2H20 V E FF 111414 endothermic absorb heat will get colder exothermic emit heat will get hotter Signature Features 4 When chemical reactions occur in closed environments the total mass of the system before and after the process remains constant takes energy abs to break bonds mass is conserved 5 Chemical processes always lead to a net energy IS released When bonds form transfer of energy between the system in which the reactions takes place and its surroundings There are chemical processes that result in a net absorption of energy from the surroundings endothermic reactions there are chemical processes that result in a net release of energy from the system to the surroundings exothermic reactions Chemical Thinking 39 log to ash irreversible Signature Features 6 Chemical reactions occur at varying speeds The speed or rate of chemical processes depends on the chemical nature of the reactants but also on factors such as temperature pressure and concentration of the reacting species 7 Chemical reactions occur with various extents Some chemical reactions go to completion This is the reactants are fully transformed into products when combined in the proper proportion However there are many other chemical systems in which the final state involves the presence of constant amounts of both reactants and products Chemical Thinking 39 The Challenge We would like to generate models that allow us to answer questions such as this What proportion of How much reactants and energy will be products are needed or 5 involved produced E E E How fast will the To what extent 392 process go 7 will the reactants g and how can I 39 be changed into E control it products C 0 Q em Chemical Thinking 39 Letps Think The atomicmolecular model of matter assumes that all substances are made up submicroscopic particles atoms ions or molecules in constant movement and interaction Discuss how this model could be used to explain the following facts 1 Mass is conserved in chemical process in closed systems 2 Chemical processes always result in energy transfer 3 The speed of chemical reactions depends on temperature 4 Not all chemical reactions go to completion Chemical Thinking 39 Modeling Chemical Reactions Assumption 1 Chemical reactions are processes in which the particles that make up the reactants are rearranged As a result of these rearrangements substances with a different chemical composition are formed Burning Methane as f in 9 gt 0 Q Q 9 CH49 029 C029 209 m E x E 39 Tc 2 E a O Chemical Equations Q How do we represent the process using formulas Reactants Products CH49 2 o g s 3029 2 209 I Stoichiometric coefficients 111414 Combustion is combining things with 02when combined with a carbon hydrogen species will always fonnvvaterandcai bonoxide The chemical equation only states the proportion in which substances react and are formed not the actual amounts in the system Chemical Thinking 39 LImItIng Reactant 02 is in excess CH4 is the limiting reactant determines the maximum amount of products What about here Same Equation CH4g 2 029 9 C02g 2 H20g Chemical Thinking 39 Mass Conservation Q a m 9 Q o 9 CH49 2 029 9 C029 2 H20g Why does the mass remain constant 0 The chemical reaction leads to a rearrangement of the atoms into new molecules 1 The mass does not change because all of the different atoms remain in the system m E x E 39 Tc 2 E a O Letps Think Oxygen gas in our atmosphere reacts with many chemical substances in the planet We take advantage of some of these reactions to generate energy through the combustion of different types of fuels However some of the products of these processes are major air pollutants Use the following submicroscopic representation of this process to derive its chemical equation 60 000 gt o oo 90 Clearly identify reactants products and the limiting reactant 111414 There39s only one carbon so only one CO2 can be formed 2 O2 is limiting reactant 5N2 202 3N2 4ND 02 is limiting reactant 3N2 did not react so 2N2 202 4N0 N202 2N0 Chemical Thinking 39 Letps Think Natural gas contains a small amount of sulfur compounds When burned they produce the pollutant 02 that further reacts with O2 in the atmosphere to produce another harmful substance Use the following submicroscopic representation of this process to derive its chemical equation 9009 09 Doc 4 9 Clearly identify reactants products and the limiting reactant Chemical Thinking 39 Let s Think The incomplete combustion of fossil fuels such as coal mainly made of carbon and ethane CZHG produce a toxic pollutant carbon monoxide CO 6 Write the balanced 3 gt chemical equations and complete the 000 0 particulate representation of each process a Which reactant is gt a limiting q a any g 4quot m E x E 39 Tc 2 E a 0 Modeling Chemical Reactions Assumption 2 Rearrangement of atoms during a chemical reaction involves electron redistribution among different atoms As a result of this process the internal potential energy of particles in the system changes Endothermic E Process In some cases products have higher Ep than 5 reactants s o T 3 i Energy Absorbed 3O 3 029 9 2 039 Reactants Products Reaction Path 111414 402 4802 202 4803 Sulfur is the limiting reactant W303 02 2802 2803 60 502 600 202 carbon is the limiting reactant 60 302 600 20 02 200 502 3C2H6 6H20 400 C2H6 22 2H 4 H2 if subtract the two positive number endothermic taKes energy to breaK bonds WWW Energy Transfer Exothermic E k Process l 39o Q In other cases products 9 0 have lower Ep than 2 H2 02 a reactants E a A Energy Released E 2 29 quot39 029 2 H20quot 2 H o 2 39 Tg Reactants Products GE Reaction Path 7 0 66 if Let s ThIn k The combustion of propane C3H8 in our stoves as well as the production of glucose C6H1206 in plants are processes in which important air components 02 C02 and H20 are involved 1 Write the chemical equations for the combustion of C3H8 in stoves and during the production of C6H1206 in plants 2 Based on your experience and knowledge about these two processes sketch their corresponding energy diagrams clearly indicating the relative internal potential energy of reactants and products 3 Are these processes as endothermic or exothermic Chemical Thinking 39 39 39 quot 39 Modeling Chemical Reactions Assumption 3 For a chemical reaction to occur particles of the different reactants must collide The more collisions per unit time the faster the process The more collisions in a given time the faster the process We can control collision frequency changing T P VandN m E x E 39 Tc 2 E a 0 111414 negative number exothermic energy released upon bond formation C3H8 502 4H20 3C02 6H20 6C02 C6H1206 602 opposite of combustion propane is exothermiccombustion giving off heat reactants top left products lower right surreunding gets het glucose is endothermic absorb energy form the sun reactants lower left products top right 111414 have to collide with proper orientation and energy Modeling Chemical Reactions Assumption 4 For a chemical reaction to occur colliding particles must be oriented in a manner that allows reacting groups to interact effectively I r O Q 1 NO Effective quot5 r V collision o IE 3 N02 02 Tu lt Ineffective I No 03 9 N02 02 collision 9 0 AB is easier C f39 39 Eff 39 on 39guratlon ecuveness NO 03 IS more complexso needed to have different The higher the of configurations that lead to effective collisions configuration effectiveness the faster the process vAnInIAnn 0 39539 ILClLIUI IO Thus in general moles of the same type of reactant is faster a the fewer the number of particles that need to interact to generate the products b the simpler the composition and structure of such particles Chemical Thinking 39 39 quot 39 the faster the process it Let 3 Think the one on the right is faster orientation doesnt matter number is the same more simple structure Reacunts Products Reactants Products 39 Which of these gt 3 gt 5 I mg to more processescanbe g 0 products IS more favorable than less products 39 expected to be faster assuming equal a 3 conditions of gt g temgeraturetprfssur 8 g 0 o the one on the rightmore of the same number of an concen ra Ian 0 E particles in the reactantsand Simpler they Just need to collide c system 0 0 z 0 2 0 o as 30 0 C 0 Chemical Thinking 39 Modeling Chemical Reactions Assumption 5 For a chemical reaction to occur colliding particles must have enough energy to reach a transition state that leads to the formation of the new products Transition The energy required to State attain the transition E NVV0V0 state is called the quot 0 0 Activation Energy Ea E Noo3 NOZ02 Reactants Products Reaction Path 111414 Chemical Thinking 39 Let s Think According to our model the rate of a reaction is determined by these three factors Collision Frequency Configuration Effectiveness Activation Energy Analyze the effect of 0 9 increasing temperature 5 concentration of reactants and 0 activation energy on the rate of 98 0 reaction 3 o A B 9 AB Go to httpwwwcbcarizonaedultppchemthinksimU4Activationhtml m E x E 39 Tc 2 E a c 0 Let s Think Temperature I Volume Activation Energy 4 of Atoms A or B y so quotquot 39 III 0 i39 Eu AB AB A E Time E 10 111414 Activation Energy Originally only the Typical n s g m Now all lhese kinetic energy distribution among particles at a fixed EEEEEELVBSZ temperature mam i new acuvallon original activation energy energy The lower Ea the more particles E R 1 have enough energy to react 3 ate Chemical Thinking 39 activation energy does not depend on temperature Temperature temp increase faster rate IS because increasmg KE The I I I kinetic energy notbecauseIowerlngactwatlfmerrergy distribution g ga mg non among particles 2 energyto react changes with changing 3 temperature The higher T the more particles T 1 R 1 have enough energy to react ate The higher the Ea the stronger the effect of T I Chemical Thinking 39 Concentration The higher the concentration of particles NIV the 0 larger the probability 2 00 they will collide 3 0 0 quotAl 1 2 NB Rate E 0 9 5 0 o 5 go a q i5 V l RateI o 953 quot 39 39 quot 3 o I TE 2 E G C 0 A NAN 1 1 NBN Rate decrease as reactants are consumed This implies that the rate should 11 111414 Let 3 Think exothermic reactions tend to be faster because they release heat and therefore KE increases The following two reactions have the same Ea A B 9 C D E 9 F everything is the sametheyre the same rate ess energy te break benels than upen bond formation more energy to break bonds than upon bond formation Reaction Coordinate Reaction Coordinate If Ao Bo Do Eo and T is held constant what reaction will be faster Chemical Thinking 39 39 39 quot 39 Modeling Chemical Reactions Assumption 6 As reactants interact to generate products in a chemical reaction particles of the products may chemically interact with each other and regenerate particles of the reactants The final state of the system is determined by the balance between these two opposite processes 53 quotE To better understand this assumption let us E explore the generic isomerization reaction 73 A 9 B E quot o Let 3 Think exothermicfavored In forvvard or product favored because activation energy IS bigger Let s imagine we En E MB 39 Sta W39thi higher percentage of forward reaction that AoAo Bo 0 backwards more B energy than A 2 A EaBgtA g What happens as quotA time passes E E When do the B reaction stops Reactants Products 2 Reaction Path 0 12 111414 034 x 06 0204 f2 Leti s ts AmolL B molL 0 1 0 Theextentofthe ATB 03969 X reaction depends on B A e 0 39 the likelihood of the 1 04 l 06 0334 0666 quot forward process A a B 04 x 06 024 39 compared to the BeA 06x03018 O X 0 6 0 backwards process 2 o34 l 055 39 39 39 Q H B 1quot 0666X03 01998 g Alt B B A e 2 E 3 l lE Imagine 60 ofA A9 3 6 goes to B but only B A e 2 30 B goes to A 4 9 9 GE When does it stop 39 39 Etc 5 reaches equilibruum when process starts to stop product favored because more products at equilibrium Kinetic Factors Eaforward It is determined by E 39 O back 0 Activation Energy 2 N029 Ea 0 Collision Effectiveness in each direction N2049 The lower Ea the more likely the transformation The higher the of effective configurations the more likely the transformation 2 N02g N204g favored by low Ea N204 9 2 N02g favored by effective configurations Chemical Thinking 39 39 39 quot 39 06 back favored by Ea g Let s Th I n k forward favored by co nflguratlons The energy diagrams for two important atmospheric reactions are shown Heed dai39a t39e39d39e39te39l m39l39 39e E E forward is favored by activation energy as well as m number of C A hi rxn i r f v r h o completion Reaction Path Reaction Path 7 Which process backward or forward is more probable Which of these reactions is likely to be more productfavored a E x E 39 Tc 2 E a 0 13 Chemical Thinking 39 Let39s apply Assess what you know 111414 Ozone Layer Ozone Good Up High Bad Nearby Ozone is a naturally occurring substance Most of it 90 is found in the lower parts of the stratosphere Ozone Layer Ozone in the Atmosphere Stratospheric Ozone Ozone Layer 12 ppm Altitude kilometers Altitude miles Ozone increases 75 from pollution Ozone concentration gt Stratospheric ozone screens out much of the Sun s harmful UV radiation How does it work Chemical Thinking 39 9amp0 Let39s apply Based on this approximate energy profile discuss whether you would expect this reaction to be product or reactant favored based on E3 and configuration effectiveness m E x E 39 Tc 2 E a 0 Predict Ozone is naturally formed in the stratosphere E M p through this chemical reaction 3 029 2 2 039 Ea 4985 lemol AH I39Xl39l 284 lemol Reaction Coordinate 14 111414 Let39sappny Light Activation 3 029 2 2 039 c 35 The reaction is I I E constantly affected The reactIons Is saId to I39E by the presence of be photolitically rather 73 UV radiation in the than thermally E stratosphere aCtWated quot 0 a0 Let39s app Represent The formation of 03 occurs in two main steps w o mmo AHm498kJImoI Iy Ea 4985 lemol 0 h Aern107 lemol Slop 2 Ea 0 lemol 0 per 03 formed Overall reaction 30 w 203 Build the Energy Profile for this process Chemical Thinking 39 Let39sapply ozone Cycle Ozone is also naturally destroyed in the stratosphere Destruction Chapman s Cycle 4 v f 02Oo00 1 Formation o 02 gt 03 0 oseozoo Destruction 03 0 9 2 02 Every day 3 x108 tons of 03 UV light Heat form and an equal mass decomposes steady state m E x E 39 Tc 2 E a 0 15 Let39sappny Ozone Protection Energy Removed by 02 l839ed Wiiyelength of mdnant eneigy 40 120 200 280 560 440 um I UV C l I UV A l UV B VWVWWV 0 1 D 20 Ozone DUkm UVB Tanning and 50 Slrntnpuuse sunburn E E 40 95 UVA Wrinkling and 207 5 I Tmpupluse 39 3 0 111414 230 Let39sapplv An ozone hole Concern about thinning of the ozone layer over Antarctica arose in the 197039 s but it was not monitored until in the 19805 Measuring Ozone in the Atmosphere Jul 01 2006 Julm l 1C0 150 200 250 300 350 400 450 Dobson Unix SDI 1 Dobson Unit 001 mm at 0 C 1 atm We know the amount of 03 is changing Observe carefully what happens over time Chemical Thinking 30 Let39sappw Some Facts N u an Average area or nzone hale m I 1 l T i l The size of the ozone hole over Antarctica N o Size million Sq km 2 increased dramatically between 19801995 a mi um nmaxim y v 19180 193985 193990 193995 2010 20105 Year 53 CORRELATION I E Radlcal Cl 5 C l N 2 CD 1 1 The source of the atmosphric Chlorine The Ozone is the mean October 0 is the World Meteorlogical Organization column thickness measured at 16 111414 Let39s apply UV C 7 as Chlorofluorocarbons CFCs Cl 7 are the main source of Cl0 ClCF gt Clo radicals in the atmosphere 139 Chlorine source gases Originally used as 0 Foarning agents for plastics 1 Solvents for cleaning 0 Aerosol propellants 0 Refrigerants 0 Fire extinguishers 8 llumann nade soars Total chlovme amounl parts per llxllonl m l E SlZU39ELS m U Chemical Thinking 0 Let39s appyl A n a lyze O3 Destruction One single Cl radical n can destroy over Ea 1quot kJ m039 100000 03 molecules Ea 21 lemol Which of these routes is faster most favored oo3 oloo3 AHrxn 161 lemol E3 04 lemol 02 CIO AHrxn 391 lemol 00 AHI39XI I Cl 230 lemol induced 202Cl Reaction Coordinate Chemical Thinking a i Identify one topic you would have liked to learn more about related to this module s topic a E x E 39 Ti 2 E CD O 17 Chemical Thinking 39 Analyzing Rate and Extent Summary The rate of a reaction is determined by these three main factors Collision frequency configuration effectiveness and activation energy The higher the first two factors and the lower the activation energy the faster the process Thus reactions tend to speed up with both increasing temperature and concentration of the reactive species Greater frac on with enou h energy to react Fraction of collisions f 111414 m E x E c 39 Tc 2 E a c 0 Analyzing Rate and Extent Summary The extent of a chemical reaction depends on the likelihood of the conversion of reactants to products compared to the backwards process The system reaches a state of chemical equilibrium when these two processes occur at the same rate The extent of a reaction is affected by temperature pressure and concentration of reactants and products Chemical equilibrium shifts in the direction of the process that becomes faster due to the change 18 Unit 4 How do we characterize chemical processes THREE MAIN MODULES M1Modeling Chemical Reactions I Identifying factors that affect chemical processes 5 0 353 M2 Understanding Proportions Determining the amount of substance formed or consumed Fquot M3 Tracking Energy Chemical Thinking 39 39 quot 39 Predicting the amount of energy absorbed or released The Challenge What amounts of reactants and products are involved gg How fast will the a process go 13 and how can I GE control it C 0 We would like to generate models that allow us to answer questions such as this How much energy will be needed or produced To what extent will the reactants be changed into products Unit4 Module 2 Understanding Proportions 5 a o a 0 3 V 3 3 a a quot 0 How do we characterize chemical processes Central goal To make quantitative predictions about the amount of substances consumed or produced during chemical reactions 112014 eeheHnic preeess energy released upon bond formation energy absorbed when bonds are breaking WWW tnkpq lpqq pnprgy tn hrpnk bonds in endothermic takes more energy to break bonds Making Predictions Determining the amounts of reactants and products involved in a chemical reaction is of central importance in many areas These calculations are simplified when chemical processes goes to completion The conversion of reactants into products is close to 100 because the probability of the forward process is much larger than the probability of the backward process during a chemical reaction AB lt CD Chemistry XXI 112014 Making Predictions Consider the combustion of octane C8H18 one of the main components of gasoline 2C8H18 25 02 gt 16 co2 18 H20 This equation has a lot of implicit information Particulate w giww705 iyt sqg ig a 3 fgawe ai fgiygq h 33 3 8 989 9amp 2 mol CaH18 25 mol 02 16 mol C02 18 mol H20 l 2284 g C8H18 8000 g 02 s 7042 g co2 3242 9 H20 Macroscopic 10284 9 s 10284 9 Chemistry XXI Making Predictions We can use the information conveyed by a chemical equation to make predictions about amounts of reactants and products Reactants 4 Products Mass Mass reactant product a l Molar Mass Molar Mass I E 1 Moles Moles W E reactant product 0 Chemical 0 Equation 112014 Amounts of Reactants For example Ethanol CZHSO is being used both as a gasoline additive and as an alternative renewable fuel Imagine for example that we want to produce ethanol biofuel by fermentation of glucose C6H1206 C6H1206s 39 2 C2H60l quot39 2 C029 How can we calculate what mass of glucose should be used per every kilogram 1000 g of ethanol that we want to produce Chemistry XXI Amounts of Reactants Substance M gmol C6H1206s 9 2 C2H60l quot39 2 C029 CeH os 1802 1000 g cznso 4607 Mass Product 9 Mol Product col 44quot nC2H6O1OOOMX 1m 1C2H60 2171m01C2H6O 4607 Mol Product 9 Mol Reactant 1m01C6H1206 nC6H12062171 2 60x 2 1085 mol C6H1206 2 6 Mol Reactant 9 Mass Reactant mC6H1206 1085 W x1802 g C6H1206 1955 g C6H1206 1m 6126 Chemistry XXI Let 3 Think 3N02 H20 2HNO3 NO Z Sei i l22125303ZZ39PSLZSEZEE i x iOAc HNO3 x 1 mi 639 16 E8 moi HNO3 phenomenonknownas acid 16E 8 mol HNO3 x 3 N022HNO3 24E8 mol N029 H20quot 9 HN03aCI N09 N02 Average concentrations of HN03 in a big city such W1 E6 9 N02 as Los Angeles are close to 10 pg 1 ug1x10396 g perm3 11 mg N02 Determine the concentration of NO2 in pigm3 that is involved in the production of HNO3 in this environment is 5 L E 0 Amounts of Products Predictions about the maximum amount of product in a chemical reaction also called the theoretical yield of the reaction require that we pay close attention to the relative mole proportions in which different reactants are present in the system Case 1 There is only one reactant or all of the reactants are excess except one Imagine for example that we were interested in estimating the maximum mass of C02g produced every time that 1 kg of octane C8H18 is burned in a combustion engine Carbon Footprint Chemistry XXI 112014 10mg Amounts of Products e 2C3H1a 25 029 9 16C029 18 H20 9 Substance M gmol caH18 1142 3 Wei 631 18 v 16 9191602 0 3200 10 8 18 X 2 1142 98141 2 W18 02 4401 Mass to Mole Mole t0 Mole H20 1801 4401 g CO2 x k 1 0 CO2 3083 gCO2 Mole to Mass The same procedure can be used to calculate any Chemistry XXI C2H6O 202 2C02 3H20 amount of reactant needed or product produced 10quot3 C2H6O x146 g 217 mol C2H6O 56 2amp2 Let s ThInk Calculate the amount of C02 produced when 10 kg of ethanol CZHGO reacts with 02 the products are C02 and H20 217 mol C2H6O x 2 mol CO21 mol C2H6O 435 mol CO2 435 mol CO2 x 44 g CO21 mol CO2 19 kg CO2 ction If 10 kg of C8H18 produces 308 kg CO2 and 10 kg of C2H60 generates 191 kg C02 what percent reduction in CO2 emissions do we get by replacing octane with ethanol is 5 L E 0 112014 Not an Easy Decision Does the previous result implies that ethanol is greener than octane Carbon Intensity of Bioethanol Production us Corn used in the UK Land Use change 0 0 for Cropland 26 for Grassland 214 for Forest land quot a years crop Production 34 E o Drying and Storage 12 2 g Feedstock Transport 1 I 390 tn 2 3 coal used to Process com to Ethanol 96 n O c n 39 n Electrlclty Produced 16 0 corn Gluten coproduct 8 0 corn on coproducz 8 3 Transport of 0 to UK E 4 Total 103 i 5 I a 450 40 20 O 20 40 60 80 100 120 S dmlaxen Irm mpzlmww dll gwukl Carbon intensity 0 I m Joule energy UK DoT gures Not an Easy Decision Sugaraaenumrea Klngdom so Wham United Kingdom 61 Wheat Ukmlne 99 When Germany 59 When France 64 Wheat Canada so Sugarcane South Alrlcz 104 Sugar cane Mozambique 21 Bioethanols Sugar cane Brazil 18 Corn France 49 Corn USA 105 Molasses UK 40 Molasses South Africa 81 Molasses Paklslan 70 Natural Gas 62 Dienl 86 20 an en 80 mnmmmm nunemnukgm m gram or camon quotiOXide r39 v Ineudjuuw 0f 1 J nqiroodslcnn39cnmsnhmaigtmtma pct Chemistry XXI Amounts of Products Case 2 When we are uncertain about which reactants are in excess and which species is the limiting reactant in a chemical process we need to take systematic steps to identify them Hydrogen gas H2g is seen as one of the most important alternative energy sources of this century PRODUCTION C2H6g 2 H20g 2 COg 5 H2g Imagine that we had a tank with a mixture of 500 g of each reactant Which substance is the limiting reactant How much hydrogen gas will be produced is 5 L E 8 o 5009 Amounts of Products 02H6g 2 H20g gt 2 609 5 H2g Mass Reactants Mole Reactants Substance M gmol 1 mol 22H6 CZH6 3007 500 g CZH6 x 1663 mol CZH6 3007 g 2sz H20 1801 CO 2801 1 mol H 0 500 g H20 x Z 2776 mol H20 H2 2016 1801 g H20 Compare ACTUAL to STOICHIOMETRIC ratios 3 Actual Ratio Stoichiometric Ratio 4 mo mo 0 27 76 I H20 1 669 H20 2 I H20 2 H20 E 1663 mol CZHG 39 C2H6 1 mol CZHG CH 0 Limiting Amounts of Products All calculations about the amount of product formed during a chemical reaction should be based on the amount of limiting reactant in the system 02H6g 2 H20g 9 2 609 5 H2g 5 mol H2 2776 mol H20 x 2 mol H20 5 2016 H 5 4 1399 g H2 2 1 mol H2 as 0 Let s Think In the search for renewable chemical paths to produce hydrogen gas for fuel consumption chemical scientists have found ways to use carbohydrates extracted from wood residues cellulosic materials to produce H2g in a process involving the following substances C6H1005s quot39 H20quot 29 Cozig What mass of hydrogen can be produced in a reactor containing a mixture of 1 kg of each reactant is 5 L E 0 112014 C6H1005 7H20 12H2 6002 10quot3 g x 1 mol C6H1005 162 g 62 mol C6H1005 10quot3 g x 1 mol H2018g 55 6mo H2O 556 H2062 C6H1005 89 ratio is 7 H20 1 C6H1005 limiting reactant is C6H1005 62 molx12H21 C6H1005 744 x 2 1492 H2 556 mol x 12 7 953 x 2 190 9 H2 112014 Let39s apply Assess what you know Chemistry XXI Reducing Pollution Given our dependence on fossil fuels we should figure out ways to reduce the pollution they generate For example We have seen that the incomplete combustion of a hydrocarbon may lead to the formation of carbon monoxide CO How can we reduce the amount of CO formed We may try to inject air and hydrocarbons into the combustion engine in the right proportions to try to ensure that all the hydrocarbon is consumed Chemistry XXI Let 3 Think 2C8H18 2502 16C02 18H20 1g C8H18 x1 mol 114 g 0009 mol If we inject 1 g of C8H18 in the engine how many moles ofOzare needed forfull combustion O IIIUI X 627 2 I I IIIUI Howmany grams is this mol X 1 mol 2 needed Now we do not inject O2 in our combustion engines We use air How much air do we need is 5 L E 0 Chemistry XXI Let s Think Air has 21 02 and 79 N2 by volume or by mol How many grams of air should be injected if we need 3503 g of 02 per gram of C8H18 in the combustion chamber What happens if we inject more or less air AirFuel Ratio The stoichiometric AirFuel ratio for Stoichiometric Ratio gasoline is actually 0 AFst 4 M Howwouldyou o expect the mol of 02 C02 and CO in theengineto 5 changewith o 5 increasing AF A I 0E Lt th kl 4quot V o 6 es In 2 I U RICH a 9 9 I x a g 0 a e I 9 E It 0 a s a b a r 39i39 9 g in 9 OPTIMAL 9 Q a Ill 4 5 1 Vb 1539 a 3 quot 3 x i 3 5 9 s a 3 e r Iii E ii 3 g 3 WV asawi g in 9amp9 9amp5 m a 395 3 3 f 9 E LEAN g a 4 3 9 E 3 4 5 1 ya I Ileana 5 g ayyd 39 g 369quot w 9 Q V w in a a in GE g wagg d in 39gyw y p 5 Q at 39 f l quot3 m a 395 3 5quot 5 quot 4 112014 3503 q 02 x1 mol 32 q 01095 mol O2 0109521 x 79 x O 412 mol N2 needed 0412 mol N2 x28g1 mol 11534 g N2 11534 3503 1504 g air ingresing AF increasing oxygen so Q2 starts increasing looks like constant up to line and then starts increasmg but CO2 and CO are limited I I I I I I CO2 look like upside down parabola with the max on the line at 147 AF CO is at a maximum on left side and starts decreasing asap because oxygen is lacking fuel at an QVF QCC I I V l v V UV right side 02 is in excess and fuel is limiting rich on left lean on right Understanding Proportions Summary During a chemical reaction new substances are formed and other are consumed If the reaction occurs in a closed container the total mass of the system is conserved Although the number of molecules of the different substances present in the system will vary the total number of atoms of each type will remain the same 6b 0 43 0 1 3939 3 300363 0 Q a 4 lt3 8 Chemistry XXI Understanding Proportions Summary A balanced chemical equation together with information about the molar masses of the reactants and products can be used to predict the mass amount of substance and number of particles that will be formed or consumed during the process Reactants products A Mass 39 product I Molar Mass Molar Mass I Chemical Equation Moles product Chemistry XXI 112014 Ammoniaexamplejolm oc N2 3H2 2NH3 nitrogen is the limiting reactant make 4NH3 and 1 H2 1 x10quot6 g NH3 x1 mol17 589 E4 mol NH3 589E4 mol NH3 x 1 N2 2 NH3 294E4 mol N2 294E4 mel N 2ee28ng meITBQ miiiien tons N2 For next class Investigate why energy is always absorbed or released during a chemical reaction In which ways the amount of energy released during chemical reactions may be related to your body weight is 5 L E 0 112014 Unit 4 How do we characterize chemical processes l THREE MAIN MODULES l chemical processes Identifying factors that affect 7 7 M1 Modeling Chemical Reactlons 55 9 go M2 Understanding Proportions Dete39m39mng the am quotquott f substance formed or consumed all M3 Tracking Energy Predicting the amount of energy I absorbed or released Chemical Thinking Unit 4 How do we model chemical change M 0d u e 3 Central goal To make qualitative and Tracking Energy quantitative predictions about the amount of energy absorbed or released during a chemical reactions based on the nature of the chemical bonds in the molecules of reactants and products Chemical Thinking The Challenge We would like to generate models that allow us to answer questions such as this What amounts of How much reactants and energy will be products are needed or an involved produced E E E How fast will the To what extent 392 process go 7 will the reactants g and how can I 39 be changed into GE control it products 0 112014 Modeling Chemical Reactions leweleeewn en Pe the srengeethe bends Assumpt 2 takes more energy to break bonds less energy Rearrangement of atoms during a chemical reaction involves electron redistribution among different released when bonds form atoms As a result of this process the internal potential energy of particles in the system changes 39 g Endothermic Exothermic Ep Process Ep Process 3 g 1 takes less energy to break bonds more energy E go T released when bonds iormwhich is why exo 5 3 02 Energy Absorbed Energy Released Z E Reactants Products Reactants Products g Remnpath Reac onpath surrounding gets colder for endo Energy Transfer Surroundings Surroundings A common signature of a chemical change is the release or absorption of energy System System The system The system releases energy q lt 0 absorbs energy q gt 0 EXOTHERMIC ENDOTHERMIC 5 T SUIT TSUquot l Heat Chemical Thinking Exothermic or Endothermic Combustion reactions are prototypical examples of exothermic processes CH4 202 e co2 2H20 y 9 ENERGY 5 g The decomposition of stable compounds such as E H20 tends to be endothermic C i5 2 H20 6 2 H2 02 How can lg 6 6 we i zlaln g ENERGY differences 0 112014 A Chemical Model During a chemical reaction some bonds are broken and some new bonds are formed o 39 o 39 o 1 0 aw l oo 7 E 31 39 The amount of energy released or absorbed in a chemical reaction is related to the nature and numbers of bonds broken or formed in the process Chemical Thinking yesnonoyes 66 Let39s think 9 Answer these questions in your groups 1 Does it take energy to break a chemical bond 0 ls energy released when a chemical bond is broken 1 Does it take energy to form a chemical bond between two atoms 0 ls energy released when a chemical bond is formed Chemical Thinking Bond Energy Energy is needed to separate a pair of bonded atoms break the bond The energy provided is transformed into potential energy of the separated atoms By convention EP 0 Ep 0 when the atoms are infinitely c 39 39 c separated on E x E 39 Tc 2 E a 0 112014 Bond Energy E1 Idmoi Bond formation corresponds to a minimum in the potential energy released CLlCK AND DRAG THE RIGHTHAND ATOM 7 a 0 Eng39 lnlemuclear dsianos pm as Potential energy is E Dissociation transformed into 392 Energy kinetic energy when g 3quot a bond is formed E so i w energy is released 0 Let 3 Explore Go to httpIlwwwchemarizonaedultpplchemthinklsim Bond Energy is g o o o O O C as 0 Zj bwmi Analyze and compare energy transformations for systems with different bond energies Discuss the questions posted on the Web page Chemical Thinking Bond Comparison C O Bond 2 Bond 0 Bond E klmoi E kJmol E kJmol E v3 a o u o o p E 100 100 100 200 r pm 1 DUI 39 10 200 I39 0quot BE 360 kJmol BE 414 kJmol BE 464 kJmol Energy Needed to Break the Bond 2 smaller Energy Released when the Bond is Formed Larger 2 E If 5 Construct these PE diagrams for ii CC CC and CEC bonds 8 39g Let 5 Discuss differences in energy transfer g Think when these bonds are brokenformed 0 Chemical Thinking Energy Transfer Let s analyze this reaction CH49 2 029 C029 2 209 What is net energy is released or absorbed 4 C 4 H 4 O 2 00 2 x 498 Energy 9 6 N released when forming bonds Energy 4CH 4x414 needed to 200 2x799 656 kJ break 15 8 kJ 4 OH 4 x 464 bonds 391 l1 0 0 3 quot 9 oco 713 1 oo 112014 Chemical Thinking Heat of Reaction Ep H CHHHHOOOO 0 1656 996 2652 CH4 2 02 Dissociation 392552 N gt 3454 1598 1856 Formation CO2 2 H20 3454 kJ Net Energy Released 802 kJ qrxn Aern Aern V BE bonds broken 2 BE bonds formed m E x E 39 Tc 2 E a c 0 Energy Conservation As with mass the total energy of a closed system undergoing a chemical change is conserved If so Ep where are the 802 kJ in this reaction coming from M 0 00 00 The main contributors are the molecules with the highest potential energy l O W It is energy that was stored as Ep in chemical bonds reactants Ep is transformed into kinetic energy or vise versa Chemical Thinking Letps Think Given the information provided build the Energy Diagram for the decomposition of water How much energy is released or absorbed 2 HOH 9 2 HH OO E Bond Dissoc quot Energy 0 KJlmol 0H 464 HH 436 OO 498 What happens to the energy we put in 112014 4x464 1 856 2 X 436 498 1370 486 kJ endothermic Chemical Thinking Letps Think Compare these two sets of bond energies Bond Energy Bond Energy Which types of kJImol kJImol bonds are easier HH 436 CH 414 to break CC 347 0H 464 00 142 C0 360 NN 163 NH 389 Based on this trend predict which of these reactions is likely to be exothermic or endothermic m E x E 39 Tc 2 E a 0 Letps Think Organic compounds made up of molecules that contain fewer hydrogen atoms and more oxygen atoms per carbon atom are said to be more oxidized Consider the following set of molecules represented in increasing order of oxidation Methane CH 4 Methanol CIH0 Formaldehyde CH 10 Formlc Acld CH 102 39i T H 0 9 H C H C C H 2 a H H H on it is stored in the bonds Ep in the product bonds in general AA bonds are weaker than AB bonds AA 33 AB exothermic ABC A B 3390 endothermic Q Letps Think Methane CH 4 Methanol CIH O Formaldehyde CH 0 Formlc Acid CH 101 Chemical Thinking r r H o o H C H H Co 39339 EL H a a H H H 0 2Hquot 802 kJmol 1 Estimate Aern for the 3039 Energy a combustion of one mole of the quotJmm 5 last three compounds 3quot 347 E 2 Build energy diagrams for each C0 360 If of the reactions cH 414 c v 3 Explain the overall trend in the 0H 464 2 value of the heat of reaction E 0 0 498 a wrth Increasing degree of 5 oxidation 3 0 799 Let39s apply Assess what you know 3amp0 Let39s apply a E x E 39 Tc 2 E a O Predict Most of the energy our body needs is supplied by the combustion of carbohydrates and lipids The most common carbohydrate is glucose Most lipids result from the combination of fatty acids such as oleic acid C6H1206 H O 1 C Predict which combustion reaction will produce more energy per mole of each substance 112014 CH4 202 CO2 2H2o 1656 996 2652 1598 1856 3454 802kJexo 2CH4O 302 2CO2 4H2O 2484 720 928 1494 3196 3712 more oxidized energy is lower so less energy is released Trom system to surroundings 5626 6908 1282 112014 a0 Let39s apply Estimate Verify your prediction by estimating the amount of energy produced by the combustion of one mole of each substance H xhnlfiwa f C6H1206 1 quot24 Bond Energy H C OH 650156 KJlmol I r 5 olvi H0 C H g zzg CC 347 C H OH 794 c c 611 39 2Q CH H C OH fjglf 414 2 m 0H 464 H C OH Q 0011 H le ZO39r 00 t I 21 t CO 799 a0 Let39s app Estimate C6H1206 6 02 e 6 co2 6 H20 cc cH CC co 0H co oo Bond Energy KJmol B 5 7 o 5 5 1 6 cc 347 F o o o o 12 12 o cc 611 C18H3402 512 02 e 18 co2 17 H20 6 414 OH 464 CC CH CC CO OH CO OO 00 493 B 16 33 1 1 1 1 255 CO 360 F 0 0 0 0 34 36 0 30 799 Remember energy is needed to break bonds energy is released when bonds form Chemical Thinking 2amp0 Let39s apply Com pare It is common to express the energy produced during fuel combustion in amount of energy per unit mass energy density leg Determine and compare the energy density in kcallg Calg for these two substances Discuss why this quantity may be more useful for purposes of fuel analysis and evaluation than its equivalent in kJImol 1 calorie 1 cal Energy required to raise the temperature of 100 g of H20 by 10 C 4184 J m E x E 39 Tc 2 E a O 112014 Are You Ready Chemical Thinking The Quest for Ammonia Ammonia NH3 is one of the most important industrial chemical substances It is widely used in the production of fertilizers pharmaceuticals refrigerants explosives and cleaning agents It ranks as one of the 10 top chemicals substances produced annually in the world Chemical Thinking The Synthesis Ammonia is mainly produced via this simple chemical reaction N29 29 2 NH39 Balance this chemical reaction complete the following particulate representation of the 3 process and identify the limiting reactant E E IE Q09 76 2 Let39s GE think 0 Chemical Thinking The Amounts Nitrogen for the synthesis of ammonia is obtained by separating it from air by liquefaction N29 3 H29 Z 2 NH39 112014 d b If approximately 100 million metric tons of 0quot ammonia are produced annually worldwide How many tons of N2 are extracted from Let39s think aIr each year 1ton1x103 kg1 x1059 Chemical Thinking Energy Transfer The optimization of the industrial production of ammonia requires a good understanding of energy transfer during the process N29 3 H29 2 2 NH39 9quot Estimate the heat of reaction for this process and determine whether the Let39s think reaction is exo or endothermic Bond Dissoc Energy KJlmol NH 389 HH 436 NEN 945 m E x E 39 Tc 2 E a 0 Let s Think N29 3 H29 Z 2 NH39 Sketch the energy profile for this process and discuss whether the forward or backward processes can be expected to be favored 10 Competing Factors 112014 c C 2 E C 39 Tv 2 9 Analyze the effect of Increasmg T and P GE on both reaction rate and reaction extent Let39s think 0 III S I m u latlo n gt 2 39 H N2 NH 2 N g o 0 0 NH3 E F u g 9e WHAT TO DO cEJ HIGH or LOW T Extent effects 6 Let39s think HIGH or LOW P Costs 11 Exam 1 Review of Key Points Through concentration effects For thermodynamic stability consider Energetic factors Internal PEenthapy Entropic factors number con gurations S Energetic factors H Bond strength chemical composition charge distribution state of matter IMF AA bonds are weaker than AB bonds Longer bonds are weaker than shorter bonds SHORT AB Bonds are favored More polar the strongerbigger diff in electronegativity Stronger the IMF lower the PE more energetic stability Coulombs law Stronger the F more attractive low PE Entropy S State of matter Molar mass Molecular complexity Number of atoms and bonds variety in the types of atoms and bonds that make up molec compound Ion size and ion charge in ionic compounds WE WANT S S increases with molar mass and molecular complexity Delt S delt H and delt G products reactants Multiply coefficients to energies Elements on compounds have no enthalpy delt H value Delt G delt H T delt S quotquotquotat diff temperatures at constant T and P G is favored Delt H delt S delt G and product favored and reactant and depends on T and P fav and low T and depends on T P fav at high T G and big value favored A LOT in product side Goes to completion mucho Equilibrium constant Kc products reactants to the exponent of stoichiometric coefficients Delt G RTnK K equotGRT Unit 5 How do we predict chemical change The central goal of this unit is to help you identify and apply the different factors that help predict the likelihood of chemical reactions Comparing the relative stability of different substances sf12 amp g M1 Analyzing Structure 397 Determining the directionality and o 339 M2 compar39ng Free Energ39es i extent ofa chemical reaction determine reaction rates M3 Understanding Mechanism 39den fying the Steps that l Chemical Thinking 39 39 39 39 Analyzing the factors that affect reaction rate M4 Measuring Rates Unit 5 How do we predict chemical change O C directionality and extent of chemical reactions Module 2 Comparing Free Energies 2 E Central goal E To quantitatively l C i determine the E E G C U 1 Transformation How do change it The Challenge Imagine that you were interested in QUANTITATIVELY determining which of two or more chemical processes could have occurred on the primitive Earth What measurable properties of the system can be used to make the prediction How could we actually quantify the directionality and extent of a chemical reaction a E x E J 39 Tc 2 E a J U some casesregardless of temperature always Relevant Factors never form products The directionality and extent of a chemical reaction depend on three main factors a C E ENERGETIC FACTORS ENTROPIC FACTORS Aern Aern 3 E 1 But 2 TEMPERATURE T how u H g What Changes Consider this chemical process 3 x E E 7 CH49 2 029 9 0029 2 H20g 0 39E g I Is this a favored process Why I 0 more stable lower PE so usu exothermic The Facts It may be favored if the process aleads to more energetically stable compounds 9 energy is released Exothermic AH lt 0 rxn Aern lt 0 39 CH49 2 029 9 0029 2 H20g a E x E r 39 Tc 2 E a r U Chemical Thinking 39 39 39 The Facts It may be favored if the process b Leads to the formation of substances with more distinguishable configurations AS gt0 rxn CH49 2 029 9 0029 2 H20g Aern gt 0 Certainly favored AHxn lt 0 exothermic AS gt o Chemical Thinking 39 39 39 Let s Think Consider the possible decomposition of CO2 in the atmosphere C029 06 029 6 an I Q as 3 cc 639 cc Analyze the sign of AH n and ASxn and decide whether the process is favored or not D E x E C I To 2 E G C U The Law In general energetic effects become less relevant at high temperature where entropic effects dominate The sign of this quantity Gibbs Free Energy G Aern Aern TAsrxn can be used to determine process directionality 2quot l Law of Thermodynamics At constant T and P AGxn lt 0 for thermodynamically favored processes WWW ALWAYS FAVORED regardless of temp spontaneous product favored 8 complex to simple AND gas to solidgas H reactant favored aqain look at PHASE first for delta 8 unfavorable endothermic rxn with decreasing entropy H S ALWAYS UNFAVORED Independent of temp When both or both completely dependent on temperature whether or not favored or unfavored high temp 8 becomes the main factor When G i favored Reaction Extent more negative number more product formed and Experimental results indicate that processes at more stable constant T and P occur in the direction in which the Gibbs free energy G of the system decreases A 9 B Aern GI3 GA lt 0 Exergonic 9 Favored C 9 D Aern GD G6 gt 0 Endergonic 9 Unfavored The more negative Aern the larger the extent of the reaction The more negative Aern the more thermodynamically stable the products than the reactants Chemical Thinking 39 39 39 39 Evaluating Likelihood The quantitative analysis of the extent of reactions is crucial to determine for example what were the likely components of our primitive atmosphere H2 is suspected to be an important component of the very early atmosphere Was CO2 or CH4 more stable in this environment How does temperature affect the answer Hydrogen constitutes 75 of the Universe 3 elemental mass Chemical Thinking 39 39 39 39 Let s explore it Let 8 Think CHANGE UNITS 2 Consider this data get a with the law so it is favored and therefore Reaction AH rxnkJ AS rxnJIK CH4 iS more favored c10291 4 H29 gt CH49 2 H20 2527 4103 Which compound ofCCO2 vs CH4 is most stable in is favored at really temperatures the presence of H2 at 25 C Hint Are reactants or products favored in the reaction O E x E C To 2 E G C U Chemical Thinking 39 39 39 39 Quantitative Data Our discussion highlights the importance of measuring or calculating Aern ASKquot and AGrxn if we want to make predictions about the thermodynamic likelihood of a reaction HOW DO WE DO IT Chemical Thinking 39 39 39 39 Entropy Change Aern The change of entropy for the reaction ASKquot under standard conditions 25 C and 1 atm can be calculated given data for the standard molar entropies of formation S f of reactants and products aAbB cCdD o o 0 AS rxn S products S reactants ASOrxn 68 C dsoo 39 aS A bSOB O E x E C To 2 E G C U Let s Think N29 3 H29 92 NH39 Substance S f Jlmol K H29 1307 NH39 1928 N29 1916 Calculate A80xn for this process blue are coefficients 3856 5837 1981 products reactants delta 8 not favored Enthalpy Change Aern Measurements of the energy absorbed or released in the form of heat during a chemical reaction are of central importance in making predictions about the extent of a chemical process Thermometer measured by quantifying changes in temperature These measurements are 3 commonly done using a i 219 21 substances in their z 392 standard state at 1 atm c I g and 25 cc This heat of 30quot l I reaction is identified as quot 3 the standard enthalpy Calorimetf change AHOW Heat transfer Is Indirectly E O Standard Enthalpy of Formation One particular useful quantity is the change in enthalpy when 1 mole of substance is formed from its constituent elements in their standard state 1 H2g 1 F2g 9 HFg AH f 2733 kJmol 1 H2g 1 Cl2g 9 HClg AH f 923 kJmol 1 H2g 1 Br2l 9 HBrg AH f 623 kJmol 1 H2g 1 l2s 9 Hlg AH f 265 kJmol The energy needed to decompose one mole of each compound is given by AH f Chemical Thinking Enthalpy Change Aern Information about AH f for reactants and products can be used to calculate the Aern Let s consider a reaction involved in the production of CO2 in the primitive atmosphere Cgraphite 2 H20l C02g 2 H2g How could we evaluate AH rxn given this information Cgraphite 029 9 0029 AH f 3935 lemol H2g 12 029 9 H20l AHof 2353 lemol m E x E J 39 Tc 2 E a J U Alternative Routes Cgraphite 2 H20l 9 COzg 2 H2g We could think of the reaction as combination of a decomposition and a combination steps 1st 2 H20l gt 2 H2g y62g AHom 2 2858 kJ the total energy transfer should be the same 2nd Cgraphite 549 9 0029 AHom 3935 kJ a 3i Cgraphite 2 H20l COzg 2 H2g E AH rxn 3935 2 2858 1781 kJ g This approach is based on the central idea E that no matter what path we follow Hess s g Law 0 57 Let s Think The following diagram represents an alternative path for a reaction likely involved in the formation of CH4g in our planet s primordial atmosphere Cs 4 28 028 Substance AH0 kJmol HZOI 2858 C0g 1105 AHo C02g 3935 C02g 21 H2g 39 CH4g 2 H20l CHAS 446 2 Determine AH rxn given the available data 2 Propose a general rule for calculating AH rxn based on the AHf of reactants and products Chemical Thinking FOR BOND ENERGY ONLY IS IT REACTSNTS PRODUCTS 0 o o AH rxn 2A fproducts 2A freactants everything plea IQ PR Reference State It is important to notice that in this relationship we only need the AH f of the chemical COMPOUNDS involved not of any elements HOW SO We are taking 01 Elements AH f 0 for I the elements in their Reactants Products Standard state 787 221 566 kJ it Let s Think 2 009 029 9 2 C029 Substance AH f lemol COg 1105 0029 3935 029 0 Calculate AH rxn for this process Chemical Thinking 39 39 Gibbs Free Energy An identical procedure can be used to calculate the AG xn given the AG f for the different chemical COMPOUNDS involved in the reaction o o 0 AG rxn 2 AG fproducts 2 AG freactants l AG f 0 for elements in their standard state m E Alternatively we can also calculate AG rxn as C I39E This route is useful 73 Aern AHorxnTAsorxn in estimating the 39g effect of T on the g value of AG 0 Summarizing aAbB cCdD ASOrxn 68 C dSOD 39 aS A bSOB AHorxn C AHof C d AHof D AHof A b AHof B AGorxn C AGof C d AGof D AGof A b AGof B These relationships provide information under standard conditions 25 C We can also use this relationship to estimate Aern at a different temperature Aern AHOrxn TAS rxn D E x E C I To 2 E G C U i Let s Think One central question in the theories about the origin of life is how complex organic compounds were synthesized from simpler molecules such as H2 N2 CH4 NH3 and H20 Consider these possibilities in the synthesis of the simplest amino acid glycine CZH5N02 2 CO9i quot39 NH39 39quot Iquot29 9 02H5N023 I J 2 CI449 quot39 NH39 quot39 2 H20 C2quotl5NOzS quot39 5 H29 1 Calculate AH rxm AS rxm and A60 for these two processes Chemical Thinking 39 39 39 39 1 5284 459 211o9 2607 H 6154 S 78 G 2 238H516S 223G g Let s ThInk Reaction AHorxn ASorxn AGOrxn kJ JI K kJ 2 009 NH39 H2g C2395N023 2 cquot449 quot39 NH39 quot39 2 H20 C2395N023 quot39 5 I429 2 Identify which reactions are productfavored and discuss the effect of temperature on the extent of each process Chemical Thinking 39 39 39 39 p Let s ThInk 2e 11 Reaction AHorxn ASorxn AGOrxn kJ JI K kJ 2 009 NH39 H2g C2395N023 2 cquot449 quot39 NH39 quot39 2 H20 C2395N023 quot39 5 I429 O E x E C To 2 E G C U First reaction is product favored because G is nega ve lowtemperkeep bengtavoredtor rstreaction high temps will start being favored for second rxn depends on the G H TS equation Reaction Extent si n of G ives us directionalit and ma nitude gives Although the change in Gibbs free energy A60 US extent allows us to make predictions about reaction directionality the actual number tells us little about the final relative amounts of product to reactants How can we better quantify reaction extent Experimental results indicate that chemical reactions tend to reach an equilibrium state in which the concentration of products and reactants does not vary over time 30quot N H SCDOSOO Time Chemical Thinking 39 39 concentration of products reactants raised to coefficients Equilibrium Constant At grandestiiiiitiiaemtbapialiawiit ratio of the aanaeairaiimrmaasmssm of r aeav teateweaueegremai steaiistant Equilibrium Constant aAbBcCdD u c d C D 5 For reaction involving 0 a b C E gases the equilibrium A B E constant may be c a I expressed in terms of K PC PD g the partial pressures p Pan E exerted by each gas A B B O AG rxn vs K The value of the equilibrium constant K is directly related to the AG rxn for the process AGO rxn AGfm RT111K e K 6 RT AHquot A50 mm rxn RT R Then K e R 8314 JK mol In general I ProductFavored Process AG rxn lt 0 K gt 1 I D E x E C I To 2 E G C U I ReactantFavored Process A60 gt 0 9 K lt 1 I 1O Predict the effect of f changing AHOrxn Asorxn and T on the value of K and the reaction Reaction pathway extent B K A Extent of reaction B 39 A Chemical Thinking 39 39 large G gives a large K value combustion rxns go to completion this sitch means that almost alllll As we have seen different reaction reach equilibrium states with different proportion of products and reactants in the system Reaction Extent For example combustion reactions tend to have very large K values CH49 029 gt 0029 2 H209 AG rxn 8009 N K 207 x 10140 at 25 c For all practical purposes this reaction fully goes to completion at all temperatures AH rxn lt 0 AS nmgt 0 CH49 029 gt 0029 2 H200 Chemical Thinking Reaction Extent But what about a reaction such as this COg H20l CH202I AG xn 129 kJ K 1820 at 25 C AH rxn lt 0 AS rxn lt 0 which may have played a central role in the formation of complex molecules in our planet Whether this reaction is product or reactantfavored depends on the actual conditions of the process Thus it is better to represent it as cogH20I lt CH202 to highlight the importance of chemical equilibrium m E x E J 39 Tc 2 E a J U 11 Chemical Thinking 39 39 0 Let39s apply Assess what you know Chemical Thinking Chemical Evolution In 1956 Urey and Miller conducted a classic experiment on the origin of life The experiment showed that biological molecules such as amino acids can form from simple reactants electrodes gas inlet CH4 NH3 direction of circulation Since then many experiments have been performed with different reactant mixtures and iheat sources of energy heat UV Xrays etc cooling m E x E J 39 Tc 2 E a J U Primitive Mixtures These are two of the reactant I H20 CH4 NH3 H2 mixtures that have been tested CO N2 H2 Results indicate that the elemental composition presence of C H N O of the mixture is more relevant than the kinds of molecules used Why No matter the mixture the concentration of species seems to be controlled by these equilibriums CH4g 2 209 0029 4 H29 Likely occurrin 0029 HA9 009 3 H209 betweeng N29 3 H203 2 NH3ggt 5004000 C39 12 Let s Think CH49 2 209pcoz9 4 29 c3029 HA9 009 H209 N29 3 29 2 NH39 39 Substance AH f kJImol S fJlmol K m H29 0 1307 5 E K f CH49 746 1863 E thjsflemparzecesges 2029 39393395 21339s h E at 500 0C COg 1105 1977 2 and 800 oC NH39 459 1928 E H20g 2418 1888 5 N29 0 1916 a i Discuss with a partner one thing you do not understand about Module 2 Make one suggestion of what a you could do b I could do to improve your understanding Chemical Thinking 39 39 39 39 Comparing Free Energies Summary At constant T and P AGrxn Aern TASrxn lt 0 for thermodynamically favored processes a A b B 69 c C d D DIRECTIONALITY m E AH xn Asom AG xn E Productfavored E Reactant favored Depends on T P fav at low T g Depends on T P fav at high T O E x E C To 2 E G C U Equilibrium Constant aAbB 9cCdD K CriD AriB EXTENT Ang RT111K 0 0 Aern Aern Ke RT R In general I ProductFavored Process AG nmlt0 9 Kgt1 I ReactantFavored Process AG nmgt0 9 Klt1 14
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